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Farey symbols of sporadic groups

Published March 20, 2008 by lieven

John Conway once wrote :

There are almost as many different constructions of $M_{24} $ as there have been mathematicians interested in that most remarkable of all finite groups.

In the inguanodon post Ive added yet another construction of the Mathieu groups $M_{12} $ and $M_{24} $ starting from (half of) the Farey sequences and the associated cuboid tree diagram obtained by demanding that all edges are odd. In this way the Mathieu groups turned out to be part of a (conjecturally) infinite sequence of simple groups, starting as follows :

$L_2(7),M_{12},A_{16},M_{24},A_{28},A_{40},A_{48},A_{60},A_{68},A_{88},A_{96},A_{120},A_{132},A_{148},A_{164},A_{196},\ldots $

It is quite easy to show that none of the other sporadics will appear in this sequence via their known permutation representations. Still, several of the sporadic simple groups are generated by an element of order two and one of order three, so they are determined by a finite dimensional permutation representation of the modular group $PSL_2(\mathbb{Z}) $ and hence are hiding in a special polygonal region of the Dedekind’s tessellation

Let us try to figure out where the sporadic with the next simplest permutation representation is hiding : the second Janko group $J_2 $, via its 100-dimensional permutation representation. The Atlas tells us that the order two and three generators act as

e:= (1,84)(2,20)(3,48)(4,56)(5,82)(6,67)(7,55)(8,41)(9,35)(10,40)(11,78)(12, 100)(13,49)(14,37)(15,94)(16,76)(17,19)(18,44)(21,34)(22,85)(23,92)(24, 57)(25,75)(26,28)(27,64)(29,90)(30,97)(31,38)(32,68)(33,69)(36,53)(39,61) (42,73)(43,91)(45,86)(46,81)(47,89)(50,93)(51,96)(52,72)(54,74)(58,99) (59,95)(60,63)(62,83)(65,70)(66,88)(71,87)(77,98)(79,80);

v:= (1,80,22)(2,9,11)(3,53,87)(4,23,78)(5,51,18)(6,37,24)(8,27,60)(10,62,47) (12,65,31)(13,64,19)(14,61,52)(15,98,25)(16,73,32)(17,39,33)(20,97,58) (21,96,67)(26,93,99)(28,57,35)(29,71,55)(30,69,45)(34,86,82)(38,59,94) (40,43,91)(42,68,44)(46,85,89)(48,76,90)(49,92,77)(50,66,88)(54,95,56) (63,74,72)(70,81,75)(79,100,83);

But as the kfarey.sage package written by Chris Kurth calculates the Farey symbol using the L-R generators, we use GAP to find those

L = e*v^-1  and  R=e*v^-2 so

L=(1,84,22,46,70,12,79)(2,58,93,88,50,26,35)(3,90,55,7,71,53,36)(4,95,38,65,75,98,92)(5,86,69,39,14,6,96)(8,41,60,72,61,17, 64)(9,57,37,52,74,56,78)(10,91,40,47,85,80,83)(11,23,49,19,33,30,20)(13,77,15,59,54,63,27)(16,48,87,29,76,32,42)(18,68, 73,44,51,21,82)(24,28,99,97,45,34,67)(25,81,89,62,100,31,94)

R=(1,84,80,100,65,81,85)(2,97,69,17,13,92,78)(3,76,73,68,16,90,71)(4,54,72,14,24,35,11)(5,34,96,18,42,32,44)(6,21,86,30,58, 26,57)(7,29,48,53,36,87,55)(8,41,27,19,39,52,63)(9,28,93,66,50,99,20)(10,43,40,62,79,22,89)(12,83,47,46,75,15,38)(23,77, 25,70,31,59,56)(33,45,82,51,67,37,61)(49,64,60,74,95,94,98)

Defining these permutations in sage and using kfarey, this gives us the Farey-symbol of the associated permutation representation

L=SymmetricGroup(Integer(100))("(1,84,22,46,70,12,79)(2,58,93,88,50,26,35)(3,90,55,7,71,53,36)(4,95,38,65,75,98,92)(5,86,69,39,14,6,96)(8,41,60,72,61,17, 64)(9,57,37,52,74,56,78)(10,91,40,47,85,80,83)(11,23,49,19,33,30,20)(13,77,15,59,54,63,27)(16,48,87,29,76,32,42)(18,68, 73,44,51,21,82)(24,28,99,97,45,34,67)(25,81,89,62,100,31,94)")

R=SymmetricGroup(Integer(100))("(1,84,80,100,65,81,85)(2,97,69,17,13,92,78)(3,76,73,68,16,90,71)(4,54,72,14,24,35,11)(5,34,96,18,42,32,44)(6,21,86,30,58, 26,57)(7,29,48,53,36,87,55)(8,41,27,19,39,52,63)(9,28,93,66,50,99,20)(10,43,40,62,79,22,89)(12,83,47,46,75,15,38)(23,77, 25,70,31,59,56)(33,45,82,51,67,37,61)(49,64,60,74,95,94,98)")

sage: FareySymbol("Perm",[L,R])

[[0, 1, 4, 3, 2, 5, 18, 13, 21, 71, 121, 413, 292, 463, 171, 50, 29, 8, 27, 46, 65, 19, 30, 11, 3, 10, 37, 64, 27, 17, 7, 4, 5], [1, 1, 3, 2, 1, 2, 7, 5, 8, 27, 46, 157, 111, 176, 65, 19, 11, 3, 10, 17, 24, 7, 11, 4, 1, 3, 11, 19, 8, 5, 2, 1, 1], [-3, 1, 4, 4, 2, 3, 6, -3, 7, 13, 14, 15, -3, -3, 15, 14, 11, 8, 8, 10, 12, 12, 10, 9, 5, 5, 9, 11, 13, 7, 6, 3, 2, 1]]

Here, the first string gives the numerators of the cusps, the second the denominators and the third gives the pairing information (where [tex[-2 $ denotes an even edge and $-3 $ an odd edge. Fortunately, kfarey also allows us to draw the special polygonal region determined by a Farey-symbol. So, here it is (without the pairing data) :

the hiding place of $J_2 $…

It would be nice to have (a) other Farey-symbols associated to the second Janko group, hopefully showing a pattern that one can extend into an infinite family as in the inguanodon series and (b) to determine Farey-symbols of more sporadic groups.

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exotic chess positions (2)

Published March 5, 2008 by lieven

Juan de Mairena linked in the comments to last post to a truly great retro-chess problem ! In the position below white is to play and mate in three!



At first this seems wrong as there is an obvious mate in two : 1. Qe2-f1, Kh1xh2 2. Rg3-h3 The ingenious point being that black claims a draw after 1. Qe2-f1 invoking the 50 moves rule which states

The game is drawn, upon a correct claim by the player having the move, if
(a) he writes on his scoresheet, and declares to the arbiter his intention to make a move which shall result in the last 50 moves having been made by each player without the movement of any pawn and without the capture of any piece, or
(b) the last 50 consecutive moves have been made by each player without the movement of any pawn and without the capture of any piece.

I’d love to have the ‘official’ solution to this problem. Here’s what I’ve come up with, spending the better part of the afternoon… It will not be optimal but hopefully isn’t too far off. The crucial part is a maneuver unlocking the lower left-hand cluster of pieces (in particular the ordering of white and black rook). All captures were done with pawns and the final pawn move was b2-b3. Immediately before it the situation might look something like the situation on the left (essential is that the white king should not be too far from its home-square as he will be needed later to block the white rook)



after b2-b3 the bisshop on c1 travels to f4 and the black rook squeezes in to block the white rook so that also the black king can come in and position himself at e2. Then the two rooks evacuate the first row, allowing the black king to go to h1 and then the white king comes in to block the white rook from checking the black king (situation on the left below).



Finally, the white rook comes in and positions itself at e2, afterwards the white king evacuates the first row via b2 and travels to the right-hand upper corner entering via g7. Meanwhile, the black rook comes to g1, the white rook then travels to a3 and the black rook to a2. Then, the white king goes to b7 allowing the bishop to unlock the rook on a8 going to g7, allowing finally the king to go to d8…

Perhaps there is a much simpler and more elegant solution, so if you know, please comment. Oh, btw. how is the original problem solved. Well white first cancels the 50-move rule by 1. Kd8xd7 to continue for example with 2. Rg3-g4, 3. Qe1

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exotic chess positions (1)

Published March 4, 2008 by lieven

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)



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