---
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 $K={\mathbb{Q}}(\sqrt{2})$.'
author:
- Gustav Hammarhjelm
bibliography:
- 'bibl.bib'
title: 'The density of visible points in the Ammann-Beenker point set'
---

Introduction {#secIntro}
============

A locally finite point set $\mathcal{P}\subset {\mathbb{R}}^d$ has an *asymptotic density* (or simply *density*) $\theta(\mathcal{P})$ if $$\lim_{R\to\infty}\frac{\#(\mathcal{P}\cap RD)}{{\mathrm{vol}}(RD)}=\theta(\mathcal{P})$$ 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 below) then it is known that $\theta(\mathcal{P})$ exists. In [@marklof2014visibility 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 [@nymann1972probability]; we sketch another proof in 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. [@baake2000diffraction 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 [@singppt1], but he has not published a proof of this result.

The density of the visible points of ${\mathbb{Z}}^d$ {#secDensZn}
=====================================================

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 $$\begin{aligned}
    \#(\widehat{{\mathbb{Z}}^d}\cap RD)&=\#\left(({\mathbb{Z}}^d_*\cap RD)\setminus \bigcup_{p\in{\mathbb{P}}}(p{\mathbb{Z}}^d_*\cap RD)\right)=\#\left(({\mathbb{Z}}^d_*\cap RD)\setminus \bigcup_{i=1}^n(p_i{\mathbb{Z}}^d_*\cap RD)\right)\\
    &=\#({\mathbb{Z}}^d_*\cap RD)+\sum_{k=1}^{m}(-1)^{k}\left(\sum_{1\leq i_1<\ldots <i_k\leq m}\#(p_{i_1}{\mathbb{Z}}^d_*\cap\cdots\cap p_{i_k}{\mathbb{Z}}^d_*\cap RD)\right).
    \end{aligned}$$ The last sum can be rewritten to $$\sum_{n\in{\mathbb{Z}}_{>0}}\mu(n)\cdot\#(n{\mathbb{Z}}^d_*\cap RD),$$ where $\mu$ is the Möbius function. Hence $$\frac{\#(\widehat{{\mathbb{Z}}^d}\cap RD)}{{\mathrm{vol}}(RD)}=\sum_{n\in{\mathbb{Z}}_{>0}}\frac{\mu(n)\cdot\#(n{\mathbb{Z}}^d_*\cap RD)}{{\mathrm{vol}}(RD)}=\sum_{n\in{\mathbb{Z}}_{>0}}\frac{\mu(n)}{n^d}\frac{\#({\mathbb{Z}}^d_*\cap {n^{-1}}RD)}{{\mathrm{vol}}({n^{-1}}RD)}.$$ 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 $$\theta(\widehat{{\mathbb{Z}}^d})=\lim_{R\to\infty}\frac{\#(\widehat{{\mathbb{Z}}^d}\cap RD)}{{\mathrm{vol}}(RD)}=1/\zeta(d).$$

Cut-and-project sets and the Ammann-Beenker point set {#secCPS}
=====================================================

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. [@baake2013aperiodic 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 [@marklof2014free Sec. 1.2]. For an introduction to cut-and-project sets, see e.g. [@baake2013aperiodic Ch. 7.2].

If ${\mathbb{R}}^n={\mathbb{R}}^d\times {\mathbb{R}}^m$, let $$\begin{aligned}
{2}
        \pi:& ~{\mathbb{R}}^n\longrightarrow {\mathbb{R}}^d  & \pi_{\mathrm{int}}: & ~ {\mathbb{R}}^n\longrightarrow {\mathbb{R}}^m \\
        & (x_1,\ldots,x_n)\longmapsto (x_1,\ldots,x_d)\hspace{1cm}& & (x_1,\ldots,x_n)\longmapsto (x_{d+1},\ldots, x_n)
    \end{aligned}$$ denote the natural projections.

\[defnCPS\] 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. [@marklof2014free 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 [@baake2013aperiodic Example 7.7] realised as $$\{x=x_1+x_2\zeta\mid x_1,x_2\in\mathcal{O}_K,x^\star\in W_8\},$$ 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 $$\mathcal{L}=\{(x,\overline{x})\mid x=(x_1,x_2)\in \mathcal{O}_K^2\}\subset {\mathbb{R}}^4$$ be the Minkowski embedding of $\mathcal{O}_K^2$ and let $$\widetilde{\mathcal{L}}=\{(x,\overline{x})\in \mathcal{L}\mid (x_1-x_2)/\sqrt{2}\in \mathcal{O}_K\}.$$ 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 .

The density of visible points of $\mathcal{A}$ {#secDensAB}
==============================================

All notation used in this section is defined in and taken from . 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 $\theta(\widehat{\mathcal{A}'})$ with $\mathcal{A}':=\sqrt{2}\mathcal{A}=\mathcal{P}(\mathcal{W}_\mathcal{A},\widetilde{\mathcal{L}})$ will give the value of $\theta(\widehat{\mathcal{A}})$. As a first step, in , 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 this result will be used to obtain $\theta(\widehat{\mathcal{A}})$.

The density of visible points of $\mathcal{B}$ {#subsecDensAB'}
----------------------------------------------

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

\[propInclExcl\] 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 $$\#(\widehat{\mathcal{P}}\cap RD)=\sum_{\substack{F\subset C\\\#F<\infty}}(-1)^{\#F}\#\left(\left(\mathcal{P}_*\cap \bigcap_{c\in F}c\mathcal{P}_*\right)\cap RD\right).$$

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 $$\begin{aligned}
        &\#(\widehat{\mathcal{P}}\cap RD)=\#\left((\mathcal{P}_*\cap RD)\setminus\bigcup_{c\in C}(\mathcal{P}_*\cap c\mathcal{P}_*\cap RD)\right)\\
        &=\#(\mathcal{P}_*\cap RD)-\#\left(\bigcup_{i=1}^n(\mathcal{P}_*\cap c_i\mathcal{P}_*\cap RD)\right),
        \end{aligned}$$ from which the result follows from the inclusion-exclusion counting formula for finite unions of finite sets.

A set $C$ as in 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 [@baake2014radial p. 477] is adapted to our situation.

\[propVisCond\] The visible points of $\mathcal{B}$ are given by $$\widehat{\mathcal{B}}=\{x=(x_1,x_2)\in \mathcal{B}\mid \gcd(x_1,x_2)=1,\lambda \overline{x}\notin \mathcal{W}_\mathcal{A}\}.$$

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 $\gcd(x_1,x_2)\neq1$. 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.

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

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

Fix $x\in\mathcal{B}\setminus\widehat{\mathcal{B}}$. As seen in the proof of 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 $\mathcal{L}_F=\{(\ell_Fx,\overline{\ell_Fx})\mid x\in \mathcal{O}_K^2\}$. 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 \[propInclExcl\], \[propOcclSet\] imply that $$\label{eqA'1}
    \#(\widehat{\mathcal{B}}\cap RD)=\sum_{\substack{F\subset C\\\#F<\infty}}(-1)^{\#F}\#\left((\mathcal{P}(m_F\mathcal{W}_\mathcal{A},\mathcal{L}_F){\setminus\{0\}})\cap RD\right).$$ Since $\ell_F\mathcal{O}_K^2\subset\pi_{\mathrm{int}}(\mathcal{L}_F)\subset{\mathbb{R}}^2$ is dense we have $$\theta(\mathcal{P}(m_F\mathcal{W}_\mathcal{A},\mathcal{L}_F){\setminus\{0\}})=\frac{{\mathrm{vol}}(m_F\mathcal{W}_\mathcal{A})}{{\mathrm{vol}}({\mathbb{R}}^4/\mathcal{L}_F)}$$ from [@marklof2014free Prop. 3.2]. Dividing by ${\mathrm{vol}}(RD)$, letting $R\to\infty$ and switching order of limit and summation (to be justified in below) we find that $$\theta(\widehat{\mathcal{B}})=\sum_{\substack{F\subset C\\\#F<\infty}}(-1)^{\#F}\frac{{\mathrm{vol}}(m_F\mathcal{W}_\mathcal{A})}{{\mathrm{vol}}({\mathbb{R}}^4/\mathcal{L}_F)}=\sum_{\substack{F\subset C\\\#F<\infty}}(-1)^{\#F}\frac{m_F^2(1+\sqrt{2})}{2N(\ell_F)^2},$$ 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 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”.

\[lemLatticeBoxBound\] 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 $\#(B\cap\mathcal{L})\leq L{\mathrm{vol}}(B)$.

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 $n=\prod_{i=1}^dn_i$ 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, $B\subset \bigcup_{m\in{\mathbb{N}}^d,0\leq m_i<n_i}(a+mc+[0,c]^d)=:B'$. 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 $\#(B\cap \mathcal{L})\leq \#(B'\cap\mathcal{L})\leq n D\leq \frac{2^dD}{c^d}{\mathrm{vol}}(B)$, so one can take $L=\frac{2^dD}{c^d}$.

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

\[lem-Estimate1\] 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 $\#F<\infty$, $$\#((\mathcal{P}(m_F\mathcal{W}_\mathcal{A},\mathcal{L}_F)\cap RD){\setminus\{0\}})\leq \frac{\widetilde{L}R^2}{N(\ell_F)^2}.$$

By definition $$\#((\mathcal{P}(m_F\mathcal{W}_\mathcal{A},\mathcal{L}_F){\setminus\{0\}})\cap RD)=\#(\{x\in \ell_F\mathcal{O}_K^2\mid\overline{x}\in m_F\mathcal{W}_\mathcal{A}\}{\setminus\{0\}})\cap RD).$$ Note that this number is independent of the choice of $\ell_F$. There is a bijection $$(\{x\in \ell_F\mathcal{O}_K^2\mid \overline{x}\in m_F\mathcal{W}_\mathcal{A}\}{\setminus\{0\}})\cap RD\longrightarrow(\{x\in\mathcal{O}_K^2\mid \overline{x}\in \tfrac{m_F\mathcal{W}_\mathcal{A}}{|\overline{\ell_F}|}\}{\setminus\{0\}})\cap \tfrac{RD}{\ell_F}$$ 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 $c'<c$ implies $(\mathcal{L}\cap(\lambda D\times c'\mathcal{W}_\mathcal{A})){\setminus\{0\}}=\emptyset$. This can be done, for otherwise $(\mathcal{L}\cap(\lambda D\times c'\mathcal{W}_\mathcal{A})){\setminus\{0\}}$ would be non-empty for each $c'>0$, 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 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_1^2m_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 $$\#\left(\left(\mathcal{L}\cap\tfrac{RD}{\ell_F}\times \left(\tfrac{\mathcal{W}}{|\overline{\ell_F}|}\right)\right){\setminus\{0\}}\right)\leq\frac{\widetilde{L}R^2}{N(\ell_F)^2}$$ with $\widetilde{L}:=16m_1^2m_2^2L$.

\[propInterchange\] The equality $$\lim_{R\to\infty}\sum_{\substack{F\subset C\\\#F<\infty}}\frac{(-1)^{\#F}\#\left((\mathcal{P}(m_F\mathcal{W}_\mathcal{A},\mathcal{L}_F){\setminus\{0\}})\cap RD\right)}{{\mathrm{vol}}(RD)}=\sum_{\substack{F\subset C\\\#F<\infty}}\frac{(-1)^{\#F}m_F^2(1+\sqrt{2})}{2N(\ell_F)^2}$$ holds for all Jordan measurable $D\subset {\mathbb{R}}^2$.

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 $$\label{eqnSwitchLimitsJustification}
    \lim\limits_{R\to\infty}\sum_{\substack{F\subset C\\\#F<\infty}}\frac{(-1)^{\#F}N(R,F)}{{\mathrm{vol}}(RD)}=\sum_{\substack{F\subset C\\\#F<\infty}}\lim\limits_{R\to\infty}\frac{(-1)^{\#F}N(R,F)}{{\mathrm{vol}}(RD)}$$ must be justified. In view of $$\sum_{\substack{F\subset C\\\#F<\infty}}\left|\frac{(-1)^{\#F}N(R,F)}{{\mathrm{vol}}(RD)}\right|\leq \frac{\widetilde{L}}{{\mathrm{vol}}(D)}\sum_{\substack{F\subset C\\\#F<\infty}}\frac{1}{N(\ell_F)^2}$$ and we note that $$\sum_{\substack{F\subset C\\\#F<\infty}}\frac{1}{N(\ell_F)^2}=\sum_{\substack{F\subset C\\\#F<\infty\\\lambda\in F}}\frac{1}{N(\ell_F)^2}+\sum_{\substack{F\subset C\\\#F<\infty\\\lambda\notin F}}\frac{1}{N(\ell_F)^2}\leq 2\sum_{I\triangleleft \mathcal{O}_K}\frac{1}{N(I)^2},$$ hence the sums of both sides of 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 $$\begin{aligned}
    &\left|\lim\limits_{R\to\infty}\sum_{\substack{F\subset C\\\#F<\infty}}\frac{(-1)^{\#F}N(R,F)}{{\mathrm{vol}}(RD)}-\sum_{\substack{F\subset C\\\#F<\infty}}\frac{m_F^2(1+\sqrt{2})}{2N(\ell_F)^2}\right|\\
    &\leq \left(\frac{\widetilde{L}}{{\mathrm{vol}}(D)}+\frac{1+\sqrt{2}}{2}\right)\sum_{\substack{F\subset C\\\#F<\infty\\|N(\ell_F)|\geq \Delta}}\frac{1}{N(\ell_F)^2}
    \end{aligned}$$ where the right hand side tends to $0$ 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 has been justified.

From it follows that $\theta(\widehat{\mathcal{B}})=\sum_{\substack{F\subset C\\\#F<\infty}}\frac{(-1)^{\#F}m_F^2(1+\sqrt{2})}{2N(\ell_F)^2}$, 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 $$\mu(I)=\begin{cases}0 & \text{ if }\exists \pi\in{\mathbb{P}}\text{ such that } I\subset \pi^2\mathcal{O}_K,\\
    (-1)^{\omega(I)} &\text{ otherwise.}
    \end{cases}$$ One verifies that $\mu(I_1I_2)=\mu(I_1)\mu(I_2)$ 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 $$\zeta_K(s)=\prod_{P\triangleleft \mathcal{O}_K,\,P\,\mathrm{prime}}\frac{1}{1-N(P)^{-s}}$$ and in analogy with the reciprocal formula for Riemann’s zeta function we have $$\label{eqnRecDedekind}
    \frac{1}{\zeta_K(s)}=\sum_{I\triangleleft \mathcal{O}_K}\frac{\mu(I)}{N(I)^s}.$$

\[thmAsDensA’\] The density of visible points of $\mathcal{B}$ is given by $$\theta(\mathcal{B})=\frac{1}{\zeta_K(2)}=\frac{48\sqrt{2}}{\pi^4}.$$

By we have $$\theta(\mathcal{B})=\sum_{\substack{F\subset C\\\#F<\infty}}\frac{(-1)^{\#F}m_F^2(1+\sqrt{2})}{2N(\ell_F)^2}.$$ 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 $$\begin{aligned}
    \theta(\mathcal{B})&=\sum_{\substack{F\subset C\\\#F<\infty\\\lambda\notin F}}\frac{(-1)^{\#F}(1+\sqrt{2})}{2N(\ell_F)^2}+\sum_{\substack{F\subset C\\\#F<\infty\\\lambda\in F}}\frac{(-1)^{\#F}|\overline{\lambda}|^2(1+\sqrt{2})}{2N(\ell_F)^2}\\
    &=\frac{(1-|\overline{\lambda}|^2)(1+\sqrt{2})}{2}\sum_{I\triangleleft\mathcal{O}_K}\frac{\mu(I)}{N(I)^2}=\frac{1}{\zeta_K(2)},
    \end{aligned}$$ last equality by . From [@washington1997introduction 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. [@washington1997introduction p. 34]) one finds that $\zeta_K(2)=\frac{\pi^4}{48\sqrt{2}}$ which proves the claim.

The density of visible points of $\mathcal{A}$ {#subsecDensAB}
----------------------------------------------

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

\[propOcclSet2\] For each $x\in\mathcal{A}'\setminus\widehat{\mathcal{A}'}$ there is $c\in C$ such that $x/c\in \mathcal{A}'$.

Since $\mathcal{A}'\subset \mathcal{B}$ we have $\mathcal{A}'\setminus\widehat{\mathcal{A}'}\subset \mathcal{B}\setminus\widehat{\mathcal{B}}$ and so for each $x\in \mathcal{A}'\setminus\widehat{\mathcal{A}'}$ 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 $x/c\in\mathcal{A}'$.

Take now $x\in \mathcal{A}'\setminus\widehat{\mathcal{A}'}$ 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 $n\geq1$. Since $x\in \mathcal{A}'\setminus\widehat{\mathcal{A}'}$ there is $c\in {\mathbb{Q}}(\sqrt{2})\cap{\mathbb{R}}_{>1}$ such that $x/c\in \mathcal{A}'$. 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 $x/\sqrt{2}\in\mathcal{A}'$.

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 $$|\overline{c}|=\frac{|\overline{\lambda}|^n}{\prod_{\pi\in {\mathbb{P}}}|\overline{\pi}|^{k(\pi)}}\leq |\overline{\lambda}|^n\leq |\overline{\lambda}|,$$ contradiction.

We have $\theta(\widehat{\mathcal{A}'})=\frac{1}{2\zeta_K(2)}$, hence $\theta(\widehat{\mathcal{A}})=\frac{1}{\zeta_K(2)}$.

Propositions \[propInclExcl\], \[propOcclSet2\] imply $$\label{eqnB1}
    \frac{\#(\widehat{\mathcal{A}'}\cap RD)}{{\mathrm{vol}}(RD)}=\sum_{\substack{F\subset C\\ \#F<\infty}}\frac{(-1)^{\#F}\#\left(\left(\mathcal{A}_*'\cap \bigcap_{c\in F}c\mathcal{A}_*'\right)\cap RD\right)}{{\mathrm{vol}}(RD)}$$ and it is straight-forward to verify that $\mathcal{A}_*'\cap \bigcap_{c\in F}c\mathcal{A}_*'=\mathcal{P}(m_F\mathcal{W}_\mathcal{A},\widetilde{\mathcal{L}}_F){\setminus\{0\}}$ with $\widetilde{\mathcal{L}}_F=\{(\ell_Fx,\overline{\ell_Fx})\mid x\in \mathcal{O}_K^2,(x_1-x_2)/\sqrt{2}\in\mathcal{O}_K\}$ a sublattice of $\mathcal{\mathcal{L}}_F$ of index $2$. Hence, by [@marklof2014free Prop. 3.2], when letting $R\to\infty$ inside the sum one obtains $$\sum_{\substack{F\subset C\\ \#F<\infty}}\frac{(-1)^{\#F}{\mathrm{vol}}(m_F\mathcal{W}_\mathcal{A})}{16N(\ell_F)},$$ whence $\theta(\widehat{\sqrt{2}\mathcal{A}})=\frac{1}{2\zeta_K(2)}$ follows by and , and the other result is immediate as $\sqrt{2}\mathcal{A}=\mathcal{A}'$.

The data of Table 2 of [@baake2014radial] 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 $$\kappa_\mathcal{A}=\lim_{R\to\infty}\frac{\#(\widehat{\mathcal{A}}\cap RD)}{\#(\mathcal{A}\cap RD)}=\frac{\theta(\widehat{\mathcal{A}})}{\theta(\mathcal{A})}=\frac{\tfrac{1}{\zeta_K(2)}}{\tfrac{2{\mathrm{vol}}(\mathcal{W}_\mathcal{A})}{16}}=\frac{2(\sqrt{2}-1)}{\zeta_K(2)}=0.5773\ldots$$
