The density of visible points in the Ammann-Beenker point set

TL;DR

This paper calculates the density of visible points in the Ammann-Beenker point set, extending known lattice results to a quasicrystal structure using Dedekind's zeta function over a quadratic field.

Contribution

It provides a formula for the density of visible points in the Ammann-Beenker set, linking it to Dedekind's zeta function, a novel extension of lattice point density results.

Findings

Density of visible points in Ammann-Beenker set is 2(√2-1)/ζ_K(2)

Connects quasicrystal point set density to algebraic number theory

Extends lattice point density results to a non-periodic structure

Abstract

The relative density of visible points of the integer lattice $\mathbb{Z}^d$ is known to be $1/\zeta(d)$ for $d\geq 2$, where $\zeta$ is Riemann's zeta function. In this paper we prove that the relative density of visible points in the Ammann-Beenker point set is given by $2(\sqrt{2}-1)/\zeta_K(2)$, where $\zeta_K$ is Dedekind's zeta function over $K=\mathbb{Q}(\sqrt{2})$.

Equations

$$\underset{R\to \infty }{\lim }\frac{\#(\mathcal{P}\cap RD)}{\mathrm{vol}(RD)}=\theta (\mathcal{P})$$

$$\#(\widehat{{\mathbb{Z}}^d}\cap RD)$$

$$=\#({\mathbb{Z}}_{\ast }^d\cap RD)+\sum _{k=1}^m(-1{)}^k\left(\sum _{1\le i_1<\dots <i_k\le m}\#(p_{i_1}{\mathbb{Z}}_{\ast }^d\cap \cdots \cap p_{i_k}{\mathbb{Z}}_{\ast }^d\cap RD)\right).$$

$$\sum _{n\in {\mathbb{Z}}_{>0}}\mu (n)\cdot \#(n{\mathbb{Z}}_{\ast }^d\cap RD),$$

$$\frac{\#(\widehat{{\mathbb{Z}}^d}\cap RD)}{\mathrm{vol}(RD)}=\sum _{n\in {\mathbb{Z}}_{>0}}\frac{\mu (n)\cdot \#(n{\mathbb{Z}}_{\ast }^d\cap RD)}{\mathrm{vol}(RD)}=\sum _{n\in {\mathbb{Z}}_{>0}}\frac{\mu (n)}{n^d}\frac{\#({\mathbb{Z}}_{\ast }^d\cap n^{-1}RD)}{\mathrm{vol}(n^{-1}RD)}.$$

$$\theta (\widehat{{\mathbb{Z}}^d})=\underset{R\to \infty }{\lim }\frac{\#(\widehat{{\mathbb{Z}}^d}\cap RD)}{\mathrm{vol}(RD)}=1/\zeta (d).$$

$$\pi :$$

$$(x_1,\dots ,x_n)⟼(x_1,\dots ,x_d)$$

$$\{x=x_1+x_2\zeta \mid x_1,x_2\in {\mathcal{O}}_K,x^{\star }\in W_8\},$$

$$\mathcal{L}=\{(x,\overline{x})\mid x=(x_1,x_2)\in {\mathcal{O}}_K^2\}\subset {\mathbb{R}}^4$$

$$\tilde{\mathcal{L}}=\{(x,\overline{x})\in \mathcal{L}\mid (x_1-x_2)/\sqrt{2}\in {\mathcal{O}}_K\}.$$

$$\#(\hat{\mathcal{P}}\cap RD)=\sum _{\begin{array}{c}F\subset C\\ \#F<\infty \end{array}}(-1{)}^{\#F}\#\left(\left({\mathcal{P}}_{\ast }\cap \bigcap _{c\in F}c{\mathcal{P}}_{\ast }\right)\cap RD\right).$$

$$\#(\hat{\mathcal{P}}\cap RD)=\#\left(({\mathcal{P}}_{\ast }\cap RD)\setminus \bigcup _{c\in C}({\mathcal{P}}_{\ast }\cap c{\mathcal{P}}_{\ast }\cap RD)\right)$$

$$=\#({\mathcal{P}}_{\ast }\cap RD)-\#\left(\bigcup _{i=1}^n({\mathcal{P}}_{\ast }\cap c_i{\mathcal{P}}_{\ast }\cap RD)\right),$$

$$\hat{\mathcal{B}}=\{x=(x_1,x_2)\in \mathcal{B}\mid \gcd (x_1,x_2)=1,\lambda \overline{x}\notin {\mathcal{W}}_{\mathcal{A}}\}.$$

$$\mathbb{P}=\{\pi \in {\mathcal{O}}_K\mid \pi \text{leavevmode}\mathrm{prime},1<\pi <\lambda \}\text{leavevmode}\text{and}\text{leavevmode}C=\mathbb{P}\cup \{\lambda \}$$

$$\#(\hat{\mathcal{B}}\cap RD)=\sum _{\begin{array}{c}F\subset C\\ \#F<\infty \end{array}}(-1{)}^{\#F}\#\left((\mathcal{P}(m_F{\mathcal{W}}_{\mathcal{A}},{\mathcal{L}}_F)\setminus \{0\})\cap RD\right).$$

$$\theta (\mathcal{P}(m_F{\mathcal{W}}_{\mathcal{A}},{\mathcal{L}}_F)\setminus \{0\})=\frac{\mathrm{vol}(m_F{\mathcal{W}}_{\mathcal{A}})}{\mathrm{vol}({\mathbb{R}}^4/{\mathcal{L}}_F)}$$

$$\theta (\hat{\mathcal{B}})=\sum _{\begin{array}{c}F\subset C\\ \#F<\infty \end{array}}(-1{)}^{\#F}\frac{\mathrm{vol}(m_F{\mathcal{W}}_{\mathcal{A}})}{\mathrm{vol}({\mathbb{R}}^4/{\mathcal{L}}_F)}=\sum _{\begin{array}{c}F\subset C\\ \#F<\infty \end{array}}(-1{)}^{\#F}\frac{m_F^2(1+\sqrt{2})}{2N({\ell }_F{)}^2},$$

$$\#((\mathcal{P}(m_F{\mathcal{W}}_{\mathcal{A}},{\mathcal{L}}_F)\cap RD)\setminus \{0\})\le \frac{\tilde{L}{R}^2}{N({\ell }_F{)}^2}.$$

$$\#((\mathcal{P}(m_F{\mathcal{W}}_{\mathcal{A}},{\mathcal{L}}_F)\setminus \{0\})\cap RD)=\#(\{x\in {\ell }_F{\mathcal{O}}_K^2\mid \overline{x}\in m_F{\mathcal{W}}_{\mathcal{A}}\}\setminus \{0\})\cap RD).$$

$$(\{x\in {\ell }_F{\mathcal{O}}_K^2\mid \overline{x}\in m_F{\mathcal{W}}_{\mathcal{A}}\}\setminus \{0\})\cap RD⟶(\{x\in {\mathcal{O}}_K^2\mid \overline{x}\in \frac{m_F{\mathcal{W}}_{\mathcal{A}}}{|\overline{{\ell }_F}|}\}\setminus \{0\})\cap \frac{RD}{{\ell }_F}$$

$$\#\left(\left(\mathcal{L}\cap \frac{RD}{{\ell }_F}\times \left(\frac{\mathcal{W}}{|\overline{{\ell }_F}|}\right)\right)\setminus \{0\}\right)\le \frac{\tilde{L}{R}^2}{N({\ell }_F{)}^2}$$

$$\underset{R\to \infty }{\lim }\sum _{\begin{array}{c}F\subset C\\ \#F<\infty \end{array}}\frac{(-1{)}^{\#F}\#\left((\mathcal{P}(m_F{\mathcal{W}}_{\mathcal{A}},{\mathcal{L}}_F)\setminus \{0\})\cap RD\right)}{\mathrm{vol}(RD)}=\sum _{\begin{array}{c}F\subset C\\ \#F<\infty \end{array}}\frac{(-1{)}^{\#F}{m}_F^2(1+\sqrt{2})}{2N({\ell }_F{)}^2}$$

$$\underset{R\to \infty }{\lim }\sum _{\begin{array}{c}F\subset C\\ \#F<\infty \end{array}}\frac{(-1{)}^{\#F}N(R,F)}{\mathrm{vol}(RD)}=\sum _{\begin{array}{c}F\subset C\\ \#F<\infty \end{array}}\underset{R\to \infty }{\lim }\frac{(-1{)}^{\#F}N(R,F)}{\mathrm{vol}(RD)}$$

$$\sum _{\begin{array}{c}F\subset C\\ \#F<\infty \end{array}}\left|\frac{(-1{)}^{\#F}N(R,F)}{\mathrm{vol}(RD)}\right|\le \frac{\tilde{L}}{\mathrm{vol}(D)}\sum _{\begin{array}{c}F\subset C\\ \#F<\infty \end{array}}\frac{1}{N({\ell }_F{)}^2}$$

$$\sum _{\begin{array}{c}F\subset C\\ \#F<\infty \end{array}}\frac{1}{N({\ell }_F{)}^2}=\sum _{\begin{array}{c}F\subset C\\ \#F<\infty \\ \lambda \in F\end{array}}\frac{1}{N({\ell }_F{)}^2}+\sum _{\begin{array}{c}F\subset C\\ \#F<\infty \\ \lambda \notin F\end{array}}\frac{1}{N({\ell }_F{)}^2}\le 2\sum _{I◃{\mathcal{O}}_K}\frac{1}{N(I{)}^2},$$

$$\left|\underset{R\to \infty }{\lim }\sum _{\begin{array}{c}F\subset C\\ \#F<\infty \end{array}}\frac{(-1{)}^{\#F}N(R,F)}{\mathrm{vol}(RD)}-\sum _{\begin{array}{c}F\subset C\\ \#F<\infty \end{array}}\frac{m_F^2(1+\sqrt{2})}{2N({\ell }_F{)}^2}\right|$$

$$\le \left(\frac{\tilde{L}}{\mathrm{vol}(D)}+\frac{1+\sqrt{2}}{2}\right)\sum _{\begin{array}{c}F\subset C\\ \#F<\infty \\ |N({\ell }_F)|\ge \Delta \end{array}}\frac{1}{N({\ell }_F{)}^2}$$

$$\mu (I)=\{\begin{array}{ll}0& \text{ if }\exists \pi \in \mathbb{P}\text{ such that }I\subset {\pi }^2{\mathcal{O}}_K,\\ (-1{)}^{\omega (I)}& \text{ otherwise.}\end{array}$$

$${\zeta }_K(s)=\prod _{P◃{\mathcal{O}}_K,P\mathrm{prime}}\frac{1}{1-N(P{)}^{-s}}$$

$$\frac{1}{{\zeta }_K(s)}=\sum _{I◃{\mathcal{O}}_K}\frac{\mu (I)}{N(I{)}^s}.$$

Full text

1 Introduction

A locally finite point set $\mathcal{P}\subset\mathbb{R}^{d}$ has an asymptotic density (or simply density) $\theta(\mathcal{P})$ if

[TABLE]

holds for all Jordan measurable $D\subset\mathbb{R}^{d}$. The density of a set can be interpreted as the asymptotic number of elements per unit volume. For instance, for a lattice $\mathcal{L}\subset\mathbb{R}^{d}$ we have $\theta(\mathcal{L})=\frac{1}{\mathrm{vol}(\mathbb{R}^{d}/\mathcal{L})}$. Let $\widehat{\mathcal{P}}=\{x\in\mathcal{P}\mid tx\notin\mathcal{P},\forall t\in(0,1)\}$ denote the subset of the visible points of $\mathcal{P}$. If $\mathcal{P}$ is a regular cut-and-project set (see 3.1 below) then it is known that $\theta(\mathcal{P})$ exists. In [5, Theorem 1], J. Marklof and A. Strömbergsson proved that $\theta(\widehat{\mathcal{P}})$ also exists and that $0<\theta(\widehat{\mathcal{P}})\leq\theta(\mathcal{P})$ if $\theta(\mathcal{P})>0$. In particular, for such $\mathcal{P}$ the relative density of visible points $\kappa_{\mathcal{P}}:=\frac{\theta(\widehat{\mathcal{P}})}{\theta(\mathcal{P})}$ exists, but is not known explicitly in most cases.

For $d\geq 2$ we have $\widehat{\mathbb{Z}^{d}}=\{(n_{1},\ldots,n_{d})\in\mathbb{Z}^{d}\mid\gcd(n_{1},\ldots,n_{d})=1\}$ and $\theta(\widehat{\mathbb{Z}^{d}})=1/\zeta(d)$ gives the probability that $d$ random integers share no common factor. This can be derived in several ways, see for instance [6]; we sketch another proof in Section 2 below. More generally, $\theta(\widehat{\mathcal{L}})=\frac{1}{\mathrm{vol}(\mathbb{R}^{d}/\mathcal{L})\zeta(d)}$ for a lattice $\mathcal{L}\subset\mathbb{R}^{d}$, see e.g. [3, Prop. 6].

A well-known point set, which can be realised both as the vertices of a substitution tiling and as a cut-and-project set, is the Ammann-Beenker point set. The goal of this paper is to prove that the relative density of visible points in the Ammann-Beenker point set is $2(\sqrt{2}-1)/\zeta_{K}(2)$. This density was computed by B. Sing in the presentation [7], but he has not published a proof of this result.

2 The density of the visible points of $\mathbb{Z}^{d}$

In this section we show that $\theta(\widehat{\mathbb{Z}^{d}})=1/\zeta(d)$. We shall see that a lot of inspiration can be drawn from this example when calculating the density of the visible points in the Ammann-Beenker point set.

Fix $R>0$, a Jordan measurable $D\subset\mathbb{R}^{d}$ and let $\mathbb{P}\subset\mathbb{Z}_{>0}$ denote the set of prime numbers. For each invisible point $n\in\mathbb{Z}^{d}\setminus\widehat{\mathbb{Z}^{d}}$, there is $p\in\mathbb{P}$ such that $\frac{n}{p}\in\mathbb{Z}^{d}$. Setting $\mathbb{Z}^{d}_{*}=\mathbb{Z}^{d}\setminus\{(0,\ldots,0)\}$ there are only finitely many $p_{1},\ldots,p_{n}\in\mathbb{P}$ such that $p_{i}\mathbb{Z}^{d}_{*}\cap RD\neq\emptyset$. By inclusion-exclusion counting we have

[TABLE]

The last sum can be rewritten to

[TABLE]

where $\mu$ is the Möbius function. Hence

[TABLE]

Letting $R\to\infty$, switching order of limit and summation (for instance justified by finding a constant $C$ depending on $D$ such that $\#(\mathbb{Z}_{*}^{d}\cap RD)\leq C\mathrm{vol}(RD)$ for all $R$), using $\theta(\mathbb{Z}^{d}_{*})=1$ and $1/\zeta(s)=\sum_{n\in\mathbb{Z}_{>0}}\frac{\mu(n)}{n^{s}}$ for $s>1$, we find that

[TABLE]

3 Cut-and-project sets and the Ammann-Beenker point set

The Ammann-Beenker point set can be obtained as the vertices of the Ammann-Beenker tiling, a substitution tiling of the plane using a square and a rhombus as tiles, see e.g. [2, Chapter 6.1]. In this paper however, the Ammann-Beenker set is realised as a cut-and-project set, a certain type of point set which we will now define. Cut-and-project sets are sometimes called (Euclidean) model sets. We will use the same notation and terminology for cut-and-project sets as in [4, Sec. 1.2]. For an introduction to cut-and-project sets, see e.g. [2, Ch. 7.2].

If $\mathbb{R}^{n}=\mathbb{R}^{d}\times\mathbb{R}^{m}$, let

[TABLE]

denote the natural projections.

Definition 3.1**.**

Let $\mathcal{L}\subset\mathbb{R}^{n}$ be a lattice and $\mathcal{W}\subset\overline{\pi_{\mathrm{int}}(\mathcal{L})}$ be a set. Then the cut-and-project set of $\mathcal{L}$ and $\mathcal{W}$ is given by $\mathcal{P}(\mathcal{W},\mathcal{L})=\{\pi(y)\mid y\in\mathcal{L},\pi_{\mathrm{int}}(y)\in\mathcal{W}\}$.

If $\partial W$ has measure zero with respect to any Haar measure on $\overline{\pi_{\mathrm{int}}(\mathcal{L})}$ we say that $\mathcal{P}(\mathcal{W},\mathcal{L})$ is regular. If the interior of $\mathcal{W}$ (the window) is non-empty, $\mathcal{P}(\mathcal{W},\mathcal{L})$ is relatively dense and if $\mathcal{W}$ is bounded, $\mathcal{P}(\mathcal{W},\mathcal{L})$ is uniformly discrete (cf. [4, Prop. 3.1]). To realise the Ammann-Beenker point set in this way, let $K$ be the number field $\mathbb{Q}(\sqrt{2})$, with algebraic conjugation $x\mapsto\overline{x}$ (we will also write $\overline{x}=(\overline{x_{1}},\ldots,\overline{x_{n}})$ for $x=(x_{1},\ldots,x_{n})\in K^{n}$) and norm $N(x)=x\overline{x}$. The ring of integers $\mathcal{O}_{K}=\mathbb{Z}[\sqrt{2}]$ of $K$ is a Euclidean domain with fundamental unit $\lambda:=1+\sqrt{2}$. With $\zeta:=e^{\tfrac{\pi i}{4}}$ and $\star:K\longrightarrow K$, $x\mapsto x^{\star}$ the automorphism generated by $\zeta\mapsto\zeta^{3}$, the Ammann-Beenker point set is in [2, Example 7.7] realised as

[TABLE]

where $W_{8}\subset\mathbb{C}$ is the regular octagon of side length $1$ centered at the origin, with sides perpendicular to the coordinate axes.

Let

[TABLE]

be the Minkowski embedding of $\mathcal{O}_{K}^{2}$ and let

[TABLE]

Then, after a straight-forward translation it is seen that the Ammann-Beenker point set $\mathcal{A}$ can be realised in $\mathbb{R}^{2}$ as $\mathcal{A}=\frac{1}{\sqrt{2}}\mathcal{P}(\mathcal{W}_{\mathcal{A}},\widetilde{\mathcal{L}})$, where $\mathcal{W}_{\mathcal{A}}:=\sqrt{2}W_{8}$, i.e. $\mathcal{A}$ is the scaling of a cut-and-project set according to 3.1.

4 The density of visible points of $\mathcal{A}$

All notation used in this section is defined in and taken from Section 3. Since, for any $\mathcal{P}\subset\mathbb{R}^{d}$ whose density exists, and any $c>0$ it holds that $\theta(c\mathcal{P})=c^{-d}\theta(\mathcal{P})$ and $c\widehat{\mathcal{P}}=\widehat{c\mathcal{P}}$, finding $\theta(\widehat{\mathcal{A}^{\prime}})$ with $\mathcal{A}^{\prime}:=\sqrt{2}\mathcal{A}=\mathcal{P}(\mathcal{W}_{\mathcal{A}},\widetilde{\mathcal{L}})$ will give the value of $\theta(\widehat{\mathcal{A}})$. As a first step, in Section 4.1, the asymptotic density of the visible points of the simpler set $\mathcal{B}=\mathcal{P}(\mathcal{W}_{\mathcal{A}},\mathcal{L})=\{x\in\mathcal{O}_{K}^{2}\mid\overline{x}\in\mathcal{W}_{\mathcal{A}}\}\subset\mathcal{O}_{K}^{2}$ will be calculated. In Section 4.2 this result will be used to obtain $\theta(\widehat{\mathcal{A}})$.

4.1 The density of visible points of $\mathcal{B}$

The following general counting formula for bounded subsets of visible points of a point set $\mathcal{P}$ will be needed. Let $\mathcal{P}_{*}=\mathcal{P}\setminus\{(0,\ldots,0)\}$.

Proposition 4.1**.**

Let $\mathcal{P}\subset\mathbb{R}^{d}$ be locally finite and fix a set $C\subset\mathbb{R}_{>1}$ such that for each $x\in\mathcal{P}\setminus\widehat{\mathcal{P}}$ there exists $c\in C$ with $x/c\in\mathcal{P}$. Let $R>0$ and a bounded set $D\subset\mathbb{R}^{d}$ be given. Then

[TABLE]

Proof.

The set $C_{R}:=\{c\in C\mid\mathcal{P}_{*}\cap c\mathcal{P}_{*}\cap RD\neq\emptyset\}$ is finite. Indeed, suppose this is not true and pick distinct $c_{1},c_{2},\ldots\in C_{R}$ and corresponding $x_{i}\in\mathcal{P}_{*}\cap c_{i}\mathcal{P}_{*}\cap RD$. Since $\mathcal{P}$ is locally finite, the sequence $x_{1},x_{2},\ldots$ contains only finitely many distinct elements. Thus, a subsequence $x_{k_{1}},x_{k_{2}},\ldots$ which is constant can be extracted, so that $x_{k_{i}}/c_{k_{i}}\in\mathcal{P}_{*}\cap\frac{RD}{c_{k_{i}}}\subset\mathcal{P}_{*}\cap RD$ are all distinct, contradiction to $\mathcal{P}$ being locally finite. Thus, we can write $C_{R}=\{c_{1},\ldots,c_{n}\}$ for some $c_{1},\ldots,c_{n}\in C$. Then

[TABLE]

from which the result follows from the inclusion-exclusion counting formula for finite unions of finite sets. ∎

A set $C$ as in 4.1 for $\mathcal{B}$ will be needed, and to this end a visibility condition for the elements of $\mathcal{B}$ is required. Given $x_{1},x_{2}\in\mathcal{O}_{K}$, let $\gcd(x_{1},x_{2})$ be a fixed generator of the ideal generated by $x_{1},x_{2}$ and write $\gcd(x_{1},x_{2})=1$ when $x_{1},x_{2}$ are relatively prime. In the following proposition a visibility condition of the complex realisation of the Ammann-Beenker point set given in [1, p. 477] is adapted to our situation.

Proposition 4.2**.**

The visible points of $\mathcal{B}$ are given by

[TABLE]

Proof.

First the necessity of the visibility conditions is established. Take $x=(x_{1},x_{2})$ and suppose that $\gcd(x_{1},x_{2})\neq 1$ so that there exists $c\in\mathcal{O}_{K}$ with $|N(c)|>1$ and $c\mid x_{1},x_{2}$. Scaling $c$ by units we may assume that $1<c<\lambda$. Suppose first that $|N(c)|=|\overline{c}|c\geq 3$, which implies $|\overline{c}|>1$. By noting that $\mathcal{W}_{\mathcal{A}}$ is star-shaped with respect to the origin and $\mathcal{W}_{\mathcal{A}}=-\mathcal{W}_{\mathcal{A}}$ it follows that $x/c\in\mathcal{B}$, so $x$ is invisible. If $|N(c)|=2$, then each prime factor of $c$ must divide $2=\sqrt{2}\cdot\sqrt{2}$, so it can be assumed that $c=\sqrt{2}$ and hence $x$ is occluded by $x/\sqrt{2}$. If $\lambda\overline{x}\in\mathcal{W}_{\mathcal{A}}$ it follows immediately that $x/\lambda\in\mathcal{B}$.

We now turn to the sufficiency of the visibility conditions. Take $x=(x_{1},x_{2})\in\mathcal{B}\setminus\widehat{\mathcal{B}}$ and $c>1$ such that $x/c\in\mathcal{B}$. As $\mathcal{B}$ is uniformly discrete, we may assume that $y:=x/c\in\widehat{\mathcal{B}}$. This implies, by necessity above, that $\gcd(y_{1},y_{2})=1$. Now, since $x_{i}=cy_{i}$ it follows that $c\in K$. Write $c=a/b$ with $a,b\in\mathcal{O}_{K}$ relatively prime. If $b$ is not a unit, $\gcd(y_{1},y_{2})=1$ is contradicted, hence $c\in\mathcal{O}_{K}$.

If $|N(c)|\neq 1$ then $\gcd(x_{1},x_{2})\neq 1$. Otherwise, $c>1$ is a unit, i.e. $c=\lambda^{k}$ for some integer $k>0$. Thus $\frac{\overline{x}}{\overline{c}}=\frac{\overline{x}}{\overline{\lambda}^{k}}\in\mathcal{W}_{\mathcal{A}}$. Since $\frac{1}{\overline{\lambda}}=-\lambda$ we get $(-\lambda)^{k}\overline{x}\in\mathcal{W}_{\mathcal{A}}$ and thus also $\lambda\overline{x}\in\mathcal{W}_{\mathcal{A}}$ as $\mathcal{W}_{\mathcal{A}}$ is star-shaped with respect to the origin and $-\mathcal{W}_{\mathcal{A}}=\mathcal{W}_{\mathcal{A}}$. This establishes sufficiency of the visibility conditions. ∎

Remark*.*

Note that the proof works just as well for more general windows, that is, $\widehat{\mathcal{P}(\mathcal{W},\mathcal{L})}=\{x\in\mathcal{P}(\mathcal{W},\mathcal{L})\mid\gcd(x_{1},x_{2})=1,\lambda\overline{x}\notin\mathcal{W}\}$ if $\mathcal{W}\subset\mathbb{R}^{2}$ is bounded with non-empty interior, star-shaped with respect to the origin and $-\mathcal{W}=\mathcal{W}$.

Let now

[TABLE]

so that $\mathbb{P}$ is a set that contains precisely one associate of every prime of $\mathcal{O}_{K}$. Then we have the following proposition.

Proposition 4.3**.**

For each $x\in\mathcal{B}\setminus\widehat{\mathcal{B}}$ there is $c\in C$ such that $x/c\in\mathcal{B}$.

Proof.

Fix $x\in\mathcal{B}\setminus\widehat{\mathcal{B}}$. As seen in the proof of 4.2 there is $c\in\mathcal{O}_{K}$, $c>1$, such that $x/c\in\mathcal{B}$. If $c$ is not a unit, fix $\pi\in\mathbb{P}$ so that $\pi\mid c$. It can be verified by hand that $\{(x,\overline{x})\mid x\in\mathcal{O}_{K}\}\cap((1,\lambda)\times(-1,1))=\emptyset$, hence $|\pi|>1$ and $x/\pi\in\mathcal{B}$. If $c$ is a unit, $x/\lambda\in\mathcal{B}$ is immediate. ∎

Given a finite set $F\subset\mathcal{O}_{K}$ let $I_{F}$ be the (principal) ideal generated by the elements of $F$ if $F\neq\emptyset$ and $I_{F}=\mathcal{O}_{K}$ otherwise. Let $\ell_{F}$ denote a fixed least common multiple of $F$, that is, a generator of the ideal $\bigcap_{c\in F}c\mathcal{O}_{K}$. Let also $m_{F}=\min\{1,\min_{c\in F}|\overline{c}|\}$ and $\mathcal{L}_{F}=\{(\ell_{F}x,\overline{\ell_{F}x})\mid x\in\mathcal{O}_{K}^{2}\}$. Write $I\triangleleft\mathcal{O}_{K}$ when $I\subset\mathcal{O}_{K}$ is an ideal and define the absolute norm $N(I)$ of $I$ by $|N(x)|$, where $x$ is any generator of $I$. Recall Dedekind’s zeta function $\zeta_{K}(s)=\sum_{I\triangleleft\mathcal{O}_{K}}\frac{1}{N(I)^{s}}$ for $s\in\mathbb{C}$ with $\mathrm{Re}(s)>1$.

Given a finite set $F\subset C$ it is verified that $\mathcal{B}_{*}\cap\bigcap_{c\in F}c\mathcal{B}_{*}=\mathcal{P}(m_{F}\mathcal{W}_{\mathcal{A}},\mathcal{L}_{F})\setminus\{0\}$. For any $R>0$ and bounded $D\subset\mathbb{R}^{2}$, Propositions 4.1, 4.3 imply that

[TABLE]

Since $\ell_{F}\mathcal{O}_{K}^{2}\subset\pi_{\mathrm{int}}(\mathcal{L}_{F})\subset\mathbb{R}^{2}$ is dense we have

[TABLE]

from [4, Prop. 3.2]. Dividing (1) by $\mathrm{vol}(RD)$, letting $R\to\infty$ and switching order of limit and summation (to be justified in 4.6 below) we find that

[TABLE]

since $\mathrm{vol}(\mathcal{W}_{\mathcal{A}})=4(1+\sqrt{2})$ and $\mathrm{vol}(\mathbb{R}^{4}/\mathcal{L}_{F})=8N(\ell_{F})^{2}$. The value of the right hand sum will be shown to be $1/\zeta_{K}(2)$ in 4.7 below. The following lemma gives a bound on the number of points in the intersection of a lattice and a box in terms of the volume of the box, provided that the box is ”not too thin”.

Lemma 4.4**.**

Let $\mathcal{L}\subset\mathbb{R}^{d}$ be a lattice and let $c>0$ be given. For any $a_{i},b_{i}\in\mathbb{R}$ with $b_{i}-a_{i}>c$ set $B=\prod_{i=1}^{d}[a_{i},b_{i}]$. Then there is a constant $L$ depending only on $\mathcal{L}$ and $c$ such that $\#(B\cap\mathcal{L})\leq L\mathrm{vol}(B)$.

Proof.

Let $n_{i}=\lceil\frac{b_{i}-a_{i}}{c}\rceil\in\mathbb{Z}_{+}$. Then $\frac{b_{i}-a_{i}}{c}\leq n_{i}<\frac{b_{i}-a_{i}}{c}+1=\frac{b_{i}-a_{i}+c}{c}<\frac{2(b_{i}-a_{i})}{c}.$ Hence, with $n=\prod_{i=1}^{d}n_{i}$ it follows that $n\leq\frac{2^{d}\mathrm{vol}(B)}{c^{d}}$. From $b_{i}\leq a_{i}+cn_{i}$ also $B\subset\prod_{i=1}^{d}[a_{i},a_{i}+cn_{i}].$ Let $a=(a_{1},\ldots,a_{d})$ and consider $-a+\prod_{i=1}^{d}[a_{i},a_{i}+cn_{i}]=\prod_{i=1}^{d}[0,cn_{i}]$. We have $\prod_{i=1}^{d}[0,cn_{i}]=\bigcup_{m\in\mathbb{N}^{d},0\leq m_{i}<n_{i}}(mc+[0,c]^{d}).$ Hence, $B\subset\bigcup_{m\in\mathbb{N}^{d},0\leq m_{i}<n_{i}}(a+mc+[0,c]^{d})=:B^{\prime}$. Find now $D>0$ depending on $\mathcal{L}$ and $c$ such that $\sup_{t\in\mathbb{R}^{d}}\#(\mathcal{L}\cap(t+[0,c]^{d}))=D$. Hence $\#(B\cap\mathcal{L})\leq\#(B^{\prime}\cap\mathcal{L})\leq nD\leq\frac{2^{d}D}{c^{d}}\mathrm{vol}(B)$, so one can take $L=\frac{2^{d}D}{c^{d}}$. ∎

The following bound will be crucial in the justification of interchanging limit and summation in (1) after division by $\mathrm{vol}(RD)$.

Lemma 4.5**.**

Let $D\subset\mathbb{R}^{2}$ be Jordan measurable. Then there is a constant $\widetilde{L}>0$ depending only on $D$ such that for every $R>0$ and $F\subset C$ with $\#F<\infty$,

[TABLE]

Proof.

By definition

[TABLE]

Note that this number is independent of the choice of $\ell_{F}$. There is a bijection

[TABLE]

given by $x\mapsto\tfrac{x}{\ell_{F}}$, so it suffices to estimate the number of elements in the latter set. Since $m_{F}\leq 1$ it follows that $\left(\mathcal{L}\cap\left(\tfrac{RD}{\ell_{F}}\times\tfrac{m_{F}\mathcal{W}_{\mathcal{A}}}{|\overline{\ell_{F}}|}\right)\right)\setminus\{0\}\subset\left(\mathcal{L}\cap\left(\tfrac{RD}{\ell_{F}}\times\tfrac{\mathcal{W}_{\mathcal{A}}}{|\overline{\ell_{F}}|}\right)\right)\setminus\{0\}$. Fix real numbers $m_{1},m_{2}>1$ so that $D\subset[-m_{1},m_{1}]^{2}=:B_{1}$ and $\mathcal{W}_{\mathcal{A}}\subset[-m_{2},m_{2}]^{2}=:B_{2}$.

Fix a number $c$ so that $c^{\prime}<c$ implies $(\mathcal{L}\cap(\lambda D\times c^{\prime}\mathcal{W}_{\mathcal{A}}))\setminus\{0\}=\emptyset$. This can be done, for otherwise $(\mathcal{L}\cap(\lambda D\times c^{\prime}\mathcal{W}_{\mathcal{A}}))\setminus\{0\}$ would be non-empty for each $c^{\prime}>0$, hence $\mathcal{L}\cap(\lambda D\times\mathcal{W}_{\mathcal{A}})$ would contain infinitely many points, contradiction, since $\mathcal{L}$ is a lattice and $\lambda D\times\mathcal{W}_{\mathcal{A}}$ is bounded.

Suppose first that $\frac{R}{|\ell_{F}\overline{\ell_{F}}|}<c$. Scale $\ell_{F}$ by units so that $1\leq\tfrac{R}{\ell_{F}}<\lambda$ which gives $\tfrac{1}{|\overline{\ell_{F}}|}<c$. Hence $\left(\mathcal{L}\cap\left(\tfrac{RD}{\ell_{F}}\times\tfrac{\mathcal{W}_{\mathcal{A}}}{|\overline{\ell_{F}}|}\right)\right)\setminus\{0\}\subset\left(\mathcal{L}\cap\lambda D\times\left(\tfrac{\mathcal{W}_{\mathcal{A}}}{|\overline{\ell_{F}}|}\right)\right)\setminus\{0\}=\emptyset$ and therefore $\#\left(\left(\mathcal{L}\cap\left(\tfrac{RD}{\ell_{F}}\times\tfrac{\mathcal{W}_{\mathcal{A}}}{|\overline{\ell_{F}}|}\right)\right)\setminus\{0\}\right)=0$.

Suppose now that $\frac{R}{|\ell_{F}\overline{\ell_{F}}|}\geq c$. Scale $\ell_{F}$ so that $\sqrt{c}\leq\tfrac{R}{\ell_{F}}<\lambda\sqrt{c}$. This implies that $\tfrac{1}{|\overline{\ell_{F}}|}\geq\tfrac{1}{\lambda}\sqrt{c}>\sqrt{c}$. Thus, $[0,\sqrt{c}]^{4}\subset\tfrac{RB_{D}}{\ell_{F}}\times\tfrac{B_{\mathcal{W}}}{|\overline{\ell_{F}}|}=:B$. From 4.4 we get a constant $L$ only depending on $\mathcal{L}$, $\sqrt{c}$ such that $\#(B\cap\mathcal{L})\leq L\mathrm{vol}(B)=L\cdot 16m_{1}^{2}m_{2}^{2}\frac{R^{2}}{N(\ell_{F})^{2}}.$ Now, since $\tfrac{RD}{\ell_{F}}\times\tfrac{\mathcal{W}_{\mathcal{A}}}{|\overline{\ell_{F}}|}\subset B$ we get that

[TABLE]

with $\widetilde{L}:=16m_{1}^{2}m_{2}^{2}L$. ∎

Proposition 4.6**.**

The equality

[TABLE]

holds for all Jordan measurable $D\subset\mathbb{R}^{2}$.

Proof.

For a finite $F\subset C$ let $N(R,F)=\#\left((\mathcal{P}(m_{F}\mathcal{W}_{\mathcal{A}},\mathcal{L}_{F})\setminus\{0\})\cap RD\right)$. We know that $\lim\limits_{R\to\infty}\frac{N(R,F)}{\mathrm{vol}(RD)}=\frac{m_{F}^{2}(1+\sqrt{2})}{2N(\ell_{F})^{2}}$ so

[TABLE]

must be justified. In view of 4.5

[TABLE]

and we note that

[TABLE]

hence the sums of both sides of (2) are absolutely convergent.

Fix $\Delta>0$. We claim that there is only a finite number of non-empty $F\subset C$, $\#F<\infty$, such that $|N(\ell_{F})|<\Delta$. Given such $F$ let $\ell_{F}=\prod_{c\in F}c>1$. Also, since $|\overline{\pi}|>1$ for all $\pi\in\mathbb{P}$ we have $|\overline{\ell_{F}}|\geq|\overline{\lambda}|$. Hence, $|N(\ell_{F})|=\ell_{F}|\overline{\ell_{F}}|\leq\Delta$ implies $\ell_{F}\leq\frac{\Delta}{|\overline{\ell_{F}}|}\leq\lambda\Delta$ and $|\overline{\ell_{F}}|\leq\frac{\Delta}{\ell_{F}}\leq\Delta<\lambda\Delta$ so $(\ell_{F},\overline{\ell_{F}})\in\{(x,\overline{x})\mid x\in\mathcal{O}_{K}\}\cap\lambda[-\Delta,\Delta]^{2}$ which is a finite set, thus elements of $F$ can only contain prime factors that occur as factors in the components of elements in this finite set, giving only finitely many possibilities for $F$.

It follows that

[TABLE]

where the right hand side tends to [math] as $\Delta\to\infty$ since $\sum\limits_{F\subset C,\#F<\infty,|N(\ell_{F})|\geq\Delta}\frac{1}{N(\ell_{F})^{2}}$ is the tail of an absolutely convergent sum, hence (2) has been justified. ∎

From 4.6 it follows that $\theta(\widehat{\mathcal{B}})=\sum_{\begin{subarray}{c}F\subset C\\ \#F<\infty\end{subarray}}\frac{(-1)^{\#F}m_{F}^{2}(1+\sqrt{2})}{2N(\ell_{F})^{2}}$, and it will now be shown that the right hand side is equal to $1/\zeta_{K}(2)$. Define the function $\omega:\mathcal{O}_{K}\longrightarrow\mathbb{C}$, $\omega(x)=\#\{\pi\in\mathbb{P}\mid x/\pi\in\mathcal{O}_{K}\}$, so that $\omega(x)$ is the number of non-associated prime divisors of $x$. Given $I\triangleleft\mathcal{O}_{K}$, let $\omega(I)=\omega(x)$ for any generator $x$ of $I$ and define a Möbius function on the ideals of $\mathcal{O}_{K}$ by

[TABLE]

One verifies that $\mu(I_{1}I_{2})=\mu(I_{1})\mu(I_{2})$ for relatively prime ideals $I_{1},I_{2}$. The function $\zeta_{K}$ can be expressed as an Euler product for $s$ with $\mathrm{Re}(s)>1$ as

[TABLE]

and in analogy with the reciprocal formula for Riemann’s zeta function we have

[TABLE]

Theorem 4.7**.**

The density of visible points of $\mathcal{B}$ is given by

[TABLE]

Proof.

By 4.6 we have

[TABLE]

Splitting the sum into two depending on whether $\lambda\in F$ or not, and using that $m_{F}$ is $1$ unless $\lambda\in F$, in which case $m_{F}=|\overline{\lambda}|=\sqrt{2}-1$, we get

[TABLE]

last equality by (3). From [8, Theorem 4.2] one can calculate $\zeta_{K}(-1)=\frac{1}{12}$ and by the functional equation for Dedekind’s zeta function (cf. e.g. [8, p. 34]) one finds that $\zeta_{K}(2)=\frac{\pi^{4}}{48\sqrt{2}}$ which proves the claim. ∎

4.2 The density of visible points of $\mathcal{A}$

Observe that $\mathcal{A}^{\prime}=\sqrt{2}\mathcal{A}\subset\mathcal{B}$. It is now shown that $C$ is also an occluding set for $\mathcal{A}^{\prime}$.

Proposition 4.8**.**

For each $x\in\mathcal{A}^{\prime}\setminus\widehat{\mathcal{A}^{\prime}}$ there is $c\in C$ such that $x/c\in\mathcal{A}^{\prime}$.

Proof.

Since $\mathcal{A}^{\prime}\subset\mathcal{B}$ we have $\mathcal{A}^{\prime}\setminus\widehat{\mathcal{A}^{\prime}}\subset\mathcal{B}\setminus\widehat{\mathcal{B}}$ and so for each $x\in\mathcal{A}^{\prime}\setminus\widehat{\mathcal{A}^{\prime}}$ there exists $c\in C$ such that $x/c\in\mathcal{B}$. If $c\neq\sqrt{2}$ then $\sqrt{2}\mid\frac{x_{1}-x_{2}}{c}$ so $x/c\in\mathcal{A}^{\prime}$.

Take now $x\in\mathcal{A}^{\prime}\setminus\widehat{\mathcal{A}^{\prime}}$ such that for all $c\in C\setminus\{\sqrt{2}\}$ we have $x/c\notin\mathcal{B}$. Then $x/\sqrt{2}\in\mathcal{B}$, hence $\gcd(x_{1},x_{2})=\sqrt{2}^{n}$ for some $n\geq 1$. Since $x\in\mathcal{A}^{\prime}\setminus\widehat{\mathcal{A}^{\prime}}$ there is $c\in\mathbb{Q}(\sqrt{2})\cap\mathbb{R}_{>1}$ such that $x/c\in\mathcal{A}^{\prime}$. Writing $c=a/b$ with $\gcd(a,b)=1$, the only possible $\pi\in\mathbb{P}$ with $\pi\mid a$ is $\pi=\sqrt{2}$. If $\sqrt{2}\mid a$, then it follows that $x/\sqrt{2}\in\mathcal{A}^{\prime}$.

It remains to check the case where $a$ is a unit, i.e. $c=\frac{\lambda^{n}}{\prod_{p\in\mathbb{P}}\pi^{m(\pi)}}$ for some $m:\mathbb{P}\longrightarrow\mathbb{Z}_{\geq 0}$ with finite support. The facts that $c>1$ and $\pi>1$ for all $\pi\in\mathbb{P}$ imply $n>0$. We have $x/\lambda\notin\mathcal{B}$, hence $\overline{x}\notin|\overline{\lambda}|\mathcal{W}_{\mathcal{A}}$. Since $x/c\in\sqrt{2}\mathcal{A}$ it follows that $\overline{x}\in|\overline{c}|\mathcal{W}_{\mathcal{A}}$ and hence $|\overline{c}|>|\overline{\lambda}|$. However

[TABLE]

contradiction. ∎

Theorem 4.9**.**

We have $\theta(\widehat{\mathcal{A}^{\prime}})=\frac{1}{2\zeta_{K}(2)}$, hence $\theta(\widehat{\mathcal{A}})=\frac{1}{\zeta_{K}(2)}$.

Proof.

Propositions 4.1, 4.8 imply

[TABLE]

and it is straight-forward to verify that $\mathcal{A}_{*}^{\prime}\cap\bigcap_{c\in F}c\mathcal{A}_{*}^{\prime}=\mathcal{P}(m_{F}\mathcal{W}_{\mathcal{A}},\widetilde{\mathcal{L}}_{F})\setminus\{0\}$ with $\widetilde{\mathcal{L}}_{F}=\{(\ell_{F}x,\overline{\ell_{F}x})\mid x\in\mathcal{O}_{K}^{2},(x_{1}-x_{2})/\sqrt{2}\in\mathcal{O}_{K}\}$ a sublattice of $\mathcal{\mathcal{L}}_{F}$ of index $2$. Hence, by [4, Prop. 3.2], when letting $R\to\infty$ inside the sum (4) one obtains

[TABLE]

whence $\theta(\widehat{\sqrt{2}\mathcal{A}})=\frac{1}{2\zeta_{K}(2)}$ follows by 4.6 and 4.7, and the other result is immediate as $\sqrt{2}\mathcal{A}=\mathcal{A}^{\prime}$. ∎

Remark*.*

The data of Table 2 of [1] shows that $\#(\widehat{\mathcal{A}}\cap RD)/\#(\mathcal{A}\cap RD)\approx 0.577$ for a particular $D$ and fairly large $R$. This agrees with our results, since

[TABLE]