Finite crystallization and Wulff shape emergence for ionic compounds in the square lattice

TL;DR

This paper studies two-dimensional ionic crystal formation on a square lattice, proving ground state configurations, Wulff shape emergence, and effects of net charge on crystallization, revealing shape sensitivity and critical charge thresholds.

Contribution

It introduces a detailed analysis of energy minimizers and shape emergence in ionic systems with net charge constraints on the square lattice, including sharp scaling laws and shape transitions.

Findings

Ground states are connected subsets with alternating atomic types.

Emergence of a square Wulff shape for large particle numbers.

Crystallization fails beyond a critical net charge, with a diamond-like shape at the threshold.

Abstract

We present two-dimensional crystallization results in the square lattice for finite particle systems consisting of two different atomic types. We identify energy minimizers of configurational energies featuring two-body short-ranged particle interactions which favor some reference distance between different atomic types and contain repulsive contributions for atoms of the same type. We first prove that ground states are connected subsets of the square lattice with alternating arrangement of the two atomic types in the crystal lattice, and address the emergence of a square macroscopic Wulff shape for an increasing number of particles. We then analyze the signed difference of the number of the two atomic types, the so-called net charge, for which we prove the sharp scaling ${\rm O}(n^{1/4})$ in terms of the particle number $n$. Afterwards, we investigate the model under prescribed net charge. We provide a characterization for the minimal energy and identify a critical net charge beyond which crystallization in the square lattice fails. Finally, for this specific net charge we prove a crystallization result and identify a diamond-like Wulff-shape of energy minimizers which illustrates the sensitivity of the macroscopic geometry on the net charge.

Tables

Table 1. Table 1. The function ϕ ​ ( n ) italic-ϕ 𝑛 \phi(n) for 1 ≤ n ≤ 32 1 𝑛 32 1\leq n\leq 32 .

$n$	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16
$\varphi \left(n\right)$	1	2	3	4	3	4	5	4	5	6	5	6	5	6	7	6
$n$	17	18	19	20	21	22	23	24	25	26	27	28	29	30	31	32
$\varphi \left(n\right)$	7	6	7	8	7	8	7	8	7	8	9	8	9	8	9	8

Equations

$$C_n=\{(x_1,q_1),\dots ,(x_n,q_n)\}\subset {\left({\mathbb{R}}^2\times \{-1,1\}\right)}^n,$$

$$\mathcal{E}(C_n)=\frac{1}{2}\sum _{\underset{q_i=q_j}{i\ne j}}{V}_{\mathrm{r}}(|x_i-x_j|)+\frac{1}{2}\sum _{\underset{q_i\ne q_j}{i\ne j}}{V}_{\mathrm{a}}(|x_i-x_j|),$$

$$[\mathrm{i}]$$

$$[\mathrm{ii}]$$

$$[\mathrm{iii}]$$

$$[\mathrm{iv}]$$

$$[\mathrm{v}]$$

$$[\mathrm{vi}]$$

$$[\mathrm{vii}]$$

$$\mathcal{N}(x_i)=\left(X_n\setminus \{x_i\}\right)\cap \{x\in {\mathbb{R}}^2:|x-x_i|\le r_0\},$$

$$b=\#\{\{x_i,x_j\}:x_j\in \mathcal{N}(x_i)\}.$$

$$\eta (C_n)=\sum _{j\ge 4}(j-4)f_j,$$

$$\mathcal{N}(x_i)\cap \left\{x_j\in X_n:q_j=q_i\right\}=\varnothing \text{ for all }i\in \{1,\dots ,n\}$$

$$\mathcal{Q}(C_n):=\sum _{i=1}^n{q}_i.$$

$$X_n^{+}:=\{x_i\in X_n:q_i=+1\},X_n^{-}:=\{x_i\in X_n:q_i=-1\}$$

$${\mathcal{E}}_{\min }^n(q_{\mathrm{net}}):=\min \left\{\mathcal{E}(C_n):C_n\subset ({\mathbb{R}}^2\times \{-1,1\}{)}^n,\mathcal{Q}(C_n)=q_{\mathrm{net}}\right\}.$$

$$\mathcal{E}(C_n)\le \mathcal{E}(C_n^{\prime })$$

$$S_n:=\left\{x\in {\mathbb{Z}}^2:x_1,x_2\ge 0,|x{|}_{\infty }\le \lfloor \sqrt{n}-1\rfloor \right\}$$

$$D_n:=\left\{x\in {\mathbb{Z}}^2:|x{|}_1\le \lfloor (-1+\sqrt{2n-1})/2\rfloor \right\}$$

$$\beta (n):=2n-2\sqrt{n}.$$

$$\mathcal{E}(X_n)=-b=-\lfloor \beta (n)\rfloor .$$

$$\min_{a\in\mathbb{R}^{2}}\#\big{(}X_{n}\triangle(a+S_{n})\big{)}\leq cn^{3/4},$$

$$\underset{j\to +\infty }{\mathrm{liminf}}{n}_j^{-1/4}|\mathcal{Q}(C_{n_j})|>0.$$

$$q_{\mathrm{sat}}^n=\min \left\{q_{\mathrm{net}}:q_{\mathrm{net}}\in {\mathcal{Q}}_{\mathrm{net}}(n)\cap [0,n],{\mathcal{E}}_{\min }^n(q_{\mathrm{net}})=-2n+2q_{\mathrm{net}}\right\}.$$

$$\varphi (n):=\{\begin{array}{ll}2\lfloor -\frac{1}{2}+\frac{1}{2}\sqrt{2n-5}\rfloor +3& \text{if }n\text{ odd},n\ge 3,\\ 2\lfloor \frac{1}{2}\sqrt{2n-4}\rfloor +2& \text{if }n\text{ even},n\ge 2.\end{array}$$

$$-2n+2|q_{\mathrm{net}}|\le {\mathcal{E}}_{\min }^n(q_{\mathrm{net}})\le -2n+2|q_{\mathrm{net}}|+4(q_{\mathrm{sat}}^n-|q_{\mathrm{net}}|{)}^{+}.$$

$$\min_{a\in\mathbb{R}^{2}}\#\big{(}X_{n}\triangle(a+D_{n})\big{)}\leq cn^{3/4}.$$

$$\mathcal{E}(C_n)=-\lfloor \beta (n)\rfloor =-\lfloor 2n-2\sqrt{n}\rfloor .$$

$$\sum _{j=1}^{2l+1}(k-l-2j)=2kl-l^2+1$$

$$\mathcal{E}(C_n)$$

$$=-2k^2+2\sqrt{k^2}-1=-\lfloor \beta (k^2)\rfloor -1=-\lfloor \beta (n)\rfloor ,$$

$$\underset{l\to +\infty }{\mathrm{liminf}}{n}_l^{-1/4}|\mathcal{Q}(C_{n_l})|>0.$$

$$\#X_n^{-}=\frac{1}{2}(n-\mathcal{Q}(C_n)).$$

Full text

1. Introduction

The question whether the ground states of particle systems for certain configurational energies arrange themselves into crystalline order is referred to as the crystallization problem [6]. Due to its paramount theoretical and applicative relevance, this mathematical issue has attracted a great deal of attention over the last decades and has led to various mathematically rigorous crystallization results for ensembles consisting of one single atomic type. For particle systems with different types of atoms, however, rigorous results appear to be scarce. The goal of this paper is to contribute to these fundamental mathematical questions by presenting a study of crystallization in the two-dimensional square lattice for finite particle systems consisting of two different atomic types.

Microscopically, crystallization can be seen as the result of interatomic interactions governed by quantum mechanics. At zero or very low temperature, atomic interactions are expected to depend only on the geometry of the atomic arrangement. In this case, configurations can be identified with the respective positions of identical atoms $\{x_{1},\ldots,x_{n}\}$. Then the crystallization problem consists in considering the minimization of a configurational energy $\mathcal{E}(\{x_{1},\ldots,x_{n}\})$ and in proving or disproving the periodicity of ground-state configurations of $\mathcal{E}$.

Various such crystallization results for different choices of the energy $\mathcal{E}$ comprising classical interaction potentials have been derived over the last decades. Here, among the vast body of literature, we only mention some of the relevant works, and refer the reader to the recent review [6] for a general perspective. Concerning results in one dimension we mention [19, 20, 29, 34], and we refer to [11, 21, 24, 26, 28, 35] for studies in dimension two for a finite number of identical particles. In particular, we highlight the work by Mainini, Piovano, and Stefanelli [24] where a comprehensive analysis of crystallization in the square lattice is performed. Besides crystallization, the authors also provide a fine characterization of ground-state geometries by proving the emergence of a square macroscopic Wulff shape for growing particle numbers. (We also refer to [2, 9, 10, 25] for similar studies for different lattices.) Under less restrictive assumptions on the potentials, various results have been obtained in the thermodynamic limit [3, 12, 13, 33], i.e., as the number of particles tends to infinity. In particular, we highlight the recent work by Bétermin, De Luca, and Petrache [3] where, for a wide class of pair-interaction potentials, crystallization in the square lattice has been obtained. The crystallization problem in three dimension seems to be very difficult and only few rigorous results [15, 16, 22, 32] are available.

For particle systems involving different types of atoms, simulations are abundant, but rigorous results seem to be limited to [4, 7, 17, 30]. To the best of our knowledge, the recent work [17] by the authors represents a first rigorous mathematical crystallization result for two-dimensional dimers, i.e., molecular compounds consisting of two atomic types. This result is inspired by problems for systems of identical particles and follows the classical molecular-mechanical frame of configurational energy minimization: configurations of $n$ particles are identified with their respective positions $\{x_{1},\ldots,x_{n}\}\subset\mathbb{R}^{2}$ and additionally with their types $\{q_{1},\ldots,q_{n}\}\in\{-1,1\}^{n}$. The goal is to determine minimizers of a corresponding interaction energy $\mathcal{E}(\{(x_{1},q_{1}),\ldots,(x_{n},q_{n})\})$ and to characterize their geometry.

More precisely, the energy $\mathcal{E}=\mathcal{E}_{\mathrm{a}}+\mathcal{E}_{\mathrm{r}}$ is assumed to consist of two short-ranged two-body interaction potentials $\mathcal{E}_{\mathrm{a}}$ and $\mathcal{E}_{\mathrm{r}}$, where $\mathcal{E}_{\mathrm{a}}$ represents the interactions between atoms of different type and $\mathcal{E}_{\mathrm{r}}$ encodes the energy contributions of atoms having the same type. The potential $\mathcal{E}_{\mathrm{a}}$ is attractive-repulsive and favors atoms sitting at some specific reference distance, whereas $\mathcal{E}_{\mathrm{r}}$ is a pure repulsive term. The main result of [17] is that, for specific quantitative assumptions on the potentials, global minimizers of the configurational energy are essentially connected subsets of the regular hexagonal lattice with the two atomic types alternating. The first main goal of the present article is to show that weaker short-ranged repulsive terms favor crystallization in the square lattice, which illustrates the sensitivity of the ground-state geometry on the precise assumptions on the potentials.

Let us mention that the choice of the interaction potentials is motivated by the modelling of ions in ionic compounds. In fact, one can interpret the two interaction energies as a (very simplified) model for the bonding of ions capturing the essential features to describe the formation of ionic crystalline solids. The shape of $\mathcal{E}_{\mathrm{a}}$ is due to electrostatic forces between ions of opposite charge as well as a small additional force due to van der Waals interactions, resulting in a long ranged attraction. The short ranged repulsive force can be explained by the Pauli exclusion principle when a pair of ions comes close enough such that their outer electron shells overlap. The balance between these forces leads to a potential energy landscape with a minimum energy when the nuclei are separated by a specific equilibrium distance. On the other hand, the repulsive energy $\mathcal{E}_{\mathrm{r}}$ between ions of same charge models a simplified Coulomb repulsive force. We will often refer to the atomic types $\{q_{1},\ldots,q_{n}\}\in\{-1,1\}^{n}$ as positive or negative charges. In this context, it is interesting to consider the net charge $\sum_{i=1}^{n}q_{i}$, i.e., the (signed) difference of the number of the two atomic types.

In [17], the free net charge problem has been addressed, i.e., the net charge is not preassigned but a fundamental part of the minimization problem. It has been shown that $\sum_{i=1}^{n}q_{i}$ is ‘almost neutral’, with a deviation from zero of order ${\rm O}(n^{1/4})$, where the scaling in terms of the particle number $n$ is sharp. Similar problems of charge distributions and net charge on Bravais lattices have been recently studied in [5]. Besides extending the results of the free net charge problem to the present setting, the second main goal of this article is to complement the aforementioned analysis by studying the prescribed net charge problem, i.e., the net charge is a given constraint in the minimization problem. This corresponds to a model of a closed physical system containing a certain number of positively and negatively charged atoms.

We now give an overview of the main results of this article representing a comprehensive analysis of finite crystallization for dimers in the two-dimensional square lattice, see Section 2.3 for details.

• (1)

We first address the free net charge problem. Under suitable assumptions on the attractive and the repulsive potentials, we characterize the ground-state energy and geometry of finite particle configurations of ions in dimension two. In particular, we prove that each global minimizer of the configurational energy is essentially a connected subset of the square lattice with alternating arrangement of the two atomic types in the crystal lattice. This characterization holds except for possibly one atom at the boundary of the configuration. Similar to [24], we identify the emergence of a square macroscopic Wulff shape for growing particle numbers.

• (2)

We provide a fine asymptotic characterization for the net charge as the number of atoms $n$ grows. More specifically, we show that the fluctuation of the net charge around zero can be at most of order ${\rm O}(n^{1/4})$, i.e., $|\sum_{i=1}^{n}q_{i}|\leq cn^{1/4}$ for some constant $c>0$ independent of $n$. By providing an explicit construction we further prove that this scaling is sharp.

• (3)

We consider the prescribed net charge problem and provide a characterization of the minimal energy in dependence of the net charge. We identify a critical net charge, the so-called saturation net charge $q_{\rm sat}$, which corresponds to the case that all atoms of the less frequent atomic type are bonded to exactly four other atoms. By way of example, see Fig. 5, we show that $q_{\rm sat}$ is critical in the sense that beyond this specific net charge we cannot expect that atoms arrange themselves in a regular lattice.

• (4)

We investigate the geometry of energy minimzers for prescribed net charge $q_{\rm sat}$ and show that also in this case optimal configurations are (essentially) subsets of the square lattice. Our interest in this specific case is twofold: (i) this problem together with the free net charge problem (which by (2) essentially corresponds to the problem with prescribed net charge zero) constitute the extreme cases for which crystallization results can be shown. (ii) for $q_{\rm sat}$, we identify a macroscopic diamond-like Wulff-shape, which illustrates the dependence of the macroscopic geometry on the prescribed net charge.

Our general proof strategy for the free net charge problem follows the induction method on bond-graph layers developed in [21, 24, 26, 28]. Let us mention that our problem is particularly related to [24], where crystallization for identical particles in the square lattice has been investigated under three-body angular potentials. Actually, the ground-state energy in the present context coincides with the one obtained there. A crucial point in the induction step is the derivation of a boundary energy estimate. The presence of repulsive instead of angular terms calls for a novel definition of the boundary energy, complementing the approach in [24] from a technical point of view, see Remark 5.2.

To prove the emergence of a square Wulff shape and the sharp scaling ${\rm O}(n^{1/4})$ for the net charge, we use the fact that in ground states the two atomic types are alternately arranged in the square lattice. This allows us to apply the $n^{3/4}$-law in [24] which states that ground states differ from a square shape by at most ${\rm O}(n^{3/4})$ atoms, or equivalently, by at most ${\rm O}(n^{1/4})$ in Hausdorff distance.

Concerning the proof of the results for the prescribed net charge problem, in principle we follow the same induction method on bond-graph layers as for the free net charge problem. The actual realization, however, is much more delicate. In fact, as a preliminary step for a crystallization result in the square lattice, a fine characterization of the saturation net charge $q_{\mathrm{sat}}$ is needed. Then, it turns out that the geometry of optimal configurations under prescribed net charge $q_{\mathrm{sat}}$ is quite flexible: by way of explicit constructions (see, e.g., Fig. 6 and Fig. 9), we observe that optimal configurations are possibly not connected or regions at the boundary are not contained in the square lattice. Fine geometric arguments are necessary to ensure that these degenerate parts consist of a controlled number of atoms only. For the identification of the global diamond-like Wulff-shape, we identify the charge of configurations with the $\infty$-perimeter of specific interpolations, and then, following an idea inspired by [8], we apply the quantitative isoperimetric inequality to obtain a bound on the deviations from a diamond.

The article is organized as follows. In Section 2 we introduce the precise mathematical setting and present the main results about the free and prescribed net charge problem. In Section 3 we construct explicitly some configurations in order to provide sharp upper bounds for the ground-state energy and the net charge. Moreover, we establish an upper bound for the saturation net charge $q_{\rm sat}$. These explicit constructions already give the right intuition for the microscopic and macroscopic geometry of optimal configurations. In Section 4 we discuss elementary geometric properties of energy minimizers. In Section 5 we give the lower bound for the ground-state energy and provide a fine characterization of the geometry of ground states. Here, we also prove the $n^{1/4}$-law for the net charge of ground states. Finally, Section 6 is devoted to the prescribed net charge problem. We first provide a lower bound for $q_{\rm sat}$ matching the upper bound derived in Section 3. Afterwards, we characterize the geometry of $q_{\rm sat}$-optimal configurations.

2. Setting and main results

In this section we first introduce our model and give some basic definitions. Afterwards, we present our main results.

2.1. Configurations and interaction energy

We consider particle systems in two dimensions consisting of two different atomic types. We model their interaction by classical potentials in the frame of Molecular Mechanics [1, 18, 23]. Let $n\in\mathbb{N}$. We indicate the configuration of $n$ particles by

[TABLE]

identified with the respective atomic positions $X_{n}=(x_{1},\ldots,x_{n})\in\mathbb{R}^{2n}$ together with their types $Q_{n}=(q_{1},\ldots,q_{n})\in\{-1,1\}^{n}$. By referring to a model for ionic dimers, we will often call $Q_{n}$ the charges of the atoms, $q=1$ representing cations and $q=-1$ representing anions. Our choice of the empirical potentials (see below) is indeed inspired by ions in ionic compounds, which are primarily held together by their electrostatic forces between the net negative charge of the anions and the net positive charge of the cations [27].

By following the setting in [17], we define the energy $\mathcal{E}:(\mathbb{R}^{2}\times\{-1,1\})^{n}$ $\to\overline{\mathbb{R}}$ of a given configuration $\{(x_{1},q_{1}),\ldots,(x_{n},q_{n})\}\in(\mathbb{R}^{2}\times\{-1,1\})^{n}$ by

[TABLE]

where $V_{\mathrm{r}},V_{\mathrm{a}}:[0,+\infty)\to\overline{\mathbb{R}}$ are a repulsive potential and an attractive-repulsive potential, respectively. The factor $1/2$ accounts for the fact that every contribution is counted twice in the sum. The two potentials are pictured schematically in Fig. 1. Let $r_{0}\in[1,(2\sin(\frac{\pi}{7}))^{-1})$ and note that $r_{0}<\sqrt{2}$. The attractive-repulsive potential $V_{\mathrm{a}}$ satisfies

[TABLE]

The distance $r=1$ represents the (unique) equilibrium distance of two atoms with opposite charge. The choice of $V_{\rm a}$ reflects a balance between a long-ranged Coulomb attraction and the short-ranged Pauli repulsion acting when a pair of ions comes too close to each other. Assumption [iii] restricts the interaction range and guarantees that the bond graph is planar, see Section 2.2.

The repulsive potential $V_{\mathrm{r}}$ satisfies

[TABLE]

The natural assumption [v] is satisfied for example for repulsive Coulomb interactions. We emphasize that some quantitative requirements of the form [vi] and [vii] are necessary to obtain a crystallization result in the square lattice. Other quantitative assumptions on the repulsive potential will favor, e.g., that the atoms arrange themselves in a hexagonal lattice, see [17].

Finally, we require the following slope conditions

• [viii]

\displaystyle V^{\prime}_{\mathrm{r},-}(\sqrt{2})<-\frac{16}{\sqrt{2}\pi},\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}\ \ \ \ \ \ \ \frac{1}{r-1}(V_{\mathrm{a}}(r)-V_{\mathrm{a}}(1))>-\frac{1}{2}V^{\prime}_{\mathrm{r},+}(1)\ \ \ \text{for all }r\in(1,r_{0}],

where the functions $V^{\prime}_{\mathrm{r},-}$, $V^{\prime}_{\mathrm{r},+}$ denote the left and right derivative, respectively. (They exist due to the convexity of $V_{\rm r}$.) These conditions are reminiscent of the soft-interaction assumption by Radin [28] and the slope condition for an angular potential by Mainini, Piovano, and Stefanelli [24]. Assumption [viii] is only needed in Lemma 5.3 where the energy contribution of atoms at the boundary of the configuration is estimated. For a more detailed discussion on the assumptions on $V_{\rm a}$ and $V_{\rm r}$ we refer the reader to [17, Section 2.1].

2.2. Basic notions

In this subsection we collect some basic notions. Consider a configuration $C_{n}\in\left(\mathbb{R}^{2}\times\{-1,1\}\right)^{n}$ with finite energy consisting of the positions $X_{n}=(x_{1},\ldots,x_{n})\in\mathbb{R}^{2n}$ and the charges $Q_{n}=(q_{1},\ldots,q_{n})\in\{-1,1\}^{n}$.

Neighborhood, bonds, angles: For each $x_{i}\in\mathbb{R}^{2}$, $i\in\{1,\ldots,n\}$, we define the neighborhood by

[TABLE]

where $r_{0}$ is defined in [iii]. If $x_{j}\in\mathcal{N}(x_{i})$, we say that $x_{i}$ and $x_{j}$ are bonded. We will say that $x_{i}$ is $k$-bonded if $\#\mathcal{N}(x_{i})=k$. Given $x_{j},x_{k}\in\mathcal{N}(x_{i})$, we define the bond-angle between $x_{j},x_{i},x_{k}$ as the angle between the two vectors $x_{k}-x_{i}$ and $x_{j}-x_{i}$. (We choose anti-clockwise orientation for definiteness.) In general, we say that it is an angle at $x_{i}$.

Bond graph: The set of atomic positions $X_{n}\subset\mathbb{R}^{2n}$ together with the set of bonds $\{\{x_{i},x_{j}\}:x_{j}\in\mathcal{N}(x_{i})\}$ forms a graph which we call the bond graph. Since for configurations with finite energy there holds $\mathrm{dist}(x_{i},X_{n}\setminus\{x_{i}\})\geq 1$ and $x_{j}\in\mathcal{N}(x_{i})$ only if $|x_{i}-x_{j}|\leq r_{0}<\sqrt{2}$, their bond graph is a planar. Indeed, given a quadrangle with all sides and one diagonal in $[1,r_{0}]$, the second diagonal is at least $\sqrt{2}>r_{0}$. If no ambiguity arises, the number of bonds in the bond graph will be denoted by $b$, i.e.,

[TABLE]

We say a configuration is connected if each two atoms are joinable through a simple path in the bond graph. In a similar fashion, we speak of connected components of a configuration. Any simple cycle of the bond graph is a polygon.

Acyclic bonds: A bond is called acyclic if it is not contained in any simple cycle of the bond graph. Among acyclic bonds we distinguish between flags and bridges. We say that an acyclic bond is a bridge if it is contained in some simple path connecting two vertices which are included in two distinct cycles. All other acyclic bonds are called flags, see Fig. 2.

Defects: By elementary polygons we denote polygons which do not contain any non-acyclic bonds in its interior region. An elementary polygon in the bond graph which is not a square is called a defect. We introduce the excess of edges $\eta(C_{n})$ by

[TABLE]

where $f_{j}$ denotes the number of elementary polygons with $j$ vertices in the bond graph. The excess of edges is a tool to quantify the number of defects in the bond graph. Note that the summation in (2.3) runs over $j\geq 4$. This is due to the fact that we use this definition only for configurations whose bond graph contains only $k$-gons with $k\geq 4$, cf. Lemma 4.2 below. If it is clear from the context, we omit the dependence on $C_{n}$ and write $\eta=\eta(C_{n})$.

In the following, we frequently refer to $C_{n}$ instead of $X_{n}$ when speaking about its bond graph, acyclic bonds, or connectedness properties.

Charges: We say that a configuration satisfying

[TABLE]

has alternating charge distribution. A configuration is called repulsion-free if $|x_{i}-x_{j}|\geq\sqrt{2}$ for all $x_{i}\neq x_{j}$ with $q_{i}=q_{j}$. The net charge of a configuration is defined as the (signed) difference of the number of the two atomic types, i.e.,

[TABLE]

We note that all possible net charges are given by $\mathcal{Q}_{\rm net}(n):=(2\mathbb{Z}+n\,{\rm mod}2)\cap[-n,n]$. If $\mathcal{Q}(C_{n})>0$, we say that the atoms with charge $+1$ represent the majority phase and the atoms with charge $-1$ the minority phase. We use a corresponding denomination if $\mathcal{Q}(C_{n})<0$. We denote by

[TABLE]

the positively and negatively charged phase, respectively.

Ground states: For $q_{\rm net}\in\mathcal{Q}_{\rm net}(n)$ we define

[TABLE]

A configuration $C_{n}$ with $\mathcal{Q}(C_{n})=q_{\rm net}$ and $\mathcal{E}(C_{n})=\mathcal{E}^{n}_{\rm min}(q_{\rm net})$ is called a $q_{\rm net}$-optimal configuration. We will often simply call $C_{n}$ an optimal configuration. Moreover, we call $C_{n}$ a ground state if $\mathcal{E}(C_{n})=\min_{q_{\rm net}\in\mathcal{Q}_{\rm net}(n)}\mathcal{E}^{n}_{\rm min}(q_{\rm net})$. In other words, $C_{n}$ is a ground state if and only if

[TABLE]

for all $C^{\prime}_{n}\subset(\mathbb{R}^{2}\times\{-1,1\})^{n}$.

Subsets of the square lattice: We denote by $\mathbb{Z}^{2}\subset\mathbb{R}^{2}$ the square lattice. We define special subsets of the square lattice representing the Wulff-shapes of optimal configurations. For $n\in\mathbb{N}$ we let

[TABLE]

the square of sidelength $\lfloor\sqrt{n}-1\rfloor$, where by $\lfloor t\rfloor$ we denote the integer part of $t\in\mathbb{R}$. By

[TABLE]

we denote the diamond of radius $\lfloor(-1+\sqrt{2n-1})/2\rfloor$. $S_{n}$ and $D_{n}$ represent the largest squares and diamonds, respectively, whose number of atoms is less or equal to $n$. In Fig. 3 some subsets of the square lattice are depicted.

Observe that these configurations can be chosen to have alternating charge distribution. By assumptions [ii], [iii], and [vii], the energy of such configurations $C_{n}$ satisfies $\mathcal{E}(C_{n})=-b$ since all atoms of the same charge have at least distance $\sqrt{2}$ and all atoms of $X_{n}$ are bonded only to atoms of opposite charge with bonds of unit length.

Equilibrated atoms: We say that an atom $x\in X_{n}$ is equilibrated if all bond-angles at $x$ lie in $\{\frac{\pi}{2},\pi,\frac{3\pi}{2}\}$. By $\mathcal{A}(X_{n})$ we denote the atoms which are not equilibrated. Note that, if $\mathcal{A}(X_{n})=\emptyset$ and $X_{n}$ is connected, then $X_{n}$ is a subset of the square lattice $\mathbb{Z}^{2}$.

2.3. Main results

In this subsection we state our main results. We will first address the free net charge problem, and characterize the energy and geometry of ground states. In particular, we will prove a rigorous planar crystallization result in the spirit of [17, 21, 24, 26, 28] and the emergence of a square macroscopic Wulff shape (cf. [2, 9, 10, 24, 31]). Then we will characterize the net charge of ground-state configurations.

Afterwards, we change the perspective and study $q_{\rm net}$-optimal configurations under prescribed net charge $q_{\rm net}$. We give an estimate for the minimal energy $\mathcal{E}^{n}_{\rm min}(q_{\rm net})$ and identify a specific net charge, the saturation net charge $q^{n}_{\rm sat}$, which corresponds to the smallest net charge where all atoms of the minority phase are $4$-bonded. Finally, we prove the emergence of a diamond-like Wulff-shape for $q^{n}_{\rm sat}$-optimal configurations which reflects the sensitivity of the Wulff-shape on the prescribed net charge.

Free net charge: Our first result characterizes the energy of ground states. For $n\in\mathbb{N}$, we introduce the function

[TABLE]

Theorem 2.1** (Ground-state energy).**

Let $n\in\mathbb{N}$. Ground states are connected and have alternating charge distribution. They do not contain any bridges. There holds

[TABLE]

Remark 2.2**.**

In view of assumptions ${\rm[ii]}$ and [vii], we have that $\mathcal{E}(C_{n})\geq-b$ with equality if and only if the configuration is repulsion-free and all bonds have unit length. In particular, Theorem 2.1 implies that ground states satisfy both properties.

The next result states that ground states are essentially subsets of the square lattice and that a square Wulff-shape emerges as $n\to\infty$. Without further notice, all following statements regarding the geometry of ground states hold up to isometry. Recall the definition of a square of sidelength $\lfloor\sqrt{n}-1\rfloor$ in (2.8).

Theorem 2.3** (Characterization of ground states).**

Let $n\in\mathbb{N}$ and let $C_{n}$ be a ground state.

• (a)

(Crystallization) Except for possibly one atom, $C_{n}$ is a subset of the square lattice.

• (b)

(Wulff-shape) For a universal constant $c>0$ independent of $n$, there holds

[TABLE]

where $\triangle$ denotes the symmetric difference of sets.

We point out that, if a ground state contains a flag, it is possibly not a subset of the square lattice, see Fig. 4. Our next result addresses the net charge (2.5) of ground-state configurations.

Theorem 2.4** (Net charge of ground states).**

The following properties for the net charge hold:

• (a)

(Net charge control) There exists a universal constant $c>0$ such that for all $n\in\mathbb{N}$ and all ground states $C_{n}$ the net charge satisfies $|\mathcal{Q}(C_{n})|\leq cn^{1/4}$.

• (b)

(Sharpness of the $n^{1/4}$-scaling) There exists an increasing sequence of integers $(n_{j})_{j}$ and ground states $(C_{n_{j}})_{j}$ such that

[TABLE]

The sharp scaling $n^{1/4}$ for the upper bound of the net charge has also been identified in a related model where ground states are subsets of the hexagonal lattice, see [17, Theorem 2.5]. The three theorems are proved in Section 5. Explicit constructions for the upper bound of the ground-state energy and Theorem 2.4(b) are given in Subsection 3.1 and Subsection 3.2, respectively.

Prescribed net charge: We now change the perspective and suppose that the net charge $q_{\rm net}$ is prescribed. We first characterize the energy of $q_{\rm net}$-optimal configurations $\mathcal{E}^{n}_{\rm min}(q_{\rm net})$, see (2.7). To this end, we introduce the saturation net charge

[TABLE]

The definition corresponds to the smallest net charge for which all atoms of the minority phase are $4$-bonded. Configurations with the latter property will be called saturated in the following. The saturation net charge can be characterized as follows.

Proposition 2.5** (Characterization of $q_{\mathrm{sat}}^{n}$).**

There holds $q_{\mathrm{sat}}^{n}=\phi(n)$, where $\phi(0)=0$, $\phi(1)=1$, and

[TABLE]

We refer to (6.8) for an equivalent characterization. The representation (2.13) shows the scaling $q_{\rm sat}^{n}\sim n^{1/2}$, which corresponds to the scaling of the number of boundary atoms of optimal configurations.

By $h^{+}:=\max(h,0)$ we denote the positive part of a function $h$. The energy of $q_{\rm net}$-optimal configurations can be characterized as follows.

Theorem 2.6** (Energy of $q_{\rm net}$-optimal configurations).**

For all $n\in\mathbb{N}$ and all $q_{\rm net}\in Q_{\rm net}(n)$ there holds

[TABLE]

We point out that the upper bound in (2.14) is consistent with Theorem 2.1, i.e., with $\mathcal{E}^{n}_{\rm min}(q_{\rm net})\geq-\lfloor\beta(n)\rfloor$ for all $q_{\rm net}^{n}$. To see this, it suffices to note that $2q_{\rm sat}^{n}\geq-\lfloor-2\sqrt{n}\rfloor$. The result states that for $|q_{\rm net}|\geq q^{n}_{\rm sat}$, the minimal energy is exactly $-2n+2|q_{\rm net}|$. This corresponds to the case that optimal configurations are saturated, i.e., each atom of the minority phase is $4$-bonded. In this sense, $q^{n}_{\rm sat}$ can be understood as a critical net charge.

It turns out that $q^{n}_{\rm sat}$ is not only a critical net charge in terms of the minimal energy $\mathcal{E}^{n}_{\rm min}(q_{\rm net})$, but also from a geometrical viewpoint: $q^{n}_{\rm sat}$ is critical in the sense that beyond $q^{n}_{\rm sat}$ no crystallization can be expected, cf. Remark 2.7. Note, however, that $q_{\mathrm{sat}}^{n}$-optimal configurations crystallize, cf. Theorem 2.8.

Remark 2.7** (Fragmentation for $q_{\rm net}>q_{\mathrm{sat}}^{n}$).**

For $q_{\rm net}>q_{\mathrm{sat}}^{n}$, $q_{\rm net}$-optimal configurations may be completely degenerate, e.g., may consist of many connected components, see Fig. 5. Their characterizing property is that atoms of the minority phase are $4$-bonded with bond angles $\frac{\pi}{2}$, cf. Lemma 4.3(a)(i),(iv).

Recall the definition of diamonds in (2.9).

Theorem 2.8** (Crystallization and Wulff-shape for $q^{n}_{\rm sat}$-optimal configurations).**

Let $n\in\mathbb{N}$ and let $C_{n}$ be a $q^{n}_{\rm sat}$-optimal configuration.

• (a)

(Crystallization) There exists a universal constant $n_{0}\in\mathbb{N}$ independent of $n$ such that $C_{n}$ is a subset of the square lattice except for at most $n_{0}$ atoms.

• (b)

(Wulff-shape) For a universal constant $c>0$ independent of $n$, there holds

[TABLE]

Complete crystallization cannot be expected for certain values of $n$, as shown for example in Fig. 6. Comparing this result to the geometry of ground states identified in Theorem 2.3, we observe that the geometry of the Wulff-shape and therefore the global geometry of optimal configurations is very sensitive to the prescribed net charge. Proposition 2.5, Theorem 2.6, and Theorem 2.8 are proved in Section 6. The upper bound for $q_{\rm sat}^{n}$ is constructed explicitly in Subsection 3.3.

3. Constructions of special subsets of the square lattice

This section is devoted to explicit constructions of sub-configurations of the square lattice with alternating charge distribution. In Subsection 3.1 we exhibit candidates for the ground-state energy which will already give the upper bound in (2.11). In Subsection 3.2 we construct ground-state configurations with net charge of order $n^{1/4}$, which will establish Theorem 2.4(b). Finally, in Subsection 3.3 we define configurations with net charge $\phi(n)$, see (2.13), for which the atoms of the minority phase are $4$-bonded. This yields the upper bound for $q_{\mathrm{sat}}^{n}$ in Proposition 2.5. We defer the lower bound on the ground-state energy and the upper bound on the net charge to Section 5. The lower bound for $q_{\mathrm{sat}}^{n}$ is addressed in Subsection 6.3.

3.1. Upper bound on the ground-state energy

This subsection is devoted to an explicit construction of configurations $C_{n}$ which maximize the number of bonds and that are subsets of the square lattice. These configurations provide a reference energy value for every $n$, namely $\mathcal{E}(C_{n})=-\lfloor\beta(n)\rfloor$, see (2.10).

By the special geometry of the square lattice, it is quite natural to give an interpretation of the two terms appearing in $\beta$. The leading order term of the energy is given by $-2n$ which corresponds to the bulk part of the energy. Its value is due to the fact that every interior atom is bonded to four other atoms of opposite charge. Furthermore, the repulsive term in the energy is zero for such configurations since the distance of two atoms with the same charge is bigger than or equal to $\sqrt{2}$. The additional lower order correction term is due to the fact that atoms on the boundary of the ground-state configuration do not have four neighbors.

The construction follows [24] and is illustrated in Fig. 7. If $n=k^{2}$, $k\in\mathbb{N}$, we arrange the atoms on the lattice points of the square $S_{n}$ (cf. (2.8)). Then for $n=k^{2}+m+1$ with $0\leq m\leq 2k-1$ we proceed as follows: for $0\leq m\leq k-1$ we recursively construct $X_{k^{2}+m+1}$ by adding the point with coordinates $(m,k)$ to $X_{k^{2}+m}$. For $k\leq m\leq 2k-1$ we construct $X_{k^{2}+m+1}$ by adding the point with coordinates $(k,m-k)$ to $X_{k^{2}+m}$. Since the bond graphs only contain cycles of even length, we can choose corresponding charges such that the resulting configurations $C_{n}$ have alternating charge distribution. One can check that $\mathcal{Q}(C_{n})\in\{-1,0,1\}$ for all $n\in\mathbb{N}$.

Proposition 3.1**.**

For all $n\in\mathbb{N}$, let $C_{n}$ be the configuration introduced above. Then there holds

[TABLE]

Proof.

The proof follows as in [24, Proposition 4.3], additionally observing that all atoms in the bond graph are bonded to particles of opposite charge only and that for such configurations, up to neglecting the charge, our energy coincides with the one considered in [24]. ∎

3.2. Squares with additional trapezoid

Recall that the above configurations have net charge in $\{-1,0,1\}$. Starting with a square and attaching a trapezoid in a suitable way, we can also construct configurations with energy $-\lfloor\beta(n)\rfloor$ having net charge of order $n^{1/4}$. The construction is inspired by related ideas [9, 17] used in connection to the derivation of the so-called $n^{3/4}$-law.

We choose $k=5l^{2}+7l+3$, $l\in 2\mathbb{N}$, and $n=k^{2}+1$. We construct a configuration $C_{n}$ as follows. We start from the square $S_{(k-l)^{2}}$. Since $k-l$ is odd, the net charge of $S_{(k-l)^{2}}$ can be chosen as $+1$. We add a new atom to the bond graph in such a way that it gets bonded to the second up-most among the rightmost atoms. Then we add descendingly atoms along the right side of the square in such a way that they are bonded to the atom in the previous step and one atom of the square. In this way, we add $(k-l-2)$ atoms. Since $k-l$ is odd, we have added $\lceil(k-l)/2\rceil-1$ atoms of charge $+1$ and $\lfloor(k-l)/2\rfloor-1$ atoms of charge $-1$. Next, we add a new column starting from the second up-most among the atoms of the previous column. In this way, we add $(k-l-4)$ atoms. We repeat this procedure until we have added $2l+1$ columns of atoms. This corresponds to

[TABLE]

added atoms. Note that in each column the number of added atoms of charge $+1$ exceeds the number of added atoms of charge $-1$ exactly by one, and that the resulting configuration consists of $n=k^{2}+1$ atoms. The construction is sketched in Fig. 8.

We now determine the energy of the configurations. Recall the definition of $\beta$ in (2.10). We observe that in a column where we add $m$ atoms we add exactly $2m-1$ bonds to the bond graph. Consequently, in view of Proposition 3.1, the energy of $C_{n}$ is given by

[TABLE]

where in the last step we used that $n=k^{2}+1$. We now determine the net charge of the configuration. Recall that the configuration $S_{(k-l)^{2}}$ has net charge equal to $+1$. As explained above, in each column the number of added atoms of charge $+1$ exceeds the number of added atoms of charge $-1$ by exactly one, i.e., $\mathcal{Q}(C_{n})=2l+2$.

We are now in the position to give the proof of Theorem 2.4(b). To this end, consider the sequence of integers $n_{l}=(5l^{2}+7l+3)^{2}+1$, $l\in 2\mathbb{N}$, and the configurations $C_{n_{l}}$ constructed above. Note that $4l\geq n_{l}^{1/4}$. Thus, we obtain $\mathcal{Q}(C_{n_{l}})=2l+2\geq\frac{1}{2}n_{l}^{1/4}.$ This yields

[TABLE]

The statement follows once we know that the ground state energy equals exactly $-\lfloor\beta(n)\rfloor$ for all $n\in\mathbb{N}$. This will be proven in Subsection 5.2.

3.3. Upper bound on $q_{\mathrm{sat}}^{n}$

We construct sub-configurations of the square lattice satisfying $\mathcal{E}(C_{n})=-2n+2|\mathcal{Q}(C_{n})|$ and $|\mathcal{Q}(C_{n})|=\phi(n)$. This shows $q_{\mathrm{sat}}^{n}\leq\phi(n)$, see (2.12)–(2.13). The main idea of the construction is to place atoms of the minority phase only at sites whose $\mathbb{Z}^{2}$ neighborhood, that is the set of points on $\mathbb{Z}^{2}$ with distance $1$ to the point, is already occupied by four points of the majority phase. This leads to configurations whose global geometry is reminiscent of a diamond. For an illustration of the construction we refer to Fig. 9.

For $1\leq n\leq 4$ we arrange $n$ atoms of charge $+1$ such that their mutual distance is bigger or equal to $\sqrt{2}$. Next, we provide the construction for $n=1+2k^{2}+2k$, $k\in\mathbb{N}$. In this case, we define $X_{n}=D_{n}$, see (2.9), and $q_{i}=(-1)^{i_{1}+i_{2}+k}$, $i=(i_{1},i_{2})\in D_{n}$. Now for $n=1+2k^{2}+2k+m$, with $1\leq m\leq 4k+3$, we recursively construct $X_{1+2k^{2}+2k+m+1}$ by adding one atom to the configuration $X_{1+2k^{2}+2k+m}$. For $m=1$ we add an atom of positive charge at position $(-(k+1),1)$. Let $2\leq m\leq 2k+2$. For $m$ even, we add an atom of positive charge at position $(-(k+1)+m/2,1+m/2)$. For $m$ odd, we add an atom of negative charge at position $(-(k+1)-\lfloor-m/2\rfloor,-\lfloor-m/2\rfloor)$. Now let $2k+3\leq m\leq 4k+3$. For $m$ odd, we add an atom of positive charge at position $(-\lfloor-m/2\rfloor-(k+1),2k+3-\lfloor m/2\rfloor)$. For $m$ even, we add an atom of negative charge at position $(m/2-(k+2),2k+3-m/2)$.

Proposition 3.2** (Upper bound for $q_{\mathrm{sat}}^{n}$).**

Let $n\in\mathbb{N}$. Then $q_{\mathrm{sat}}^{n}\leq\phi(n)$.

Proof.

We check that for every $n\in\mathbb{N}$ the constructed configuration $C_{n}$ satisfies $\mathcal{E}(C_{n})=-2n+2\mathcal{Q}(C_{n})$ and $\mathcal{Q}(C_{n})=\phi(n)$. This implies $q_{\mathrm{sat}}^{n}\leq\phi(n)$.

Let us first confirm that, for all $n\in\mathbb{N}$, $C_{n}$ satisfies $\mathcal{E}(C_{n})=-2n+2\mathcal{Q}(C_{n})$. Recall definition (2.6). Since $\#X_{n}^{+}+\#X_{n}^{-}=n$, $\#X_{n}^{+}-\#X_{n}^{-}=\mathcal{Q}(C_{n})$, and $X_{n}^{-}\leq X_{n}^{+}$ we get

[TABLE]

The construction shows that each atom with negative charge has exactly four neighbors of positive charge since we place atoms of negative charge only on lattice sites whose $\mathbb{Z}^{2}$-neighbourhood is already occupied by four atoms of positive charge. Moreover, we observe that the distance of atoms with the same charge is at least $\sqrt{2}$. Thus, by [ii], [vii], and (3.1) we get $\mathcal{E}(C_{n})=-4\#X_{n}^{-}=-2n+2\mathcal{Q}(C_{n})$.

It remains to prove that $\mathcal{Q}(C_{n})=\phi(n)$. To this end, it is convenient to use a different representation of $\phi$ given by

[TABLE]

see (6.8) in Subsection 6.3 below. Let $n=1+2k^{2}+2k=1+2(k-1)^{2}+2(k-1)+4(k-1)+4$. In this case, $C_{n}$ is a union of $2k+1$ rows for each of which the number of atoms of charge $+1$ exceeds the number of atoms of charge $-1$ by exactly one. We therefore have $\mathcal{Q}(C_{n})=1+2k=\phi(n)$.

For $n=1+2k^{2}+2k+m$ with $m\in\{1,2\}$, we add $m$ atoms of charge $+1$ to the configuration $C_{1+2k^{2}+2k}$. Consequently, we get $\mathcal{Q}(C_{n})=\mathcal{Q}(C_{1+2k^{2}+2k})+m=1+2k+m=\phi(n)$.

For $3\leq m\leq 2k+2$, we add alternatingly first an atom of charge $-1$ and then an atom of charge $+1$. We therefore obtain $\mathcal{Q}(C_{n})=\mathcal{Q}(C_{1+2k^{2}+2k})+1=1+2k+1=\phi(n)$ if $m$ is odd and $\mathcal{Q}(C_{n})=\mathcal{Q}(C_{1+2k^{2}+2k})+2=1+2k+2=\phi(n)$ if $m$ is even. For $2k+3\leq m\leq 4k+3$, we add alternatingly first an atom of charge $+1$ and then an atom of charge $-1$. We therefore obtain $\mathcal{Q}(C_{n})=\mathcal{Q}(C_{1+2k^{2}+4k+2})+1=1+2k+3=\phi(n)$ if $m$ is odd and \mathcal{Q}(C_{n})=\mathcal{Q}(C_{1+2k^{2}+4k+2})=2k+3\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}=\phi(n) if $m$ is even. ∎

4. Elementary properties of optimal configurations

In this section we prove elementary geometric properties of optimal configurations. For their definition, we refer to the paragraph below (2.7).

Lemma 4.1**.**

Let $C_{n}$ be an optimal configuration. Then $C_{n}$ has alternating charge distribution, all bond angles satisfy

[TABLE]

and

[TABLE]

Proof.

In this proof, we will use the following convention: we say that we relocate $(x,q)\in C_{n}$, and write $C_{n}-\{(x,q)\}$, by considering the configuration $(C_{n}\cup(x+\tau,q))\setminus(x,q)$, where $\tau\in\mathbb{R}^{2}$ is chosen such that

[TABLE]

Since $C_{n}$ is a $q_{\rm net}$-optimal configuration, there holds $\mathcal{E}(C_{n})<+\infty$. Thus, by [i] and [iv]

[TABLE]

For brevity, we define $\mathcal{N}_{\rm rep}(x_{i})=\mathcal{N}(x_{i})\cap\left\{x_{j}\in X_{n}:q_{j}=q_{i}\right\}$ for all $i\in\{1,\ldots,n\}$. We give the proof of the statement in two steps.

Step 1: $\#\mathcal{N}_{\rm rep}(x_{i})=0\text{ for all }i\in\{1,\ldots,n\}$. First, (4.3) and $r_{0}<(2\sin(\frac{\pi}{7}))^{-1}$ entail by an elementary argument that

[TABLE]

Indeed, if $\#\mathcal{N}(x_{i})\geq 7$, two neighbors of $x_{i}$ would necessarily have distance smaller than $1$. Now assume by contradiction that $\#\mathcal{N}_{\rm rep}(x_{i})\geq 1$. Note that every bond between atoms of different charge contributes at least $-1$ to the energy by $[\mathrm{ii}]$. This along with $V_{\rm r}\geq 0$, see [iv], and the fact that the energy per neighbor of same charge exceeds $6$ (see $[\mathrm{iv}],[\mathrm{v}]$, and $[\mathrm{vi}]$) allows us to relocate $(x_{i},q_{i})$: by (4.4) we obtain

[TABLE]

This contradicts the fact that $C_{n}$ is a $q_{\rm net}$-optimal configuration. Thus, Step 1 is proved.

*Step 2: $\tfrac{2\pi}{5}\leq\theta\leq\tfrac{8\pi}{5}.$ * Assume by contradiction that there exists a bond angle such that $\theta<\tfrac{2\pi}{5}$. Denote by $x_{1},x_{0},x_{2}$, $x_{1},x_{2}\in\mathcal{N}(x_{0})$, the three atoms forming the bond angle $\theta$. By Step 1 we get that $q_{0}=-q_{1}=-q_{2}$. Since $\sin(x)$ is increasing for $x\in[0,\frac{\pi}{2}]$, we have $|x_{1}-x_{2}|\leq 2r_{0}\sin\left(\theta/2\right)\leq 2r_{0}\sin\left(\pi/5\right)$. By using $[\mathrm{ii}]$, $[\mathrm{v}]$, [vi], $V_{\rm r}\geq 0$ (cf. $[\mathrm{iv}]$), and (4.4) we get

[TABLE]

If a bond angle $\theta$ satisfies $\theta>\tfrac{8\pi}{5}$, consider the opposite angle formed by the three points, i.e., $2\pi-\theta<\tfrac{2\pi}{5}$. In both cases, this contradicts the fact that $C_{n}$ is an optimal configuration and concludes the proof of Step 2.

Step 3: $\#\mathcal{N}(x_{i})\leq 4$ for all $i\in\{1,\ldots,n\}$. Assume by contradiction that there exists $i\in\{1,\ldots,n\}$ such that $\#\mathcal{N}(x_{i})\geq 5$. By Step 1 we may suppose that there exist $\{x_{0},\ldots,x_{4}\}\subset\mathcal{N}(x_{i})$ with $q_{j}=-q_{i}$ $j=0,\ldots,4$. We let $\theta_{j}\in[0,2\pi)$ be the angle between $x_{j},x_{i},x_{j+1}$. (Here and in the following the indices have to be understood modulo $4$.) We can choose $j_{0}\in\{0,\ldots,4\}$ such that

[TABLE]

Now (4.1) and (4.2) follow from Step 2 and 3, respectively. The property of alternating charge (see (2.4)) follows from Step 1. This concludes the proof of Lemma 4.1. ∎

Recall the definition of non-equilibrated atoms $\mathcal{A}(X_{n})$ at the end of Subsection 2.2.

Lemma 4.2** (Bond-angles and polygons).**

Let $C_{n}$ be a configuration.

1. (a)

If $C_{n}$ is an optimal configuration, then the bond graph consists only of polygons of even length. For each polygon $P\subset X_{n}$ satisfying $\mathcal{A}(X_{n})\cap P\neq\emptyset$ there holds

[TABLE] 2. (b)

If $C_{n}$ is such that $\mathcal{E}(C_{n})=-b$, then all bonds have unit length, the configuration is repulsion-free, and all bond angles $\theta$ satisfy

[TABLE]

In particular, all squares in the bond graph are regular.

Proof.

Proof of (a). Let $C_{n}$ be an optimal configuration. By Lemma 4.1 the configuration has alternating charge distribution. Hence, the bond graph consists only of elementary polygons of even length. Let $P=\{x_{1},\ldots,x_{k}\}\subset X_{n}$ be a polygon, and denote the interior angles by $\{\theta_{1},\ldots,\theta_{k}\}$. Assume by contradiction that $\#(\mathcal{A}(X_{n})\cap P)=1$. Without loss of generality $x_{1}\in\mathcal{A}(X_{n})$, i.e., $\theta_{1}\notin\frac{\pi}{2}\mathbb{N}$. We have $\theta_{1}=\pi(k-2)-\sum_{i=2}^{k}\theta_{i}$. Now by assumption $\theta_{i}\in\frac{\pi}{2}\mathbb{N}$ for all $i\in\{2,\ldots,k\}$ and so the right hand side is an integer multiple of $\frac{\pi}{2}$. This contradicts the fact that $x_{1}\in\mathcal{A}(X_{n})$.

Proof of (b): Given a configuration $C_{n}$ such that $\mathcal{E}(C_{n})=-b$, all bonds are necessarily of unit length and the configuration is repulsion-free, see Remark 2.2. We show that all bond angles $\theta$ satisfy (4.5). In fact, suppose that $x_{1}$, $x_{0}$, $x_{2}$ form the angle $\theta$. Observe that $C_{n}$ has alternating charge distribution since it is repulsion-free. As $x_{1}$ and $x_{2}$ are neighbors of $x_{0}$, we thus have $q_{1}=q_{2}$. Then we get $|x_{1}-x_{0}|=|x_{2}-x_{0}|=1$ as well as $|x_{1}-x_{2}|\geq\sqrt{2}$ by [vii]. A simple geometric argument yields $\pi/2\leq\theta\leq 3\pi/2$. By (4.5) and the fact that the four interior angles of a square sum up to $2\pi$ we deduce that all squares are regular. ∎

We now investigate the relation of net charge and energy. Without restriction we consider configurations with non-negative net charge in order to simplify notation.

Lemma 4.3** (Net charge controls energy).**

Let $C_{n}$ be a configuration with $\mathcal{Q}(C_{n})\geq 0$.

(a) There holds

[TABLE]

with equality only if

• (i)

All atoms with charge $-1$ are $4$-bonded, and all bonds are of unit length,

• (ii)

The configuration is repulsion-free,

• (iii)

$\mathcal{Q}(C_{n})\geq q_{\mathrm{sat}}^{n}$,

• (iv)

All $4$-bonded atoms have only bond angles $\frac{\pi}{2}$,

• (v)

If $x_{i}$ satisfies $\#\mathcal{N}(x_{i})=3$, then $x_{i}\notin\mathcal{A}(X_{n})$.

• (vi)

For all connected components $C_{m}\subset C_{n}$ there holds $\mathcal{Q}(C_{m})\geq q_{\mathrm{sat}}^{m}$.

(b) Conversely, if properties (i) and (ii) are satisfied, then equality holds in (4.6).

Proof.

We first show (4.6). To this end, it is not restrictive to assume that $C_{n}$ is an optimal configuration. Recall (2.6). Since $\#X_{n}^{+}+\#X_{n}^{-}=n$ and $\#X_{n}^{+}-\#X_{n}^{-}=\mathcal{Q}(C_{n})$, we get $\min\{\#X_{n}^{+},\#X_{n}^{-}\}=(n-|\mathcal{Q}(C_{n})|)/2$. As $C_{n}$ is an optimal configuration, (4.2) yields $\mathcal{N}(x_{i})\leq 4$ for all $i=1,\ldots,n$. We therefore obtain by [ii] and $V_{\rm r}\geq 0$

[TABLE]

This shows (4.6) since by assumption $|\mathcal{Q}(C_{n})|=\mathcal{Q}(C_{n})$. In a similar fashion, to see (b), it suffices to note that (i) and (ii) imply that all inequalities in (4.7) are actually equalities.

Now we assume that equality holds in (4.6). (In particular, this implies that $C_{n}$ is an optimal configuration and satisfies $\mathcal{E}(C_{n})=-b$.) We confirm (i)–(vi).

Proof of (i), (ii): First, by Lemma 4.2(b) we get that all bonds have unit length and that the configuration is repulsion-free. Suppose now that there exists an atom with charge $-1$ such that $\mathcal{N}(x_{i})<4$. This implies strict inequality in the second inequality in (4.7). This contradicts the equality in (4.6).

Proof of (iii): The inequality follows directly from definition (2.12).

Proof of (iv), (v): Now, (iv) follows from (4.5) and the fact that the bond angles at each atom sum up to $2\pi$. To see (v), let $x_{0}$ be a $3$-bonded atom. By (i) it necessarily has charge $+1$ and it is only bonded to atoms with charge $-1$. These atoms are $4$-bonded by (i). Denote the bond angles at $x_{0}$ by $\theta_{1},\theta_{2},\theta_{3}$. Without loss of generality, suppose that $\pi/2\leq\theta_{1}\leq 2\pi/3\leq 3\pi/4$, see (4.5). Assume by contradiction that $\theta_{1}>\pi/2$. Denote by $x_{1},x_{2}$ the two $4$-bonded atoms whose bonds enclose $\theta_{1}$. Denote by $z_{1},z_{2}$ the two atoms $z_{1}\in\mathcal{N}(x_{1})\setminus\{x_{0}\}$, $z_{2}\in\mathcal{N}(x_{2})\setminus\{x_{0}\}$ that have minimal distance to each other. Observe that $z_{1}$ and $z_{2}$ have charge $+1$. We proceed to show that $z_{1}\neq z_{2}$ and $|z_{1}-z_{2}|<\sqrt{2}$ which contradicts (ii). By (iv) and the fact that all bonds are of unit length we get $|z_{1}-x_{0}|=|z_{2}-x_{0}|=\sqrt{2}$. Moreover, the angle enclosed by $z_{1}$, $x_{0}$, $z_{2}$ is equal to $\theta_{1}-2\pi/4$, where $0<\theta_{1}-2\pi/4\leq\pi/4$. Thus, we obtain

[TABLE]

This yields $z_{1}\neq z_{2}$ and contradicts (ii). Therefore, we have shown $\theta_{1}=\pi/2$. One of the remaining two bond angles, say $\theta_{2}$, then also satisfies $\theta_{2}\leq 3\pi/4$. By the same argument we have that $\theta_{2}=\pi/2$. Now $\theta_{3}=\pi$ and the claim is proved.

Proof of (vi): Assume that there exist two connected components $C_{m}$ and $C_{n-m}$ with no bonds between them. We have that

[TABLE]

By (4.7) applied to $C_{m}$ and $C_{n-m}$ there holds

[TABLE]

and therefore

[TABLE]

We now have equality everywhere in (4.10) and therefore necessarily also in (4.9). Moreover, in view of (4.8), equality also implies that there holds $\mathcal{Q}(C_{m})\geq 0$ and $\mathcal{Q}(C_{n-m})\geq 0$. By equality in (4.9) and (iii) (applied on $C_{m}$ and $C_{n-m}$) we get $\mathcal{Q}(C_{m})\geq q_{\mathrm{sat}}^{m}$ and $\mathcal{Q}(C_{n-m})\geq q_{\mathrm{sat}}^{n-m}$. ∎

5. Characterization of ground states

This section is devoted to the proofs of Theorem 2.1, Theorem 2.3, and Theorem 2.4.

5.1. Boundary energy

This subsection is devoted to the concept of boundary energy and a corresponding estimate which will be instrumental for the characterization of ground states and their energy in Subsections 5.2 and 5.3. It is convenient to first introduce an auxiliary energy, sorted by the contributions of single atoms. We then define the so-called boundary energy in terms of this auxiliary energy.

Reduced energy: Let $C_{n}$ be a configuration and let $x_{i}\in X_{n}$. We set

[TABLE]

where $\mathcal{N}(x_{i})$ is defined in (2.2). We define the reduced energy $\mathcal{R}:(\mathbb{R}^{2}\times\{-1,1\})^{n}\to\overline{\mathbb{R}}$ by

[TABLE]

Lemma 5.1** (Relation of $\mathcal{E}$ and $\mathcal{R}$).**

If $C_{n}$ has alternating charge distribution and satisfies $\#\mathcal{N}(x_{i})\leq 4$ for all $i=1,\ldots,n$, then

[TABLE]

with equality if $C_{n}$ is repulsion-free.

Proof.

Since $C_{n}$ has alternating charge distribution, the contributions of $V_{\rm a}$ in (2.1) and (5.2) coincide. In the case that $C_{n}$ is repulsion-free, the contributions of $V_{\rm r}$ vanish and thus there indeed holds $\mathcal{E}(C_{n})=\mathcal{R}(C_{n})$. In the general case, we observe that in (2.1) each pair $\{x_{i},x_{j}\}$ with $q_{i}=q_{j}$ contributes $V_{\rm r}(|x_{i}-x_{j}|)$ to the energy. In (5.2), however, $\{x_{i},x_{j}\}$ contributes at most $V_{\rm r}(|x_{i}-x_{j}|)$ since the pair appears at most eight times in the sum. This is due to double counting and the fact that each atom can be bonded to at most four other atoms. ∎

By Lemma 4.1, ground states $C_{n}$ satisfy the assumptions of the lemma. Later we will also see that ground states are repulsion-free which will imply $\mathcal{E}(C_{n})=\mathcal{R}(C_{n})$.

Boundary atoms, boundary energy: Within the bond graph, we say that an atom is a boundary atom if it is not contained in the interior region of any simple cycle. Otherwise, we call it bulk atom. We denote the union of the boundary atoms by $\partial X_{n}$ and let $d=\#\partial X_{n}$. A boundary bond is a bond containing a boundary atom. All other bonds are called bulk bonds. Given $C_{n}$, we define its bulk, denoted by $C_{n}^{\mathrm{bulk}}$, as the sub-configuration obtained by dropping all boundary atoms (and the corresponding charges). Similarly, the particle positions are indicated by $X_{n}^{\mathrm{bulk}}$. With the above definition, we have that the bulk is an $(n-d)$-atom configuration. There are two contributions to the energy of $C_{n}$, namely $\mathcal{R}^{\mathrm{bnd}}$ and $\mathcal{R}^{\mathrm{bulk}}$, defined by

[TABLE]

Remark 5.2** (Boundary energy).**

Since $C_{n}^{\mathrm{bulk}}$ does not contain boundary bonds of $C_{n}$, each boundary bond $\{x_{i},x_{j}\}$ with $q_{i}\neq q_{j}$ contributes $V_{\rm a}(|x_{i}-x_{j}|)$ to $\mathcal{R}^{\mathrm{bnd}}(C_{n})$, and each boundary bond $\{x_{i},x_{j}\}$ with $q_{i}=q_{j}$ contributes at least $\frac{1}{4}V_{\rm r}(|x_{i}-x_{j}|)$ to $\mathcal{R}^{\mathrm{bnd}}(C_{n})$. Note that $\mathcal{R}^{\mathrm{bnd}}(C_{n})$ contains also pair interactions of certain bulk bonds, namely if the corresponding bulk atoms are neighbors of the same boundary atom, see Fig. 10. We point out that, when the boundary energy in (5.3) is defined with $\mathcal{E}$ in place of $\mathcal{R}$ (see, e.g., [17, 24, 26]), such pair interactions do not contribute to the boundary energy. Our definition, slightly different in comparison to [17], is necessary from a technical point of view in order to derive the ‘correct’ boundary energy estimate (5.4) and to obtain (5.6)-(5.7) as necessary conditions for equality in (5.4).

Maximal polygon: We introduce an additional notion in the case that $C_{n}$ is connected and does not contain acyclic bonds. In this case, the bond graph is delimited by a simple cycle which we call the maximal polygon. We denote the atoms of the maximal polygon by $\{x_{1},\ldots,x_{d}\}$ and the interior angle at $x_{i}\in\partial X_{n}$ by $\theta_{i}$. Moreover, for $2\leq k\leq 4$, we indicate by

[TABLE]

the set of $k$-bonded boundary atoms. For ground states $C_{n}$ there holds $\#I_{2}+\#I_{3}+\#I_{4}=d$ by (4.2).

We now provide an estimate for the boundary energy $\mathcal{R}^{\rm bnd}$. Its proof is inspired by [24, Lemma 3.1]. The precise estimates, however, deviate significantly from [24] due to the presence of the repulsive potential $V_{\rm r}$ instead of an angular potential. We defer the proof to Appendix A.

Lemma 5.3**.**

Let $n\geq 4$ and let $C_{n}$ be a connected ground state with no acyclic bonds. Then

[TABLE]

with equality only if the following conditions are satisfied:

[TABLE]

Remark 5.4**.**

Observe by (5.6) that equality in (5.4) implies that $2d-4$ bonds contribute to the boundary energy. Thus, equality in (5.4) together with [ii] and [vii] imply that for all boundary atoms $x_{i}$ one has $\min\{|x_{i}-x_{j}|:\ j\in\{1,\ldots,n\},j\neq i,\ q_{j}=q_{i}\}\geq\sqrt{2}$. Furthermore, note that (5.5)-(5.7) together with (5.2)-(5.3) imply that for ground-state configurations $C_{n}$ such that there holds equality in (5.4) the atoms

[TABLE]

are subset of the same square lattice. We point out that each $x_{i}\in I_{k},k=3,4$, has exactly $k-2$ neighbors in $X_{n}^{\mathrm{bulk}}$ and that the interior bond angles are exactly $\pi/2$.

Recall the excess of edges $\eta=\sum_{j\geq 4}(j-4)f_{j}$, introduced in (2.3), where $f_{j}$ denotes the number of elementary polygons with $j$ vertices in the bond graph. Clearly, $\eta=0$ if and only if the bond graph consists of squares only. We also recall that $b$ denotes the number of bonds in the bond graph.

Lemma 5.5** (Cardinality of the bulk).**

Let $C_{n}$ be a connected ground state with no acyclic bonds. Then

[TABLE]

Proof.

Let $f_{j}$ be the number elementary $j$-gons in the bond graph and let $f$ be the number of elementary polygons in the bond graph. From Lemma 4.2(a) we obtain

[TABLE]

since by the summation on the left all bonds contained in the maximal polygon are counted only once whereas all other bonds are counted twice. By (2.3) we get $4f=2b-d-\eta$. This along with Euler’s formula $n-b+f=1$ (omitting the exterior face) yields $n-d=2b+4-3n+\eta$. ∎

5.2. Energetic characterization

We start with the proof of Theorem 2.1. We use the following properties of the function $\beta$ defined in (2.10).

Lemma 5.6**.**

The function $\beta:\mathbb{N}\to\mathbb{R}$ satisfies

• (i)

$\lfloor\beta(n-m)\rfloor+\lfloor\beta(m)\rfloor+1<\lfloor\beta(n)\rfloor$* for all $n\geq 4$ and $1\leq m,n-m\leq n$,*

• (ii)

$\lfloor\beta(n)\rfloor+1\leq\lfloor\beta(n+1)\rfloor$* for all $n\geq 1$.*

Proof.

See [24, Proposition 4.1, Proposition 4.2]. ∎

Lemma 5.7**.**

Let $j,n,m\in\mathbb{N}$ and let $x\in\mathbb{R}$ satisfy

[TABLE]

Then $x\geq-2n+j-4+2\sqrt{m+n+8-2j}$.

Proof.

The proof is elementary: we note that the function

[TABLE]

is strictly increasing and vanishes for $x=-2n+j-4+2\sqrt{m+n+8-2j}$. ∎

Proof of Theorem 2.1.

We start by noting that every ground state $C_{n}$ has alternating charge distribution and every atom has at most four neighbors by Lemma 4.1. By Proposition 3.1 and Lemma 5.1 the ground-state energy satisfies

[TABLE]

We prove that, if $C_{n}$ is a ground state, then $C_{n}$ is connected, does not contain bridges, and satisfies

[TABLE]

The same statement then holds also for $\mathcal{E}(C_{n})$. In fact, due to (5.9)-(5.10), we have

[TABLE]

This implies $\mathcal{E}(C_{n})=\mathcal{R}(C_{n})=-b=-\lfloor\beta(n)\rfloor$.

Let $n\geq 4$. We proceed by induction. Suppose that the statement has been proven for all $m<n$ (for $1\leq m\leq 3$ this is elementary). We first show connectedness of the ground state and the non-existence of bridges (Step 1). Afterwards, we prove the energy equality (Step 2).

Step 1: $C_{n}$ is connected and does not contain bridges. Assume by contradiction that $C_{n}$ consists of two sub-configurations $C_{m}$ and $C_{n-m}$ that are connected by at most one bond. The energy contribution of this bond, if it exists, is greater or equal to $-1$. Apart from that, we can estimate the sum of the energy contributions of both components separately. In both cases, as $m,n-m<n$ $n\geq 4$, using the induction assumption, Lemma 5.6(i), and $V_{\rm r}\geq 0$ (cf. [iv]), we get

[TABLE]

This contradicts (5.9) and shows that $C_{n}$ is connected and its bond graph does not contain any bridges.

Step 2: Energy equality $\mathcal{R}(C_{n})=-b=-\lfloor\beta(n)\rfloor$. We divide the proof into three steps. We first treat the case that $C_{n}$ contains acyclic bonds (Step 2.1). Afterwards, we consider only configurations $C_{n}$ without acyclic bonds and show $\mathcal{R}(C_{n})=-b$ (Step 2.2) as well as $\mathcal{R}(C_{n})=-\lfloor\beta(n)\rfloor$ (Step 2.3).

Step 2.1: $C_{n}$ contains acyclic bonds. By Step 1, $C_{n}$ does not contain bridges. If there exist flags, we can find an atom $x_{i}$ such that removing $x_{i}$ removes exactly one flag. We can count the energy contribution of this flag by at least $-1$ and we estimate the energy of the rest of the configuration by induction. By Lemma 5.6(ii) we get

[TABLE]

Equality also shows that $C_{n}\setminus\{(x_{i},q_{i})\}$ has $\lfloor\beta(n-1)\rfloor$ bonds by induction and $C_{n}$ has $\lfloor\beta(n-1)\rfloor+1=\lfloor\beta(n)\rfloor$ bonds.

Step 2.2: $\mathcal{R}(C_{n})=-b$ for connected $C_{n}$ with no acyclic bonds. Assume by contradiction that $\mathcal{R}(C_{n})>-b$, i.e., there exist $x_{1},x_{2}\in X_{n}$ such that $q_{1}=q_{2}$ and $|x_{1}-x_{2}|<\sqrt{2}$ or there exists a bond between $x_{1},x_{2}\in X_{n}$ such that $q_{1}=-q_{2}$ and $|x_{1}-x_{2}|>1$. Now if $x_{1}\in\partial X_{n}$ or $x_{2}\in\partial X_{n}$ we have by (5.4), (5.5), and Remark 5.4

[TABLE]

Moreover, by (5.3), the induction hypothesis, and (5.10) there holds

[TABLE]

On the other hand, if $x_{1},x_{2}\notin\partial X_{n}$, by Lemma 5.3 we get $\mathcal{R}^{\mathrm{bnd}}(C_{n})\geq-2d+4$. By (5.3) and the induction assumption we obtain

[TABLE]

In both cases, by (5.3) and (2.10) there holds $\mathcal{R}(C_{n})>-\lfloor 2n-2\sqrt{(n-d)}\rfloor+4$. Since the right hand side is an integer, we obtain

[TABLE]

In a similar fashion, the assumption $\mathcal{R}(C_{n})>-b$ implies $-\lfloor-\mathcal{R}(C_{n})\rfloor-1\geq-b$. Now by Lemma 5.5 we obtain $n-d\geq 2(\lfloor-\mathcal{R}(C_{n})\rfloor+1)+4-3n$. This along with (5.11) yields

[TABLE]

Note that $-(\lfloor-\mathcal{R}(C_{n})\rfloor+1)\leq 2-3n/2$. By using Lemma 5.7 with $j=4,m=0$, and $x=-(\lfloor-\mathcal{R}(C_{n})\rfloor+1)$ we obtain

[TABLE]

Since the left hand side is an integer, we get by (2.10)

[TABLE]

This contradicts (5.9).

Step 2.3: $\mathcal{R}(C_{n})=-\lfloor\beta(n)\rfloor$ for connected $C_{n}$ with no acyclic bonds. Due to (5.9), it suffices to prove $\mathcal{R}(C_{n})\geq-\lfloor\beta(n)\rfloor$. We again proceed by induction. By Lemma 5.3, (5.3), and the induction hypothesis we obtain

[TABLE]

By (2.10) and (5.3) there holds $\mathcal{R}(C_{n})\geq-2n+2\sqrt{n-d}+4$. By Lemma 5.5 and Step 2.2 we obtain $n-d\geq-2\mathcal{R}(C_{n})+4-3n$. This yields

[TABLE]

By applying Lemma 5.7 with $j=4$, $m=0$, and $x=\mathcal{R}(C_{n})$ we obtain $\mathcal{R}(C_{n})\geq-\beta(n)$. Finally, since $\mathcal{R}(C_{n})$ is an integer due to Step 2.2, we conclude $\mathcal{R}(C_{n})\geq-\lfloor\beta(n)\rfloor$. ∎

5.3. Geometric Characterization

We now proceed with the proof of Theorem 2.3. To this end, we first prove two lemmas about flags and non-equilibrated atoms.

Lemma 5.8** (Flags).**

Let $n\geq 4$ and let $C_{n}$ be a ground state. Then the bond graph of $C_{n}$ contains at most one flag.

Proof.

Assume by contradiction that there exist at least two flags. We can choose two flags such that removing the two flags increases the energy of the configuration at most by $2$. By applying Thereom 2.1 to the sub-configuration after removing the two flags, we obtain $\mathcal{E}(C_{n})\geq-2-\lfloor\beta(n-2)\rfloor$. Therefore, we get by (2.10)

[TABLE]

We have that $2-\frac{4}{\sqrt{n-2}+\sqrt{n}}\geq 1$ for all $n\geq 6$. With the above estimate this implies $\mathcal{E}(C_{n})>-\lfloor\beta(n)\rfloor$. This gives a contradiction to Theorem 2.1 in the cases $n\geq 6$. For $n=4,5$ it can be checked directly that $-2-\lfloor\beta(n-2)\rfloor>-\lfloor\beta(n)\rfloor$. ∎

Recall the definition of non-equilibrated atoms in Subsection 2.2. By $\mathcal{A}_{\rm bulk}(X_{n})\subset\mathcal{A}(X_{n})$ we additionally denote the bulk atoms which are not equilibrated.

Lemma 5.9** (Non-equilibrated atoms).**

Let $C_{n}$ be a ground state and assume that its bond graph does not contain any acyclic bonds. If $\mathcal{A}(X_{n})\neq\emptyset$, then $\#\mathcal{A}(X_{n})\geq 2$ and $\eta\geq 2$. If $\mathcal{A}_{\mathrm{bulk}}(X_{n})\neq\emptyset$, then $\#\mathcal{A}(X_{n})\geq 2$ and $\eta\geq 4$.

Proof.

We split the proof into several steps. First, we prove that $\theta\in[\frac{\pi}{2},\frac{3\pi}{2}]$ for all bond angles $\theta$. Secondly, we show that, if $\mathcal{A}(X_{n})\neq\emptyset$, then $\#\mathcal{A}(X_{n})\geq 2$. Then we prove $\eta\geq 2$ if $\mathcal{A}(X_{n})\neq\emptyset$. Finally, we confirm that, if $\mathcal{A}_{\mathrm{bulk}}(X_{n})\neq\emptyset$, then $\eta\geq 4$. These statements show the thesis of the lemma.

Step 1: $\theta\in[\frac{\pi}{2},\frac{3\pi}{2}]$ for all bond angles $\theta$. By Theorem 2.1 there holds $\mathcal{E}(C_{n})=-b$. Thus, Lemma 4.2(b) implies that $\theta\in[\frac{\pi}{2},\frac{3\pi}{2}]$ for all bond angles $\theta$.

Step 2: If $\mathcal{A}(X_{n})\neq\emptyset$, then $\#\mathcal{A}(X_{n})\geq 2$. Since the bond graph of $C_{n}$ does not contain any acyclic bonds and $\mathcal{A}(X_{n})\neq\emptyset$, there holds $\mathcal{A}(X_{n})\cap P\neq\emptyset$ for some polygon $P$ in the bond graph. By Lemma 4.2(a) it follows that $\#(\mathcal{A}(X_{n})\cap P)\geq 2$.

Step 3: If $\mathcal{A}(X_{n})\neq\emptyset$, then $\eta\geq 2$. Assume by contradiction that $\eta=0$. By the definition of $\eta$ this implies that all elementary polygons in the bond graph are squares. Furthermore, since the bond graph of $C_{n}$ does not contain any acyclic bonds, all atoms are contained in some elementary polygon, i.e., in some square. Lastly, we note that, due to Theorem 2.1, there holds $\mathcal{E}(C_{n})=-b$ and therefore by Lemma 4.2(b) all squares are regular. In particular, all their atoms are equilibrated. This contradicts the fact that $\mathcal{A}(X_{n})\neq\emptyset$.

Step 4: If $\mathcal{A}_{\mathrm{bulk}}(X_{n})\neq\emptyset$, then $\eta\geq 4$. Let $x\in\mathcal{A}_{\rm bulk}(X_{n})$ and denote by $\theta_{1},\ldots,\theta_{k}$ the bond angles at $x$. By (4.2) we have that $k\leq 4$. We have $k\in\{2,3,4\}$. We conclude once we have shown that there exist at least two bond angles at $x$ that are not equilibrated. In fact, since no acyclic bonds are present in the bond graph and $x\in\mathcal{A}_{\rm bulk}(X_{n})$, all the bond angles at $x$ belong to different elementary polygons. Then, there exist at least two non squares in the bond graph and hence $\eta\geq 4$.

It thus remains to prove that there exist $i_{1},i_{2}\in\{1,\ldots,k\}$, $i_{1}\neq i_{2}$, such that $\theta_{i_{1}},\theta_{i_{2}}\notin\{\frac{\pi}{2},\pi,\frac{3\pi}{2}\}$. Recall $x\in\mathcal{A}(X_{n})$. Without restriction we suppose that $\theta_{1}\notin\{\frac{\pi}{2},\pi,\frac{3\pi}{2}\}$. Now assume that $\theta_{j}\in\{\frac{\pi}{2},\pi,\frac{3\pi}{2}\}$ for all $j=2,\ldots,k$. Since the bond angles at $x$ need to sum to $2\pi$, we obtain

[TABLE]

By Step 1 we get $\theta_{1}\in\frac{\pi}{2}\mathbb{N}\cap[\frac{\pi}{2},\frac{3\pi}{2}]$. This contradicts the fact that $\theta_{1}\notin\{\frac{\pi}{2},\pi,\frac{3\pi}{2}\}$. ∎

We are now in the position to prove Theorem 2.3(a).

Proof of Theorem 2.3(a).

We prove the statement by induction. For $=1,2,3$ the statement is clearly true, and for $n=4$ it follows from Lemma 4.2(b). Let $n\geq 5$. By Theorem 2.1 and Lemma 5.8 the bond graph of ground states does not contain bridges and at most one flag. Therefore, in view of Lemma 5.6(ii), up to removing a flag, it is not restrictive to assume that $C_{n}$ is a ground state with no acyclic bonds. We will use the following fact several times: by Remark 2.2, $C_{n}$ is repulsion-free. Thus Lemma 5.1 implies

[TABLE]

where $C_{n}^{\mathrm{bulk}}$ is defined before (5.3). By the definition of $\mathcal{A}(X_{n})$ and due to the fact that ground states are connected (cf. Theorem 2.1), it suffices to prove that $\mathcal{A}(X_{n})=\emptyset$. We divide the proof into two steps. We first prove that $\mathcal{A}(X_{n})\setminus\mathcal{A}_{\mathrm{bulk}}(X_{n})=\emptyset$ and secondly we show that $\mathcal{A}_{\mathrm{bulk}}(X_{n})=\emptyset$.

Step 1: $\mathcal{A}(X_{n})\setminus\mathcal{A}_{\mathrm{bulk}}(X_{n})=\emptyset$. Assume by contradiction that $\mathcal{A}(X_{n})\setminus\mathcal{A}_{\mathrm{bulk}}(X_{n})\neq\emptyset$. By Lemma 5.3 this implies $\mathcal{R}^{\mathrm{bnd}}(C_{n})>-2d+4$. Then (5.12) and Theorem 2.1 (applied for $C_{n}^{\mathrm{bulk}}$) yield

[TABLE]

This along with $\mathcal{R}^{\mathrm{bnd}}(C_{n})>-2d+4$, (5.3), and again Lemma 5.1 gives

[TABLE]

By Theorem 2.1, $\mathcal{E}(C_{n})$ is an integer and consequently we derive

[TABLE]

Now by using Lemma 5.5 and $\mathcal{E}(C_{n})=-b$ (see Theorem 2.1) we obtain

[TABLE]

Observe that $\eta\geq 2$ by Lemma 5.9. By applying Lemma 5.7 with $j=5,m=\eta$ and $x=\mathcal{E}(C_{n})$ we conclude $\mathcal{E}(C_{n})\geq-\beta(n)+1>-\lfloor\beta(n)\rfloor$. This contradicts Theorem 2.1.

Step 2: $\mathcal{A}_{\mathrm{bulk}}(X_{n})=\emptyset$. Assume by contradiction that $\mathcal{A}_{\mathrm{bulk}}(X_{n})\neq\emptyset$. Note that $\eta\geq 4$ by Lemma 5.9 and thus $n\geq 8$. There are two cases two consider: a) $n-d\leq 3$ and b) $n-d\geq 4$.

Proof for $n-d\leq 3$. Using Lemma 5.5 we have that

[TABLE]

By Theorem 2.1 we have that $\mathcal{E}(C_{n})=-b=-\lfloor 2n-2\sqrt{n}\rfloor$. This together with (5.14) leads to

[TABLE]

where the last step follows from $\eta\geq 4$. Inequality (5.15) is violated for all $n\geq 8$. This yields a contradiction and concludes the proof in this case.

Proof for $n-d\geq 4$. We first observe that $C_{n}^{\mathrm{bulk}}$ is clearly not a subset of the square lattice if $\mathcal{A}(X_{n}^{\mathrm{bulk}})\neq\emptyset$. In addition, we notice that $\mathcal{A}_{\mathrm{bulk}}(X_{n})\setminus\mathcal{A}(X_{n}^{\mathrm{bulk}})\neq\emptyset$ implies that $C_{n}^{\mathrm{bulk}}$ is not subset of the square lattice. To see this, assume that there exists $x\in\mathcal{A}_{\mathrm{bulk}}(X_{n})\setminus\mathcal{A}(X_{n}^{\mathrm{bulk}})$ and denote by $y\in\partial X_{n}$ one of its neighbors. If $y$ is $4$-bonded, (5.8) and $x\in\mathcal{A}(X_{n})$ yield that $C_{n}^{\mathrm{bulk}}$ is not a subset of the square lattice. On the other hand, if $y$ is $3$-bonded, (5.8) together with a simple geometric consideration implies that $C_{n}$ is not repulsion-free, which contradicts Remark 2.2.

First, assume that the bond graph of $C_{n}^{\mathrm{bulk}}$ does not contain a flag. By induction hypothesis, ground states with less than $n$ atoms without flags are subsets of the square lattice. Thus, by the assumption $\mathcal{A}_{\mathrm{bulk}}(X_{n})\neq\emptyset$, $C_{n}^{\mathrm{bulk}}$ cannot be a ground state. This along with (5.3) and (5.12) yields

[TABLE]

By Lemma 5.3 we also have $\mathcal{R}^{\mathrm{bnd}}(C_{n})\geq-2d+4$. The two inequalities together with (5.3) lead to the strict inequality (5.13), and we may proceed exactly as in Step 1 of the proof to obtain a contradiction to the fact that $C_{n}$ is a ground state.

Now assume that the bond graph of $C_{n}^{\mathrm{bulk}}$ contains a flag. Without restriction we can assume that $C_{n}^{\mathrm{bulk}}$ is a ground state since otherwise the strict inequality (5.16) holds, and we obtain a contradiction exactly as before. Since $n-d\geq 4$, the bond graph of $C_{n}^{\mathrm{bulk}}$ contains exactly one flag by Lemma 5.8. There holds $\#\mathcal{A}(X_{n})\geq 2$ by Lemma 5.9 and $\mathcal{A}(X_{n})\setminus\mathcal{A}_{\mathrm{bulk}}(X_{n})=\emptyset$ by Step 1. Therefore, $\#\mathcal{A}_{\mathrm{bulk}}(X_{n})\geq 2$. After removing the flag from $C_{n}^{\mathrm{bulk}}$, we get a configuration $C_{n-d-1}$ which does not contain flags, but there still holds \mathcal{A}(X_{n})\cap X_{n-d-1}\neq\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}\emptyset. By induction hypothesis, $C_{n-d-1}$ can thus not be a ground state and we get \mathcal{E}(C_{n-d-1})>\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}-\lfloor\beta(n-d-1)\rfloor. By counting the contribution of the flag to the energy by at least $-1$ we obtain

[TABLE]

where we also used (5.12). This along with $\mathcal{R}^{\mathrm{bnd}}(C_{n})\geq-2d+4$ (see Lemma 5.3) and (5.3) gives

[TABLE]

Since $\mathcal{E}(C_{n})=-b$ is an integer by Theorem 2.1, we obtain $\mathcal{E}(C_{n})\geq-2n+6+2\sqrt{n-d-1}$. Now using Lemma 5.5 and $\mathcal{E}(C_{n})=-b$ we obtain

[TABLE]

Recall that $\eta\geq 4$ by Lemma 5.9. By applying Lemma 5.7 with $j=6,m=\eta-1$ and $x=\mathcal{E}(C_{n})$ we obtain

[TABLE]

Recall that $n\geq 8$. Therefore, we get $2-\frac{2}{\sqrt{n}+\sqrt{n-1}}\geq 1$ and thus

[TABLE]

This contradicts the fact that $C_{n}$ is a ground state and concludes the proof. ∎

We now address the proof of Theorem 2.3(b). To this end, we use the following result.

Theorem 5.10** (Deviation from Wulff-shape).**

Let $n\in\mathbb{N}$ and let $C_{n}$ be a ground state with no acyclic bonds. Then, possibly after translation we find two squares $S_{k_{1}^{2}}\subset\mathbb{Z}^{2}$ and $S_{k_{2}^{2}}\subset\mathbb{Z}^{2}$, $k_{1},k_{2}\in\mathbb{N}$, with $S_{k_{1}^{2}}\subset X_{n}\subset S_{k_{2}^{2}}$ such that

[TABLE]

where $c>0$ is a universal constant independent of $n$ and $C_{n}$.

Proof.

Let $n\geq 4$. From Theorem 2.3(a) and the fact that $C_{n}$ does not contain acyclic bonds we get that $C_{n}$ is a subset of the square lattice. Moreover, $C_{n}$ is repulsion-free, see Remark 2.2. Therefore, our energy on ground-state configurations coincides (up to distributing alternating charge or neglecting it) with the one considered in [24]. The statement then follows from [24, Theorem 8.1]. ∎

Proof of Theorem 2.3(b).

By Theorem 2.1 and Lemma 5.8 we know that the bond graph contains at most one flag and no other acyclic bonds. If it contains a flag, one can remove the flag and the remaining configuration is still a ground state by Lemma 5.6(i). We can therefore assume that there are no acyclic bonds in the bond graph of $C_{n}$. The statement follows from Theorem 5.10. In fact, observe that Theorem 5.10 also implies that the diameter of a ground state is of order $\sqrt{n}$. ∎

5.4. Characterization of the net charge

We now prove Theorem 2.4. Part (b) of the statement has already been addressed by an explicit construction in Subsection 3.2. Thus, it remains to prove part (a).

Definition 5.11** (Line segment).**

A tuple $(x_{0},\ldots,x_{m})\subset\mathbb{Z}^{2}$ is called a line segment if there exists $v\in\{(1,0),(0,1)\}$ such that $x_{k+1}=x_{k}+v$ for all $k\in\{0,\ldots,m-1\}$.

Proposition 5.12** (Convexity of ground states).**

Let $n\in\mathbb{N}$ and let $C_{n}$ be a ground state with no acyclic bonds. Then each line segment $(x_{0},\ldots,x_{m})$ with $x_{0},x_{m}\in X_{n}$ satisfies $x_{i}\in X_{n}$ for $i=0,\ldots,m$.

Proof.

For the proof we refer to [24, Proposition 6.3], where this property is called convexity by rows and columns. The result in [24, Proposition 6.3] is applicable due to the fact that our energy on ground-state configurations coincides with the one considered in [24], see the proof of Theorem 5.10. ∎

Proof of Theorem 2.4(a).

In view of Theorem 2.1 and Lemma 5.8, it suffices to treat the case that the bond graph of $C_{n}$ does not contain any acyclic bonds. We apply Theorem 5.10 to find two squares $S_{k_{1}^{2}}\subset X_{n}\subset S_{k_{2}^{2}}$. By Proposition 5.12 it is elementary to see that $X_{n}\setminus S_{k_{1}^{2}}$ can be written as the union of at most $4(k_{2}-k_{1})$ line segments. Recall that $C_{n}$ has alternating charge distribution and therefore the net charge of each line segment is in $\{-1,0,1\}$. Also recall from Subsection 3.1 that squares have charge in $\{-1,0,1\}$. This implies that the net charge of the configuration $C_{n}$ satisfies

[TABLE]

The statement follows from the fact that $k_{2}-k_{1}\leq cn^{1/4}$, see Theorem 5.10. ∎

6. Characterization of optimal configurations for Prescribed charge

In this section we prove Proposition 2.5, Theorem 2.6, and Theorem 2.8.

6.1. Energy of optimal configurations

In this short subsection we prove Theorem 2.6.

Proof of Theorem 2.6:.

Without restriction we may suppose that $q_{\rm net}\geq 0$ since the proof for $q_{\mathrm{net}}\leq 0$ follows analogously. We proceed in two steps. First, we prove the statement for $q_{\mathrm{net}}\geq q_{\mathrm{sat}}^{n}$ and then for $0\leq q_{\mathrm{net}}<q_{\mathrm{sat}}^{n}$. The proof is performed by an induction argument.

Step 1: $q_{\mathrm{net}}\geq q_{\mathrm{sat}}^{n}$. Our goal is to show

[TABLE]

For $q_{\rm net}=q_{\mathrm{sat}}^{n}$ the statement is clearly true, see (2.12). Suppose that the statement holds for $q_{\mathrm{net}}\geq q_{\mathrm{sat}}^{n}$ with $q_{\mathrm{net}}\leq n-2$. We show (6.1) for $q_{\mathrm{net}}+2$. By Lemma 4.3(a), for all configurations $C_{n}$ satisfying $\mathcal{Q}(C_{n})=q_{\mathrm{net}}+2$ there holds $\mathcal{E}(C_{n})\geq-2n+2(q_{\mathrm{net}}+2)$.

It thus suffices to construct a configuration $\tilde{C}_{n}$ with $\mathcal{Q}(\tilde{C}_{n})=q_{\mathrm{net}}+2$ and $\mathcal{E}(\tilde{C}_{n})=-2n+2(q_{\mathrm{net}}+2)$. To this end, let $C_{n}$ be a configuration with $\mathcal{Q}(C_{n})=q_{\mathrm{net}}$ and $\mathcal{E}(C_{n})=-2n+2q_{\mathrm{net}}$, which exists by the induction hypothesis.

Choose $x_{i}\in C_{n}$ with $q_{i}=-1$ and modify $C_{n}$ as follows: remove $x_{i}$ and add a new atom $\tilde{x}_{i}$ with charge $+1$ to the configuration such that $|\tilde{x}_{i}-x_{j}|\geq\sqrt{2}$ for all $j=1,\ldots,n$. Denote this configuration by $\tilde{C}_{n}$. We have \mathcal{Q}(\tilde{C}_{n}\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0})=q_{\mathrm{net}}+2. By Lemma 4.3(a)(i), the relocated atom was $4$-bonded and the bond lengths were of unit length. Hence, we obtain by [ii]

[TABLE]

Thus, (6.1) is proven for $q_{\rm net}+2$.

Step 2: 0\leq\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}q_{\mathrm{net}}<\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}q_{\mathrm{sat}}^{n}. As before, the lower bound in (2.14) follows from Lemma 4.3(a). Therefore, it remains to show that

[TABLE]

For $q_{\rm net}=q_{\mathrm{sat}}^{n}$ the statement is clearly true, see (2.12). Assume that (6.2) holds for $2\leq q_{\mathrm{net}}\leq q_{\mathrm{sat}}^{n}$. We show (6.2) for $q_{\mathrm{net}}-2$. To this end, we have to construct a configuration $\tilde{C}_{n}$ with $\mathcal{Q}(\tilde{C}_{n})=q_{\mathrm{net}}-2$ and $\mathcal{E}(\tilde{C}_{n})\leq-2n+4q_{\mathrm{sat}}^{n}-2(q_{\mathrm{net}}-2)$.

Let $C_{n}$ be a configuration with $\mathcal{Q}(C_{n})=q_{\mathrm{net}}$ and $\mathcal{E}(C_{n})=\mathcal{E}^{n}_{\mathrm{min}}(q_{\mathrm{net}})$. Choose an atom $x_{i}$ with $q_{i}=1$ and modify $C_{n}$ in the following way: remove $x_{i}$ and add a new atom $\tilde{x}_{i}$ with charge $-1$ to the configuration such that $|\tilde{x}_{i}-x_{j}|\geq\sqrt{2}$ for all $j=1,\ldots,n$. Denote this configuration by $\tilde{C}_{n}$. First, note that $\mathcal{Q}(\tilde{C}_{n})=q_{\mathrm{net}}-2$. By [ii] and $\#\mathcal{N}(x_{i})\leq 4$, see (4.2), we get

[TABLE]

As (6.2) holds for $q_{\mathrm{net}}$ by induction hypothesis, we obtain

[TABLE]

This shows (6.2) for $q_{\rm net}-2$ and concludes the proof. ∎

6.2. Boundary and interior net charge

In the following, we will consider without restriction configurations which satisfy $\mathcal{Q}(C_{n})\geq 0$, i.e., the $+1$ phase is the majority phase. This can indeed always be achieved by interchanging the roles of the positive and negative charges. We define the notion of a bridging atom. An atom $x_{0}$ is called bridging atom if it is $2$-bonded and not contained in any simple cycle, see Fig. 11. Denote by $C_{m}$ and $C_{n-m-1}$ the two connected components of $C_{n}\setminus\{x_{0},q_{0}\}$. We set

[TABLE]

In the following we say that $C_{m}$ and $C_{m-n}$ are connected through a bridging atom. Recall the definition of acyclic bonds in Subsection 2.2. In a similar fashion, we say that an atom $x_{i}$ is an acyclic atom if it is not contained in any cycle. We denote the union of the acyclic atoms by $I_{\rm ac}$.

Lemma 6.1** (Bridging atom).**

Consider $C_{n}$ with $0\leq\mathcal{Q}(C_{n})$ and $\mathcal{E}(C_{n})=-2n+2\mathcal{Q}(C_{n})$.

(a) If there exists a bridging atom connecting two configurations $C_{m}$ and $C_{n-m}$, there holds

[TABLE]

Moreover, $C_{n-m}$ is an optimal configuration and there holds $m,n-m\geq 4$.

(b) If $n\geq 6$ and $C_{n}$ is connected without bridging atoms, then $I_{\mathrm{ac}}\cap X_{n}^{-}=\emptyset$ and $I_{\mathrm{ac}}\cap X_{n}^{+}$ consists of $1$-bonded atoms.

Proof.

(a) Suppose that a bridging atom $x_{0}$ exists. As $\mathcal{E}(C_{n})=-2n+2\mathcal{Q}(C_{n})$, Lemma 4.3(a)(i) implies that $x_{0}$ has charge $q_{0}=+1$. This clearly gives

[TABLE]

Define by $C_{n+1}$ a configuration given by the union of the two connected components $C_{n-m}$ and $C_{m+1}$, where $C_{m+1}$ is translated in such a way that $\mathrm{dist}(X_{n-m},X_{m+1})\geq\sqrt{2}$. We observe that the atoms of charge $-1$ are $4$-bonded by construction since $C_{n}$ satisfies this property. In a similar fashion, $C_{n+1}$ is repulsion-free and its bonds have unit length by Lemma 4.3(a)(i),(ii). Therefore, in view of Lemma 4.3(b), for $C_{n+1}$ there holds the equality $\mathcal{E}(C_{n+1})=-2(n+1)+2\mathcal{Q}(C_{n+1})$. Then applying Lemma 4.3(a)(vi) on the two connected components $C_{m+1}$ and $C_{n-m}$ we find $\mathcal{Q}(C_{m+1})\geq q_{\mathrm{sat}}^{m+1}$ and $\mathcal{Q}(C_{n-m})\geq q_{\mathrm{sat}}^{n-m}$. The first part of the claim now follows from (6.4).

In view of Lemma 4.3(b), for $C_{n-m}$ there holds the equality $\mathcal{E}(C_{n-m})=-2(n-m)+2\mathcal{Q}(C_{n-m})$. Thus, $C_{n-m}$ is an optimal configuration by (4.6). Finally, since atoms of charge $-1$ are $4$-bonded, each of the two components contains at least four atoms, i.e., $m,n-m\geq 4$.

(b) Now assume that $n\geq 6$ and that $C_{n}$ is connected without bridging atoms. By $\mathcal{E}(C_{n})=-2n+2\mathcal{Q}(C_{n})$ and (4.6) we have that $C_{n}$ is optimal. First, suppose by contradiction that there exists some $x_{i}\in I_{\mathrm{ac}}\cap X_{n}^{-}\neq\emptyset$. Since $C_{n}$ is connected with alternating charge distribution (see Lemma 4.1) and $n\geq 6$, there exist $x_{k}\in X_{n}^{-}$, $x_{k}\neq x_{i}$, and $x_{j}\in\mathcal{N}(x_{i})\cap\mathcal{N}(x_{k})$. Since $C_{n}$ does not contain bridging atoms and $x_{i}\in I_{\rm ac}$, $x_{j}$ is at least $3$-bonded. By Lemma 4.3(a)(iv),(v) all bond angles of $x_{j}$ are integer multiples of $\frac{\pi}{2}$. Thus, as sketched in Fig. 11, $x_{i}$ is contained in a square. This contradicts $x_{i}\in I_{\mathrm{ac}}$. In a similar fashion, Fig. 11 shows that some $x_{j}\in X_{n}^{+}$ which is $3$-bonded or $4$-bonded is contained in a square. This along with (4.2) and the fact that by assumption there are no bridging atoms, i.e., no $2$-bonded atoms in $I_{\rm ac}$, shows that $I_{\rm ac}\cap X_{n}^{+}$ consists of $1$-bonded atoms only. ∎

Sub-configurations: Similar to the definition of $I_{\rm ac}$, we say that an atom $x_{i}$ is an exterior acyclic atom if it is not contained in any cycle and not contained in the interior region of any cycle. We denote the union of the exterior acyclic atoms by $I_{\rm ac}^{\rm ext}$. Clearly, there holds $I_{\rm ac}^{\rm ext}\subset I_{\rm ac}$.

Let $n\geq 6$ and let $C_{n}$ be a connected, optimal configuration without bridging atoms satisfying $\mathcal{Q}(C_{n})\geq q_{\rm sat}^{n}$. In particular, there holds $\mathcal{E}(C_{n})=-2n+2\mathcal{Q}(C_{n})$ by Theorem 2.6. We denote by $C_{n}^{a}$ the configuration without the exterior acyclic atoms and their charges. We observe that $C_{n}^{a}$ is still connected. This follows from Lemma 6.1(b). By $\partial C_{n}^{a}$ we indicate the maximal polygon of $C_{n}^{a}$, i.e, the simple cycle which delimits the bond graph, see also Subsection 5.1. The cardinality of $\partial C_{n}^{a}$ is denoted by $d$. Furthermore, for $k=2,3,4$, we set

[TABLE]

Finally, we set $C_{n}^{a,\mathrm{bulk}}=C_{n}^{a}\setminus\partial C_{n}^{a}$. In a similar fashion, we denote by $X_{n}^{a}$, $\partial X_{n}^{a}$, and $X_{n}^{a,\mathrm{bulk}}$ the atomic positions of the sub-configurations. Recall the definition of $\eta$ in (2.3).

Lemma 6.2** (Cardinality of $n-d$).**

Let $n\geq 6$ and let $C_{n}$ be a connected, optimal configuration which satisfies $\mathcal{Q}(C_{n})\geq q_{\mathrm{sat}}^{n}$ and does not contain a bridging atom. Then

[TABLE]

Proof.

Let $f_{j}$ be the number elementary $j$-gons in the bond graph and let $f$ be the number of elementary polygons. There holds

[TABLE]

where $b_{\mathrm{ac}}$ denotes the cardinality of the acyclic bonds. In fact, by the summation on the left the bonds of the maximal polygon are counted once, the acyclic bonds are not counted, and all other bonds are counted twice. By Lemma 4.2(a) we obtain $4f+\eta=2b-d-2b_{\rm ac}$. This along with Euler’s formula $n-b+f=1$ (omitting the exterior face) and the fact that $b_{\rm ac}\geq\#I_{\rm ac}^{\rm ext}$ yields

[TABLE]

As $b=2n-2\mathcal{Q}(C_{n})$, see Remark 2.2, Theorem 2.6, and Lemma 4.3(a)(i),(ii), the claim follows. ∎

Recall the definition of the non-equilibrated atoms $\mathcal{A}(X_{n})$ in Subsection 2.2. The following estimate for the net charge of the boundary and the interior configuration will be instrumental for our analysis. Its proof will be given in Appendix A, along with the proof of the boundary energy estimate (Lemma 5.3).

Lemma 6.3** (Net charge of the interior and boundary).**

Let $n\geq 6$ and let $C_{n}$ be a connected $q_{\mathrm{sat}}^{n}$-optimal configuration without bridging atoms.

(a) Then there exists $m\geq 4$ such that

[TABLE]

If $\#\mathcal{A}(X_{n})\geq\#\mathcal{A}(X_{n}^{\mathrm{a},\mathrm{bulk}})+2$, then $m\geq 6$.

(b) Suppose that (6.6) holds. Then $\#I_{\mathrm{ac}}^{\mathrm{ext}}\geq m$ and there exists an optimal configuration $C_{n-d-m}$ satisfying $\mathcal{Q}(C_{n-d-m})\geq q_{\mathrm{sat}}^{n-d-m}$ and

[TABLE]

Moreover, $X_{n-d-m}$ is a subset of $X_{n}$ up to [math]-bonded atoms and $\mathcal{A}(X_{n-d-m})=\mathcal{A}(X_{n}^{\mathrm{a},\mathrm{bulk}})$.

Roughly speaking, the configurations $C_{n-d-m}$ are constructed by removing $\partial X_{n}^{a}$ and $I_{\mathrm{ac}}^{\rm ext}$ from $C_{n}$. In this sense, $\mathcal{Q}(C_{n-d-m})$ can be regarded as the net charge of the interior. The statement then shows that the boundary net charge can be controlled from below by $4$, and by at least 6 if the number of non-equilibrated atoms decreases by at least 2 when the boundary is removed.

6.3. Characterization of $q_{\mathrm{sat}}^{n}$

The goal of this subsection is to prove Proposition 2.5. In view of the construction in Subsection 3.3, see Proposition 3.2, it remains to show the lower bound $q_{\mathrm{sat}}^{n}\geq\phi(n)$, where the function $\phi$ is defined in (2.13). As a preparation, we first provide an equivalent representation of $\phi$, and state some monotonicity and subadditivity properties.

Lemma 6.4** (Representation of $\phi$).**

For $n\geq 2$ there holds

[TABLE]

where $k\in\mathbb{N}_{0}$.

Proof.

We divide the proof into two steps. First, we prove the statement in the case that $n$ is even and then in the case that $n$ is odd.

$n$ even. Let $n=1+2k^{2}+2k+m$ with $n$ even, i.e., $1\leq m\leq 4k+4$ and $m$ odd. By (2.13) we have

[TABLE]

It is elementary to check that

[TABLE]

This along with (6.9) shows the desired equality of $\phi(n)$ with the expression on the right hand side of (6.8), in the case that $n$ is even.

$n$ odd. Let $n=1+2k^{2}+2k+m$ with $n$ odd, i.e., $1\leq m\leq 4k+4$ and $m$ even. By (2.13) we have

[TABLE]

It is again elementary to see that

[TABLE]

This together with (6.10) shows the desired equality of $\phi(n)$ with the expression on the right hand side of (6.8), in the case that $n$ is odd. ∎

Lemma 6.5** (Properties of $\phi$).**

The following properties hold true:

• (i)

$\phi(n)\leq\phi(n-1)+1$* and $\phi(n)\leq\phi(n+1)+1$ for all $n\in\mathbb{N}$,*

• (ii)

$\phi(n)\leq\phi(n+2)$* for all $n\in\mathbb{N}$,*

• (iii)

$\phi(n)\leq\phi(n-3)+1$* for all $n\geq 5$,*

• (iv)

$\phi(n)\leq\phi(n-5)+1$* for all $n\geq 11$,*

• (v)

$\phi(n_{1}-t)+\phi(n_{2}+t)\leq\phi(n_{1})+\phi(n_{2})+6$* for all $t\leq n_{1}\leq n_{2}$.*

• (vi)

$\phi(n)+2\leq\phi(m)+\phi(n-m)$* for all $3\leq m,n-m\leq n,\ \ n\geq 11$,*

• (vii)

There exists $m_{0}\in\mathbb{N}$ such that $\phi(n)+52\leq\phi(m)+\phi(n-m)$ for all $m_{0}\leq m,n-m\leq n$.

Proof.

Properties (i)-(v) are elementary and can be checked by using (6.8): for (ii), we use the monotonicity of $\phi$ when restricted to even and odd numbers, respectively. Property (i) follows by looking closely at (6.8). To see (iii)-(iv), we denote by $k(n)\in\mathbb{N}_{0}$ and $1\leq m(n)\leq 4k(n)+4$ the numbers such that $n=1+2(k(n))^{2}+2k(n)+m(n)$. We also set $n_{3}=7$ and $n_{5}=15$. It is elementary to check that for $t\in\{3,5\}$ and for all $n\geq n_{t}$ there holds either $k(n)=k(n-t)$ or

[TABLE]

This along with a careful inspection of (6.8) implies (iii)-(iv) for $n\geq 7$ and $n\geq 15$ respectively. The remaining cases can be checked by Table 1. To see (v), we first note that for $0\leq m\leq n$ there holds $\phi(n)-\phi(m)\leq 2(k(n)-k(m))+2$ and $\phi(m)-\phi(n)\leq 2(k(m)-k(n))+2$. This can be seen by careful inspection of (6.8). Additionally, we use that for $t\leq n_{1}\leq n_{2}$ there holds $k(n_{1}-t)-k(n_{1})+k(n_{2}+t)-k(n_{2})\leq 1$.

We proceed with (vi). The case $m=3$ follows from (iii) and $\phi(3)=3$, see Table 1 below. The case $m=4$ follows by taking additionally (i) and $\phi(4)=4$ into account. For $m=5$, we use (iv) and the fact that $\phi(5)=3$. Now suppose that $n\geq 11$ and $6\leq m,n-m\leq n$.

We will use the following property: let $k_{1},k_{2},m_{0}\in\mathbb{N}$, with $m_{0}\geq k_{1}\geq k_{2}$. Since the function $x\mapsto\sqrt{x}$ is concave and increasing, the function

[TABLE]

attains its minimum for $m=m_{0}$. Moreover, we will use that $\lfloor a+b\rfloor\leq\lfloor a\rfloor+\lfloor b\rfloor+1$. We will work directly with the definition of $\phi$, see (2.13). Consulting Table 1, we note that the cases $11\leq n\leq 32$ can be checked directly by comparing the values of $\phi(n-m)+\phi(m)$ and $\phi(n)$. We can therefore assume that $n\geq 33$. There are four cases to consider: (a) $n$ even, $m$ even. (b) $n$ even, $m$ odd. (c) $n$ odd, $m$ even. (d) $n$ odd, $m$ odd.

(a) $n$ even, $m$ even. By (6.11) for $k_{1}=k_{2}=4$ and $m_{0}=6$, we get

[TABLE]

Now for $n\geq 32$ we have that $\sqrt{2}-6(\sqrt{2n-4}+\sqrt{2n-16})^{-1}\geq 1$, which indeed yields $\phi(m)+\phi(n-m)\geq 2\lfloor\frac{1}{2}\sqrt{2n-4}\rfloor+2+2=\phi(n)+2$.

(b) $n$ even, $m$ odd. Observe that in this case we have $m,n-m\geq 7$ and thus $n\geq 14$. By (6.11) for $k_{1}=k_{2}=5$ and $m_{0}=7$, we obtain

[TABLE]

We check that for all $n\geq 34$ (note $n$ is even) there holds $1-15(\sqrt{2n-4}+\sqrt{2n-19})^{-1}\geq 0$, and thus $\phi(m)+\phi(n-m)\geq\phi(n)+2$.

(c) $n$ odd, $m$ even. By (6.11) for $k_{1}=5$, $k_{2}=4$ and $m_{0}=6$, we obtain

[TABLE]

For $n\geq 33$ we have that $\sqrt{2}-6(\sqrt{2n-5}+\sqrt{2n-17})^{-1}\geq 1$, and thus $\phi(m)+\phi(n-m)\geq\phi(n)+2$.

(d) $n$ odd, $m$ odd. We proceed as in (c) by interchanging the roles of $m$ and $n-m$.

Finally, to see (vii), one may follow the lines of the proof of (vi). We sketch only the case where $n$ and $m$ are even. By repeating the argument in (a) for general $m_{0}$ we find

[TABLE]

One can check that for m_{0}=3945\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0} we have $\phi(m)+\phi(n-m)\geq\phi(n)+52$ for all $n\geq 2m_{0}$. ∎

We now are in a position to establish the lower bound for $q_{\mathrm{sat}}^{n}$. This lower bound together with the upper bound in Proposition 3.2 shows Proposition 2.5.

Proposition 6.6** (Lower bound for $q_{\mathrm{sat}}^{n}$).**

Let $n\in\mathbb{N}$. Then $q_{\mathrm{sat}}^{n}\geq\phi(n)$.

Proof.

Let $C_{n}$ be a $q_{\mathrm{sat}}^{n}$-optimal configuration, i.e., $\mathcal{Q}(C_{n})=q_{\mathrm{sat}}^{n}$ and $\mathcal{E}(C_{n})=\mathcal{E}_{\mathrm{min}}^{n}(q_{\mathrm{sat}}^{n})=-2n+2q_{\mathrm{sat}}^{n}$. We proceed to show that $q_{\mathrm{sat}}^{n}=\mathcal{Q}(C_{n})\geq\phi(n)$. We prove the claim by induction. In view of Table 1, it is elementary to check that for $1\leq n\leq 10$ the claim holds. Indeed, for $1\leq n\leq 4$ the cardinality of the minority phase is zero, for $5\leq n\leq 7$ the cardinality of the minority phase is at most one, and for $8\leq n\leq 10$ the cardinality of the minority phase is at most two. This is due to the fact that by Lemma 4.3(a)(i),(iv) each atom of the minority phase is $4$-bonded with bond angles $\pi/2$ and thus two atoms of the minority phase can share at most two neighbors. Finally, three such atoms can only have one neighbor in common.

Let $n\geq 11$. We assume that the statement holds for all $m\in\mathbb{N}$ with $1\leq m<n$, and prove the statement for $n$. We proceed in three steps. First, we show that the claim holds true if $C_{n}$ is not connected. Then, we treat the case where $C_{n}$ is connected and contains a bridging atom. Finally, we address the case of connected $C_{n}$ without any bridging atoms.

Step 1: $C_{n}$ is not connected. We assume that $C_{n}$ is not connected. Denote by $C_{m}$ and $C_{n-m}$, $1\leq m,n-m\leq n$, two sub-configurations consisting of $m$ and $n-m$ atoms, respectively, with no bonds between them. Lemma 4.3(a)(vi) implies that $\mathcal{Q}(C_{m})\geq q_{\mathrm{sat}}^{m}$ and $\mathcal{Q}(C_{n-m})\geq q_{\mathrm{sat}}^{n-m}$. Suppose that $m\leq 2$ or $n-m\leq 2$, without restriction say $m\leq 2$. We can apply the induction hypothesis, Table 1, and Lemma 6.5(i) ($m$ times) to obtain

[TABLE]

On the other hand, if $m,n-m\geq 3$, the induction hypothesis and Lemma 6.5(vi) (recall $n\geq 11$) yield

[TABLE]

This yields the claim in the case that $C_{n}$ is not connected.

Step 2: $C_{n}$ is connected and contains bridging atoms. Assume that the bond graph contains a bridging atom. Denote the two configurations that a connected through are bridging atom by $C_{m}$ and $C_{n-m}$, see the definition before (6.3). By Lemma 6.1(a) and the induction hypothesis we have $n\geq m,n-m\geq 4$ and

[TABLE]

By using Lemma 6.5(i),(vi) we conclude

[TABLE]

Step 3: $C_{n}$ is connected and does not contain any bridging atoms. Since $C_{n}$ does not contain any bridging atoms, Lemma 6.3 along with the induction hypothesis yields

[TABLE]

for some $m\geq 4$. By Lemma 6.2 and Lemma 6.5(i),(ii) we then obtain

[TABLE]

By Lemma 6.5(ii), \#I_{\rm ac}^{\rm ext}\geq m\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}\geq 4, and the fact that $\eta$ is even (see Lemma 4.2(a)) we get

[TABLE]

Suppose first that $n+8-4q_{\mathrm{sat}}^{n}\leq-1$. Then by Table 1, (2.13), and the fact that $n-q_{\mathrm{sat}}^{n}$ is even it is elementary to check that $q_{\mathrm{sat}}^{n}\geq\lceil(n+9)/4\rceil\geq\phi(n)-1$. Otherwise, the claim follows from (6.12) and Lemma 6.7 below for $x=q_{\mathrm{sat}}^{n}$, where we again use that $n-q_{\mathrm{sat}}^{n}$ is always even. ∎

In the previous proof we have used the following lemma.

Lemma 6.7**.**

Let $n\geq 7$. Let $x\in\mathbb{N}$ with $x\leq n/4+2$, $n-x$ even, and

[TABLE]

Then there holds $x\geq\phi(n)$.

Proof.

If $4+\phi\left(8+n-4x\right)\geq\phi(n)$, the statement follows from (6.13) and if $x\geq\phi(n)$ there is nothing to prove. We now assume by contradiction that

[TABLE]

Since $\phi(8+n-4x)-\phi(n)$ is even by (2.13), and $x-\phi(n)$ is even by assumption and (2.13), (6.14) yields

[TABLE]

Let $n=1+2k^{2}+2k+m$, $m\in\mathbb{N}$, $1\leq m\leq 4k+4$. By (6.15)(i) there holds $\phi(n)\geq 6$. Using Table 1, we observe that $n\geq 10$ and thus also $k\geq 1$. We distinguish three cases: (a) $2\leq m\leq 4k+4$ and $m$ even. (b) $1\leq m\leq 2k+1$, $m$ odd. (c) $2k+3\leq m\leq 4k+3$, $m$ odd.

(a) $2\leq m\leq 4k+4$ and $m$ even. By (6.8) there holds $\phi(n)=3+2k$, and (6.15)(ii) thus implies

[TABLE]

One can also check that the difference of the first and the last expression is even. By Lemma 6.5(ii), (6.15)(i), and $\phi(n)=3+2k$ we then obtain

[TABLE]

For $n\geq 14$ and thus $k\geq 2$, there holds $\phi(1+2(k-2)^{2}+2(k-2)+2)=3+2(k-2)$ by (6.8). This yields the contradiction $5+2k\leq 3+2k$. A contradiction in the cases $n=11,13$, i.e., $k=1$, can be obtained by noting $1+2(k-2)^{2}+2(k-2)+2=3$ and $\phi(3)=3$, cf. Table 1.

(b) $1\leq m\leq 2k+1$ and $m$ odd. By (6.8) there holds $\phi(n)=2+2k$ and (6.15)(ii) thus yields

[TABLE]

The difference of the first and the last expression is even. By Lemma 6.5(ii), (6.15)(i), and $\phi(n)=2+2k$ we obtain

[TABLE]

For $n\geq 14$ and thus $k\geq 2$, there holds $\phi(1+2(k-2)^{2}+2(k-2)+1)=2+2(k-2)$ by (6.8). This yields the contradiction $4+2k\leq 2+2k$. A contradiction in the cases $n=10,12$, i.e., $k=1$, can be obtained by noting $1+2(k-2)^{2}+2(k-2)+1=2$ and $\phi(2)=2$, cf. Table 1.

(c) $2k+3\leq m\leq 4k+3$ and $m$ odd. By (6.8) there holds $\phi(n)=4+2k$ and (6.15)(ii) thus implies

[TABLE]

The difference of the first and the last expression is even. By Lemma 6.5(ii), (6.15)(i), and $\phi(n)=4+2k$ we obtain

[TABLE]

For $n\geq 14$ and thus $k\geq 2$, there holds $\phi(1+2(k-2)^{2}+2(k-2)+2(k-2)+3)=4+2(k-2)$ by (6.8). This yields the contradiction $6+2k\leq 4+2k$. A contradiction in the cases $n=10,12$, i.e., $k=1$, can be obtained by noting $1+2(k-2)^{2}+2(k-2)+2(k-2)+3=2$ and $\phi(2)=2$, cf. Table 1. ∎

6.4. Crystallization result for $q_{\mathrm{sat}}^{n}$-optimal configurations

This subsection is devoted to the proof of Theorem 2.8(a). In Lemma 6.8 we first show that $q^{n}_{\rm sat}$-optimal configurations are connected and do not contain bridging atoms after removing a finite number of atoms independently of $n$. Afterwards, we control the number of non-equilibrated atoms (Lemma 6.9). This then allows us to show Theorem 2.8(a).

Recall that in Proposition 2.5 we have shown that $\phi(n)=q_{\rm sat}^{n}$ for all $n\in\mathbb{N}$. In the following, we will use this equality without further notice. As before, it is not restrictive to consider configurations $C_{n}$ with $\mathcal{Q}(C_{n})\geq 0$.

Lemma 6.8** (Connectedness, bridging atoms).**

Let $C_{n}$ be a $q_{\mathrm{sat}}^{n}$-optimal configuration.

(a) If $n\geq 11$ and $C_{n}$ is not connected, we can remove $m\in\{1,2\}$ [math]-bonded atoms from $C_{n}$ to obtain a connected $q_{\rm sat}^{n-m}$-optimal configuration.

(b) Let $n\geq 26$ and let $m_{0}$ be the constant from Lemma 6.5(vii). If $C_{n}$ contains a bridging atom, there holds \mathcal{Q}(C_{n})\geq\phi(n-26\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0})+2. We can remove $m\in\{1,\ldots,m_{0}\}$ atoms from $C_{n}$ to obtain a configuration $C_{n-m}$ which is $q_{\rm sat}^{n-m}$-optimal and does not contain bridging atoms.

We defer the proof and continue with the next ingredient for the proof of Theorem 2.8(a). We show that the number of non-equilibrated atoms $\mathcal{A}(X_{n})$ can be controlled. We remark that the bound on $\mathcal{A}(X_{n})$ is not sharp and could be improved at the expense of more elaborated methods. As our focus lies on a qualitative description of the geometry of optimal configurations, we refrain from entering into finer estimates.

Lemma 6.9** (Control on $\#\mathcal{A}(X_{n})$).**

Let $C_{n}$ be a $q_{\mathrm{sat}}^{n}$-optimal configuration without bridging atoms. Then $\#\mathcal{A}(X_{n})\leq 50$.

We again defer the proof and proceed to show Theorem 2.8(a).

Proof of Theorem 2.8(a).

Let $m_{0}$ be the constant from Lemma 6.5(vii) and define $n_{0}=52m_{0}+2$. Let $C_{n}$ be a $q_{\mathrm{sat}}^{n}$-optimal configuration. If $n<n_{0}$, the statement is trivial, we therefore suppose that $n\geq n_{0}$. The goal is to prove that, after removing at most $2m_{0}+2$ atoms, the remaining configurations is a subset of the square lattice. In view of Lemma 6.8, we can remove $m\leq m_{0}+2$ atoms from $C_{n}$ to obtain a connected configuration $C_{n-m}$ without bridging atoms which is a $q_{\mathrm{sat}}^{n-m}$-optimal configuration. Consequently, it suffices to consider a $q_{\mathrm{sat}}^{n}$-configuration $C_{n}$, n\geq 51\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}m_{0}, which is connected without bridging atoms and to show that, after removing at most $m_{0}$ atoms, it is a subset of the square lattice.

Step 1: Proof for connected configurations without bridging atoms. We introduce

[TABLE]

Note that, if $X_{m}\in\mathcal{X}(X_{n})$, then up to isometry $X_{m}\subset\mathbb{Z}^{2}$. Choose $X_{n}^{\mathrm{max}}\in\mathcal{X}(X_{n})$ as a maximal element, that is $\#X_{m}\leq\#X_{n}^{\mathrm{max}}$ for all $X_{m}\in\mathcal{X}(X_{n})$. Denote its cardinality by $n_{\mathrm{max}}\leq n$. As $C_{n}$ is connected and \#\mathcal{A}(C_{n})\leq 50\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}, it is elementary to see that $C_{n}$ consists of at most 51\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0} sub-configuration each of which subset of a (different) square lattice. Since n\geq 51\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}m_{0}, this implies $n_{\rm max}\geq m_{0}$. Additionally, we set $C_{n}^{\mathrm{max}}=\{(x_{i},q_{i}):x_{i}\in X_{n}^{\mathrm{max}}\}.$ Our goal is now to prove that

[TABLE]

The main ingredient for the proof is the estimate

[TABLE]

We defer the proof of (6.17) to Step 2 below and first show (6.16). Assume by contradiction that (6.16) does not hold true, i.e., $n-n_{\mathrm{max}}>m_{0}$. By (6.17) we obtain

[TABLE]

Since also $n_{\rm max}\geq m_{0}$, we then derive by Lemma 6.5(ii),(vii)

[TABLE]

This yields $\mathcal{Q}(C_{n})\geq q_{\mathrm{sat}}^{n}+2$ and contradicts $\mathcal{Q}(C_{n})=q_{\mathrm{sat}}^{n}$.

Step 2: Proof of (6.17). To conclude the proof, it remains to confirm (6.17). As a preparation, we introduce

[TABLE]

where $\mathcal{N}(x_{i})$ (see (2.2)) is defined with respect to $X_{n}$. First, we prove that

[TABLE]

To see this, let $x_{i}\in\mathrm{d}X_{n}^{\mathrm{max}}$ and choose $x_{j}\in\mathcal{N}(x_{i})\setminus X_{n}^{\mathrm{max}}$. Suppose by contradiction that $x_{i}\notin\mathcal{A}(X_{n})$ or $x_{i}\in X_{n}^{-}$. In both cases, this implies $x_{i}\notin\mathcal{A}(X_{n})$ by Lemma 4.3(a)(i),(iv). Then all bond angles at $x_{i}$ are integer multiples of $\frac{\pi}{2}$, and thus the configuration $X_{n}^{\mathrm{max}}\cup\{x_{j}\}$ is such that $\mathcal{A}(X_{n}^{\mathrm{max}}\cup\{x_{j}\})=\emptyset$. This implies $X_{n}^{\mathrm{max}}\cup\{x_{j}\}\in\mathcal{X}(X_{n})$ and contradicts the maximality of $\#X_{n}^{\rm max}$.

We now show (6.17) and begin with (i). By (6.18) and Lemma 4.3(a)(i) there holds $\#(\mathcal{N}(x_{i})\cap X_{n}^{\mathrm{max}})=\#\mathcal{N}(x_{i})=4$ for all $x_{i}\in X_{n}^{\rm max}$ with $q_{i}=-1$, i.e., all negatively charged atoms of $X_{n}^{\mathrm{max}}$ are $4$-bonded. Moreover, since $X_{n}^{\mathrm{max}}\subset X_{n}$, $X_{n}^{\mathrm{max}}$ is repulsion-free and all bonds have unit length by Lemma 4.3(a)(i),(ii). By Lemma 4.3(b) this implies the equality $\mathcal{E}(C_{n}^{\mathrm{max}})=-2n_{\rm max}+2\mathcal{Q}(C_{n}^{\mathrm{max}})$. Then Lemma 4.3(a)(iii) yields $\mathcal{Q}(C_{n}^{\mathrm{max}})\geq\phi(n_{\mathrm{max}})$, as desired.

We now prove (6.17)(ii). We set $\tilde{X}_{n}:=(X_{n}\setminus X_{n}^{\mathrm{max}})\cup\mathrm{d}X_{n}^{\mathrm{max}}$ and

[TABLE]

Since $\mathrm{d}X_{n}^{\mathrm{max}}\subset X_{n}^{+}$ by (6.18), we obtain

[TABLE]

We now proceed as in the proof of (6.17)(i): let x_{i}\in\tilde{X}_{n}^{-}\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}. By (6.18) and Lemma 4.3(a)(i) there holds $\#(\mathcal{N}(x_{i})\cap\tilde{X}_{n})=\#\mathcal{N}(x_{i})=4$. Since $\tilde{X}_{n}\subset X_{n}$, $\tilde{X}_{n}$ is repulsion-free and all bonds are of unit length by Lemma 4.3(a)(i),(ii). By Lemma 4.3(b) this implies the equality $\mathcal{E}(\tilde{C}_{n})=-2\#\tilde{X}_{n}+2\mathcal{Q}(\tilde{C}_{n})$. Then Lemma 4.3(a)(iii) implies $\mathcal{Q}(\tilde{C}_{n})\geq\phi(\#\tilde{X}_{n})$.

By (6.18) and Lemma 6.9 we obtain \#\mathrm{d}X_{n}^{\mathrm{max}}\leq\#\mathcal{A}(X_{n})\leq 50\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}. By using (6.19), $\#\tilde{X}_{n}=n-n_{\mathrm{max}}+\#\mathrm{d}X_{n}^{\mathrm{max}}$, and applying Lemma 6.5(i) (50\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}-\#\mathrm{d}X_{n}^{\mathrm{max}})-times, we obtain

[TABLE]

This concludes the proof of (6.17)(ii). ∎

Remark 6.10** (Maximal component).**

For later purposes, we observe that the configuration $C_{n}^{\mathrm{max}}$ identified in the proof is a connected subset of the square lattice and that it is saturated, i.e., the atoms with charge $-1$ are $4$-bonded. We also note that $\mathcal{Q}(C_{n}^{\mathrm{max}})\leq\phi(n_{\mathrm{max}})+49$. In fact, otherwise by (6.17)(ii) and Lemma 6.5(ii),(vi) we would get the contradiction $\mathcal{Q}(C_{n})\geq\phi(n_{\rm max})+50+\phi(n-n_{\mathrm{max}}+50)-50\geq\phi(n)+2$.

We proceed with the proofs of Lemma 6.8 and Lemma 6.9.

Proof of Lemma 6.8.

(a) Let $n\geq 11$ and assume by contradiction that $C_{n}$ is not connected with two connected components of at least three atoms. Denote by $C_{m}$ and $C_{n-m}$, $3\leq m,n-m\leq n$, two sub-configurations consisting of $m$ and $n-m$ atoms, respectively, with no bonds between them. By Lemma 4.3(a)(vi) we get $\mathcal{Q}(C_{m})\geq\phi(m)$ and $\mathcal{Q}(C_{n-m})\geq\phi(n-m)$. By Lemma 6.5(vi) we obtain

[TABLE]

This contradicts $\mathcal{Q}(C_{n})=q_{\mathrm{sat}}^{n}$. Observe that two single atoms may indeed exist, see Fig. 9. We also note that these (at most two) atoms are [math]-bonded since they have the same charge $+1$ and $C_{n}$ has alternating charge distribution.

Thus, we can remove $m\in\{1,2\}$ atoms of charge $+1$ to obtain a connected configuration $C_{n-m}$ with $\mathcal{Q}(C_{n-m})=\mathcal{Q}(C_{n})-m=q_{\rm sat}^{n}-m$. By Lemma 6.5(i) this implies $\mathcal{Q}(C_{n-m})\leq q_{\rm sat}^{n-m}$. On the other hand, by construction, $C_{n-m}$ satisfies

[TABLE]

Therefore, we get $\mathcal{Q}(C_{n-m})\geq q_{\rm sat}^{n-m}$, see Lemma 4.3(a)(iii). Thus, $C_{n-m}$ is $q_{\rm sat}^{n-m}$-optimal.

(b) Step 1: There exists at most one bridging atom. Assume by contradiction that there exist two bridging atoms. Denote the two components connected through the first bridging atom by $C_{m_{1}},C_{n-m_{1}}$, and denote the two components of $C_{n-m_{1}}$ connected through the second bridging atom by $C_{m_{2}}$ and $C_{n-m_{1}-m_{2}}$, see (6.3). By Lemma 6.1(a) we get $n\geq n-m_{1}-m_{2},m_{1},m_{2}\geq 4$ and

[TABLE]

By using Lemma 6.5(ii) and Lemma 6.5(vi) twice we derive

[TABLE]

This contradicts $\mathcal{Q}(C_{n})=q_{\mathrm{sat}}^{n}$.

Step 2: If there exists a bridging atom, then one of the components contains at most $m_{0}$ atoms. Assume by contradiction that $C_{n}$ consists of two components $C_{m}$, $C_{n-m}$ connected through a bridging atom with $m,n-m\geq m_{0}+1$. By Lemma 6.1(a) we get $\mathcal{Q}(C_{m})\geq\phi(m+1)-1$ and $\mathcal{Q}(C_{n-m})\geq\phi(n-m)$. Then Lemma 6.5(i),(vii) yield

[TABLE]

Thus, $\mathcal{Q}(C_{n})>q_{\mathrm{sat}}^{n}$. This contradicts $\mathcal{Q}(C_{n})=q_{\mathrm{sat}}^{n}$.

Step 3: Conclusion. In view of Step 1 and Step 2, we can remove $m\in\{4,\ldots,m_{0}\}$, $n-m\geq 4$, atoms to obtain a configuration $C_{n-m}$ without bridging atoms. We denote the sub-configuration consisting of the removed atoms by $C_{m}$. By Lemma 6.1(a) we get that $C_{n-m}$ is an optimal configuration and $\mathcal{Q}(C_{n-m})\geq q_{\rm sat}^{n-m}$. Now suppose by contradiction that $\mathcal{Q}(C_{n-m})>q_{\rm sat}^{n-m}$, i.e., $\mathcal{Q}(C_{n-m})\geq q_{\rm sat}^{n-m}+2$. Recall by Lemma 6.1(a) that $\mathcal{Q}(C_{m})\geq\phi(m+1)-1$. Therefore, since $n\geq m,n-m\geq 4$, by Lemma 6.5(i),(vi) we derive

[TABLE]

This contradicts $\mathcal{Q}(C_{n})=q_{\mathrm{sat}}^{n}$ and shows $\mathcal{Q}(C_{n-m})=q_{\rm sat}^{n-m}$. It remains to prove \mathcal{Q}(C_{n}\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0})\geq\phi(n-26\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0})+2. To see this, we recall by Lemma 6.1(a) that

[TABLE]

In view of Lemma 6.5(i),(ii), for 4\leq m\leq 26\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0} we derive \mathcal{Q}(C_{n})\geq\phi(m+1)-1+\phi(n-26\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0})-m\,{\rm mod}\,2. Then the result follows by checking Table 1. If m\geq 27\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}, we suppose without restriction that $m+1\leq n-m$, and we use Lemma 6.5(v) for $n_{1}=m+1$, $n_{2}=n-m$, t=n_{1}-27\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0} to get

[TABLE]

where the last step follows from $\phi(27)=9$, see Table 1. In view of (6.22), this concludes the proof. ∎

Proof of Lemma 6.9.

We prove the statement by induction. For $n\leq 50$ the statement is clearly true. Now let $n\geq 51$. We assume that the statement holds for all $m<n$ and we proceed to prove the statement for $n$. Let $C_{n}$ be a $q_{\mathrm{sat}}^{n}$-optimal configuration without bridging atoms. If $C_{n}$ is not connected, we can apply Lemma 6.8(a) to remove $k\in\{1,2\}$ [math]-bonded atoms to obtain a $q_{\mathrm{sat}}^{n-k}$-optimal configuration. By the induction hypothesis, this new configuration contains at most $50$ non-equilibrated atoms and therefore also $\#\mathcal{A}(X_{n})\leq 50$. Therefore, it is not restrictive to assume that $C_{n}$ is connected. We divide the proof into several steps. Recall the definition of $I_{\rm ac}$ below (6.3).

Step 1: $\mathcal{A}(X_{n})\cap I_{\mathrm{ac}}=\emptyset$ and $\#\mathcal{A}(X_{n})\leq 3\eta/2$. We first prove that $\mathcal{A}(X_{n})\cap I_{\rm ac}=\emptyset$. By Lemma 6.1(b) there holds $I_{\mathrm{ac}}\cap X_{n}^{-}=\emptyset$ and $I_{\mathrm{ac}}\cap X_{n}^{+}$ consists of $1$-bonded atoms only. Since $1$-bonded atoms do not have bond angles, this implies $I_{\mathrm{ac}}\cap\mathcal{A}(X_{n})=\emptyset$.

Next, we prove $\#\mathcal{A}(X_{n})\leq 3\eta/2$. By Theorem 2.6, Remark 2.2, Lemma 4.2(b), and Lemma 4.3(a)(i),(ii) all squares are regular. Thus, if x\in\mathcal{A}(X_{n})\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0} is contained in a polygon, this polygon cannot be a square. Moreover, $x\in X_{n}^{+}$ by Lemma 4.3(a)(i),(iv). Denote now by $f_{j}$ the number of elementary $j$-gons in the bond graph and recall that $f_{j}=\emptyset$ for $j$ odd or $j\leq 3$, see Lemma 4.2(a). As \mathcal{A}(X_{n})\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}\cap I_{\rm ac}=\emptyset, each x\in\mathcal{A}(X_{n})\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0} is contained in at least one $j$-gon, $j\geq 6$. We also observe that, due to alternating charge distribution, there can be at most $j/2$ atoms of positive charge in each $j$-gon. This implies by (2.3)

[TABLE]

Step 2: Preliminaries. From now on, we suppose by contradiction that $\#\mathcal{A}(X_{n})\geq 51$. This implies $\eta\geq 34$ by Step 1. Note by Lemma 6.3 that $\#I_{2}+2\#I_{3}+3\#I_{4}=2d-m$ and $\#I_{\mathrm{ac}}^{\mathrm{ext}}\geq m$ for some $m\geq 4$. We will use the following fact several times: the estimate

[TABLE]

along with $\eta\geq 34$ leads to a contradiction: in fact, by Lemma 6.2 and Lemma 6.3 there holds $n-d\in n+4-4\mathcal{Q}(C_{n})+\eta+2\#I_{\mathrm{ac}}^{\mathrm{ext}}+2\mathbb{N}_{0}$ with $\eta\geq 34$ and $\#I_{\mathrm{ac}}^{\mathrm{ext}}\geq m\geq 4$. Using Lemma 6.5(ii) we obtain

[TABLE]

In view of Lemma 6.7 applied for $x=\mathcal{Q}(C_{n})-2$, we get the contradiction $\phi(n)-2=\mathcal{Q}(C_{n})-2\geq\phi(n)$, see (6.12) for a similar argument.

We denote by $C_{n-d-m}$ the optimal configuration obtained from Lemma 6.3(b) satisfying $\mathcal{A}(X_{n-d-m})=\mathcal{A}(X_{n}^{\mathrm{a},\mathrm{bulk}})$, $X_{n-d-m}\subset X_{n}$ up to [math]-bonded atoms, and

[TABLE]

Note by $n-4\phi(n)\geq-16$ for $n\geq 51$, (2.13), Lemma 6.2, and Lemma 6.3 that $n-d-m\geq n-4\phi(n)+4+\eta+m\geq 26$.

Step 3: $C_{n-d-m}$ satisfies $\mathcal{Q}(C_{n-d-m})=\phi(n-d-m)$. Assume by contradiction that $\mathcal{Q}(C_{n-d-m})\geq\phi(n-d-m)+2$. This together with (6.24), $m\geq 4$, and Lemma 6.5(ii) leads to $\mathcal{Q}(C_{n})\geq\phi(n-d-m)+6\geq\phi(n-d-26-m)+6$ which yields a contradiction by (6.23). In view of Lemma 6.8(a), up to removing at most two [math]-bonded atoms, we can thus assume that $C_{n-d-m}$ is also connected, and that $X_{n-d-m}\subset X_{n}$.

Step 4: $\#\mathcal{A}(X_{n})\leq\#\mathcal{A}(X_{n}^{\mathrm{a},\mathrm{bulk}})+1$, in particular $\mathcal{A}(X_{n})\cap\partial X_{n}^{\mathrm{a}}=\emptyset$. Recall the definition of $\partial X_{n}^{a}$ and $X_{n}^{a,\mathrm{bulk}}$ below (6.5). Suppose by contradiction $\#\mathcal{A}(X_{n})\geq\#\mathcal{A}(X_{n}^{\mathrm{a},\mathrm{bulk}})+2$. By Lemma 6.3(a) we have $m\geq 6$. Thus, by (6.24) and Lemma 6.5(ii) we get $\mathcal{Q}(C_{n})\geq\phi(n-d-m)+6\geq\phi(n-d-26-m)+6$ which yields a contradiction by (6.23). In particular, this also shows $\mathcal{A}(X_{n})\cap\partial X_{n}^{\mathrm{a}}=\emptyset$. Indeed, otherwise we would get $\#(\mathcal{A}(X_{n})\cap\partial X_{n}^{\mathrm{a}})\geq 2$ by Lemma 4.2 which contradicts $\#\mathcal{A}(X_{n})\leq\#\mathcal{A}(X_{n}^{\mathrm{a},\mathrm{bulk}})+1$.

Step 5: $C_{n-d-m}$ contains a bridging atom. Suppose by contradiction that $C_{n-d-m}$ does not contain a bridging atom. By Step 3 and Step 4 we get that

[TABLE]

and $\mathcal{Q}(C_{n-d-m})=\phi(n-d-m)$. (The equality in (6.25) follows from Lemma 6.3(b).) By the induction hypothesis applied on $C_{n-d-m}$, we get $\#\mathcal{A}(X_{n-d-m})\leq 50$. This implies equality in (6.25), i.e.,  50=\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}\#\mathcal{A}(X_{n})-1=\#\mathcal{A}(X_{n-d-m})\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}. Note that this together with Step 1 yields $\eta(C_{n-d-m})\geq 34$. By Step 4 (applied for $X_{n-d-m}$ in place of $X_{n}$; note that $C_{n-d-m}$ is connected, $q_{\rm sat}^{n-d-m}$-optimal, without bridging atoms, and $\eta(C_{n-d-m})\geq 34$) we also get $\mathcal{A}(X_{n-d-m})\cap\partial X_{n-d-m}^{\mathrm{a}}=\emptyset$.

Consider the unique $x\in\mathcal{A}(X_{n})\setminus\mathcal{A}(X_{n}^{\mathrm{a},\mathrm{bulk}})$. Since $X_{n}$ does not contain bridging atoms and all elements in $\mathcal{A}(X_{n})$ are at least $2$-bonded, $x$ is contained in an elementary polygon $P$. By Lemma 4.2 we find some $y\in\mathcal{A}(X_{n})\cap(P\setminus\{x\})$. We now show that $y$ is a bridging atom in $X_{n-d-m}$. Note first that $y\in\mathcal{A}(X_{n-d-m})$ since otherwise there would hold strict inequality in (6.25). Observe by Lemma 4.3(a)(iv),(v) that $x$ is $2$-bonded. Then, it is also clear that $P\cap X_{n-d-m}$ is contained in the boundary of the exterior face of $X_{n-d-m}$. Therefore, $y$ has to be an acyclic atom since otherwise we obtain the contradiction $\mathcal{A}(X_{n-d-m})\cap\partial X_{n-d-m}^{\mathrm{a}}\neq\emptyset$. As also $y$ is is $2$-bonded, see Lemma 4.3(a)(iv),(v), this shows that the configuration $C_{n-d-m}$ contains the bridging atom $y$. This contradicts our assumption that $C_{n-d-m}$ does not contain bridging atoms. Summarizing, $C_{n-d-m}$ needs to contain a bridging atom.

Step 6: $\#\mathcal{A}(X_{n})\leq 50$. Recall that we have supposed by contradiction that $\#\mathcal{A}(X_{n})\geq 51$ which implies $\eta\geq 34$ by Step 1. By Lemma 6.8(b) we obtain $\mathcal{Q}(C_{n-d-m})\geq\phi(n-d-m-26)+2$. This along with (6.24) and $m\geq 4$ gives

[TABLE]

which leads to a contradiction by (6.23). ∎

Remark 6.11** (Theorem 2.8(a) without bridging atoms).**

By increasing $n_{0}$ independently of $n$, the configuration in Theorem 2.8(a) can be constructed in such a way that it is a connected, saturated subset of the square lattice without bridging atoms: the configuration $C_{n}^{\mathrm{max}}$ in Remark 6.10 may contain at most $49$ bridging atoms. In fact, otherwise one could apply Lemma 6.1(a) and Lemma 6.5(vi) $50$ times to obtain an estimate of the form (see (6.4) for a similar argument)

[TABLE]

which contradicts $\mathcal{Q}(C_{n}^{\mathrm{max}})\leq\phi(n_{\mathrm{max}})+49$. In a similar fashion, repeating the arguments in (6.21), one of the components connected through a bridging atom has at most $m_{0}$ atoms as otherwise (6.26) holds. Thus, it suffices to remove at most $49m_{0}$ atoms to get the desired configuration without bridging atoms.

6.5. Wulff-shape emergence for $q_{\mathrm{sat}}^{n}$-optimal configurations

This subsection is devoted to the proof of Theorem 2.8(b). Our strategy is to apply an anisotropic quantitative isoperimetric inequality on suitable interpolations of the configurations. (A similar idea can be found in [8].) We first state two lemmas which relate the $\infty$-perimeter of the interpolation to the net charge and then prove the main result. Recall that configurations are called saturated if the atoms of the minority phase are $4$-bonded. We also recall (2.3).

Lemma 6.12** (Control on $\eta$).**

Let $N_{1}\in\mathbb{N}$. Let $C_{n}$ be a connected and saturated sub-configuration of the square lattice without bridging atoms satisfying

[TABLE]

Then there exists $N_{2}=N_{2}(N_{1})>0$ such that we can add k\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}\leq N_{2} atoms to $C_{n}$ to obtain a connected, saturated sub-configuration C_{n+k\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}} of the square lattice without bridging atoms satisfying \mathcal{Q}(C_{n+k\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}})\geq 0 and \eta(C_{n+k\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}})=0.

We defer the proof to the end of the subsection. A key ingredient for the proof of Theorem 2.8(b) is that, for (special) saturated configurations, the net charge coincides with the $\infty$-perimeter of a suitable interpolation of the configuration. We define $D(x):=\left\{|x-\cdot|_{1}\leq 1\right\}$ for $x\in\mathbb{Z}^{2}$. For a saturated configuration $C_{n}$ with $\mathcal{Q}(C_{n})\geq 0$ we introduce the interpolation

[TABLE]

and define the $\infty$-perimeter by

[TABLE]

Here, $\nu_{\partial A(C_{n})}$ is the outer unit normal to the set $A(C_{n})$. The second identity in (6.29) is elementary since by definition there holds $|(\nu_{\partial A(C_{n})})_{1}(x)|=|(\nu_{\partial A(C_{n})})_{2}(x)|=1/\sqrt{2}$ for $\mathcal{H}^{1}$-almost every point $x\in\partial A(C_{n})$.

Lemma 6.13**.**

Let $C_{n}$ be a saturated, connected sub-configuration of the square lattice which does not contain bridging atoms and satisfies $\eta=0$ as well as $\mathcal{Q}(C_{n})\geq 0$. Then $\mathrm{Per}_{\infty}(A(C_{n}))=2\mathcal{Q}(C_{n})-2$.

We again defer the proof and proceed to show Theorem 2.8(b).

Proof of Theorem 2.8(b).

Our strategy is to apply an anisotropic quantitative isoperimetric inequality on $A(C_{n})$, defined in (6.28). To this end, we will modify a given $q_{\rm sat}^{n}$-optimal configuration such that Lemma 6.13 is applicable. We split the proof into three steps. In the following, $c$ denotes a universal constant which may vary from line to line, but is independent of $n$.

Step 1: Some elementary facts. For $n\in\mathbb{N}$ we set

[TABLE]

A careful inspection of (6.8) yields for all $n\in\mathbb{N}$

[TABLE]

Let $C^{\prime}_{n}$ be a saturated configuration with $\mathcal{Q}(C_{n}^{\prime})\geq 0$. We find $\mathcal{L}^{2}(A(C^{\prime}_{n}))=2\#\{q_{i}=-1\}$ since $\mathcal{L}^{2}(D(x))=2$ for all $x\in\mathbb{Z}^{2}$. Thus, $\mathcal{L}^{2}(A(C^{\prime}_{n}))=n-\mathcal{Q}(C_{n}^{\prime})$. We let $W_{r}:=\{|\cdot|_{1}\leq r\}$ and note that

[TABLE]

Step 2: Modification of $q_{\rm sat}^{n}$-optimal configurations. Let $C_{n}$ be a $q_{\rm sat}^{n}$-optimal configuration. By Remark 6.11, after removing at most $n_{0}$ atoms, we get a saturated, connected sub-configuration of the square lattice $C_{n-m}$, $0\leq m\leq n_{0}$, whose bond graph does not contain any bridging atoms. By Lemma 6.5(i) there holds

[TABLE]

By Lemma 6.12 we can add at most $N_{2}(2n_{0})$ atoms to obtain a saturated, connected sub-configuration of the square lattice without bridging atoms satisfying $\eta=0$. For simplicity, we denote the configuration again by $C_{n}$ and observe that

[TABLE]

In Step 3 below we will apply Lemma 6.13 on $C_{n}$. The newly constructed configuration and the original configuration differ in cardinality and net charge by a finite constant independent of $n$, i.e., it suffices to prove the statement for this new configuration.

Step 3: Application of the quantitative isoperimetric inequality. Recall that the diamond $D_{\underline{n}}$ with alternating charge distribution is a $q_{\mathrm{sat}}^{\underline{n}}$-optimal configuration, see (2.9) and Subsection 3.3. By (6.30)(ii), (6.31), and (6.32) this implies $r(D_{\underline{n}})^{2}-c\leq r(C_{n})^{2}\leq r(D_{\underline{n}})^{2}+c\sqrt{n}$. Therefore, we get

[TABLE]

where for brevity we have set $r_{n}=r(C_{n})$. Using the definition r_{n}=\sqrt{\tfrac{1}{2}\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}(n-\mathcal{Q}(C_{n}))}, (2.13), and (6.32) one can also check that for $n$ sufficiently large there holds

[TABLE]

By the quantitative anisotropic isoperimetric inequality [14, Theorem 1.1] (applied for the convex set $K=W_{1}$) there exists a translation $\tau\in\mathbb{R}^{2}$ such that

[TABLE]

As by (6.29), (6.31), and (6.34) there holds \mathcal{L}^{2}(W_{r_{n}})=2r_{n}^{2}\leq n\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0} and \mathrm{Per}_{\infty}(W_{r_{n}})=4r_{n}\geq 2\sqrt{n/2}\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}, we get

[TABLE]

To estimate the right hand side, we apply Lemma 6.13 (on both $C_{n}$ and $D_{\underline{n}}$), (6.30)(i), (6.32), (6.33)(i), and $\phi(\underline{n})=\mathcal{Q}(D_{\underline{n}})$ to get

[TABLE]

After translation we may suppose that $X_{n}+\tau$ and $D_{\underline{n}}$ are both contained in $\mathbb{Z}^{2}$. By the fact that $\mathcal{L}^{2}(D(x))=2$ for all $x\in\mathbb{Z}^{2}$, (6.28), (6.33)(ii), (6.35), and (6.5) we get

[TABLE]

It remains to note that $\#(D_{\underline{n}}\triangle(X_{n}+\tau))\leq 5\,\#(D_{\underline{n}}^{-}\triangle(X^{-}_{n}+\tau))$ since $C_{n}$ is saturated. ∎

It remains to prove Lemma 6.12 and Lemma 6.13.

Proof of Lemma 6.12.

We first construct the configuration $C_{n+k}$. By $X_{n}^{\mathrm{int}}$ we denote the union of the points $\mathbb{Z}^{2}\setminus X_{n}$ which are contained in the interior of a simple cycle of the bond graph of $X_{n}$. Let $C_{n}^{\mathrm{int}}=(X_{n}^{\mathrm{int}},Q^{\mathrm{int}})$, where $Q^{\mathrm{int}}$ is chosen such that $C_{n}\cup C_{n}^{\mathrm{int}}$ has alternating charge distribution. This is possible since $X_{n}\cup X_{n}^{\mathrm{int}}\subset\mathbb{Z}^{2}$. Define $k:=\#X_{n}^{\mathrm{int}}$ and $C_{n+k}:=C_{n}\cup C_{n}^{\mathrm{int}}$. Note that $C_{n+k}$ is still connected, saturated and does not contain bridging atoms. Moreover, $C_{n+k}$ has alternating charge distribution, is subset of the square lattice, and $\eta(C_{n+k})=0$. Therefore, the configuration satisfies Lemma 4.3(a)(i),(ii). Hence, Lemma 4.3(b),(a)(iii) implies

[TABLE]

It remains to control $k$. Choose $N_{2}\in\mathbb{N}$ depending only on $N_{1}$ such that

[TABLE]

We now show $k\leq N_{2}$. Assume by contradiction that $k\geq N_{2}+1$. We claim that

[TABLE]

We postpone the proof of this estimate to the end and proceed to establish a contradiction. Using (6.27) and (6.37)-(6.39) we obtain

[TABLE]

Now by $k\geq N_{2}+1$, Lemma 6.5(i),(ii), and (6.38) there holds $\phi(k)\geq\phi(N_{2}+1)-1\geq N_{1}+2$. This together with (6.40) leads to $\phi(n+k)\leq\phi(n)-2$ which contradicts Lemma 6.5(i),(ii).

It remains to prove (6.39). By construction there holds $X_{n}^{\mathrm{int}}\subset\mathbb{Z}^{2}$. In particular, the configuration is repulsion-free and all bonds have unit length. In view of Lemma 4.3(b),(a)(iii) (note that it is applicable even though the configuration has negative net charge: consider $(X_{n}^{\mathrm{int}},-Q^{\mathrm{int}})$), it remains to show that each $x\in(X_{n}^{\mathrm{int}})^{+}$ has four neighbors in $X_{n}^{\mathrm{int}}$. If there existed some $x\in(X_{n}^{\mathrm{int}})^{+}$ which is at most $3$-bonded, then there would exist y\in X^{-}_{n}\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0} such that $|x-y|=1$. Since every point on the square lattices has at most four neighboring points on the square lattice, this would imply that $y$ is not $4$-bonded in $X_{n}$ and thus $X_{n}$ would not have been a saturated configuration: a contradiction. ∎

Proof of Lemma 6.13.

We prove the claim by induction over $k=\#X_{n}^{-}$. Let $k=1$. As $C_{n}$ is connected, this implies $n=5$ and therefore $\mathcal{Q}(C_{n})=3$ as well as $\mathrm{Per}_{\infty}(A(C_{n}))=4$. Thus, the claim is true for $k=1$. Now assume that we have proved the statement for all $l<k$. We proceed to prove the claim for $k$. Consider the leftmost of the uppermost negatively charged atoms and denote it by $x_{0}$. There are three cases to consider (see Fig. 12):

• (a)

$D(x_{0})$ shares exactly one side with another square $D(x_{1})$, $x_{1}\in X^{-}_{n}$.

• (b)

$D(x_{0})$ shares exactly two adjacent sides with two other squares $D(x_{1}),D(x_{2})$, $x_{1},x_{2}\in X^{-}_{n}$. The atom which is contained in both of these sides is contained in four squares.

• (c)

$D(x_{0})$ shares exactly two adjacent sides with two other squares $D(x_{1}),D(x_{2})$, $x_{1},x_{2}\in X^{-}_{n}$. The atom which is contained in both of these sides is contained in only three squares.

This is a consequence of the choice of $x_{0}$, the absence of bridging atoms, the connectedness of $C_{n}$, and $\eta=0$. In fact, by the choice of the point $x_{0}$, the top corner cannot be contained in another square. Next, at least one side of $D(x_{0})$ has to be shared with another square since otherwise the configuration would not be connected or there would be a 2-bonded atom. The latter in turn would be a bridging atom or would lead to $\eta\geq 2$. Finally, if $D(x_{0})$ shares exactly two sides with other squares, the bottom corner is either contained in three or four squares.

Proof of (a): We remove the two corners of positive charge not contained in any other square as well as $x_{0}$ to obtain a configuration $C_{n-3}$ such that $\#X_{n-3}^{-}=k-1$. Note that

[TABLE]

Since we have removed three sides of the square $D(x_{0})$ but count a side of the square $D(x_{1})$ to the perimeter, we obtain

[TABLE]

It is elementary to check that the new configuration is still saturated, connected, satisfies $\eta(C_{n-3})=0$, and does not contain any bridging atom. Therefore, we can apply the induction hypothesis together with (6.41) and (6.42) to obtain

[TABLE]

Proof of (b): We remove the top corner of positive charge as well as $x_{0}$ to obtain a configuration $C_{n-2}$ such that $\#X_{n-2}^{-}=k-1$. Again, the new configuration is still saturated, connected, satisfies $\eta(C_{n-2})=0$, and does not contain any bridging atom. We get $\mathcal{Q}(C_{n-2})=\mathcal{Q}(C_{n})$ and, since we have removed two sides of the square $D(x_{0})$ but count one side of each square $D(x_{1})$, $D(x_{2})$ to the perimeter, we obtain $\mathrm{Per}_{\infty}(A(C_{n-2}))=\mathrm{Per}_{\infty}(A(C_{n}))$. This along with the induction hypothesis yields

[TABLE]

Proof of (c): We remove the top corner of positive charge as well as $x_{0}$ to obtain two sub-configurations $C_{m_{1}}$,$C_{m_{2}}$ which share only the bottom atom of $D(x_{0})$. This implies $m_{1}+m_{2}=n-1$ and

[TABLE]

We again observe that the new configurations are saturated, connected, satisfy $\eta(C_{m_{1}})=\eta(C_{m_{2}})=0$, and do not contain any bridging atom. (Note that the configuration obtained by removing the top corner and $x_{0}$ would contain one bridging atom.) We have removed two sides of the square $D(x_{0})$, but count one side of $D(x_{1})$ to the perimeter of $C_{m_{1}}$ and count one side of $D(x_{2})$ to the perimeter of $C_{m_{2}}$. This implies

[TABLE]

Note that $\#X_{m_{1}}^{-},\#X_{m_{2}}^{-}<k$. Thus, we can use the induction hypothesis along with (6.43)-(6.44) to obtain

[TABLE]

This concludes the proof. ∎

Appendix A Proof of the boundary energy and net charge estimates

This section is devoted to the proofs of Lemma 5.3 and Lemma 6.3.

Proof of Lemma 5.3.

For convenience, we decompose the proof into three steps.

Step 1: Preliminary estimate. Assume that $C_{n}$ is a connected ground state with no acyclic bonds. Suppose that $\{x_{1},\ldots,x_{d}\}$ are ordered such that $x_{i}\in\mathcal{N}(x_{i+1})$, $i=1,\ldots,d$. Here and in the following, we use the identification $x_{d+1}=x_{1}$ and $x_{0}=x_{d}$. For a 3-bonded atom $x_{i}$, denote by $x_{i}^{b}\in X_{n}\setminus\{x_{i-1},x_{i+1}\}$ the atom that is connected to $x_{i}$ with the third bond. In a similar fashion, for a 4-bonded atom $x_{i}$, denote by $x_{i}^{b,1},x_{i}^{b,2}\in X_{n}\setminus\{x_{i-1},x_{i+1}\}$ the two atoms which are bonded to $x_{i}$. Recall (5.1)-(5.3) and Remark 5.2. We first observe that the boundary energy can be estimated by

[TABLE]

For a $3$-bonded atom $x_{i}$, denote by $\theta_{i}^{1},\theta_{i}^{2}\in[0,2\pi]$ the angle between $x_{i-1},x_{i},x_{i}^{b}$ and the angle between $x_{i}^{b},x_{i},x_{i+1}$ respectively. For a $4$-bonded atom $x_{i}$, denote by $\theta_{i}^{1},\theta_{i}^{2},\theta_{i}^{3}\in[0,2\pi]$ the angle between $x_{i-1},x_{i},x_{i}^{b,1}$, the angle between $x_{i}^{b,1},x_{i},x_{i}^{b,2}$ and the angle between $x_{i}^{b,2},x_{i},x_{i+1}$, respectively. Finally, we define $\delta:=(\#I_{2}+2\#I_{3}+3\#I_{4})/d$ and note that $\delta\in[1,3]$ since $\#I_{2}+\#I_{3}+\#I_{4}=d$. We will prove that

[TABLE]

where the first inequality is strict if not all lengths of boundary bonds are equal to $1$. We defer the proof of (A) to Step 3. At this stage, let us only point out that the specific definition of the boundary energy including also bulk bonds, see Remark 5.2 and Fig. 10, ensures that also $\theta_{i}^{2}$, $x_{i}\in I_{4}$, gives a contribution.

Step 2: Proof of the statement. We now show that (A) implies the statement of the lemma. First, we introduce

[TABLE]

and observe that estimate (A) can be written as

[TABLE]

We obtain (5.4) by minimizing the right hand side of (A.3) with respect to $\delta$. To see this, set $\delta_{0}=2-\frac{4}{d}$. For 1\leq\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}\delta\leq\delta_{0}, we have $\alpha(\delta_{0})=\frac{\pi}{4}\leq\alpha(\delta)\leq\frac{\pi}{2}$. By $[\mathrm{vii}]$ we get V_{\mathrm{r}}(2\sin(\alpha(\delta)))=\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}0 for all 1\leq\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}\delta\leq\delta_{0}. Therefore, we find

[TABLE]

and we obtain (5.4) for 1\leq\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}\delta\leq\delta_{0}. Now for $\delta_{0}<\delta\leq 3$, we have $\alpha(\delta)<\alpha(\delta_{0})$. By $[\mathrm{v}]$ we get

[TABLE]

Then by the fact that $\sin(\theta)$ is concave for $\theta\in[0,\pi]$, $V_{\rm r}(\sqrt{2})=0$ by [vii], $V_{\mathrm{r},-}^{\prime}(\sqrt{2})<-16/(\sqrt{2}\pi)<0$ by $[\mathrm{viii}]$, $\alpha(\delta_{0})=\frac{\pi}{4}$, $\sin(\alpha(\delta_{0}))=\cos(\alpha(\delta_{0}))=\frac{1}{2}\sqrt{2}$, and $\alpha(\delta)-\alpha(\delta_{0})<0$ we derive

[TABLE]

From the previous calculation and (A.3), estimate (5.4) follows also for $\delta_{0}<\delta\leq 3$.

We now show that we have strict inequality in (5.4) if one of the conditions (5.5)-(5.7) is violated. First, if a boundary bond is not of unit length, we have strict inequality in (A) and thus also in (A.3). If (5.6) is violated, we find $\delta\neq\delta_{0}$ after a short computation. Then we get strict inequalities from (A.4) or (A.5), respectively. Finally, let use assume that (5.7) is violated. We can suppose that $\delta=\delta_{0}$ and (5.5)-(5.6) hold as otherwise the inequality in (5.4) is strict. If equality holds in (5.4), then equality also holds in (A). As $V_{\rm r}(2\sin(\alpha(\delta_{0})))=0$, this implies

[TABLE]

In view of ${\rm[vii]}$, this gives

[TABLE]

Under the assumption that (5.7) is violated, in view of (A.6), $\theta_{i}=\theta_{i}^{1}+\theta_{i}^{2}$ for $x_{i}\in I_{3}$ and $\theta_{i}=\theta_{i}^{1}+\theta_{i}^{2}+\theta_{i}^{3}$ for $x_{i}\in I_{4}$, we obtain

[TABLE]

where the first equality follows from the fact that the maximal polygon has $d$ vertices, and the last equality holds by (5.6). This is a contradiction and shows that strict inequality in (5.4) holds if (5.7) is violated.

Step 3: Proof of (A). To complete the proof, it remains to show (A) and its strict version. We follow the lines of the proof of [17, Lemma 5.1]. In the case of a $2$-bonded $x_{i}$, define $r^{1}_{i}=|x_{i}-x_{i-1}|$ and $r^{2}_{i}=|x_{i}-x_{i+1}|$. In the case of a $3$-bonded $x_{i}$, define additionally $r^{3}_{i}=|x_{i}^{b}-x_{i}|$. In the case of a $4$-bonded $x_{i}$, define $r^{3}_{i}=|x_{i}^{b,1}-x_{i}|$ and $r^{4}_{i}=|x_{i}^{b,2}-x_{i}|$. By the cosine rule we obtain

[TABLE]

where we have used the shorthand

[TABLE]

We want to prove that for every boundary atom $x_{i}$ its contribution to the energy can be controlled by the energy contribution in a modified configuration which has the same angles but unit bond lengths instead of $r_{i}^{1},r_{i}^{2}$. Recall that by $[\mathrm{viii}]$ we have for all $r\in(1,r_{0}]$

[TABLE]

Let $\theta\in[2\pi/5,8\pi/5]$, $r_{1},r_{2}\in[1,r_{0}]$. Then $\ell(\theta,r_{1},r_{2})\geq 1$. Note that $V_{\mathrm{r},+}^{\prime}(r)\leq 0$ for $r\geq 1$ and $V_{\mathrm{r},+}^{\prime}$ is monotone increasing due to the convexity assumption in $(\mathrm{v})$ on $V_{\mathrm{r}}$. Moreover, we have $\partial_{r_{1}}\ell(\theta,r_{1},r_{2})\leq 1$ by an elementary computation. We therefore obtain

[TABLE]

By integration from $1$ to $r$ in the variable $s$ we get

[TABLE]

By applying this estimate in the second as well as in the third component of $\ell(\theta,r_{1},r_{2})$ with $1$ and $r_{2}$ respectively, we derive

[TABLE]

Due to (4.1), we have that all angles in (A) lie in $[2\pi/5,8\pi/5]$ and all bond lengths lie in $[1,r_{0}]$. Now for all $2$-bonded $x_{i}$, using (A.10) with $r_{i}^{1}$, $r_{i}^{2}$, and $\theta_{i}$, we have

[TABLE]

For all $3$-bonded $x_{i}$, using (A.10) twice, we get

[TABLE]

Finally, for all $4$-bonded $x_{i}$, using (A.10) three times, we have

[TABLE]

By applying (A), (A), (A.11)-(A.13), $V_{\rm a}(1)=-1$, and $V_{\rm r}\geq 0$ we obtain

[TABLE]

For later purposes, we remark that this inequality is strict if one bond does not have unit length. This follows from the strict inequality in (A.9).

Recall $\delta=(\#I_{2}+2\#I_{3}+3\#I_{4})/d$ and note that $\ell(\theta,1,1)=2\sin(\theta/2)$ by (A.8). By using $\theta_{i}=\theta_{i}^{1}+\theta_{i}^{2}$ for $x_{i}\in I_{3}$, $\theta_{i}=\theta_{i}^{1}+\theta_{i}^{2}+\theta_{i}^{3}$ for $x_{i}\in I_{4}$, and (A) we obtain

[TABLE]

This yields the first inequality in (A). We note that this inequality is strict if one of the bonds does not have unit length since then (A) is strict. The second inequality in (A) follows by a convexity argument: since $V_{\mathrm{r}}$ is convex and non-increasing by ${\rm[v]}$, and $\sin(\theta/2)$ is concave for $\theta\in[0,2\pi]$, we observe that $\theta\mapsto V_{\mathrm{r}}(2\sin(\frac{\theta}{2}))$ is a convex function for $\theta\in[0,2\pi]$, see [17, Proof of Lemma 5.1] for details. This along with the fact that $\#I_{2}+2\#I_{3}+3\#I_{4}=\delta d$ and

[TABLE]

yields the second inequality in (A). This concludes the proof. ∎

We now proceed with the proof of Lemma 6.3. Let $C_{n}$ be a connected $q_{\mathrm{sat}}^{n}$-optimal configuration without bridging atoms. Recall the decomposition $X_{n}=X_{n}^{a,\mathrm{bulk}}\cup\partial X_{n}^{a}\cup I_{\mathrm{ac}}^{\rm ext}$ introduced in Subsection 6.2. For $2\leq k\leq 4$ we define

[TABLE]

Moreover, we set $I^{\pm}=\bigcup_{k=2}^{4}I_{k}^{\pm}$. Note that $I_{k}=I_{k}^{+}\cup I_{k}^{-}$, where $I_{k}$ is given by (6.5). By Lemma 4.1, $C_{n}$ has alternating charge distribution. Thus, $\partial C_{n}^{a}$ is a cycle of even length $d\in 2\mathbb{N}$ and, by (4.2) there holds $\#I^{-}=\#I^{+}=d/2$. We denote the interior angle of the polygon $\partial X_{n}^{a}$ at $x_{i}$ by $\theta_{i}$. We start with some preliminary properties.

Lemma A.1**.**

Let $n\geq 6$ and let $C_{n}$ be a connected $q_{\mathrm{sat}}^{n}$-optimal configuration. Then

• (i)

$\#I_{4}^{+}=0$.

• (ii)

Let $m\geq 4$ be such that $\#I_{2}+2\#I_{3}+3\#I_{4}\leq 2d-m$. Then $2\#I_{2}^{-}+\#I_{3}^{-}\geq m+\#I_{3}^{+}$.

Proof.

Proof of (i). Assume by contradiction that there exists $x_{i}\in I_{4}^{+}$. Since $C_{n}$ is a $q_{\mathrm{sat}}^{n}$-optimal configuration, all atoms of charge $-1$ are $4$-bonded with bond angles $\frac{\pi}{2}$, see Lemma 4.3(a)(i),(iv). Hence, $x_{i}$ is contained in four squares and thus not part of $\partial X_{n}^{a}$.

Proof of (ii). As each atom is at most $4$-bonded, see (4.2), we have $\#I_{2}+\#I_{3}+\#I_{4}=d$. This along with (i) and the assumption of (ii) shows

[TABLE]

This implies

[TABLE]

Since there holds $\#I^{-}=\#I^{+}$ due to alternating charge distribution and $\#I_{4}^{+}=0$ by property (i), property (ii) follows. ∎

Proof of Lemma 6.3(a).

By Theorem 2.6, Lemma 4.3(a)(i),(ii), and Remark 2.2 there holds $\mathcal{E}(C_{n})=-b$. Therefore, $\theta_{i}\geq(k-1)\frac{\pi}{2}\text{ if }x_{i}\in I_{k}$ by Lemma 4.2(b). This implies

[TABLE]

where in the last step we used that the sum of the interior angles equals $\pi(d-2)$. This shows (6.6).

We now show the additional statement. To this end, suppose that $\#\mathcal{A}(X_{n})\geq\#\mathcal{A}(X_{n}^{\mathrm{a},\mathrm{bulk}})+2$. We claim that

[TABLE]

We defer the proof of (A.15) and first conclude the proof of the statement. The fact that the sum of the interior angles equals $\pi(d-2)$, (A.15), and $\theta_{i}\geq(k-1)\frac{\pi}{2}$ for $x_{i}\in I_{k}$ imply

[TABLE]

Since the left and the right hand side are integers, we find $2+\#I_{2}+2\#I_{3}+3\#I_{4}\leq 2d-4$. This shows that $m\geq 6$ in (6.6). It remains to prove (A.15). As a preparation, we first show that

[TABLE]

By Lemma 4.3(a)(i),(iv),(v) we observe that $x$ is $2$-bonded and has charge $+1$. Denote the two neighbors of x\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}\in\mathcal{A}(X_{n}) by $z_{1}$ and $z_{2}$. Since $z_{1}$ and $z_{2}$ have charge $-1$, they are $4$-bonded. Also note that $(\mathcal{N}(z_{1})\cap\mathcal{N}(z_{2}))\setminus\{x\}=\emptyset$. In fact, if there was an atom y\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}\in(\mathcal{N}(z_{1})\cap\mathcal{N}(z_{2}))\setminus\{x\}, the four atoms $z_{1}$, $x$, $z_{2}$, and $y$ would form a square. Due to Lemma 4.2(b), this square is regular which contradicts the fact that $x\in\mathcal{A}(X_{n})$. Now, if we had that a bond-angle at $x$ does not lie in $[5\pi/6,7\pi/6]$, it is elementary to see that $\mathrm{dist}(\mathcal{N}(z_{1})\setminus\{x\},\mathcal{N}(z_{2})\setminus\{x\})<\sqrt{2}$. This would contradict the fact that $C_{n}$ is repulsion-free.

We now show that (A.15) holds. Recall the assumption $\#\mathcal{A}(X_{n})\geq\#\mathcal{A}(X_{n}^{\mathrm{a},\mathrm{bulk}})+2$. We consider the two cases (a) $\mathcal{A}(X_{n})\cap\partial X_{n}^{\mathrm{a}}\neq\emptyset$ and (b) $\mathcal{A}(X_{n})\cap\partial X_{n}^{\mathrm{a}}=\emptyset$.

Proof of Case $\mathrm{(a)}$. Observe that $\#(\partial X_{n}^{a}\cap\mathcal{A}(X_{n}))\geq 2$ by Lemma 4.2(a). Additionally, by (A.16) we get $\partial X_{n}^{a}\cap\mathcal{A}(X_{n})\subset I_{2}^{+}$ and $\theta_{i}\geq 5\pi/6$ for $x_{i}\in\partial X_{n}^{a}\cap\mathcal{A}(X_{n})$. Hence, (A.15) holds.

Proof of Case $\mathrm{(b)}$. We first show that for each $x\in\mathcal{A}(X_{n})\setminus\mathcal{A}(X_{n}^{\mathrm{a},\mathrm{bulk}})$ we can find

[TABLE]

By (A.16) we get that $x\in X_{n}^{+}$, that it is $2$-bonded in $X_{n}$ and, since $x\in\mathcal{A}(X_{n})\setminus\mathcal{A}(X_{n}^{\mathrm{a},\mathrm{bulk}})$, it is $1$-bonded in $X_{n}^{\mathrm{a},\mathrm{bulk}}$. Thus, there exists exactly one atom in $\mathcal{N}(x)\cap\partial X_{n}^{\mathrm{a}}$, denoted by $x_{i}$. By $x_{i-1},x_{i+1}$ we denote the two neighbors of $x_{i}$ in $\partial X_{n}^{\mathrm{a}}$. At least one of these two atoms, say without restriction $x_{i-1}$, satisfies $|x_{i-1}-x|=\sqrt{2}$, see Fig. 13. We will identify $x_{i-1}$ as the atom $z_{x}$ in (A.17). Recall the assumption $\partial X_{n}^{a}\cap\mathcal{A}(X_{n})=\emptyset$, which implies that the bond-angles at $x_{i-1}$ lie in $\{\pi/2,\pi,3\pi/2\}$. Let $y\in\mathcal{N}(x_{i-1})\setminus\{x_{i}\}$. Suppose $y$, $x_{i}$, and $x_{i-1}$ would form an angle in $\{\pi/2,3\pi/2\}$, i.e., $|y-x_{i}|=\sqrt{2}$. Then either (i) $x_{i-1},y,x,x_{i}$ or (ii) $x_{i-1},y,z,x_{i}$ form a square, where $z\in\mathcal{N}(x_{i})\setminus\{x,x_{i-1}\}$, see Fig. 13. In view of (A.16), case (i) contradicts $x\in\mathcal{A}(X_{n})$, and case (ii) contradicts $x_{i-1}\in\partial X_{n}^{\mathrm{a}}$. This shows that $y,x_{i-1},x_{i}$ form an angle $\pi$, and this also yields that $x_{i-1}\in I^{+}_{2}$. We denote $x_{i-1}$ by $z_{x}$.

Since $\#\mathcal{A}(X_{n})\geq\#\mathcal{A}(X_{n}^{\mathrm{a},\mathrm{bulk}})+2$ by assumption, we find $x\neq y\in\mathcal{A}(X_{n})\setminus\mathcal{A}(X_{n}^{\mathrm{a},\mathrm{bulk}})$. We now show that $z_{x}\neq z_{y}$, where $z_{x}$ and $z_{y}$ are the atoms identified in (A.17). This will imply (A.15). Assume by contradiction that $z_{x}=z_{y}$. Denote $z_{x}=z_{y}$ by $x_{i}\in I_{2}^{+}$ and recall $\theta_{i}=\pi$. Denote its two neighbors by $x_{i-1},x_{i+1}$, and recall that they are $4$-bonded, in particular bonded to $x$ or $y$, respectively. Consider the unique atoms $w_{x}\in\mathcal{N}(x)\setminus\partial X_{n}^{\mathrm{a}}$ and $w_{y}\in\mathcal{N}(y)\setminus\partial X_{n}^{\mathrm{a}}$. Observe that they are $4$-bonded. First, note that \mathcal{N}(w_{x}\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0})\cap\mathcal{N}(w_{y}\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0})=\emptyset since the contrary would imply that $x,y$ would be contained in an octagon with bond angles $\pi,\pi/2$ respectively, contradicting the fact that $x,y\in\mathcal{A}(X_{n})$. Since by (A.16) the bond angles at $x,y\in\mathcal{A}(X_{n})$ lie in $[5\pi/6,7\pi/6]$, an elementary argument shows that \mathrm{dist}(\mathcal{N}(w_{x}\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}),\mathcal{N}(w_{y}\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}))<\sqrt{2}, see Fig. 13. This contradicts the fact that $C_{n}$ is repulsion-free and concludes the proof of case (b). ∎

Proof of Lemma 6.3(b).

We proceed in several steps. We first introduce an auxiliary configuration $C_{n-d}^{\prime}$ which arises from $C_{n}$ by removing the boundary and adding again the $3$-bonded boundary atoms of charge $+1$ (Step 1). We collect some properties about the net charge of $C_{n-d}^{\prime}$ whose proofs are deferred to Steps 3 and 4. In Step 2 we define the configuration $C_{n-d-m}$ by suitably adding atoms of charge $+1$ to $C_{n-d}^{\prime}$, and we prove the statement of the lemma.

Step 1: An auxiliary configuration. We recall the decomposition $X_{n}=X_{n}^{a,\mathrm{bulk}}\cup\partial X_{n}^{a}\cup I_{\mathrm{ac}}^{\rm ext}$ introduced in Subsection 6.2. By Lemma 4.2(a) and Lemma 4.1, $\partial C_{n}^{a}$ is a cycle of even length $d$ with alternating charge distribution, and therefore $\mathcal{Q}(\partial C_{n}^{a})=0$. Moreover, all exterior acyclic bonds have charge $+1$ by Lemma 6.1(b). This implies

[TABLE]

We introduce an auxiliary configuration $C_{n-d}^{\prime}$ by

[TABLE]

In a similar fashion, we denote the atomic positions by $X_{n-d}^{\prime}$. It is clear that $\mathcal{Q}(C_{n-d}^{\prime})=\mathcal{Q}(C_{n}^{a,\mathrm{bulk}})+\#I_{3}^{+}$. By construction there holds $\mathcal{A}(X_{n}^{\mathrm{a},\mathrm{bulk}})=\mathcal{A}(X_{n-d}^{\prime})$ since all the atoms added from $I_{3}^{+}$ are $1$-bonded in $C^{\prime}_{n-d}$, more precisely, bonded to an atom which does not lie in $\mathcal{A}(X_{n})$. By (A.18) we get

[TABLE]

The goal is to estimate the contributions of the different terms in (A.20) separately: we will show that

[TABLE]

where $m\geq 4$ is given in (6.6). We defer the proof of (A.21)–(A.22) to Steps 3 and 4 below and proceed to show the statement of the lemma.

Step 2: Proof of the statement: First, (A.22) implies $\#I_{\mathrm{ac}}^{\rm ext}\geq\#I_{3}^{+}+m\geq m$. Next, we construct the configuration $C_{n-d-m}$. A difficulty arises from the fact that the auxiliary configuration $C_{n-d}^{\prime}$ possibly does not satisfy the net charge equality (6.7) and therefore we need to add atoms of charge $+1$.

We set $k_{m}:=\#I_{\mathrm{ac}}^{\rm ext}-\#I_{3}^{+}-m$ and observe that $k_{m}\geq 0$ by (A.22). We define

[TABLE]

where the positions are chosen such that $\min\{\mathrm{dist}(z_{i},X_{n}),\mathrm{dist}(z_{i},z_{j}):i\neq j\}\geq\sqrt{2}$. In view of (A.19), we observe that $C_{n-d-m}$ indeed consists of

[TABLE]

atoms. Since the added atoms are all [math]-bonded, in view of (A.19), $X_{n-d-m}$ is a subset of $X_{n}$ up to [math]-bonded atoms. Moreover, we clearly have $\mathcal{A}(X_{n-d-m})=\mathcal{A}(X_{n-d}^{\prime})$ and thus $\mathcal{A}(X_{n-d-m})=\mathcal{A}(X_{n}^{\mathrm{a},\mathrm{bulk}})$. As we have added $k_{m}$ [math]-bonded atoms of charge $+1$ to $C_{n-d}^{\prime}$, $C_{n-d-m}$ still satisfies Lemma 4.3(a)(i),(ii), and thus by Lemma 4.3(b) there holds equality in (4.6). Therefore, $C_{n-d-m}$ is optimal and satisfies $\mathcal{Q}(C_{n-d-m})\geq q_{\mathrm{sat}}^{n-d-m}$ by Lemma 4.3(a)(iii). Moreover, we clearly have $\mathcal{Q}(C_{n-d-m})=\mathcal{Q}(C_{n-d}^{\prime})+k_{m}$. By (A.20) and the definition $k_{m}=\#I_{\mathrm{ac}}^{\rm ext}-\#I_{3}^{+}-m$ we observe

[TABLE]

This shows (6.7) and concludes the proof of the statement. To conclude the proof, it thus remains to show (A.21) and (A.22).

Step 3: Proof of (A.21). We first recall that $C_{n}^{a,\mathrm{bulk}}$ consists of $n-d-\#I_{\mathrm{ac}}^{\rm ext}$ atoms and thus $C_{n-d}^{\prime}$ consists of $\#X_{n-d}^{\prime}:=n-d-(\#I_{\mathrm{ac}}^{\rm ext}-\#I^{+}_{3})$ atoms. Thus, in view of Lemma 4.3(a)(iii) it suffices to prove

[TABLE]

By Lemma 4.3(a)(i),(ii) we find that $C_{n}$ is repulsion-free with unit bond lengths and all atoms of charge $-1$ are $4$-bonded. Since $C_{n-d}^{\prime}\subset C_{n}$, also $C_{n-d}^{\prime}$ is repulsion-free with unit bond lengths. We now show that each atom of $C_{n-d}^{\prime}$ with charge $-1$ is $4$-bonded. Let $x_{i}\in X_{n-d}^{\prime}$ with $q_{i}=-1$. We show that \mathcal{N}(x_{i})\cap(X_{n}\setminus X_{n-d}^{\prime}\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0})=\emptyset. In fact, in view of (A.19) and $q_{i}=-1$, we have

[TABLE]

We first recall that $\#I_{4}^{+}=\emptyset$, see Lemma A.1(i). Moreover, atoms in \mathcal{N}(x_{i})\cap(X_{n}\setminus X_{n-d}^{\prime}\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}) cannot lie in $I_{2}$ since atoms in $I_{2}$ are only bonded to other atoms in $\partial X_{n}^{a}$ or to atoms in $I^{\rm ext}_{\mathrm{ac}}$, see (6.5). Then (A.24) shows \mathcal{N}(x_{i})\cap(X_{n}\setminus X_{n-d}^{\prime}\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0})=\emptyset.

This implies that the neighborhood of $x_{i}$ in $X_{n-d}^{\prime}$ coincides with the one in $X_{n}$, and thus $x_{i}$ is still $4$-bonded, cf. Lemma 4.3(a)(i). Consequently, we have shown that $C_{n-d}^{\prime}$ satisfies Lemma 4.3(a)(i),(ii). Then (A.23) follows from Lemma 4.3(b).

Step 4: Proof of (A.22). Since $n\geq 6$ and $C_{n}$ does not contain bridging atoms, we can apply Lemma 6.1(b). We get that $I_{\rm ac}^{\rm ext}=I_{\rm ac}^{\rm ext}\cap X_{n}^{+}$ and that each atom in $I_{\rm ac}^{\rm ext}$ is $1$-bonded. For all $x_{i}\in I_{3}^{-}$ there exists exactly one $x_{j}\in\mathcal{N}(x_{i})\cap I_{\mathrm{ac}}^{\rm ext}$, whereas for $x_{i}\in I_{2}^{-}$ there exist exactly two $x_{j},x_{k}\in\mathcal{N}(x_{i})\cap I_{\mathrm{ac}}^{\rm ext}$. As atoms in $I_{\rm ac}^{\rm ext}$ are $1$-bonded, it is clear that no atom in $I_{\rm ac}^{\rm ext}$ is bonded to two different atoms in $I_{2}^{-}\cup I_{3}^{-}$. This yields

[TABLE]

Then (A.22) follows from (6.6) and Lemma A.1(ii). ∎

Acknowledgements

This work was funded by the Deutsche Forschungsgemeinschaft (DFG, German Research Foundation) under Germany’s Excellence Strategy EXC 2044 -390685587, Mathematics Münster: Dynamics–Geometry–Structure.