Conway’s musical sequences (2)

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 $\frac{1}{\tau }$ (with $\tau$ the golden ratio) and assign to any integer $a\in \mathbb{Z}$ a tile
$$P_c(a)=\{\begin{array}{l}S\\ L\end{array}\text{iff}\lceil c+(a+1)\frac{1}{\tau }\rceil –\lceil c+a\frac{1}{\tau }\rceil =\{\begin{array}{l}0\\ 1\end{array}$$
(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 $c+a\frac{1}{\tau }$ and $c+(a+1)\frac{1}{\tau }$, because $2\frac{1}{\tau }\approx 1.236$ there must be an interval in the preceding and the following $\frac{1}{\tau }$-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 $\frac{1}{\tau }$-intervals have an integer in them, the next interval cannot contain an integer as $3\frac{1}{\tau }\approx 1.854<2$.

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 $\frac{1}{\tau }$ and $L^{\perp }$ the line perpendicular it.

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

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

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 $\frac{1}{\sqrt{1+{\tau }^2}}$ and long intervals of length $\frac{\tau }{\sqrt{1+{\tau }^2}}$.

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 $1+\frac{1}{\tau }=\tau \approx 1.618$ so there cannot be two consecutive short intervals in the projection.

Ready for the punchline?

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
$$\{\begin{array}{l}{\pi }^{\perp }(c+1,-\tau c)=(\frac{(1+{\tau }^2)c+1}{1+{\tau }^2},-\tau \frac{(1+{\tau }^2)c+1}{1+{\tau }^2})\\ {\pi }^{\perp }(c,-\tau c+1)=(\frac{(1+{\tau }^2)c-\tau }{1+{\tau }^2},-\tau \frac{(1+{\tau }^2)c-\tau }{1+{\tau }^2})\end{array}$$
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.