Tag: Mathieu

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‘

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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Generators of modular subgroups

Published July 12, 2007 by lievenlb

In older NeverEndingBooks-posts (and here) proofs were given that the modular group $\Gamma = PSL_2(\mathbb{Z}) $ is the group free product $C_2 \ast C_3 $, so let’s just skim over details here. First one observes that $\Gamma $ is generated by (the images of) the invertible 2×2 matrices

$U= \begin{bmatrix} 0 & -1 \\\ 1 & 0 \end{bmatrix} $ and $V= \begin{bmatrix} 0 & 1 \\\ -1 & 1 \end{bmatrix} $

A way to see this is to consider X=U.V and Y=V.U and notice that multiplying with powers of X adds multiples of the second row to the first (multiply on the left) or multiples of the first column to the second (multiply on the right) and the other cases are handled by taking multiples with powers of Y. Use this together with the fact that matrices in $GL_2(\mathbb{Z}) $ have their rows and columns made of coprime numbers to get any such matrix by multiplication on the left or right by powers of X and Y into the form

$\begin{bmatrix} \pm 1 & 0 \\\ 0 & \pm 1 \end{bmatrix} $ and because $U^2=V^3=\begin{bmatrix} -1 & 0 \\\ 0 & -1 \end{bmatrix} $

we see that $\Gamma $ is an epimorphic image of $C_2 \ast C_3 $. To prove isomorphism one can use the elegant argument due to Roger Alperin considering the action of the Moebius transformations $u(z) = -\frac{1}{z} $ and $v(z) = \frac{1}{1-z} $ (with $v^{-1}(z) = 1-\frac{1}{z} $) induced by the generators U and V on the sets $\mathcal{P} $ and $\mathcal{N} $ of all positive (resp. negative) irrational real numbers. Observe that

$u(\mathcal{P}) \subset \mathcal{N} $ and $v^{\pm}(\mathcal{N}) \subset \mathcal{P} $

Hence, if $w $ is a word in $u $ and $v^{\pm} $ of off length we either have $w(\mathcal{P}) \subset \mathcal{N} $ or $w(\mathcal{N}) \subset \mathcal{P} $ so $w $ can never be the identity. If the length is even we can conjugate $w $ such that it starts with $v^{\pm} $. If it starts with $v $ then $w(\mathcal{P}) \subset v(\mathcal{N}) $ is a subset of positive rationals less than 1 whereas if it starts with $v^{-1} $ then $w(\mathcal{P}) \subset v^{-1}(\mathcal{N}) $ is a subset of positive rationals greater than 1, so again it cannot be the identity. Done!

By a result of Aleksandr Kurosh it follows that every modular subgroup is the group free product op copies of $C_2, C_3 $ or $C_{\infty} $ and we would like to determine the free generators explicitly for a cofinite subgroup starting from its associated Farey code associated to a special polygon corresponding to the subgroup.

To every even interval [tex]\xymatrix{x_i = \frac{a_i}{b_i} \ar@{-}[r]_{\circ} & x_{i+1}= \frac{a_{i+1}}{b_{i+1}}}[/tex] in the Farey code one associates the generator of a $C_2 $ component

$A_i = \begin{bmatrix} a_{i+1}b_{i+1}+ a_ib_i & -a_i^2-a_{i+1}^2 \\\ b_i^2+b_{i+1}^2 & -a_{i+1}b_{i+1}-a_ib_i \end{bmatrix} $

to every odd interval [tex]\xymatrix{x_i = \frac{a_i}{b_i} \ar@{-}[r]_{\bullet} & x_{i+1} = \frac{a_{i+1}}{b_{i+1}}}[/tex] in the Farey code we associate the generator of a $C_3 $ component

$B_i = \begin{bmatrix} a_{i+1}b_{i+1}+a_ib_{i+1}+a_ib_i & -a_i^2-a_ia_{i+1}-a_{i+1}^2 \\\ b_i^2+b_ib_{i+1} + b_{i+1}^2 & -a_{i+1}b_{i+1} – a_{i+1}b_i – a_i b_i \end{bmatrix} $

and finally, to every pair of free intervals [tex]\xymatrix{x_k \ar@{-}[r]_{a} & x_{k+1}} \ldots \xymatrix{x_l \ar@{-}[r]_{a} & x_{l+1}}[/tex] we associate the generator of a $C_{\infty} $ component

$C_{k,l} = \begin{bmatrix} a_l & -a_{l+1} \\\ b_l & – b_{l+1} \end{bmatrix} \begin{bmatrix} a_{k+1} & a_k \\\ b_{k+1} & b_k \end{bmatrix}^{-1} $

Kulkarni’s result states that these matrices are free generators of the cofiniite modular subgroup determined by the Farey code. For example, for the M(12) special polygon on the left (bounded by the thick black geodesics), the Farey-code for this Mathieu polygon is

[tex]\xymatrix{\infty \ar@{-}[r]_{1} & 0 \ar@{-}[r]_{\bullet} & \frac{1}{3} \ar@{-}[r]_{\bullet} & \frac{1}{2} \ar@{-}[r]_{\bullet} & 1 \ar@{-}[r]_{1} & \infty}[/tex]

Therefore, the structure of the subgroup must be $C_{\infty} \ast C_3 \ast C_3 \ast C_3 $ with the generator of the infinite factor being

$\begin{bmatrix} -1 & 1 \\\ -1 & 0 \end{bmatrix} $ and those of the cyclic factors of order three

$\begin{bmatrix} 3 & -1 \\\ 13 & -4 \end{bmatrix}, \begin{bmatrix} 7 & -3 \\\ 19 & 8 \end{bmatrix} $ and $\begin{bmatrix} 4 & -3 \\\ 7 & -5 \end{bmatrix} $

This approach also gives another proof of the fact that $\Gamma = C_2 \ast C_3 $ because the Farey code to the subgroup of index 1 is [tex]\xymatrix{\infty \ar@{-}[r]_{\circ} & 0 \ar@{-}[r]_{\bullet} & \infty}[/tex] corresponding to the fundamental domain on the left. This finishes (for now) this thread on Kulkarni’s paper (or rather, part of it). On the Lost? page I will try to list threads in a logical ordering when they materialize.

Reference

Ravi S. Kulkarni, “An arithmetic-geometric method in the study of the subgroups of the modular group”, Amer. J. Math 113 (1991) 1053-1133

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Farey codes

Published July 1, 2007 by lievenlb

John Farey (1766-1826) was a geologist of sorts. Eyles, quoted on the math-biographies site described his geological work as

“As a geologist Farey is entitled to respect for the work which he carried out himself, although it has scarcely been noticed in the standard histories of geology.”

That we still remember his name after 200 years is due to a short letter he wrote in 1816 to the editor of the Philosophical Magazine

“On a curious Property of vulgar Fractions.
By Mr. J. Farey, Sen. To Mr. Tilloch

Sir. – On examining lately, some very curious and elaborate Tables of “Complete decimal Quotients,” calculated by Henry Goodwyn, Esq. of Blackheath, of which he has printed a copious specimen, for private circulation among curious and practical calculators, preparatory to the printing of the whole of these useful Tables, if sufficient encouragement, either public or individual, should appear to warrant such a step: I was fortunate while so doing, to deduce from them the following general property; viz.

If all the possible vulgar fractions of different values, whose greatest denominator (when in their lowest terms) does not exceed any given number, be arranged in the order of their values, or quotients; then if both the numerator and the denominator of any fraction therein, be added to the numerator and the denominator, respectively, of the fraction next but one to it (on either side), the sums will give the fraction next to it; although, perhaps, not in its lowest terms.

For example, if 5 be the greatest denominator given; then are all the possible fractions, when arranged, 1/5, 1/4, 1/3, 2/5, 1/2, 3/5, 2/3, 3/4, and 4/5; taking 1/3, as the given fraction, we have (1+1)/(5+3) = 2/8 = 1/4 the next smaller fraction than 1/3; or (1+1)/(3+2) = 2/5, the next larger fraction to 1/3. Again, if 99 be the largest denominator, then, in a part of the arranged Table, we should have 15/52, 28/97, 13/45, 24/83, 11/38, &c.; and if the third of these fractions be given, we have (15+13)/(52+45) = 28/97 the second: or (13+11)/(45+38) = 24/83 the fourth of them: and so in all the other cases.

I am not acquainted, whether this curious property of vulgar fractions has been before pointed out?; or whether it may admit of any easy or general demonstration?; which are points on which I should be glad to learn the sentiments of some of your mathematical readers; and am

Sir, Your obedient humble servant,
J. Farey. Howland-street.”

So, if we interpolate “childish addition of fractions” $\frac{a}{b} \oplus \frac{c}{d} = \frac{a+c}{b+d} $ and start with the numbers $0 = \frac{0}{1} $ and $\infty = \frac{1}{0} $ we get the binary Farey-tree above. For a fixed natural number n, if we stop the interpolation whenever the denominator of the fraction would become larger than n and order the obtained fractions (smaller or equal to one) we get the Farey sequence F(n). For example, if n=3 we start with the sequence $ \frac{0}{1},\frac{1}{1} $. The next step we get $\frac{0}{1},\frac{1}{2},\frac{1}{1} $ and the next step gives

$\frac{0}{1},\frac{1}{3},\frac{1}{2},\frac{2}{3},\frac{1}{1} $

and as all the denomnators of childish addition on two consecutive fractions will be larger than 3, the above sequence is F(3). A remarkable feature of the series F(n) is that if $\frac{a}{b} $ and $\frac{c}{d} $ are consecutive terms in F(n), then

$det \begin{bmatrix} a & c \\\ b & d \end{bmatrix} = -1 $

and so these two fractions are the endpoints of an even geodesic in the Dedekind tessellation.

A generalized Farey series is an ordered collection of fractions $\infty,x_0,x_1,\cdots,x_n,\infty $ such that $x_0 $ and $x_n $ are integers and some $x_i=0 $. Moreover, writing $x_i = \frac{a_i}{b_i} $ we have that

$det \begin{bmatrix} a_i & a_{i+1} \\\ b_i & b_{i+1} \end{bmatrix} = -1 $

A Farey code is a generalized Farey sequence consisting of all the vertices of a special polygon that lie in $\mathbb{R} \cup \{ \infty \} $ together with side-pairing information. If two consecutive terms are such that the complete geodesic between $x_i $ and $x_{i+1} $ consists of two sides of the polygon which are paired we denote this fact by
[tex]\xymatrix{x_i \ar@{-}[r]_{\circ} & x_{i+1}}[/tex]. If they are the endpoints of two odd sides of the polygon which are paired we denote this by [tex]\xymatrix{x_i \ar@{-}[r]_{\bullet} & x_{i+1}}[/tex]. Finally, if they are the endpoints of a free side which is paired to another free side determined by $x_j $ and $x_{j+1} $ we denote this fact by marking both edges [tex]\xymatrix{x_i \ar@{-}[r]_{k} & x_{i+1}}[/tex] and [tex]\xymatrix{x_j \ar@{-}[r]_{k} & x_{j+1}}[/tex] with the same number.

For example, for the M(12) special polygon on the left (bounded by the thick black geodesics), the only vertices in $\mathbb{R} \cup \{ \infty \} $ are $\infty,0,\frac{1}{3},\frac{1}{2},1 $. The two vertical lines are free sides and are paired, whereas all other sides of the polygon are odd. Therefore the Farey-code for this Mathieu polygon is

[tex]\xymatrix{\infty \ar@{-}[r]_{1} & 0 \ar@{-}[r]_{\bullet} & \frac{1}{3} \ar@{-}[r]_{\bullet} & \frac{1}{2} \ar@{-}[r]_{\bullet} & 1 \ar@{-}[r]_{1} & \infty}[/tex]

Conversely, to a Farey-code we can associate a special polygon by first taking the hyperbolic convex hull of all the terms in the sequence (the region bounded by the vertical lines and the bottom red circles in the picture on the left) and adding to it for each odd interval [tex]\xymatrix{x_i \ar@{-}[r]_{\bullet} & x_{i+1}}[/tex] the triangle just outside the convex hull consisting of two odd edges in the Dedekind tessellation (then we obtain the region bounded by the black geodesics). Again, the side-pairing of the obained special polygon can be obtained from that of the Farey-code.

This correspondence gives a natural one-to-one correspondence special polygons <---> Farey-codes . _Later_ we will see how the Farey-code determines the group structure of the corresponding finite index subgroup of the modular group $\Gamma = PSL_2(\mathbb{Z}) $.

Reference

Ravi S. Kulkarni, “An arithmetic-geometric method in the study of the subgroups of the modular group”, Amer. J. Math 113 (1991) 1053-1133

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Hyperbolic Mathieu polygons

Published June 27, 2007 by lievenlb

Today we will link modular quilts (via their associated cuboid tree diagrams) to special hyperbolic polygons. The above drawing gives the hyperbolic polygon (the gray boundary) associated to the M(24) tree diagram (the black interior graph). In general, the correspondence goes as follows.

Recall that a cuboid tree diagram is a tree such that all internal vertices are 3-valent and have a specified ordering on the incident edges (given by walking counterclockwise around the vertex) and such that all leaf-vertices are tinted blue or red, the latter ones are paired via an involution (indicated by giving paired red vertices the same label). Introduce a new 2-valent vertex on all edges joining two internal vertices or a blue vertex to an internal vertex. So, the picture on the right corresponds to the tree diagram on the left. Equip this extended tree with a metric such that every edge has length equal to an f-edge in the Dedekind tessellation. Fix an edge having a red vertex and develop this isometrically onto the f-edge connecting $i $ to $\rho $ in the tessellation. Then, the extended tree develops uniquely along the f-edges of the tessellation and such that the circled black and blue vertices correspond to odd vertices, the circled red and added uncircled vertices correspond to even vertices in the tessellation. Starting from the above tree (and choosing the upper-left edge to start the embedding), we obtain the picture on the left (we have removed the added 2-valent vertices)

We will now associate a special hyperbolic polygon to this tree. At a red vertex take the even line going through the vertex. If under the involution the red vertex is send to itself, the even edges will be paired. Otherwise, the line is a free side and will be paired to the free side containing the red vertex corresponding under the involution. At a blue vertex, take the two odd edges making an angle of $\frac{\pi}{3} $ with the tree-edge containing the blue vertex. These odd edges will be paired. If we do this procedure for all blue and red vertices, we obtain a special polygon (see the picture on the right, the two vertical lines are paired).
Conversely, suppose we start with a special polygon such as the one on the left below

and consider all even and odd vertices on the boundary (which are tinted red, respectively blue) together with all odd vertices in the interior of the special polygon. These are indicated in the picture on the right above. If we connect these vertices with the geodesics in the polygon we get a cuboid tree diagram. This correspondence special polygons —>> tree diagrams is finite to one as we have made a choice in the starting red vertex and edge. If we would have taken the other edge containing a red vertex we would end up with the following special polygon

It is no accident that these two special polygons consist of exactly 24 triangles of the Dedekind tessellation as they correspond to the index 12 subgroup of the modular group $\Gamma $ determining the 12-dimensional permutation representation of the Mathieu group $M_{12} $. Similarly, the top drawing has 48 hyperbolic triangles and corresponds to the 24-dimensional permutation representation of $M_{24} $. Another time we will make the connection with Farey series which will allow us to give free generators of finite index subgroups.

Reference

Ravi S. Kulkarni, “An arithmetic-geometric method in the study of the subgroups of the modular group”, Amer. J. Math. 113 (1991) 1053-1133

One Comment

Modular quilts and cuboid tree diagrams

Published June 25, 2007 by lievenlb

Conjugacy classes of finite index subgroups of the modular group $\Gamma = PSL_2(\mathbb{Z}) $ are determined by a combinatorial gadget : a modular quilt. By this we mean a finite connected graph drawn on a Riemann surface such that its vertices are either black or white. Moreover, every edge in the graph connects a black to a white vertex and the valency (that is, the number of edges incodent to a vertex) of a black vertex is either 1 or 2, that of a white vertex is either 1 or 3. Finally, for every white vertex of valency 3, there is a prescribed cyclic order on the edges incident to it.

On the left a modular quilt consisting of 18 numbered edges (some vertices and edges re-appear) which gives a honeycomb tiling on a torus. All white vertices have valency 3 and the order of the edges is given by walking around a point in counterclockwise direction. For example, the order of the edges at the top left vertex (which re-appears at the middle right vertex) can be represented by the 3-cycle (6,11,14), that around the central vertex gives the 3-cycle (2,7,16).

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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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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?

NeverEndingBooks-groups

Published June 11, 2007 by lievenlb

Here a collection of pdf-files of NeverEndingBooks-posts on groups, in reverse chronological order.

bloomsday end

Published June 9, 2007 by lievenlb

From time to time you may see here a message that NeverEndingBooks ends on Bloomsday (June 16th). Soon after, I hope to restart with another blog at the same URL. For starters, Neverendingbooks refers to my never-ending bookproject on noncommutative geometry started in 1999, a millenium ago… Today I\’m correcting the proofs and have even seen the cover-design of the book, supposed to be published in the fall. So, it should be really EndingBook(s), finally. From time to time it is good to start afresh. The next project is still pretty vague to me but it will be a lot more focussed and center around topics like Moonshine, the Monster, the Mathieu groups, Modular forms and group etc. Suggestions for a blogtitle are welcome (M-theory is already taken…). Besides there are technical problems with the machine running the blog, a new one is expected around June 16th. As I will not be able to clone between the two (one PPC, the new one Intel) I decided to start again from scratch. Anyway, Ive made a database-dump of NeverEndingBooks and will make it available to anyone interested in reading old posts (even the ones with a private-status). Finally, there are other reasons, better kept private. Give me a couple of weeks to resurface. For now, all the best.

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