Topological densities of quasiperiodic structures

Remark 1.

It is important to distinguish topological density defined above from the metrical density that can be defined for the set of vertices of the graph embedded into R d as

where Λ is the set of graph vertices and B r d 0 is a d-dimensional ball of radius r centered at zero point. Particularly, topological density depends only on graph G but metrical density also depends on the embedding of G into R d . However, it is quite interesting to note that our methods of calculation of the topological density use some embedding of the graph.

Proposition 1.

If the topological density exists, it does not depend on the choice of the initial vertex.

Proof.

Note that for any vertex y there exists some constant C (depending on y) such that

Therefore

Using the existence of the topological density t d G , x we have

Similar arguments also allow to prove the equality lim n → ∞ ∑ k = n − С + 1 n e k x n d = 0 . Hence, Proposition 1 is proved.

In view of Proposition 1, we will denote the topological density as t d G .

Remark 2.

The topological density can also be determined for tilings by considering the tiles of the tiling instead of the vertices of the graph and supposing the tiles adjacent if they have a common boundary of a nonzero (d-1)-dimensional volume. Obviously, the topological density of the tiling is equal to the topological density of its dual graph.

Theorem 1.

The following asymptotic formula for the coordination numbers of the vertex graph of the Ammann-Beenker tiling holds:

where the function C(x) for x ∈ [ 0 ; 1 is

Theorem 2.

The topological density td(AB) of the vertex graph of the Ammann-Beenker tiling is 8 − 4 2 .

Proof.

From the Theorem 1 we have

And, hence,

Since 2 2 is irrational, the sequence 2 2 k − 1 2 is uniformly distributed modulo one. From the general theory of the uniform distribution of sequences ¹⁴ it is known that for any sequence {x _n } uniformly distributed modulo one and any Riemann integrable function f the following asymptotic formula holds

Using summation by parts, we obtain

Substituting this in (1) we find

So, to complete the proof of Theorem 2, it only remains to calculate the integral.

We now turn to the study of the vertex graph of the Penrose tiling. We use the construction from ref. 13]. Consider two projections π 1 : Z 4 → C and π 2 : Z 4 → C × C 5 (C ₅ is a fifth-order cyclic group) defined as

π 2 h , j , k , l = h + j ζ 5 2 + k ζ 5 4 + l ζ 5 , h + j + k + l mod 5 where ζ 5 = e 2 π i 5 is a complex fifth root of unity. We define a set W⊂С × С₅ as follows: W = ∪ i = 0 4 Ω i , i , where Ω₀ = {0}, Ω₁ = P, Ω₂ = −τP, Ω₃ = τP, Ω₄ = −P, and P is a regular pentagon on a complex plane with vertices 1 , ζ 5 , ζ 5 2 , ζ 5 3 , ζ 5 4   a n d   τ = 1 + 5 2 . Here we assume that the boundaries of the pentagons Ω₂, Ω₄ and vertices of the pentagon Ω₃ are excluded from W. Then the set of vertices of the Penrose tiling is

Again, two vertices are connected by and edge if and only if the distance between them is unity. A fragment of the Penrose tiling is represented in Figure 2.

Note that, with the exception of the zero point, the set of vertices of the Penrose tiling can also be represented in the form V = ∪ i = 1 4 V i , V i = π 1 x : x ∈ L + i , 0 , 0 , 0 , π 2 ′ x ∈ Ω i where π 2 ′ h , j , k , l = h + j ζ 5 2 + k ζ 5 4 + l ζ 5 and the lattice L is L = { a , b , c , d ∈  Z ⁴ : a + b + c + d mod   5 = 0 .

In ref. 11] the following asymptotic formula for the coordination numbers of the vertex graph of the Penrose tiling was proved.

Theorem 3.

For the coordination numbers of the vertex graph of the Penrose tiling the following asymptotic formula holds:

1. C n = 10 τ − 2 + 5 2 τ − 1 , if n is odd and n − 1 2 τ − 2 ∈ [ 0 , τ − 1 )

2. C(n) = 10τ ⁻², if n is odd and n − 1 2 τ − 2 ∈ [ τ − 1 , 1 )

3. C n = 5 τ − 2 + 25 2 τ − 3 , if n is even and n − 2 2 τ − 2 ∈ [ 0 , τ − 3 )

4. C n = 5 − 5 2 τ − 3 n − 2 2 τ − 2 , if n is even and n − 2 2 τ − 2 ∈ [ τ − 3 , τ − 1

5. C n = 5 − 5 τ − 4 + 5 2 τ − 3 n − 2 2 τ − 2 , if n is even and n − 2 2 τ − 2 ∈ [ τ − 1 , 1 ) .

Arguing similarly to the proof of Theorem 2, but with a separate summation over even and odd n and calculating the resulting integrals, we obtain the following result.

Theorem 4.

The topological density td(Penr) of the vertex graph of the Penrose tiling is 25 4 τ − 2 .

In ref. 15] were introduced a fractal set which was named the Rauzy fractal. To define it, consider a substitution σ

It can be proved that σ ⁿ (1) always begins with σ ⁿ⁻¹ (1). Therefore, we can consider an infinite sequence u = σ ^∞ (1) that is a fixed point of this substitution. Let r _n (i) be a number of occurrences of the symbol i between the first n symbols of the sequence u. Let ς be the unique real root of the equation ς ³+ς ²+ς = 1. Assume

Then the set T = δ n : n ∈ N ‾ is the Rauzy fractal. Alternative approaches to define the Rauzy fractal and its generalizations can be found in ref. 16]. Assume

Then we have a partition of the Rauzy fractal into three similar sets (see Figure 3)

Hence, using the inflation-deflation method we obtain a quasiperiodic tiling of the plane known as Rauzy tiling. Its fragment is represented in Figure 4.

In ref. 17] the following conjecture about coordination numbers of the Rauzy tilings was proposed.

Conjecture 1.

For the coordination numbers of the Rauzy tiling the following asymptotic formula holds

An obvious calculation using the definition of topological density and asymptotic ∑ k = 1 n k ∼ n 2 2 shows that the Conjecture 1 implies that the topological density of the Rauzy tiling is 1 2 5 + 2 ς + 5 ς 2 .

Remark 3.

The above definition is a special case of the more general concept of a model set ¹⁸ in which the phase space may not be Euclidean, but an arbitrary locally compact Abelian group.

The set of vertices of the Ammann-Beenker tiling will be a cut-and-project set if we identify the set of complex numbers with the real plane. The definition of the set of vertices of the Penrose tiling, which we used earlier, is not formally covered by this construction (it is only a model set). Nevertheless, it is shown above that the set of vertices of Penrose tiling can be represented as a union of one point (zero point) and four sets obtained by translation from the cut-and-project sets. So, it possible to use the methods discussed below for an alternative calculation of the topological density of the vertex graph of the Penrose tiling. In the case of the Rauzy tiling, it was shown in ref. 19] that in each tile of the Rauzy tiling one can select a special point (so called Rauzy point) in such way that the set of all Rauzy points will be a cut-and-project set with L = Z 3 , π 1 ( ( a , b , c ) ) = ( ( 4 ς − 2 ) a + ( 2 ς − 1 ) b + ( ς − 1 ) c , ( 4 ς 2 − 1 ) a + ( 2 ς 2 − 1 ) + ς 2 c ) ,

Theorem 5.

The growth form of the vertex graph of the Ammann-Beenker tiling is a regular octagon whose vertices on the complex plane are R e k π i 4 , 0 ≤ k ≤ 7, R = 2 2 − 1 .

Theorem 6.

The growth form of the vertex graph of the Penrose tiling is a regular decagon whose vertices on the complex plane are R e 2 π i k 10 , 0 ≤ k ≤ 9, R = τ − 1 sin 2 π 5 + τ − 2 sin π 5 i .

In the case of the Rauzy tiling we only have the following conjecture. ¹⁷

Conjecture 2.

The growth form of the Rauzy tiling is an octagon whose vertices have coordinates ± ( 1 − 2 ς + 2 ς 2 2 − 3 ς − 3 ς 2 ) , ± ( − 2 + 3 ς + 3 ς 2 2 − 4 ς + 2 ς 2 ) , ± ( − 0.5 + ς + 0.5 ς 2 1 − ς + ς 2 ) , ± ( − 0.5 + ς − 0.5 ς 2 ς + ς 2 ) .

Growth forms for some infinite family of graphs whose vertices can be obtained by cut-and-project method were found in ref. 23].

Analysis of the proofs of known results on growth forms and coordination sequences shows that finding the growth form of some graph is a much simpler problem than finding the asymptotic formula for its coordination sequence. Moreover, all known approaches to the study of coordination sequences in the non-periodic case include finding the growth form as one of the steps.

Consider some graph whose set of vertices is a cut-and-project set. Such graphs are studied in ref. 24]. Suppose that we know its growth form. Now we show how to find its topological density without any knowledge about its coordination sequence.

Theorem 7.

Let G be some graph whose set of vertices is a cut-and-project set. Suppose that its growth form exists and is equal to pol _G . Then the topological density of the graph G is

where M _L is a matrix which rows are base vectors of the lattice L and the map π : R m + d → R m + d is π(x)=(π ₁(x),π ₂(x)).

We give two proofs of this theorem.

First proof.

Consider the set E q n x = ∪ k = 1 n e q k x . Let E _n (x) be the number of graph vertices in Eq _n (x). The existence of a growth form pol _G implies that there exists a sequence {c _n } such that lim n → ∞ c n n = 0 and

Here Pol _G is a set with boundary pol _G . Further, consider the following cylinders in R d + m

and Cyl = Cyl (1). From our definitions it follows that

Since images of the lattice L under projections π ₁ and π ₂ are dense in physical and phase spaces respectively, we see that hyperplanes π ₁ (x) = 0 and π ₂ (x) = 0 are totally irrational relatively to the lattice L. Recall that, the m-dimensional volume of the boundary of the set W is zero. So, we have

Since vol _d + m (Cyl (n)) = n ^d vol _d + m (Cyl), we can rewrite (4) as

Further, lim n → ∞ c n n = 0 , and, hence, n ± с n d ∼ n d . In combination with the inclusion (3) and the definition of the topological density, this gives

and finding the topological density is reduced to calculating the volume vol _d+m (Cyl).

Note that π maps cylinder Cyl to the set Pol _G  ⊕ W = {(x, y) : x ∈ Pol _G , y ∈ W}. So, vol _d+m (Pol _G  ⊕ W) = vol _d (Pol _G ) vol _m (W) and vol _d+m (Pol _G  ⊕ W) = |det π| vol _d+m (Cyl). Therefore, we have  v o l d + m C y l = v o l d P o l G v o l m W det π as was required.

Second proof.

Define the metrical density of the set Λ _W as d e n s Λ W = lim r → ∞ # Λ W ∩ B r d 0 v o l d B r d 0 , where B r d 0 is a d-dimensional ball of radius r centered at zero point. If the metrical density exists, then for any vertex x ∈ Λ _W and any convex body Ω we have an asymptotic formula

Choosing Ω = Pol _G and arguing as in the previous proof, we see that

For any vertex x∈Λ _W the map ^* : x → π 2 π 1 − 1 x from the physical to the phase space is correctly defined. Moreover, the set

is a lattice in R m + d . Consider the projections π ₃ ((x, x ^*)) = x and π ₄ ((x, x ^*)) = x ^*. Then the hyperplanes π ₃ (x) = 0 and π ₄ (x) = 0 are orthogonal. Therefore, we get

Note that in the case of orthogonal hyperplanes π ₃ (x) = 0 and π ₄ (x) = 0 it is known, ¹³ that

if the (m-1)-dimensional volume of the boundary of the set W is zero. As a result, we obtain

It is easy to see that L = π L and, hence, det M L = det M L det π . This transform (6) to (2).

Let us show, how we can use Theorem 7 to calculate the topological densities for the examples considered above.

In the case of the vertex graph of the Ammann-Beenker tiling we have v o l 2 W = 2 1 + 2 and det M _L  = det E = 1. From Theorem 5, we get v o l 2 P o l G = 8 3 2 − 4 . The matrix of the map π has form

So, det π = 4 and direct calculation leads to the same result as in the Theorem 2.

In the case of the vertex graph of the Penrose tiling we need separate calculations for each of the sets of vertices V _i . Therefore, we need to rewrite (4) as t d = v o l 2 P o l G ∑ i = 1 4 v o l 2 Ω i det M L det π . Further, we have vol ₂ (Ω₄) = vol ₂ (Ω₁) and vol ₂ (Ω₂) = vol ₂ (Ω₃) = τ ² vol ₂ (Ω₁). Furthermore, det M _L  = 5 and

det π = 5 4 5 . Hence, t d = 4 25 5 2 + 2 τ 2 v o l 2 P o l G v o l 2 Ω 1 . By direct calculation, we have v o l 2 Ω 1 = 5 2 5 + 5 8 and, from Theorem 6, we obtain that

Further, direct calculation leads to the same result as in Theorem 4.

In the case of the Rauzy tiling we have det M _L  = 1, vol ₁ (W) = 1−ς ² and the map π has the matrix

with the determinant det π = ς−5ς ³. If we apply (5) we obtain that t d = 1 2 5 + 2 ς + 5 ς 2 if the Conjecture 2 is true.

Remark 4.

Note that formula (5) give a connection between topological and metrical densities. Metrical densities of some important some graph whose set of vertices is a cut-and-project set were previously known (see, for example, important paper ²⁵ on this topic). Particularly, in ref. 25] it was shown that metrical density of the set of vertices of the Penrose tiling is 125 8 τ 4 2 − τ 2 tan 2 π 5 . Combining this with (5) and (7) we obtain yet another proof that topological density in the Penrose case is 25 4 τ − 2 .