Introduction
A Delone set in R d \mathbb{R}^{d} R d rectifiable if it is bi-Lipschitz (BL) equivalent to Z d \mathbb{Z}^{d} Z d Z d \mathbb{Z}^{d} Z d [9 ] , motivated by quasi-isometries of groups, and Hillel Furstenberg, motivated by Kakutani equivalence of R d \mathbb{R}^{d} R d R d \mathbb{R}^{d} R d [17 Section 3] for the history). For d = 1 d=1 d = 1 [5 ] and, independently, Curtis McMullen [15 ] showed that for d ≥ 2 d \ge2 d ≥ 2 ρ \rho ρ ρ \rho ρ
Sets produced in this way are deliberately irregular, and much of the work since then has concerned Delone sets coming from aperiodic order, which are described by a finite amount of data. Every Delone set arising from a primitive substitution tiling is rectifiable [18 ] , and so is every linearly repetitive Delone set [2 ] . On the other hand, Cortez and Navas [7 ] constructed repetitive subsets of Z 2 \mathbb{Z}^{2} Z 2 [4 ] found such sets with close to optimal repetitivity functions. In both cases non-rectifiability comes from a version of the Burago–Kleiner encoding of a non-realisable density.
The question of whether every cut-and-project set (also called a model set) is BL equivalent to a lattice has remained open. Burago and Kleiner [6, Introduction, 2020 Mathematics Subject Classification . 52C23; 51F30; 11J71.
Key words and phrases. Cut-and-project sets, model sets, Delone sets, separated nets, bi-Lipschitz equivalence. Problem 3] considered the following construction: let P ⊆ R n P \subseteq\mathbb{R}^n P ⊆ R n B ⊆ R n B \subseteq\mathbb{R}^n B ⊆ R n P P P z ∈ Z n z \in\mathbb{Z}^n z ∈ Z n ( z + B ) ∩ P ≠ ∅ (z+B) \cap P \ne\varnothing ( z + B ) ∩ P = ∅ Z 2 \mathbb{Z}^2 Z 2 [6 ] settles the case n = 3 n=3 n = 3 [6 ] ), under the condition that the orthogonal projection of B B B P P P [6 ] and what follows from the machinery of the proof: their theorem is stated for arbitrary bounded B B B Z 3 \mathbb{Z}^3 Z 3 P P P B B B P P P B B B Z 3 \mathbb{Z}^3 Z 3
Haynes, Kelly, and Weiss [12 Theorem 1.1] showed, in all dimensions, that for almost every subspace the answer to the BL equivalence question is yes, whenever the boundary of the window has upper Minkowski dimension smaller than the dimension of the internal space. Later, Koivusalo and Lagacé [13 ] found sharp bounds for the discrepancy of cut-and-project sets, which indicate that this is close to the largest class to which the known sufficient criteria for rectifiability apply.
The BL problem for cut and project sets appears again, for windows which are bounded, have nonempty interior, and have boundary of measure zero, as Problem 2.8.2 in [1 ] , where it is pointed out that if the window is only required to be bounded then it is easy to produce non-rectifiable examples. Most recently it is listed by Smilansky and Solomon as Open Problem 7.2 in their survey [17 ] . To our knowledge, no example of a non-rectifiable cut-and-project set with window satisfying these conditions was previously known.
To state our main result, for α = ( α 1 , α 2 ) ∈ R 2 \alpha=(\alpha_1,\alpha_2)\in\mathbb{R}^2 α = ( α 1 , α 2 ) ∈ R 2
V α = { z ∈ R 3 : z 3 = α 1 z 1 + α 2 z 2 } , n α = ( α 1 , α 2 , − 1 ) , V_\alpha=\{z\in\mathbb{R}^3:z_3=\alpha_1z_1+\alpha_2z_2\},\qquad n_\alpha=(\alpha_1,\alpha_2,-1), V α = { z ∈ R 3 : z 3 = α 1 z 1 + α 2 z 2 } , n α = ( α 1 , α 2 , − 1 ) , and let π α \pi_\alpha π α R 3 \mathbb{R}^3 R 3 V α V_\alpha V α W ⊆ R W\subseteq\mathbb{R} W ⊆ R
Y α ( W ) = { π α ( z ) : z ∈ Z 3 , ⟨ z , n α ⟩ ∈ W } . (1) Y_\alpha(W)=\{\pi_\alpha(z):z\in\mathbb{Z}^3,\ \langle z,n_\alpha\rangle\in W\}. \tag*{(1)} Y α ( W ) = { π α ( z ) : z ∈ Z 3 , ⟨ z , n α ⟩ ∈ W } . (1) This is the cut-and-project set with physical space V α V_\alpha V α V α ⊥ V_\alpha^\perp V α ⊥ R \mathbb{R} R w ↦ ⟨ w , n α ⟩ w\mapsto\langle w,n_\alpha\rangle w ↦ ⟨ w , n α ⟩ Z 3 \mathbb{Z}^3 Z 3 W W W
Theorem 1. There exist α 1 , α 2 ∈ R \alpha_1,\alpha_2\in\mathbb{R} α 1 , α 2 ∈ R 1 , α 1 1,\alpha_1 1 , α 1 α 2 \alpha_2 α 2 Q \mathbb{Q} Q W ⊆ R W\subseteq\mathbb{R} W ⊆ R
(i) There is no z ∈ Z 3 z\in\mathbb{Z}^3 z ∈ Z 3 ⟨ z , n α ⟩ ∈ ∂ W \langle z,n_\alpha\rangle\in\partial W ⟨ z , n α ⟩ ∈ ∂ W
(ii) The set Y α ( W ) Y_{\alpha}(W) Y α ( W ) V α V_{\alpha} V α V α V_{\alpha} V α
Since 1 1 1 α 1 \alpha_1 α 1 α 2 \alpha_2 α 2 Q \mathbb{Q} Q V α V_{\alpha} V α R 3 / Z 3 \mathbb{R}^3/\mathbb{Z}^3 R 3 / Z 3 π α \pi_{\alpha} π α Z 3 \mathbb{Z}^3 Z 3 α 1 \alpha_1 α 1 Z 3 \mathbb{Z}^3 Z 3 Z α 1 + Z α 2 + Z \mathbb{Z}\alpha_1+\mathbb{Z}\alpha_2+\mathbb{Z} Z α 1 + Z α 2 + Z R \mathbb{R} R [3 ] , Y α ( W ) Y_{\alpha}(W) Y α ( W ) [1 Problem 2.8.2] , so Theorem 1 gives a negative answer to that problem when the ambient space has dimension 3 and the physical space has dimension 2. It also answers Open Problem 7.2 of [17 ] . To see that it answers Burago and Kleiner’s question for n = 3 n=3 n = 3 B = { − s n α / ∣ n α ∣ 2 : s ∈ W } + D B=\{-s n_{\alpha}/\lvert n_{\alpha}\rvert^2:s\in W\}+D B = { − s n α / ∣ n α ∣ 2 : s ∈ W } + D D D D V α V_{\alpha} V α z ∈ Z 3 z\in\mathbb{Z}^3 z ∈ Z 3 ( z + B ) ∩ V α ≠ ∅ (z+B)\cap V_{\alpha}\ne\varnothing ( z + B ) ∩ V α = ∅ ⟨ z , n α ⟩ ∈ W \langle z,n_{\alpha}\rangle\in W ⟨ z , n α ⟩ ∈ W B B B Y α ( W ) Y_{\alpha}(W) Y α ( W ) V α ⊥ V_{\alpha}^{\perp} V α ⊥ R \mathbb{R} R B B B V α V_{\alpha} V α { − s : s ∈ W } \{-s:s\in W\} { − s : s ∈ W } § 6 , at the beginning of the proof of part (ii) of Theorem 3 ).
Our construction works for every irrational α 1 \alpha_1 α 1 α 1 = 2 \alpha_1=\sqrt{2} α 1 = 2 [6 ] (in our notation, condition (1.1) concerns α 1 \alpha_1 α 1 [6 Theorem 1.2] imposes no condition on α 2 \alpha_2 α 2 Theorem 1 also shows that the property of B B B [6 Theorem 1.2] , that its projection to the normal line of P P P
The window in Theorem 1 cannot be replaced by an interval. In a companion paper [11 ] we show that if W W W α 1 \alpha_1 α 1 α 2 \alpha_2 α 2 Y α ( W ) Y_{\alpha}(W) Y α ( W ) Z 2 \mathbb{Z}^2 Z 2 α 1 \alpha_1 α 1 [6 ] , so any such example would have to use an α 1 \alpha_1 α 1
Theorem 1 will be deduced in § 2 from Theorem 3 , which is a statement about the set of integer points m ∈ Z 2 m\in\mathbb{Z}^2 m ∈ Z 2 α 1 m 1 + α 2 m 2 \alpha_1m_1+\alpha_2m_2 α 1 m 1 + α 2 m 2 W + Z W+\mathbb{Z} W + Z Z 2 \mathbb{Z}^2 Z 2 [10 ] on codimension one cut-and-project sets. There, after moving the window along the physical space to a segment parallel to a coordinate axis, the problem was reduced to one about a set of integer points of exactly this form, which was then analyzed one row at a time, through the return times of an irrational rotation to the window.
Our set also has a property which bears on a different question. Following [4 ] (see also [17 Section 3.3] ), a measurable function ρ : [ 0 , 1 ] 2 → ( 0 , ∞ ) \rho: [0,1]^2 \to(0,\infty) ρ : [ 0 , 1 ] 2 → ( 0 , ∞ ) non-realisable if there is no map Φ : [ 0 , 1 ] 2 → R 2 \Phi: [0,1]^2 \to\mathbb{R}^2 Φ : [ 0 , 1 ] 2 → R 2 ∣ det D Φ ∣ = ρ |\det D\Phi|=\rho ∣ det D Φ∣ = ρ Λ ⊆ R 2 \Lambda\subseteq\mathbb{R}^2 Λ ⊆ R 2 encodes ρ \rho ρ r k → ∞ r_k \to\infty r k → ∞ Q k ⊆ R 2 Q_k \subseteq\mathbb{R}^2 Q k ⊆ R 2 r k r_k r k φ k : [ 0 , 1 ] 2 → Q k \varphi_k : [0,1]^2 \to Q_k φ k : [ 0 , 1 ] 2 → Q k
μ k = 1 r k 2 ∑ x ∈ Λ ∩ Q k δ φ k − 1 ( x ) \mu_k=\frac{1}{r_k^2}\sum_{x\in\Lambda\cap Q_k}\delta_{\varphi_k^{-1}(x)} μ k = r k 2 1 x ∈ Λ ∩ Q k ∑ δ φ k − 1 ( x ) converge weakly to ρ ( u ) d u \rho(u)\,du ρ ( u ) d u [5 Lemma 2.1] ; see [17 Proposition 3.13] ), and this is the mechanism behind all of the examples of non-rectifiable Delone sets mentioned above. Bhat and Dymond [4 ] asked whether the converse holds, that is, whether every non-rectifiable Delone set encodes a non-realisable density, and this question is recorded as Open Problem 3.14 in [17 ] . In what follows, the area of a square in V α V_\alpha V α R 3 \mathbb{R}^3 R 3 V α V_\alpha V α R 2 \mathbb{R}^2 R 2
Corollary 2. Let α \alpha α W W W Theorem 1 , and let δ = ∣ W ∣ / ∣ n α ∣ \delta=|W|/|\mathfrak{n}_\alpha| δ = ∣ W ∣/∣ n α ∣ ϵ > 0 \epsilon>0 ϵ > 0 R ϵ > 0 R_\epsilon>0 R ϵ > 0 Q ⊆ V α Q\subseteq V_\alpha Q ⊆ V α R ≥ R ϵ R\geq R_\epsilon R ≥ R ϵ
∣ # ( Y α ( W ) ∩ Q ) − δ R 2 ∣ ≤ ϵ R 2 . \left|\#\left(Y_\alpha(W)\cap Q\right)-\delta R^2\right|\leq\epsilon R^2. # ( Y α ( W ) ∩ Q ) − δ R 2 ≤ ϵ R 2 . Consequently, up to equality almost everywhere, the only density encoded by Y α ( W ) Y_\alpha(W) Y α ( W ) δ \delta δ Y α ( W ) Y_\alpha(W) Y α ( W )
The constant function δ \delta δ u ↦ δ 1 / 2 u u\mapsto\delta^{1/2}u u ↦ δ 1/2 u Corollary 2 therefore gives a negative answer to the question of Bhat and Dymond. We point out that Burago and Kleiner remark in [6 Remark 1.5] , without giving details, that the technique of [5 ] can be used to produce separated nets which have uniform asymptotic density but are not BL equivalent to Z 2 \mathbb{Z}^2 Z 2 Corollary 2 , and by the second part of the proof of that corollary a net with this property does not encode any non-realisable density. To our knowledge, however, no proof of their remark has appeared, and the question is recorded as open in [17 ] . The uniformity of the density is a general feature of regular model sets (cf. [16 ] ), and in our setting it follows from an elementary counting lemma along rows of lattice points (Lemma 10), which we need anyway for the proof of Theorem 1 .
The first main ingredient of the proof is a finite-scale version of the argument of Burago and Kleiner [5 Section 3] based on stretched pairs of points (Proposition 7 ). Some modification of their argument is needed, since it obtains a contradiction from a limit of rescaled copies of the set, and by Corollary 2 every such limit of our set has constant density (this is presumably the situation which Burago and Kleiner have in mind in [6 Remark 1.5] ). For each L ≥ 1 L\geq1 L ≥ 1 c > 0 c>0 c > 0 L L L c c c L L L Z 2 \mathbb{Z}^2 Z 2 c c c contrast . We deduce this from a statement about BL homeomorphisms of the plane, in which the hypothesis of Burago and Kleiner on the Jacobian is replaced by a lower bound for the differences between the areas of the images of adjacent squares.
The second ingredient is the window. Since the internal space is one-dimensional, windows of a particular form correspond to simple patterns in the physical space. For an integer Q ≥ 1 Q \ge1 Q ≥ 1 x ∈ R / Z x \in\mathbb{R}/\mathbb{Z} x ∈ R / Z ∥ Q x − γ ∥ < w / 2 \lVert Qx-\gamma\rVert< w/2 ∥ Q x − γ ∥ < w /2 Q Q Q R 2 \mathbb{R}^2 R 2 Q α − μ Q\alpha-\mu Q α − μ μ ∈ Z 2 \mu\in\mathbb{Z}^2 μ ∈ Z 2 Lemma 11 ). We fix α 1 \alpha_1 α 1 α 2 \alpha_2 α 2 Proposition 7 for any given L L L ∣ A L ∣ |A_L| ∣ A L ∣ A L A_L A L A L A_L A L L → ∞ L \to\infty L → ∞ L L L L L L α 1 \alpha_1 α 1 α 2 \alpha_2 α 2 ∣ A L ∣ |A_L| ∣ A L ∣ α \alpha α ⟨ α , m ⟩ \langle\alpha,m\rangle ⟨ α , m ⟩ m ∈ Z 2 m \in\mathbb{Z}^2 m ∈ Z 2 Theorem 1 .
We have restricted our attention to planes in R 3 \mathbb{R}^3 R 3 d ≥ 2 d \ge2 d ≥ 2 α 1 \alpha_1 α 1 α 1 = 2 \alpha_1 = \sqrt{2} α 1 = 2 α 2 \alpha_2 α 2 V α V_\alpha V α [12 ] , so that V α V_\alpha V α [12 Theorem 1.1] applies to every Diophantine subspace, and it follows that the boundary of our window necessarily has upper Minkowski dimension equal to 1. This can also be proven directly from our construction.
This paper is organized as follows: In § 2 we reduce Theorem 1 to a statement about a subset of Z 2 \mathbb{Z}^2 Z 2 § 3 we prove the stretched-pair lemma for BL homeomorphisms and use it to show that no BL homeomorphism of the plane can match the area pattern of our configurations. In § 4 we transfer this conclusion to Delone sets, proving Proposition 7 . § 5 contains the counting lemmas, from which we deduce Corollary 2 , and the construction of strip families with prescribed normals. Finally, in § 6 we construct the window and complete the proofs.
Notation: For x , y ∈ R n x,y\in\mathbb{R}^n x , y ∈ R n ⟨ x , y ⟩ \langle x,y\rangle ⟨ x , y ⟩ ∣ x ∣ |x| ∣ x ∣ x ∈ R x\in\mathbb{R} x ∈ R ∥ x ∥ \|x\| ∥ x ∥ x x x A A A ∣ A ∣ |A| ∣ A ∣ A A A # A \#A # A f f f L L L
L − 1 ∣ x − y ∣ ≤ ∣ f ( x ) − f ( y ) ∣ ≤ L ∣ x − y ∣ L^{-1}|x-y|\leq|f(x)-f(y)|\leq L|x-y| L − 1 ∣ x − y ∣ ≤ ∣ f ( x ) − f ( y ) ∣ ≤ L ∣ x − y ∣ for all x x x y y y Λ ⊆ R 2 \Lambda\subseteq\mathbb{R}^2 Λ ⊆ R 2 inf { ∣ x − y ∣ : x , y ∈ Λ , x ≠ y } > 0 \inf\{|x-y|:x,y\in\Lambda,\ x\ne y\}>0 inf { ∣ x − y ∣ : x , y ∈ Λ , x = y } > 0 sup u ∈ R 2 inf x ∈ Λ ∣ u − x ∣ \sup_{u\in\mathbb{R}^2}\inf_{x\in\Lambda}|u-x| sup u ∈ R 2 inf x ∈ Λ ∣ u − x ∣ x ∈ R x\in\mathbb{R} x ∈ R U ⊆ R / Z U\subseteq\mathbb{R}/\mathbb{Z} U ⊆ R / Z x ∈ U x\in U x ∈ U x x x R / Z \mathbb{R}/\mathbb{Z} R / Z U U U
Disclosure of AI use: AI was used in this research, as well as to assist in the preparation of this manuscript. However, the final details of all proofs, as well as every statement, citation, and claim in the paper, has been checked by the author.
Reduction to a problem about subsets of Z 2 \mathbb{Z}^2 Z 2
For α ∈ R 2 \boldsymbol{\alpha}\in\mathbb{R}^2 α ∈ R 2 W ⊆ R W\subseteq\mathbb{R} W ⊆ R
Λ α ( W ) = { m ∈ Z 2 : ⟨ α , m ⟩ ∈ W + Z } . (2) \Lambda_{\boldsymbol{\alpha}}(W)=\{m\in\mathbb{Z}^2:\langle\boldsymbol{\alpha},m\rangle\in W+\mathbb{Z}\}. \tag*{(2)} Λ α ( W ) = { m ∈ Z 2 : ⟨ α , m ⟩ ∈ W + Z } . (2) It is helpful to picture this set one row at a time. On the horizontal line m 2 = k m_2=k m 2 = k Λ α ( W ) \Lambda_{\boldsymbol{\alpha}}(W) Λ α ( W ) m 1 m_1 m 1 m 1 α 1 + k α 2 m_1\alpha_1+k\alpha_2 m 1 α 1 + k α 2 k α 2 k\alpha_2 k α 2 α 1 \alpha_1 α 1 W W W W W W Y α ( W ) Y_{\boldsymbol{\alpha}}(W) Y α ( W ) Λ α ( W ) \Lambda_{\boldsymbol{\alpha}}(W) Λ α ( W ) Theorem 1 will then follow from the following result, which is what we will actually prove.
Theorem 3. There exist α 1 , α 2 ∈ R \alpha_1,\alpha_2\in\mathbb{R} α 1 , α 2 ∈ R 1 , α 1 1,\alpha_1 1 , α 1 α 2 \alpha_2 α 2 Q \mathbb{Q} Q W ⊆ R W\subseteq\mathbb{R} W ⊆ R
(i) For every m ∈ Z 2 m\in\mathbb{Z}^2 m ∈ Z 2 ⟨ α , m ⟩ ∉ ∂ W + Z \langle\boldsymbol{\alpha},m\rangle\notin\partial W+\mathbb{Z} ⟨ α , m ⟩ ∈ / ∂ W + Z
(ii) The set Λ α ( W ) \Lambda_{\boldsymbol{\alpha}}(W) Λ α ( W ) Z 2 \mathbb{Z}^2 Z 2
The proof of Theorem 3 occupies § 3 -§ 6 . To compare Y α ( W ) Y_{\boldsymbol{\alpha}}(W) Y α ( W ) Λ α ( W ) \Lambda_{\boldsymbol{\alpha}}(W) Λ α ( W ) pr : V α → R 2 \operatorname{pr}:V_{\boldsymbol{\alpha}}\to\mathbb{R}^2 pr : V α → R 2 pr ( z ) = ( z 1 , z 2 ) \operatorname{pr}(z)=(z_1,z_2) pr ( z ) = ( z 1 , z 2 ) V α V_{\boldsymbol{\alpha}} V α ( z 1 , z 2 ) ↦ α 1 z 1 + α 2 z 2 (z_1,z_2)\mapsto\alpha_1z_1+\alpha_2z_2 ( z 1 , z 2 ) ↦ α 1 z 1 + α 2 z 2 pr \operatorname{pr} pr 1 / ∣ n α ∣ 1/|\mathbf{n}_{\boldsymbol{\alpha}}| 1/∣ n α ∣
Lemma 4. Let α ∈ R 2 \alpha\in\mathbb{R}^2 α ∈ R 2 W ⊆ R W \subseteq\mathbb{R} W ⊆ R s W = sup { ∣ s ∣ : s ∈ W } s_W=\sup\{|s|:s\in W\} s W = sup { ∣ s ∣ : s ∈ W } Ψ : Y α ( W ) → Λ α ( W ) \Psi:Y_\alpha(W)\to\Lambda_\alpha(W) Ψ : Y α ( W ) → Λ α ( W )
∣ pr ( y ) − Ψ ( y ) ∣ ≤ s W for all y ∈ Y α ( W ) . |\operatorname{pr}(y)-\Psi(y)|\le s_W \qquad\text{for all } y\in Y_\alpha(W). ∣ pr ( y ) − Ψ ( y ) ∣ ≤ s W for all y ∈ Y α ( W ) . Proof. Write n = n α \mathbf{n}=\mathbf{n}_\alpha n = n α π = π α \pi=\pi_\alpha π = π α K = { z ∈ Z 3 : ⟨ z , n ⟩ ∈ W } K=\{z\in\mathbb{Z}^3:\langle z,\mathbf{n}\rangle\in W\} K = { z ∈ Z 3 : ⟨ z , n ⟩ ∈ W } Y α ( W ) = π ( K ) Y_\alpha(W)=\pi(K) Y α ( W ) = π ( K ) z ∈ K z\in K z ∈ K s z = ⟨ z , n ⟩ s_z=\langle z,\mathbf{n}\rangle s z = ⟨ z , n ⟩ π ( z ) = z − s z n / ∣ n ∣ 2 \pi(z)=z-s_z\mathbf{n}/|\mathbf{n}|^2 π ( z ) = z − s z n /∣ n ∣ 2
First we show that π \pi π K K K z , z ′ ∈ K z,z'\in K z , z ′ ∈ K π ( z ) = π ( z ′ ) \pi(z)=\pi(z') π ( z ) = π ( z ′ ) z − z ′ = t n z-z'=t\mathbf{n} z − z ′ = t n t ∈ R t\in\mathbb{R} t ∈ R t ∈ Z t\in\mathbb{Z} t ∈ Z t ∣ n ∣ 2 = s z − s z ′ t|\mathbf{n}|^2=s_z-s_{z'} t ∣ n ∣ 2 = s z − s z ′ ∣ n ∣ ≥ 1 |\mathbf{n}|\ge1 ∣ n ∣ ≥ 1 t = 0 t=0 t = 0
Next, the map z ↦ ( z 1 , z 2 ) z\mapsto(z_1,z_2) z ↦ ( z 1 , z 2 ) K K K Λ α ( W ) \Lambda_\alpha(W) Λ α ( W ) m ∈ Z 2 m\in\mathbb{Z}^2 m ∈ Z 2 ⟨ ( m 1 , m 2 , z 3 ) , n ⟩ = ⟨ α , m ⟩ − z 3 \langle(m_1,m_2,z_3),\mathbf{n}\rangle=\langle\alpha,m\rangle-z_3 ⟨( m 1 , m 2 , z 3 ) , n ⟩ = ⟨ α , m ⟩ − z 3 z 3 z_3 z 3 ( m 1 , m 2 , z 3 ) ∈ K (m_1,m_2,z_3)\in K ( m 1 , m 2 , z 3 ) ∈ K ⟨ α , m ⟩ − W \langle\alpha,m\rangle-W ⟨ α , m ⟩ − W ⟨ α , m ⟩ ∈ W + Z \langle\alpha,m\rangle\in W+\mathbb{Z} ⟨ α , m ⟩ ∈ W + Z
Combining these two facts, the map Ψ ( π ( z ) ) = ( z 1 , z 2 ) \Psi(\pi(z))=(z_1,z_2) Ψ ( π ( z )) = ( z 1 , z 2 ) Y α ( W ) Y_\alpha(W) Y α ( W ) Λ α ( W ) \Lambda_\alpha(W) Λ α ( W ) z ∈ K z\in K z ∈ K
pr ( π ( z ) ) − Ψ ( π ( z ) ) = − s z ∣ n ∣ 2 ( α 1 , α 2 ) , \operatorname{pr}(\pi(z))-\Psi(\pi(z))=-\frac{s_z}{|\mathbf{n}|^2}(\alpha_1,\alpha_2), pr ( π ( z )) − Ψ ( π ( z )) = − ∣ n ∣ 2 s z ( α 1 , α 2 ) , and the norm of the right-hand side is at most s W s_W s W ∣ ( α 1 , α 2 ) ∣ ≤ ∣ n ∣ |(\alpha_1,\alpha_2)|\le|\mathbf{n}| ∣ ( α 1 , α 2 ) ∣ ≤ ∣ n ∣
We remark that the bijection Ψ \Psi Ψ α \alpha α
Proof of Theorem 1, assuming Theorem 3: Let α \alpha α W W W Theorem 3 , and write n = n α \mathbf{n}=\mathbf{n}_\alpha n = n α z ∈ Z 3 z\in\mathbb{Z}^3 z ∈ Z 3 m = ( z 1 , z 2 ) m=(z_1,z_2) m = ( z 1 , z 2 ) ⟨ z , n ⟩ = ⟨ α , m ⟩ − z 3 \langle z,\mathbf{n}\rangle=\langle\alpha,m\rangle-z_3 ⟨ z , n ⟩ = ⟨ α , m ⟩ − z 3 ⟨ z , n ⟩ ∈ ∂ W \langle z,\mathbf{n}\rangle\in\partial W ⟨ z , n ⟩ ∈ ∂ W ⟨ α , m ⟩ ∈ ∂ W + Z \langle\alpha,m\rangle\in\partial W+\mathbb{Z} ⟨ α , m ⟩ ∈ ∂ W + Z Theorem 3 . This proves part (i) of Theorem 1 .
For part (ii), note that Y α ( W ) Y_\alpha(W) Y α ( W ) V α V_\alpha V α V α ⊥ V_\alpha^\perp V α ⊥ Z 3 \mathbb{Z}^3 Z 3 Theorem 1 , the projection π α \pi_\alpha π α Z 3 \mathbb{Z}^3 Z 3 Z 3 \mathbb{Z}^3 Z 3 Y α ( W ) Y_\alpha(W) Y α ( W ) V α V_\alpha V α [3 Chapter 7] .
Now suppose, by way of contradiction, that there is a BL bijection from Y α ( W ) Y_\alpha(W) Y α ( W ) Γ ⊂ V α \Gamma\subset V_\alpha Γ ⊂ V α pr ( y ) ↦ Ψ ( y ) \operatorname{pr}(y)\mapsto\Psi(y) pr ( y ) ↦ Ψ ( y ) pr ( Y α ( W ) ) \operatorname{pr}(Y_\alpha(W)) pr ( Y α ( W )) Λ α ( W ) \Lambda_\alpha(W) Λ α ( W ) [17 Proposition 1.8] . The set pr ( Γ ) \operatorname{pr}(\Gamma) pr ( Γ ) R 2 \mathbb{R}^2 R 2 Z 2 \mathbb{Z}^2 Z 2 pr \operatorname{pr} pr pr − 1 \operatorname{pr}^{-1} pr − 1 Λ α ( W ) \Lambda_\alpha(W) Λ α ( W ) Z 2 \mathbb{Z}^2 Z 2 Theorem 3 . This completes the proof of Theorem 1 , assuming Theorem 3 .
Stretched pairs
In this section we prove that no BL homeomorphism of the plane can reproduce a certain hierarchical pattern of areas. The argument is that of Burago and Kleiner [5 Section 3] (see Lemmas 3.2 and 3.7 there), and our only real change is to the hypothesis. Burago and Kleiner assume that the Jacobian of the map is equal to a prescribed checkerboard function outside a set of small measure. We will only assume that the images of any two adjacent squares of a checkerboard differ in area by a definite amount.
Let e = ( 1 , 0 ) e=(1,0) e = ( 1 , 0 ) N ≥ 2 N \ge2 N ≥ 2
R N = [ 0 , 1 ] × [ 0 , 1 / N ] , S i = [ i − 1 N , i N ] × [ 0 , 1 N ] ( 1 ≤ i ≤ N ) . R_N=[0,1]\times[0,1/N],\qquad S_i=\left[\frac{i-1}{N},\frac{i}{N}\right]\times\left[0,\frac{1}{N}\right]\qquad(1\le i\le N). R N = [ 0 , 1 ] × [ 0 , 1/ N ] , S i = [ N i − 1 , N i ] × [ 0 , N 1 ] ( 1 ≤ i ≤ N ) . so that R N R_N R N N N N S i S_i S i N − 2 N^{-2} N − 2 M ≥ 1 M\ge1 M ≥ 1 marked points of R N R_N R N
x p q i = ( p + M ( i − 1 ) N M , q N M ) , 1 ≤ i ≤ N , 0 ≤ p , q ≤ M , x_{pq}^{i}=\left(\frac{p+M(i-1)}{NM},\frac{q}{NM}\right),\qquad1\le i\le N,\ 0\le p,q\le M, x pq i = ( N M p + M ( i − 1 ) , N M q ) , 1 ≤ i ≤ N , 0 ≤ p , q ≤ M , which form a square grid of mesh 1 / ( N M ) 1/(NM) 1/ ( N M ) R N R_N R N x M q i = x 0 q i + 1 x_{Mq}^{i}=x_{0q}^{i+1} x M q i = x 0 q i + 1 R N R_N R N R × { q / ( N M ) } \mathbb{R}\times\{q/(NM)\} R × { q / ( N M )} 0 ≤ q ≤ M 0\le q\le M 0 ≤ q ≤ M Σ ( u ) = ξ + ℓ u \Sigma(u)=\xi+\ell u Σ ( u ) = ξ + ℓ u ξ ∈ R 2 \xi\in\mathbb{R}^{2} ξ ∈ R 2 ℓ > 0 \ell>0 ℓ > 0 Σ ( R N ) \Sigma(R_N) Σ ( R N ) Σ ( S i ) \Sigma(S_i) Σ ( S i ) Σ ( x p q i ) \Sigma(x_{pq}^{i}) Σ ( x pq i ) Σ ( R N ) \Sigma(R_N) Σ ( R N ) built on the segment Σ ( [ 0 , 1 ] × { 0 } ) \Sigma([0,1]\times\{0\}) Σ ([ 0 , 1 ] × { 0 }) x ≠ y x\ne y x = y A A A f f f ∣ f ( x ) − f ( y ) ∣ ≥ A ∣ x − y ∣ |f(x)-f(y)|\ge A|x-y| ∣ f ( x ) − f ( y ) ∣ ≥ A ∣ x − y ∣
Lemma 5. Let L ≥ 1 L\ge1 L ≥ 1 c ∈ ( 0 , 1 ] c\in(0,1] c ∈ ( 0 , 1 ] M ≥ 1 M\ge1 M ≥ 1 β > 0 \beta>0 β > 0 N 0 ≥ 2 N_0\ge2 N 0 ≥ 2 L L L c c c N ≥ N 0 N\ge N_0 N ≥ N 0 Σ ( u ) = ξ + ℓ u \Sigma(u)=\xi+\ell u Σ ( u ) = ξ + ℓ u ℓ > 0 \ell>0 ℓ > 0 f : R 2 → R 2 f:\mathbb{R}^{2}\to\mathbb{R}^{2} f : R 2 → R 2 L L L
∣ f ( Σ ( S i + 1 ) ) ∣ − ∣ f ( Σ ( S i ) ) ∣ ≥ c ∣ Σ ( S i ) ∣ for 1 ≤ i ≤ N − 1. (3) \left|f(\Sigma(S_{i+1}))\right|-\left|f(\Sigma(S_i))\right|\ge c\left|\Sigma(S_i)\right|\qquad\text{for }1\le i\le N-1. \tag*{(3)} ∣ f ( Σ ( S i + 1 )) ∣ − ∣ f ( Σ ( S i )) ∣ ≥ c ∣ Σ ( S i ) ∣ for 1 ≤ i ≤ N − 1. (3) If the pair Σ ( 0 , 0 ) , Σ ( 1 , 0 ) \Sigma(0,0),\Sigma(1,0) Σ ( 0 , 0 ) , Σ ( 1 , 0 ) A A A f f f A > 0 A>0 A > 0 x x x y y y Σ ( R N ) \Sigma(R_N) Σ ( R N ) ℓ / ( N M ) \ell/(NM) ℓ / ( N M ) x , y x,y x , y ( 1 + β ) A (1+\beta)A ( 1 + β ) A f f f
The integer M M M β \beta β N 0 N_0 N 0 L L L c c c M M M M + 1 M+1 M + 1
Proof. Let g ( u ) = ℓ − 1 f ( Σ ( u ) ) g(u)=\ell^{-1}f(\Sigma(u)) g ( u ) = ℓ − 1 f ( Σ ( u )) L L L R 2 \mathbb{R}^{2} R 2 Ω ⊆ R 2 \Omega\subseteq\mathbb{R}^{2} Ω ⊆ R 2 ∣ g ( Ω ) ∣ = ℓ − 2 ∣ f ( Σ ( Ω ) ) ∣ |g(\Omega)|=\ell^{-2}|f(\Sigma(\Omega))| ∣ g ( Ω ) ∣ = ℓ − 2 ∣ f ( Σ ( Ω )) ∣ ∣ Σ ( Ω ) ∣ = ℓ 2 ∣ Ω ∣ |\Sigma(\Omega)|=\ell^{2}|\Omega| ∣Σ ( Ω ) ∣ = ℓ 2 ∣Ω∣ g g g Σ \Sigma Σ x ≠ y x\ne y x = y g g g x , y x,y x , y f f f Σ ( x ) , Σ ( y ) \Sigma(x),\Sigma(y) Σ ( x ) , Σ ( y ) Σ \Sigma Σ R N R_N R N Σ ( R N ) \Sigma(R_N) Σ ( R N ) f f f g g g Σ \Sigma Σ
η = c 128 π ( L + 2 ) , M = ⌈ L η ⌉ , λ = ( η 3 L ) 2 , \eta=\frac{c}{128\pi(L+2)},\qquad M=\left\lceil\frac{L}{\eta}\right\rceil,\qquad\lambda=\left(\frac{\eta}{3L}\right)^2, η = 128 π ( L + 2 ) c , M = ⌈ η L ⌉ , λ = ( 3 L η ) 2 , and
β = λ 16 ( M + 1 ) , N 0 = ⌈ 32 L 2 ( M + 1 ) λ ⌉ . \beta= \frac{\lambda}{16(M+1)}, \qquad N_0 = \left\lceil\frac{32L^2(M+1)}{\lambda} \right\rceil. β = 16 ( M + 1 ) λ , N 0 = ⌈ λ 32 L 2 ( M + 1 ) ⌉ . Note that β ≤ λ ≤ 1 \beta\le\lambda\le1 β ≤ λ ≤ 1 N 0 ≥ 2 N_0 \ge2 N 0 ≥ 2
After composing f f f R 2 \mathbb{R}^2 R 2 f ( 0 , 0 ) = ( 0 , 0 ) f(0,0)=(0,0) f ( 0 , 0 ) = ( 0 , 0 ) f ( 1 , 0 ) = ( z , 0 ) f(1,0)=(z,0) f ( 1 , 0 ) = ( z , 0 ) z > 0 z>0 z > 0 z ≥ A z\ge A z ≥ A f f f L L L 1 / L ≤ z ≤ L 1/L\le z\le L 1/ L ≤ z ≤ L x , y x,y x , y ( 1 + β ) z (1+\beta)z ( 1 + β ) z
∣ f ( x ) − f ( y ) ∣ < ( 1 + β ) z ∣ x − y ∣ , (4) |f(x)-f(y)| < (1+\beta)z|x-y|, \tag*{(4)} ∣ f ( x ) − f ( y ) ∣ < ( 1 + β ) z ∣ x − y ∣ , (4) for all distinct marked points x , y x,y x , y x x x y y y (4) is equivalent to the same inequality for adjacent marked points only. This is why it suffices to contradict (4) .
For 1 ≤ i ≤ N − 1 1\le i\le N-1 1 ≤ i ≤ N − 1 0 ≤ p , q ≤ M 0\le p,q\le M 0 ≤ p , q ≤ M
v p q i = f ( x p q i + 1 ) − f ( x p q i ) , v ‾ = ( z / N , 0 ) . v_{pq}^{i}=f(x_{pq}^{i+1})-f(x_{pq}^{i}), \qquad\overline{v}=(z/N,0). v pq i = f ( x pq i + 1 ) − f ( x pq i ) , v = ( z / N , 0 ) . Since x p q i + 1 = x p q i + e / N x_{pq}^{i+1}=x_{pq}^{i}+e/N x pq i + 1 = x pq i + e / N (4) that ∣ v p q i ∣ < ( 1 + β ) z / N |v_{pq}^{i}|<(1+\beta)z/N ∣ v pq i ∣ < ( 1 + β ) z / N v p q i v_{pq}^{i} v pq i ( 1 − λ ) z / N (1-\lambda)z/N ( 1 − λ ) z / N S i S_i S i i ≤ N − 1 i\le N-1 i ≤ N − 1 ( M + 1 ) 2 (M+1)^2 ( M + 1 ) 2 v p q i v_{pq}^{i} v pq i
Step 1: a regular square exists. Suppose not, and for each 1 ≤ i ≤ N − 1 1\le i\le N-1 1 ≤ i ≤ N − 1 p i , q i p_i,q_i p i , q i v p i q i i v_{p_iq_i}^{i} v p i q i i i i i q i q_i q i i i i q ∈ { 0 , … , M } q\in\{0,\ldots,M\} q ∈ { 0 , … , M } i 1 < i 2 < ⋯ < i t i_1<i_2<\cdots<i_t i 1 < i 2 < ⋯ < i t q i j = q q_{i_j}=q q i j = q j j j
t ≥ N − 1 2 ( M + 1 ) ≥ N 4 ( M + 1 ) . t\ge\frac{N-1}{2(M+1)}\ge\frac{N}{4(M+1)}. t ≥ 2 ( M + 1 ) N − 1 ≥ 4 ( M + 1 ) N . Write p j = p i j p_j=p_{i_j} p j = p i j
( 0 , 0 ) , x 0 q 1 , x p 1 q i 1 , x p 1 q i 1 + 1 , x p 2 q i 2 , x p 2 q i 2 + 1 , … , x p t q i t , x p t q i t + 1 , x M q N , ( 1 , 0 ) . (0,0),\ x_{0q}^{1},\ x_{p_1q}^{i_1},\ x_{p_1q}^{i_1+1},\ x_{p_2q}^{i_2},\ x_{p_2q}^{i_2+1},\ \ldots,\ x_{p_tq}^{i_t},\ x_{p_tq}^{i_t+1},\ x_{Mq}^{N},\ (1,0). ( 0 , 0 ) , x 0 q 1 , x p 1 q i 1 , x p 1 q i 1 + 1 , x p 2 q i 2 , x p 2 q i 2 + 1 , … , x p t q i t , x p t q i t + 1 , x M q N , ( 1 , 0 ) . All of these points except the first and last lie on the row q q q i j + 1 ≥ i j + 2 i_{j+1}\ge i_j+2 i j + 1 ≥ i j + 2 x p j q i j + 1 x_{p_jq}^{i_{j+1}} x p j q i j + 1 M ( i j + 1 ) / ( N M ) M(i_j+1)/(NM) M ( i j + 1 ) / ( N M ) x p j + 1 q i j + 1 x_{p_{j+1}q}^{i_{j+1}} x p j + 1 q i j + 1 f f f z z z
(i) Each of the t t t x p j q i j x_{p_jq}^{i_j} x p j q i j x p j q i j + 1 x_{p_jq}^{i_j+1} x p j q i j + 1 v p j q i j v_{p_jq}^{i_j} v p j q i j ( 1 − λ ) z / N (1-\lambda)z/N ( 1 − λ ) z / N
(ii) The remaining horizontal pieces join marked points on the row q q q 1 − t / N 1-t/N 1 − t / N (4) they contribute at most ( 1 + β ) z ( 1 − t / N ) (1+\beta)z(1-t/N) ( 1 + β ) z ( 1 − t / N )
(iii) The two vertical pieces contribute at most L / N L/N L / N f f f L L L
z ≤ ( 1 − λ ) z t N + ( 1 + β ) z ( 1 − t N ) + 2 L N , z \le(1-\lambda)z\frac{t}{N}+(1+\beta)z\left(1-\frac{t}{N}\right)+\frac{2L}{N}, z ≤ ( 1 − λ ) z N t + ( 1 + β ) z ( 1 − N t ) + N 2 L , which, after dividing by z z z z ≥ 1 / L z \ge1/L z ≥ 1/ L
( λ + β ) t N ≤ β + 2 L 2 N . (\lambda+\beta)\frac{t}{N}\le\beta+\frac{2L^2}{N}. ( λ + β ) N t ≤ β + N 2 L 2 . The left-hand side is at least λ / ( 4 ( M + 1 ) ) \lambda/(4(M+1)) λ / ( 4 ( M + 1 )) β \beta β N ≥ N 0 N\ge N_0 N ≥ N 0 λ / ( 8 ( M + 1 ) ) \lambda/(8(M+1)) λ / ( 8 ( M + 1 )) S i S_i S i i ≤ N − 1 i\le N-1 i ≤ N − 1
Step 2: regular vectors are close to v ˉ \bar{v} v ˉ v = v p q i = ( v 1 , v 2 ) v=v^i_{pq}=(v_1,v_2) v = v pq i = ( v 1 , v 2 )
( 1 − λ ) z / N < v 1 ≤ ∣ v ∣ < ( 1 + β ) z / N , (1-\lambda)z/N<v_1\le\lvert v\rvert<(1+\beta)z/N, ( 1 − λ ) z / N < v 1 ≤ ∣ v ∣ < ( 1 + β ) z / N , so ∣ v 1 − z / N ∣ ≤ λ z / N \lvert v_1-z/N\rvert\le\lambda z/N ∣ v 1 − z / N ∣ ≤ λ z / N β ≤ λ \beta\le\lambda β ≤ λ
v 2 2 = ∣ v ∣ 2 − v 1 2 < ( ( 1 + β ) 2 − ( 1 − λ ) 2 ) z 2 N 2 = ( β + λ ) ( 2 + β − λ ) z 2 N 2 ≤ 4 λ z 2 N 2 . v_2^2=\lvert v\rvert^2-v_1^2<\left((1+\beta)^2-(1-\lambda)^2\right)\frac{z^2}{N^2}=(\beta+\lambda)(2+\beta-\lambda)\frac{z^2}{N^2}\le\frac{4\lambda z^2}{N^2}. v 2 2 = ∣ v ∣ 2 − v 1 2 < ( ( 1 + β ) 2 − ( 1 − λ ) 2 ) N 2 z 2 = ( β + λ ) ( 2 + β − λ ) N 2 z 2 ≤ N 2 4 λ z 2 . It follows that ∣ v − v ˉ ∣ 2 ≤ ( λ 2 + 4 λ ) z 2 / N 2 ≤ 5 λ L 2 / N 2 \lvert v-\bar{v}\rvert^2\le(\lambda^2+4\lambda)z^2/N^2\le5\lambda L^2/N^2 ∣ v − v ˉ ∣ 2 ≤ ( λ 2 + 4 λ ) z 2 / N 2 ≤ 5 λ L 2 / N 2
∣ v − v ˉ ∣ ≤ 3 L λ N = η N . \lvert v-\bar{v}\rvert\le\frac{3L\sqrt{\lambda}}{N}=\frac{\eta}{N}. ∣ v − v ˉ ∣ ≤ N 3 L λ = N η . Step 3: the images of a regular square and the next square have nearly equal area. Let S i S_i S i θ = ( η + L / M ) / N \theta=(\eta+L/M)/N θ = ( η + L / M ) / N
∣ f ( S i + 1 ) ∣ − ∣ f ( S i ) ∣ ≤ 16 π θ ( L N + θ ) . (5) \lvert f(S_{i+1})\rvert-\lvert f(S_i)\rvert\le16\pi\theta\left(\frac{L}{N}+\theta\right). \tag*{(5)} ∣ f ( S i + 1 )∣ − ∣ f ( S i )∣ ≤ 16 π θ ( N L + θ ) . (5) Parametrize ∂ S i \partial S_i ∂ S i γ 0 ( x ) = f ( x ) \gamma_0(x)=f(x) γ 0 ( x ) = f ( x ) γ 1 ( x ) = f ( x + e / N ) − v ˉ \gamma_1(x)=f(x+e/N)-\bar{v} γ 1 ( x ) = f ( x + e / N ) − v ˉ e / N e/N e / N ∂ S i \partial S_i ∂ S i ∂ S i + 1 \partial S_{i+1} ∂ S i + 1 γ 1 \gamma_1 γ 1 f ( S i + 1 ) − v ˉ f(S_{i+1})-\bar{v} f ( S i + 1 ) − v ˉ S i S_i S i 1 / ( N M ) 1/(NM) 1/ ( N M ) x ∈ ∂ S i x\in\partial S_i x ∈ ∂ S i 1 / ( 2 N M ) 1/(2NM) 1/ ( 2 N M ) x p q i ∈ ∂ S i x^i_{pq}\in\partial S_i x pq i ∈ ∂ S i x + e / N x+e/N x + e / N 1 / ( 2 N M ) 1/(2NM) 1/ ( 2 N M ) x p q i + 1 x^{i+1}_{pq} x pq i + 1
∣ γ 1 ( x ) − γ 0 ( x ) ∣ ≤ ∣ f ( x + e / N ) − f ( x p q i + 1 ) ∣ + ∣ v p q i − v ˉ ∣ + ∣ f ( x p q i ) − f ( x ) ∣ ≤ L 2 N M + η N + L 2 N M = θ . \begin{aligned}
\lvert\gamma_1(x)-\gamma_0(x)\rvert\le\lvert f(x+e/N)-f(x^{i+1}_{pq})\rvert+\lvert v^i_{pq}-\bar{v}\rvert+\lvert f(x^i_{pq})-f(x)\rvert\\
&\le\frac{L}{2NM}+\frac{\eta}{N}+\frac{L}{2NM}=\theta.
\end{aligned} ∣ γ 1 ( x ) − γ 0 ( x )∣ ≤ ∣ f ( x + e / N ) − f ( x pq i + 1 )∣ + ∣ v pq i − v ˉ ∣ + ∣ f ( x pq i ) − f ( x )∣ ≤ 2 N M L + N η + 2 N M L = θ . Now we use the fact that f f f R 2 \mathbb{R}^2 R 2 f ( ∂ S i ) f(\partial S_i) f ( ∂ S i ) f ( int S i ) f(\operatorname{int} S_i) f ( int S i ) f ( R 2 ∖ S i ) f(\mathbb{R}^2\setminus S_i) f ( R 2 ∖ S i ) f ( ∂ S i ) f(\partial S_i) f ( ∂ S i ) f ( int S i ) f(\operatorname{int} S_i) f ( int S i ) R 2 ∖ f ( ∂ S i ) \mathbb{R}^2\setminus f(\partial S_i) R 2 ∖ f ( ∂ S i ) γ 0 \gamma_0 γ 0 y ∉ f ( ∂ S i ) y\notin f(\partial S_i) y ∈ / f ( ∂ S i ) ± 1 \pm1 ± 1 y ∈ f ( int S i ) y\in f(\operatorname{int} S_i) y ∈ f ( int S i ) [8 Proposition 5.20] , and it is 0 0 0 y y y γ 1 \gamma_1 γ 1 f ( int S i + 1 ) − v ˉ f(\operatorname{int} S_{i+1})-\bar{v} f ( int S i + 1 ) − v ˉ y y y θ \theta θ f ( ∂ S i ) f(\partial S_i) f ( ∂ S i ) ( 1 − t ) γ 0 + t γ 1 (1-t)\gamma_0+t\gamma_1 ( 1 − t ) γ 0 + t γ 1 0 ≤ t ≤ 1 0\le t\le1 0 ≤ t ≤ 1 y y y γ 0 \gamma_0 γ 0 γ 1 \gamma_1 γ 1 y y y [8 Chapter 3] ). Therefore, for such y y y y ∈ f ( S i ) y\in f(S_i) y ∈ f ( S i ) y ∈ f ( S i + 1 ) − v ˉ y \in f(S_{i+1}) - \bar{v} y ∈ f ( S i + 1 ) − v ˉ f ( S i ) f(S_i) f ( S i ) f ( S i + 1 ) − v ˉ f(S_{i+1}) - \bar{v} f ( S i + 1 ) − v ˉ θ \theta θ N θ ( f ( ∂ S i ) ) N_\theta(f(\partial S_i)) N θ ( f ( ∂ S i )) f ( ∂ S i ) f(\partial S_i) f ( ∂ S i )
∣ ∣ f ( S i + 1 ) ∣ − ∣ f ( S i ) ∣ ∣ ≤ ∣ N θ ( f ( ∂ S i ) ) ∣ . \left| |f(S_{i+1})| - |f(S_i)| \right| \leq|N_\theta(f(\partial S_i))|. ∣ ∣ f ( S i + 1 ) ∣ − ∣ f ( S i ) ∣ ∣ ≤ ∣ N θ ( f ( ∂ S i )) ∣. To estimate ∣ N θ ( f ( ∂ S i ) ) ∣ |N_\theta(f(\partial S_i))| ∣ N θ ( f ( ∂ S i )) ∣ M ′ = ⌈ L / ( N θ ) ⌉ M' = \lceil L/(N\theta) \rceil M ′ = ⌈ L / ( N θ )⌉ S i S_i S i M ′ M' M ′ 1 / ( N M ′ ) 1/(NM') 1/ ( N M ′ ) 4 M ′ 4M' 4 M ′ L / ( N M ′ ) ≤ θ L/(NM') \leq\theta L / ( N M ′ ) ≤ θ θ \theta θ N θ ( f ( ∂ S i ) ) N_\theta(f(\partial S_i)) N θ ( f ( ∂ S i )) 4 M ′ 4M' 4 M ′ 2 θ 2\theta 2 θ
∣ N θ ( f ( ∂ S i ) ) ∣ ≤ 16 π M ′ θ 2 ≤ 16 π θ 2 ( L N θ + 1 ) = 16 π θ ( L N + θ ) . |N_\theta(f(\partial S_i))| \leq16\pi M'\theta^2 \leq16\pi\theta^2\left(\frac{L}{N\theta}+1\right) = 16\pi\theta\left(\frac{L}{N}+\theta\right). ∣ N θ ( f ( ∂ S i )) ∣ ≤ 16 π M ′ θ 2 ≤ 16 π θ 2 ( N θ L + 1 ) = 16 π θ ( N L + θ ) . This proves (5) .
Conclusion. Our choices of η \eta η M M M L / M ≤ η L/M \leq\eta L / M ≤ η N θ ≤ 2 η = c / ( 64 π ( L + 2 ) ) ≤ 1 N\theta\leq2\eta= c/(64\pi(L+2)) \leq1 N θ ≤ 2 η = c / ( 64 π ( L + 2 )) ≤ 1 (5) we have that
16 π N 2 ( N θ ) ( L + N θ ) ≤ 16 π N 2 ⋅ c 64 π ( L + 2 ) ⋅ ( L + 1 ) ≤ c 4 N 2 < c ∣ S i ∣ . \frac{16\pi}{N^2}(N\theta)(L+N\theta) \leq\frac{16\pi}{N^2}\cdot\frac{c}{64\pi(L+2)}\cdot(L+1) \leq\frac{c}{4N^2} < c|S_i|. N 2 16 π ( N θ ) ( L + N θ ) ≤ N 2 16 π ⋅ 64 π ( L + 2 ) c ⋅ ( L + 1 ) ≤ 4 N 2 c < c ∣ S i ∣. However, this contradicts (3) . Therefore (4) is false, and we conclude that there is a pair of adjacent marked points on a common row which is ( 1 + β ) z (1+\beta)z ( 1 + β ) z
Next we iterate Lemma [5 ] . For integers M ≥ 1 M \geq1 M ≥ 1 N ≥ 2 N \geq2 N ≥ 2 J ≥ 1 J \geq1 J ≥ 1 C = C ( M , N , J ) \mathcal{C} = \mathcal{C}(M,N,J) C = C ( M , N , J ) j j j R N R_N R N [ 0 , 1 ] × { 0 } [0,1] \times\{0\} [ 0 , 1 ] × { 0 } R = ξ + ℓ R N R = \xi+ \ell R_N R = ξ + ℓ R N j < J j < J j < J x x x x + ℓ e / ( N M ) x+\ell e/(NM) x + ℓ e / ( N M ) R R R
x + ℓ N M R N x+\frac{\ell}{NM}R_N x + N M ℓ R N built on the segment between them is a rectangle of level j + 1 j+1 j + 1 C \mathcal{C} C 1 , … , J 1,\ldots,J 1 , … , J j j j ( N M ) 1 − j (NM)^{1-j} ( N M ) 1 − j N − 1 ( N M ) 1 − j N^{-1}(NM)^{1-j} N − 1 ( N M ) 1 − j
Proposition 6. Let L ≥ 1 L \geq1 L ≥ 1 c ∈ ( 0 , 1 ] c \in(0,1] c ∈ ( 0 , 1 ] M M M β \beta β N 0 N_0 N 0 [5 ] . Suppose that N ≥ N 0 N \geq N_0 N ≥ N 0 J ≥ 1 J \geq1 J ≥ 1 ( 1 + β ) J > L 2 (1+\beta)^J > L^2 ( 1 + β ) J > L 2 L L L f : R 2 → R 2 f:\mathbb{R}^2 \to\mathbb{R}^2 f : R 2 → R 2 C ( M , N , J ) \mathcal{C}(M,N,J) C ( M , N , J ) S S S S ′ S' S ′
∣ ∣ f ( S ′ ) ∣ − ∣ f ( S ) ∣ ∣ ≥ c ∣ S ∣ . (6) \left| |f(S')| - |f(S)| \right| \geq c|S|. \tag*{(6)} ∣ ∣ f ( S ′ ) ∣ − ∣ f ( S ) ∣ ∣ ≥ c ∣ S ∣. (6) Proof. Suppose that f f f ( 0 , 0 ) , ( 1 , 0 ) (0,0),(1,0) ( 0 , 0 ) , ( 1 , 0 ) L − 1 L^{-1} L − 1 f f f f f f L L L 1 ≤ j ≤ J 1 \leq j \leq J 1 ≤ j ≤ J R R R j j j ( 1 + β ) j − 1 L − 1 (1+\beta)^{j-1}L^{-1} ( 1 + β ) j − 1 L − 1 f f f (6) , the map f f f (3) for R R R [5 ] gives a pair of adjacent marked points of R R R ( 1 + β ) j L − 1 (1+\beta)^jL^{-1} ( 1 + β ) j L − 1 j < J j<J j < J C \mathcal{C} C j + 1 j+1 j + 1 J J J ( 1 + β ) J L − 1 (1+\beta)^JL^{-1} ( 1 + β ) J L − 1 ( 1 + β ) J L − 1 > L (1+\beta)^JL^{-1} > L ( 1 + β ) J L − 1 > L f f f L L L
Proposition 6 applies equally to every translated and dilated copy x + s C x+sC x + s C x ∈ R 2 x \in\mathbb{R}^2 x ∈ R 2 s > 0 s>0 s > 0 L L L f f f (6) for every pair of adjacent squares of every rectangle in x + s C x+sC x + s C
g ( u ) = s − 1 f ( x + s u ) g(u)=s^{-1}f(x+su) g ( u ) = s − 1 f ( x + s u ) is again an L L L R 2 \mathbb{R}^2 R 2 S S S C C C
∣ g ( S ) ∣ = s − 2 ∣ f ( x + s S ) ∣ . |g(S)|=s^{-2}|f(x+sS)|. ∣ g ( S ) ∣ = s − 2 ∣ f ( x + s S ) ∣. Since ∣ x + s S ∣ = s 2 ∣ S ∣ |x+sS|=s^2|S| ∣ x + s S ∣ = s 2 ∣ S ∣ g g g (6) for every pair of adjacent squares of every rectangle in C C C N N N N ≥ N 0 N\geq N_0 N ≥ N 0
From homeomorphisms to Delone sets
In this section we transfer Proposition 6 from homeomorphisms to Delone sets, by replacing the area of the image of a square with the number of points of the set in a large dilate of that square. The idea is that, viewed from far away, an L L L Λ \Lambda Λ Z 2 \mathbb{Z}^2 Z 2 L L L f f f S S S s s s Λ \Lambda Λ s S sS s S s 2 s^2 s 2 f ( S ) f(S) f ( S ) Λ \Lambda Λ C ( M , N , J ) C(M,N,J) C ( M , N , J ) f f f C ( M , N , J ) C(M,N,J) C ( M , N , J )
To make this precise we use the compactness argument from the proof of [5 Lemma 2.1] (see also [17 Proposition 3.13] ), with one change. Burago and Kleiner apply the argument to a single Delone set, rescaled by a sequence of factors tending to infinity, and they draw their conclusion from the limit. For our construction we need a conclusion at one finite scale, which holds for every set whose counts on adjacent squares differ in this way at that scale, so that the scale can be fixed before the set is known. We therefore argue by contradiction. If there were no such scale, then there would be sets Λ n \Lambda_n Λ n L L L Z 2 \mathbb{Z}^2 Z 2 s n C ( M , N , J ) s_nC(M,N,J) s n C ( M , N , J ) s n s_n s n n n n L L L
Proposition 7. Let L ≥ 1 L\geq1 L ≥ 1 c ∈ ( 0 , 1 ] c\in(0,1] c ∈ ( 0 , 1 ] M M M β \beta β N 0 N_0 N 0 N ≥ N 0 N\geq N_0 N ≥ N 0 J ≥ 1 J\geq1 J ≥ 1 ( 1 + β ) J > L 2 (1+\beta)^J>L^2 ( 1 + β ) J > L 2 b > 0 b>0 b > 0 s ∗ > 0 s_*>0 s ∗ > 0 L L L c c c N N N J J J b b b Λ ⊂ R 2 \Lambda\subset\mathbb{R}^2 Λ ⊂ R 2 x ∈ R 2 x\in\mathbb{R}^2 x ∈ R 2 s ≥ s ∗ s\geq s_* s ≥ s ∗ x + s C ( M , N , J ) x+sC(M,N,J) x + s C ( M , N , J ) S S S S ′ S' S ′
∣ # ( Λ ∩ S ′ ) − # ( Λ ∩ S ) ∣ ≥ c ∣ S ∣ . (7) \left|\#(\Lambda\cap S')-\#(\Lambda\cap S)\right|\geq c|S|. \tag*{(7)} ∣ # ( Λ ∩ S ′ ) − # ( Λ ∩ S ) ∣ ≥ c ∣ S ∣. (7) Then there is no L L L Λ \Lambda Λ Z 2 \mathbb{Z}^2 Z 2
Proof. Suppose, by way of contradiction, that the proposition fails for some b > 0 b > 0 b > 0 n ∈ N n \in\mathbb{N} n ∈ N Λ n ⊆ R 2 \Lambda_n \subseteq\mathbb{R}^2 Λ n ⊆ R 2 b b b x n ∈ R 2 x_n \in\mathbb{R}^2 x n ∈ R 2 s n ≥ n s_n \ge n s n ≥ n L L L g n : Λ n → Z 2 g_n : \Lambda_n \to\mathbb{Z}^2 g n : Λ n → Z 2 (7) holds for Λ n \Lambda_n Λ n x n + s n C x_n + s_n C x n + s n C C = C ( M , N , J ) C = C(M,N,J) C = C ( M , N , J ) g n g_n g n L L L Λ n \Lambda_n Λ n 1 / L 1/L 1/ L y n ∈ Λ n y_n \in\Lambda_n y n ∈ Λ n ∣ y n − x n ∣ ≤ b |y_n-x_n| \le b ∣ y n − x n ∣ ≤ b Λ n \Lambda_n Λ n Λ n − x n \Lambda_n-x_n Λ n − x n g n g_n g n y ↦ g n ( y + x n ) − g n ( y n ) y \mapsto g_n(y+x_n)-g_n(y_n) y ↦ g n ( y + x n ) − g n ( y n ) L L L Z 2 \mathbb{Z}^2 Z 2 x n = 0 x_n=0 x n = 0 ∣ y n ∣ ≤ b |y_n| \le b ∣ y n ∣ ≤ b g n ( y n ) = 0 g_n(y_n)=0 g n ( y n ) = 0
Rescaling and passing to a limit. Let X n = s n − 1 Λ n X_n=s_n^{-1}\Lambda_n X n = s n − 1 Λ n Z n = s n − 1 Z 2 Z_n=s_n^{-1}\mathbb{Z}^2 Z n = s n − 1 Z 2 G n : X n → Z n G_n:X_n\to Z_n G n : X n → Z n G n ( u ) = s n − 1 g n ( s n u ) G_n(u)=s_n^{-1}g_n(s_nu) G n ( u ) = s n − 1 g n ( s n u ) G n G_n G n L L L H n : Z n → X n H_n:Z_n\to X_n H n : Z n → X n X n X_n X n b / s n b/s_n b / s n Z n Z_n Z n 1 / s n 1/s_n 1/ s n u n = s n − 1 y n u_n=s_n^{-1}y_n u n = s n − 1 y n ∣ u n ∣ ≤ b / s n |u_n|\le b/s_n ∣ u n ∣ ≤ b / s n G n ( u n ) = 0 G_n(u_n)=0 G n ( u n ) = 0 H n ( 0 ) = u n H_n(0)=u_n H n ( 0 ) = u n G n G_n G n H n H_n H n R 2 \mathbb{R}^2 R 2 L L L inf x ( ϕ ( x ) + L ∣ u − x ∣ ) \inf_x(\phi(x)+L|u-x|) inf x ( ϕ ( x ) + L ∣ u − x ∣ ) ϕ \phi ϕ G ~ n , H ~ n : R 2 → R 2 \widetilde{G}_n,\widetilde{H}_n:\mathbb{R}^2\to\mathbb{R}^2 G n , H n : R 2 → R 2 2 L \sqrt{2}L 2 L G n G_n G n X n X_n X n H n H_n H n Z n Z_n Z n G ~ n ( u n ) = 0 \widetilde{G}_n(u_n)=0 G n ( u n ) = 0 H ~ n ( 0 ) = u n \widetilde{H}_n(0)=u_n H n ( 0 ) = u n G ~ n → f \widetilde{G}_n\to f G n → f H ~ n → h \widetilde{H}_n\to h H n → h f , h : R 2 → R 2 f,h:\mathbb{R}^2\to\mathbb{R}^2 f , h : R 2 → R 2
The limit is a BL homeomorphism. Let u , u ′ ∈ R 2 u,u'\in\mathbb{R}^2 u , u ′ ∈ R 2 x n , x n ′ ∈ X n x_n,x'_n\in X_n x n , x n ′ ∈ X n ∣ x n − u ∣ , ∣ x n ′ − u ′ ∣ ≤ b / s n |x_n-u|,|x'_n-u'|\le b/s_n ∣ x n − u ∣ , ∣ x n ′ − u ′ ∣ ≤ b / s n
∣ G n ( x n ) − f ( u ) ∣ ≤ ∣ G ~ n ( x n ) − G ~ n ( u ) ∣ + ∣ G ~ n ( u ) − f ( u ) ∣ ≤ 2 L b s n + ∣ G ~ n ( u ) − f ( u ) ∣ → 0 , |G_n(x_n)-f(u)| \le|\widetilde{G}_n(x_n)-\widetilde{G}_n(u)|+|\widetilde{G}_n(u)-f(u)|
\le\frac{\sqrt{2}Lb}{s_n}+|\widetilde{G}_n(u)-f(u)|\to0, ∣ G n ( x n ) − f ( u ) ∣ ≤ ∣ G n ( x n ) − G n ( u ) ∣ + ∣ G n ( u ) − f ( u ) ∣ ≤ s n 2 L b + ∣ G n ( u ) − f ( u ) ∣ → 0 , and similarly G n ( x n ′ ) → f ( u ′ ) G_n(x'_n)\to f(u') G n ( x n ′ ) → f ( u ′ )
L − 1 ∣ x n − x n ′ ∣ ≤ ∣ G n ( x n ) − G n ( x n ′ ) ∣ ≤ L ∣ x n − x n ′ ∣ L^{-1}|x_n-x'_n|\le|G_n(x_n)-G_n(x'_n)|\le L|x_n-x'_n| L − 1 ∣ x n − x n ′ ∣ ≤ ∣ G n ( x n ) − G n ( x n ′ ) ∣ ≤ L ∣ x n − x n ′ ∣ for every n n n f f f L L L G n ( x n ) ∈ Z n G_n(x_n)\in Z_n G n ( x n ) ∈ Z n
H ~ n ( G n ( x n ) ) = H n ( G n ( x n ) ) = x n . \widetilde{H}_n(G_n(x_n))=H_n(G_n(x_n))=x_n. H n ( G n ( x n )) = H n ( G n ( x n )) = x n . Since the maps H ~ n \widetilde{H}_n H n h h h h ( f ( u ) ) h(f(u)) h ( f ( u )) u u u h ∘ f h\circ f h ∘ f Z n Z_n Z n w ∈ R 2 w\in\mathbb{R}^2 w ∈ R 2 f ∘ h f\circ h f ∘ h f f f L L L R 2 \mathbb{R}^2 R 2 h h h
Counts converge to areas. Let S S S C C C 0 ≤ θ ≤ 1 0\le\theta\le1 0 ≤ θ ≤ 1 S θ − S_\theta^- S θ − S S S θ \theta θ R 2 ∖ S \mathbb{R}^2\setminus S R 2 ∖ S S θ + S_\theta^+ S θ + R 2 \mathbb{R}^2 R 2 θ \theta θ S S S
lim n → ∞ # ( Λ n ∩ s n S ) s n 2 = ∣ f ( S ) ∣ . (8) \lim_{n\to\infty}\frac{\#(\Lambda_n\cap s_nS)}{s_n^2}=|f(S)|. \tag*{(8)} n → ∞ lim s n 2 # ( Λ n ∩ s n S ) = ∣ f ( S ) ∣. (8) Since H n H_n H n Z n \mathbb{Z}_n Z n X n X_n X n
# ( Λ n ∩ s n S ) = # ( X n ∩ S ) = # { w ∈ Z n : H n ( w ) ∈ S } . \#(\Lambda_n \cap s_nS)=\#(X_n \cap S)=\#\{w\in\mathbb{Z}_n:H_n(w)\in S\}. # ( Λ n ∩ s n S ) = # ( X n ∩ S ) = # { w ∈ Z n : H n ( w ) ∈ S } . If H n ( w ) ∈ S H_n(w)\in S H n ( w ) ∈ S w = G n ( H n ( w ) ) w=G_n(H_n(w)) w = G n ( H n ( w )) ∣ G n ( H n ( w ) ) ∣ ≤ L ( ∣ H n ( w ) ∣ + ∣ u n ∣ ) \lvert G_n(H_n(w))\rvert\le L(\lvert H_n(w)\rvert+\lvert u_n\rvert) ∣ G n ( H n ( w ))∣ ≤ L (∣ H n ( w )∣ + ∣ u n ∣) w w w n n n B B B f ( S θ + ) f(S_\theta^+) f ( S θ + )
Now fix θ ∈ ( 0 , 1 ] \theta\in(0,1] θ ∈ ( 0 , 1 ] H ~ n → h \widetilde{H}_n\to h H n → h B B B n n n ∣ H n ( w ) − h ( w ) ∣ < θ \lvert H_n(w)-h(w)\rvert<\theta ∣ H n ( w ) − h ( w )∣ < θ w ∈ Z n ∩ B w\in\mathbb{Z}_n\cap B w ∈ Z n ∩ B n n n h ( w ) ∈ S θ − h(w)\in S_\theta^- h ( w ) ∈ S θ − H n ( w ) ∈ S H_n(w)\in S H n ( w ) ∈ S H n ( w ) ∈ S H_n(w)\in S H n ( w ) ∈ S h ( w ) ∈ S θ + h(w)\in S_\theta^+ h ( w ) ∈ S θ + h ( w ) ∈ A h(w)\in A h ( w ) ∈ A w ∈ f ( A ) w\in f(A) w ∈ f ( A ) f ( S θ − ) ⊆ B f(S_\theta^-)\subseteq B f ( S θ − ) ⊆ B
Z n ∩ f ( S θ − ) ⊆ { w ∈ Z n : H n ( w ) ∈ S } ⊆ Z n ∩ f ( S θ + ) . \mathbb{Z}_n\cap f(S_\theta^-)\subseteq\{w\in\mathbb{Z}_n:H_n(w)\in S\}\subseteq\mathbb{Z}_n\cap f(S_\theta^+). Z n ∩ f ( S θ − ) ⊆ { w ∈ Z n : H n ( w ) ∈ S } ⊆ Z n ∩ f ( S θ + ) . The sets f ( S θ ± ) f(S_\theta^\pm) f ( S θ ± ) f ( ∂ S θ ± ) f(\partial S_\theta^\pm) f ( ∂ S θ ± ) K ⊆ R 2 K\subseteq\mathbb{R}^2 K ⊆ R 2 Z n = s n − 1 Z 2 \mathbb{Z}_n=s_n^{-1}\mathbb{Z}^2 Z n = s n − 1 Z 2 K K K s n 2 s_n^2 s n 2 ∣ K ∣ \lvert K\rvert ∣ K ∣ s n − 1 s_n^{-1} s n − 1 Z n \mathbb{Z}_n Z n ∂ K \partial K ∂ K ∣ ∂ K ∣ = 0 \lvert\partial K\rvert=0 ∣ ∂ K ∣ = 0 K K K
∣ f ( S θ − ) ∣ ≤ lim inf n → ∞ # ( Λ n ∩ s n S ) s n 2 ≤ lim sup n → ∞ # ( Λ n ∩ s n S ) s n 2 ≤ ∣ f ( S θ + ) ∣ . \lvert f(S_\theta^-)\rvert\le\liminf_{n\to\infty}\frac{\#(\Lambda_n\cap s_nS)}{s_n^2}\le\limsup_{n\to\infty}\frac{\#(\Lambda_n\cap s_nS)}{s_n^2}\le\lvert f(S_\theta^+)\rvert. ∣ f ( S θ − )∣ ≤ n → ∞ lim inf s n 2 # ( Λ n ∩ s n S ) ≤ n → ∞ lim sup s n 2 # ( Λ n ∩ s n S ) ≤ ∣ f ( S θ + )∣ . Finally, f ( S θ − ) ⊆ f ( S ) ⊆ f ( S θ + ) f(S_\theta^-)\subseteq f(S)\subseteq f(S_\theta^+) f ( S θ − ) ⊆ f ( S ) ⊆ f ( S θ + ) ∣ f ( S θ + ) ∖ f ( S θ − ) ∣ ≤ 2 π L 2 ∣ S θ + ∖ S θ − ∣ \lvert f(S_\theta^+)\setminus f(S_\theta^-)\rvert\le2\pi L^2\lvert S_\theta^+\setminus S_\theta^-\rvert ∣ f ( S θ + ) ∖ f ( S θ − )∣ ≤ 2 π L 2 ∣ S θ + ∖ S θ − ∣ θ \theta θ
Conclusion. Applying (7) to Λ n \Lambda_n Λ n s n S s_nS s n S s n S ′ s_nS' s n S ′ S S S S ′ S' S ′ C \mathcal{C} C s n 2 s_n^2 s n 2
∣ f ( S ′ ) ∣ − ∣ f ( S ) ∣ ≥ c ∣ S ∣ . \lvert f(S')\rvert-\lvert f(S)\rvert\ge c\lvert S\rvert. ∣ f ( S ′ )∣ − ∣ f ( S )∣ ≥ c ∣ S ∣ . This holds for every pair of adjacent squares of every rectangle in C \mathcal{C} C f f f L L L R 2 \mathbb{R}^2 R 2 □ \square □
We emphasize that s ∗ s_* s ∗ L L L c c c N N N J J J b b b Λ \Lambda Λ c c c N N N J J J L L L C \mathcal{C} C s ∗ s_* s ∗ b = L b=L b = L
Counting along rows, and strip families
In this section we collect the facts about the sets { m ∈ Z 2 : ⟨ α , m ⟩ ∈ U } \{m\in\mathbb{Z}^2:\langle\alpha,m\rangle\in U\} { m ∈ Z 2 : ⟨ α , m ⟩ ∈ U } U ⊆ R / Z U\subseteq\mathbb{R}/\mathbb{Z} U ⊆ R / Z α 1 \alpha_1 α 1 α 2 \alpha_2 α 2 U U U α 2 \alpha_2 α 2
Our counting estimates are based on the following uniform version of Weyl’s equidistribution theorem for irrational rotations (cf. [14 Chapter 1] ). We include the short proof, to make clear that the bound depends only on α 1 \alpha_1 α 1 ϵ \epsilon ϵ
Lemma 8. Let α 1 ∈ R ∖ Q \alpha_1 \in\mathbb{R} \setminus\mathbb{Q} α 1 ∈ R ∖ Q ϵ ∈ ( 0 , 1 ) \epsilon\in(0,1) ϵ ∈ ( 0 , 1 ) N ϵ ≥ 1 N_\epsilon\ge1 N ϵ ≥ 1 N ≥ N ϵ N \ge N_\epsilon N ≥ N ϵ ζ ∈ R \zeta\in\mathbb{R} ζ ∈ R I ⊆ R / Z I \subseteq\mathbb{R}/\mathbb{Z} I ⊆ R / Z
∣ # { 0 ≤ j < N : ζ + j α 1 ∈ I } − N ∣ I ∣ ∣ ≤ ϵ N . \left|\#\{0 \le j < N : \zeta+ j\alpha_1 \in I\} - N|I|\right| \le\epsilon N. ∣ # { 0 ≤ j < N : ζ + j α 1 ∈ I } − N ∣ I ∣ ∣ ≤ ϵ N . Proof. Choose an integer q ≥ 1 q \ge1 q ≥ 1 0 < ∥ q α 1 ∥ < ϵ / 2 0 < \lVert q\alpha_1\rVert< \epsilon/2 0 < ∥ q α 1 ∥ < ϵ /2 κ = ∥ q α 1 ∥ \kappa= \lVert q\alpha_1\rVert κ = ∥ q α 1 ∥ { 0 , … , N − 1 } \{0,\ldots,N-1\} { 0 , … , N − 1 } q q q K ≤ N / q + 1 K \le N/q+1 K ≤ N / q + 1 j 0 , j 0 + q , … , j 0 + ( K − 1 ) q j_0,j_0+q,\ldots,j_0+(K-1)q j 0 , j 0 + q , … , j 0 + ( K − 1 ) q 1 1 1 ζ + j α 1 \zeta+j\alpha_1 ζ + j α 1 ± κ \pm\kappa ± κ κ \kappa κ x 0 + i κ x_0+i\kappa x 0 + iκ 0 ≤ i < K 0 \le i < K 0 ≤ i < K X = ( K − 1 ) κ X=(K-1)\kappa X = ( K − 1 ) κ I + Z I+\mathbb{Z} I + Z ∣ I ∣ ≤ 1 |I| \le1 ∣ I ∣ ≤ 1 X ∣ I ∣ X|I| X ∣ I ∣ X + 2 X+2 X + 2 ℓ \ell ℓ ℓ / κ − 1 \ell/\kappa-1 ℓ / κ − 1 ℓ / κ + 1 \ell/\kappa+1 ℓ / κ + 1 I I I ( K − 1 ) ∣ I ∣ (K-1)|I| ( K − 1 ) ∣ I ∣ 1 / κ + X + 2 1/\kappa+X+2 1/ κ + X + 2 K ∣ I ∣ K|I| K ∣ I ∣ 1 / κ + K κ + 3 1/\kappa+K\kappa+3 1/ κ + K κ + 3 q q q q ( 1 / κ + 3 ) + N κ q(1/\kappa+3)+N\kappa q ( 1/ κ + 3 ) + N κ κ < ϵ / 2 \kappa<\epsilon/2 κ < ϵ /2 ϵ N \epsilon N ϵ N N ≥ N ϵ = ⌈ 2 q ( 1 / κ + 3 ) / ϵ ⌉ N \ge N_\epsilon=\lceil2q(1/\kappa+3)/\epsilon\rceil N ≥ N ϵ = ⌈ 2 q ( 1/ κ + 3 ) / ϵ ⌉
The next lemma is the main counting estimate. For a finite union of arcs it holds uniformly in α 2 \alpha_2 α 2 ν \nu ν ν \nu ν R / Z \mathbb{R}/\mathbb{Z} R / Z
Lemma 9. Let α 1 ∈ R ∖ Q \alpha_1 \in\mathbb{R} \setminus\mathbb{Q} α 1 ∈ R ∖ Q ϵ ∈ ( 0 , 1 ) \epsilon\in(0,1) ϵ ∈ ( 0 , 1 ) ν ≥ 1 \nu\ge1 ν ≥ 1 r ≥ 1 r \ge1 r ≥ 1 α 1 \alpha_1 α 1 ϵ \epsilon ϵ ν \nu ν α 2 ∈ R \alpha_2 \in\mathbb{R} α 2 ∈ R U U U ν \nu ν K ⊆ R 2 K \subseteq\mathbb{R}^2 K ⊆ R 2 r r r
∣ # { m ∈ Z 2 ∩ K : ⟨ α , m ⟩ ∈ U } − ∣ U ∣ ∣ K ∣ ∣ ≤ ϵ ∣ K ∣ . \left|\#\{m \in\mathbb{Z}^2 \cap K : \langle\alpha,m\rangle\in U\}-|U||K|\right| \le\epsilon|K|. # { m ∈ Z 2 ∩ K : ⟨ α , m ⟩ ∈ U } − ∣ U ∣∣ K ∣ ≤ ϵ ∣ K ∣. Proof. If U = R / Z U=\mathbb{R}/\mathbb{Z} U = R / Z U ≠ R / Z U\ne\mathbb{R}/\mathbb{Z} U = R / Z U U U ν \nu ν N 1 = N ϵ / ( 4 ν ) N_1=N_{\epsilon/(4\nu)} N 1 = N ϵ / ( 4 ν ) ϵ / ( 4 ν ) \epsilon/(4\nu) ϵ / ( 4 ν ) ϵ \epsilon ϵ h h h K K K n K = # ( Z 2 ∩ K ) n_K=\#(\mathbb{Z}^2\cap K) n K = # ( Z 2 ∩ K )
First we record two geometric facts. Since K K K r r r K K K h / 2 h/2 h /2 r h / 2 rh/2 r h /2 per ( K ) \operatorname{per}(K) per ( K ) per ( K ) ≤ 2 ∣ K ∣ / r \operatorname{per}(K)\le2|K|/r per ( K ) ≤ 2∣ K ∣/ r ∣ K ∣ |K| ∣ K ∣ ∂ K \partial K ∂ K K K K r r r
h ≤ 2 ∣ K ∣ r and per ( K ) ≤ 2 ∣ K ∣ r . (9) h \le\frac{2|K|}{r}
\qquad\text{and}\qquad
\operatorname{per}(K) \le\frac{2|K|}{r}. \tag*{(9)} h ≤ r 2∣ K ∣ and per ( K ) ≤ r 2∣ K ∣ . (9) For k ∈ Z k \in\mathbb{Z} k ∈ Z n k n_k n k Z 2 ∩ K \mathbb{Z}^2 \cap K Z 2 ∩ K m 2 = k m_2=k m 2 = k K K K ( a k + j , k ) (a_k+j,k) ( a k + j , k ) 0 ≤ j < n k 0 \leq j<n_k 0 ≤ j < n k a k a_k a k ⟨ α , m ⟩ \langle\alpha,m\rangle ⟨ α , m ⟩ ζ + j α 1 \zeta+j\alpha_1 ζ + j α 1 ζ = a k α 1 + k α 2 \zeta=a_k\alpha_1+k\alpha_2 ζ = a k α 1 + k α 2 n k ≥ N 1 n_k \geq N_1 n k ≥ N 1 Lemma 8 to each component of U U U ⟨ α , m ⟩ ∈ U \langle\alpha,m\rangle\in U ⟨ α , m ⟩ ∈ U ϵ n k / 4 \epsilon n_k/4 ϵ n k /4 ∣ U ∣ n k |U|n_k ∣ U ∣ n k m 2 = k m_2=k m 2 = k 0 < n k < N 1 0<n_k<N_1 0 < n k < N 1 K K K N 1 + 1 N_1+1 N 1 + 1 K K K y y y y y y h h h 2 r 2r 2 r 2 r 2r 2 r y y y h h h ( N 1 + 1 ) h / ( 2 r ) (N_1+1)h/(2r) ( N 1 + 1 ) h / ( 2 r ) ( N 1 + 1 ) h / r + 2 (N_1+1)h/r+2 ( N 1 + 1 ) h / r + 2 N 1 N_1 N 1 ∣ U ∣ ≤ 1 |U|\leq1 ∣ U ∣ ≤ 1 ∣ U ∣ n K |U|n_K ∣ U ∣ n K (9) , and using ∣ K ∣ ≥ π r 2 |K|\geq\pi r^2 ∣ K ∣ ≥ π r 2
∣ # { m ∈ Z 2 ∩ K : ⟨ α , m ⟩ ∈ U } − ∣ U ∣ n K ∣ ≤ ϵ n K 4 + N 1 ( ( N 1 + 1 ) h 2 r + 2 ) ≤ ϵ n K 4 + 2 N 1 ( N 1 + 2 ) r 2 ∣ K ∣ . \begin{aligned}
\left|\#\{m\in\mathbb{Z}^2\cap K:\langle\alpha,m\rangle\in U\}-|U|n_K\right|\leq\frac{\epsilon n_K}{4}+N_1\left(\frac{(N_1+1)h}{2r}+2\right) \\
&\leq\frac{\epsilon n_K}{4}+\frac{2N_1(N_1+2)}{r^2}|K|.
\end{aligned} # { m ∈ Z 2 ∩ K : ⟨ α , m ⟩ ∈ U } − ∣ U ∣ n K ≤ 4 ϵ n K + N 1 ( 2 r ( N 1 + 1 ) h + 2 ) ≤ 4 ϵ n K + r 2 2 N 1 ( N 1 + 2 ) ∣ K ∣. Finally we compare n K n_K n K ∣ K ∣ |K| ∣ K ∣ Z 2 \mathbb{Z}^2 Z 2 ∂ K \partial K ∂ K 2 \sqrt{2} 2 ∂ K \partial K ∂ K 2 2 per ( K ) + 2 π 2\sqrt{2}\operatorname{per}(K)+2\pi 2 2 per ( K ) + 2 π K K K (9) and ∣ K ∣ ≥ π r 2 |K|\geq\pi r^2 ∣ K ∣ ≥ π r 2
∣ n K − ∣ K ∣ ∣ ≤ 2 2 per ( K ) + 2 π ≤ ( 4 2 r + 2 r 2 ) ∣ K ∣ ≤ 8 ∣ K ∣ r . |n_K-|K||\leq2\sqrt{2}\operatorname{per}(K)+2\pi\leq\left(\frac{4\sqrt{2}}{r}+\frac{2}{r^2}\right)|K|\leq\frac{8|K|}{r}. ∣ n K − ∣ K ∣∣ ≤ 2 2 per ( K ) + 2 π ≤ ( r 4 2 + r 2 2 ) ∣ K ∣ ≤ r 8∣ K ∣ . Putting everything together, the difference in the statement is at most
( ϵ 4 ( 1 + 8 r ) + 2 N 1 ( N 1 + 2 ) r 2 + 8 r ) ∣ K ∣ , \left(\frac{\epsilon}{4}\left(1+\frac{8}{r}\right)+\frac{2N_1(N_1+2)}{r^2}+\frac{8}{r}\right)|K|, ( 4 ϵ ( 1 + r 8 ) + r 2 2 N 1 ( N 1 + 2 ) + r 8 ) ∣ K ∣ , which is at most ϵ ∣ K ∣ \epsilon|K| ϵ ∣ K ∣ r ≥ max ( 32 / ϵ , ( 8 N 1 ( N 1 + 2 ) / ϵ ) 1 / 2 ) r\geq\max\left(32/\epsilon,(8N_1(N_1+2)/\epsilon)^{1/2}\right) r ≥ max ( 32/ ϵ , ( 8 N 1 ( N 1 + 2 ) / ϵ ) 1/2 )
The next lemma, which we use to prove Corollary 2 , and again in § 6 , follows from Lemma 9 by approximating U U U
Lemma 10. Let α 1 ∈ R ∖ Q \alpha_1\in\mathbb{R}\setminus\mathbb{Q} α 1 ∈ R ∖ Q α 2 ∈ R \alpha_2\in\mathbb{R} α 2 ∈ R U ⊆ R / Z U\subseteq\mathbb{R}/\mathbb{Z} U ⊆ R / Z ∣ ∂ U ∣ = 0 |\partial U|=0 ∣ ∂ U ∣ = 0 ϵ > 0 \epsilon>0 ϵ > 0 r > 0 r>0 r > 0 K ⊆ R 2 K\subseteq\mathbb{R}^2 K ⊆ R 2 r r r
∣ # { m ∈ Z 2 ∩ K : ⟨ α , m ⟩ ∈ U } − ∣ U ∣ ∣ K ∣ ∣ ≤ ϵ ∣ K ∣ . \left|\#\{m\in\mathbb{Z}^2\cap K:\langle\alpha,m\rangle\in U\}-|U||K|\right|\leq\epsilon|K|. # { m ∈ Z 2 ∩ K : ⟨ α , m ⟩ ∈ U } − ∣ U ∣∣ K ∣ ≤ ϵ ∣ K ∣. Proof. Assume without loss of generality that ϵ < 1 \epsilon<1 ϵ < 1 ∂ U = ∅ \partial U=\varnothing ∂ U = ∅ U U U R / Z \mathbb{R}/\mathbb{Z} R / Z Lemma 9 with ν = 1 \nu=1 ν = 1 ∂ U \partial U ∂ U O O O ∣ O ∣ < ϵ / 4 |O|<\epsilon/4 ∣ O ∣ < ϵ /4 ( R / Z ) ∖ O (\mathbb{R}/\mathbb{Z})\setminus O ( R / Z ) ∖ O ∂ U \partial U ∂ U U U U U U U U − U_{-} U − U U U U + = U − ∪ O U_{+}=U_{-}\cup O U + = U − ∪ O U − ⊆ U ⊆ U + U_{-}\subseteq U\subseteq U_{+} U − ⊆ U ⊆ U + U − U_{-} U − U + U_{+} U + ∣ U + ∣ − ∣ U − ∣ ≤ ∣ O ∣ < ε / 4 |U_{+}|-|U_{-}|\leq|O|<\varepsilon/4 ∣ U + ∣ − ∣ U − ∣ ≤ ∣ O ∣ < ε /4 Lemma 9 to U − U_{-} U − U + U_{+} U + ε / 4 \varepsilon/4 ε /4 ε \varepsilon ε □ \square □
Proof of Corollary 2, assuming Theorem 3: Let α \alpha α W W W Theorem 3 , write Y = Y α ( W ) Y=Y_{\alpha}(W) Y = Y α ( W ) Λ = Λ α ( W ) \Lambda=\Lambda_{\alpha}(W) Λ = Λ α ( W ) Ψ \Psi Ψ Lemma 4 , and let U ⊆ R / Z U\subseteq\mathbb{R}/\mathbb{Z} U ⊆ R / Z W W W R / Z \mathbb{R}/\mathbb{Z} R / Z Λ = { m ∈ Z 2 : ⟨ α , m ⟩ ∈ U } \Lambda=\{m\in\mathbb{Z}^{2}:\langle\alpha,m\rangle\in U\} Λ = { m ∈ Z 2 : ⟨ α , m ⟩ ∈ U } W W W 1 1 1 ∣ U ∣ = ∣ W ∣ |U|=|W| ∣ U ∣ = ∣ W ∣ ∣ ∂ U ∣ = ∣ ∂ W ∣ = 0 |\partial U|=|\partial W|=0 ∣ ∂ U ∣ = ∣ ∂ W ∣ = 0 ∂ W \partial W ∂ W r > 0 r>0 r > 0 A ⊆ R 2 A\subseteq\mathbb{R}^{2} A ⊆ R 2 N r ( A ) N_{r}(A) N r ( A ) r r r A A A
Let Q ⊆ V α Q\subseteq V_{\alpha} Q ⊆ V α R ≥ 1 R\geq1 R ≥ 1 P = pr ( Q ) P=\operatorname{pr}(Q) P = pr ( Q ) P P P ∣ P ∣ = R 2 / ∣ n α ∣ |P|=R^{2}/|n_{\alpha}| ∣ P ∣ = R 2 /∣ n α ∣ c 1 R c_{1}R c 1 R c 2 R c_{2}R c 2 R c 1 , c 2 > 0 c_{1},c_{2}>0 c 1 , c 2 > 0 α \alpha α Lemma 4 , if y ∈ Y y\in Y y ∈ Y Ψ ( y ) \Psi(y) Ψ ( y ) s W s_{W} s W pr ( y ) \operatorname{pr}(y) pr ( y ) y ∈ Y ∩ Q y\in Y\cap Q y ∈ Y ∩ Q Ψ ( y ) ∉ P \Psi(y)\notin P Ψ ( y ) ∈ / P Ψ ( y ) ∈ N s W ( ∂ P ) \Psi(y)\in N_{s_{W}}(\partial P) Ψ ( y ) ∈ N s W ( ∂ P ) m ∈ Λ ∩ P m\in\Lambda\cap P m ∈ Λ ∩ P Ψ − 1 ( m ) ∉ Q \Psi^{-1}(m)\notin Q Ψ − 1 ( m ) ∈ / Q m ∈ N s W ( ∂ P ) m\in N_{s_{W}}(\partial P) m ∈ N s W ( ∂ P )
∣ # ( Y ∩ Q ) − # ( Λ ∩ P ) ∣ ≤ # ( Z 2 ∩ N s W ( ∂ P ) ) ≤ c 3 R , \left|\#(Y\cap Q)-\#(\Lambda\cap P)\right|\leq\#\left(\mathbb{Z}^{2}\cap N_{s_{W}}(\partial P)\right)\leq c_{3}R, ∣ # ( Y ∩ Q ) − # ( Λ ∩ P ) ∣ ≤ # ( Z 2 ∩ N s W ( ∂ P ) ) ≤ c 3 R , where c 3 c_{3} c 3 α \alpha α W W W ε > 0 \varepsilon>0 ε > 0 ε / 2 \varepsilon/2 ε /2 ε \varepsilon ε R R R
∣ # ( Λ ∩ P ) − ∣ U ∣ ∣ P ∣ ∣ ≤ ε ∣ P ∣ 2 ≤ ε R 2 2 . \left|\#(\Lambda\cap P)-|U||P|\right|\leq\frac{\varepsilon|P|}{2}\leq\frac{\varepsilon R^{2}}{2}. ∣ # ( Λ ∩ P ) − ∣ U ∣∣ P ∣ ∣ ≤ 2 ε ∣ P ∣ ≤ 2 ε R 2 . Since ∣ U ∣ ∣ P ∣ = ∣ W ∣ R 2 / ∣ n α ∣ = δ R 2 |U||P|=|W|R^{2}/|n_{\alpha}|=\delta R^{2} ∣ U ∣∣ P ∣ = ∣ W ∣ R 2 /∣ n α ∣ = δ R 2
For the second statement, identify V α V_{\alpha} V α R 2 \mathbb{R}^{2} R 2 r k r_{k} r k Q k Q_{k} Q k ϕ k \phi_{k} ϕ k S ⊆ [ 0 , 1 ] 2 S\subseteq[0,1]^{2} S ⊆ [ 0 , 1 ] 2 1 / T 1/T 1/ T T ≥ 1 T\geq1 T ≥ 1 ϕ k \phi_{k} ϕ k r k r_{k} r k ϕ k ( S ) \phi_{k}(S) ϕ k ( S ) r k / T r_{k}/T r k / T
μ k ( S ) = # ( Y ∩ ϕ k ( S ) ) r k 2 ⟶ δ T 2 = δ ∣ S ∣ as k → ∞ . \mu_{k}(S)=\frac{\#(Y\cap\phi_{k}(S))}{r_{k}^{2}}\longrightarrow\frac{\delta}{T^{2}}=\delta|S|\quad\text{as }k\to\infty. μ k ( S ) = r k 2 # ( Y ∩ ϕ k ( S )) ⟶ T 2 δ = δ ∣ S ∣ as k → ∞. Since Y Y Y Y Y Y ϕ k ( S ) \phi_{k}(S) ϕ k ( S ) S S S T 2 T^{2} T 2 1 / T 1/T 1/ T [ 0 , 1 ] 2 [0,1]^{2} [ 0 , 1 ] 2 O T ( r k ) O_{T}(r_{k}) O T ( r k ) [ 0 , 1 ] 2 [0,1]^{2} [ 0 , 1 ] 2 μ k \mu_{k} μ k δ d u \delta\,du δ d u Y Y Y δ \delta δ u ↦ δ 1 / 2 u u\mapsto\delta^{1/2}u u ↦ δ 1/2 u
Now we consider the patterns in the plane produced by windows of a particular kind. For an integer Q ≥ 1 Q\geq1 Q ≥ 1 w ∈ ( 0 , 1 ) w\in(0,1) w ∈ ( 0 , 1 ) γ ∈ R \gamma\in\mathbb{R} γ ∈ R
U ( Q , w , γ ) = { x ∈ R / Z : ∣ ∥ Q x − γ ∥ ∣ < w / 2 } , U(Q,w,\gamma)=\{x\in\mathbb{R}/\mathbb{Z}:|\|Qx-\gamma\||<w/2\}, U ( Q , w , γ ) = { x ∈ R / Z : ∣∥ Q x − γ ∥∣ < w /2 } , U > ( Q , w , γ ) = { x ∈ R / Z : ∣ ∥ Q x − γ ∥ ∣ > w / 2 } . U^{>}(Q,w,\gamma)=\{x\in\mathbb{R}/\mathbb{Z}:|\|Qx-\gamma\||>w/2\}. U > ( Q , w , γ ) = { x ∈ R / Z : ∣∥ Q x − γ ∥∣ > w /2 } . The first of these is a union of Q Q Q w / Q w/Q w / Q stage with modulus Q Q Q w w w γ \gamma γ ω ∈ R 2 ∖ { 0 } \boldsymbol{\omega} \in\mathbb{R}^{2} \setminus\{0\} ω ∈ R 2 ∖ { 0 }
G ( ω , w , γ ) = { x ∈ R 2 : ∥ ⟨ ω , x ⟩ − γ ∥ < w / 2 } , G > ( ω , w , γ ) = { x ∈ R 2 : ∥ ⟨ ω , x ⟩ − γ ∥ > w / 2 } . G(\boldsymbol{\omega},w,\gamma)=\{x\in\mathbb{R}^{2}:\lVert\langle\boldsymbol{\omega},x\rangle-\gamma\rVert<w/2\}, \\
G^{>}(\boldsymbol{\omega},w,\gamma)=\{x\in\mathbb{R}^{2}:\lVert\langle\boldsymbol{\omega},x\rangle-\gamma\rVert>w/2\}. G ( ω , w , γ ) = { x ∈ R 2 : ∥⟨ ω , x ⟩ − γ ∥ < w /2 } , G > ( ω , w , γ ) = { x ∈ R 2 : ∥⟨ ω , x ⟩ − γ ∥ > w /2 } . The set G ( ω , w , γ ) G(\boldsymbol{\omega},w,\gamma) G ( ω , w , γ ) ω \boldsymbol{\omega} ω 1 / ∣ ω ∣ 1/|\boldsymbol{\omega}| 1/∣ ω ∣ w / ∣ ω ∣ w/|\boldsymbol{\omega}| w /∣ ω ∣ G > ( ω , w , γ ) G^{>}(\boldsymbol{\omega},w,\gamma) G > ( ω , w , γ ) strip families .
Lemma 11. Let α ∈ R 2 \alpha\in\mathbb{R}^{2} α ∈ R 2 Q ≥ 1 Q\geq1 Q ≥ 1 Q α = μ + ω Q\alpha=\boldsymbol{\mu}+\boldsymbol{\omega} Q α = μ + ω μ ∈ Z 2 \boldsymbol{\mu}\in\mathbb{Z}^{2} μ ∈ Z 2 ω ∈ R 2 ∖ { 0 } \boldsymbol{\omega}\in\mathbb{R}^{2}\setminus\{0\} ω ∈ R 2 ∖ { 0 } w ∈ ( 0 , 1 ) w\in(0,1) w ∈ ( 0 , 1 ) γ ∈ R \gamma\in\mathbb{R} γ ∈ R m ∈ Z 2 m\in\mathbb{Z}^{2} m ∈ Z 2 ⟨ α , m ⟩ ∈ U ( Q , w , γ ) \langle\alpha,m\rangle\in U(Q,w,\gamma) ⟨ α , m ⟩ ∈ U ( Q , w , γ ) m ∈ G ( ω , w , γ ) m\in G(\boldsymbol{\omega},w,\gamma) m ∈ G ( ω , w , γ ) ⟨ α , m ⟩ ∈ U > ( Q , w , γ ) \langle\alpha,m\rangle\in U^{>}(Q,w,\gamma) ⟨ α , m ⟩ ∈ U > ( Q , w , γ ) m ∈ G > ( ω , w , γ ) m\in G^{>}(\boldsymbol{\omega},w,\gamma) m ∈ G > ( ω , w , γ ) U ⊆ R / Z U\subseteq\mathbb{R}/\mathbb{Z} U ⊆ R / Z E ⊆ R 2 E\subseteq\mathbb{R}^{2} E ⊆ R 2
{ m ∈ Z 2 : ⟨ α , m ⟩ ∈ U } = Z 2 ∩ E . \{m\in\mathbb{Z}^{2}:\langle\alpha,m\rangle\in U\}=\mathbb{Z}^{2}\cap E. { m ∈ Z 2 : ⟨ α , m ⟩ ∈ U } = Z 2 ∩ E . Proof. We have that Q ⟨ α , m ⟩ = ⟨ μ , m ⟩ + ⟨ ω , m ⟩ Q\langle\alpha,m\rangle=\langle\boldsymbol{\mu},m\rangle+\langle\boldsymbol{\omega},m\rangle Q ⟨ α , m ⟩ = ⟨ μ , m ⟩ + ⟨ ω , m ⟩ ⟨ μ , m ⟩ ∈ Z \langle\boldsymbol{\mu},m\rangle\in\mathbb{Z} ⟨ μ , m ⟩ ∈ Z ∥ Q ⟨ α , m ⟩ − γ ∥ = ∥ ⟨ ω , m ⟩ − γ ∥ \lVert Q\langle\alpha,m\rangle-\gamma\rVert=\lVert\langle\boldsymbol{\omega},m\rangle-\gamma\rVert ∥ Q ⟨ α , m ⟩ − γ ∥ = ∥⟨ ω , m ⟩ − γ ∥ m m m □ \square □
The representation Q α = μ + ω Q\alpha=\boldsymbol{\mu}+\boldsymbol{\omega} Q α = μ + ω ω \boldsymbol{\omega} ω α 1 \alpha_{1} α 1 α 2 \alpha_{2} α 2 γ \gamma γ α \alpha α
Lemma 12. Let α 1 ∈ R ∖ Q \alpha_{1}\in\mathbb{R}\setminus\mathbb{Q} α 1 ∈ R ∖ Q v = 1 , 2 , … v=1,2,\ldots v = 1 , 2 , … t v ∈ R 2 ∖ { 0 } \mathbf{t}_{v}\in\mathbb{R}^{2}\setminus\{0\} t v ∈ R 2 ∖ { 0 } ϵ v ∈ ( 0 , 1 ) \epsilon_{v}\in(0,1) ϵ v ∈ ( 0 , 1 ) ϵ v ∣ t v ∣ < 1 \epsilon_{v}|\mathbf{t}_{v}|<1 ϵ v ∣ t v ∣ < 1 Q v ≥ 1 Q_{v}\geq1 Q v ≥ 1 α 2 ∈ R \alpha_{2}\in\mathbb{R} α 2 ∈ R 1 , α 1 1,\alpha_{1} 1 , α 1 α 2 \alpha_{2} α 2 Q \mathbb{Q} Q v v v μ v ∈ Z 2 \boldsymbol{\mu}_{v}\in\mathbb{Z}^{2} μ v ∈ Z 2
∣ ω v − t v ∣ ≤ ϵ v ∣ t v ∣ , where ω v = Q v α − μ v . |\boldsymbol{\omega}_{v}-\mathbf{t}_{v}|\leq\epsilon_{v}|\mathbf{t}_{v}|,\qquad\text{where }\boldsymbol{\omega}_{v}=Q_{v}\alpha-\boldsymbol{\mu}_{v}. ∣ ω v − t v ∣ ≤ ϵ v ∣ t v ∣ , where ω v = Q v α − μ v . Proof. Enumerate the countable set Q + Q α 1 \mathbb{Q}+\mathbb{Q}\alpha_{1} Q + Q α 1 ρ 1 , ρ 2 , … \rho_{1},\rho_{2},\ldots ρ 1 , ρ 2 , … I 0 = [ 0 , 1 ] I_{0}=[0,1] I 0 = [ 0 , 1 ] I 0 ⊇ I 1 ⊇ I 2 ⊇ ⋯ I_{0}\supseteq I_{1}\supseteq I_{2}\supseteq\cdots I 0 ⊇ I 1 ⊇ I 2 ⊇ ⋯ I v − 1 I_{v-1} I v − 1 η = ϵ v ∣ t v ∣ / 2 \eta=\epsilon_{v}|\mathbf{t}_{v}|/2 η = ϵ v ∣ t v ∣/2 Q v Q_{v} Q v
Q v ≥ 2 ∣ I v − 1 ∣ and ∥ Q v α 1 − t v , 1 ∥ < η , Q_{v}\geq\frac{2}{|I_{v-1}|}\qquad\text{and}\qquad\lVert Q_{v}\alpha_{1}-t_{v,1}\rVert<\eta, Q v ≥ ∣ I v − 1 ∣ 2 and ∥ Q v α 1 − t v , 1 ∥ < η , where t v = ( t v , 1 , t v , 2 ) \mathbf{t}_{v}=(t_{v,1},t_{v,2}) t v = ( t v , 1 , t v , 2 ) y ∈ R y\in\mathbb{R} y ∈ R ∥ Q v y − t v , 2 ∥ ≤ η / 2 \lVert Q_{v}y-t_{v,2}\rVert\leq\eta/2 ∥ Q v y − t v , 2 ∥ ≤ η /2 η / Q v \eta/Q_{v} η / Q v ( t v , 2 + k ) / Q v (t_{v,2}+k)/Q_{v} ( t v , 2 + k ) / Q v k ∈ Z k\in\mathbb{Z} k ∈ Z ∣ I v − 1 ∣ ≥ 2 / Q v |I_{v-1}|\geq2/Q_{v} ∣ I v − 1 ∣ ≥ 2/ Q v I v − 1 I_{v-1} I v − 1 I v I_{v} I v ρ v \rho_{v} ρ v α 2 \alpha_2 α 2 ⋂ v I v \bigcap_v I_v ⋂ v I v ∣ I v ∣ ≤ η / Q v ≤ 1 / Q v |I_v| \le\eta/Q_v \le1/Q_v ∣ I v ∣ ≤ η / Q v ≤ 1/ Q v Q v ≥ 2 / ∣ I v − 1 ∣ Q_v \ge2/|I_{v-1}| Q v ≥ 2/∣ I v − 1 ∣ v v v μ v , 1 ∈ Z \mu_{v,1} \in\mathbb{Z} μ v , 1 ∈ Z ∣ Q v α 1 − μ v , 1 − t v , 1 ∣ < η |Q_v\alpha_1-\mu_{v,1}-t_{v,1}|<\eta ∣ Q v α 1 − μ v , 1 − t v , 1 ∣ < η μ v , 2 ∈ Z \mu_{v,2}\in\mathbb{Z} μ v , 2 ∈ Z ∣ Q v α 2 − μ v , 2 − t v , 2 ∣ ≤ η / 2 |Q_v\alpha_2-\mu_{v,2}-t_{v,2}|\le\eta/2 ∣ Q v α 2 − μ v , 2 − t v , 2 ∣ ≤ η /2 ∣ ω v − t v ∣ < ( 5 / 4 ) 1 / 2 η < 2 η = ϵ v ∣ t v ∣ |\omega_v-t_v|<(5/4)^{1/2}\eta<2\eta=\epsilon_v|t_v| ∣ ω v − t v ∣ < ( 5/4 ) 1/2 η < 2 η = ϵ v ∣ t v ∣ α 2 ≠ ρ v \alpha_2\ne\rho_v α 2 = ρ v v v v α 2 ∉ Q + Q α 1 \alpha_2\notin\mathbb{Q}+\mathbb{Q}\alpha_1 α 2 ∈ / Q + Q α 1 α 1 ∉ Q \alpha_1\notin\mathbb{Q} α 1 ∈ / Q 1 , α 1 1,\alpha_1 1 , α 1 α 2 \alpha_2 α 2 Q \mathbb{Q} Q
The window: proof of Theorem 3
Throughout this section we fix an irrational number α 1 \alpha_1 α 1 m ∈ Z 2 m\in\mathbb{Z}^2 m ∈ Z 2 t m t_m t m ⟨ α , m ⟩ \langle\alpha,m\rangle ⟨ α , m ⟩ R / Z \mathbb{R}/\mathbb{Z} R / Z L ≥ 1 L\ge1 L ≥ 1 L L L m m m t m t_m t m Proposition 7 , the area pattern which rules out L L L L L L
Geometry of the configuration. Let M ≥ 2 M\ge2 M ≥ 2 N ≥ 1000 N\ge1000 N ≥ 1000 N N N J ≥ 1 J\ge1 J ≥ 1 s > 0 s>0 s > 0 s C ( M , N , J ) s\,\mathcal{C}(M,N,J) s C ( M , N , J ) § 3 . For j ≥ 1 j\ge1 j ≥ 1
ℓ j = s ( N M ) 1 − j and σ j = ℓ j / N , \ell_j=s(NM)^{1-j}\qquad\text{and}\qquad\sigma_j=\ell_j/N, ℓ j = s ( N M ) 1 − j and σ j = ℓ j / N , so that the rectangles of level j j j ℓ j \ell_j ℓ j σ j \sigma_j σ j σ j \sigma_j σ j ℓ j + 1 = σ j / M \ell_{j+1}=\sigma_j/M ℓ j + 1 = σ j / M σ j + 1 = σ j / ( N M ) \sigma_{j+1}=\sigma_j/(NM) σ j + 1 = σ j / ( N M ) reg \operatorname{reg} reg s C ( M , N , J ) s\mathcal{C}(M,N,J) s C ( M , N , J )
Lemma 13. With the notation above, the following statements hold.
(i) Every square of a rectangle of level j j j [ k σ j , ( k + 1 ) σ j ] × [ y , y + σ j ] [k\sigma_j,(k+1)\sigma_j]\times[y,y+\sigma_j] [ k σ j , ( k + 1 ) σ j ] × [ y , y + σ j ] k ∈ Z k\in\mathbb{Z} k ∈ Z y ∈ ℓ j Z y\in\ell_j\mathbb{Z} y ∈ ℓ j Z i i i i ≡ k + 1 ( m o d 2 ) i\equiv k+1\pmod2 i ≡ k + 1 ( mod 2 ) j j j { x 2 ∈ ℓ j + 1 Z } \{x_2\in\ell_{j+1}\mathbb{Z}\} { x 2 ∈ ℓ j + 1 Z }
(ii) Every rectangle of level j ′ > j j'>j j ′ > j { x ∈ R 2 : dist ( x 2 , ℓ j + 1 Z ) ≤ 1.001 σ j + 1 } \{x\in\mathbb{R}^2:\operatorname{dist}(x_2,\ell_{j+1}\mathbb{Z})\le1.001\sigma_{j+1}\} { x ∈ R 2 : dist ( x 2 , ℓ j + 1 Z ) ≤ 1.001 σ j + 1 }
(iii) Every x ∈ reg x\in\operatorname{reg} x ∈ reg ∣ x ∣ ≤ 1.001 s |x|\le1.001s ∣ x ∣ ≤ 1.001 s
Proof. By induction on j j j j j j ℓ j Z × ℓ j Z \ell_j\mathbb{Z}\times\ell_j\mathbb{Z} ℓ j Z × ℓ j Z j = 1 j=1 j = 1 j j j j j j ℓ j + 1 Z × ℓ j + 1 Z \ell_{j+1}\mathbb{Z}\times\ell_{j+1}\mathbb{Z} ℓ j + 1 Z × ℓ j + 1 Z ℓ j = N M ℓ j + 1 \ell_j=NM\ell_{j+1} ℓ j = N M ℓ j + 1 i i i ( x 0 , y 0 ) (x_0,y_0) ( x 0 , y 0 ) k = x 0 / σ j + i − 1 k=x_0/\sigma_j+i-1 k = x 0 / σ j + i − 1 x 0 / σ j ∈ N Z x_0/\sigma_j\in N\mathbb{Z} x 0 / σ j ∈ N Z
Each rectangle of level j + 1 j+1 j + 1 σ j + 1 \sigma_{j+1} σ j + 1 R R R j + 1 j+1 j + 1 R R R R R R
σ j + 1 ( 1 + ( N M ) − 1 + ( N M ) − 2 + ⋯ ) ≤ 1.001 σ j + 1 \sigma_{j+1}\left(1+(NM)^{-1}+(NM)^{-2}+\cdots\right)\le1.001\sigma_{j+1} σ j + 1 ( 1 + ( N M ) − 1 + ( N M ) − 2 + ⋯ ) ≤ 1.001 σ j + 1 above it, using N M ≥ 2000 NM \ge2000 N M ≥ 2000 R R R x 2 ∈ ℓ j + 1 Z x_2 \in\ell_{j+1}\mathbb{Z} x 2 ∈ ℓ j + 1 Z j ′ > j j' > j j ′ > j R R R reg ⊆ [ 0 , s ] × [ 0 , 1.001 σ 1 ] \operatorname{reg} \subseteq[0,s] \times[0,1.001\,\sigma_1] reg ⊆ [ 0 , s ] × [ 0 , 1.001 σ 1 ] σ 1 = s / N ≤ s / 1000 \sigma_1 = s/N \le s/1000 σ 1 = s / N ≤ s /1000 □ \square □
Strip families for one configuration. Keeping the notation above, suppose that for each 1 ≤ j ≤ J 1 \le j \le J 1 ≤ j ≤ J S t j = G ( ω , 1 / 2 , γ ) \mathrm{St}_j = G(\omega,1/2,\gamma) St j = G ( ω , 1/2 , γ ) 1 ≤ j < J 1 \le j < J 1 ≤ j < J B d j + 1 = G ( ω ′ , 5 / ( 2 N ) , γ ′ ) \mathrm{Bd}_{j+1} = G(\omega',5/(2N),\gamma') Bd j + 1 = G ( ω ′ , 5/ ( 2 N ) , γ ′ ) e = ( 1 , 0 ) e=(1,0) e = ( 1 , 0 ) e ⊥ = ( 0 , 1 ) e^\perp=(0,1) e ⊥ = ( 0 , 1 )
∣ ω − e 2 σ j ∣ ≤ σ j 250 s ⋅ 1 2 σ j , ∣ γ − 3 4 ∣ ≤ 1 1000 N , \left|\omega-\frac{e}{2\sigma_j}\right| \le\frac{\sigma_j}{250s}\cdot\frac{1}{2\sigma_j},\qquad\left|\gamma-\frac{3}{4}\right| \le\frac{1}{1000N}, ω − 2 σ j e ≤ 250 s σ j ⋅ 2 σ j 1 , γ − 4 3 ≤ 1000 N 1 , ∣ ω ′ − e ⊥ ℓ j + 1 ∣ ≤ ℓ j + 1 500 N s ⋅ 1 ℓ j + 1 , ∣ γ ′ ∣ ≤ 1 1000 N . (10) \left|\omega'-\frac{e^\perp}{\ell_{j+1}}\right| \le\frac{\ell_{j+1}}{500Ns}\cdot\frac{1}{\ell_{j+1}},\qquad|\gamma'| \le\frac{1}{1000N}. \tag*{(10)} ω ′ − ℓ j + 1 e ⊥ ≤ 500 N s ℓ j + 1 ⋅ ℓ j + 1 1 , ∣ γ ′ ∣ ≤ 1000 N 1 . (10) The stripes S t j \mathrm{St}_j St j j j j bands B d j + 1 \mathrm{Bd}_{j+1} Bd j + 1 B d ‾ j + 1 = R 2 ∖ B d j + 1 \overline{\mathrm{Bd}}_{j+1} = \mathbb{R}^2 \setminus\mathrm{Bd}_{j+1} Bd j + 1 = R 2 ∖ Bd j + 1 B d j + 1 > = G > ( ω ′ , 5 / ( 2 N ) , γ ′ ) \mathrm{Bd}^{>}_{j+1} = G^{>}(\omega',5/(2N),\gamma') Bd j + 1 > = G > ( ω ′ , 5/ ( 2 N ) , γ ′ ) B d ‾ J + 1 = ∅ \overline{\mathrm{Bd}}_{J+1} = \varnothing Bd J + 1 = ∅ B d J + 1 > = R 2 \mathrm{Bd}^{>}_{J+1} = \mathbb{R}^2 Bd J + 1 > = R 2 pattern associated with these families is
E = ⋃ j = 1 J ( B d 2 ∩ ⋯ ∩ B d j ∩ B d j + 1 > ∩ S t j ) , (11) E = \bigcup_{j=1}^{J}\left(\mathrm{Bd}_2 \cap\cdots\cap\mathrm{Bd}_j \cap\mathrm{Bd}^{>}_{j+1} \cap\mathrm{St}_j\right), \tag*{(11)} E = j = 1 ⋃ J ( Bd 2 ∩ ⋯ ∩ Bd j ∩ Bd j + 1 > ∩ St j ) , (11) where for j = 1 j=1 j = 1 R 2 \mathbb{R}^2 R 2 Figure 1 shows this set near a rectangle of level 1, for small values of M M M N N N
Figure 1. Schematic of the pattern E E E N = 8 N=8 N = 8 M = 3 M=3 M = 3 J = 3 J=3 J = 3 E E E B d 2 \mathrm{Bd}_2 Bd 2 S t 1 \mathrm{St}_1 St 1 S t 2 \mathrm{St}_2 St 2 S t 2 \mathrm{St}_2 St 2 S t 3 \mathrm{St}_3 St 3 B d 3 \mathrm{Bd}_3 Bd 3
Lemma 14. With the notation above, let S S S i i i j j j C ( M , N , J ) \mathcal{C}(M,N,J) C ( M , N , J ) σ = σ j \sigma=\sigma_j σ = σ j
(i) Let S ∘ ⊆ S S^\circ\subseteq S S ∘ ⊆ S S S S σ / 100 \sigma/100 σ /100 S S S S l ⊆ S S_l \subseteq S S l ⊆ S σ / 100 \sigma/100 σ /100 S S S i i i S ∘ ⊆ S t j S^\circ\subseteq\mathrm{St}_j S ∘ ⊆ St j i i i S ∩ S t j ⊆ S l S \cap\mathrm{St}_j \subseteq S_l S ∩ St j ⊆ S l
(ii) We have that S ⊆ B d 2 ∩ ⋯ ∩ B d j S \subseteq\mathrm{Bd}_2\cap\cdots\cap\mathrm{Bd}_j S ⊆ Bd 2 ∩ ⋯ ∩ Bd j j < J j<J j < J S ∩ B d ‾ j + 1 S\cap\overline{\mathrm{Bd}}_{j+1} S ∩ Bd j + 1 B n d \mathrm{Bnd} Bnd M + 1 M+1 M + 1 σ \sigma σ 1.26 σ j + 1 1.26\,\sigma_{j+1} 1.26 σ j + 1 2.52 σ j + 1 2.52\,\sigma_{j+1} 2.52 σ j + 1 6 ∣ S ∣ / N 6|S|/N 6∣ S ∣/ N
(iii) We have that E ∩ S ∖ B d ‾ j + 1 = S t j ∩ S ∖ B d ‾ j + 1 E\cap S\setminus\overline{\mathrm{Bd}}_{j+1}=\mathrm{St}_j\cap S\setminus\overline{\mathrm{Bd}}_{j+1} E ∩ S ∖ Bd j + 1 = St j ∩ S ∖ Bd j + 1
Proof. By Lemma 13 (iii), all points x x x ∣ x ∣ ≤ 1.001 s |x|\le1.001\,s ∣ x ∣ ≤ 1.001 s x x x ∣ ⟨ ω , x ⟩ − x 1 / ( 2 σ j ) ∣ ≤ 1.001 / 500 |\langle\omega,x\rangle-x_1/(2\sigma_j)|\le1.001/500 ∣ ⟨ ω , x ⟩ − x 1 / ( 2 σ j ) ∣ ≤ 1.001/500 γ \gamma γ ⟨ ω , x ⟩ − γ \langle\omega,x\rangle-\gamma ⟨ ω , x ⟩ − γ x 1 / ( 2 σ j ) − 3 / 4 x_1/(2\sigma_j)-3/4 x 1 / ( 2 σ j ) − 3/4 1 / 200 1/200 1/200 ∥ x 1 / ( 2 σ j ) − 3 / 4 ∥ < 1 / 4 \|x_1/(2\sigma_j)-3/4\|<1/4 ∥ x 1 / ( 2 σ j ) − 3/4∥ < 1/4 x 1 x_1 x 1 [ k σ j , ( k + 1 ) σ j ] [k\sigma_j,(k+1)\sigma_j] [ k σ j , ( k + 1 ) σ j ] k k k x 1 x_1 x 1 σ j / 100 \sigma_j/100 σ j /100 σ j Z \sigma_j\mathbb{Z} σ j Z x 1 / ( 2 σ j ) − 3 / 4 x_1/(2\sigma_j)-3/4 x 1 / ( 2 σ j ) − 3/4 1 / 200 1/200 1/200 ± 1 / 4 + Z \pm1/4+\mathbb{Z} ± 1/4 + Z x x x S t j \mathrm{St}_j St j k k k Lemma 13 (i), k k k i i i
In the same way, the conditions on ω ′ \omega' ω ′ γ ′ \gamma' γ ′ ∣ x ∣ ≤ 1.001 s |x|\le1.001s ∣ x ∣ ≤ 1.001 s ⟨ ω ′ , x ⟩ − γ ′ \langle\omega',x\rangle-\gamma' ⟨ ω ′ , x ⟩ − γ ′ x 2 / ℓ j + 1 x_2/\ell_{j+1} x 2 / ℓ j + 1 1 / ( 100 N ) 1/(100N) 1/ ( 100 N ) ℓ j + 1 / N = σ j + 1 \ell_{j+1}/N=\sigma_{j+1} ℓ j + 1 / N = σ j + 1 reg \operatorname{reg} reg dist ( x 2 , ℓ j + 1 Z ) < 1.24 σ j + 1 \operatorname{dist}(x_2,\ell_{j+1}\mathbb{Z})<1.24\sigma_{j+1} dist ( x 2 , ℓ j + 1 Z ) < 1.24 σ j + 1 B d j + 1 \mathrm{Bd}_{j+1} Bd j + 1 reg \operatorname{reg} reg dist ( x 2 , ℓ j + 1 Z ) > 1.26 σ j + 1 \operatorname{dist}(x_2,\ell_{j+1}\mathbb{Z})>1.26\sigma_{j+1} dist ( x 2 , ℓ j + 1 Z ) > 1.26 σ j + 1 B d j + 1 > \mathrm{Bd}^{>}_{j+1} Bd j + 1 > j ′ < j j'<j j ′ < j S S S j > j ′ j>j' j > j ′ Lemma 13 (ii) it lies in B d j ′ + 1 \mathrm{Bd}_{j'+1} Bd j ′ + 1 Lemma 13 (i) the bottom side of S S S x 2 = y x_2=y x 2 = y y ∈ ℓ j Z y\in\ell_j\mathbb{Z} y ∈ ℓ j Z S S S σ j = M ℓ j + 1 \sigma_j=M\ell_{j+1} σ j = M ℓ j + 1 x 2 ∈ ℓ j + 1 Z x_2\in\ell_{j+1}\mathbb{Z} x 2 ∈ ℓ j + 1 Z 1.26 σ j + 1 1.26\sigma_{j+1} 1.26 σ j + 1 S S S M + 1 M+1 M + 1 x 2 = y + q ℓ j + 1 x_2=y+q\ell_{j+1} x 2 = y + q ℓ j + 1 0 ≤ q ≤ M 0\le q\le M 0 ≤ q ≤ M ℓ j + 1 = N σ j + 1 \ell_{j+1}=N\sigma_{j+1} ℓ j + 1 = N σ j + 1 S S S S S S 1.26 σ j + 1 1.26\sigma_{j+1} 1.26 σ j + 1 M + 1 M+1 M + 1 ( M + 1 ) ⋅ 2.52 σ j + 1 σ j ≤ 6 ∣ S ∣ / N (M+1)\cdot2.52\sigma_{j+1}\sigma_j\le6|S|/N ( M + 1 ) ⋅ 2.52 σ j + 1 σ j ≤ 6∣ S ∣/ N σ j + 1 = σ j / ( N M ) \sigma_{j+1}=\sigma_j/(NM) σ j + 1 = σ j / ( N M )
For (iii), consider the terms in (11) . If j ′ < j j'<j j ′ < j j ′ j' j ′ B d j ′ + 1 > \mathrm{Bd}^{>}_{j'+1} Bd j ′ + 1 > B d j ′ + 1 ⊇ S \mathrm{Bd}_{j'+1}\supseteq S Bd j ′ + 1 ⊇ S j ′ > j j'>j j ′ > j j ′ j' j ′ B d j + 1 ⊆ B d ‾ j + 1 \mathrm{Bd}_{j+1}\subseteq\overline{\mathrm{Bd}}_{j+1} Bd j + 1 ⊆ Bd j + 1 S ∖ B d ‾ j + 1 S\setminus\overline{\mathrm{Bd}}_{j+1} S ∖ Bd j + 1 B d j + 1 > \mathrm{Bd}^{>}_{j+1} Bd j + 1 > B d 2 ∩ ⋯ ∩ B d j \mathrm{Bd}_2\cap\cdots\cap\mathrm{Bd}_j Bd 2 ∩ ⋯ ∩ Bd j j j j S t j \mathrm{St}_j St j
The construction. For each integer L ≥ 1 L \ge1 L ≥ 1
a L = 400 − L 4 , c L = a L 2 , τ L = a L 100 . a_L = \frac{400^{-L}}{4}, \qquad c_L = \frac{a_L}{2}, \qquad\tau_L = \frac{a_L}{100}. a L = 4 40 0 − L , c L = 2 a L , τ L = 100 a L . Let M L M_L M L β L \beta_L β L N 0 , L N_{0,L} N 0 , L Lemma 5 for L L L c L c_L c L M L ≥ 2 M_L \ge2 M L ≥ 2 N L N_L N L max ( N 0 , L , 1000 ) \max(N_{0,L},1000) max ( N 0 , L , 1000 ) J L J_L J L ( 1 + β L ) J L > L 2 (1+\beta_L)^{J_L}>L^2 ( 1 + β L ) J L > L 2 s ∗ , L s_{*,L} s ∗ , L Proposition 7 for L L L c L c_L c L N L N_L N L J L J_L J L b = L b=L b = L L L L
The piece of the window for L L L 2 J L − 1 2J_L-1 2 J L − 1 § 5 , one for each of the strip families S t 1 , B d 2 , S t 2 , … , B d J L , S t J L \mathrm{St}_1,\mathrm{Bd}_2,\mathrm{St}_2,\ldots,\mathrm{Bd}_{J_L},\mathrm{St}_{J_L} St 1 , Bd 2 , St 2 , … , Bd J L , St J L (11) with J = J L J=J_L J = J L hierarchy L L L v ≥ 1 v\ge1 v ≥ 1 1 1 1 2 2 2 Q v Q_v Q v v v v Lemma 12 , so when we come to choose the data for hierarchy L L L 1 , … , L − 1 1,\ldots,L-1 1 , … , L − 1
ν L = 1 + ∑ L ′ < L ( 2 + ∑ v 2 Q v ) . \nu_L=1+\sum_{L'<L}\left(2+\sum_v 2Q_v\right). ν L = 1 + L ′ < L ∑ ( 2 + v ∑ 2 Q v ) . where the inner sum is over the indices v v v L ′ L' L ′ L ′ < L L'<L L ′ < L
Let r L r_L r L Lemma 9 for α 1 \alpha_1 α 1 ϵ = τ L \epsilon=\tau_L ϵ = τ L ν = ν L \nu=\nu_L ν = ν L
s L ≥ max ( s ∗ , L , 200 r L N L ( N L M L ) J L − 1 ) , s_L\ge\max\left(s_{*,L},\,200r_LN_L(N_LM_L)^{J_L-1}\right), s L ≥ max ( s ∗ , L , 200 r L N L ( N L M L ) J L − 1 ) , so that the squares of the finest level of s L C ( M L , N L , J L ) s_L\mathcal{C}(M_L,N_L,J_L) s L C ( M L , N L , J L ) σ J L ≥ 200 r L \sigma_{J_L}\ge200r_L σ J L ≥ 200 r L s = s L s=s_L s = s L ℓ j \ell_j ℓ j σ j \sigma_j σ j v v v L L L w v w_v w v t v t_v t v ϵ v \epsilon_v ϵ v Lemma 12 . For the stage corresponding to S t j \mathrm{St}_j St j
w v = 1 2 , t v = e 2 σ j , ϵ v = σ j 250 s L , w_v=\frac{1}{2}, \qquad t_v=\frac{e}{2\sigma_j}, \qquad\epsilon_v=\frac{\sigma_j}{250s_L}, w v = 2 1 , t v = 2 σ j e , ϵ v = 250 s L σ j , and for the stage corresponding to B d j + 1 \mathrm{Bd}_{j+1} Bd j + 1
w v = 5 2 N L , t v = e ⊥ ℓ j + 1 , ϵ v = ℓ j + 1 500 N L s L . w_v=\frac{5}{2N_L}, \qquad t_v=\frac{e^\perp}{\ell_{j+1}}, \qquad\epsilon_v=\frac{\ell_{j+1}}{500N_Ls_L}. w v = 2 N L 5 , t v = ℓ j + 1 e ⊥ , ϵ v = 500 N L s L ℓ j + 1 . The construction alternates between the choices above and the steps of Lemma 12 : once the data for hierarchy L L L Lemma 12 for the stages of hierarchy L L L L + 1 L+1 L + 1 ϵ v ∣ t v ∣ \epsilon_v|t_v| ϵ v ∣ t v ∣ 1 / ( 500 s L ) 1/(500s_L) 1/ ( 500 s L ) 1 / ( 500 N L s L ) 1/(500N_Ls_L) 1/ ( 500 N L s L ) 1 1 1 L L L Lemma 12 then gives a number α 2 \alpha_2 α 2 1 1 1 α 1 \alpha_1 α 1 α 2 \alpha_2 α 2 Q \mathbb{Q} Q v v v ω v = Q v α − μ v \omega_v=Q_v\alpha-\mu_v ω v = Q v α − μ v ω \omega ω ω ′ \omega' ω ′ (10) .
Now that α \alpha α T = { t m : m ∈ Z 2 } T=\{t_m:m\in\mathbb{Z}^2\} T = { t m : m ∈ Z 2 } R / Z \mathbb{R}/\mathbb{Z} R / Z v v v L L L w v w_v w v γ v 0 \gamma_v^0 γ v 0 γ v \gamma_v γ v ∣ γ v − γ v 0 ∣ ≤ 1 / ( 1000 N L ) |\gamma_v-\gamma_v^0|\leq1/(1000N_L) ∣ γ v − γ v 0 ∣ ≤ 1/ ( 1000 N L )
γ v ∉ { ⟨ ω v , m ⟩ ± w v / 2 : m ∈ Z 2 } + Z . (12) \gamma_v\notin\{\langle\omega_v,m\rangle\pm w_v/2:m\in\mathbb{Z}^2\}+\mathbb{Z}. \tag*{(12)} γ v ∈ / {⟨ ω v , m ⟩ ± w v /2 : m ∈ Z 2 } + Z . (12) This is possible because the excluded set is countable. Then choose p ∈ R p\in\mathbb{R} p ∈ R p p p p + 400 − L / 2 p+400^{-L}/2 p + 40 0 − L /2 p + 400 − L / 2 + a L p+400^{-L}/2+a_L p + 40 0 − L /2 + a L L ≥ 1 L\geq1 L ≥ 1 T T T
I L = ( p , p + 400 − L ] and A L = ( p + 400 − L 2 , p + 400 − L 2 + a L ) I_L=(p,p+400^{-L}]
\qquad\text{and}\qquad
A_L=\left(p+\frac{400^{-L}}{2},p+\frac{400^{-L}}{2}+a_L\right) I L = ( p , p + 40 0 − L ] and A L = ( p + 2 40 0 − L , p + 2 40 0 − L + a L ) (as subsets of R / Z \mathbb{R}/\mathbb{Z} R / Z U v = U ( Q v , w v , γ v ) U_v=U(Q_v,w_v,\gamma_v) U v = U ( Q v , w v , γ v ) U v > = U > ( Q v , w v , γ v ) U_v^{>} = U^{>}(Q_v,w_v,\gamma_v) U v > = U > ( Q v , w v , γ v ) H L ⊆ R / Z H_L\subseteq\mathbb{R}/\mathbb{Z} H L ⊆ R / Z L L L (11) , and let
W ∘ = ⋃ L ≥ 1 ( A L ∩ H L ) . W^\circ=\bigcup_{L\geq1}(A_L\cap H_L). W ∘ = L ≥ 1 ⋃ ( A L ∩ H L ) . We define W ⊆ R W\subseteq\mathbb{R} W ⊆ R W ∘ W^\circ W ∘ ( p , p + 1 / 400 ) (p,p+1/400) ( p , p + 1/400 )
Note that A L ⊆ I L ∖ I L + 1 A_L\subseteq I_L\setminus I_{L+1} A L ⊆ I L ∖ I L + 1 A L A_L A L A L A_L A L I L + 1 I_{L+1} I L + 1 ∣ I L + 1 ∣ = 400 − L − 1 = a L / 100 |I_{L+1}|=400^{-L-1}=a_L/100 ∣ I L + 1 ∣ = 40 0 − L − 1 = a L /100 Lemma 11 , for each L L L E L E_L E L (11) from the strip families G ( ω v , w v , γ v ) G(\omega_v,w_v,\gamma_v) G ( ω v , w v , γ v ) L L L
{ m ∈ Z 2 : t m ∈ A L ∩ H L } = { m ∈ Z 2 ∩ E L : t m ∈ A L } . (13) \{m\in\mathbb{Z}^2:t_m\in A_L\cap H_L\}=\{m\in\mathbb{Z}^2\cap E_L:t_m\in A_L\}. \tag*{(13)} { m ∈ Z 2 : t m ∈ A L ∩ H L } = { m ∈ Z 2 ∩ E L : t m ∈ A L } . (13) By our choice of phases and Lemma 12 , the strip families of hierarchy L L L (10) with s = s L s=s_L s = s L M = M L M=M_L M = M L N = N L N=N_L N = N L J = J L J=J_L J = J L Lemma 14 applies to them.
Properties of the window. Each set A L ∩ H L A_L\cap H_L A L ∩ H L ∂ A L \partial A_L ∂ A L L L L x ∈ W ∘ ∖ W ∘ ‾ x\in W^\circ\setminus\overline{W^\circ} x ∈ W ∘ ∖ W ∘ x ≠ p x\neq p x = p A L A_L A L p p p x x x x x x A L ∩ H L A_L\cap H_L A L ∩ H L ∂ W ∘ \partial W^\circ ∂ W ∘ p p p A L A_L A L W ∘ W^\circ W ∘ ∂ W ∘ ⊆ ∂ W ∘ ‾ \partial W^\circ\subseteq\partial\overline{W^\circ} ∂ W ∘ ⊆ ∂ W ∘ W W W W W W W ∘ W^\circ W ∘ W W W A L ∩ H L A_L\cap H_L A L ∩ H L W W W
Proof of part (i) of Theorem 3 . We show that T T T ∂ W ∘ \partial W^\circ ∂ W ∘ ∂ W \partial W ∂ W R / Z \mathbb{R}/\mathbb{Z} R / Z p p p T T T p p p A L A_L A L U v U_v U v x x x Q v x ∈ γ v ± w v / 2 + Z Q_vx\in\gamma_v\pm w_v/2+\mathbb{Z} Q v x ∈ γ v ± w v /2 + Z x = t m x=t_m x = t m Q v ⟨ α , m ⟩ = ⟨ μ v , m ⟩ + ⟨ ω v , m ⟩ Q_v\langle\alpha,m\rangle=\langle\mu_v,m\rangle+\langle\omega_v,m\rangle Q v ⟨ α , m ⟩ = ⟨ μ v , m ⟩ + ⟨ ω v , m ⟩ ⟨ μ v , m ⟩ ∈ Z \langle\mu_v,m\rangle\in\mathbb{Z} ⟨ μ v , m ⟩ ∈ Z γ v ∈ ⟨ ω v , m ⟩ ∓ w v / 2 + Z \gamma_v\in\langle\omega_v,m\rangle\mp w_v/2+\mathbb{Z} γ v ∈ ⟨ ω v , m ⟩ ∓ w v /2 + Z (12) . This proves part (i). It also shows that
Λ α ( W ) = { m ∈ Z 2 : t m ∈ W ∘ } . (14) \Lambda_\alpha(W)=\{m\in\mathbb{Z}^2:t_m\in W^\circ\}. \tag*{(14)} Λ α ( W ) = { m ∈ Z 2 : t m ∈ W ∘ } . (14) Counting points in the squares of one configuration. Fix L L L a = a L a=a_L a = a L τ = τ L \tau=\tau_L τ = τ L r = r L r=r_L r = r L Λ = Λ α ( W ) \Lambda=\Lambda_\alpha(W) Λ = Λ α ( W )
U b g = ⋃ L ′ < L ( A L ′ ∩ H L ′ ) . U^{\mathrm{bg}}=\bigcup_{L'<L}(A_{L'}\cap H_{L'}). U bg = L ′ < L ⋃ ( A L ′ ∩ H L ′ ) . We claim that every square S S S j j j s L C ( M L , N L , J L ) s_LC(M_L,N_L,J_L) s L C ( M L , N L , J L )
# ( Λ ∩ S ) = ∣ U b g ∣ ∣ S ∣ + χ S a ∣ S ∣ + e S , with ∣ e S ∣ ≤ 0.07 a ∣ S ∣ , (15) \#(\Lambda\cap S)=|U^{\mathrm{bg}}||S|+\chi_Sa|S|+e_S,\qquad\text{with } |e_S|\leq0.07a|S|, \tag*{(15)} # ( Λ ∩ S ) = ∣ U bg ∣∣ S ∣ + χ S a ∣ S ∣ + e S , with ∣ e S ∣ ≤ 0.07 a ∣ S ∣ , (15) where χ S = 1 \chi_S=1 χ S = 1 S S S χ S = 0 \chi_S=0 χ S = 0
To see this, note first that the pieces A L ′ ∩ H L ′ A_{L'}\cap H_{L'} A L ′ ∩ H L ′ L ′ < L L'<L L ′ < L A L ∩ H L A_L\cap H_L A L ∩ H L L ′ ′ > L L''>L L ′′ > L U b g U^{\mathrm{bg}} U bg A L A_L A L I L + 1 I_{L+1} I L + 1 [14 ] and [13 ] , # ( Λ ∩ S ) = X 1 + X 2 + X 3 \#(\Lambda\cap S)=X_1+X_2+X_3 # ( Λ ∩ S ) = X 1 + X 2 + X 3
X 1 = # { m ∈ Z 2 ∩ S : t m ∈ U b g } , X 2 = # { m ∈ Z 2 ∩ S ∩ E L : t m ∈ A L } , X_1=\#\{m\in\mathbb{Z}^2\cap S:t_m\in U^{\mathrm{bg}}\},\qquad X_2=\#\{m\in\mathbb{Z}^2\cap S\cap E_L:t_m\in A_L\}, X 1 = # { m ∈ Z 2 ∩ S : t m ∈ U bg } , X 2 = # { m ∈ Z 2 ∩ S ∩ E L : t m ∈ A L } , and
0 ≤ X 3 ≤ # { m ∈ Z 2 ∩ S : t m ∈ I L + 1 } . 0\leq X_3\leq\#\{m\in\mathbb{Z}^2\cap S:t_m\in I_{L+1}\}. 0 ≤ X 3 ≤ # { m ∈ Z 2 ∩ S : t m ∈ I L + 1 } . All of the rectangles to which we apply Lemma 9 below are compact and convex, and contain a closed disc of radius r r r S S S S ∘ S^\circ S ∘ 0.98 σ j 0.98\sigma_j 0.98 σ j σ j / 100 ≥ 2 r \sigma_j/100\geq2r σ j /100 ≥ 2 r 1.26 σ j + 1 ≥ 2 r 1.26\sigma_{j+1}\geq2r 1.26 σ j + 1 ≥ 2 r j < J L j<J_L j < J L σ J ≥ 200 r \sigma_J\geq200r σ J ≥ 200 r
The set U b g U^{\mathrm{bg}} U bg R / Z \mathbb{R}/\mathbb{Z} R / Z A L ′ A_{L'} A L ′ L ′ < L L'<L L ′ < L A L ′ A_{L'} A L ′ Q v Q_v Q v 2 Q v 2Q_v 2 Q v ν L \nu_L ν L U b g U^{\mathrm{bg}} U bg ν L \nu_L ν L U b g U^{\mathrm{bg}} U bg ν L \nu_L ν L Lemma 9 with ϵ = τ \epsilon=\tau ϵ = τ
X 1 = ∣ U b g ∣ ∣ S ∣ ± τ ∣ S ∣ and X 3 ≤ ( ∣ I L + 1 ∣ + τ ) ∣ S ∣ = a 50 ∣ S ∣ . X_1=|U^{\mathrm{bg}}||S|\pm\tau|S|\qquad\text{and}\qquad X_3\leq(|I_{L+1}|+\tau)|S|=\frac{a}{50}|S|. X 1 = ∣ U bg ∣∣ S ∣ ± τ ∣ S ∣ and X 3 ≤ ( ∣ I L + 1 ∣ + τ ) ∣ S ∣ = 50 a ∣ S ∣. For X 2 X_2 X 2
cnt ( R ) = # { m ∈ Z 2 ∩ R : t m ∈ A L } . \operatorname{cnt}(R)=\#\{m\in\mathbb{Z}^2\cap R:t_m\in A_L\}. cnt ( R ) = # { m ∈ Z 2 ∩ R : t m ∈ A L } . If R R R S S S S ∘ S^\circ S ∘ A L A_L A L Lemma 9 gives cnt ( R ) = a ∣ R ∣ ± τ ∣ R ∣ \operatorname{cnt}(R)=a|R|\pm\tau|R| cnt ( R ) = a ∣ R ∣ ± τ ∣ R ∣ Lemma 14 (iii), X 2 X_2 X 2 cnt ( S ∩ S t j ) \operatorname{cnt}(S\cap\mathrm{St}_j) cnt ( S ∩ St j ) B d ‾ j + 1 \overline{\mathrm{Bd}}_{j+1} Bd j + 1 cnt ( B n d ) \operatorname{cnt}(\mathrm{Bnd}) cnt ( Bnd ) Lemma 14 (ii) is at most ( a + τ ) ⋅ 6 ∣ S ∣ / N L ≤ 0.007 a ∣ S ∣ (a+\tau)\cdot6|S|/N_L\leq0.007a|S| ( a + τ ) ⋅ 6∣ S ∣/ N L ≤ 0.007 a ∣ S ∣ j = J L j=J_L j = J L S S S Lemma 14 (i), cnt ( S ∘ ) ≤ cnt ( S ) \operatorname{cnt}(S^\circ)\leq\operatorname{cnt}(S) cnt ( S ∘ ) ≤ cnt ( S ) ∣ S ∘ ∣ = 0.98 ∣ S ∣ |S^\circ|=0.98|S| ∣ S ∘ ∣ = 0.98∣ S ∣ cnt ( S ∩ S t j ) = a ∣ S ∣ ± ( 0.02 a + τ ) ∣ S ∣ \operatorname{cnt}(S\cap\mathrm{St}_j)=a|S|\pm(0.02a+\tau)|S| cnt ( S ∩ St j ) = a ∣ S ∣ ± ( 0.02 a + τ ) ∣ S ∣ S S S 0 ≤ cnt ( S ∩ S t j ) ≤ cnt ( S l ) ≤ ( a + τ ) ∣ S ∣ / 50 ≤ 0.03 a ∣ S ∣ 0\leq\operatorname{cnt}(S\cap\mathrm{St}_j)\leq\operatorname{cnt}(\mathrm{Sl})\leq(a+\tau)|S|/50\leq0.03a|S| 0 ≤ cnt ( S ∩ St j ) ≤ cnt ( Sl ) ≤ ( a + τ ) ∣ S ∣/50 ≤ 0.03 a ∣ S ∣ X 2 = χ S a ∣ S ∣ ± 0.037 a ∣ S ∣ X_2=\chi_Sa|S|\pm0.037a|S| X 2 = χ S a ∣ S ∣ ± 0.037 a ∣ S ∣ τ = a / 100 \tau=a/100 τ = a /100 (15) .
Proof of part (ii) of Theorem 3 . First we record the nonemptiness properties mentioned above. Fix L L L S S S s L C ( M L , N L , J L ) s_LC(M_L,N_L,J_L) s L C ( M L , N L , J L ) X 2 = a ∣ S ∣ ± 0.037 a ∣ S ∣ X_2=a|S|\pm0.037a|S| X 2 = a ∣ S ∣ ± 0.037 a ∣ S ∣ (15) , there is an m ∈ Z 2 ∩ S m\in\mathbb{Z}^2\cap S m ∈ Z 2 ∩ S t m ∈ A L ∩ H L t_m\in A_L\cap H_L t m ∈ A L ∩ H L A L ∩ H L A_L\cap H_L A L ∩ H L A L A_L A L W W W
It also follows that W ∘ W^\circ W ∘ W W W I I I U U U I I I R / Z \mathbb{R}/\mathbb{Z} R / Z ϵ = ∣ I ∣ / 2 \epsilon=|I|/2 ϵ = ∣ I ∣/2 r 0 > 0 r_0>0 r 0 > 0 r 0 r_0 r 0 R 2 \mathbb{R}^2 R 2 Λ α ( I ) \Lambda_\alpha(I) Λ α ( I ) Λ \Lambda Λ Λ \Lambda Λ b Λ b_\Lambda b Λ
Suppose, by way of contradiction, that there is an L 0 L_0 L 0 g : Λ → Z 2 g:\Lambda\to\mathbb{Z}^2 g : Λ → Z 2 L ≥ max ( L 0 , b Λ ) L\geq\max(L_0,b_\Lambda) L ≥ max ( L 0 , b Λ ) g g g L L L Λ \Lambda Λ L L L S S S S ′ S' S ′ sLC ( M L , N L , J L ) \operatorname{sLC}(M_L,N_L,J_L) sLC ( M L , N L , J L ) (15) ,
∣ # ( Λ ∩ S ′ ) − # ( Λ ∩ S ) ∣ ≥ a L ∣ S ∣ − 0.14 a L ∣ S ∣ > c L ∣ S ∣ . \left|\#(\Lambda\cap S')-\#(\Lambda\cap S)\right|\geq a_L|S|-0.14a_L|S|>c_L|S|. ∣ # ( Λ ∩ S ′ ) − # ( Λ ∩ S ) ∣ ≥ a L ∣ S ∣ − 0.14 a L ∣ S ∣ > c L ∣ S ∣. Since s L ≥ s ∗ , L s_L\geq s_{*,L} s L ≥ s ∗ , L Proposition 7 , applied with x = 0 x=0 x = 0 s = s L s=s_L s = s L b = L b=L b = L L L L Λ \Lambda Λ Z 2 \mathbb{Z}^2 Z 2 Theorem 3 . Together with the properties of W W W 1 , α 1 1,\alpha_1 1 , α 1 α 2 \alpha_2 α 2 Theorem 3 , and therefore also the proofs of Theorem 1 and Corollary 2 .
We conclude with a remark. The construction gives each hierarchy a fixed contrast c L = a L / 2 c_L=a_L/2 c L = a L /2 0 0 0 L → ∞ L\to\infty L → ∞ Λ \Lambda Λ c > 0 c>0 c > 0 (7) cannot hold at arbitrarily large scales. The price is that the depth J L J_L J L L L L β L \beta_L β L c L 3 L − 6 c_L^3L^{-6} c L 3 L − 6 Lemma 5 ). Department of Mathematics University of Houston Houston, TX, United States haynes@math.uh.edu
