what does the monster see?

The Monster is the largest of the 26 sporadic simple groups and has order

808 017 424 794 512 875 886 459 904 961 710 757 005 754 368 000 000 000

= 2^46 3^20 5^9 7^6 11^2 13^3 17 19 23 29 31 41 47 59 71.

It is not so much the size of its order that makes it hard to do actual calculations in the monster, but rather the dimensions of its smallest non-trivial irreducible representations (196 883 for the smallest, 21 296 876 for the next one, and so on).

In characteristic two there is an irreducible representation of one dimension less (196 882) which appears to be of great use to obtain information. For example, Robert Wilson used it to prove that The Monster is a Hurwitz group. This means that the Monster is generated by two elements g and h satisfying the relations

$g^2 = h^3 = (gh)^7 = 1 $

Geometrically, this implies that the Monster is the automorphism group of a Riemann surface of genus g satisfying the Hurwitz bound 84(g-1)=#Monster. That is,

g=9619255057077534236743570297163223297687552000000001=42151199 * 293998543 * 776222682603828537142813968452830193

Or, in analogy with the Klein quartic which can be constructed from 24 heptagons in the tiling of the hyperbolic plane, there is a finite region of the hyperbolic plane, tiled with heptagons, from which we can construct this monster curve by gluing the boundary is a specific way so that we get a Riemann surface with exactly 9619255057077534236743570297163223297687552000000001 holes. This finite part of the hyperbolic tiling (consisting of #Monster/7 heptagons) we’ll call the empire of the monster and we’d love to describe it in more detail.



Look at the half-edges of all the heptagons in the empire (the picture above learns that every edge in cut in two by a blue geodesic). They are exactly #Monster such half-edges and they form a dessin d’enfant for the monster-curve.

If we label these half-edges by the elements of the Monster, then multiplication by g in the monster interchanges the two half-edges making up a heptagonal edge in the empire and multiplication by h in the monster takes a half-edge to the one encountered first by going counter-clockwise in the vertex of the heptagonal tiling. Because g and h generated the Monster, the dessin of the empire is just a concrete realization of the monster.

Because g is of order two and h is of order three, the two permutations they determine on the dessin, gives a group epimorphism $C_2 \ast C_3 = PSL_2(\mathbb{Z}) \rightarrow \mathbb{M} $ from the modular group $PSL_2(\mathbb{Z}) $ onto the Monster-group.

In noncommutative geometry, the group-algebra of the modular group $\mathbb{C} PSL_2 $ can be interpreted as the coordinate ring of a noncommutative manifold (because it is formally smooth in the sense of Kontsevich-Rosenberg or Cuntz-Quillen) and the group-algebra of the Monster $\mathbb{C} \mathbb{M} $ itself corresponds in this picture to a finite collection of ‘points’ on the manifold. Using this geometric viewpoint we can now ask the question What does the Monster see of the modular group?

To make sense of this question, let us first consider the commutative equivalent : what does a point P see of a commutative variety X?



Evaluation of polynomial functions in P gives us an algebra epimorphism $\mathbb{C}[X] \rightarrow \mathbb{C} $ from the coordinate ring of the variety $\mathbb{C}[X] $ onto $\mathbb{C} $ and the kernel of this map is the maximal ideal $\mathfrak{m}_P $ of
$\mathbb{C}[X] $ consisting of all functions vanishing in P.

Equivalently, we can view the point $P= \mathbf{spec}~\mathbb{C}[X]/\mathfrak{m}_P $ as the scheme corresponding to the quotient $\mathbb{C}[X]/\mathfrak{m}_P $. Call this the 0-th formal neighborhood of the point P.

This sounds pretty useless, but let us now consider higher-order formal neighborhoods. Call the affine scheme $\mathbf{spec}~\mathbb{C}[X]/\mathfrak{m}_P^{n+1} $ the n-th forml neighborhood of P, then the first neighborhood, that is with coordinate ring $\mathbb{C}[X]/\mathfrak{m}_P^2 $ gives us tangent-information. Alternatively, it gives the best linear approximation of functions near P.
The second neighborhood $\mathbb{C}[X]/\mathfrak{m}_P^3 $ gives us the best quadratic approximation of function near P, etc. etc.

These successive quotients by powers of the maximal ideal $\mathfrak{m}_P $ form a system of algebra epimorphisms

$\ldots \frac{\mathbb{C}[X]}{\mathfrak{m}_P^{n+1}} \rightarrow \frac{\mathbb{C}[X]}{\mathfrak{m}_P^{n}} \rightarrow \ldots \ldots \rightarrow \frac{\mathbb{C}[X]}{\mathfrak{m}_P^{2}} \rightarrow \frac{\mathbb{C}[X]}{\mathfrak{m}_P} = \mathbb{C} $

and its inverse limit $\underset{\leftarrow}{lim}~\frac{\mathbb{C}[X]}{\mathfrak{m}_P^{n}} = \hat{\mathcal{O}}_{X,P} $ is the completion of the local ring in P and contains all the infinitesimal information (to any order) of the variety X in a neighborhood of P. That is, this completion $\hat{\mathcal{O}}_{X,P} $ contains all information that P can see of the variety X.

In case P is a smooth point of X, then X is a manifold in a neighborhood of P and then this completion
$\hat{\mathcal{O}}_{X,P} $ is isomorphic to the algebra of formal power series $\mathbb{C}[[ x_1,x_2,\ldots,x_d ]] $ where the $x_i $ form a local system of coordinates for the manifold X near P.

Right, after this lengthy recollection, back to our question what does the monster see of the modular group? Well, we have an algebra epimorphism

$\pi~:~\mathbb{C} PSL_2(\mathbb{Z}) \rightarrow \mathbb{C} \mathbb{M} $

and in analogy with the commutative case, all information the Monster can gain from the modular group is contained in the $\mathfrak{m} $-adic completion

$\widehat{\mathbb{C} PSL_2(\mathbb{Z})}_{\mathfrak{m}} = \underset{\leftarrow}{lim}~\frac{\mathbb{C} PSL_2(\mathbb{Z})}{\mathfrak{m}^n} $

where $\mathfrak{m} $ is the kernel of the epimorphism $\pi $ sending the two free generators of the modular group $PSL_2(\mathbb{Z}) = C_2 \ast C_3 $ to the permutations g and h determined by the dessin of the pentagonal tiling of the Monster’s empire.

As it is a hopeless task to determine the Monster-empire explicitly, it seems even more hopeless to determine the kernel $\mathfrak{m} $ let alone the completed algebra… But, (surprise) we can compute $\widehat{\mathbb{C} PSL_2(\mathbb{Z})}_{\mathfrak{m}} $ as explicitly as in the commutative case we have $\hat{\mathcal{O}}_{X,P} \simeq \mathbb{C}[[ x_1,x_2,\ldots,x_d ]] $ for a point P on a manifold X.

Here the details : the quotient $\mathfrak{m}/\mathfrak{m}^2 $ has a natural structure of $\mathbb{C} \mathbb{M} $-bimodule. The group-algebra of the monster is a semi-simple algebra, that is, a direct sum of full matrix-algebras of sizes corresponding to the dimensions of the irreducible monster-representations. That is,

$\mathbb{C} \mathbb{M} \simeq \mathbb{C} \oplus M_{196883}(\mathbb{C}) \oplus M_{21296876}(\mathbb{C}) \oplus \ldots \ldots \oplus M_{258823477531055064045234375}(\mathbb{C}) $

with exactly 194 components (the number of irreducible Monster-representations). For any $\mathbb{C} \mathbb{M} $-bimodule $M $ one can form the tensor-algebra

$T_{\mathbb{C} \mathbb{M}}(M) = \mathbb{C} \mathbb{M} \oplus M \oplus (M \otimes_{\mathbb{C} \mathbb{M}} M) \oplus (M \otimes_{\mathbb{C} \mathbb{M}} M \otimes_{\mathbb{C} \mathbb{M}} M) \oplus \ldots \ldots $




and applying the formal neighborhood theorem for formally smooth algebras (such as $\mathbb{C} PSL_2(\mathbb{Z}) $) due to Joachim Cuntz (left) and Daniel Quillen (right) we have an isomorphism of algebras

$\widehat{\mathbb{C} PSL_2(\mathbb{Z})}_{\mathfrak{m}} \simeq \widehat{T_{\mathbb{C} \mathbb{M}}(\mathfrak{m}/\mathfrak{m}^2)} $

where the right-hand side is the completion of the tensor-algebra (at the unique graded maximal ideal) of the $\mathbb{C} \mathbb{M} $-bimodule $\mathfrak{m}/\mathfrak{m}^2 $, so we’d better describe this bimodule explicitly.

Okay, so what’s a bimodule over a semisimple algebra of the form $S=M_{n_1}(\mathbb{C}) \oplus \ldots \oplus M_{n_k}(\mathbb{C}) $? Well, a simple S-bimodule must be either (1) a factor $M_{n_i}(\mathbb{C}) $ with all other factors acting trivially or (2) the full space of rectangular matrices $M_{n_i \times n_j}(\mathbb{C}) $ with the factor $M_{n_i}(\mathbb{C}) $ acting on the left, $M_{n_j}(\mathbb{C}) $ acting on the right and all other factors acting trivially.

That is, any S-bimodule can be represented by a quiver (that is a directed graph) on k vertices (the number of matrix components) with a loop in vertex i corresponding to each simple factor of type (1) and a directed arrow from i to j corresponding to every simple factor of type (2).

That is, for the Monster, the bimodule $\mathfrak{m}/\mathfrak{m}^2 $ is represented by a quiver on 194 vertices and now we only have to determine how many loops and arrows there are at or between vertices.

Using Morita equivalences and standard representation theory of quivers it isn’t exactly rocket science to determine that the number of arrows between the vertices corresponding to the irreducible Monster-representations $S_i $ and $S_j $ is equal to

$dim_{\mathbb{C}}~Ext^1_{\mathbb{C} PSL_2(\mathbb{Z})}(S_i,S_j)-\delta_{ij} $

Now, I’ve been wasting a lot of time already here explaining what representations of the modular group have to do with quivers (see for example here or some other posts in the same series) and for quiver-representations we all know how to compute Ext-dimensions in terms of the Euler-form applied to the dimension vectors.

Right, so for every Monster-irreducible $S_i $ we have to determine the corresponding dimension-vector $~(a_1,a_2;b_1,b_2,b_3) $ for the quiver

$\xymatrix{ & & & &
\vtx{b_1} \\ \vtx{a_1} \ar[rrrru]^(.3){B_{11}} \ar[rrrrd]^(.3){B_{21}}
\ar[rrrrddd]_(.2){B_{31}} & & & & \\ & & & & \vtx{b_2} \\ \vtx{a_2}
\ar[rrrruuu]_(.7){B_{12}} \ar[rrrru]_(.7){B_{22}}
\ar[rrrrd]_(.7){B_{23}} & & & & \\ & & & & \vtx{b_3}} $

Now the dimensions $a_i $ are the dimensions of the +/-1 eigenspaces for the order 2 element g in the representation and the $b_i $ are the dimensions of the eigenspaces for the order 3 element h. So, we have to determine to which conjugacy classes g and h belong, and from Wilson’s paper mentioned above these are classes 2B and 3B in standard Atlas notation.

So, for each of the 194 irreducible Monster-representations we look up the character values at 2B and 3B (see below for the first batch of those) and these together with the dimensions determine the dimension vector $~(a_1,a_2;b_1,b_2,b_3) $.

For example take the 196883-dimensional irreducible. Its 2B-character is 275 and the 3B-character is 53. So we are looking for a dimension vector such that $a_1+a_2=196883, a_1-275=a_2 $ and $b_1+b_2+b_3=196883, b_1-53=b_2=b_3 $ giving us for that representation the dimension vector of the quiver above $~(98579,98304,65663,65610,65610) $.

Okay, so for each of the 194 irreducibles $S_i $ we have determined a dimension vector $~(a_1(i),a_2(i);b_1(i),b_2(i),b_3(i)) $, then standard quiver-representation theory asserts that the number of loops in the vertex corresponding to $S_i $ is equal to

$dim(S_i)^2 + 1 – a_1(i)^2-a_2(i)^2-b_1(i)^2-b_2(i)^2-b_3(i)^2 $

and that the number of arrows from vertex $S_i $ to vertex $S_j $ is equal to

$dim(S_i)dim(S_j) – a_1(i)a_1(j)-a_2(i)a_2(j)-b_1(i)b_1(j)-b_2(i)b_2(j)-b_3(i)b_3(j) $

This data then determines completely the $\mathbb{C} \mathbb{M} $-bimodule $\mathfrak{m}/\mathfrak{m}^2 $ and hence the structure of the completion $\widehat{\mathbb{C} PSL_2}_{\mathfrak{m}} $ containing all information the Monster can gain from the modular group.

But then, one doesn’t have to go for the full regular representation of the Monster. Any faithful permutation representation will do, so we might as well go for the one of minimal dimension.

That one is known to correspond to the largest maximal subgroup of the Monster which is known to be a two-fold extension $2.\mathbb{B} $ of the Baby-Monster. The corresponding permutation representation is of dimension 97239461142009186000 and decomposes into Monster-irreducibles

$S_1 \oplus S_2 \oplus S_4 \oplus S_5 \oplus S_9 \oplus S_{14} \oplus S_{21} \oplus S_{34} \oplus S_{35} $

(in standard Atlas-ordering) and hence repeating the arguments above we get a quiver on just 9 vertices! The actual numbers of loops and arrows (I forgot to mention this, but the quivers obtained are actually symmetric) obtained were found after laborious computations mentioned in this post and the details I’ll make avalable here.

Anyone who can spot a relation between the numbers obtained and any other part of mathematics will obtain quantities of genuine (ie. non-Inbev) Belgian beer…

the McKay-Thompson series

Monstrous moonshine was born (sometime in 1978) the moment John McKay realized that the linear term in the j-function

$j(q) = \frac{1}{q} + 744 + 196884 q + 21493760 q^2 + 864229970 q^3 + \ldots $

is surprisingly close to the dimension of the smallest non-trivial irreducible representation of the monster group, which is 196883. Note that at that time, the Monster hasn’t been constructed yet, and, the only traces of its possible existence were kept as semi-secret information in a huge ledger (costing 80 pounds…) kept in the Atlas-office at Cambridge. Included were 8 huge pages describing the character table of the monster, the top left fragment, describing the lower dimensional irreducibles and their characters at small order elements, reproduced below

If you look at the dimensions of the smallest irreducible representations (the first column) : 196883, 21296876, 842609326, … you will see that the first, second and third of them are extremely close to the linear, quadratic and cubic coefficient of the j-function. In fact, more is true : one can obtain these actual j-coefficients as simple linear combination of the dimensions of the irrducibles :

$\begin{cases} 196884 &= 1 + 196883 \\
21493760 &= 1 + 196883 + 21296876 \\
864229970 &= 2 \times 1 + 2 \times 196883 + 21296876 + 842609326
\end{cases} $

Often, only the first relation is attributed to McKay, whereas the second and third were supposedly discovered by John Thompson after MKay showed him the first. Marcus du Sautoy tells a somewhat different sory in Finding Moonshine :

McKay has also gone on to find these extra equations, but is was Thompson who first published them. McKay admits that “I was a bit peeved really, I don’t think Thompson quite knew how much I knew.”

By the work of Richard Borcherds we now know the (partial according to some) explanation behind these numerical facts : there is a graded representation $V = \oplus_i V_i $ of the Monster-group (actually, it has a lot of extra structure such as being a vertex algebra) such that the dimension of the i-th factor $V_i $ equals the coefficient f $q^i $ in the j-function. The homogeneous components $V_i $ being finite dimensional representations of the monster, they decompose into the 194 irreducibles $X_j $. For the first three components we have the decompositions

$\begin{cases} V_1 &= X_1 \oplus X_2 \\
V_2 &= X_1 \oplus X_2 \oplus X_3 \\
V_3 &= X_1^{\oplus 2 } \oplus X_2^{\oplus 2} \oplus X_3 \oplus X_4
\end{cases} $

Calculating the dimensions on both sides give the above equations. However, being isomorphisms of monster-representations we are not restricted to just computing the dimensions. We might as well compute the character of any monster-element on both sides (observe that the dimension is just the character of the identity element). Characters are the traces of the matrices describing the action of a monster-element on the representation and these numbers fill the different columns of the character-table above.

Hence, the same integral combinations of the character values of any monster-element give another q-series and these are called the McKay-Thompson series. John Conway discovered them to be classical modular functions known as Hauptmoduln.

In most papers and online material on this only the first few coefficients of these series are documented, which may be just too little information to make new discoveries!

Fortunately, David Madore has compiled the first 3200 coefficients of all the 172 monster-series which are available in a huge 8Mb file. And, if you really need to have more coefficients, you can always use and modify his moonshine python program.

In order to reduce bandwidth, here a list containing the first 100 coefficients of the j-function

jfunct=[196884, 21493760, 864299970, 20245856256, 333202640600, 4252023300096, 44656994071935, 401490886656000, 3176440229784420, 22567393309593600, 146211911499519294, 874313719685775360, 4872010111798142520, 25497827389410525184, 126142916465781843075, 593121772421445058560, 2662842413150775245160, 11459912788444786513920, 47438786801234168813250, 189449976248893390028800, 731811377318137519245696, 2740630712513624654929920, 9971041659937182693533820, 35307453186561427099877376, 121883284330422510433351500, 410789960190307909157638144, 1353563541518646878675077500, 4365689224858876634610401280, 13798375834642999925542288376, 42780782244213262567058227200, 130233693825770295128044873221, 389608006170995911894300098560, 1146329398900810637779611090240, 3319627709139267167263679606784, 9468166135702260431646263438600, 26614365825753796268872151875584, 73773169969725069760801792854360, 201768789947228738648580043776000, 544763881751616630123165410477688, 1452689254439362169794355429376000, 3827767751739363485065598331130120, 9970416600217443268739409968824320, 25683334706395406994774011866319670, 65452367731499268312170283695144960, 165078821568186174782496283155142200, 412189630805216773489544457234333696, 1019253515891576791938652011091437835, 2496774105950716692603315123199672320, 6060574415413720999542378222812650932, 14581598453215019997540391326153984000, 34782974253512490652111111930326416268, 82282309236048637946346570669250805760, 193075525467822574167329529658775261720, 449497224123337477155078537760754122752, 1038483010587949794068925153685932435825, 2381407585309922413499951812839633584128, 5421449889876564723000378957979772088000, 12255365475040820661535516233050165760000, 27513411092859486460692553086168714659374, 61354289505303613617069338272284858777600, 135925092428365503809701809166616289474168, 299210983800076883665074958854523331870720, 654553043491650303064385476041569995365270, 1423197635972716062310802114654243653681152, 3076095473477196763039615540128479523917200, 6610091773782871627445909215080641586954240, 14123583372861184908287080245891873213544410, 30010041497911129625894110839466234009518080, 63419842535335416307760114920603619461313664, 133312625293210235328551896736236879235481600, 278775024890624328476718493296348769305198947, 579989466306862709777897124287027028934656000, 1200647685924154079965706763561795395948173320, 2473342981183106509136265613239678864092991488, 5070711930898997080570078906280842196519646750, 10346906640850426356226316839259822574115946496, 21015945810275143250691058902482079910086459520, 42493520024686459968969327541404178941239869440, 85539981818424975894053769448098796349808643878, 171444843023856632323050507966626554304633241600, 342155525555189176731983869123583942011978493364, 679986843667214052171954098018582522609944965120, 1345823847068981684952596216882155845897900827370, 2652886321384703560252232129659440092172381585408, 5208621342520253933693153488396012720448385783600, 10186635497140956830216811207229975611480797601792, 19845946857715387241695878080425504863628738882125, 38518943830283497365369391336243138882250145792000, 74484518929289017811719989832768142076931259410120, 143507172467283453885515222342782991192353207603200, 275501042616789153749080617893836796951133929783496, 527036058053281764188089220041629201191975505756160, 1004730453440939042843898965365412981690307145827840, 1908864098321310302488604739098618405938938477379584, 3614432179304462681879676809120464684975130836205250, 6821306832689380776546629825653465084003418476904448, 12831568450930566237049157191017104861217433634289960, 24060143444937604997591586090380473418086401696839680, 44972195698011806740150818275177754986409472910549646, 83798831110707476912751950384757452703801918339072000]

This information will come in handy when we will organize our Monstrous Easter Egg Race, starting tomorrow at 6 am (GMT)…

Farey symbols of sporadic groups

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.

exotic chess positions (2)

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

exotic chess positions (1)

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)