Planar Rosa: a family of quasiperiodic substitution discrete plane tilings with 2n-fold rotational symmetry

Abstract

We present Planar Rosa, a family of rhombus tilings with a 2n-fold rotational symmetry that are generated by a primitive substitution and that are also discrete plane tilings, meaning that they are obtained as a projection of a higher dimensional discrete plane. The discrete plane condition is a relaxed version of the cut-and-project condition. We also prove that the Sub Rosa substitution tilings with 2n-fold rotational symmetry defined by Kari and Rissanen do not satisfy even the weaker discrete plane condition. We prove these results for all even $n\geqslant 4$. This completes our previously published results for odd values of n.

Technical details

Lemma 34

(Decompposition of $\mathbb {R}^n$ as the orthogonal direct sum of the spaces $\mathcal {E}_n^k$) For any even integer n. Define the planes $\mathcal {E}_n^k$ for $0\leqslant k < \tfrac{n}{2}$ as

$$\begin{aligned} \mathcal {E}_n^k := \left\langle \left( \cos \frac{(2k+1)i\pi }{n}\right) _{0\leqslant i< n}, \left( \sin \frac{(2k+1)i\pi }{n}\right) _{0\leqslant i < n} \right\rangle . \end{aligned}$$

The planes $\mathcal {E}_n^k$ are pairwise orthogonal and their direct sum is $\mathbb {R}^n$.

Proof

Remark that the fact that the direct sum of the spaces $\mathcal {E}_n^k$ is $\mathbb {R}^n$ is a consequence of the fact that they are pairwise orthogonal. Indeed if they are pairwise orthogonal then their sum is direct (trivial intersection) and the dimension of the sum is the sum of the dimensions so the dimension is n and the sum is $\mathbb {R}^n$.

Now we prove that they are pairwise orthogonal. Take $0\leqslant k_1< k_2 < \tfrac{n}{2}$. $\mathcal {E}_n^{k_1}$ and $\mathcal {E}_n^{k_2}$ are orthogonal, indeed :

• $\left\langle \left( \cos \frac{(2k_1+1)i\pi }{n}\right) _{0\leqslant i< n} \bigg | \left( \cos \frac{(2k_2+1)i\pi }{n}\right) _{0\leqslant i < n} \right\rangle = 0$, indeed we have

  $$\begin{aligned}&\left\langle \left( \cos \frac{(2k_1+1)i\pi }{n}\right) _{0\leqslant i< n} \bigg | \left( \cos \frac{(2k_2+1)i\pi }{n}\right) _{0\leqslant i < n} \right\rangle \\&\qquad = \sum \limits _{i=0}^{n-1} \cos \frac{(2k_1+1)i\pi }{n}\cos \frac{(2k_2+1)i\pi }{n}\\&\qquad = \sum \limits _{i=0}^{n-1}\frac{1}{2}\left( \cos \frac{2i(k_1+k_2+1)\pi }{n}+ \cos \frac{2i(k_2-k_1)\pi }{n}\right) \\&\qquad = \left( \sum \limits _{i=0}^{n-1}\frac{1}{2} \cos \frac{2i(k_1+k_2+1)\pi }{n}\right) + \left( \sum \limits _{i=0}^{n-1}\frac{1}{2}\cos \frac{2i(k_2-k_1)\pi }{n}\right) \\&\qquad = 0 \end{aligned}$$

  Note that the important hypothesis is $0\leqslant k_1< k_2 < \tfrac{n}{2}$ so $0< (k_2 - k_1), (k_1+k_2+1) < n$ and $\sum \limits _{i=0}^{n-1}\frac{1}{2} \cos \frac{2i(k_1+k_2+1)\pi }{n} = 0$ (and same with $(k_2-k_1$)).

• similarly $\left\langle \left( \sin \frac{(2k_1+1)i\pi }{n}\right) _{0\leqslant i< n} \bigg | \left( \sin \frac{(2k_2+1)i\pi }{n}\right) _{0\leqslant i < n} \right\rangle = 0$ with

  $$\begin{aligned} 2\sin \frac{(2k_1+1)i\pi }{n}\sin \frac{(2k_2+1)i\pi }{n} = \cos \frac{2i(k_1+k_2+1)\pi }{n}- \cos \frac{2i(k_2-k_1)\pi }{n}. \end{aligned}$$

• similarly we have $\left\langle \left( \cos \frac{(2k_1+1)i\pi }{n}\right) _{0\leqslant i< n} \bigg | \left( \sin \frac{(2k_2+1)i\pi }{n}\right) _{0\leqslant i < n} \right\rangle = 0$ with

  $$\begin{aligned} 2\cos \frac{(2k_1+1)i\pi }{n}\sin \frac{(2k_2+1)i\pi }{n} = \sin \frac{2i(k_1+k_2+1)\pi }{n}- \sin \frac{2i(k_2-k_1)\pi }{n}. \end{aligned}$$

• similarly we have $\left\langle \left( \sin \frac{(2k_1+1)i\pi }{n}\right) _{0\leqslant i< n} \bigg | \left( \cos \frac{(2k_2+1)i\pi }{n}\right) _{0\leqslant i < n} \right\rangle = 0$.

Lemma 35

(The billiard word $w$ of slope $\gamma$ is well-defined) Let n be an even integer at least 4. Let $\gamma :=(\cos \tfrac{i\pi }{n})_{0\leqslant i < n/2}$. Let $\Gamma _\frac{1}{2} := \langle \gamma \rangle + (\frac{1}{2} ,\frac{1}{2} , \dots ,\frac{1}{2} ) = \{ (\frac{1}{2} + t\cos \tfrac{i\pi }{n})_{0\leqslant i < n/2} |\ t \in \mathbb {R}^+\}$. Let $H_{j,k}:= \{ x \in \mathbb {R}^{n/2}|\ x_j = k\}$ with $j<\frac{n}{2}$ and $k> 1$.

The billiard word $w$ of line $\Gamma _\frac{1}{2}$ is well-defined , i.e., the line $\Gamma _\frac{1}{2}$ never intersects two distinct hyperplanes at once, i.e.,

for any $j,j'\in \{0,1, \dots , \frac{n}{2} -1\}$, $k,k' \in \mathbb {N}$, if $(j,k)\ne (j',k')$ then $\Gamma _\frac{1}{2} \cap H_{j,k} \cap H_{j',k'} = \emptyset$.

Proof

If $j=j'$ the result is trivial because in that case $k\ne k'$ and therefore $H_{j,k} \cap H_{j',k'} = \emptyset$.

If $j\ne j'$, without loss of generality we assume $j<j'$. For contradiction, assume that the intersection is non-empty, i.e., assume that there exists $x\in \Gamma _\frac{1}{2} \cap H_{j,k} \cap H_{j',k'}$. By definition of $\Gamma _\frac{1}{2}$, $x = t\gamma + (\frac{1}{2} ,\frac{1}{2} ,\dots ,\frac{1}{2} )$ for some $t \in \mathbb {R}^+$. By definition of $H_{j,k}$ and $H_{j',k'}$, $x_j = k$ and $x_{j'} = k'$.

So we get $t = (k-\frac{1}{2} )/\cos (\tfrac{j\pi }{n}) = (k'-\frac{1}{2} )/\cos (\tfrac{j'\pi }{n})$ which implies $(k-\frac{1}{2} )\cos (\tfrac{j'\pi }{n}) - (k'-\frac{1}{2} )\cos (\tfrac{j\pi }{n}) = 0$. Since the angles are rational with $\pi$ with $0\leqslant \tfrac{j\pi }{n}< \tfrac{j'\pi }{n} < \tfrac{\pi }{2}$ and the coefficients are rational, this is a trigonometric diophantine equation with two terms so we can apply known results on trigonometric diophantine equations Conway and Jones (1976). Using this known result we obtain that the only possible case is $\tfrac{j\pi }{n}=0$ and $\tfrac{j'\pi }{n}=\tfrac{\pi }{3}$. We now have $(k-\frac{1}{2} )\cdot \frac{1}{2} - (k'-\frac{1}{2} ) = 0$, i.e., $\tfrac{k}{2} + \tfrac{1}{4} = k'$ which is impossible since k and $k'$ are integers. We get the expected contradiction so this means that the intersection is empty.

Lemma 36

(The eigenvalue matrix $N_{n}$ is almost orthogonal) Let n be an even integer greater than 2. Let $N_{n}$ be defined as

$$\begin{aligned} N_{n} := \left( \lambda _{n,j,i} \right) _{0\leqslant i,j< \frac{n}{2}} = \left( \eta _j \cos \tfrac{(2i+1)j\pi }{n}\right) _{0\leqslant i,j < \frac{n}{2}}, \end{aligned}$$

with $\eta _0:=1$ and $\eta _j:= 2$ for $0<j<\tfrac{n}{2}$ (Definition 12).

The matrix $N_{n}$ is almost orthogonal in the sense that $N_{n}\cdot D^\textsf{T}= D\cdot N_{n}^\textsf{T}= \frac{n}{2}\textrm{Id}_{\frac{n}{2}}$ with the matrix D defined below.

In particular, $N_{n}\cdot \gamma ^\textsf{T}= (\frac{n}{2},0,0,\dots 0)^\textsf{T}$ with $\gamma := \left( \cos (\tfrac{i\pi }{n})\right) _{0\leqslant i < \frac{n}{2}}$ the optimal rhombus frequency vector (Definition 15).

Proof

Let us first recall that here n is an even integer.

Definitions: We define four matrices A, B, C and D by:

$$\begin{aligned} A&:= \left( a_i \cos \frac{i(2j+1)\pi }{2n}\right) _{0\leqslant i,j< n} \qquad \qquad \qquad a_i = {\left\{ \begin{array}{ll} \sqrt{\frac{1}{n}} \text { if } i=0 \\ \sqrt{\frac{2}{n}} \text { otherwise}\end{array}\right. }\\ B&:= \left( b_i \cos \frac{i(2j+1)\pi }{n}\right) _{0\leqslant i< \frac{n}{2},0\leqslant j< n} \qquad \qquad b_i = {\left\{ \begin{array}{ll} \sqrt{\frac{1}{n}} \text { if } i=0 \\ \sqrt{\frac{2}{n}} \text { otherwise}\end{array}\right. }\\ C&:= \left( c_i \cos \frac{i(2j+1)\pi }{n}\right) _{0\leqslant i,j< \frac{n}{2}} \qquad \qquad \qquad c_i = {\left\{ \begin{array}{ll} \sqrt{\frac{2}{n}} \text { if } i=0 \\ \frac{2}{\sqrt{n}} \text { otherwise}\end{array}\right. }\\ D&:= \left( \cos \frac{(2i+1)j\pi }{n}\right) _{0\leqslant i,j< \frac{n}{2}} \end{aligned}$$

Overview of the proof: the matrix A is a Discrete Cosine Transform matrix, sometimes called DCT-III, which is know to be orthogonal. From this we prove that B is semi-orthogonal and C is orthogonal. From the fact that C is orthogonal we get $N_{n}\cdot D^\textsf{T}= D\cdot N_{n}^\textsf{T}= \tfrac{n}{2}\textrm{Id}_{\frac{n}{2}}$ because $(N_{n}\cdot D^\textsf{T})_{i,k} = \tfrac{n}{2}(C\cdot C^\textsf{T})_{i,k}$.

Details and computations: We consider as known the fact that A (DCT-III) is orthogonal. As B is a rectangular matrix and not a square matrix, it cannot be orthogonal, however we say it is semi-orthogonal for $B\cdot B^\textsf{T}= \textrm{Id}_{\frac{n}{2}}$. This holds due to the fact that A is orthogonal and

$$\begin{aligned} (B\cdot B^\textsf{T})_{i,k} = \sum \limits _{j=0}^{n-1}B_{i,j}B_{k,j} = \sum \limits _{j=0}^{n-1}A_{2i,j}A_{2k,j} = (A\cdot A^\textsf{T})_{2i,2k}. \end{aligned}$$

Now remark that $B_{i,j}=B_{i,n-1-j}$ because

$$\begin{aligned} \cos \left( \frac{i(2(n-1-j)+1)\pi }{n}\right) = \cos \left( 2i\pi - \frac{i(2j+1)\pi }{n}\right) = \cos \left( \frac{i(2j+1)\pi }{n}\right) , \end{aligned}$$

So $(B\cdot B^\textsf{T})_{i,k} = (C\cdot C^\textsf{T})_{i,k}$ because $C_{i,j}=\sqrt{2}B_{i,j}$ and

$$\begin{aligned} (B\cdot B^\textsf{T})_{i,k} = \sum \limits _{j=0}^{n-1} B_{i,j}B_{k,j} = 2\sum \limits _{j=0}^{\frac{n}{2}-1} B_{i,j}B_{k,j} = \sum \limits _{j=0}^{\frac{n}{2}-1} \sqrt{2}B_{i,j}\sqrt{2}B_{k,j} = \sum \limits _{j=0}^{\frac{n}{2}-1} C_{i,j}C_{k,j}. \end{aligned}$$

So C is orthogonal. Now remark that C is somewhat similar to $N_{n}^\textsf{T}$.

We actually have $(C^\textsf{T}\cdot C)_{i,j} = \tfrac{2}{n} (N_{n}\cdot D^\textsf{T})_{i,j}$ , for any $0\leqslant i,j < n$, indeed

$$\begin{aligned} \left( C^\textsf{T}\cdot C\right) _{i,j}&= \sum \limits _{k=0}^{\frac{n}{2}-1} C_{k,i}C_{k,j} = \sum \limits _{k=0}^{\frac{n}{2}-1} c_k^2 \cos \left( \frac{k(2i+1)\pi }{n}\right) \cos \left( \frac{k(2j+1)\pi }{n}\right) \\&= \sum \limits _{k=0}^{\frac{n}{2}-1} \eta _k\frac{2}{n} \cos \left( \frac{k(2i+1)\pi }{n}\right) \cos \left( \frac{k(2j+1)\pi }{n}\right) = \tfrac{2}{n} \left( N_{n}\cdot D^\textsf{T}\right) _{i,j}, \end{aligned}$$

recall that $N_{n}=(\eta _j \cos \tfrac{(2i+1)j\pi }{n})_{0\leqslant i,j < \frac{n}{2}}$ with $\eta _0 = 1$ and $\eta _j= 2$ for $j>0$, so we indeed have $c_j^2 = \tfrac{2 \eta _j}{n}$.

In particular we have $N_{n}\cdot \gamma ^\textsf{T}= (\tfrac{n}{2},0,0\dots , 0)^\textsf{T}$.