Tag: games

The odd knights of the round table

Published January 28, 2010 by lievenlb

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).

Leave a Comment

Olivier Messiaen & Mathieu 12

Published December 31, 2009 by lievenlb

To mark the end of 2009 and 6 years of blogging, two musical compositions with a mathematical touch to them. I wish you all a better 2010!

Remember from last time that we identified Olivier Messiaen as the ‘Monsieur Modulo’ playing the musical organ at the Bourbaki wedding. This was based on the fact that his “modes à transposition limitée” are really about epimorphisms between modulo rings Z/12Z→Z/3Z and Z/12Z→Z/4Z.

However, Messiaen had more serious mathematical tricks up his sleeve. In two of his compositions he did discover (or at least used) one of the smaller sporadic groups, the Mathieu group $M_{12} $ of order 95040 on which we have based a whole series of Mathieu games two and a half years ago.

Messiaen’s ‘Ile de fey 2’ composition for piano (part of Quatre études de rythme (“Four studies in rhythm”), piano (1949–50)) is based on two concurrent permutations. The first is shown below, with the underlying motive rotational permutation shown.

This gives the permutation (1,7,10,2,6,4,5,9,11,12)(3,8). A second concurrent permutation is based on the permutation (1,6,9,2,7,3,5,4,8,10,11) and both of them generate the Mathieu group $M_{12} $. This can be seen by realizing the two permutations as the rotational permutations

and identifying them with the Mongean shuffles generating $M_{12} $. See for example, Dave Benson’s book “Music: A Mathematical Offering”, freely available online.

Clearly, Messiaen doesn’t use all of its 95040 permutations in his piece! Here’s how it sounds. The piece starts 2 minutes into the clip.

The second piece is “Les Yeux dans les Roues” (The Eyes in the Wheels), sixth piece from the “Livre d’Orgue” (1950/51).

According to Hauptwerk, the piece consists of a melody/theme in the pedal, accompanied by two fast-paced homorhythmic lines in the manuals. The pedal presents a sons-durées theme which is repeated six times, in different permutations. Initially it is presented in its natural form. Afterwards, it is presented alternatively picking notes from each end of the original form. Similar transformations are applied each time until the sixth, which is the retrograde of the first. The entire twelve-tone analysis (pitch only, not rhythm) of the pedal is shown below:

That is we get the following five permutations which again generate Mathieu 12 :

a=(2,3,5,9,8,10,6,11,4,7,12)
b=(1,2,4,8,9,7,11,3,6,12)(5,10)=e*a
c=(1,12,11,9,5,4,6,2,10,7)(3,8)=e*d
d=(1,11,10,8,4,5,3,7,2,9,6)
e=(1,12)(2,11)(3,10)(4,9)(5,8)(6,7)

Here’s the piece performed on organ :

Considering the permutations $X=d.a^{-1} $ and $Y=(a.d^2.a.d^3)^{-1} $ one obtains canonical generators of $M_{12} $, that is, generators satisfying the defining equations of this sporadic group

$X^2=Y^3=(XY)^{11}=[X,Y]^6=(XYXYXY^{-1})^6=1 $

I leave you to work out the corresponding dessin d’enfant tonight after a couple of glasses of champagne! It sure has a nice form. Once again, a better 2010!

5 Comments

The artist and the mathematician

Published October 19, 2009 by lievenlb

Over the week-end I read The artist and the mathematician (subtitle : The story of Nicolas Bourbaki, the genius mathematician who never existed) by Amir D. Aczel.

Whereas the central character of the book should be Bourbaki, it focusses more on two of Bourbaki’s most colorful members, André Weil and Alexander Grothendieck, and the many stories and myths surrounding them.

The opening chapter (‘The Disappearance’) describes the Grothendieck’s early years (based on the excellent paper by Allyn Jackson Comme Appelé du Néant ) and his disappearance in the Pyrenees in the final years of last century. The next chapter (‘An Arrest in Finland’) recount the pre-WW2 years of Weil and the myth of his arrest in Finland and his near escape from execution (based on Weil’s memoires The Apprenticeship of a Mathematician). Chapter seven (‘The Café’) describes the first 10 proto-Bourbaki meetings following closely the study ‘A Parisian Café and Ten Proto-Bourbaki Meetings (1934-1935)‘ by Liliane Beaulieu. Etc. etc.

All the good ‘Bourbaki’-stories get a place in this book, not always historically correct. For example, on page 90 it is suggested that all of the following jokes were pulled at the Besse-conference, July 1935 : the baptizing of Nicolas, the writing of the Comptes-Rendus paper, the invention of the Bourbaki-daughter Betti and the printing of the wedding invitation card. In reality, all of these date from much later, the first two from the autumn of 1935, the final two no sooner than april 1939…

One thing I like about this book is the connection it makes with other disciplines, showing the influence of Bourbaki’s insistence on ‘structuralism’ in fields as different as philosophy, linguistics, anthropology and literary criticism. One example being Weil’s group-theoretic solution to the marriage-rules problem in tribes of Australian aborigines studied by Claude Lévi-Strauss, another the literary group Oulipo copying Bourbaki’s work-method.

Another interesting part is Aczel’s analysis of Bourbaki’s end. In the late 50ties, Grothendieck tried to convince his fellow Bourbakis to redo their work on the foundations of mathematics, changing these from set theory to category theory. He failed as others felt that the foundations had already been laid and there was no going back. Grothendieck left, and Bourbaki would gradually decline following its refusal to accept new methods. In Grothendieck’s own words (in “Promenade” 63, n. 78, as translated by Aczel) :

“Additionally, since the 1950s, the idea of structure has become passé, superseded by the influx of new ‘categorical’ methods in certain of the most dynamical areas of mathematics, such as topology or algebraic geometry. (Thus, the notion of ‘topos’ refuses to enter into the ‘Bourbaki sack’ os structures, decidedly already too full!) In making this decision, in full cognizance, not to engage in this revision, Bourbaki has itself renounced its initial ambition, which has been to furnish both the foundations and the basic language for all of modern mathematics.”

Finally, it is interesting to watch Aczel’s own transformation throughout the book, from slavishly copying the existing Weil-myths and pranks at the beginning of the book, to the following harsh criticism on Weil, towards the end (p. 209) :

“From other information in his autobiography, one gets the distinct impression that Weil was infatuated with the childish pranks of ‘inventing’ a person who never existed, creating for him false papers and a false identity, complete with a daughter, Betti, who even gets married, parents and relatives, and membership in a nonexistent Academy of Sciences of the nonexistent nation of Polvedia (sic).
Weil was so taken with these activities that he even listed, as his only honor by the time of his death ‘Member, Poldevian Academy of Sciences’. It seems that Weil could simply not go beyond these games: he could not grasp the deep significance and power of the organization he helped found. He was too close, and thus unable to see the great achievements Bourbaki was producing and to acknowledge and promote these achievements. Bourbaki changed the way we do mathematics, but Weil really saw only the pranks and the creation of a nonexistent person.”

Judging from my own reluctance to continue with the series on the Bourbaki code, an overdose reading about Weil’s life appears to have this effect on people…

Leave a Comment

On2 : Conway’s nim-arithmetics

Published January 26, 2009 by lievenlb

Last time we did recall Cantor’s addition and multiplication on ordinal numbers. Note that we can identify an ordinal number $\alpha $ with (the order type of) the set of all strictly smaller ordinals, that is, $\alpha = { \alpha’~:~\alpha’ < \alpha } $. Given two ordinals $\alpha $ and $\beta $ we will denote their Cantor-sums and products as $[ \alpha + \beta] $ and $[\alpha . \beta] $.

The reason for these square brackets is that John Conway constructed a well behaved nim-addition and nim-multiplication on all ordinals $\mathbf{On}_2 $ by imposing the ‘simplest’ rules which make $\mathbf{On}_2 $ into a field. By this we mean that, in order to define the addition $\alpha + \beta $ we must have constructed before all sums $\alpha’ + \beta $ and $\alpha + \beta’ $ with $\alpha’ < \alpha $ and $\beta’ < \beta $. If + is going to be a well-defined addition on $\mathbf{On}_2 $ clearly $\alpha + \beta $ cannot be equal to one of these previously constructed sums and the ‘simplicity rule’ asserts that we should take $\alpha+\beta $ the least ordinal different from all these sums $\alpha’+\beta $ and $\alpha+\beta’ $. In symbols, we define

$\alpha+ \beta = \mathbf{mex} { \alpha’+\beta,\alpha+ \beta’~|~\alpha’ < \alpha, \beta’ < \beta } $

where $\mathbf{mex} $ stands for ‘minimal excluded value’. If you’d ever played the game of Nim you will recognize this as the Nim-addition, at least when $\alpha $ and $\beta $ are finite ordinals (that is, natural numbers) (to nim-add two numbers n and m write them out in binary digits and add without carrying). Alternatively, the nim-sum n+m can be found applying the following two rules :

the nim-sum of a number of distinct 2-powers is their ordinary sum (e.g. $8+4+1=13 $, and,
the nim-sum of two equal numbers is 0.

So, all we have to do is to write numbers n and m as sums of two powers, scratch equal terms and add normally. For example, $13+7=(8+4+1)+(4+2+1)=8+2=10 $ (of course this is just digital sum without carry in disguise).

Here’s the beginning of the nim-addition table on ordinals. For example, to define $13+7 $ we have to look at all values in the first 7 entries of the row of 13 (that is, ${ 13,12,15,14,9,8,11 } $) and the first 13 entries in the column of 7 (that is, ${ 7,6,5,4,3,2,1,0,15,14,13,12,11 } $) and find the first number not included in these two sets (which is indeed $10 $).

In fact, the above two rules allow us to compute the nim-sum of any two ordinals. Recall from last time that every ordinal can be written uniquely as as a finite sum of (ordinal) 2-powers :
$\alpha = [2^{\alpha_0} + 2^{\alpha_1} + \ldots + 2^{\alpha_k}] $, so to determine the nim-sum $\alpha+\beta $ we write both ordinals as sums of ordinal 2-powers, delete powers appearing twice and take the Cantor ordinal sum of the remaining sum.

Nim-multiplication of ordinals is a bit more complicated. Here’s the definition as a minimal excluded value

$\alpha.\beta = \mathbf{mex} { \alpha’.\beta + \alpha.\beta’ – \alpha’.\beta’ } $

for all $\alpha’ < \alpha, \beta’ < \beta $. The rationale behind this being that both $\alpha-\alpha’ $ and $\beta – \beta’ $ are non-zero elements, so if $\mathbf{On}_2 $ is going to be a field under nim-multiplication, their product should be non-zero (and hence strictly greater than 0), that is, $~(\alpha-\alpha’).(\beta-\beta’) > 0 $. Rewriting this we get $\alpha.\beta > \alpha’.\beta+\alpha.\beta’-\alpha’.\beta’ $ and again the ‘simplicity rule’ asserts that $\alpha.\beta $ should be the least ordinal satisfying all these inequalities, leading to the $\mathbf{mex} $-definition above. The table gives the beginning of the nim-multiplication table for ordinals. For finite ordinals n and m there is a simple 2 line procedure to compute their nim-product, similar to the addition-rules mentioned before :

the nim-product of a number of distinct Fermat 2-powers (that is, numbers of the form $2^{2^n} $) is their ordinary product (for example, $16.4.2=128 $), and,
the square of a Fermat 2-power is its sesquimultiple (that is, the number obtained by multiplying with $1\frac{1}{2} $ in the ordinary sense). That is, $2^2=3,4^2=6,16^2=24,… $

Using these rules, associativity and distributivity and our addition rules it is now easy to work out the nim-multiplication $n.m $ : write out n and m as sums of (multiplications by 2-powers) of Fermat 2-powers and apply the rules. Here’s an example

$5.9=(4+1).(4.2+1)=4^2.2+4.2+4+1=6.2+8+4+1=(4+2).2+13=4.2+2^2+13=8+3+13=6 $

Clearly, we’d love to have a similar procedure to calculate the nim-product $\alpha.\beta $ of arbitrary ordinals, or at least those smaller than $\omega^{\omega^{\omega}} $ (recall that Conway proved that this ordinal is isomorphic to the algebraic closure $\overline{\mathbb{F}}_2 $ of the field of two elements). From now on we restrict to such ‘small’ ordinals and we introduce the following special elements :

$\kappa_{2^n} = [2^{2^{n-1}}] $ (these are the Fermat 2-powers) and for all primes $p > 2 $ we define
$\kappa_{p^n} = [\omega^{\omega^{k-1}.p^{n-1}}] $ where $k $ is the number of primes strictly smaller than $p $ (that is, for p=3 we have k=1, for p=5, k=2 etc.).

Again by associativity and distributivity we will be able to multiply two ordinals $< \omega^{\omega^{\omega}} $ if we know how to multiply a product

$[\omega^{\alpha}.2^{n_0}].[\omega^{\beta}.2^{m_0}] $ with $\alpha,\beta < [\omega^{\omega}] $ and $n_0,m_0 \in \mathbb{N} $.

Now, $\alpha $ can be written uniquely as $[\omega^t.n_t+\omega^{t-1}.n_{t-1}+\ldots+\omega.n_2 + n_1] $ with t and all $n_i $ natural numbers. Write each $n_k $ in base $p $ where $p $ is the $k+1 $-th prime number, that is, we have for $n_0,n_1,\ldots,n_t $ an expression

$n_k=[\sum_j p^j.m(j,k)] $ with $0 \leq m(j,k) < p $

The point of all this is that any of the special elements we want to multiply can be written as a unique expression as a decreasing product

$[\omega^{\alpha}.2^{n_0}] = [ \prod_q \kappa_q^m(q) ] $

where $q $ runs over all prime powers. The crucial fact now is that for this decreasing product we have a rule similar to addition of 2-powers, that is Conway-products coincide with the Cantor-products

$[ \prod_q \kappa_q^m(q) ] = \prod_q \kappa_q^m(q) $

But then, using associativity and commutativity of the Conway-product we can ‘nearly’ describe all products $[\omega^{\alpha}.2^{n_0}].[\omega^{\beta}.2^{m_0}] $. The remaining problem being that it may happen that for some q we will end up with an exponent $m(q)+m(q’)>p $. But this can be solved if we know how to take p-powers. The rules for this are as follows

$~(\kappa_{2^n})^2 = \kappa_{2^n} + \prod_{1 \leq i < n} \kappa_{2^i} $, for 2-powers, and,

$~(\kappa_{p^n})^p = \kappa_{p^{n-1}} $ for a prime $p > 2 $ and for $n \geq 2 $, and finally

$~(\kappa_p)^p = \alpha_p $ for a prime $p > 2 $, where $\alpha_p $ is the smallest ordinal $< \kappa_p $ which cannot be written as a p-power $\beta^p $ with $\beta < \kappa_p $. Summarizing : if we will be able to find these mysterious elements $\alpha_p $ for all prime numbers p, we are able to multiply in $[\omega^{\omega^{\omega}}]=\overline{\mathbb{F}}_2 $.

Let us determine the first one. We have that $\kappa_3 = \omega $ so we are looking for the smallest natural number $n < \omega $ which cannot be written in num-multiplication as $n=m^3 $ for $m < \omega $ (that is, also $m $ a natural number). Clearly $1=1^3 $ but what about 2? Can 2 be a third root of a natural number wrt. nim-multiplication? From the tabel above we see that 2 has order 3 whence its cube root must be an element of order 9. Now, the only finite ordinals that are subfields of $\mathbf{On}_2 $ are precisely the Fermat 2-powers, so if there is a finite cube root of 2, it must be contained in one of the finite fields $[2^{2^n}] $ (of which the mutiplicative group has order $2^{2^n}-1 $ and one easily shows that 9 cannot be a divisor of any of the numbers $2^{2^n}-1 $, that is, 2 doesn’t have a finte 3-th root in nim! Phrased differently, we found our first mystery number $\alpha_3 = 2 $. That is, we have the marvelous identity in nim-arithmetic

$\omega^3 = 2 $

Okay, so what is $\alpha_5 $? Well, we have $\kappa_5 = [\omega^{\omega}] $ and we have to look for the smallest ordinal which cannot be written as a 5-th root. By inspection of the finite nim-table we see that 1,2 and 3 have 5-th roots in $\omega $ but 4 does not! The reason being that 4 has order 15 (check in the finite field [16]) and 25 cannot divide any number of the form $2^{2^n}-1 $. That is, $\alpha_5=4 $ giving another crazy nim-identity

$~(\omega^{\omega})^5 = 4 $

And, surprises continue to pop up… Conway showed that $\alpha_7 = \omega+1 $ giving the nim-identity $~(\omega^{\omega^2})^7 = \omega+1 $. The proof of this already uses some clever finite field arguments. Because 7 doesn’t divide any number $2^{2^n}-1 $, none of the finite subfields $[2^{2^n}] $ contains a 7-th root of unity, so the 7-power map is injective whence surjective, so all finite ordinal have finite 7-th roots! That is, $\alpha_7 \geq \omega $. Because $\omega $ lies in a cubic extension of the finite field [4], the field generated by $\omega $ has 64 elements and so its multiplicative group is cyclic of order 63 and as $\omega $ has order 9, it must be a 7-th power in this field. But, as the only 7th powers in that field are precisely the powers of $\omega $ and by inspection $\omega+1 $ is not a 7-th power in that field (and hence also not in any field extension obtained by adjoining square, cube and fifth roots) so $\alpha_7=\omega +1 $.

Conway did stop at $\alpha_7 $ but I’ve always been intrigued by that one line in ONAG p.61 : “Hendrik Lenstra has computed $\alpha_p $ for $p \leq 43 $”. Next time we will see how Lenstra managed to do this and we will use sage to extend his list a bit further, including the first open case : $\alpha_{47}= \omega^{\omega^7}+1 $.

For an enjoyable video on all of this, see Conway’s MSRI lecture on Infinite Games. The nim-arithmetic part is towards the end of the lecture but watching the whole video is a genuine treat!

Leave a Comment

On2 : transfinite number hacking

Published January 8, 2009 by lievenlb

In ONAG, John Conway proves that the symmetric version of his recursive definition of addition and multiplcation on the surreal numbers make the class On of all Cantor’s ordinal numbers into an algebraically closed Field of characteristic two : On2 (pronounced ‘Onto’), and, in particular, he identifies a subfield
with the algebraic closure of the field of two elements. What makes all of this somewhat confusing is that Cantor had already defined a (badly behaving) addition, multiplication and exponentiation on ordinal numbers.

Over the last week I’ve been playing a bit with sage to prove a few exotic identities involving ordinal numbers. Here’s one of them ($\omega $ is the first infinite ordinal number, that is, $\omega={ 0,1,2,\ldots } $),

$~(\omega^{\omega^{13}})^{47} = \omega^{\omega^7} + 1 $

answering a question in Hendrik Lenstra’s paper Nim multiplication.

However, it will take us a couple of posts before we get there. Let’s begin by trying to explain what brought this on. On september 24th 2008 there was a meeting, intended for a general public, called a la rencontre des dechiffeurs, celebrating the 50th birthday of the IHES.

One of the speakers was Alain Connes and the official title of his talk was “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). Instead, he talked about a seemingly trivial problem : what is the algebraic closure of $\mathbb{F}_2 $, the field with two elements? My only information about the actual content of the talk comes from the following YouTube-blurb

Alain argues that we do not have a satisfactory description of $\overline{\mathbb{F}}_2 $, the algebraic closure of $\mathbb{F}_2 $. Naturally, it is the union (or rather, limit) of all finite fields $\mathbb{F}_{2^n} $, but, there are too many non-canonical choices to make here.

Recall that $\mathbb{F}_{2^k} $ is a subfield of $\mathbb{F}_{2^l} $ if and only if $k $ is a divisor of $l $ and so we would have to take the direct limit over the integers with respect to the divisibility relation… Of course, we can replace this by an increasing sequence of a selection of cofinal fields such as

$\mathbb{F}_{2^{1!}} \subset \mathbb{F}_{2^{2!}} \subset \mathbb{F}_{2^{3!}} \subset \ldots $

But then, there are several such suitable sequences! Another ambiguity comes from the description of $\mathbb{F}_{2^n} $. Clearly it is of the form $\mathbb{F}_2[x]/(f(x)) $ where $f(x) $ is a monic irreducible polynomial of degree $n $, but again, there are several such polynomials. An attempt to make a canonical choice of polynomial is to take the ‘first’ suitable one with respect to some natural ordering on the polynomials. This leads to the so called Conway polynomials.

Conway polynomials for the prime $2 $ have only been determined up to degree 400-something, so in the increasing sequence above we would already be stuck at the sixth term $\mathbb{F}_{2^{6!}} $…

So, what Alain Connes sets as a problem is to find another, more canonical, description of $\overline{\mathbb{F}}_2 $. The problem is not without real-life interest as most finite fields appearing in cryptography or coding theory are subfields of $\overline{\mathbb{F}}_2 $.

(My guess is that Alain originally wanted to talk about the action of the Galois group on the roots of unity, which would be the corresponding problem over the field with one element and would explain the title of the talk, but decided against it. If anyone knows what ‘coupling-problem’ he is referring to, please drop a comment.)

Surely, Connes is aware of the fact that there exists a nice canonical recursive construction of $\overline{\mathbb{F}}_2 $ due to John Conway, using Georg Cantor’s ordinal numbers.

In fact, in chapter 6 of his book On Numbers And Games, John Conway proves that the symmetric version of his recursive definition of addition and multiplcation on the surreal numbers make the class $\mathbf{On} $ of all Cantor’s ordinal numbers into an algebraically closed Field of characteristic two : $\mathbf{On}_2 $ (pronounced ‘Onto’), and, in particular, he identifies a subfield

$\overline{\mathbb{F}}_2 \simeq [ \omega^{\omega^{\omega}} ] $

with the algebraic closure of $\mathbb{F}_2 $. What makes all of this somewhat confusing is that Cantor had already defined a (badly behaving) addition, multiplication and exponentiation on ordinal numbers. To distinguish between the Cantor/Conway arithmetics, Conway (and later Lenstra) adopt the convention that any expression between square brackets refers to Cantor-arithmetic and un-squared ones to Conway’s. So, in the description of the algebraic closure just given $[ \omega^{\omega^{\omega}} ] $ is the ordinal defined by Cantor-exponentiation, whereas the exotic identity we started out with refers to Conway’s arithmetic on ordinal numbers.

Let’s recall briefly Cantor’s ordinal arithmetic. An ordinal number $\alpha $ is the order-type of a totally ordered set, that is, if there is an order preserving bijection between two totally ordered sets then they have the same ordinal number (or you might view $\alpha $ itself as a totally ordered set, namely the set of all strictly smaller ordinal numbers, so e.g. $0= \emptyset,1= { 0 },2={ 0,1 },\ldots $).

For two ordinals $\alpha $ and $\beta $, the addition $[\alpha + \beta ] $ is the order-type of the totally ordered set $\alpha \sqcup \beta $ (the disjoint union) ordered compatible with the total orders in $\alpha $ and $\beta $ and such that every element of $\beta $ is strictly greater than any element from $\alpha $. Observe that this definition depends on the order of the two factors. For example,$ [1 + \omega] = \omega $ as there is an order preserving bijection ${ \tilde{0},0,1,2,\ldots } \rightarrow { 0,1,2,3,\ldots } $ by $\tilde{0} \mapsto 0,n \mapsto n+1 $. However, $\omega \not= [\omega + 1] $ as there can be no order preserving bijection ${ 0,1,2,\ldots } \rightarrow { 0,1,2,\ldots,0_{max} } $ as the first set has no maximal element whereas the second one does. So, Cantor’s addition has the bad property that it may be that $[\alpha + \beta] \not= [\beta + \alpha] $.

The Cantor-multiplication $ \alpha . \beta $ is the order-type of the product-set $\alpha \times \beta $ ordered via the last differing coordinate. Again, this product has the bad property that it may happen that $[\alpha . \beta] \not= [\beta . \alpha] $ (for example $[2 . \omega ] \not=[ \omega . 2 ] $). Finally, the exponential $\beta^{\alpha} $ is the order type of the set of all maps $f~:~\alpha \rightarrow \beta $ such that $f(a) \not=0 $ for only finitely many $a \in \alpha $, and ordered via the last differing function-value.

Cantor’s arithmetic allows normal-forms for ordinal numbers. More precisely, with respect to any ordinal number $\gamma \geq 2 $, every ordinal number $\alpha \geq 1 $ has a unique expression as

$\alpha = [ \gamma^{\alpha_0}.\eta_0 + \gamma^{\alpha_1}.\eta_1 + \ldots + \gamma^{\alpha_m}.\eta_m] $

for some natural number $m $ and such that $\alpha \geq \alpha_0 > \alpha_1 > \ldots > \alpha_m \geq 0 $ and all $1 \leq \eta_i < \gamma $. In particular, taking the special cases $\gamma = 2 $ and $\gamma = \omega $, we have the following two canonical forms for any ordinal number $\alpha $

$[ 2^{\alpha_0} + 2^{\alpha_1} + \ldots + 2^{\alpha_m}] = \alpha = [ \omega^{\beta_0}.n_0 + \omega^{\beta_1}.n_1 + \ldots + \omega^{\beta_k}.n_k] $

with $m,k,n_i $ natural numbers and $\alpha \geq \alpha_0 > \alpha_1 > \ldots > \alpha_m \geq 0 $ and $\alpha \geq \beta_0 > \beta_1 > \ldots > \beta_k \geq 0 $. Both canonical forms will be important when we consider the (better behaved) Conway-arithmetic on $\mathbf{On}_2 $, next time.

One Comment

5 years blogging

Published December 31, 2008 by lievenlb

Here’s a 5 move game from $\mathbb{C} $, the complex numbers game, annotated by Hendrik Lenstra in Nim multiplication.

$\begin{matrix} & \text{White} & \text{Black} \\ 1. & 3-2i & { 3_{\mathbb{R}} } \\ 2. & 3_{\mathbb{R}} & (22/7)_{\mathbb{Q}} \\ 3. & (-44_{\mathbb{Z}},-14_{\mathbb{Z}})? & { -44_{\mathbb{Z}} } \\ 4. & -44_{\mathbb{Z}} & ( 0_{\mathbb{N}},44_{\mathbb{N}} )! \\ 5. & \text{Resigns} & \\ \end{matrix} $

He writes : “The following 5 comments will make the rules clear.

1 : White selected a complex numbers. Black knows that $\mathbb{C} = \mathbb{R} \times \mathbb{R} $ by $a+bi = (a,b) $, and remembers Kuratowski’s definition of an ordered pair: $~(x,y) = { { x }, { x,y } } $. Thus black must choose an element of ${ { 3_{\mathbb{R}} }, { 3_{\mathbb{R}},-2_{\mathbb{R}} } } $. The index $\mathbb{R} $ here, and later $\mathbb{Q},\mathbb{Z} $ and $\mathbb{N} $, serve to distinguish between real numbers, rational numbers, integers and natural numbers usually denoted by the same symbol. Black’s move leaves White a minimum of choice, but it is not the best one.

2 : White has no choice. The Dedekind definition of $\mathbb{R} $ which the players agreed upon identifies a real number with the set of all strictly larger rational numbers; so Black’s move is legal.

3 : A rational number is an equivalence class of pairs of integers $~(a,b) $ with $b \not= 0 $; here $~(a,b) $ represents the rational number $a/b $. The question mark denotes that White’s move is a bad one.

4 : The pair $~(a,b) $ of natural numbers represents the integer $a-b $. Black’s move is the only winning one.

5 : White resigns, since he can choose between ${ 0_{\mathbb{N}} } $ and ${ 0_{\mathbb{N}},44_{\mathbb{N}} } $. In both cases Black will reply by $0_{\mathbb{N}} $, which is the empty set” (and so wins because White has no move left).

These rules make it clear what we mean by the natural numbers $\mathbb{N} $ game, the $\mathbb{Z} $-game and the $\mathbb{Q} $ and $\mathbb{R} $ games. A sum of games is defined as usual (players are allowed to move in exactly one of the component games).

Here’s a 5 term exercise from Lenstra’s paper : Determine the unique winning move in the game $\mathbb{N} + \mathbb{Z} + \mathbb{Q} + \mathbb{R} + \mathbb{C} $

It will take you less than 5 minutes to solve this riddle. Some of the other ‘exercises’ in Lenstra’s paper may take you a lot longer, if not forever…

Exactly 5 years ago I wrote : “As it is probably better to run years behind than to stand eternally still, I’ll try out how much of a blogger I am in 2004.”

5 months ago this became : “from january 1st 2009, I’ll be moving out of here. I will leave the neverendingbooks-site intact for some time to come, so there is no need for you to start archiving it en masse, yet.”

5 minutes before the deadline, this will be my last post….

of 2008

less entropy in 2009!

5 Comments

sporadic simple games

Published July 13, 2008 by lievenlb

About a year ago I did a series of posts on games associated to the Mathieu sporadic group $M_{12} $, starting with a post on Conway’s puzzle M(13), and, continuing with a discussion of mathematical blackjack. The idea at the time was to write a book for a general audience, as discussed at the start of the M(13)-post, ending with a series of new challenging mathematical games. I asked : “What kind of puzzles should we promote for mathematical thinking to have a fighting chance to survive in the near future?”

Now, Scientific American has (no doubt independently) taken up this lead. Their July 2008 issue features the article Rubik’s Cube Inspired Puzzles Demonstrate Math’s “Simple Groups” written by Igor Kriz and Paul Siegel.

By far the nicest thing about this article is that it comes with three online games based on the sporadic simple groups, the Mathieu groups $M_{12} $, $M_{24} $ and the Conway group $.0 $.

the M(12) game

Scrambles to an arbitrary permutation in $M_{12} $ and need to use the two generators $INVERT=(1,12)(2,11)(3,10)(4,9)(5,8)(6,7) $ and $MERGE=(2,12,7,4,11,6,10,8,9,5,3) $ to return to starting position.

Here is the help-screen :

They promise the solution by july 27th, but a few-line GAP-program cracks the puzzle instantly.

the M(24) game

Similar in nature, again using two generators of $M_{24} $. GAP-solution as before.

This time, they offer this help-screen :

the .0 game

Their most original game is based on Conway’s $.0 $ (dotto) group. Unfortunately, they offer only a Windows-executable version, so I had to install Bootcamp and struggle a bit with taking screenshots on a MacBook to show you the game’s starting position :

Dotto:

Dotto, our final puzzle, represents the Conway group Co0, published in 1968 by mathematician John H. Conway of Princeton University. Co0 contains the sporadic simple group Co1 and has exactly twice as many members as Co1. Conway is too modest to name Co0 after himself, so he denotes the group “.0” (hence the pronunciation “dotto”).

In Dotto, there are four moves. This puzzle includes the M24 puzzle. Look at the yellow/blue row in the bottom. This is, in fact, M24, but the numbers are arranged in a row instead of a circle. The R move is the “circle rotation to the right”: the column above the number 0 stays put, but the column above the number 1 moves to the column over the number 2 etc. up to the column over the number 23, which moves to the column over the number 1. You may also click on a column number and then on another column number in the bottom row, and the “circle rotation” moving the first column to the second occurs. The M move is the switch, in each group of 4 columns separated by vertical lines (called tetrads) the “yellow” columns switch and the “blue” columns switch. The sign change move (S) changes signs of the first 8 columns (first two tetrads). The tetrad move (T) is the most complicated: Subtract in each row from each tetrad 1/2 times the sum of the numbers in that tetrad. Then in addition to that, reverse the signs of the columns in the first tetrad.

Strategy hints: Notice that the sum of squares of the numbers in each row doesn’t change. (This sum of squares is 64 in the first row, 32 in every other row.) If you manage to get an “8”in the first row, you have almost reduced the game to M24 except those signs. To have the original position, signs of all numbers on the diagonal must be +. Hint on signs: if the only thing wrong are signs on the diagonal, and only 8 signs are wrong, those 8 columns can be moved to the first 8 columns by using only the M24 moves (M,R).

Leave a Comment

Surreal numbers & chess

Published April 8, 2008 by lievenlb

Most chess programs are able to give a numerical evaluation of a position. For example, the position below is considered to be worth +8.7 with white to move, and, -0.7 with black to move (by a certain program). But, if one applies combinatorial game theory as in John Conway’s ONAG and the Berlekamp-Conway-Guy masterpiece Winning Ways for your Mathematical Plays it will turn out that the position can be proved to have an infinitesimal advantage for white…

So, what do we mean by this? First some basic rules of combinatorial game theory. To start, we evaluate a position without knowing which player has the move. A zero-game is by definition a position in which neither player has a good move, that is, any move by either player quickly leads to losing the game. Hence, a zero-game is a position in which the second player to move wins.

What is the chess-equivalent of a zero-position game? A position in which neither player has a good move is called a Mutual Zugzwang in chess literature. An example is given by the above position, if we restrict attention only to the 4 pieces in the upper right-hand corner and forget the rest. We don’t know who has the move, but, White cannot move at all and Black cannot move the King or Bishop without losing the Bishop and allowing White to promote the pawn and win quickly. In CGT-parlance, the upper-right position has value $\{ \emptyset | \emptyset \} = 0 $ where the left options denote the White moves and the right options the Black moves.

All other values are determined by recursion. For example, consider a position in which White has just one move left before the sitution is again a Mutual Zugzwang, and, Black has no good move whatsoever. After white’s move, the position will again be a zero-position and Black has no options, so the value of this position would be denoted by $\{ 0 | \emptyset \} $ and we call the value of this position to be $+1 $. Similarly, if white has no options and black has one final move to make, the position would be considered to have value $\{ \emptyset | 0 \}= -1 $.

Clearly, these are just the three easiest game-values to have and the real kick comes further down the road when one can prove by recursion that some games have non-integer values (such as $\{ 0 | 1 \} = \frac{1}{2} $ for a position in which white has one move to get to a mutual zugzwang and black has a move leading to a position of value $+1 $ (defined as before)), or non-number values such as $\ast = \{ 0 | 0 \} $ where both white and black’s best move is to get to a mutal zugzwang. Game-values such as $\ast $ are called fuzzy (or confused with zero) and are defined by the property that the first player to move wins.

Similarly, positive game-values are those positions where White wins, independent of who has the move and negatives are those that Black wins. There is a whole menagery of game-values and the WinningWays-booklets give an example based introduction to this fascinating theory.

Brief as this introduction was, it will allow us to determine the exact value of the position in the above diagram. We know already that we can forget about the right-hand upper corner (as this is a zero-position) and concentrate attention to the left-hand side of the board.

It is easy to see that neither Knight can move without loosing quickly, nor can the pawns on a5 and b7. That is, white has just 2 options : either c3-c4 (quickly loosing after d5xc4 2. d3xc4,d4-d3 3. Nc1xd3,Na1-b3) or, and this is the only valid option c3xd4 leading to the position on the left below. Black has only one valid move : d4xc3 leading to the position on the right below.

Clearly, the left-diagram has value 0 as it is a mutual Zugzwang. The position on the right takes a moment’s thought : White has one move left d3-d4 leading to a 0-position, whereas black has one move d5-d4 leading to a position of value -1 (as black still has one move left d6-d5, whereas white has none). That is, the CGT-value of the right-hand position is $\{ 0 | -1 \} $ and therefore, the value of the starting position is precisely equal to

\[ \{ 0 | \{ 0 | -1 \} \} = +_{1} \]

(called tiny-one among ONAGers)

It can be shown that $+_1 $ has a positive value (that is, White wins independently of who has the first move) but smaller than any positive number-valued games!

Noam Elkies has written a beautiful paper On numbers and endgames: Combinatorial game theory in chess endgames containing many interesting examples (the example above is an adaptation of his diagram9).

2 Comments

exotic chess positions (1)

Published March 4, 2008 by lievenlb

Ever tried a chess problem like : White to move, mate in two! Of course you have, and these are pretty easy to solve : you only have to work through the finite list of white first moves and decide whether or not black has a move left preventing mate on the next white move. This is even a (non-optimal) fool-proof algorithm to find the solution to this kind of chess problems. Right?

Wrong! There exist concrete positions, provable mate in two in which it is NOT possible to determine the winning first move for white! So, what’s wrong with the argument above? We did assume that, given the position, it is possible to determine all legal moves for the two players. So?

Well, some moves are legal only depending on the history of the game. For example, you are only allowed to do a castling if your king nor your rook made a prior move. Further, you can only make an en-passant-capture on the next move.

But surely all this is just theoretical? No-one ever constructed a provable 2-mate with impossible winning move. Wrong again. The logician Raymond Smullyan did precisely that in his retro-chess puzzle book Chess mysteries of Sherlock Holmes. Here’s the position :

Presumably every chess player goes for the mate : 1. Kf5-e6 2. g7-g8D But what if black counters your first move with a castling 1. … 0-0-0 ? Surely he isnt allowed to do this. Why not, is there any clue in the position to prove that either the king or the rook must have moved before?

Well, what was black previous move then? It cannot be the pawn move e6-e5 as before that move the white King would be in check, so what was it? Just one possibility left : it must have been e7-e5.

This offers then another winning strategy for white, as white can capture en-passant. 1. d5xe6 e.p. and then if black castles 1. … 0.0.0, 2. b6-b7 or is black does any other move : 2. g7-g8.

Hence, whatever the games’ history, white has a mate in two! However, looking ONLY at the given position, it is impossible for him to judge whether Kf5-e6 will do the trick!

Anyone seen similar constructions?

UPDATE

According to the wikipedia page on Retrograde chess analysis, the Smullyan-idea is an adaptation of a much older problem due to W. Langstaff in the Chess Amateur of 1922. Here’s the situation (the solution is the same as above)

4 Comments

Writing & Blogging

Published February 27, 2008 by lievenlb

Terry Tao is reworking some of his better blogposts into a book, to be published by the AMS (here’s a preliminary version of the book “What’s New?”)

After some thought, I decided not to transcribe all of my posts from last year (there are 93 of them!), but instead to restrict attention to those articles which (a) have significant mathematical content, (b) are not announcements of material that will be published elsewhere, and (c) are not primarily based on a talk given by someone else. As it turns out, this still leaves about 33 articles from 2007, leading to a decent-sized book of a couple hundred pages in length.

If you have a blog and want to turn it into a LaTeX-book, there’s no need to transcribe or copy every single post, thanks to the WPTeX tool. Note that this is NOT a WP-plugin, but a (simple at that) php-program which turns all posts into a bookcontent.tex file. This file can then be edited further into a proper book.

Unfortunately, the present version chokes on LaTeXrender-code (which is easy enough to solve doing a global ‘find-and-replace’ of the tex-tags by dollar-signs) but worse, on Markdown-code… But then, someone fluent in php-regex will have no problems extending the libs/functions.php file (I hope…).

At the moment I’m considering turning the Mathieu-games-posts into a booklet. A possible title might be Mathieumatical Games. Rereading them (and other posts) I regret to be such an impatient blogger. Often I’m interested in something and start writing posts about it without knowing where or when I’ll land. This makes my posts a lot harder to get through than they might have been, if I would blog only after having digested the material myself… Typical recent examples are the tori-crypto-posts and the Bost-Connes algebra posts.

So, I still have a lot to learn from other bloggers I admire, such as Jennifer Ouellette who maintains the Coctail Party Physics blog. At the moment, Jennifer is resident blogger-journalist at the Kavli Institute where she is running a “Journal Club” workshop giving ideas on how to write better about science.

But the KITP is also committed to fostering scientific communication. That’s where I come in. Each Friday through April 26th, I’ll be presiding over a “Journal Club” meeting focusing on some aspect of communicating science.

Her most recent talk was entitled To Blog or Not to Blog? That is the Question and you can find the slides as well as a QuickTime movie of her talk. They even plan to set up a blog for the participants of the workshop. I will surely follow the rest of her course with keen interest!

3 Comments