Tag: Conway

How to play Nimbers?

Published January 7, 2011 by lievenlb

Nimbers is a 2-person game, winnable only if you understand the arithmetic of the finite fields $\mathbb{F}_{2^{2^n}} $ associated to Fermat 2-powers.

It is played on a rectangular array (say a portion of a Go-board, for practical purposes) having a finite number of stones at distinct intersections. Here’s a typical position

The players alternate making a move, which is either

removing one stone, or
moving a stone to a spot on the same row (resp. the same column) strictly to the left (resp. strictly lower), and if there’s already a stone on this spot, both stones are removed, or
adding stones to the empty corners of a rectangle having as its top-right hand corner a chosen stone and removing stones at the occupied corners

Here we illustrate two possible moves from the above position, in the first we add two new stones and remove 2 existing stones, in the second we add three new stones and remove only the top right-hand stone.

As always, the last player able to move wins the game!

Note that Nimbers is equivalent to Lenstra’s ‘turning corners’-game (as introduced in his paper Nim-multiplication or mentioned in Winning Ways Chapter 14, page 473).

If all stones are placed on the left-most column (or on the bottom row) one quickly realizes that this game reduces to classical Nim with Nim-heap sizes corresponding to the stones (for example, the left-most stone corresponds to a heap of size 3).

Nim-addition $n \oplus m $ is defined inductively by

$n \oplus m = mex(n’ \oplus m,n \oplus m’) $

where $n’ $ is any element of ${ 0,1,\ldots,n-1 } $ and $m’ $ any element of ${ 0,1,\ldots,m-1 } $ and where ‘mex’ stands for Minimal EXcluded number, that is the smallest natural number which isn’t included in the set. Alternatively, one can compute $n \oplus m $ buy writing $n $ and $m $ in binary and add these binary numbers without carrying-over. It is well known that a winning strategy for Nim tries to shorten one Nim-heap such that the Nim-addition of the heap-sizes equals zero.

This allows us to play Nimber-endgames, that is, when all the stones have been moved to the left-column or the bottom row.

To evaluate general Nimber-positions it is best to add another row and column, the coordinate axes of the array

and so our stones lie at positions (1,3), (4,7), (6,4), (10,3) and (14,8). In this way all legal moves follow the rectangle-rule when we allow rectangles to contain corners on the added coordinate axes. For example, removing a stone is achieved by taking a rectangle with two sides on the added axes, and, moving a stone to the left (or the bottom) is done by taking a rectangle with one side at the x-axes (resp. the y-axes)

However, the added stones on the coordinate axes are considered dead and may be removed from the game. This observation allows us to compute the Grundy number of a stone at position (m,n) to be

$G(m,n)=mex(G(m’,n’) \oplus G(m’,n) \oplus G(m,n’)~:~0 \leq m’ < m, 0 \leq n’ < n) $

and so by induction these Grundy numbers are equal to the Nim-multiplication $G(m,n) = m \otimes n $ where

$m \otimes n = mex(m’ \otimes n’ \oplus m’ \otimes n \oplus m \otimes n’~:~0 \leq m’ < m, 0 \leq n’ < n) $

Thus, we can evaluate any Nimbers-position with stone-coordinates smaller than $2^{2^n} $ by calculating in a finite field using the identification (as for example in the odd Knights of the round table-post) $\mathbb{F}_{2^{2^n}} = \{ 0,1,2,\ldots,2^{2^n}-1 \} $

For example, when all stones lie in a 15×15 grid (as in the example above), all calculations can be performed using

Here, we’ve identified the non-zero elements of $\mathbb{F}_{16} $ with 15-th roots of unity, allowing us to multiply, and we’ve paired up couples $(n,n \oplus 1) $ allowing u to reduce nim-addition to nim-multiplication via

$n \oplus m = (n \otimes \frac{1}{m}) \otimes (m \oplus 1) $

In particular, the stone at position (14,8) is equivalent to a Nim-heap of size $14 \otimes 8=10 $. The nim-value of the original position is equal to 8

Suppose your opponent lets you add one extra stone along the diagonal if you allow her to start the game, where would you have to place it and be certain you will win the game?

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Seating the first few billion Knights

Published December 30, 2010 by lievenlb

The odd Knight of the round table problem asks for a consistent placement of the n-th Knight in the row at an odd root of unity, compatible with the two different realizations of the algebraic closure of the field with two elements.

The first identifies the multiplicative group of its non-zero elements with the group of all odd complex roots of unity, under complex multiplication. The second uses Conway’s ‘simplicity rules’ in ONAG to define an addition and multiplication on the set of all ordinal numbers.

Here’s the seating arrangement for the first 15 knights. Conway proved that all finite ordinals smaller than $2^{2^i} $ form a subfield of $\overline{\mathbb{F}}_2 $. The first non-trivial one being $\{ 0,1,2,3 \} $ with smallest multiplicative generator $2 $, whence we place Knight 2 at $e^{2 i \pi/3} $ and as $2^2=3 $ we know where to place the third Knight.

The next subfield is made of the numbers $\{ 0,1,2,\ldots,13,14,15 \} $ and its non-zero numbers form a cyclic group of order 15. Hence we need to find the smallest generator of this group satisfying the additional property of being compatible with the earlier seating, that is, its fifth power must equal to 2. Checking the multiplication table reproduced here one verifies that the wanted generator is 4 and so we place Knight 4 at $e^{\frac{2 \pi i}{15}} $ and as all the ordinals smaller than 16 are powers of 4 this tells us where to place the Knights until we get to the 15th in the row.

In february we were able to seat the first few thousand Knights by showing by hand that 32 is the smallest ordinal such that its 15-th power is equal to 4 and using SAGE that 1051 is the smallest ordinal whose 257-th power equals 32. These calculations enabled us to seat the Knights until we get to the 65536-th in the row.

Today I managed to show that 1361923 is the smallest ordinal such that its 65537-th power equals 1051. You can verify this statement in SAGE using the method explained in the february post

sage: R.< x,y,z,t,u >=GF(2)[]

sage: S.< a,b,c,d,e > =
R.quotient((x^2+x+1,y^2+y+x,z^2+z+x*y,t^2+t+x*y*z,u^2+u+x*y*z*t))

sage: (c*e+b*e+a*b*c*d+b*c*d+a*b*d+a+1)^65537
c^2 + b*d + a + 1

(It takes a bit longer to check minimality of 1361923). That is, we have to seat Knight 1361923 at $e^{\frac{2 \pi i}{4294967295}} $ and because all the numbers smaller than 4294967296 are powers of 1361923 we have seating arrangements for the first 4294967295 Knights!

I did try the same method in february but ran into time- and memory-problems on my 2.4Ghz 2Gb MacBook. Today I upgraded from Sage 3.3 to Sage 4.6 and this version is a lot faster (using the 64-bit architecture) and also appears to be much better at memory-management. Thank you, Sage-community!

Wishing you all a lot of mathematical (and other) fun in the prime-number year 2011.

atb :: lieven.

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So, who did discover the Leech lattice?

Published December 20, 2010 by lievenlb

For the better part of the 30ties, Ernst Witt (1) did hang out with the rest of the ‘Noetherknaben’, the group of young mathematicians around Emmy Noether (3) in Gottingen.

In 1934 Witt became Helmut Hasse‘s assistent in Gottingen, where he qualified as a university lecturer in 1936. By 1938 he has made enough of a name for himself to be offered a lecturer position in Hamburg and soon became an associate professor, the down-graded position held by Emil Artin (2) until he was forced to emigrate in 1937.

A former fellow student of him in Gottingen, Erna Bannow (4), had gone earlier to Hamburg to work with Artin. She continued her studies with Witt and finished her Ph.D. in 1939. In 1940 Erna Bannow and Witt married.

So, life was smiling on Ernst Witt that sunday january 28th 1940, both professionally and personally. There was just one cloud on the horizon, and a rather menacing one. He was called up by the Wehrmacht and knew he had to enter service in february. For all he knew, he was spending the last week-end with his future wife… (later in february 1940, Blaschke helped him to defer his military service by one year).

Still, he desperately wanted to finish his paper before entering the army, so he spend most of that week-end going through the final version and submitted it on monday, as the published paper shows.

In the 70ties, Witt suddenly claimed he did discover the Leech lattice $ {\Lambda} $ that sunday. Last time we have seen that the only written evidence for Witt’s claim is one sentence in his 1941-paper Eine Identität zwischen Modulformen zweiten Grades. “Bei dem Versuch, eine Form aus einer solchen Klassen wirklich anzugeben, fand ich mehr als 10 verschiedene Klassen in $ {\Gamma_{24}} $.”

But then, why didn’t Witt include more details of this sensational lattice in his paper?

Ina Kersten recalls on page 328 of Witt’s collected papers : “In his colloquium talk “Gitter und Mathieu-Gruppen” in Hamburg on January 27, 1970, Witt said that in 1938, he had found nine lattices in $ {\Gamma_{24}} $ and that later on January 28, 1940, while studying the Steiner system $ {S(5,8,24)} $, he had found two additional lattices $ {M} $ and $ {\Lambda} $ in $ {\Gamma_{24}} $. He continued saying that he had then given up the tedious investigation of $ {\Gamma_{24}} $ because of the surprisingly low contribution

$ \displaystyle | Aut(\Lambda) |^{-1} < 10^{-18} $

to the Minkowski density and that he had consented himself with a short note on page 324 in his 1941 paper.”

In the last sentence he refers to the fact that the sum of the inverse orders of the automorphism groups of all even unimodular lattices of a given dimension is a fixed rational number, the Minkowski-Siegel mass constant. In dimension 24 this constant is

$ \displaystyle \sum_{L} \frac{1}{| Aut(L) |} = \frac {1027637932586061520960267}{129477933340026851560636148613120000000} \approx 7.937 \times 10^{-15} $

That is, Witt was disappointed by the low contribution of the Leech lattice to the total constant and concluded that there might be thousands of new even 24-dimensional unimodular lattices out there, and dropped the problem.

If true, the story gets even better : not only claims Witt to have found the lattices $ {A_1^{24}=M} $ and $ {\Lambda} $, but also enough information on the Leech lattice in order to compute the order of its automorphism group $ {Aut(\Lambda)} $, aka the Conway group $ {Co_0 = .0} $ the dotto-group!

Is this possible? Well fortunately, the difficulties one encounters when trying to compute the order of the automorphism group of the Leech lattice from scratch, is one of the better documented mathematical stories around.

The books From Error-Correcting Codes through Sphere Packings to Simple Groups by Thomas Thompson, Symmetry and the monster by Mark Ronan, and Finding moonshine by Marcus du Sautoy tell the story in minute detail.

It took John Conway 12 hours on a 1968 saturday in Cambridge to compute the order of the dotto group, using the knowledge of Leech and McKay on the properties of the Leech lattice and with considerable help offered by John Thompson via telephone.

But then, John Conway is one of the fastest mathematicians the world has known. The prologue of his book On numbers and games begins with : “Just over a quarter of a century ago, for seven consecutive days I sat down and typed from 8:30 am until midnight, with just an hour for lunch, and ever since have described this book as “having been written in a week”.”

Conway may have written a book in one week, Ernst Witt did complete his entire Ph.D. in just one week! In a letter of August 1933, his sister told her parents : “He did not have a thesis topic until July 1, and the thesis was to be submitted by July 7. He did not want to have a topic assigned to him, and when he finally had the idea, he started working day and night, and eventually managed to finish in time.”

So, if someone might have beaten John Conway in fast-computing the dottos order, it may very well have been Witt. Sadly enough, there is a lot of circumstantial evidence to make Witt’s claim highly unlikely.

For starters, psychology. Would you spend your last week-end together with your wife to be before going to war performing an horrendous calculation?

Secondly, mathematical breakthroughs often arise from newly found insight. At that time, Witt was also working on his paper on root lattices “Spiegelungsgrupen and Aufzähling halbeinfacher Liescher Ringe” which he eventually submitted in january 1941. Contained in that paper is what we know as Witt’s lemma which tells us that for any integral lattice the sublattice generated by vectors of norms 1 and 2 is a direct sum of root lattices.

This leads to the trick of trying to construct unimodular lattices by starting with a direct sum of root lattices and ‘adding glue’. Although this gluing-method was introduced by Kneser as late as 1967, Witt must have been aware of it as his 16-dimensional lattice $ {D_{16}^+} $ is constructed this way.

If Witt wanted to construct new 24-dimensional even unimodular lattices in 1940, it would be natural for him to start off with direct sums of root lattices and trying to add vectors to them until he got what he was after. Now, all of the Niemeier-lattices are constructed this way, except for the Leech lattice!

I’m far from an expert on the Niemeier lattices but I would say that Witt definitely knew of the existence of $ {D_{24}^+} $, $ {E_8^3} $ and $ {A_{24}^+} $ and that it is quite likely he also constructed $ {(D_{16}E_8)^+, (D_{12}^2)^+, (A_{12}^2)^+, (D_8^3)^+} $ and possibly $ {(A_{17}E_7)^+} $ and $ {(A_{15}D_9)^+} $. I’d rate it far more likely Witt constructed another two such lattices on sunday january 28th 1940, rather than discovering the Leech lattice.

Finally, wouldn’t it be natural for him to include a remark, in his 1941 paper on root lattices, that not every even unimodular lattices can be obtained from sums of root lattices by adding glue, the Leech lattice being the minimal counter-example?

If it is true he was playing around with the Steiner systems that sunday, it would still be a pretty good story he discovered the lattices $ {(A_2^{12})^+} $ and $ {(A_1^{24})^+} $, for this would mean he discovered the Golay codes in the process!

Which brings us to our next question : who discovered the Golay code?

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Who discovered the Leech lattice?

Published November 30, 2010 by lievenlb

The Leech lattice was, according to wikipedia, ‘originally discovered by Ernst Witt in 1940, but he did not publish his discovery’ and it ‘was later re-discovered in 1965 by John Leech’. However, there is very little evidence to support this claim.

The facts

What is certain is that John Leech discovered in 1965 an amazingly dense 24-dimensional lattice $ {\Lambda} $ having the property that unit balls around the lattice points touch, each one of them having exactly 196560 neighbors. The paper ‘Notes on sphere packings’ appeared in 1967 in the Canad. J. Math. 19, 251-267.

Compare this to the optimal method to place pennies on a table, leading to the hexagonal tiling, each penny touching exactly 6 others. Similarly, in dimension 8 the densest packing is the E8 lattice in which every unit ball has exactly 240 neighbors.

The Leech lattice $ {\Lambda} $ can be characterized as the unique unimodular positive definite even lattice such that the length of any non-zero vector is at least two.

The list of all positive definite even unimodular lattices, $ {\Gamma_{24}} $, in dimension 24 was classified later by Hans-Volker Niemeier and are now known as the 24 Niemeier lattices.

For the chronology below it is perhaps useful to note that, whereas Niemeier’s paper did appear in 1973, it was submitted april 5th 1971 and is just a minor rewrite of Niemeier’s Ph.D. “Definite quadratische Formen der Dimension 24 und Diskriminante 1” obtained in 1968 from the University of Göttingen with advisor Martin Kneser.

The claim

On page 328 of Ernst Witt’s Collected Papers Ina Kersten recalls that Witt gave a colloquium talk on January 27, 1970 in Hamburg entitled “Gitter und Mathieu-Gruppen” (Lattices and Mathieu-groups). In this talk Witt claimed to have found nine lattices in $ {\Gamma_{24}} $ as far back as 1938 and that on January 28, 1940 he found two additional lattices $ {M} $ and $ {\Lambda} $ while studying the Steiner system $ {S(5,8,24)} $.

On page 329 of the collected papers is a scan of the abstract Witt wrote in the colloquium book in Bielefeld where he gave a talk “Uber einige unimodularen Gitter” (On certain unimodular lattices) on January 28, 1972

Here, Witt claims that he found three new lattices in $ {\Gamma_{24}} $ on January 28, 1940 as the lattices $ {M} $, $ {M’} $ and $ {\Lambda} $ ‘feiern heute ihren 32sten Gebursttag!’ (celebrate today their 32nd birthday).

He goes on telling that the lattices $ {M} $ and $ {\Lambda} $ were number 10 and 11 in his list of lattices in $ {\Gamma_{24}} $ in his paper “Eine Identität zwischen Modulformen zweiten Grades” in the Abh. Math. Sem. Univ. Hamburg 14 (1941) 323-337 and he refers in particular to page 324 of that paper.

He further claims that he computed the orders of their automorphism groups and writes that $ {\Lambda} $ ‘wurde 1967 von Leech wieder-entdeckt’ (was re-discovered by Leech in 1967) and that its automorphism group $ {G(\Lambda)} $ was studied by John Conway. Recall that Conway’s investigations of the automorphism group of the Leech lattice led to the discovery of three new sporadic groups, the Conway groups $ {Co_1,Co_2} $ and $ {Co_3} $.

However, Witt’s 1941-paper does not contain a numbered list of 24-dimensional lattices. In fact, apart from $ {E_8+E_8+E_8} $ is does not contain a single lattice in $ {\Gamma_{24}} $. The only relevant paragraph is indeed on page 324

He observes that Mordell already proved that there is just one lattice in $ {\Gamma_8} $ (the $ {E_8} $-lattice) and that the main result of his paper is to prove that there are precisely two even unimodular 16-dimensional lattices : $ {E_8+E_8} $ and another lattice, now usually called the 16-dimensional Witt-lattice.

He then goes on to observe that Schoeneberg knew that $ {\# \Gamma_{24} > 1} $ and so there must be more lattices than $ {E_8+E_8+E_8} $ in $ {\Gamma_{24}} $. Witt concludes with : “In my attempt to find such a lattice, I discovered more than 10 lattices in $ {\Gamma_{24}} $. The determination of $ {\# \Gamma_{24}} $ does not seem to be entirely trivial.”

Hence, it is fair to assume that by 1940 Ernst Witt had discovered at least 11 of the 24 Niemeier lattices. Whether the Leech lattice was indeed lattice 11 on the list is anybody’s guess.

Next time we will look more closely into the historical context of Witt’s 1941 paper.

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Seating the first few thousand Knights

Published February 3, 2010 by lievenlb

The Knight-seating problems asks for a consistent placing of n-th Knight at an odd root of unity, compatible with the two different realizations of the algebraic closure of the field with two elements.
The first identifies the multiplicative group of its non-zero elements with the group of all odd complex roots of unity, under complex multiplication. The second uses Conway’s ‘simplicity rules’ to define an addition and multiplication on the set of all ordinal numbers.

The odd Knights of the round table-problem asks for a specific one-to-one correspondence between two realizations of ‘the’ algebraic closure $\overline{\mathbb{F}_2} $ of the field of two elements.

The first identifies the multiplicative group of its non-zero elements with the group of all odd complex roots of unity, under complex multiplication. The addition on $\overline{\mathbb{F}_2} $ is then recovered by inducing an involution on the odd roots, pairing the one corresponding to x to the one corresponding to x+1.

The second uses Conway’s ‘simplicity rules’ to define an addition and multiplication on the set of all ordinal numbers. Conway proves in ONAG that this becomes an algebraically closed field of characteristic two and that $\overline{\mathbb{F}_2} $ is the subfield of all ordinals smaller than $\omega^{\omega^{\omega}} $. The finite ordinals (the natural numbers) form the quadratic closure of $\mathbb{F}_2 $.

On the natural numbers the Conway-addition is binary addition without carrying and Conway-multiplication is defined by the properties that two different Fermat-powers $N=2^{2^i} $ multiply as they do in the natural numbers, and, Fermat-powers square to its sesquimultiple, that is $N^2=\frac{3}{2}N $. Moreover, all natural numbers smaller than $N=2^{2^{i}} $ form a finite field $\mathbb{F}_{2^{2^i}} $. Using distributivity, one can write down a multiplication table for all 2-powers.

The Knight-seating problems asks for a consistent placing of n-th Knight $K_n $ at an odd root of unity, compatible with the two different realizations of $\overline{\mathbb{F}_2} $. Last time, we were able to place the first 15 Knights as below, and asked where you would seat $K_{16} $

$K_4 $ was placed at $e^{2\pi i/15} $ as 4 was the smallest number generating the ‘Fermat’-field $\mathbb{F}_{2^{2^2}} $ (with multiplicative group of order 15) subject to the compatibility relation with the generator 2 of the smaller Fermat-field $\mathbb{F}_2 $ (with group of order 15) that $4^5=2 $.

To include the next Fermat-field $\mathbb{F}_{2^{2^3}} $ (with multiplicative group of order 255) consistently, we need to find the smallest number n generating the multiplicative group and satisfying the compatibility condition $n^{17}=4 $. Let’s first concentrate on finding the smallest generator : as 2 is a generator for 1st Fermat-field $\mathbb{F}_{2^{2^1}} $ and 4 a generator for the 2-nd Fermat-field $\mathbb{F}_{2^{2^2}} $ a natural conjecture might be that 16 is a generator for the 3-rd Fermat-field $\mathbb{F}_{2^{2^3}} $ and, more generally, that $2^{2^i} $ would be a generator for the next field $\mathbb{F}_{2^{2^{i+1}}} $.

However, an “exercise” in the 1978-paper by Hendrik Lenstra Nim multiplication asks : “Prove that $2^{2^i} $ is a primitive root in the field $\mathbb{F}_{2^{2^{i+1}}} $ if and only if i=0 or 1.”

I’ve struggled with several of the ‘exercises’ in Lenstra’s paper to the extend I feared Alzheimer was setting in, only to find out, after taking pen and paper and spending a considerable amount of time calculating, that they are indeed merely exercises, when looked at properly… (Spoiler-warning : stop reading now if you want to go through this exercise yourself).

In the picture above I’ve added in red the number $x(x+1)=x^2+1 $ to each of the involutions. Clearly, for each pair these numbers are all distinct and we see that for the indicated pairing they make up all numbers strictly less than 8.

By Conway’s simplicity rules (or by checking) the pair (16,17) gives the number 8. In other words, the equation
$x^2+x+8 $ is an irreducible polynomial over $\mathbb{F}_{16} $ having as its roots in $\mathbb{F}_{256} $ the numbers 16 and 17. But then, 16 and 17 are conjugated under the Galois-involution (the Frobenius $y \mapsto y^{16} $). That is, we have $16^{16}=17 $ and $17^{16}=16 $ and hence $16^{17}=8 $. Now, use the multiplication table in $\mathbb{F}_{16} $ given in the previous post (or compute!) to see that 8 is of order 5 (and NOT a generator). As a consequence, the multiplicative order of 16 is 5×17=85 and so 16 cannot be a generator in $\mathbb{F}_{256} $.
For general i one uses the fact that $2^{2^i} $ and $2^{2^i}+1 $ are the roots of the polynomial $x^2+x+\prod_{j<i} 2^{2^j} $ over $\mathbb{F}_{2^{2^i}} $ and argues as before.

Right, but then what is the minimal generator satisfying $n^{17}=4 $? By computing we see that the pairings of all numbers in the range 16…31 give us all numbers in the range 8…15 and by the above argument this implies that the 17-th powers of all numbers smaller than 32 must be different from 4. But then, the smallest candidate is 32 and one verifies that indeed $32^{17}=4 $ (use the multiplication table given before).

Hence, we must place Knight $K_{32} $ at root $e^{2 \pi i/255} $ and place the other Knights prior to the 256-th at the corresponding power of 32. I forgot the argument I used to find-by-hand the requested place for Knight 16, but one can verify that $32^{171}=16 $ so we seat $K_{16} $ at root $e^{342 \pi i/255} $.

But what about Knight $K_{256} $? Well, by this time I was quite good at squaring and binary representations of integers, but also rather tired, and decided to leave that task to the computer.

If we denote Nim-addition and multiplication by $\oplus $ and $\otimes $, then Conway’s simplicity results in ONAG establish a field-isomorphism between $~(\mathbb{N},\oplus,\otimes) $ and the field $\mathbb{F}_2(x_0,x_1,x_2,\ldots ) $ where the $x_i $ satisfy the Artin-Schreier equations

$x_i^2+x_i+\prod_{j < i} x_j = 0 $

and the i-th Fermat-field $\mathbb{F}_{2^{2^i}} $ corresponds to $\mathbb{F}_2(x_0,x_1,\ldots,x_{i-1}) $. The correspondence between numbers and elements from these fields is given by taking $x_i \mapsto 2^{2^i} $. But then, wecan write every 2-power as a product of the $x_i $ and use the binary representation of numbers to perform all Nim-calculations with numbers in these fields.

Therefore, a quick and dirty way (and by no means the most efficient) to do Nim-calculations in the next Fermat-field consisting of all numbers smaller than 65536, is to use sage and set up the field $\mathbb{F}_2(x_0,x_1,x_2,x_3) $ by

R.< x,y,z,t > =GF(2)[]
S.< a,b,c,d >=R.quotient((x^2+x+1,y^2+y+x,z^2+z+x*y,t^2+t+x*y*z))

To find the smallest number generating the multiplicative group and satisfying the additional compatibility condition $n^{257}=32 $ we have to find the smallest binary number $i_1i_2 \ldots i_{16} $ (larger than 255) satisfying

(i1*a*b*c*t+i2*b*c*t+i3*a*c*t+i4*c*t+i5*a*b*t+i6*b*t+
i7*a*t+i8*t+i9*a*b*c+i10*b*c+i11*a*c+i12*c+i13*a*b+
i14*b+i15*a+i16)^257=a*c

It takes a 2.4GHz 2Gb-RAM MacBook not that long to decide that the requested generator is 1051 (killing another optimistic conjecture that these generators might be 2-powers). So, we seat Knight
$K_{1051} $ at root $e^{2 \pi i/65535} $ and can then arrange seatings for all Knight queued up until we reach the 65536-th! In particular, the first Knight we couldn’t place before, that is Knight $K_{256} $, will be seated at root $e^{65826 \pi i/65535} $.

If you’re lucky enough to own a computer with more RAM, or have the patience to make the search more efficient and get the seating arrangement for the next Fermat-field, please drop a comment.

I’ll leave you with another Lenstra-exercise which shouldn’t be too difficult for you to solve now : “Prove that $x^3=2^{2^i} $ has three solutions in $\mathbb{N} $ for each $i \geq 2 $.”

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The odd knights of the round table

Published January 28, 2010 by lievenlb

Here’s a tiny problem illustrating our limited knowledge of finite fields : “Imagine an infinite queue of Knights ${ K_1,K_2,K_3,\ldots } $, waiting to be seated at the unit-circular table. The master of ceremony (that is, you) must give Knights $K_a $ and $K_b $ a place at an odd root of unity, say $\omega_a $ and $\omega_b $, such that the seat at the odd root of unity $\omega_a \times \omega_b $ must be given to the Knight $K_{a \otimes b} $, where $a \otimes b $ is the Nim-multiplication of $a $ and $b $. Which place would you offer to Knight $K_{16} $, or Knight $K_n $, or, if you’re into ordinals, Knight $K_{\omega} $?”

What does this have to do with finite fields? Well, consider the simplest of all finite field $\mathbb{F}_2 = { 0,1 } $ and consider its algebraic closure $\overline{\mathbb{F}_2} $. Last year, we’ve run a series starting here, identifying the field $\overline{\mathbb{F}_2} $, following John H. Conway in ONAG, with the set of all ordinals smaller than $\omega^{\omega^{\omega}} $, given the Nim addition and multiplication. I know that ordinal numbers may be intimidating at first, so let’s just restrict to ordinary natural numbers for now. The Nim-addition of two numbers $n \oplus m $ can be calculated by writing the numbers n and m in binary form and add them without carrying. For example, $9 \oplus 1 = 1001+1 = 1000 = 8 $. Nim-multiplication is slightly more complicated and is best expressed using the so-called Fermat-powers $F_n = 2^{2^n} $. We then demand that $F_n \otimes m = F_n \times m $ whenever $m < F_n $ and $F_n \otimes F_n = \frac{3}{2}F_n $. Distributivity wrt. $\oplus $ can then be used to calculate arbitrary Nim-products. For example, $8 \otimes 3 = (4 \otimes 2) \otimes (2 \oplus 1) = (4 \otimes 3) \oplus (4 \otimes 2) = 12 \oplus 8 = 4 $. Conway’s remarkable result asserts that the ordinal numbers, equipped with Nim addition and multiplication, form an algebraically closed field of characteristic two. The closure $\overline{\mathbb{F}_2} $ is identified with the subfield of all ordinals smaller than $\omega^{\omega^{\omega}} $. For those of you who don’t feel like going transfinite, the subfield $~(\mathbb{N},\oplus,\otimes) $ is identified with the quadratic closure of $\mathbb{F}_2 $.

The connection between $\overline{\mathbb{F}_2} $ and the odd roots of unity has been advocated by Alain Connes in his talk before a general public at the IHES : “L’ange de la géométrie, le diable de l’algèbre et le corps à un élément” (the angel of geometry, the devil of algebra and the field with one element). He describes its content briefly in this YouTube-video

At first it was unclear to me which ‘coupling-problem’ Alain meant, but this has been clarified in his paper together with Caterina Consani Characteristic one, entropy and the absolute point. The non-zero elements of $\overline{\mathbb{F}_2} $ can be identified with the set of all odd roots of unity. For, if x is such a unit, it belongs to a finite subfield of the form $\mathbb{F}_{2^n} $ for some n, and, as the group of units of any finite field is cyclic, x is an element of order $2^n-1 $. Hence, $\mathbb{F}_{2^n}- { 0 } $ can be identified with the set of $2^n-1 $-roots of unity, with $e^{2 \pi i/n} $ corresponding to a generator of the unit-group. So, all elements of $\overline{\mathbb{F}_2} $ correspond to an odd root of unity. The observation that we get indeed all odd roots of unity may take you a couple of seconds (( If m is odd, then (2,m)=1 and so 2 is a unit in the finite cyclic group $~(\mathbb{Z}/m\mathbb{Z})^* $ whence $2^n = 1 (mod~m) $, so the m-roots of unity lie within those of order $2^n-1 $ )).

Assuming we succeed in fixing a one-to-one correspondence between the non-zero elements of $\overline{\mathbb{F}_2} $ and the odd roots of unity $\mu_{odd} $ respecting multiplication, how can we recover the addition on $\overline{\mathbb{F}_2} $? Well, here’s Alain’s coupling function, he ties up an element x of the algebraic closure to the element s(x)=x+1 (and as we are in characteristic two, this is an involution, so also the element tied up to x+1 is s(x+1)=(x+1)+1=x. The clue being that multiplication together with the coupling map s allows us to compute any sum of two elements as $x+y=x \times s(\frac{y}{x}) = x \times (\frac{y}{x}+1) $.
For example, all information about the finite field $\mathbb{F}_{2^4} $ is encoded in this identification with the 15-th roots of unity, together with the pairing s depicted as

Okay, we now have two identifications of the algebraic closure $\overline{\mathbb{F}_2} $ : the smaller ordinals equipped with Nim addition and Nim multiplication and the odd roots of unity with complex-multiplication and the Connes-coupling s. The question we started from asks for a general recipe to identify these two approaches.

To those of you who are convinced that finite fields (LOL, even characteristic two!) are objects far too trivial to bother thinking about : as far as I know, NOBODY knows how to do this explicitly, even restricting the ordinals to merely the natural numbers!

Please feel challenged! To get you started, I’ll show you how to place the first 15 Knights and give you a procedure (though far from explicit) to continue. Here’s the Nim-picture compatible with that above

To verify this, and to illustrate the general strategy, I’d better hand you the Nim-tables of the first 16 numbers. Here they are

It is known that the finite subfields of $~(\mathbb{N},\oplus,\otimes) $ are precisely the sets of numbers smaller than the Fermat-powers $F_n $. So, the first one is all numbers smaller than $F_1=4 $ (check!). The smallest generator of the multiplicative group (of order 3) is 2, so we take this to correspond to the unit-root $e^{2 \pi i/3} $. The next subfield are all numbers smaller than $F_2 = 16 $ and its multiplicative group has order 15. Now, choose the smallest integer k which generates this group, compatible with the condition that $k^{\otimes 5}=2 $. Verify that this number is 4 and that this forces the identification and coupling given above.

The next finite subfield would consist of all natural numbers smaller than $F_3=256 $. Hence, in this field we are looking for the smallest number k generating the multiplicative group of order 255 satisfying the extra condition that $k^{\otimes 17}=4 $ which would fix an identification at that level. Then, the next level would be all numbers smaller than $F_4=65536 $ and again we would like to find the smallest number generating the multiplicative group and such that the appropriate power is equal to the aforementioned k, etc. etc.

Can you give explicit (even inductive) formulae to achieve this? I guess even the problem of placing Knight 16 will give you a couple of hours to think about… (to be continued).

Olivier Messiaen & Mathieu 12

Published December 31, 2009 by lievenlb

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

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

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

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

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

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

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

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

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

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

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

Here’s the piece performed on organ :

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

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

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

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E(8) from moonshine groups

Published May 15, 2009 by lievenlb

Are the valencies of the 171 moonshine groups are compatible, that is, can one construct a (disconnected) graph on the 171 vertices such that in every vertex (determined by a moonshine group G) the vertex-valency coincides with the valency of the corresponding group? Duncan describes a subset of 9 moonshine groups for which the valencies are compatible. These 9 groups are characterized as those moonshine groups G
having width 1 at the cusp and such that their intersection with the modular group is big.

Time to wrap up this series on John Duncan‘s paper Arithmetic groups and the affine E8 Dynkin diagram in which he gives a realization of the extended E(8)-Dynkin diagram (together with its isotropic root vector) from the moonshine groups, compatible with McKay’s E(8)-observation.

In the previous post we have described all 171 moonshine groups using Conway’s big picture. This description will allow us to associate two numbers to a moonshine group $G \subset PSL_2(\mathbb{R}) $.
Recall that for any such group we have a positive integer $N $ such that

$\Gamma_0(N) \subset G \subset \Gamma_0(h,\frac{N}{h})+ $

where $h $ is the largest divisor of 24 such that $h^2 | N $. Let us call $n_G=\frac{N}{h} $ the dimension of $G $ (Duncan calls this number the ‘normalized level’) as it will give us the dimension component at the vertex determined by $G $.

We have also seen last time that any moonshine group is of the form $G = \Gamma_0(n_G || h)+e,f,g $, that is, $G/\Gamma_0(n_G ||h) $ is an elementary abelian group $~(\mathbb{Z}/2\mathbb{Z})^m $ generated by Atkin-Lehner involutions. Let’s call $v_G=m+1 $ the valency of the group $G $ as it will give s the valency of the vertex determined by $G $.

It would be nice to know whether the valencies of the 171 moonshine groups are compatible, that is, whether one can construct a (disconnected) graph on the 171 vertices such that in each vertex (determined by a moonshine group $G $) the vertex-valency coincides with the valency of the corresponding group.

Duncan describes a subset of 9 moonshine groups for which the valencies are compatible. These 9 groups are characterized as those moonshine groups $G $
having width 1 at the cusp and such that their intersection with the modular group $\Gamma = PSL_2(\mathbb{Z}) $ is big, more precisely the index $[\Gamma : \Gamma \cap G] \leq 12 $ and $[\Gamma : \Gamma \cap G]/[G : \Gamma \cap G] \leq 3 $.

They can be described using the mini-moonshine picture on the right. They are :

The modular group itself $1=\Gamma $, being the stabilizer of the lattice 1. This group has clearly dimension and valency equal to one.

The modular subgroup $2=\Gamma_0(2) $ being the point-wise stabilizer of the lattices 1 and 2 (so it has valency one and dimension two, and, its normalizer $2+ =\Gamma_0(2)+ $ which is the set-wise stabilizer of the lattices 1 and 2 and the one Atkin-Lehner involution interchanges both. So, this group has valency two (as we added one involution) as well as dimension two.

Likewise, the groups $3+=\Gamma_0(3)+ $ and $5+=\Gamma_0(5)+ $ are the stabilzer subgroups of the red 1-cell (1,3) resp. the green 1-cell (1,5) and hence have valency two (as we add one involution) and dimensions 3 resp. 5.

The group $4+=\Gamma_0(4)+ $ stabilizes the (1|4)-thread and as we add one involution must have valency 2 and dimension 4.

On the other hand, the group $6+=\Gamma_0(6)+ $ stabilizes the unique 2-cell in the picture (having lattices 1,2,3,6) so this time we will add three involutions (horizontal and vertical switches and their product the antipodal involution). Hence, for this group the valency is three and its dimension is equal to six.

Remain the two groups connected to the mini-snakes in the picture. The red mini-snake (top left hand) is the ball with center 3 and hyperdistance 3 and determines the group $3||3=\Gamma_0(3||3) $ which has valency one (we add no involutions) and dimension 3. The blue mini-snake (the extended D(5)-Dynkin in the lower right corner) determines the group $4||2+=\Gamma(4||2)+ $ which has valency two and dimension 4.

The valencies of these 9 moonshine groups are compatible and they can be arranged in the extended E(8) diagram depicted below

Moreover, the dimensions of the groups give the exact dimension-components of the isotropic root of the extended E(8)-diagram. Further, the dimension of the group is equal to the order of the elements making up the conjugacy class of the monster to which exactly the given groups correspond via monstrous moonshine and hence compatible with John McKay‘s original E(8)-observation!

Once again, I would love to hear when someone has more information on the cell-decomposition of the moonshine picture or if someone can extend the moonshine E(8)-graph, possibly to include all 171 moonshine groups.

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looking for the moonshine picture

Published May 11, 2009 by lievenlb

We have seen that Conway’s big picture helps us to determine all arithmetic subgroups of $PSL_2(\mathbb{R}) $ commensurable with the modular group $PSL_2(\mathbb{Z}) $, including all groups of monstrous moonshine.

As there are exactly 171 such moonshine groups, they are determined by a finite subgraph of Conway’s picture and we call the minimal such subgraph the moonshine picture. Clearly, we would like to determine its structure.

On the left a depiction of a very small part of it. It is the minimal subgraph of Conway’s picture needed to describe the 9 moonshine groups appearing in Duncan’s realization of McKay’s E(8)-observation. Here, only three primes are relevant : 2 (blue lines), 3 (reds) and 5 (green). All lattices are number-like (recall that $M \frac{g}{h} $ stands for the lattice $\langle M e_1 + \frac{g}{h} e_2, e_2 \rangle $).

We observe that a large part of this mini-moonshine picture consists of the three p-tree subgraphs (the blue, red and green tree starting at the 1-lattice $1 = \langle e_1,e_2 \rangle $. Whereas Conway’s big picture is the product over all p-trees with p running over all prime numbers, we observe that the mini-moonshine picture is a very small subgraph of the product of these three subtrees. In fact, there is just one 2-cell (the square 1,2,6,3).

Hence, it seems like a good idea to start our investigation of the full moonshine picture with the determination of the p-subtrees contained in it, and subsequently, worry about higher dimensional cells constructed from them. Surely it will be no major surprise that the prime numbers p that appear in the moonshine picture are exactly the prime divisors of the order of the monster group, that is p=2,3,5,7,11,13,17,19,23,29,31,41,47,59 or 71. Before we can try to determine these 15 p-trees, we need to know more about the 171 moonshine groups.

Recall that the proper way to view the modular subgroup $\Gamma_0(N) $ is as the subgroup fixing the two lattices $L_1 $ and $L_N $, whence we will write $\Gamma_0(N)=\Gamma_0(N|1) $, and, by extension we will denote with $\Gamma_0(X|Y) $ the subgroup fixing the two lattices $L_X $ and $L_Y $.

As $\Gamma_0(N) $ fixes $L_1 $ and $L_N $ it also fixes all lattices in the (N|1)-thread, that is all lattices occurring in a shortest path from $L_1 $ to $L_N $ (on the left a picture of the (200|1)-thread).

If $N=p_1^{a_1} p_2^{a_2} \ldots p_k^{a_k} $, then the (N|1)-thread has $2^k $ involutions as symmetries, called the Atkin-Lehner involutions. For every exact divisor $e || N $ (that is, $e|N $ and $gcd(e,\frac{N}{e})=1 $ we have an involution $W_e $ which acts by sending each point in the thread-cell corresponding to the prime divisors of $e $ to its antipodal cell-point and acts as the identity on the other prime-axes. For example, in the (200|1)-thread on the left, $W_8 $ is the left-right reflexion, $W_{25} $ the top-bottom reflexion and $W_{200} $ the antipodal reflexion. The set of all exact divisors of N becomes the group $~(\mathbb{Z}/2\mathbb{Z})^k $ under the operation $e \ast f = \frac{e \times f}{gcd(e,f)^2} $.

Most of the moonshine groups are of the form $\Gamma_0(n|h)+e,f,g,… $ for some $N=h.n $ such that $h | 24 $ and $h^2 | N $. The group $\Gamma_0(n|h) $ is then conjugate to the modular subgroup $\Gamma_0(\frac{n}{h}) $ by the element $\begin{bmatrix} h & 0 \ 0 & 1 \end{bmatrix} $. With $\Gamma_0(n|h)+e,f,g,… $ we mean that the group $\Gamma_0(n|h) $ is extended with the involutions $W_e,W_f,W_g,… $. If we simply add all Atkin-Lehner involutions we write $\Gamma_0(n|h)+ $ for the resulting group.

Finally, whenever $h \not= 1 $ there is a subgroup $\Gamma_0(n||h)+e,f,g,… $ which is the kernel of a character $\lambda $ being trivial on $\Gamma_0(N) $ and on all involutions $W_e $ for which every prime dividing $e $ also divides $\frac{n}{h} $, evaluating to $e^{\frac{2\pi i}{h}} $ on all cosets containing $\begin{bmatrix} 1 & \frac{1}{h} \ 0 & 1 \end{bmatrix} $ and to $e^{\pm \frac{2 \pi i }{h}} $ for cosets containing $\begin{bmatrix} 1 & 0 \ n & 0 \end{bmatrix} $ (with a + sign if $\begin{bmatrix} 0 & -1 \ N & 0 \end{bmatrix} $ is present and a – sign otherwise). Btw. it is not evident at all that this is a character, but hard work shows it is!

Clearly there are heavy restrictions on the numbers that actually occur in moonshine. In the paper On the discrete groups of moonshine, John Conway, John McKay and Abdellah Sebbar characterized the 171 arithmetic subgroups of $PSL_2(\mathbb{R}) $ occuring in monstrous moonshine as those of the form $G = \Gamma_0(n || h)+e,f,g,… $ which are

(a) of genus zero, meaning that the quotient of the upper-half plane by the action of $G \subset PSL_2(\mathbb{R}) $ by Moebius-transformations gives a Riemann surface of genus zero,
(b) the quotient group $G/\Gamma_0(nh) $ is a group of exponent 2 (generated by some Atkin-Lehner involutions), and
(c) every cusp can be mapped to $\infty $ by an element of $PSL_2(\mathbb{R}) $ which conjugates the group to one containing $\Gamma_0(nh) $.

Now, if $\Gamma_0(n || h)+e,f,g,… $ is of genus zero, so is the larger group $\Gamma_0(n | h)+e,f,g,… $, which in turn, is conjugated to the group $\Gamma_0(\frac{n}{h})+e,f,g,… $. Therefore, we need a list of all groups of the form $\Gamma_0(\frac{n}{h})+e,f,g,… $ which are of genus zero. There are exactly 123 of them, listed on the right.

How does this help to determine the structure of the p-subtree of the moonshine picture for the fifteen monster-primes p? Look for the largest p-power $p^k $ such that $p^k+e,f,g… $ appears in the list. That is for p=2,3,5,7,11,13,17,19,23,29,31,41,47,59,71 these powers are resp. 5,3,2,2,1,1,1,1,1,1,1,1,1,1,1. Next, look for the largest p-power $p^l $ dividing 24 (that is, 3 for p=2, 1 for p=3 and 0 for all other primes). Then, these relevant moonshine groups contain the modular subgroup $\Gamma_0(p^{k+2l}) $ and are contained in its normalizer in $PSL_2(\mathbb{R}) $ which by the Atkin-Lehner theorem is precisely the group $\Gamma_0(p^{k+l}|p^l)+ $.

Right, now the lattices fixed by $\Gamma_0(p^{k+2l}) $ (and permuted by its normalizer), that is the lattices in our p-subtree, are those that form the $~(p^{k+2l}|1) $-snake in Conway-speak. That is, the lattices whose hyper-distance to the $~(p^{k+l}|p^l) $-thread divides 24. So for all primes larger than 2 or 3, the p-tree is just the $~(p^l|1) $-thread.

For p=3 the 3-tree is the (243|1)-snake having the (81|3)-thread as its spine. It contains the following lattices, all of which are number-like.

Depicting the 2-tree, which is the (2048|1)-snake may take a bit longer… Perhaps someone should spend some time figuring out which cells of the product of these fifteen trees make up the moonshine picture!

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Conway’s big picture

Published May 2, 2009 by lievenlb

Conway and Norton showed that there are exactly 171 moonshine functions and associated two arithmetic subgroups to them. We want a tool to describe these and here’s where Conway’s big picture comes in very handy. All moonshine groups are arithmetic groups, that is, they are commensurable with the modular group. Conway’s idea is to view several of these groups as point- or set-wise stabilizer subgroups of finite sets of (projective) commensurable 2-dimensional lattices.

Expanding (and partially explaining) the original moonshine observation of McKay and Thompson, John Conway and Simon Norton formulated monstrous moonshine :

To every cyclic subgroup $\langle m \rangle $ of the Monster $\mathbb{M} $ is associated a function

$f_m(\tau)=\frac{1}{q}+a_1q+a_2q^2+\ldots $ with $q=e^{2 \pi i \tau} $ and all coefficients $a_i \in \mathbb{Z} $ are characters at $m $ of a representation of $\mathbb{M} $. These representations are the homogeneous components of the so called Moonshine module.

Each $f_m $ is a principal modulus for a certain genus zero congruence group commensurable with the modular group $\Gamma = PSL_2(\mathbb{Z}) $. These groups are called the moonshine groups.

Conway and Norton showed that there are exactly 171 different functions $f_m $ and associated two arithmetic subgroups $F(m) \subset E(m) \subset PSL_2(\mathbb{R}) $ to them (in most cases, but not all, these two groups coincide).

Whereas there is an extensive literature on subgroups of the modular group (see for instance the series of posts starting here), most moonshine groups are not contained in the modular group. So, we need a tool to describe them and here’s where Conway’s big picture comes in very handy.

All moonshine groups are arithmetic groups, that is, they are subgroups $G $ of $PSL_2(\mathbb{R}) $ which are commensurable with the modular group $\Gamma = PSL_2(\mathbb{Z}) $ meaning that the intersection $G \cap \Gamma $ is of finite index in both $G $ and in $\Gamma $. Conway’s idea is to view several of these groups as point- or set-wise stabilizer subgroups of finite sets of (projective) commensurable 2-dimensional lattices.

Start with a fixed two dimensional lattice $L_1 = \mathbb{Z} e_1 + \mathbb{Z} e_2 = \langle e_1,e_2 \rangle $ and we want to name all lattices of the form $L = \langle v_1= a e_1+ b e_2, v_2 = c e_1 + d e_2 \rangle $ that are commensurable to $L_1 $. Again this means that the intersection $L \cap L_1 $ is of finite index in both lattices. From this it follows immediately that all coefficients $a,b,c,d $ are rational numbers.

It simplifies matters enormously if we do not look at lattices individually but rather at projective equivalence classes, that is $~L=\langle v_1, v_2 \rangle \sim L’ = \langle v’_1,v’_2 \rangle $ if there is a rational number $\lambda \in \mathbb{Q} $ such that $~\lambda v_1 = v’_1, \lambda v_2=v’_2 $. Further, we are of course allowed to choose a different ‘basis’ for our lattices, that is, $~L = \langle v_1,v_2 \rangle = \langle w_1,w_2 \rangle $ whenever $~(w_1,w_2) = (v_1,v_2).\gamma $ for some $\gamma \in PSL_2(\mathbb{Z}) $.
Using both operations we can get any lattice in a specific form. For example,

$\langle \frac{1}{2}e_1+3e_2,e_1-\frac{1}{3}e_2 \overset{(1)}{=} \langle 3 e_1+18e_2,6e_1-2e_2 \rangle \overset{(2)}{=} \langle 3 e_1+18 e_2,38 e_2 \rangle \overset{(3)}{=} \langle \frac{3}{38}e_1+\frac{9}{19}e_2,e_2 \rangle $

Here, identities (1) and (3) follow from projective equivalence and identity (2) from a base-change. In general, any lattice $L $ commensurable to the standard lattice $L_1 $ can be rewritten uniquely as $L = \langle Me_1 + \frac{g}{h} e_2,e_2 \rangle $ where $M $ a positive rational number and with $0 \leq \frac{g}{h} < 1 $.

Another major feature is that one can define a symmetric hyper-distance between (equivalence classes of) such lattices. Take $L=\langle Me_1 + \frac{g}{h} e_2,e_2 \rangle $ and $L’=\langle N e_1 + \frac{i}{j} e_2,e_2 \rangle $ and consider the matrix

$D_{LL’} = \begin{bmatrix} M & \frac{g}{h} \\ 0 & 1 \end{bmatrix} \begin{bmatrix} N & \frac{i}{j} \\ 0 & 1 \end{bmatrix}^{-1} $ and let $\alpha $ be the smallest positive rational number such that all entries of the matrix $\alpha.D_{LL’} $ are integers, then

$\delta(L,L’) = det(\alpha.D_{LL’}) \in \mathbb{N} $ defines a symmetric hyperdistance which depends only of the equivalence classes of lattices (hyperdistance because the log of it behaves like an ordinary distance).

Conway’s big picture is the graph obtained by taking as its vertices the equivalence classes of lattices commensurable with $L_1 $ and with edges connecting any two lattices separated by a prime number hyperdistance. Here’s part of the 2-picture, that is, only depicting the edges of hyperdistance 2.

The 2-picture is an infinite 3-valent tree as there are precisely 3 classes of lattices at hyperdistance 2 from any lattice $L = \langle v_1,v_2 \rangle $ namely (the equivalence classes of) $\langle \frac{1}{2}v_1,v_2 \rangle~,~\langle v_1, \frac{1}{2} v_2 \rangle $ and $\langle \frac{1}{2}(v_1+v_2),v_2 \rangle $.

Similarly, for any prime hyperdistance p, the p-picture is an infinite p+1-valent tree and the big picture is the product over all these prime trees. That is, two lattices at square-free hyperdistance $N=p_1p_2\ldots p_k $ are two corners of a k-cell in the big picture!
(Astute readers of this blog (if such people exist…) may observe that Conway’s big picture did already appear here prominently, though in disguise. More on this another time).

The big picture presents a simple way to look at arithmetic groups and makes many facts about them visually immediate. For example, the point-stabilizer subgroup of $L_1 $ clearly is the modular group $PSL_2(\mathbb{Z}) $. The point-stabilizer of any other lattice is a certain conjugate of the modular group inside $PSL_2(\mathbb{R}) $. For example, the stabilizer subgroup of the lattice $L_N = \langle Ne_1,e_2 \rangle $ (at hyperdistance N from $L_1 $) is the subgroup

${ \begin{bmatrix} a & \frac{b}{N} \\ Nc & d \end{bmatrix}~|~\begin{bmatrix} a & b \\ c & d \end{bmatrix} \in PSL_2(\mathbb{Z})~} $

Now the intersection of these two groups is the modular subgroup $\Gamma_0(N) $ (consisting of those modular group element whose lower left-hand entry is divisible by N). That is, the proper way to look at this arithmetic group is as the joint stabilizer of the two lattices $L_1,L_N $. The picture makes it trivial to compute the index of this subgroup.

Consider the ball $B(L_1,N) $ with center $L_1 $ and hyper-radius N (on the left, the ball with hyper-radius 4). Then, it is easy to show that the modular group acts transitively on the boundary lattices (including the lattice $L_N $), whence the index $[ \Gamma : \Gamma_0(N)] $ is just the number of these boundary lattices. For N=4 the picture shows that there are exactly 6 of them. In general, it follows from our knowledge of all the p-trees the number of all lattices at hyperdistance N from $L_1 $ is equal to $N \prod_{p | N}(1+ \frac{1}{p}) $, in accordance with the well-known index formula for these modular subgroups!

But, there are many other applications of the big picture giving a simple interpretation for the Hecke operators, an elegant proof of the Atkin-Lehner theorem on the normalizer of $\Gamma_0(N) $ (the whimsical source of appearances of the number 24) and of Helling’s theorem characterizing maximal arithmetical groups inside $PSL_2(\mathbb{C}) $ as conjugates of the normalizers of $\Gamma_0(N) $ for square-free N.
J.H. Conway’s paper “Understanding groups like $\Gamma_0(N) $” containing all this material is a must-read! Unfortunately, I do not know of an online version.