Canonical Projection Continued Fractions

A flexible substitution rule

Consider the family of topologically conjugate substitution rules and action on the window given below. Mouse over to see the relationship between window regions and tiles in the substitution.

Edge ratio b : a

01

They take the set of canonical projection tilings whose 4x4 projection matrices (defining the physical and internal spaces) commute with:

$$R = \begin{pmatrix} 0&0&-1&0\\ 0&0&0&-1\\ 1&0&0&0\\ 0&1&0&0 \end{pmatrix} M = \begin{pmatrix} 0&0&1&0\\ 0&1&0&0\\ 1&0&0&0\\ 0&0&0&-1 \end{pmatrix}$$

This family has two pairs of generators a,c and b,d that are each at right angles to each other, and the two pairs are at 45 degrees to each other. The degree of freedom adjusted above is the ratio of the two lengths. Assuming the projections of $a = (1,0,0,0)$ and $c = (0,0,1,0)$ are horizontal and vertical of length $1$ and longer than the projections of the other two (at 45 degrees) $b = (0,1,0,0)$ and $d = (0,0,0,1)$.

A continued fraction on this family

Depending on the initial length $x$ of the $b$ and $d$ the rule above finds a unique predecessor for any such canonical projection tiling (the central octagon). The partition of the window shows that this is a predecessor to the tiling, as the substitution rule takes the subwindow back to the full window, and every region of the full window is uniquely covered away from the partition boundaries. The classic challenges of dealing with situations where points in the projection lie on the boundary of the window apply here.

The action is either the inverse of the action on the edges:

$$S_1 = \begin{pmatrix} 1&0&0&0\\ 1&1&1&0\\ 0&0&1&0\\ -1&0&1&1 \end{pmatrix}$$

when $x$ is less than $\frac{1}{\sqrt{2}}$ (so $1-\sqrt{2}x$ is positive) or (the inverse of):

$$S_2 = \begin{pmatrix} 1&1&0&-1\\ 1&1&1&0\\ 0&1&1&1\\ -1&0&1&1 \end{pmatrix}$$

otherwise. It is easiest to see the effects on the length if we consider the two lengths $(a,b)$ we have:

$$(a,b) \;\longmapsto\; \begin{cases} (a-\sqrt{2}\,b,\; b) & \text{if } a \ge \sqrt{2}\,b,\\[2pt] (\sqrt{2}\,b-a,\; \sqrt{2}\,a-b) & \text{if } a \ge b,\\[2pt] (b,\; a) & \text{otherwise.} \end{cases}$$

This is related to continued fractions on $(2,4,\infty)$. On the ratio $t = a/b$ the three branches act as the Mobius transforms:

$$s_1=\begin{pmatrix} 1 & -\sqrt{2} \\ 0 & 1 \end{pmatrix}, \qquad s_2=\begin{pmatrix} -1 & \sqrt{2} \\ \sqrt{2} & -1 \end{pmatrix}, \qquad r=\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$$

The first is parabolic. The other two have determinant $-1$ so reverse orientation: the third is the reflection in the unit geodesic, and the second is a glide reflection along the geodesic joining its fixed points $\pm 1$ (the fixed point $t = 1$ is the Ammann-Beenker tiling). Together the three generate the $(2,4,\infty)$ group with reflections. So it is different to the Rosen continued fraction, which is orientation preserving.

This system is periodic exactly when there is a substitution rule for the tilings in this family, see The octagonal system catches all substitution tilings.

Relationship to billiards on $(2,4,\infty)$

The continued fraction is in fact a cross-section of the billiard flow in the $(2,4,\infty)$ triangle. Using David Fried, “Symbolic dynamics for triangle groups”, Invent. Math. 125 (1996), 487–521 gives the map:

$$F(x)= \begin{cases} -x & \text{if } x < 0,\\[2pt] 1/x & \text{if } 0 < x < \tfrac{\sqrt{2}}{2},\\[2pt] \sqrt{2}-x & \text{if } x > \tfrac{\sqrt{2}}{2}, \end{cases}$$ $$\qquad f_1=\begin{pmatrix}-1&0\\0&1\end{pmatrix},\quad r=\begin{pmatrix}0&1\\1&0\end{pmatrix},\quad f_2=\begin{pmatrix}-1&\sqrt{2}\\0&1\end{pmatrix}.$$

Here $f_1$, $r$ and $f_2$ are the reflections in the three mirrors of the $(2,4,\infty)$ triangle with corners at $i$, $e^{i\pi/4}$ and the cusp at $\infty$. Each application of $F$ is a single reflection, so $F$ codes the billiard trajectory by bounces. The octagonal system is a recoding of these trajectories: reading right to left, each branch is a fixed collection of bounces, with $s_1 = f_1 f_2$ and $s_2 = r\,f_1 f_2\,r\,f_2$. So a subtraction step crosses the cusp strip with two bounces, and the glide reflection fixing the Ammann-Beenker tiling is a five-bounce word.

The $\pi/3$ system (Eisenstein case)

I have not yet done the work on the substitution rule, but a similar situation might work for the case, where the projection commutes with:

$$R_6 = \begin{pmatrix} 0&0&-1&0\\ 0&0&0&-1\\ 1&0&1&0\\ 0&1&0&1 \end{pmatrix} M_6 = \begin{pmatrix} 0&0&1&0\\ 0&1&0&1\\ 1&0&0&0\\ 0&0&0&-1 \end{pmatrix}$$

Which is two pairs of edges at 60 degrees to each other and 30 degree between the pairs.

On the edges the action is again given by inverses, now of the three matrices. They are all elements of $Gl_4(\Z)$ and have positive minors so give susbstitution rules.

$$N_1 = \begin{pmatrix} 1&0&0&0\\ 2&1&1&0\\ 0&0&1&0\\ -1&0&1&1 \end{pmatrix} N_2 = \begin{pmatrix} 2&1&0&-1\\ 2&1&1&0\\ 0&1&2&2\\ -1&0&1&1 \end{pmatrix} N_3 = \begin{pmatrix} 2&1&0&-1\\ 2&2&1&0\\ 0&1&2&2\\ -1&0&1&2 \end{pmatrix}$$

Each commutes with $R_6$ and $M_6$ and has determinant $1$. The corresponding action on the two lengths has four branches, the first three the inverses of $N_1$, $N_2$ and $N_3$ and the last swapping the two pairs:

$$(a,b) \;\longmapsto\; \begin{cases} (a-\sqrt{3}\,b,\; b) & \text{if } a \ge \sqrt{3}\,b,\\[2pt] (\sqrt{3}\,b-a,\; \sqrt{3}\,a-2b) & \text{if } a \ge \tfrac{2}{\sqrt{3}}\,b,\\[2pt] (2a-\sqrt{3}\,b,\; 2b-\sqrt{3}\,a) & \text{if } a \ge b,\\[2pt] (b,a) & \text{otherwise.} \end{cases}$$

On the ratio $t = a/b$ the branches act as the Mobius transforms:

$$n_1=\begin{pmatrix} 1 & -\sqrt{3} \\ 0 & 1 \end{pmatrix}, \qquad n_2=\begin{pmatrix} -1 & \sqrt{3} \\ \sqrt{3} & -2 \end{pmatrix}, \qquad n_3=\begin{pmatrix} 2 & -\sqrt{3} \\ -\sqrt{3} & 2 \end{pmatrix}$$

The first is parabolic and $r$ is the reflection in the unit geodesic as before. The second has determinant $-1$ and is a glide reflection. The third is new: it is hyperbolic with determinant $1$, translating along the geodesic joining its fixed points $\pm 1$ with eigenvalue $2+\sqrt{3}$ (the fixed point $t = 1$ is the equal length tiling). Together the four generate the $(2,6,\infty)$ group with reflections.

The relationship to billiards runs as for the octagon. Fried’s map for $(2,6,\infty)$ has the same shape as before with $\sqrt{3} = 2\cos(\pi/6)$ in place of $\sqrt{2}$, with reflections $h_1$, $r$ and $h_2$ in the mirrors of the triangle with corners at $i$ (order $2$), $e^{i\pi/6}$ (order $6$) and the cusp at $\infty$. In these letters the branches are:

$$n_1 = h_1 h_2, \qquad n_2 = r\,h_1 h_2\,r h_2, \qquad n_3 = h_1 h_2\,r h_2\,r h_2,$$

with $r$ itself a single bounce off the mirror joining the two cone points. So $n_2$ is the same five bounce word as the octagonal glide reflection $s_2$, the hyperbolic $n_3$ is a six bounce word, and the continued fraction is again a recoding of the billiard trajectories, now in the $(2,6,\infty)$ triangle.