Tag: Conway

Mathieu’s blackjack (2)

Published July 17, 2007 by lievenlb

(continued from part one). Take twelve cards and give them values 0,1,2,…,11 (for example, take the jack to have value 11 and the queen to have value 0). The hexads are 6-tuples of cards having the following properties. When we star their values by the scheme on the left below and write a 0 below a column if it has just one star at the first row or two stars on rows two and three (a + if the unique star is at the first row or two stars in the other columns, and a – if the unique star in on the second row or two stars in rows one and two) or a ? if the column has 3 or 0 stars, we get a tetracodeword where we are allowed to replace a ? by any digit. Moreover, we want that the stars are NOT distributed over the four columns such that all of the possible outcomes 0,1,2,3 appear once. For example, the card-pile { queen, 3, 4, 7, 9, jack } is an hexad as is indicated on the right below and has column-distributions (1,1,2,2).

$\begin{array}{|c|ccc|} \hline 6 & 3 & 0 & 9 \\ 5 & 2 & 7 & 10 \\ 4 & 1 & 8 & 11 \\ \hline & & & \end{array} $ $\begin{array}{|c|ccc|} \hline & \ast & \ast & \ast \\ & & \ast & \\ \ast & & & \ast \\ \hline – & 0 & – & + \end{array} $

The hexads form a Steiner-system S(5,6,12), meaning that every 5-pile of cards is part of a unique hexad. The permutations on these twelve cards, having the property that they send every hexad to another hexad, form the sporadic simple group $M_{12} $, the _Mathieu group_ of order 95040. For now, we assume these facts and deduce from them the Conway-Ryba winning strategy for Mathieu’s blackjack : the hexads are exactly the winning positions and from a non-hexad pile of total value at least 21 there is always a legal (that is, total value decreasing) move to an hexad by replacing one card in the pile by a card from the complement.

It seems that the first proof of this strategy consisted in calculating the Grundy values of all 905 legal positions in Mathieu’s blackjack. Later Joseph Kahane and Alex Ryba gave a more conceptual proof, that we will try to understand.

Take a non-hexad 6-pile such that the total value of its cards is at least 21, then removing any one of the six cards gives a 5-pile and is by the Steiner-property contained in a unique hexad. Hence we get 6 different hexads replacing one card from the non-hexad pile by a card not contained in it. We claim that at least one of these operations is a legal move, meaning that the total value of the cards decreases. Let us call a counterexample a misfit and record some of its properties until we can prove its non-existence.

A misfit is a non-hexad with total value at least 21 such that all 6 hexads, obtained from it by replacing one card by a card from its complement, have greater total value

A misfit must contain the queen-card. If not, we could get an hexad replacing one misfit-card (value > 0) by the queen (value zero) so this would be a legal move. Further, the misfit cannot contain the jack-card for otherwise replacing it by a lower-valued card to obtain an hexad is a legal move.

A misfit contains at least three cards from {queen,1,2,3,4}. If not, three of these cards are the replacements of misfit-cards to get an hexad, but then at least one of the replaced cards has a greater value than the replacement, giving a legal move to an hexad.

A misfit contains more than three cards from {queen=0, 1,2,3,4}. Assume there are precisely three $\{ c_1,c_2,c_3 \} $ from this set, then the complement of the misfit in the hexad {queen,1,2,3,4,jack} consists of three elements $\{ d_1,d_2,d_3 \} $ (a misfit cannot contain the jack). The two leftmost columns of the value-scheme (left above) form the hexad {1,2,3,4,5,6} and because the Mathieu group acts 5-transitively there is an element of $M_{12} $ taking $\{ 0,1,2,3,4,11 \} \rightarrow \{ 1,2,3,4,5,6 \} $ and we may even assume that it takes $\{ c_1,c_2,c_3 \} \rightarrow \{ 4,5,6 \} $. But then, in the new value-scheme (determined by that $M_{12} $-element) the two leftmost columns of the misfit look like

$\begin{array}{|c|ccc|} \hline \ast & . & ? & ? \\ \ast & . & ? & ? \\ \ast & . & ? & ? \\ \hline ? & ? & & \end{array} $

and the column-distribution of the misfit must be either (3,0,2,1) or (3,0,1,2) (it cannot be (3,0,3,0) or (3,0,0,3) otherwise the (image of the) misfit would be an hexad). Let {i,j} be the two misfit-values in the 2-starred column. Replacing either of them to get an hexad must have the replacement lying in the second column (in order to get a valid column distribution (3,1,1,1)). Now, the second column consists of two small values (from {0,1,2,3,4}) and the large jack-value (11). So, at least one of {i,j} is replaced by a smaller valued card to get an hexad, which cannot happen by the misfit-property.

Now, if the misfit shares four cards with {queen,1,2,3,4} then it cannot contain the 10-card. Otherwise, the replacement to get an hexad of the 10-card must be the 11-card (by the misfit-property) but then there would be another hexads containing five cards from {queen,0,1,2,3,jack} which cannot happen by the Steiner-property. Right, let’s summarize what we know so far about our misfit. Its value-scheme looks like

$\begin{array}{|c|ccc|} \hline 6 & III & \ast & 9 \\ 5 & II & 7 & . \\ IV & I & 8 & . \\ \hline & & & \end{array} $ and it must contain three of the four Romans. At this point Kahane and Ryba claim that the two remaining cards (apart from the queen and the three romans) must be such that there is exactly one from {5,6} and exactly one from {7,8,9}. They argue this follows from duality where the dual pile of a card-pile $\{ x_1,x_2,\ldots,x_6 \} $ is the pile $\{ 11-x_1,11-x_2,\ldots,11-x_6 \} $. This duality acts on the hexads as the permutation $~(0,11)(1,10)(2,9)(3,8)(4,7)(5,6) \in M_{12} $. Still, it is unclear to me how they deduce from it the above claim (lines 13-15 of page 4 of their paper). I’d better have some coffee and work around this (to be continued…)

If you want to play around a bit with hexads and the blackjack game, you’d better first download SAGE (if you haven’t done so already) and then get David Joyner’s hexad.sage file and put it in a folder under your sage installation (David suggests ‘spam’ himself…). You can load the routines into sage by typing from the sage-prompt attach ‘spam/hexad.sage’. Now, you can find the hexad from a 5-pile via the command find_hexad([a1,a2,a3,a4,a5],minimog_shuffle) and you can get the winning move for a blackjack-position via blackjack_move([a1,a2,a3,a4,a5,a6],minimog_shuffle). More details are in the Joyner-Casey(Luers) paper referenced last time.

Reference

Joseph Kahane and Alexander J. Ryba, ‘The hexad game‘

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Mathieu’s blackjack (1)

Published July 16, 2007 by lievenlb

Mathieu’s blackjack is a two-person combinatorial game played with 12 cards of values 0,1,2,…,11. For example take from any deck the numbered cards together with the jack (value 11) and the queen (value 0) (btw. if you find this PI by all means replace the queen by a zero-valued king). Shuffle the cards and divide them into two piles of 6 cards (all of them face up on the table) : the main-pile and the other-pile. The rules of the game are

players alternate moves
a move consists of exchanging a card of the main-pile with a lower-valued card from the other-pile
the player whose move makes the sum of all cards in the main-pile under 21 looses the game

For example, the starting main-pile might consist of the six cards

This pile has total value 3+4+7+8+9+11=42. A move replaces one of these cards with a lowever vlued one not in the pile. So for example, replacing 8 with 5 or 1 or 2 or the queen are all valid moves. A winning move from this situation is for example replacing 8 by the queen (value 0) decreasing the value from 42 to 34

But there are otthers, such as replacing 11 by 5, 9 by 1 or 4 by 2. To win this game you need to know the secrets of the tetracode and the MINIMOG.

The tetracode is a one-error correcting code consisting of the following nine words of length four over $\mathbb{F}_3 = { 0,+,- } $

$~\begin{matrix} 0~0 0 0 & 0~+ + + & 0~- – – \\ +~0 + – & +~+ – 0 & +~- 0 + \\ -~0 – + & -~+ 0 – & -~- + 0 \end{matrix} $

The first element (which is slightly offset from the rest) is the slope s of the words, and the other three digits cyclically increase by s (in the field $\mathbb{F}_3 $). Because the Hamming-distance is 3 (the minimal number of different digits between two codewords), the tetracode can correct one error, meaning that if at most one of the four digits gets distorted by the channel one can detect and correct this. For example, if you would receive the word $+~++- $ (which is not a codeword) and if you would know that at most one digits went wrong, you can deduce that the word $+~0+- $ was sent. Thus, one can solve the 4-problem for the tetracode : correctt a tetracodeword given all 4 of its digits, one of which may be mistaken.

Another easy puzzle is the 2-problem for the tertracode : complete a tetracodeword from any 2 of its digits. For example, given the incomplete word $?~?0+ $ you can decide that the slope should be + and hence that the complete word must be $+~-0+ $.

We will use the MINIMOG here as a way to record the blackjack-position. It is a $4 \times 3 $ array where the 12 boxes correspond to the card-values by the following scheme

$\begin{array}{|c|ccc|} \hline 6 & 3 & 0 & 9 \\ 5 & 2 & 7 & 10 \\ 4 & 1 & 8 & 11 \\ \hline \end{array} $

and given a blackjack-position we place a star in the corresponding box, so the above start-position (resp. after the first move) corresponds to

$~\begin{array}{|c|ccc|} \hline & \ast & & \ast \\ & & \ast & \\ \ast & & \ast & \ast \\ \hline – & 0 & 0 & + \end{array}~ $ respectively $\begin{array}{|c|ccc|} \hline & \ast & \ast & \ast \\ & & \ast & \\ \ast & & & \ast \\ \hline – & 0 & – & + \end{array} $

In the final row we have added elements of $\mathbb{F}_3 $ indicating wher ethe stars are placed in that column (if there is just one star, we write the row-number of the star (ordered 0,+,- from top to bottom), if there are two stars we record the row-number of the empty spot. If we would have three or no stars in a column we would record a wild-card character : ?

Observe that the final row of the start position is $-~00+ $ which is NOT a tetracodeword, whereas that of the winning position $-~0-+ $ IS a tetracodeword! This is the essence of the _Conway-Ryba winning strategy_ for Mathieu’s blackjack. There are precisely 132 winning positions forming the Steiner-system S(5,6,12). By an S(5,6,12) we mean a collection of 6-element subsets (our card-piles) from a 12-element set (the deck minus the king) having the amazing property that for EVERY 5-tuple from the 12-set there is a UNIQUE 6-element set containing this 5-tuple. Hence, there are exactly $\begin{pmatrix} 12 \\\ 5 \end{pmatrix}/6 = 132 $ elements in a Steiner S(5,6,12) system. The winning positions are exactly those MINOMOGs having 6 stars such that the final row is a tetracodeword (or can be extended to a tetracodeword replacing the wildcards ? by suitable digits) and such that the distribution of the stars over the columns is NOT (3,2,1,0) in any order.

Provided the given blackjack-position is not in this Steiner-system (and there is only a 1/7 chance that it is), the strategy is clear : remove one of the stars to get a 5-tuple and determine the unique 6-set of the Steiner-system containing this 5-tuple. If the required extra star corresponds to a value less than the removed star you have a legal and winning move (if not, repeat this for another star). Finding these winning positions means solving 2- and 4-problems for the tetracode. _Another time_ we will say more about this Steiner system and indicate the relation with the Mathieu group $M_{12} $.

References

J.H. Conway and N.J.A. Sloane, ‘The Golay codes and the Mathieu groups’, chp. 10 of “Sphere Packings, Lattices and Groups“

David Joyner and Ann Casey-Luers, ‘Kittens, S(5,6,12) and Mathematical blackjack in SAGE‘

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The Mathieu groupoid (1)

Published June 20, 2007 by lievenlb

Conway’s puzzle M(13) is a variation on the 15-puzzle played with the 13 points in the projective plane $\mathbb{P}^2(\mathbb{F}_3) $. The desired position is given on the left where all the counters are placed at at the points having that label (the point corresponding to the hole in the drawing has label 0). A typical move consists in choosing a line in the plane going through the point where the hole is, choose one of the three remaining points on this line and interchange the counter on it for the hole while at the same time interchanging the counters on the other two points. In the drawing on the left, lines correspond to the little-strokes on the circle and edges describe which points lie on which lines. For example, if we want to move counter 5 to the hole we notice that both of them lie on the line represented by the stroke just to the right of the hole and this line contains also the two points with counters 1 and 11, so we have to replace these two counters too in making a move. Today we will describe the groupoid corresponding to this slide-puzzle so if you want to read on, it is best to play a bit with Sebastian Egner’s M(13) Java Applet to see the puzzle in action (and to use it to verify the claims made below). Clicking on a counter performs the move taking the counter to the hole.

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The 15-puzzle groupoid (1)

Published June 17, 2007 by lievenlb

Before we go deeper into Conway’s M(13) puzzle, let us consider a more commonly known sliding puzzle: the 15-puzzle. A heated discussion went on a couple of years ago at sci-physics-research, starting with this message. Lubos Motl argued that group-theory is sufficient to analyze the problem and that there is no reason to resort to groupoids (‘The human(oids) who like groupoids…’ and other goodies, in pre-blog but vintage Motl-speak) whereas ‘Jason’ defended his viewpoint that a groupoid is the natural symmetry for this puzzle.

I’m mostly with Lubos on this. All relevant calculations are done in the symmetric group $S_{16} $ and (easy) grouptheoretic results such as the distinction between even and odd permutations or the generation of the alternating groups really crack the puzzle. At the same time, if one wants to present this example in class, one has to be pretty careful to avoid confusion between permutations encoding positions and those corresponding to slide-moves. In making such a careful analysis, one is bound to come up with a structure which isn’t a group, but is precisely what some people prefer to call a groupoid (if not a 2-group…).

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Conway’s puzzle M(13)

Published June 16, 2007 by lievenlb

Recently, I’ve been playing with the idea of writing a book for the general public. Its title is still unclear to me (though an idea might be “The disposable science”, better suggestions are of course wellcome) but I’ve fixed the subtitle as “Mathematics’ puzzling fall from grace”. The book’s concept is simple : I would consider the mathematical puzzles creating an hype over the last three centuries : the 14-15 puzzle for the 19th century, Rubik’s cube for the 20th century and, of course, Sudoku for the present century.

For each puzzle, I would describe its origin, the mathematics involved and how it can be used to solve the puzzle and, finally, what the differing quality of these puzzles tells us about mathematics’ changing standing in society over the period. Needless to say, the subtitle already gives away my point of view. The final part of the book would then be more optimistic. What kind of puzzles should we promote for mathematical thinking to have a fighting chance to survive in the near future?

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The miracle of 163

Published June 14, 2007 by lievenlb

On page 227 of Symmetry and the Monster, Mark Ronan tells the story of Conway and Norton computing the number of independent _mini j-functions_ (McKay-Thompson series) arising from the Moonshine module. There are 194 distinct characters of the monster (btw. see the background picture for the first page of the character table as given in the Atlas), but some of them give the same series reducing the number of series to 171. But, these are not all linearly independent. Mark Ronan writes :

“Conway recalls that, ‘As we went down into the 160s, I said let’s guess what number we will reach.’ They guessed it would be 163 – which has a very special property in number theory – and it was!
There is no explanation for this. We don’t know whether it is merely a coincidence, or something more. The special property of 163 in number theory has intruiging consequences, among which is the fact that
$e^{\pi \sqrt{163}} = 262537412640768743.99999999999925… $
is very close to being a whole number.”

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group think 2

Published May 23, 2007 by lievenlb

Someone from down under commented on the group think post yesterday :

Nice post, but I might humbly suggest that there’s not much in it that anyone would disagree with. I’d be interested in your thoughts on the following:
1. While many doomed research programs have the seven symptoms you mention, so do some very promising research programs. For instance, you could argue that Grothendieck’s school did. While it did eventually explode, it remains one of the high points in the history of mathematics. But at the time, many people (Mordell, Siegel) thought it was all garbage. Indeed there was even doubt into the late eighties. Is there anything close to a necessary and sufficient condition that an outsider can use to get some idea of whether a research group is doing work that will last?
2. Pretty much everyone thinks they’re underappreciated. It’s easy to advise them to pull a Perelman because it costs you nothing. But most unappreciated researchers are unappreciated for a good reason. How can unappreciated researchers decide whether their ideas really are good or not before spending ten years of their lives finding out?

First the easy bit : the ‘do a Perelman’-sentence seems to have been misread by several people (probably due to my inadequate English). I never suggested ‘unappreciated researchers’ to pull a Perelman but rather the key figures in seemingly successful groups making outrageous claims for power-reasons. Here is what I actually wrote

An aspect of these groupthinking science-groups that worries me most of all is their making of exagerated claims to potential applications, not supported (yet) by solid proof. Short-time effect may be to attract more people to the subject and to keep doubting followers on board, but in the long term (at least if the claimed results remain out of reach) this will destroy the subject itself (and, sadly enough, also closeby subjects making no outrageous claims!). My advice to people making such claims is : do a Perelman! Rather than doing a PR-job, devote yourself for as long as it takes to prove your hopes, somewhere in splendid isolation and come back victoriously. I have a spare set of keys if you are in search for the perfect location!
Before I will try to answer both questions let me stress that this is just my personal opinion to which I attach no particular value. Sure, I will forget things and will over-stress others. You can always leave a comment if you think I did, but I will not enter a discussion. I think it is important that a person develops his or her own scientific ethic and tries to live by it. 1. Is there anything close to a necessary and sufficient condition that an outsider can use to get some idea of whether a research group is doing work that will last? Clearly, the short answer to this is “no”. Still, there are some signs an outsider might pick up to form an opinion. – What is the average age of the leading people in the group? (the lower, the better) – The percentage of talks given by young people at a typical conference of the group (the higher, the better) – The part of a typical talk in the subject spend setting up notation, referring to previous results and namedropping (the lower, the better) – The number of group-outsiders invited to speak at a typical conference (the higher, the better) – The number of self-references in a typical paper (the lower, the better) – The number of publications by the group in non-group controlled journals (the higher, the better) – The number of group-controlled journals (the lower, the better) – The readablity of survey papers and textbooks on the subject (the higher, the better) – The complexity of motivating examples not covered by competing theories (the lower, the better) – The number of subject-gurus (the higher, the better) – The number of phd-students per guru (the lower, the better) – The number of main open problems (the higher, the better) – The Erdoes-like number of a typical group-member wrt. John Conway (the lower, the better) Okay, Im starting to drift but I hope you get the point. It is not that difficult to set up your own tools to measure the amount to which a scientific group suffers from group think. Whether the group will make a long-lasting contribution is another matter which is much harder to predict. Here, I would go for questions like : – Does the theory offer a new insight into classical & central mathematical objects such as groups, curves, modular forms, Dynkin diagrams etc. ? – Does the theory offer tools to reduce the complexity of a problem or does is instead add a layer of technical complexity? That is, are they practicing mathematics or obscurification? 2. How can unappreciated researchers decide whether their ideas really are good or not before spending ten years of their lives finding out? Here is my twofold advice to all the ‘unappreciated’ : (1) be at least as critical to your own work as you are to that of others (it is likely you will find out that you are rightfully under-appreciated compared to others) and (2) enjoy the tiny tokens of appreciation because they are likely all that you will ever get. Speaking for myself, I do not feel unappreciated compared to what I did. I did prove a couple of good results to which adequate reference is given and I had a couple of crazy ideas which were ridiculed by some at the time. A silly sense of satisfaction comes from watching the very same people years later fall over each other trying to reclaim some of the credit for these ideas. Okay, it may not have the same status of recognition as a Fields medal or a plenary talk at the ICM but it is enough to put a smile on my face from time to time and to continue stubbornly with my own ideas.

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devilish symmetries

Published April 2, 2007 by lievenlb

In another post we introduced
Minkowski’s question-mark function, aka the devil’s straircase
and related it to
Conways game of _contorted fractions_. Side remark : over at Good Math, Bad Math Mark Chu-Carroll is running
a mini-series on numbers&games, so far there is a post on surreal numbers,
surreal arithmetic and the connection with
games but
probably this series will go on for some time.

About a year ago I had
an email-exchange with Linas Vepstas because I was
intrigued by one of his online publications linking the fractal
symmetries of the devil’s staircase to the modular group. Unfortunately,
his paper contained some inaccuracies and I’m happy some of my comments
made it into his rewrite The Minkowski question mark, GL(2,Z) and the
modular group. Still, several
mistakes remain so read this paper only modulo his own caveat

XXXX This paper is unfinished. Although this version
corrects a number of serious errors in the previous drafts, it is still
misleading and confusing in many ways. The second half, in particular
must surely contain errors and mis-statements! Caveat emptor! XXXX

For example, on page 15 of the march 24-version he claims
that the third braid group $B_3 \simeq SL_2(\mathbb{Z}) $ which
would make life, mathematics and even physics a lot easier, but
unfortunately is not true. Recall that Artin’s defining relation for the
3-string braid group is $\sigma_1 \sigma_2 \sigma_1 = \sigma_2
\sigma_1 \sigma_2 $ as can be seen because the 3-strings below can
be transformed into each other
But from this
relation it follows that $c=(\sigma_1 \sigma_2 \sigma_1)^2 $ is
a central element in $B_3 $ and it is not difficult to verify
that indeed $B_3/ \langle c \rangle \simeq PSL_2(\mathbb{Z}) $
and $B_3/ \langle c^2 \rangle \simeq SL_2(\mathbb{Z}) $ An easy
way to see that the third braid group and the modular group are quite
different is to look at their one-dimensional representations. Any
group-map $B_3 \rightarrow \mathbb{C}^_ $ is determined by
non-zero complex numbers x and y satisfying $x^2y=y^2x $ so are
parametrized by the torus $\mathbb{C}^_ $ whereas there are only
6 one-dimensional representations of $PSL_2(\mathbb{Z}) = C_2 \ast
C_3 $ (and similarly, there are only 12 one-dimensional
$SL_2(\mathbb{Z}) $-representations). Btw. for those still
interested in noncommutative geometry : $(P)SL_2(\mathbb{Z}) $
are noncommutative manifolds whereas $B_3 $ is definitely
singular, if I ever get to the definitions of all of this… Still,
there is a gem contained in Linas’ paper and here’s my reading of it :
the fractal symmetries of the devil’s staircase form a generating
sub-semigroup $C_2 \ast \mathbb{N} $ of
$GL_2(\mathbb{Z}) $ . To begin, let us recall that the
question-mark function is defined in terms of continued fraction
expressions. So, what group of symmetries may be around the corner?
Well, if $a = \langle a_0;a_1,a_2,\ldots \rangle $ is the
continued fraction of a (see this
post for details) then if we
look at the n-th approximations $\frac{p_n}{q_n} $ (that is, the
rational numbers obtained after breaking off the continued fraction at
step n) it is failrly easy to show that $\begin{bmatrix} p_n &
p_{n-1} \\ q_n & q_{n-1} \end{bmatrix} \in GL_2(\mathbb{Z}) $ and
recall (again) that this group acts on
$\mathbb{P}^1_{\mathbb{C}} $ via Moebius transformations
$\begin{bmatrix} a & b \ c & d \end{bmatrix} $ via $z
\mapsto \frac{az+b}{cz+d} $ One of the symmetries is easy to spot
(reflexion along the 1/2-axis) That is, $?(x-1) = 1 – ?(x) $ Observe that the left-hand
side transformation is given by the Moebius transformation determined by
the matrix $r = \begin{bmatrix} -1 & 1 \\ 0 & 1 \end{bmatrix} \in
GL_2(\mathbb{Z}) $ Other symmetries are harder to see as they are
_fractal symmetries_, that is they are self-symmetries but at different
scales. For example, let us blow-up the ?-function at the interval
[1/3,1/2] and compare it with the function at the interval [1/2,1]
which has the same graph, while halving the function value. More
generally, substituting the ?-function definition using continued
fraction expressions one verifies that $?(\frac{x}{x+1}) =
\frac{1}{2} ?(x) $ and this time the left-hand transformation is
determined by the matrix $g = \begin{bmatrix} 1 & 0 \\ 1 & 1
\end{bmatrix} \in GL_2(\mathbb{Z}) $ We obtain a semi-group $S
= \langle r,g \rangle $ of fractal symmetries which are induced (the
right hand sides of the above expressions) via a 2-dimensional
representation of S $S \rightarrow GL_2(\mathbb{C})~\qquad r
\mapsto \begin{bmatrix} 1 & 0 \\ 1 & -1 \end{bmatrix}~\qquad g \mapsto
\begin{bmatrix} 1 & 0 \\ 0 & \frac{1}{2} \end{bmatrix} $ acting
via left-multiplication on the two-dimensional vectorspace
$\mathbb{C}1+\mathbb{C}x $. We claim that S is the free
semi-group $C_2 \ast \mathbb{N} $. Clearly, $r^2=1 $ and
g is of infinite order, but we have to show that no expression of the
form $rg^{i_1}rg^{i_2}r \ldots rg^{i_l}r $ can be the identity
in S. We will prove this by computing its action on the continued
fraction expression of $a = \langle 0;a_0,a_1,\ldots \rangle $.
It is a pleasant exercise to show that $g. \langle 0;a_1,a_2,\ldots
\rangle = \langle 0;a_1+1,a_2,\ldots \rangle $ whence by induction
$g^n. \langle 0;a_1,a_2,\ldots \rangle = \langle 0;a_1+n,a_2,\ldots
\rangle $ Moreover, the action on r is given by $r. \langle
0;a_1,a_2,\ldots \rangle = \langle 0;1,a_1-1,a_2,\ldots \rangle $ if
$a_1 \not= 1 $ whereas $r. \langle 0;1,a_2,a_3,\ldots
\rangle = \langle 0;a_2+1,a_3,\ldots \rangle $ But then, as a
consequence we have that $g^{n-1}rg . \langle 0;a_1,a_2,\ldots
\rangle = \langle 0;n,a_1,a_2,\ldots \rangle $ and iterating this
procedure gives us finally that an expression $g^{j-1} r g^k r g^l
r \ldots g^z r g = (g^{j-1} r g)(g^{k-1} r g)(g^{l-1} r g) \ldots
(g^{z-1} r g) $ acts on $a = \langle 0;a_1,a_2,\ldots
\rangle $ by sending it to $\langle
0;j,k,l,\ldots,z,a_1,a_2,\ldots \rangle $ whence such an expression
can never act as the identity element, proving that indeed $S \simeq
C_2 \ast \mathbb{N} $. As for the second claim, recall from this
post that
$GL_2(\mathbb{Z}) $ is generated by the matrices $U =
\begin{bmatrix} 0 & -1 \ 1 & 0 \end{bmatrix}~\quad V = \begin{bmatrix}
0 & 1 \ -1 & 1 \end{bmatrix}~\quad R = \begin{bmatrix} 0 & 1 \ 1 & 0
\end{bmatrix} $ and a straightforward verification shows that
$r = RV,~\quad g = VU $ and $R = g^{-1}rg,~\quad
V=g^{-1}rgr,\quad U=rg^{-1}rg^2 $ whence, indeed, the semi-group S
generates the whole of $GL_2(\mathbb{Z}) $!

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the father of all beamer talks

Published March 2, 2007 by lievenlb

Who was the first mathematician to give a slide show talk? I don’t have the
definite answer to this question, but would like to offer a strong
candidate : Hermann Minkowski gave the talk “Zur Geometrie der Zahlen” (On the
geometry of numbers) before the third ICM in 1904 in Heidelberg and even
the title page of his paper in the proceedings indicates that he did
present his talk using slides (Mit Projektionsbildern auf einer
Doppeltafel)

Seven
of these eight slides would be hard to improve using LaTeX

What concerns
us today is the worst of all slides, the seventh, where Minkowski tries
to depict his famous questionmark function $?(x) $, sometimes also called
the _devil’s staircase_

The devil’s
staircase is a fractal curve and can be viewed as a mirror (taking a
point on the horizontal axis to the point on the vertical axis through
the function value) having magical simplifying properties : – it takes
rational numbers to _dyadic numbers_, that is those of the form
$n.2^{-m}$ with $n,m \in \mathbb{Z} $. – it takes quadratic
_irrational_ numbers to rational numbers. So, iterating this
mirror-procedure, the devil’s staircase is a device solving the main
problem of Greek Mathematics : which lengths can be constructed using
ruler and compass? These _constructible numbers_ are precisely those
real numbers which become after a finite number of devil-mirrors a
dyadic number. The proofs of these facts are not very difficult but
they involve a piece of long-forgotten mathematical technology :
_continued fractions_. By repeted approximations using the
floor-function (the largest natural number less than or equal to the real
number), every positive real number can be written as

$a = a_0 +
\frac{1}{a_1 + \frac{1}{a_2 + \frac{1}{a_3 + \frac{1}{\dots}}}} $

with all $a_i $ natural numbers. So, let us just denote from now on this
continued fraction of a by the expression

$a = \langle
a_0;a_1,a_2,a_3,\dots \rangle $

Clearly, a is a rational number if
(and also if but this requires a small argument using the Euclidian
algorithm) the above description has a tail of zeroes at the end and
(slightly more difficult) $a$ is a real quadratic irrational number
(that is, an element of a quadratic extension field
$\mathbb{Q}\sqrt{n} $) if and only if the continued fraction-expression
has a periodic tail. There is a lot more to say about
continued-fraction expressions and I’ll do that in another
‘virtual-course-post’ (those prepended with a (c): sign). For the
impatient let me just say that two real numbers will lie in the same
$GL_2(\mathbb{Z}) $-orbit (under the action via Moebius-transformations)
if and only if their continued fraction expressions have the same tails
eventually (which has applications in noncommutative geometry as in the
work of Manin and Marcolli but maybe I’ll come to this in the (c):
posts).

Right, now we can define the mysterious devil-stair function
$?(x) $. If a is in the real interval $[0,1] $ and if $a \in
\mathbb{Q} $ then $a = \langle 0;a_1,a_2,\dots,a_n,0,0,\dots
\rangle $ and we define $?(a) = 2 \sum_{k=1}^{n} (-1)^k
2^{-(a_1+a_2+\dots+a_k)} $ and if a is irrational with continued
fraction expression $a = \langle 0;a_1,a_2,a_3,\dots \rangle $, then

$?(a) = 2 \sum_{k=1}^{\infty} (-1)^{k+1} 2^{-(a_1+a_2+\dots+a_k)} $

A
perhaps easier description is that with the above continued-fraction
expression, the _binary_ expansion of $?(a) $ has the following form

$?(a) = 0,0 \dots 01 \dots 1 0 \dots 0 1 \dots 1 0 \dots 0 1 \dots
1 0 \dots $

where the first batch of zeroes after the comma has length
$a_1-1 $, the first batch of ones has length $a_2 $ the next batch of
zeroes length $a_3 $ and so on.

It is a pleasant exercise to verify that
this function does indeed have the properties we claimed before. A
recent incarnation of the question mark function is in Conway’s game of
_contorted fractions_. A typical position consists of a finite number of
boxed real numbers, for example the position might be

$\boxed{\pi} + \boxed{\sqrt{2}} + \boxed{1728} +
\boxed{-\frac{1}{3}} $

The Rules of the game are : (1) Both
players L and R take turns modifying just one of the numbers such that
the denominator becomes strictly smaller (irrational numbers are
supposed to have $\infty$ as their ‘denominator’). And if the boxed
number is already an integer, then its absolute value must decrease.
(2) Left must always _decrease_ the value of the boxed number, Right
must always increase it. (3) The first player unable to move looses
the game. To decide who wins a particular game, one needs to compute
the value of a position $\boxed{x} $ according to the rules of
combinatorial game theory (see for example the marvelous series of four
books Winning Ways for your Mathematical Plays. It turns out that this CG-value is no other than $?(x)$
… And, Conway has a much improved depiction of the devil-staircase in
his book On Numbers And Games

One Comment

symmetry and the monster

Published June 2, 2006 by lievenlb

Mark
Ronan has written a beautiful book intended for the general public
on Symmetry and the Monster. The
book’s main theme is the classification of the finite simple groups. It
starts off with the introduction of groups by Galois, gives the
classifivcation of the finite Lie groups, the Feit-Thompson theorem and
the construction of several of the sporadic groups (including the
Mathieu groups, the Fischer and Conway groups and clearly the
(Baby)Monster), explains the Leech lattice and the Monstrous Moonshine
conjectures and ends with Richard Borcherds proof of them using vertex
operator algebras. As in the case of Music of the
Primes it is (too) easy to be critical about notation. For example,
whereas groups are just called symmetry groups, I don’t see the point of
calling simple groups ‘atoms of symmetry’. But, unlike du Sautoy,
Mark Ronan stays close to mathematical notation, lattices are just
lattices, characer-tables are just that, j-function is what it is etc.
And even when he simplifies established teminology, for example
‘cyclic arithmetic’ for modular arithmetic, ‘cross-section’
for involution centralizer, ‘mini j-functions’ for Hauptmoduln
etc. there are footnotes (as well as a glossary) mentioning the genuine
terms. Group theory is a topic with several colourful people
including the three Johns John Leech, John
McKay and John Conway
and several of the historical accounts in the book are a good read. For
example, I’ve never known that the three Conway groups were essentially
discovered in just one afternoon and a few telephone exchanges between
Thompson and Conway. This year I’ve tried to explain some of
monstrous moonshine to an exceptionally good second year of
undergraduates but failed miserably. Whereas I somehow managed to give
the construction and proof of simplicity of Mathieu 24, elliptic and
modular functions were way too difficult for them. Perhaps I’ll give it
another (downkeyed) try using ‘Symmetry and the Monster’ as
reading material. Let’s hope Oxford University Press will soon release a
paperback (and cheaper) version.

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