# Tag: Conway

Here’s the upper part of Kneser‘s neighbourhood graph of the Niemeier lattices:

The Leech lattice has a unique neighbour, that is, among the $23$ remaining Niemeier lattices there is a unique one, $(A_1^{24})^+$, sharing an index two sub-lattice with the Leech.

How would you try to construct $(A_1^{24})^+$, an even unimodular lattice having the same roots as $A_1^{24}$?

The root lattice $A_1$ is $\sqrt{2} \mathbb{Z}$. It has two roots $\pm \sqrt{2}$, determinant $2$, its dual lattice is $A_1^* = \tfrac{1}{\sqrt{2}} \mathbb{Z}$ and we have $A_1^*/A_1 \simeq C_2 \simeq \mathbb{F}_2$.

Thus, $A_1^{24}= \sqrt{2} \mathbb{Z}^{\oplus 24}$ has $48$ roots, determinant $2^{24}$, its dual lattice is $(A_1^{24})^* = \tfrac{1}{\sqrt{2}} \mathbb{Z}^{\oplus 24}$ and the quotient group $(A_1^{24})^*/A_1^{24}$ is $C_2^{24}$ isomorphic to the additive subgroup of $\mathbb{F}_2^{\oplus 24}$.

A larger lattice $A_1^{24} \subseteq L$ of index $k$ gives for the dual lattices an extension $L^* \subseteq (A_1^{24})^*$, also of index $k$. If $L$ were unimodular, then the index has to be $2^{12}$ because we have the situation
$A_1^{24} \subseteq L = L^* \subseteq (A_1^{24})^*$
So, Kneser’s glue vectors form a $12$-dimensional subspace $\mathcal{C}$ in $\mathbb{F}_2^{\oplus 24}$, that is,
$L = \mathcal{C} \underset{\mathbb{F}_2}{\times} (A_1^{24})^* = \{ \tfrac{1}{\sqrt{2}} \vec{v} ~|~\vec{v} \in \mathbb{Z}^{\oplus 24},~v=\vec{v}~mod~2 \in \mathcal{C} \}$
Because $L = L^*$, the linear code $\mathcal{C}$ must be self-dual meaning that $v.w = 0$ (in $\mathbb{F}_2$) for all $v,w \in \mathcal{C}$. Further, we want that the roots of $A_1^{24}$ and $L$ are the same, so the minimal number of non-zero coordinates in $v \in \mathcal{C}$ must be $8$.

That is, $\mathcal{C}$ must be a self-dual binary code of length $24$ with Hamming distance $8$.

Marcel Golay (1902-1989) – Photo Credit

We now know that there is a unique such code, the (extended) binary Golay code, $\mathcal{C}_{24}$, which has

• one vector of weight $0$
• $759$ vectors of weight $8$ (called ‘octads’)
• $2576$ vectors of weight $12$ (called ‘dodecads’)
• $759$ vectors of weight $16$
• one vector of weight $24$

The $759$ octads form a Steiner system $S(5,8,24)$ (that is, for any $5$-subset $S$ of the $24$-coordinates there is a unique octad having its non-zero coordinates containing $S$).

Witt constructed a Steiner system $S(5,8,24)$ in his 1938 paper “Die $5$-fach transitiven Gruppen von Mathieu”, so it is not unthinkable that he checked the subspace of $\mathbb{F}_2^{\oplus 24}$ spanned by his $759$ octads to be $12$-dimensional and self-dual, thereby constructing the Niemeier-lattice $(A_1^{24})^+$ on that sunday in 1940.

John Conway classified all nine self-dual codes of length $24$ in which the weight
of every codeword is a multiple of $4$. Each one of these codes $\mathcal{C}$ gives a Niemeier lattice $\mathcal{C} \underset{\mathbb{F}_2}{\times} (A_1^{24})^*$, all but one of them having more roots than $A_1^{24}$.

Vera Pless and Neil Sloan classified all $26$ binary self-dual codes of length $24$.

Around 1975 Sir Roger Penrose discovered his aperiodic P2 tilings of the plane, using only two puzzle pieces: Kites (K) and Darts (D)

The inner angles of these pieces are all multiples of $36^o = \tfrac{180^o}{5}$, the short edges have length $1$, and the long edges have length $\tau = \tfrac{1+\sqrt{5}}{2}$, the golden ratio. These pieces must be joined together so that the colours match up (if we do not use this rule it is easy to get periodic tilings with these two pieces).

There is plenty of excellent online material available:

• The two original Martin Gardner Scientific American articles on Penrose tiles have been made available by the MAA, and were reprinted in “Penrose Tiles to Trapdoor Ciphers”. They contain most of Conway’s early discoveries about these tilings, but without proofs.
• A JavaScript application by Kevin Bertman to play around with these tilings. You can deflate and inflate tilings, find forced tiles and much more. Beneath the app-window there’s a detailed explanation of all the basics, including inflation and deflation of the P2-tiles, the seven types of local vertex configurations (naming by Conway, of course),

proofs of aperiodicity (similar to the one for Conway’s musical sequences), that every tile lies within an ace (similar to the LSL-subword in musical sequences) with application to local isomorphism (again similar to the $1$-dimensional case).
• Course notes of an Oxford masterclass in geometry Lectures on Penrose tilings by Alexander Ritter, again with proofs of all of the above and a discussion about the Cartwheel tilings (similar to that in the post on musical sequences), giving an algorithm to decide whether or not a partial tiling can be extended to the entire plane, or not.

There’s no point copying this material here. Rather, I’d like to use some time in this GoV series of posts to talk about de Bruijn’s pentagrid results. For this reason, I now need to make the connection with Penrose’s ‘other’ tilings, the P3 tiles of ‘thin’ and ‘thick’ rhombi (sometimes called ‘skinny’ and ‘fat’ rhombi).

Every Penrose P2-tiling can be turned into a P3-rhombic tiling, and conversely.

From kites and darts to rhombi: divide every kite in two halves along its line of reflection. Then combine darts and half-kites into rhombi where a fat rhombus consists of a dart and two half-kites, joined at a long edge, and a skinny rhombus is made of two half-kites, joined at a short edge. This can always be done preserving the gluing conditions. It suffices to verify this for the deflated kite and dart (on the left below) and we see that the matching colour-conditions are those of the rhombi.

From rhombi to kites and darts: divide every fat rhombus into a dart (placed at the red acute angle) and two half-kites, joined at a long edge, and divide every skinny rhombus into two half-kites along its short diagonal.

All results holding for Kites and Darts tilings have therefore their versions for Rhombic tilings. For example, we have rhombic deflation and inflation rules

A Rhombic tiling can be seen as an intermediate step in the inflation process of a Penrose tiling $P$. Start by dividing all kites in two halves along their long diagonal and all darts in two halves along their short diagonal (the purple lines below).

If we consider all original black lines together with the new purple ones, we get a tiling of the plane by triangles and we call this the $A$-tiling. There are two triangles, s small triangle $S_A$ (the whiter ones, with two small edges ofd length $1$ and one edge of length $\tau$) and a large triangle $L_A$ (the greyer ones, with two edges of length $\tau$ and one edge of length $1$).

Next, we remove the black lines joining an $S_A$-triangle with an $L_A$-triangle, and get another tiling with triangles, the $B$-tiling, with two pieces, a large triangle $L_B$ with two sides of length $\tau$ and one side of length $1+\tau$ (the white ones) (the union of an $L_A$ and a $S_A$), and a small triangle $S_B$ which has the same form as $L_A$ (the greyer ones). If we now join two white triangles $L_B$ along their common longer edge, and two greyer triangles $S_B$ along their common smaller edge, we obtain the Rhombic tiling $R$ corresponding to $P$.

We can repeat this process starting with the Rhombic tiling $R$. We divide all fat rhombi in two along their long diagonals, and the skinny rhombi in two along their short diagonals (the purple lines). We obtain a tiling of the plane by triangles, which is of course just the $B$-tiling above.

Remove the edge joining a small $S_B$ with a large $L_B$ triangle, then we get a new tiling by triangles, the $\tau A$-tiling consisting of large triangles $L_{\tau A}$ (the white ones) with two long edges of length $1+\tau$ and one short edge of length $\tau$ (note that $L_{\tau A}$ is $\tau$ times the triangle $L_A$), and a smaller one $S_{\tau A}$ (the grey ones) having two short edges of length $\tau$ and one long edge of length $1+\tau$ (again $S_{\tau A}$ is $\tau$ times the triangle $S_A$). If we join two $L_{\tau A}$-triangles sharing a common long edge we obtain a Kite ($\tau$-times larger than the original Kite) and if we joint two $S_{\tau A}$-triangles along their common small edge we get a Dart ($\tau$ times larger than the original Dart). The Penrose tiling we obtain is the inflation $inf(P)$ of the original $P$.

If we repeat the whole procedure starting from $inf(P)$ instead of $P$ we get, in turn, triangle tilings of the plane, subsequently the $\tau B$-tiling, the $\tau^2 A$-tiling, the $\tau^2 B$-tiling and so on, the triangles in a $\tau^n A$-tiling being of size $\tau^2$-times those of the $A$-tiling, and those in the $\tau^n B$-tiling $\tau^n$-times as large as those of the $B$-tiling.

The upshot of this is that we can associate to a Penrose tiling $P$ of the plane a sequence of $0$’s and $1$’s. Starting from $P$ we have an infinite sequence of associated triangle tilings
$A,~B,~\tau A,~\tau B,~\tau^2 A, \dots,~\tau^n A,~\tau^n B,~\tau^{n+1} A, \dots$
with larger and larger triangle tiles. Let $p$ be a point of the plane lying in the interior of an $A$-tile, then we define its index sequence
$i(P,p) = (x_0,x_1,x_2,\dots ) \quad x_{2n} = \begin{cases} 0~\text{if p \in L_{\tau^n A}} \\ 1~\text{if p \in S_{\tau^n A}} \end{cases}~ x_{2n+1} = \begin{cases} 0~\text{if p \in L_{\tau^n B}} \\ 1~\text{if p \in S_{\tau^n B}} \end{cases}$
That is, $x_0=1$ if $p$ lies in a small triangle of the $A$-tiling and $x_0=1$ if it lies in a large triangle, $x_1=1$ if $p$ lies in a small triangle of the $B$-tiling and is $0$ if it lies in a large $B$-triangle, and so on.

The beauty of this is that every infinite sequence $(x_0,x_1,x_2,\dots )$ whose terms are $0$ or $1$, and in which no two consecutive terms are equal to $1$, is the index sequence $i(P,p)$ of some Penrose tiling $P$ in some point $p$ of the plane.

From the construction of the sequence of triangle-tilings it follows that a small triangle is part of a large triangle in the next tiling. For this reason an index sequence can never have two consecutive ones. An index sequence gives explicit instructions as to how the Penrose tiling is constructed. In each step add a large or small triangle as to fit the sequence, together with the matching triangle (the other half of a Penrose or Rhombic tile). Next, look at the patches $P_i$ of Kites and Darts (in the $\tau^i A$ tiling), for $i=0,1,\dots$, then $P_{i-k}$ is contained in the $k$-times inflated $P_i$, $i^k(P_i)$ (without rescaling), for each $i$ and all $1 \leq k \leq i$. But then, we have a concentric series of patches
$P_0 \subset i(P_1) \subset i^2(P_2) \subset \dots$
filling the entire plane.

The index sequence depends on the choice of point $p$ in the plane. If $p’$ is another point, then as the triangle tiles increase is size at each step, $p$ and $p’$ will lie in the same triangle-tile at some stage, and after that moment their index sequences will be the same.

As a result, there exists an uncountable infinity of distinct Penrose tilings of the plane.

From the discussions we see that a Penrose tiling is determined by the index-sequence in any point in the plane $(x_0,x_1,\dots )$ consisting of $0$’s and $1$’s having no two consecutive $1$’s, and that another such sequence $(x’_0,x’_1,\dots )$ determines the same Penrose tilings if $x_n=x’_n$ for all $n \geq N$ for some number $N$. That is, the Penrose tiling is determined by the the equivalence class of such sequences, and it is easy to see that there are uncountably many such equivalence classes.

If you want to play a bit with Penrose tiles, you can order the P2 tiles or the P3 tiles from
Cherry Arbor Design.

A Conway musical sequence is an infinite word in $L$ and $S$, containing no two consecutive $S$’s nor three consecutive $L$’s, such that all its inflations remain musical sequences.

We’ve seen that such musical sequences encode an aperiodic tiling of the line in short ($S$) and long ($L$) intervals, and that such tilings are all finite locally isomorphic.

But, apart from the middle $C$-sequences (the one-dimensional cartwheel tilings) we gave no examples of such tilings (or musical sequences). Let’s remedy this!

Take any real number $c$ as long as it is not an integral combination of $1$ and $\tfrac{1}{\tau}$ (with $\tau$ the golden ratio) and assign to any integer $a \in \mathbb{Z}$ a tile
$P_c(a) = \begin{cases} S \\ L \end{cases} ~\text{iff}~\lceil c+(a+1)\frac{1}{\tau} \rceil – \lceil c+a \frac{1}{\tau} \rceil = \begin{cases} 0 \\ 1 \end{cases}$
(instead of ceilings we might have taken floors, because of the restriction on $c$).

With a little bit of work we see that the deflated word determined by $P_c$ is again of this type, more precisely $def(P_c) = P_{-(c-\lfloor c \rfloor)\frac{1}{\tau}}$. But then it also follows that inflated words are of this type, meaning that all $P_c$ define a musical sequence.

Let’s just check that these sequences satisfy the gluing restrictions. If there is no integer between $c+a\tfrac{1}{\tau}$ and $c+(a+1)\tfrac{1}{\tau}$, because $2 \tfrac{1}{\tau} \approx 1.236$ there must be an interval in the preceding and the following $\tfrac{1}{\tau}$-interval, showing that an $S$ in the sequence has an $L$ on its left and right, so there are no two consecutive $S$’s in the sequences.

Similarly, if two consecutive $\tfrac{1}{\tau}$-intervals have an integer in them, the next interval cannot contain an integer as $3 \tfrac{1}{\tau} \approx 1.854 < 2$.

Now we come to the essential point: these sequences can be obtained by the cut-and-project method.

Take the line $L$ through the origin with slope $\tfrac{1}{\tau}$ and $L^{\perp}$ the line perpendicular it.

Consider the unit square $H$ and $H_{\vec{\gamma}}=H + \vec{\gamma}$ its translation under a shift vector $\vec{\gamma}=(\gamma_x,\gamma_y)$ and let $\pi$ (or $\pi^{\perp}$) be the orthogonal projection of the plane onto $L$ (or onto $L^{\perp}$). One quickly computes that
$\pi(a,b) = (\frac{\tau^2 a + \tau b}{1+\tau^2},\frac{\tau a + b}{1+\tau^2}) \quad \text{and} \quad \pi^{\perp}(a,b) = (\frac{a-\tau b}{1+\tau^2},\frac{\tau^2b-\tau a}{1+\tau^2})$
In the picture, we take $\vec{\gamma}=(c,-\tau c)$.

The window $W$ will be the strip, parallel with $L$ with basis $\pi^{\perp}(H_{\vec{\gamma}})$.

We cut the standard lattice $\mathbb{Z}^2$, of all points with integer coordinates in the plane, by retricting to the window $\mathcal{P}=\mathbb{Z}^2 \cap W$.

Next, we project $\mathcal{P}$ onto the line $L$, and we get a set of endpoints of intervals which divide the line $L$ into short intervals of length $\tfrac{1}{\sqrt{1+\tau^2}}$ and long intervals of length $\tfrac{\tau}{\sqrt{1+\tau^2}}$.

For $(a,b) \in W$, the interval will be short if $(a,b+1) \in W$ and long if $(a+1,b) \in W$.

Because these intervals differ by a factor $\tau$ in length, we get a tiling of the line by short intervals $S$ and long intervals $L$. It is easy to see that they satisfy the gluing restrictions (remember, no two consecutive short intervals and no three consecutive long intervals): the horizontal width of the window $W$ is $1+\tau \approx 2.618$ (so there cannot be three consecutive long intervals in the projection) and the vertical width of the window $W$ is $1+\tfrac{1}{\tau} = \tau \approx 1.618$ so there cannot be two consecutive short intervals in the projection.

The sequence obtained from projecting $\mathcal{P}$ is equal to the sequence $P_{(1+\tau^2)c}$. So, we get all musical sequences of this form from the cut-and-project method!

On $L^{\perp}$ the two end-points of the window are
$\begin{cases} \pi^{\perp}(c+1,-\tau c) = (\frac{(1+\tau^2)c+1}{1+\tau^2},- \tau \frac{(1+\tau^2)c+1}{1+\tau^2}) \\ \pi^{\perp}(c,-\tau c+1) = (\frac{(1+\tau^2)c-\tau}{1+\tau^2},-\tau \frac{(1+\tau^2)c-\tau}{1+\tau^2}) \end{cases}$
Therefore, a point $(a,b) \in \mathbb{Z}^2$ lies in the window $W$ if and only if
$(1+\tau^2)c-\tau < a-\tau b < (1+\tau^2)c+1$ or equivalently, if $(1+\tau^2)c+(b-1)\tau < a < (1+\tau^2)c+b \tau + 1$ Observe that $\lceil (1+\tau^2)c + b\tau \rceil - \lceil (1+\tau^2)c+(b-1)\tau \rceil = P_{(1+\tau^2)c}(b-1) + 1 \in \{ 1,2 \}$ We separate the two cases: (1) : If $\lceil (1+\tau^2)c + (b+1)\tau \rceil - \lceil (1+\tau^2)c+b \tau \rceil =1$, then there must be an integer $a$ such that $(1+\tau^2)c +(b+1) \tau -1 < a < (1+\tau^2) b+1$, and this forces $\lceil (1+\tau^2)c + (b+2)\tau \rceil - \lceil (1+\tau^2)c+(b+1)\tau \rceil =2$. With $b_i = (1+\tau^2)c+(b+i)\tau$ and $d_i = b_i+1$ we have the situation

and from the inequalities above this implies that both $(a+1,b+1)$ and $(a+1,b+2)$ are in $W$, giving a short interval $S$ in the projection.

(2) : If $\lceil (1+\tau^2)c + (b+1)\tau \rceil – \lceil (1+\tau^2)c+b \tau \rceil =1$, then there must be an integer $a$ such that $(1+\tau^2)c+b \tau < a < (1+\tau^2)cv + (b+1)\tau -1$, giving the situation

giving from the inequalities that both $(a+1,b+1)$ and $(a+2,b+1)$ are in $W$, giving a long interval $L$ in the projection, finishing the proof.

Whenever I visit someone’s YouTube or Twitter profile page, I hope to see an interesting banner image. Here’s the one from Richard Borcherds’ YouTube Channel.

Not too surprisingly for Borcherds, almost all of these numbers are related to the monster group or its moonshine.

Let’s try to decode them, in no particular order.

196884

John McKay’s observation $196884 = 1 + 196883$ was the start of the whole ‘monstrous moonshine’ industry. Here, $1$ and $196883$ are the dimensions of the two smallest irreducible representations of the monster simple group, and $196884$ is the first non-trivial coefficient in Klein’s j-function in number theory.

$196884$ is also the dimension of the space in which Robert Griess constructed the Monster, following Simon Norton’s lead that there should be an algebra structure on the monster-representation of that dimension. This algebra is now known as the Griess algebra.

Here’s a recent talk by Griess “My life and times with the sporadic simple groups” in which he tells about his construction of the monster (relevant part starting at 1:15:53 into the movie).

1729

1729 is the second (and most famous) taxicab number. A long time ago I did write a post about the classic Ramanujan-Hardy story the taxicab curve (note to self: try to tidy up the layout of some old posts!).

Recently, connections between Ramanujan’s observation and K3-surfaces were discovered. Emory University has an enticing press release about this: Mathematicians find ‘magic key’ to drive Ramanujan’s taxi-cab number. The paper itself is here.

“We’ve found that Ramanujan actually discovered a K3 surface more than 30 years before others started studying K3 surfaces and they were even named. It turns out that Ramanujan’s work anticipated deep structures that have become fundamental objects in arithmetic geometry, number theory and physics.”

Ken Ono

24

There’s no other number like $24$ responsible for the existence of sporadic simple groups.

24 is the length of the binary Golay code, with isomorphism group the sporadic Mathieu group $M_24$ and hence all of the other Mathieu-groups as subgroups.

24 is the dimension of the Leech lattice, with isomorphism group the Conway group $Co_0 = .0$ (dotto), giving us modulo its center the sporadic group $Co_1=.1$ and the other Conway groups $Co_2=.2, Co_3=.3$, and all other sporadics of the second generation in the happy family as subquotients (McL,HS,Suz and $HJ=J_2$)

24 is the central charge of the Monster vertex algebra constructed by Frenkel, Lepowski and Meurman. Most experts believe that the Monster’s reason of existence is that it is the symmetry group of this vertex algebra. John Conway was one among few others hoping for a nicer explanation, as he said in this interview with Alex Ryba.

24 is also an important number in monstrous moonshine, see for example the post the defining property of 24. There’s a lot more to say on this, but I’ll save it for another day.

60

60 is, of course, the order of the smallest non-Abelian simple group, $A_5$, the rotation symmetry group of the icosahedron. $A_5$ is the symmetry group of choice for most viruses but not the Corona-virus.

3264

3264 is the correct solution to Steiner’s conic problem asking for the number of conics in $\mathbb{P}^2_{\mathbb{C}}$ tangent to five given conics in general position.

Steiner himself claimed that there were $7776=6^5$ such conics, but realised later that he was wrong. The correct number was first given by Ernest de Jonquières in 1859, but a rigorous proof had to await the advent of modern intersection theory.

Eisenbud and Harris wrote a book on intersection theory in algebraic geometry, freely available online: 3264 and all that.

248

248 is the dimension of the exceptional simple Lie group $E_8$. $E_8$ is also connected to the monster group.

If you take two Fischer involutions in the monster (elements of conjugacy class 2A) and multiply them, the resulting element surprisingly belongs to one of just 9 conjugacy classes:

1A,2A,2B,3A,3C,4A,4B,5A or 6A

The orders of these elements are exactly the dimensions of the fundamental root for the extended $E_8$ Dynkin diagram.

This is yet another moonshine observation by John McKay and I wrote a couple of posts about it and about Duncan’s solution: the monster graph and McKay’s observation, and $E_8$ from moonshine groups.

163

163 is a remarkable number because of the ‘modular miracle’
$e^{\pi \sqrt{163}} = 262537412640768743.99999999999925…$
This is somewhat related to moonshine, or at least to Klein’s j-function, which by a result of Kronecker’s detects the classnumber of imaginary quadratic fields $\mathbb{Q}(\sqrt{-D})$ and produces integers if the classnumber is one (as is the case for $\mathbb{Q}(\sqrt{-163})$).

The details are in the post the miracle of 163, or in the paper by John Stillwell, Modular Miracles, The American Mathematical Monthly, 108 (2001) 70-76.

His description of the $j$-function (at 4:13 in the movie) is simply hilarious!

Borcherds connects $163$ to the monster moonshine via the $j$-function, but there’s another one.

The monster group has $194$ conjugacy classes and monstrous moonshine assigns a ‘moonshine function’ to each conjugacy class (the $j$-function is assigned to the identity element). However, these $194$ functions are not linearly independent and the space spanned by them has dimension exactly $163$.

Before we’ll come to applications of quasicrystals to viruses it is perhaps useful to illustrate essential topics such as deflation, inflation, aperiodicity, local isomorphism and the cut-and project method in the simplest of cases, that of $1$-dimensional tilings.

We want to tile the line $\mathbb{R}^1$ with two kinds of tiles, short ($S$) and ($L$) long intervals, differing by a golden ratio factor $\tau=\tfrac{1}{2}(1+\sqrt{5}) \approx 1.618$.

Clearly, no two tiles may overlap and we impose a gluing restriction: there can be no two consecutive $S$-intervals in the tiling, and no three consecutive $L$-intervals.

The code of a tiling is a doubly infinite word in $S$ and $L$ such that there are no two consecutive $S$’s nor three consecutive $L$’s. For example
$\sigma = \dots LSLLSLS\underline{L}LSLLSL \dots$
We underline one tile to distinguish the sequence from shifts of it.

Conway’s musical sequences will be special codes (or tilings), allowing for the inverse operations of inflation and deflation, terms coined by John Conway in relation to Penrose tilings. The musical sequences are important to understand Conway’s worms (sometimes called “wormholes”) which are strings of Long and Short bow ties in a Penrose tiling, and to measure the distances between Amman bars. In fact, many of the properties of Penrose tilings and $3$-dimensional quasicrystals (for example, local isomorphism) have their counterparts for tilings having a musical sequence as code.

Conway’s investigations of Penrose tiles held up work on the ATLAS-project and caused some problems at home:

“In pursuing his investigations, he unsurped some of his wife Eileen’s territory, covering the dining table with an infinite nuisance of tiles. He cut them out himself, causing his right hand to hurt with cramps for days. To Eileen’s dismay, he studied the dining table mosaic for a year, relegating family meals to the kitchen and prohibiting dinner parties.”

From “Genius at Play – The curious mind of John Horton Conway” by Siobhan Roberts

Let’s investigate inflation and deflation of these tilings.

The point of the golden factor $\tau$ is to allow for deflation. That is, we can replace a tiling by another one with tiles $S$ and $L$ both a factor $\tfrac{1}{\tau}=\tau-1 \approx 0.618$ smaller than the original tiles. If the original $S$-tile has length $a$ (and the $L$-tile length $\tau a$), then the new tile $S$ wil have length $\tfrac{1}{\tau}a$ and the new $L$-tile length $a$.
We do this by replacing each old $S$-tile by a new $L$-tile, and to break up any old $L$-tile in a new $L$ and new $S$-tile, as $\tau a = a + \tfrac{1}{\tau}a$ (note that $\tau^2=\tau+1$)

To get the code of the deflated tiling we replace each letter $S$ by a letter $L$ and each $L$ by $LS$. The underlined letter will be the first letter of the deflated underlined letter in the original sequence. The deflated sequence of the one above is
$def(\sigma) = \dots LSLLSLSLLSL\underline{L}SLSLLSLSLLS \dots$
and it is easy to see that the deflated tiling satisfies again the gluing condition.

Certain of these tilings (not all!) allow for an inverse to deflation, called inflation, increasing the size of the tiles by a factor $\tau$.

Starting from a tiling we divide each $L$-tile in half and these mid-points will be end-points of the tiles in the new tiling, erasing all endpoints of the original one. The inflated tiling will have two sorts of tiles, a new short one $S$ of length $\tau a$ obtained from the end-half of an original $L$-tile, followed b the start-half of an original $L$-tile, and a new long tile $L$ of length $\tau^2 a = (\tau+1) a$, made of the end-half of an original $L$, followed by an original $S$, followed by the start-half of an original $L$.

We get the code of the inflated tiling by replacing first each $L$ by $ll$ and subsequently replace each word $lSl$ by a letter $L$ and each $ll$ by $S$. An example,
$\sigma = \dots LSLLSLSLLSL\underline{L}SLSLLSLSLLS \dots \\ \dots llSllllSllSllllSlll\underline{l}SllSllllSllSllllS \dots \\ inf(\sigma) = \dots (l)LSLLSLS\underline{L}LSLLS(lS) \dots$

But, the inflated tiling may no longer satisfy the gluing condition. An example
$\dots LSLSLSL \dots \mapsto \dots llSllSllSll \dots \mapsto \dots (l)LLL(l) \dots$

A Conway musical sequence is the code of a tiling $\sigma$ such that all its consecutive inflations $inf^n(\sigma)$ satisfy the gluing condition. For the corresponding Conway tiling $\sigma$ we have that
$def(inf(\sigma))=\sigma=inf(def(\sigma))$

Let’s construct at least two such Conway tilings (later we’ll see that there are uncountably many). Take $C_n=def^n(LS\underline{L})$ and write it in a special form to highlight symmetries.
\begin{eqnarray*}
C_0 =& (L.S)\underline{L} \\
C_1 =& L(S.L)\underline{L}S \\
C_2 =& LSL(L.S)\underline{L}SL \\
C_3 =& LSLLSL(S.L)\underline{L}SLLS \\
C_4 =& LSLLSLSLLSL(L.S)\underline{L}SLLSLSL
\end{eqnarray*}

The even terms have middle-part $(L.S)$ and the odd ones $(S.L)$. The remaing left and right parts are each others reflexion (or part of it). This is easily seen by induction as are the inclusions
$C_0 \subset C_2 \subset C_4 \subset \dots \subset C_{even} \quad \text{and} \quad C_1 \subset C_3 \subset C_5 \subset \dots \subset C_{odd}$

$C_{even}$ and $C_{odd}$ are special Conway musical sequences, called the middle $C$-sequences, and are each others inflation and deflation. If you are familiar with Penrose tilings, these are the $1$-dimensional counterparts of the cartwheel Penrose tiling (here with the $10$ Conway worms emanating from the center, and with the borders of the first few cartwheels drawn).

A direct consequence of inflation on Conway’s musical sequences is that the corresponding tiling is aperiodic, that is, it has no translation symmetry.

For, inflation only depends on the local configuration of tiles, so if translation by $R$ is a symmetry of a musical sequence $\sigma$ then it is also a symmetry of $inf(\sigma)$, and so also of $inf^n(\sigma)$. But for large $n$ we will have that $R < \tau^n a$ (with $a$ the size of the tiles in $\sigma$). But then a tile in $inf^n(\sigma)$ and its translation by $R$ must overlap which is impossible if $+R$ is a translation symmetry of $inf^n(\sigma)$. Done!

Returning to the middle C-sequences, what was the point of starting with $C_0 = LSL$? Well, it follows directly from the gluing restrictions that any letter in a musical sequence is part of a subword $LSL$ of $\sigma$. But then, every finite subword $W$ of $\sigma$ is also a subword of $C_{2n}$ for some large $n$.

For, let $d$ be the length of the interval corresponding to $W$ and choose $n$ such that $d > \tau^{2n} a$ then the interval of the line corresponding to $W$ is contained in a single tile in $inf^{2n}(\sigma)$ and this tile belongs to a subword $LSL$ of $inf^{2n}(\sigma)$. But then $W$ will be a subword of the $2n$-th deflation of that interval $LSL \subset inf^{2n}(\sigma)$, which is $C_{2n}$.

Or, as Conway would phrase it with respect to Penrose tilings (quote again from Siobhan Roberts’ book)

“Every points is in the cartwheel somewhere. If you jab your finger anywhere, on any point anywhere on teh pattern, you are part of a cartwheel. The whole ting is overlapping cartwheels.”

An immediate consequence is the local isomorphism theorem: Every subword of a musical sequence $\sigma$ appears infinitely many times as subword of any other musical sequence. That is, one cannot distinguish two tilings of the line with musical sequence codes from each other by looking at finite intervals!

The argument is similar to the one above. The finite interval corresponding to the subword lies in a unique tile of $inf^n(\sigma)$ for $n$ large enough. Now, take another musical sequence $\mu$ and consider any of the infinitely many tiles of the same type in $inf^n(\mu)$, then $def^n$ of such a tile will contain the subword in $\phi$.

Another time, we’ll see that musical sequences can be produced by the ‘cut-and-project’-method (what I called the ‘windows’-method before).
This time we will project parts of the standard $2$-dimensional lattice $\mathbb{Z}^2$ onto the line, which is a lot easier to visualise than de Bruijn’s projection from $\mathbb{R}^5$ to produce Penrose tilings or the projection from six dimensional space to harvest quasicrystals.

I’m on vacation, and re-reading two ‘metabiographies’:

and

Siobhan Roberts : Genius At Play: The Curious Mind of John Horton Conway

.

Siobhan Roberts’ book is absolutely brilliant! I’m reading it for the n-th time, first on Kindle, then hardcopy, and now I’m just flicking through its pages, whenever I want to put a smile on my face.

So, here’s today’s gem of a Conway quote (on page 150):

Pure mathematicians usually don’t found companies and deal with the world in an aggressive way. We sit in our ivory towers and think.

(Conway complains his words were taken out of context, in an article
featuring Stephen Wolfram.)

If only university administrations worldwide would accept the ‘sitting in an ivory tower and think’-bit as the job description, and evaluation criterium, for their pure mathematicians.

Sadly… they prefer managers to thinkers.

This reminds me of another brilliant text, perhaps not receiving the attention it deserves:

Daniel J. Woodhouse : An open letter to the mathematical community.

Woodhouse offers a reaction to the ‘neoliberal upper management and bloated administration’ of universities:

Within the sphere of pure mathematics — the oldest and most successful of humanity’s intellectual endeavors — I believe our best chance at preserving the integrity and dignity of our tradition is to return to our Pythagorean roots. We should become a cult.

How?

Let us seclude ourselves in mountain caves and daub mysterious equations in blood across rock-faces to ward off outsiders. Let us embrace our most impenetrable mathematical texts as sacred and requiring divinely distributed revelation.

Why?

I am convinced that the current system has dulled our understanding of the value we offer through our instruction. Modern mathematical techniques are the foundation of modern science, medicine, and technology, and we should be the literal, rather than metaphorical, high priests of this temple. Only by withholding our insights will we be able to reassert the intrinsic worth of our knowledge.

I hope these few paragraphs have wetted your appetite to read the manifesto in full, and then take action!

Let’s try to identify the $\Psi(n) = n \prod_{p|n}(1+\frac{1}{p})$ points of $\mathbb{P}^1(\mathbb{Z}/n \mathbb{Z})$ with the lattices $L_{M \frac{g}{h}}$ at hyperdistance $n$ from the standard lattice $L_1$ in Conway’s big picture.

Here are all $24=\Psi(12)$ lattices at hyperdistance $12$ from $L_1$ (the boundary lattices):

You can also see the $4 = \Psi(3)$ lattices at hyperdistance $3$ (those connected to $1$ with a red arrow) as well as the intermediate $12 = \Psi(6)$ lattices at hyperdistance $6$.

The vertices of Conway’s Big Picture are the projective classes of integral sublattices of the standard lattice $\mathbb{Z}^2=\mathbb{Z} e_1 \oplus \mathbb{Z} e_2$.

Let’s say our sublattice is generated by the integral vectors $v=(v_1,v_2)$ and $w=(w_1.w_2)$. How do we determine its class $L_{M,\frac{g}{h}}$ where $M \in \mathbb{Q}_+$ is a strictly positive rational number and $0 \leq \frac{g}{h} < 1$?

Here’s an example: the sublattice (the thick dots) is spanned by the vectors $v=(2,1)$ and $w=(1,4)$

Well, we try to find a basechange matrix in $SL_2(\mathbb{Z})$ such that the new 2nd base vector is of the form $(0,z)$. To do this take coprime $(c,d) \in \mathbb{Z}^2$ such that $cv_1+dw_1=0$ and complete with $(a,b)$ satisfying $ad-bc=1$ via Bezout to a matrix in $SL_2(\mathbb{Z})$ such that
$\begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} v_1 & v_2 \\ w_1 & w_2 \end{bmatrix} = \begin{bmatrix} x & y \\ 0 & z \end{bmatrix}$
then the sublattice is of class $L_{\frac{x}{z},\frac{y}{z}~mod~1}$.

In the example, we have
$\begin{bmatrix} 0 & 1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 2 & 1 \\ 1 & 4 \end{bmatrix} = \begin{bmatrix} 1 & 4 \\ 0 & 7 \end{bmatrix}$
so this sublattice is of class $L_{\frac{1}{7},\frac{4}{7}}$.

Starting from a class $L_{M,\frac{g}{h}}$ it is easy to work out its hyperdistance from $L_1$: let $d$ be the smallest natural number making the corresponding matrix integral
$d. \begin{bmatrix} M & \frac{g}{h} \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} u & v \\ 0 & w \end{bmatrix} \in M_2(\mathbb{Z})$
then $L_{M,\frac{g}{h}}$ is at hyperdistance $u . w$ from $L_1$.

Now that we know how to find the lattice class of any sublattice of $\mathbb{Z}^2$, let us assign a class to any point $[c:d]$ of $\mathbb{P}^1(\mathbb{Z}/n\mathbb{Z})$.

As $gcd(c,d)=1$, by Bezout we can find a integral matrix with determinant $1$
$S_{[c:d]} = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$
But then the matrix
$\begin{bmatrix} a.n & b.n \\ c & d \end{bmatrix}$
has determinant $n$.

Working backwards we see that the class $L_{[c:d]}$ of the sublattice of $\mathbb{Z}^2$ spanned by the vectors $(a.n,b.n)$ and $(c,d)$ is of hyperdistance $n$ from $L_1$.

This is how the correspondence between points of $\mathbb{P}^1(\mathbb{Z}/n\mathbb{Z})$ and classes in Conway’s big picture at hyperdistance $n$ from $L_1$ works.

Let’s do an example. Take the point $[7:3] \in \mathbb{P}^1(\mathbb{Z}/12\mathbb{Z})$ (see last time), then
$\begin{bmatrix} -2 & -1 \\ 7 & 3 \end{bmatrix} \in SL_2(\mathbb{Z})$
so we have to determine the class of the sublattice spanned by $(-24,-12)$ and $(7,3)$. As before we have to compute
$\begin{bmatrix} -2 & -7 \\ 7 & 24 \end{bmatrix} \begin{bmatrix} -24 & -12 \\ 7 & 3 \end{bmatrix} = \begin{bmatrix} -1 & 3 \\ 0 & -12 \end{bmatrix}$
giving us that the class $L_{[7:3]} = L_{\frac{1}{12}\frac{3}{4}}$ (remember that the second term must be taken $mod~1$).

If you do this for all points in $\mathbb{P}^1(\mathbb{Z}/12\mathbb{Z})$ (and $\mathbb{P}^1(\mathbb{Z}/6\mathbb{Z})$ and $\mathbb{P}^1(\mathbb{Z}/3 \mathbb{Z})$) you get this version of the picture we started with

You’ll spot that the preimages of a canonical coordinate of $\mathbb{P}^1(\mathbb{Z}/m\mathbb{Z})$ for $m | n$ are the very same coordinate together with ‘new’ canonical coordinates in $\mathbb{P}^1(\mathbb{Z}/n\mathbb{Z})$.

To see that this correspondence is one-to-one and that the index of the congruence subgroup
$\Gamma_0(n) = \{ \begin{bmatrix} p & q \\ r & s \end{bmatrix}~|~n|r~\text{and}~ps-qr=1 \}$
in the full modular group $\Gamma = PSL_2(\mathbb{Z})$ is equal to $\Psi(n)$ it is useful to consider the action of $PGL_2(\mathbb{Q})^+$ on the right on the classes of lattices.

The stabilizer of $L_1$ is the full modular group $\Gamma$ and the stabilizer of any class is a suitable conjugate of $\Gamma$. For example, for the class $L_n$ (that is, of the sublattice spanned by $(n,0)$ and $(0,1)$, which is of hyperdistance $n$ from $L_1$) this stabilizer is
$Stab(L_n) = \{ \begin{bmatrix} a & \frac{b}{n} \\ c.n & d \end{bmatrix}~|~ad-bc = 1 \}$
and a very useful observation is that
$Stab(L_1) \cap Stab(L_n) = \Gamma_0(n)$
This is the way Conway likes us to think about the congruence subgroup $\Gamma_0(n)$: it is the joint stabilizer of the classes $L_1$ and $L_n$ (as well as all classes in the ‘thread’ $L_m$ with $m | n$).

On the other hand, $\Gamma$ acts by rotations on the big picture: it only fixes $L_1$ and maps a class to another one of the same hyperdistance from $L_1$.The index of $\Gamma_0(n)$ in $\Gamma$ is then the number of classes at hyperdistance $n$.

To see that this number is $\Psi(n)$, first check that the classes at hyperdistance $p^k$ for $p$ a prime number and for all $k$ for the $p+1$ free valent tree with root $L_1$, so there are exactly $p^{k-1}(p+1)$ classes as hyperdistance $p^k$.

To get from this that the number of hyperdistance $n$ classes is indeed $\Psi(n) = \prod_{p|n}p^{v_p(n)-1}(p+1)$ we have to use the prime- factorisation of the hyperdistance (see this post).

The fundamental domain for the action of $\Gamma_0(12)$ by Moebius tranfos on the upper half plane must then consist of $48=2 \Psi(12)$ black or white hyperbolic triangles

Next time we’ll see how to deduce the ‘monstrous’ Grothendieck dessin d’enfant for $\Gamma_0(12)$ from it

Dedekind’s Psi-function $\Psi(n)= n \prod_{p |n}(1 + \frac{1}{p})$ pops up in a number of topics:

• $\Psi(n)$ is the index of the congruence subgroup $\Gamma_0(n)$ in the modular group $\Gamma=PSL_2(\mathbb{Z})$,
• $\Psi(n)$ is the number of points in the projective line $\mathbb{P}^1(\mathbb{Z}/n\mathbb{Z})$,
• $\Psi(n)$ is the number of classes of $2$-dimensional lattices $L_{M \frac{g}{h}}$ at hyperdistance $n$ in Conway’s big picture from the standard lattice $L_1$,
• $\Psi(n)$ is the number of admissible maximal commuting sets of operators in the Pauli group of a single qudit.

The first and third interpretation have obvious connections with Monstrous Moonshine.

Conway’s big picture originated from the desire to better understand the Moonshine groups, and Ogg’s Jack Daniels problem
asks for a conceptual interpretation of the fact that the prime numbers such that $\Gamma_0(p)^+$ is a genus zero group are exactly the prime divisors of the order of the Monster simple group.

Here’s a nice talk by Ken Ono : Can’t you just feel the Moonshine?

For this reason it might be worthwhile to make the connection between these two concepts and the number of points of $\mathbb{P}^1(\mathbb{Z}/n\mathbb{Z})$ as explicit as possible.

Surely all of this is classical, but it is nicely summarised in the paper by Tatitscheff, He and McKay “Cusps, congruence groups and monstrous dessins”.

The ‘monstrous dessins’ from their title refers to the fact that the lattices $L_{M \frac{g}{h}}$ at hyperdistance $n$ from $L_1$ are permuted by the action of the modular groups and so determine a Grothendieck’s dessin d’enfant. In this paper they describe the dessins corresponding to the $15$ genus zero congruence subgroups $\Gamma_0(n)$, that is when $n=1,2,3,4,5,6,7,8,9,10,12,13,16,18$ or $25$.

Here’s the ‘monstrous dessin’ for $\Gamma_0(6)$

But, one can compute these dessins for arbitrary $n$, describing the ripples in Conway’s big picture, and try to figure out whether they are consistent with the Riemann hypothesis.

We will get there eventually, but let’s start at an easy pace and try to describe the points of the projective line $\mathbb{P}^1(\mathbb{Z}/n \mathbb{Z})$.

Over a field $k$ the points of $\mathbb{P}^1(k)$ correspond to the lines through the origin in the affine plane $\mathbb{A}^2(k)$ and they can represented by projective coordinates $[a:b]$ which are equivalence classes of couples $(a,b) \in k^2- \{ (0,0) \}$ under scalar multiplication with non-zero elements in $k$, so with points $[a:1]$ for all $a \in k$ together with the point at infinity $[1:0]$. When $n=p$ is a prime number we have $\# \mathbb{P}^1(\mathbb{Z}/p\mathbb{Z}) = p+1$. Here are the $8$ lines through the origin in $\mathbb{A}^2(\mathbb{Z}/7\mathbb{Z})$

Over an arbitrary (commutative) ring $R$ the points of $\mathbb{P}^1(R)$ again represent equivalence classes, this time of pairs
$(a,b) \in R^2~:~aR+bR=R$
with respect to scalar multiplication by units in $R$, that is
$(a,b) \sim (c,d)~\quad~\text{iff}~\qquad \exists \lambda \in R^*~:~a=\lambda c, b = \lambda d$
For $\mathbb{P}^1(\mathbb{Z}/n \mathbb{Z})$ we have to find all pairs of integers $(a,b) \in \mathbb{Z}^2$ with $0 \leq a,b < n$ with $gcd(a,b)=1$ and use Cremona’s trick to test for equivalence:
$(a,b) = (c,d) \in \mathbb{P}^1(\mathbb{Z}/n \mathbb{Z})~\quad \text{iff}~\quad ad-bc \equiv 0~mod~n$
The problem is to find a canonical representative in each class in an efficient way because this is used a huge number of times in working with modular symbols.

Perhaps the best algorithm, for large $n$, is sketched in pages 145-146 of Bill Stein’s Modular forms: a computational approach.

For small $n$ the algorithm in $\S 1.3$ in the Tatitscheff, He and McKay paper suffices:

• Consider the action of $(\mathbb{Z}/n\mathbb{Z})^*$ on $\{ 0,1,…,n-1 \}=\mathbb{Z}/n\mathbb{Z}$ and let $D$ be the set of the smallest elements in each orbit,
• For each $d \in D$ compute the stabilizer subgroup $G_d$ for this action and let $C_d$ be the set of smallest elements in each $G_d$-orbit on the set of all elements in $\mathbb{Z}/n \mathbb{Z}$ coprime with $d$,
• Then $\mathbb{P}^1(\mathbb{Z}/n\mathbb{Z})= \{ [c:d]~|~d \in D, c \in C_d \}$.

Let’s work this out for $n=12$ which will be our running example (the smallest non-squarefree non-primepower):

• $(\mathbb{Z}/12\mathbb{Z})^* = \{ 1,5,7,11 \} \simeq C_2 \times C_2$,
• The orbits on $\{ 0,1,…,11 \}$ are
$\{ 0 \}, \{ 1,5,7,11 \}, \{ 2,10 \}, \{ 3,9 \}, \{ 4,8 \}, \{ 6 \}$
and $D=\{ 0,1,2,3,4,6 \}$,
• $G_0 = C_2 \times C_2$, $G_1 = \{ 1 \}$, $G_2 = \{ 1,7 \}$, $G_3 = \{ 1,5 \}$, $G_4=\{ 1,7 \}$ and $G_6=C_2 \times C_2$,
• $1$ is the only number coprime with $0$, giving us $[1:0]$,
• $\{ 0,1,…,11 \}$ are all coprime with $1$, and we have trivial stabilizer, giving us the points $[0:1],[1:1],…,[11:1]$,
• $\{ 1,3,5,7,9,11 \}$ are coprime with $2$ and under the action of $\{ 1,7 \}$ they split into the orbits
$\{ 1,7 \},~\{ 3,9 \},~\{ 5,11 \}$
giving us the points $[1:2],[3:2]$ and $[5:2]$,
• $\{ 1,2,4,5,7,8,10,11 \}$ are coprime with $3$, the action of $\{ 1,5 \}$ gives us the orbits
$\{ 1,5 \},~\{ 2,10 \},~\{ 4,8 \},~\{ 7,11 \}$
and additional points $[1:3],[2:3],[4:3]$ and $[7:3]$,
• $\{ 1,3,5,7,9,11 \}$ are coprime with $4$ and under the action of $\{ 1,7 \}$ we get orbits
$\{ 1,7 \},~\{ 3,9 \},~\{ 5,11 \}$
and points $[1:4],[3:4]$ and $[5,4]$,
• Finally, $\{ 1,5,7,11 \}$ are the only coprimes with $6$ and they form a single orbit under $C_2 \times C_2$ giving us just one additional point $[1:6]$.

This gives us all $24= \Psi(12)$ points of $\mathbb{P}^1(\mathbb{Z}/12 \mathbb{Z})$ (strangely, op page 43 of the T-H-M paper they use different representants).

One way to see that $\# \mathbb{P}^1(\mathbb{Z}/n \mathbb{Z}) = \Psi(n)$ comes from a consequence of the Chinese Remainder Theorem that for the prime factorization $n = p_1^{e_1} … p_k^{e_k}$ we have
$\mathbb{P}^1(\mathbb{Z}/n \mathbb{Z}) = \mathbb{P}^1(\mathbb{Z}/p_1^{e_1} \mathbb{Z}) \times … \times \mathbb{P}^1(\mathbb{Z}/p_k^{e_k} \mathbb{Z})$
and for a prime power $p^k$ we have canonical representants for $\mathbb{P}^1(\mathbb{Z}/p^k \mathbb{Z})$
$[a:1]~\text{for}~a=0,1,…,p^k-1~\quad \text{and} \quad [1:b]~\text{for}~b=0,p,2p,3p,…,p^k-p$
which shows that $\# \mathbb{P}^1(\mathbb{Z}/p^k \mathbb{Z}) = (p+1)p^{k-1}= \Psi(p^k)$.

Next time, we’ll connect $\mathbb{P}^1(\mathbb{Z}/n \mathbb{Z})$ to Conway’s big picture and the congruence subgroup $\Gamma_0(n)$.