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Substitution Discrete Plane Tilings with 2n-Fold Rotational Symmetry for Odd n


We study substitution tilings that are also discrete plane tilings, that is, satisfy a relaxed version of cut-and-projection. We prove that the Sub Rosa substitution tilings with a 2n-fold rotational symmetry for odd \(n>5\) defined by Kari and Rissanen are not discrete planes—and therefore not cut-and-project tilings either. We then define new Planar Rosa substitution tilings with a 2n-fold rotational symmetry for any odd n, and show that these satisfy the discrete plane condition. The tilings we consider are edge-to-edge rhombus tilings. We give an explicit construction for the 10-fold case, and provide a construction method for the general case of any odd n. Our methods are to lift the tilings and substitutions to \({\mathbb {R}}^{{n}}\) using the lift operator first defined by Levitov, and to study the planarity of substitution tilings in \({\mathbb {R}}^{{n}}\) using mainly linear algebra, properties of circulant matrices, and trigonometric sums. For the construction of the Planar Rosa substitutions we additionally use the Kenyon criterion and a result on De Bruijn multigrid dual tilings.

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We wish to thank Thierry Monteil and Nicolas Bédaride for their help regarding billiard words, and also Thomas Fernique for his help and proofreading.

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Correspondence to Victor H. Lutfalla.

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Lemma A.1

(eigenvalues of the elementary matrices) Let n be an odd integer and \(i\in \{0,\dots , \lfloor {n}/{2}\rfloor -1\}\). The elementary matrix \(M_i(n)\) defined in Definition 4.1 has eigenspaces \(\Delta \) and \({\mathcal {E}}_{{j}}\) for \(0\leqslant j<\lfloor n/2\rfloor \) with eigenvalue \(\lambda _\Delta =0\) and

$$\begin{aligned} \lambda _{i,j}(n)= 2\cos \frac{(2j+1)(2i+1)\pi }{2n}\,e^{-{(2j+1) \pi \mathrm{i}}/({2n})}. \end{aligned}$$


Let z be a complex number such that \(z^n=1\) and Z be the vector \((z^k)_{0\leqslant k < n}\). We have

$$\begin{aligned} Z\cdot M_i(n) = (-1)^i(z^{I({i})} - z^{J({i})})\cdot Z. \end{aligned}$$


$$\begin{aligned} z={\mathrm{exp}\,{\frac{2(2j+1)\pi \mathrm{i}}{n}}} \end{aligned}$$

we get that \({\mathcal {E}}_{{j}}\) is eigenspace of \(M_i(n)\) with eigenvalue \(\lambda _{i,j}(n)\),

$$\begin{aligned} \lambda _{i,j}(n)&=(-1)^i\biggl (\mathrm{exp}\,{\frac{2(2j+1) I({i})\pi \mathrm{i}}{n}}-\mathrm{exp}\,{\frac{2(2j+1)J({i}) \pi \mathrm{i}}{n}}\biggr )\\&= (-1)^i\mathrm{exp}\,\biggl (\frac{2(2j+1) i\pi \mathrm{i}}{n} \cdot \frac{n+1}{2}\biggr ) \\&\quad + (-1)^{i+1}\mathrm{exp} \,\biggl (\frac{-2(2j+1)(i+1)\pi \mathrm{i}}{n}\cdot \frac{n+1}{2}\biggr ) \\&= (-1)^i\mathrm{exp}\,{\frac{(2j+1) i(n+1)\pi \mathrm{i}}{n}} + (-1)^{i+1} \mathrm{exp}\,{\frac{-(2j+1)(i+1)(n+1)\pi \mathrm{i}}{n}} \\&= (-1)^i\mathrm{exp}\,{((2j+1) i\pi \mathrm{i})}\,\mathrm{exp}\,{\frac{(2j+1) i\pi \mathrm{i}}{n}} \\&\qquad +(-1)^{i+1}\,\mathrm{exp}\,{(-(2j+1)(i+1) \pi \mathrm{i})}\,\mathrm{exp}\,{\frac{-(2j+1)(i+1)\pi \mathrm{i}}{n}}\\&= (-1)^i(-1)^i\mathrm{exp}\,{\frac{(2j+1) i\pi \mathrm{i}}{n}} +(-1)^{i+1}(-1)^{i+1}\mathrm{exp}\,{\mathrm{i}\frac{-(2j+1)(i+1)\pi }{n}}\\&= \mathrm{exp}\,{\frac{(2j+1) i\pi \mathrm{i}}{n}} +\mathrm{exp} \,{\frac{-(2j+1)(i+1)\pi \mathrm{i}}{n}}\\&=2\cos \frac{(2j+1)(2i+1)\pi }{2n}\,\mathrm{exp} \,{\frac{-(2j+1)\pi \mathrm{i}}{2n}}. \end{aligned}$$

With \(z=1\) we get that \(\Delta \) is an eigenspace with eigenvalue 0. \(\square \)

Lemma A.2

(trigonometric manipulations) Let k be an integer and \(i\in \{0,\dots , k-1\}\). Let us define \(C_{j,k}\) by

$$\begin{aligned} C_{j,k}:=\sum \limits _{i=0}^{k-1}4(k-i)\cos {(2i+1)\theta _{j,k}}, \end{aligned}$$

with \(\theta _{j,k}:={(2j+1)\pi }/({2(2k+1)})\). We have

$$\begin{aligned} C_{j,k}\sin ^2\theta _{j,k}= \cos \theta _{j,k}. \end{aligned}$$


Let us write \(\theta \) for \(\theta _{j,k}\) for the sake of simplicity.

$$\begin{aligned}&C_{j,k} \sin ^2\theta \\&\quad =\sum \limits _{i=0}^{k-1}4(k-i)\cos \frac{(2j+1) (2i+1)\pi }{2(2k+1)}\sin ^2\theta \\&\quad =\sum \limits _{i=0}^{k-1}4(k-i) \frac{e^{\mathrm{i}(2i+1)\theta } + e^{-\mathrm{i}(2i+1)\theta }}{2} \biggl (\frac{e^{\mathrm{i}\theta } - e^{-\mathrm{i}\theta }}{2i}\biggr )^{2}\\&\quad =\sum \limits _{i=0}^{k-1}\frac{k-i}{2} \bigl (e^{\mathrm{i}(2i+1)\theta } + e^{-\mathrm{i}(2i+1)\theta }\bigr ) \bigl ( 2 - e^{\mathrm{i}2\theta } - e^{-\mathrm{i}2\theta }\bigr )\\&\quad = \sum \limits _{i=0}^{k-1}\frac{k-i}{2} \bigl ( 2e^{\mathrm{i}(2i+1)\theta } + 2e^{-\mathrm{i}(2i+1)\theta } -e^{\mathrm{i}(2i+3)\theta } \\&\qquad \qquad \qquad \qquad \qquad \qquad \quad -e^{-\mathrm{i}(2i-1)\theta } -e^{\mathrm{i}(2i-1)\theta } -e^{-\mathrm{i}(2i+3)\theta }\bigr )\\&\quad = \sum \limits _{i=0}^{k-1}(k-i)( 2\cos {(2i+1)\theta } - \cos {(2i+3)\theta } - \cos {(2i-1)\theta })\\&\quad = \sum \limits _{i=0}^{k-1}(k-i)2\cos (2i+1)\theta - \sum \limits _{i=0}^{k-1}(k-i)\cos (2i+3)\theta \\&\qquad -\sum \limits _{i=0}^{k-1}(k-i)\cos (2i-1)\theta \\&\quad = \sum \limits _{i=0}^{k-1}(k-i)2\cos {(2i+1)\theta } - \sum \limits _{i=1}^{k}(k+1-i)\cos {(2i+1)\theta }\\&\qquad -\sum \limits _{i=-1}^{k-2}(k-1-i)\cos {(2i+1)\theta }\\&\quad = \sum \limits _{i=0}^{k-1}(k-i)2\cos {(2i+1)\theta }\\&\qquad - \sum \limits _{i=0}^{k-1}(k+1-i)\cos {(2i+1)\theta } +(k+1)\cos \theta - \cos \frac{(2j+1)\pi }{2} \\&\qquad -\sum \limits _{i=0}^{k-1}(k-1-i)\cos {(2i+1)\theta } + 0\cos {(2k-1)\theta } - k\cos \theta \\&\quad = \sum \limits _{i=0}^{k-1}(2(k-i)-(k+1-i) -(k-1-i))\cos {(2i+1)\theta }\\&\qquad +(k+1-k)\cos \theta = \cos \theta . \end{aligned}$$

\(\square \)

Lemma A.3

(the eigenvalue matrix \(N_{{n}}\) is orthogonal up to a scalar) Let n be an odd integer, then the matrix \(N_{{n}}\) from Definition 4.4 is orthogonal up to a scalar. More precisely, \((1/\sqrt{n})N_{{n}}\) is an orthogonal matrix, i.e.,

$$\begin{aligned} \frac{1}{\sqrt{n}}N_{{n}}\cdot \frac{1}{\sqrt{n}} N_{{n}}^\top = \mathrm{Id}_{\lfloor {{n}/{2}}\rfloor }. \end{aligned}$$


Let us first recall that n is an odd integer. Let us define three matrices A, B, and C by:

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

Let us first remark that A is a Discrete Cosine Transform matrix, sometimes called DCT-III, which is known to be orthogonal. From this we will prove that \(B\cdot B^\top = \mathrm{Id}_{\rfloor {n/2}\rfloor }\) and then that C is orthogonal.

Let us prove \(B\cdot B^\top = \mathrm{Id}_{\lfloor {{n}/{2}}\rfloor }\). Let us look at the jk coefficient of \(B\cdot B^\top \) for \(0\leqslant j,k <n/2\):

$$\begin{aligned} (B\cdot B^\top )_{j,k}&= \sum \limits _{i=0}^{n-1} \sqrt{\frac{2}{n}}\cos \frac{(2j+1)(2i+1)\pi }{2n} \cdot \sqrt{\frac{2}{n}}\cos \frac{(2k+1)(2i+1)\pi }{2n}\\&=(A\cdot A^\top )_{2j+1,2k+1}= {\left\{ \begin{array}{ll} 1 &{}\text { if } j = k ,\\ 0 &{}\text { otherwise}. \end{array}\right. } \end{aligned}$$

Now from the fact that \(B\cdot B^\top = \mathrm{Id}_{\lfloor {{n}/ {2}}\rfloor }\) let us prove that C is orthogonal. Let us first remark that for \(0\leqslant i<\lfloor {{n}/{2}}\rfloor \) we have

$$\begin{aligned} B_{i,\lfloor {{n}/{2}}\rfloor } = \sqrt{\frac{2}{n}} \cos \frac{(2i+1)\pi }{2}=0. \end{aligned}$$

This is due to the fact that n is odd, which implies that \(n=2\lfloor {{n}/{2}}\rfloor +1\). Let us also remark that for \(0\leqslant i,j<\lfloor {{n}/{2}}\rfloor \) we have

$$\begin{aligned} \sqrt{2} B_{i,j} = C_{i,j} \qquad \text { and } \qquad B_{i,n-1-j} = -B_{i, j}. \end{aligned}$$

The first equality is just a reformulation of the definition, and for the second equality let us develop \(B_{i,n-1-j}\) as follows:

$$\begin{aligned} B_{i,n-1-j}&= \sqrt{\frac{2}{n}} \cos \frac{(2i+1)(2(n-1-j)+1)\pi }{2n}\\&= \sqrt{\frac{2}{n}} \cos \frac{(2i+1)(2n - 2 - 2j + 1)\pi }{2n}\\&= \sqrt{\frac{2}{n}} \cos \frac{(2i+1)(2n - (2j + 1))\pi }{2n}\\&= \sqrt{\frac{2}{n}} \cos {\biggl ((2i+1)\pi -\frac{(2i+1)(2j + 1)\pi }{2n}\biggr )}\\&=\sqrt{\frac{2}{n}} \cos {\biggl ( \pi -\frac{(2i+1)(2j + 1)\pi }{2n}\biggr )}\\&= -\sqrt{\frac{2}{n}} \cos \frac{(2i+1) (2j + 1)\pi }{2n}= -B_{i,j}. \end{aligned}$$

Now let us prove that \((C\cdot C^\top )_{j,k} =(B\cdot B^\top )_{j,k}\) for \(0\leqslant j,k < \lfloor {{n}/{2}}\rfloor \).

$$\begin{aligned} (B\cdot B^\top )_{j,k}&= \sum \limits _{i=0}^{n-1} B_{j,i}B_{k,i} \\&=\sum \limits _{i=0}^{\lfloor {{n}/{2}}\rfloor -1} B_{j,i}B_{k,i} + B_{j,\lfloor {{n}/{2}}\rfloor }B_{k,\lfloor {{n}/{2}}\rfloor } +\sum \limits _{i=\lfloor {{n}/{2}}\rfloor +1}^{n-1} B_{j,i}B_{k,i} \\&=\sum \limits _{i=0}^{\lfloor {{n}/{2}}\rfloor -1} B_{j,i} B_{k,i} + 0 + \sum \limits _{i=\lfloor {{n}/{2}}\rfloor +1}^{n-1} B_{j,i}B_{k,i} \\&=\sum \limits _{i=0}^{\lfloor {{n}/{2}}\rfloor -1} B_{j,i}B_{k,i} +\sum \limits _{i=0}^{\lfloor {{n}/{2}}\rfloor -1} B_{j,n-1-i} B_{k,n-1-i} \\&= \sum \limits _{i=0}^{\lfloor {{n}/{2}}\rfloor -1}( B_{j,i} B_{k,i} + (-B_{j,i})\cdot (-B_{k,i}))\\&= \sum \limits _{i=0}^{\lfloor {{n}/{2}}\rfloor -1} 2B_{j,i} B_{k,i}= \sum \limits _{i=0}^{\lfloor {{n}/{2}}\rfloor -1} \sqrt{2}\,B_{j,i}\sqrt{2}B_{k,i}\\&= \sum \limits _{i=0}^{\lfloor {{n}/{2}}-1} C_{j,i} C_{k,i}=( C\cdot C^\top )_{j,k}. \end{aligned}$$

Hence C is orthogonal and \(N_{{n}}\) is orthogonal up to a scalar. \(\square \)

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Kari, J., Lutfalla, V.H. Substitution Discrete Plane Tilings with 2n-Fold Rotational Symmetry for Odd n. Discrete Comput Geom 69, 349–398 (2023).

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  • Substitution tilings
  • Discrete planes
  • Cut-and-project tiling
  • n-Fold symmetric tiling
  • Quasiperiodic tilings
  • Rhombus tiling

Mathematics Subject Classification

  • 52C20
  • 37B50