Tiling Spaces: Quasicrystals & Geometry
MA4054 Mathematics Project: Project Dissertation
Thesis submitted at the University of Leicester in partial ful lment of the requirements for the degree of Master of Mathematics
by
Daniel Bragg Department of Mathematics University of Leicester
May 2014 Contents
Declaration iii
Abstract iv
Introduction 1
1 Lattices & Voronoi Cells 3
1.1 Lattices ...... 3
1.1.1 Lattices in one-dimension ...... 4
1.1.2 Lattices in two-dimensions ...... 4
1.1.3 Lattices in three-dimensions ...... 5
1.2 Crystallographic Restriction Theorem ...... 7
1.2.1 De nition of a Crystal ...... 9
1.3 Projection Lattices ...... 10
1.3.1 Voronoi Cells ...... 10
1.3.2 The Projection Method ...... 13
1.4 Dual Lattices ...... 17
1.4.1 Properties of the Dual Lattice ...... 19
1.4.2 Examples of Dual Lattices ...... 19
2 Penrose Tilings 21
2.1 Tiles and Tilings ...... 21
2.2 Penrose Tilings of the Plane ...... 24
2.2.1 Matching Rules ...... 24
i 2.3 Construction Techniques ...... 31
2.3.1 Voronoi Decompositions ...... 31
2.3.2 Substitution ...... 31
2.3.3 Multigrids ...... 33
2.3.4 Pentagrids ...... 36
3 Projection Method for Constructing Tilings of the Plane 41
3.1 The Idea ...... 41
3.2 Penrose Tilings as Projections ...... 43
3.2.1 I5 and the Unit Hypercube Q5 ...... 45
3.2.2 Projecting I5 ...... 47 3.2.3 Projected Plane as Complex Plane ...... 50
3.2.4 Shift Vector −→γ ...... 55
4 Diraction Geometry 58
4.1 Far Field Diraction Geometry ...... 59
4.1.1 One Point Diraction ...... 60
4.1.2 Two Point Diraction ...... 61
4.1.3 N-Point Diraction ...... 64
4.2 Fourier Transforms and the Diraction Condition ...... 65
4.2.1 The Fourier Transform ...... 66
4.2.2 Dirac Deltas and the Diraction Condition ...... 69
4.3 Topological Bragg Peaks - An Alternate Approach ...... 77
5 One Dimensional Crystals 80
5.1 One-Dimensional Delone Sets ...... 80
5.2 Fibonacci Sequences ...... 82
5.3 Projected One-Dimensional Crystals ...... 85
Concluding Remarks 91
ii Declaration
All sentences or passages quoted in this project dissertation from other people's work have been speci cally acknowledged by clear cross referencing to author, work and page(s). I understand that failure to do this amounts to plagiarism and will be considered grounds for failure in this module and the degree examination as a whole.
Name: Daniel Bragg
Signed:
Date:
iii Abstract
Throughout the years a clear link has been established between the structure of crystals and the mathematical concepts of tiling spaces. Until recently crystals were thought to be periodic and ordered structures which can be related to a set of points in the n - dimensional Euclidean space En known as a lattice, where these points act as vertices to the building blocks of crystals. It was known that crystals could only be periodic and ordered if their corresponding lattice had either one, two, three, four or six fold rotational symmetry. However in the mid-20th century quasicrystals were discovered which were ordered and aperiodic but possessed rotational symmetries not listed above. This sparked into life the new area of study known as quasi-crystallography.
This report delves into the mathematics behind these interesting structures. First considered are the lattice structures related to the structure of a crystal. I consider the dierent lattice types in various dimensions and then look at the properties such as the
Crystallographic Restriction Theorem described above. I then introduce the idea of a dual lattice, Voronoi cell construction and the canonical projection method - all of which are important tools to be used later. Tiling spaces are then considered. In particular I will be exploring the properties of the rhomb Penrose tiling which are derived from the famous kite and dart Penrose tiling. These tilings are aperiodic so therefore provide a good model for quasicrystal structures. With all of this knowledge in place, I will then explain the projection method for constructing tilings of the plane from ve dimensional space. I then digress somewhat and explore the mathematics behind the diraction patterns created by crystals and quasicrystals. This is so that the diraction condition can be introduced, which de nes the diraction patterns that represent a crystalline structure. A new method of
iv constructing these diraction patterns known as topological Bragg peaks is also discussed.
Finally I discuss the possibility of having one dimensional crystals. In particular we look at the Fibonacci sequence and show how they produce crystalline structures.
v Introduction
There is a rich history of experimentation, misjudgments and discoveries in the eld of crystallography which presents it as an interesting area of research for both mathematicians and solid state scientists. Although there was no science directly linked to crystals before the seventeenth century, there is evidence of ideas which are related to crystallography as far back as the ancient Greeks, where Plato assigned each of the `four elements' to a regular polyhedra. In 1984 it was announced that there existed crystals with icosahedral symmetry, made from an alloy of Aluminium and Manganese. This was something which was thought to have been impossible for over 200 years prior to this discovery. This breakthrough opened up the door to new areas of research within geometry and tiling spaces as it was now suggested that crystals need no longer have a periodic atomic pattern. Therefore the de nition of what a crystal actually is has been constantly changing throughout all of history.
Throughout my research one book has been of particular interest to me - Marjorie
Senechal's Quasicrystals and Geometry. The main goal of this project was to try and mathematically relate to crystallography, a subject which is predominantly associated with chemists and physicists, and Senechal's book does exactly that. Therefore the main part of my research below is based on this book and is cited with the notation [1]. A lot of the work carried out on the projection method for constructing Penrose tilings of the plane was by the late N.G De Bruijn (1918-2012). One of his papers, Algebraic Theory of Penrose's non-periodic tiling of the plane, was of great use to me in Chapter 3 and is sited with notation [16]. Johannes Kellendonk's work on topological Bragg peaks has completely changed the way we look at diraction patterns. His paper, entitled Topological Bragg
1 Peaks And How They Characterize Point Sets, views diraction patterns as topological in nature rather than statistical. This is sited with notation [12].
I feel that I have deviated away slightly from the goals I set myself in the project description document. I feel that my naivety of the subject led me to believe that some of my chosen areas of research would take up more time than they actually did. However by pursuing these initial objectives I found that the subject branched o so that I could easily study other areas that I didn't even know existed. This has been one of the main factors as to why I've enjoyed researching crystallography so much - the research possibilities are endless!
The content of this project dissertation has successfully covered all the areas which I set as goals to explore in my interim report and therefore feel that this gives a well rounded view into the mathematics of crystalline structures and I feel like I've achieved a great deal over the past eight months. Although following the material set out by the papers above, original material in this dissertation comes in the form of proofs left for the reader to prove, such as the proof of Proposition 5.3.1 in Chapter 5.
2 Chapter 1
Lattices & Voronoi Cells
In this chapter we are going to describe some of the basic de nitions and notions which I will use throughout this paper. To begin with we will consider lattices in Euclidean space and will then go on to applying the ideas behind such tiling spaces to crystallography.
1.1 Lattices
n Consider the n-dimensional Euclidean space E with orthonormal basis ~e1, ..., ~en. Points in our Euclidean space are denoted by lower-case letters (x, y, z etc.) and vectors will always be vectors from the origin 0 ∈ En to the corresponding lower-case letter in space.
n De nition. 1.2.1 [1] Any set of vectors ~u1, ..., ~uk ∈ E generates a countably in nite group under addition and is called a Z- module; it's elements are vectors of the form
~x = m1 ~u1 + ... + mk ~uk
where mi are integers.
A Z- module is the building block for any type of lattice. Before we proceed any further, let's refresh our memory on a particular area of group theory. Let G be a transformation group acting on an object M, where M could be anything from the whole space En to G itself. Let x ∈ M.
De nition. 1.2.2 [1] The orbit of x under G is the set: OG(x) = {gx ∈ M : g ∈ G}
3 Figure 1.1: A lattice in one-dimension is generated by one vector −→u
Using this notion of orbits we can generate lattices.
n De nition. 1.2.3 [1] A Z- module in E is a lattice (of dimension n ) if it is generated by n linearly independent vectors. We denote a lattice by L, and any orbit of L by Lp.
We sometimes refer to lattices of this kind as Bravais lattices, after Auguste Bravais
(1850), who studied lattices in great detail. As we will see very shortly, there are classi - cations for Bravais lattices in n-dimensional space.
De nition. 1.2.4 [1] Any set of n linearly independent lattice vectors that generates L is called a basis for L.
It is important to note here that our choice in basis is not unique. This is due to the periodic nature of Bravais lattices. We're assuming that our basis vectors start at the origin, but if they were to originate from another point in space we could still produce the same lattice because there will exist a linear combination of these new basis vectors which will include our original origin.
1.1.1 Lattices in one-dimension
Lattices in one dimension are described by one vector only. Therefore a lattice is generated by R = m1 ~u1 where m1 ∈ Z. This implies that there is only one possible type of lattice in one-dimension, namely the line with regularly spaced points along it (Figure 1.1).
1.1.2 Lattices in two-dimensions
[2] Analogous to the one-dimensional lattice above, our lattice is now generated by R = m1 ~u1 + m2 ~u2 where m1, m2 ∈ Z. Bravais discovered that in two-dimensions there are 5
4 classi cations of lattice (Figure 1.2):
1. Square Lattices: The two basis vectors have the same magnitude (i.e. | ~u1| = | ~u2|) and the angle between them, π θ = 2
2. Rectangular Lattices: The two basis vectors have dierent magnitudes (i.e. | ~u1| 6= ), but the angle between them, π | ~u2| θ = 2
3. Oblique Lattices: The two basis vectors have dierent magnitudes and the angle
between them, π θ < 2
4. Rhombic (or centered rectangular) Lattices: The two basis vectors have dierent
magnitudes and the angle between them, π . This value of is smaller than that θ < 2 θ of the oblique lattice.
5. Hexagonal Lattices: The two basis vectors have equal magnitudes and the angle
between them, 2π θ = 3
1.1.3 Lattices in three-dimensions
[2] Our lattice is now generated by R = m1 ~a1 +m2 ~a2 +m3 ~a3 where m1, m2, m3 ∈ Z. There are 14 classi cations of Bravais lattices in three-dimensions (Figure 1.3). There are seven basic shapes of lattice along with dierent versions of each cell. For example if we take the orthorhombic lattice there are 4 dierent versions of this:
1. Orthorhombic (P) - This is the original, or primitive orthorhombic lattice shape in
three-dimensions, which is a unit cell built on the basis of the lattice.
2. Orthorhombic (I) - This is the primitive orthorhombic lattice shape with an extra
vertex in the center of it. All vertices surrounding this point will connect to this
vertex.
3. Orthorhombic (C) - Again we have the primitive orthorhombic lattice shape but this
time we have a vertex at the center of each side face. There will only be two extra
vertices in this case (one face and its opposite face). These new vertices will connect
to each vertex of the face they are positioned on.
5 −→ −→ Figure 1.2: The ve lattice types in two-dimensions generated by the vectors u1 and u2
6 Figure 1.3: The fourteen lattice types in three-dimensional space generated by vectors −→ −→ −→ a1, a2 and a3 [3]
4. Orthorhombic (F) - This again has the primitive orthorhombic lattice shape except
this time we have a vertex at the center of each face of the shape. Similar to (C),
these face vertices will connect to each vertex of the face they are positioned on.
The other 10 classi cations are described in Figure 1.4. The angles αij described in this
gure are the angles between basis vectors ai and aj. It is also interesting to add that there are 64 Bravais lattices in four-dimensions. Out of these 64, 23 of them are primitive lattices and 41 are centered. [2]
1.2 Crystallographic Restriction Theorem
If we consider two or three dimensional lattices, you may be surprised that there is as fewer arrangements as there are. This is down to the periodicity of the lattice. In order for
7 our lattice to be periodic it must have a certain rotational symmetry about it. If we take a point in our lattice , rotate about this point by an angle 2π and our lattice x ∈ L θ = N coincides with itself, we can say that the lattice has N- fold rotational symmetry. Without this symmetry we would not see sharp bright spots in the diraction pattern of the crystal which is represented by the lattice. The following theorem tells us what values of N we can use in order to maintain N- fold symmetry.
Theorem. 1.2.1 [1](The Crystallographic Restriction): Rotational symmetries of order greater than six, and also ve-fold rotational symmetry, are impossible for a periodic pattern in the plane or in three-dimensional space.
Proof. We will rst show that periodicity is only possible for N-fold symmetry, where
N ≤ 6. Assume we have a lattice that is both periodic and invariant under N-fold rotation and let x ∈ L be a rotation center. Since the lattice is periodic, we have a countable in nity of such points and there is a minimum distance d between them. Assume y ∈ L is another such point and that . A counter-clockwise rotation through 2π radians about |x − y| = d N x carries y to another rotation point y0and we must have that |y − y0| ≥ d. This is only possible when N ≤ 6 (Figure 1.4 (a))
If N = 5 then, from the argument above, we know that |y − y0| > d, but we then encounter another problem. Since y is also a rotation point, a clockwise rotation about y must take x to another point x0. After looking at (Figure 1.4 (b)) we can see that
2π d1 d2 2π 2π 2π cos( 5 ) = d = d ⇒ d1 + d2 = 2d cos( 5 ) ⇒ d3 = d − 2d cos( 5 ) = d(1 − 2 cos( 5 )) and as 2π it implies that and 0 0 . Therefore periodicity is 1 − 2 cos( 5 ) < 1 d3 < d |x − y | < d incompatible with ve-fold symmetry.
This theorem therefore restricts the number of possible lattices in our space En which in turn suggests that there is only a very small number of crystal formations in these spaces. But as we are about to see quasicrystal theory changed all of this.
8 Figure 1.4: (a) If the angle of rotation is less than π/3 then the edge connecting y and y0 will be less than d. (b) using basic trigonometry we can see that the edge connecting x0 and y0 is less than d.
1.2.1 De nition of a Crystal
The de nition of a crystal has changed on several occasions throughout history. The working informal de nition of a crystal is that it is any solid with an essentially discrete diraction diagram [1]. This cleverly incorporates both crystals and quasicrystals. So what are the dierences between a crystal and a quasicrystal? This brief subsection will provide the de nitions of the two entities to clear up any confusion as to what they are.
De nition. 1.2.2 [1] An n - dimensional periodic crystal is an n - dimensional regular system of points which admits translations in n independent directions. An n - dimensional aperiodic crystal is one with no translational symmetry.
The reason for providing this de nition is that quasicrystals are a special class of aperiodic crystals
De nition. 1.2.3 [1] An n - dimensional quasicrystal is a crystal whose diraction pattern exhibits symmetries forbidden by the crystallographic restriction theorem for En.
9 This means that the quasicrystal phenomenon actually begins in E2 because there are no non-crystallographic rotational symmetries of the line.
1.3 Projection Lattices
We have seen that, due to the Crystallographic Restriction, periodic lattices can only have one, two, three, four or six fold rotational symmetry in order to produce sharp diraction patterns. However since the discovery of quasicrystals lattices with ve, eight, ten or twelve fold symmetry have been discovered which have sharp diraction patterns. This suggests that the only way quasicrystals can be structured is if they are ordered and aperiodic. The best way to gain a better understanding of their structure is to leave three-dimensions and consider higher dimensional structures. The Crystallographic restriction only applies to two or three-dimensional structures, so by moving to higher dimensions, we discover that we have lattices with the forbidden symmetries I have described above. By considering a higher dimensional structure and projecting it down to two or three-dimensions it is possible to preserve rotational symmetry. In this section I will describe this process, but to begin with we need to de ne some mathematical tools which will help us carry this out.
1.3.1 Voronoi Cells
In crystallography it is often the case that lattices are associated with tiling spaces. The tiles in question are called unit cells and they are chosen so that their edges run parallel to the basis vectors of the lattice. However the unit cell does have its limitations as a carrier of visual information as it's not always chosen to display the symmetry of the lattice.
Luckily there is another type of cell which we can use to display things such as sym- metry, namely the Voronoi cell.
De nition. 1.3.1 [1] A point set Λ ∈ En is said to be discrete if there exists a positive real number r0 such that, for every x, y ∈ Λ, |x − y| ≥ 2r0. 1.3.2 [1] A point set Λ ∈ En is relatively dense in En if there is a positive real number
R0 such that every sphere of radius greater than R0 contains at least one point of Λ in its interior.
10 Figure 1.5: The construction of a Voronoi cell about a point X. 1. Connect X to all of the nearest points around it with line segments.
1.3.3 [1] A point set Λ ∈ En is called a Delone Set if it is discrete and relatively dense.
These three de nitions allow us to now de ne a Voronoi Cell
De nition. 1.3.4 [1] Let Λ ∈ En be any Delone set. The V oronoi Cell of a point x ∈ Λ is the set of points of En that lie at least as close to x as to any other point of Λ. That is
V (x) = {u ∈ En : |x − u| ≤ |y − u| ∀y ∈ Λ}
Construction:
Take a point x in the lattice. Connect to x each of the points in the lattice which are closest by straight line segments (Figure 1.5). From this we construct the (n − 1)- dimensional perpendicular bisector of each of these straight line segments (Figure 1.6).
The Voronoi cell V (x) is the smallest convex region about x which is bounded by these perpendicular bisectors (Figure 1.7).
When performed on each point x ∈ Λ we obtain a partition of En into cells. This partition is called the Voronoi tessellation induced by Λ. The (n − 1)- dimensional faces of a Voronoi cell V (x) are called its facets and the vectors joining x to another lattice point whose Voronoi cell shares a facet with V (x) are called facet vectors. Therefore the lines connecting the point X to all of its nearest points are in fact facet vectors.
11 Figure 1.6: The construction of a Voronoi cell about a point X. 2. Take the perpendicular bisector of these line segments.
Figure 1.7: The construction of a Voronoi cell about a point X. 3. The Voronoi cell is the smallest convex region bounded by these bisectors containing X.
12 1.3.2 The Projection Method
We now look at the method of projecting a higher dimensional lattice onto a lower dimen- sional plane. We will also consider the properties of this new lattice to show that it is in fact aperiodic.
De nition. 1.3.5 [1] Let L be a lattice and ε be any k- dimensional subspace of En, where
0 < k < n. If εTL = {0} then ε is said to be totally irrational.
Let L be a lattice in En, ε be a totally irrational d- dimensional subspace of En and let
ε⊥ be its orthogonal complement (this orthogonal complement may or may not be totally irrational). We now de ne an orthogonal projection function Π which projects the points x ∈ L onto ε and let Π⊥ be the orthogonal projection function which maps the points of the lattice onto ε⊥. The rst step of our projection method is to x a compact subset
K ⊂ ε⊥ with a non-empty interior. From now on we are going to call K our acceptance domain for the projection. We now have a selection criterion:
[1] Project x ∈ L orthogonally onto ε if and only if Π⊥(x) lies in the acceptance domain K.
An immediate question which arises is what is our choice for the acceptance domain
K? The answer to this question is the `canonical' choice of K is Π⊥(V (0)) where V (0) is the Voronoi cell of the origin of L. So there we have it. We now have a method for projecting a higher dimensional lattice onto a lower dimensional plane. The best way to see this is if we consider the example where we are projecting a two-dimensional lattice onto a one-dimensional plane (the case where n = 2 and d = 1). Consider the integer lattice and draw through L ∈ E2 the totally irrational one-dimensional subspace, ε , which only intersects L at one point. We call this point the origin of L. From this we can immediately draw the orthogonal plane ε⊥. For convenience we will draw this orthogonal plane through the origin as well. We then construct the Voronoi cell of the origin and we will project, onto ε, the points x ∈ L such that Π⊥(x) ∈ K. These will be the points de ned by the cylinder C (Figure 1.8) which, by construction, is given by C = K L ε. We then proceed to project all points in this C onto our subspace ε (Figure 1.9).
13 Figure 1.8: Construction of the acceptance domain, K, and the cylinder of points, C, which we shall project onto ε
Figure 1.9: Projection onto the one-dimensional subspace
14 Now that we have constructed a projected lattice, we need to ask questions about the properties of this lattice. Is it a Delone set? Is it aperiodic? The following results will answer these questions.
Proposition. 1.3.6 [1] De ne X = C TL. Then Π(X) is a Delone set.
Proof. In order to prove this we must show that Π(X) is both uniformly discrete and relatively dense in ε. The most challenging thing to prove here is discreteness, as relative density immediately follows from the fact that L is relatively dense in En by de nition and hence also in the cylinder C. To prove discreteness, we must show that there is a neighborhood of the origin in ε that contains no other points of Π(X). Let x ∈ L and assume that Π(x) lies in B0(c), a ball of radius c > 0 in ε which is centered at the origin. 2 2 2 We know that |x| = |Π(x)| + Π⊥(x) and we also know that Π⊥(x) is bounded by assumption ⇒ x must lie in some sphere of radius m > 0 which is centered at the origin. T Since L is discrete, L B0(m) is a nite subset of points of the lattice U ⊂ L ⇒ Π(U) is T also a nite set of points and for su ciently small r > 0 we have that Π(U) B0(r) = {0}⇒ Π(X) is uniformly discrete and therefore a Delone set.
In order to show that Π(X) is a non-periodic lattice, we need to prove some properties of the orbits of Z- modules.
k n Theorem. 1.3.7 Let Π: R −→ R be a surjective linear mapping, where n < k. Then n n there are subspaces V,W of R such that R = V ⊕ W and k k T k T (i) Π(Z ) = Π(Z ) W + Π(Z ) V , k T (ii) Π(Z ) W is a discrete point lattice in W , k T (iii) Π(Z ) V is a dense subgroup of V .
k Proof. [1] We shall assume that Π(Z ) w Z has an accumulation point. We will go on to k show that in this case V is non trivial. Since Π(Z ) is a group, 0 is an accumulation point. n For each r > 0 and v ∈ R we let Ur(v) be the open ball of radius r about v. Since 0 is T k an accumulation point we have that for all r > 0, {0} ⊂ Ur(0) Π(Z ). Let Vr be the R - linear span of T k . Then and implies . Set T . Ur(0) Π(Z ) Vr 6= 0 r > s Vs ⊂ Vr V = r>0 Vr T k For all su ciently small r, Vr = V and Ur(0) Π(Z ) spans V .
15 −1 T k k k Let K := Π (V ) Z = {x ∈ Z : Π(x) ∈ V }. Then K is a subgroup of Z and hence k there is a basis e1, ..., ek of Z and positive integers a1, ..., am for some m ≤ k such that
{a1e1, ..., akek} is a basis of K. By the de nition of K we see that aiei ∈ K implies that
k ek ∈ K so K = Ze1 ⊕ ... ⊕ Zen. We de ne L := Zen+1 + ... + Zek so Z = L ⊕ K. n n n Let π : R → R /V be the natural map and let η : L → R /V such that η = π ◦(Π |L). Then
(i) η is injective since L T Π−1(V ) = 0,
n k m (ii) η(L) spans R /V since Π(Z ) spans R , (iii) η(L) is discrete.
In fact, if we suppose that {li} is a sequence of elements of L such that {η(li)} converges in n . We can assume that converges to . Write n 0 and 0 R /V {η(li)} 0 R = V ⊕V Π(li) = vi+vi for all such that 0 0 and . We can identify n and 0. The hypothesis i vi ∈ V vi ∈ V R /V V is then that {vi} −→ 0. Since Π(K) is dense in V we can choose ji ∈ K such that
|vi − Π(ji)| < 1/(2i) . Then
0 0 |Π(li − ji)| = |Π(li) − Π(ji)| = vi + vi − Π(ji) ≤ vi + |vi − Π(ki)| −→ 0
T Thus for positive i, Π(li −ji) ∈ V and li −ji ∈ K. Then li ∈ K L = 0 and η(li) = 0. This
n proves discreteness. From (ii) and (iii), η(L) is a lattice of rank =dim R −dimV . From (i), L has the same rank. Thus k − j = n − dimV . De ne W to be the real span of Π(L).
n Since L = Zen+1 + ... + Zek, dimW ≤ k − j = n − dimV and since R = W + V we have n that dimW = m − dimV and R = W ⊕ V . Since η(L) is discrete and π is continuous, we have that Π(L) is discrete also and it has the same rank as L. So L ' Π(L) and Π(L) is a lattice in W . Finally we have that
k \ k \ Π(L) = Π(Z ) W, Π(K) = Π(Z ) V since
16 k Π(Z ) = Π(L) + Π(K)
Π(L) ⊂ W, Π(K) ⊂ V
This allows us to prove the following proposition.
Proposition. 1.3.8 If ε is totally irrational, then Π(X) is non periodic
Proof. [1] By using Theorem 1.3.7 above, we can decompose ε⊥ into subspaces V and
W . Since Π⊥(L) is one-to-one and dimε⊥ < n, V is assumed to be non trivial and the restriction of Π to (V ⊕ ε) T L is one-to-one since Π⊥((V ⊕ ε) T L) is dense in V
⇒since K has a non empty interior, (K T V ) T Π⊥(L) is dense in K T V . We write T T XV = ((K V ) ⊕ ε) L. It is easy to show that if Π(XV ) is periodic then XV is also periodic. However this is impossible because K is compact and thus cannot contain all of the integer multiples of any vector. Since (K T V ) T Π⊥(L) is nite ⇒ X cannot be periodic and neither can Π(X).
1.4 Dual Lattices
We will end this chapter with some discussion on dual lattices. We will need these construc- tions to understand ideas in later chapters, for example when we consider the multigrid construction of tiling spaces in chapter 2.
Suppose we have a three dimensional lattice Λ which is generated by the vectors −→ −→a , b , −→c ∈ E3. Since Λ is an orbit of the group generated by the corresponding trans- −→ lations, we can label each lattice point by a linear combination u−→a + v b + w−→c with u, v, w ∈ R. A lattice plane through the origin has the equation of the form hx+ky+lz = 0 3 where h, k, l ∈ Z with no common factors. This implies that for any triple (u, v, w) ∈ Z : hu + kv + lw = c where c ∈ Z. What we have here is an equation of a plane parallel to the
rst one. Every lattice point will satisfy this equation for some c ∈ Z which will rely on the
17 Figure 1.10: A three dimensional lattice decomposed into parallel planes [10] appropriate choices of (u, v, w). We can therefore interpret our lattice as a decomposition of stacks of parallel planes by the choices of (h, k, l) (Figure 1.10)
If we were to consider the Cartesian coordinate system, the expression hu + kv + lw =
(h, k, l) · (u, v, w) is a dot product so we can interpret (h, k, l) as the Cartesian coordinates of a normal vector to the stack of parallel planes. This is generally not a lattice vector.
If we want to think of it as a linear combination of basis vectors, the vectors will not be −→ −→ −→ −→ −→a , b and −→c but some other triple of vectors a∗, b∗ and c∗ such that:
−→ −→ −→ −→ (ha∗ + kb∗ + lc∗ ) · (u−→a + v b + w−→c ) = hu + kv + lw
−→ −→ −→ −→ −→ −→ −→ −→ This is only possible when a∗ · −→a = b∗ · b = c∗ · −→c = 1 and b∗ · −→a = c∗ · −→a = a∗ · b = −→ −→ −→ −→ c∗ · b = a∗ · −→c = b∗ · −→c = 0 The integral linear combination of these three vectors form a translation group whose orbit is also a lattice which we will denote by Λ∗. This is what we call the dual lattice of
Λ. We can equivalently de ne the dual lattice as the set of all vectors (h, k, l) such that, for each (u, v, w) in Λ we have:
(h, k, l) · (u, v, w) = c where c ∈ Z [10] However, possibly the most formal de nition of a dual lattice is given below
18 ∗ n De nition. 1.4.1 For a lattice Λ we de ne the dual lattice: Λ = {y ∈ R | ∀x ∈ Λ: hx, yi ∈ Z} [11]
1.4.1 Properties of the Dual Lattice
One of the main properties of the dual lattice is that the dual of the dual lattice is the original lattice (i.e. Λ∗∗ = Λ). This can be proven from the fact that the vectors of Λ∗ are orthogonal to the lattice planes of Λ and the vectors of Λ are also orthogonal to the lattice planes of Λ∗. In particular we can say that the faces of the Voronoi cells of a point in Λ are parallel to the lattice planes of Λ∗ and vice versa.
Consider, for example, two successive planes of Λ whose equations are hx + ky + lz = 0
(a) and hx + ky + lz = 1 (b), and suppose that d is the distance between these two planes.
Let (u, v, w) be a lattice point in the plane of (b). The length of the projection of the vector (u, v, w) onto the line containing (h, k, l) is equal to d and this projection is given by:
(u, v, w) · (h, k, l) hu + kv + lw 1 = = = d |(h, k, l)| |(h, k, l)| |(h, k, l)|
Hence the length of (h, k, l) is the reciprocal to the distance between the lattice planes, i.e |(h, k, l)| = 1/d . This implies that if we want to nd a stack of lattice planes which has the largest inter-planar spacing, all we have to do is nd the shortest dual lattice vector.
Since the unit cells of a lattice all have the same volume and since the height of a unit cell is the inter-planar spacing, we can conclude that
Corollary. 1.4.2 |(h, k, l)|is proportional to the density of the plane lattice to which is it orthogonal [10]
1.4.2 Examples of Dual Lattices
n ∗ n 1. The lattice of integer points satis es (Z ) = Z . Hence it is often called self dual.
2. Let n=2. Then we can show 2 ∗ 1 2 (Figure 1.11). This is obvious when you (2Z ) = 2 Z take a look at the de nition of a dual lattice. Take a point of the original lattice, say
19 2 Figure 1.11: Section of the lattice of 2Z (in blue) and it's dual (in green)
2 (2, 2). What we want to nd is all of the points in (x, y) ∈ R such that when we take the dot product of these two vectors we will have an integer.
20 Chapter 2
Penrose Tilings
We are now going to consider some of the basic ideas of tiles and tiling spaces as well as various construction techniques of such spaces. Once we have these ideas in place we will then look into a speci c type of Penrose tiling of the plane - that is the Penrose tiling with two rhombs. We are going to consider this tiling in particular because these rhombs t in well with the standard crystallographic ideas; they look like two-dimensional versions of the unit cells of periodic crystals and because they are convex they are more suited to the models for crystal structure compared to other Penrose tilings (such as the kite & dart tiling, which we will see is just another form of the rhomb tiling). We will then go onto look at dierent ways to construct these Penrose tilings, with the aim of eventually using the ideas from the previous chapter to consider Penrose tilings as a projection from higher dimensional spaces.
2.1 Tiles and Tilings
A tiling is a subdivision of space into a countable number of tiles. Tilings arise in various contexts in real life situations. For example we may consider the arrangement of cells during cell division or, more relevantly, one may study how crystals grow as atomic units arrange themselves into some sort of pattern.
De nition. 2.1.1 [1] A tiling T of the space En is a countable family of closed sets called tiles:
21 Figure 2.1: An example of a tiling which has two prototiles [4]
T = {T1,T2,...}
such that T Ø if and S∞ n. int(Ti) int(Tj) = i 6= j i=1 Ti = E
To keep in with the context of this report, we will be considering tilings whose tiles are a copy of a nite set of `prototiles'. So for example, if we consider a chess board, we have two prototiles: one black square and one white square (Figure 2.1)
De nition. [1] 2.1.2 Let {T1,T2,...} be the tiles of a tiling T , partitioned into the set of equivalence classes by some criterion M. A set P of representatives of these classes is called the protoset for T
When I talk about equivalence classes here, all I mean is that each prototile may be repeated in multiple places within the space. It is simply a way of talking about these tiles collectively. M may be a transformation group or decorations of the tiles, or both. It is also reasonable to assume that, for the purposes of crystallography, the shapes of the tiles we will be considering are of reasonable shape and size. Speci cally we will assume that for a xed protoset P of a tiling T ⊂ E n : 1) Each tile in our space is homeomorphic to the n-dimensional ball.
22 Figure 2.2: An example of a non-periodic tiling [1]
T 2) The intersection of any pair of tiles Ti Tj is a connected set (the empty set is connected)
Tiles do not need to be convex or even polytopes, but for the cases we will consider (in crystallography) they will be.
De nition. 2.1.3 [1] A tiling of En is said to be periodic if it admits translations in n linearly independent directions.
The chess board example in Figure 2.1 is a periodic tiling. We say that periodic tilings can be partitioned into unit cells.
De nition. 2.1.4 [1] A tiling of En is said to be non-periodic if it admits no translations.
For example Figure 2.2 is non-periodic because the point from which the spiral origi- nates is unique which means that if we were to translate this shape by a vector, we could
nd no such vector so that the tiling would ` t' back into its original pattern
De nition. 2.1.5 [1] A protoset is said to be aperiodic if it admits only non-periodic tilings.
Tilings with aperiodic protosets are called aperiodic tilings.
The most famous aperiodic protosets are the sets of polygons discovered by Penrose
(i.e. The Penrose tiles) which is the main topic of this chapter.
23 2.2 Penrose Tilings of the Plane
2.2.1 Matching Rules
A Penrose tiling is a non-periodic tiling generated by an aperiodic set of prototiles. These tilings are named after the mathematician Roger Penrose who investigated these tilings throughout the 1970's [5]. Penrose tilings are self-similar, so the patterns occur at larger and larger scales. If a Penrose tiling was to be implemented into a physical structure it would produce sharp bright spots on a diraction pattern, implying that it is a quasicrystal.
We will focus on one speci c type of Penrose tiling which consists of two rhombs, one
'thick` and one 'thin`, with arrows on their edges which must be matched as the tiles are
tted together. These rhombs t in well with the standard crystallographic ideas: They look like two-dimensional versions of the unit cells of periodic crystals and because they are convex they are more suited as models for crystal structure than any other Penrose protoset.
These rhomb tilings can be constructed through various methods such as substitutions and projection methods, which is something I will be working towards explaining throughout this project.
One of Penrose's famous tilings is the kite and dart tiling which consists of two quadri- laterals which t together to form a rhombus. However matching rules deny such a pairing of the two shapes. One of the matching rules which can be applied is to colour the vertices of the kite and dart with two colours and require that in order for two tiles to t together the colours must match. [5]. The rhomb tilings which we desire can be derived from the kite and dart tilings in the following way: We bisect each kite and dart in the tiling into isosceles triangles and we then merge these triangles so that we create a tiling of 'thick` rhombs and 'thin` rhombs. We denote these rhombs by T and t respectively. The T rhombs have angles of size and and the t rhombs have angles of size and where π 2θ 3θ θ 4θ θ = 5 (Figure 2.3).
.
We now need matching rules for our rhomb tiling. There are many dierent matching rules which will tile the plane, but the one which we will focus on is the one where the edges are marked with either single or double arrows (Figure 2.4). These matching rules
24 Figure 2.3: A 'T` and 't` rhomb
Figure 2.4: Matching rule for the rhombs. Edges are allowed to meet if and only if the arrows match. [1] not only guarantee non-periodicity - they also force any tiling which obeys them to be a
Penrose tiling.
Even though matching tiles with these rules does produce Penrose tilings, it cannot be guaranteed that every con guration of tiles will produce a Penrose tiling automatically.
There are many so called 'locally legal` con gurations which are not 'globally legal`. If a tiling contains any of these globally illegal con gurations then it is not a Penrose tiling.
In other words it is essential that the tiling obeys the matching rules described in Figure
2.4. An example of some of these con gurations can be seen in Figure 2.5.
It has been discovered that there are a nite number of 'globally legal` con gurations,
25 Figure 2.5: An example of some con gurations which are 'globally illegal` [1] in fact there are seven con gurations which are shown in Figure 2.6. We characterise these con gurations by labeling the vertices with integers from the set {1,2,3,4}. Which number we choose to label the vertex with depends on the sum of the angles (as a multiple of θ) that meet there. We can see from the gure that there are two ways of displaying our matching rules for the (2,2,2,2,2) con guration. We now have enough knowledge to tackle the following theorem:
Theorem. 2.2.1 [1] Any tiling of the plane by T and t rhombs whose stars are restricted to the seven types shown is a Penrose tiling, provided that no two stars that share a rhomb are related by a half turn about the center of the rhomb.
Proof. In order to prove this theorem we will show that every tiling satisfying these condi- tions obeys Penrose's matching rules. This implies we need to prove that there is never an inconsistency when these stars are extended in accordance with the vertex atlas. We will therefore take each star in turn and inspect the vertices. Each boundary vertex is part of an incomplete con guration so we need to show that, when completed, there won't be a contradiction. To prove this for each boundary vertex of each star will be very laborious so I will prove it for a select number of vertices of a select number of stars to show the
26 Figure 2.6: Atlas of the seven globally legal vertex stars with their vertex cycles [1]
27 Figure 2.7: By looking at the vertex with partial string 1,3 on the (2,2,2,4) star, we can see that the only star which ts this vertex is the (1,2,1,3,3) star and after closer inspection we can see it doesn't lead to any globally illegal con gurations. [1] methods we would use to prove the rest.
1) Consider the star with vertex cycle (2,2,2,4) - This has eight boundary vertices, whose angles (beginning with the top vertex) are given by 2; 3, 3; 2; 1, 3; 4; 1, 3; 2; 3, 3 (as multiples of θ). Take the vertex with angles 1,3. By looking at the star atlas and their vertex cycles, we can see that the only possible cycle which would complete this would be
(1,2,1,3,3) and we can see by inspection that there can be no contradiction (Figure 2.7).
Take each of the other boundary vertices and test in a similar way.
2) Consider the star with vertex cycle (2,2,2,2,2) - This has ten boundary vertices with partial strings 3,3 and 2. A vertex with string 3,3 can be completed in two ways:
a) (1,2,1,3,3) - Figure 2.8 shows that after completing this cycle we must next complete a cycle with string 2,4. There are two ways to do this: (2,4,4) and (2,2,2,4). The former of these ts in with the con guration evident on Figure 2.8, however the latter does not as it forces a vertex with cycle (3,3,3,1) which is not in our atlas. We discard this option and
28 Figure 2.8: Completing the 3,3 partial string on the (2,2,2,2,2) star with a (1,2,1,3,3) star. [1] are left with (2,4,4) which forces every 3,3 vertex to be completed with (1,2,1,3,3). This is consistent with (2,2,2,2,2) (a) of Figure 2.6.
b) (3,3,4) - This choice forces the cycles of the neighbouring vertices to contain the string
1,2. There are three ways of completing this: (1,2,1,3,3), (1,1,2,2,2,2) and (1,1,2,1,1,2,2).
By looking at Figure 2.9 we can see that the rst one of these works with no inconsistency in the arrowing, the second one gives rise to the forbidden (1,3,3,3) cycle and the third one can be realised in two ways: one of which works and one that doesn't.
We can look at the boundary vertices with partial string 2 in a similar manner.
We have considered two of the vertex stars here but we can approach the other ve in a similar way to show that this theorem does indeed hold.
There are many methods of constructing Penrose tilings, but we will focus only one of them which we will see later on in this chapter, namely the pentagrid construction method.
29 Figure 2.9: Completing the 3,3 partial string of the (2,2,2,2,2) star with a (3,3,4) star. [1]
30 2.3 Construction Techniques
In this section we discuss various techniques for constructing non-periodic tilings of the space En.
2.3.1 Voronoi Decompositions
As we have seen in the previous chapter, every Delone set Λ ⊂ En induces an associated tiling V (Λ), namely the Voronoi tessellation of En. If we have a Delone set which is non-periodic this means that V (Λ) will also be non-periodic.
We construct such a tiling by constructing the Voronoi cell of each point of Λ as explained previously. The tiles are convex and share whole facets.
2.3.2 Substitution
De nition. 2.3.1 [1] A tiling is said to be hierarchical if its tiles can be merged or composed to form a tiling on a larger scale with a nite protoset, and these tiles can then be composed to form a tiling on a still larger scale with a nite protoset, and so on ad in nitum.
We will only consider the case where the prototiles of the larger scale tiling are rescaled versions of the smaller ones.
To a protoset P we can assign a composition matrix M = (mij)i,j=1,...,m where m is the number of prototiles and mij is the number of a particular prototile in the composition. If M happens to be primitive, by which we mean all of its entries are non-negative and for some power all of its entries are positive, we call the resultant tiling a substitution tiling.
[1]
A simple example of a substitution tiling is the so-called chair tiling (Figure 2.10). The chair prototile consists of three unit squares and each chair can be decomposed into four smaller chairs. We can then enlarge this tiling until each of the four chairs is the same size as the original prototile, then these chairs can be decomposed into 4 even smaller chairs and so on ad in nitum. The limit of this procedure gives us a substitution tiling of the plane. The tiles of the original tiling are tiles of level 0, the tiles of the rst hierarchical level are tiles of level 1, and so forth.
31 Figure 2.10: The chair tiling of the plane is an example of a substitution tiling [6]
De nition. 2.3.2 [1] A hierarchical tiling T is said to be uniquely hierarchical if its level
-m tiles can be composed into level -(m + 1) tiles in only one way, for all m = 0, 1, 2, ... An example of a hierarchical tiling which is not uniquely hierarchical is the tiling of the plane by unit squares. We can choose many dierent substitution rules which would all tile the plane in the same way but will group squares in dierent ways.
One way of creating a substitution tiling is by a method called the up-down generation.
We start o with one prototile, say , and any level-1 prototile 0 in which it appears. T1 Tj Next build 0 with copies of level-0 tiles. After this we choose any prototile at level-2 in Tj which 0 appears and construct it using tiles of level-1; then subdivide the level-1 tiles into Tj level-0 tiles. If we repeat this process ad in nitum we will obtain a tiling of the space En. We will use this method later on.
Theorem. 2.3.3 [1] A uniquely hierarchical tiling is non-periodic
Proof. Since we have a nite protoset, inscribe a circle of radius c in the level-0 tiles, a circle of radius λc in the level-1 tiles and a circle of radius λkc in the level-k tiles, λ > 1. If the tiling had translational symmetry, we would be able to shift the whole tiling by some vector −→u of nite length u and it would ` t' back into the original tiling. This is only possible when u > λkc, otherwise the level-k tiles would overlap their original positions.
At some hierarchical level we will have that λkc > u and the translational vector will lie
32 entirely inside the circle of the level-k tiles. Therefore −→u is not a translational symmetry operation for the su ciently large kth hierarchical level and it is not a symmetry operation for the original tiling either. Thus the tiling is non-periodic.
2.3.3 Multigrids
It was de Bruijn [16] who coined this term and introduced this method for creating Penrose tilings of the plane. He noticed that any edge to edge tiling of the plane by rhombs contains
'ribbons` made of rhombs linked by parallel edges. If we were to choose one particular rhomb in a diagram and label its sets of parallel edges with {a,a'} and {b,b'}, we can take one set of parallel edges (say {a,a'}) and we notice that each rhombs adjacent to our rhomb along the edges a and a' has another edge parallel to this set, as do these new rhombs adjacents and so on. The ribbon de ned by the set {a,a'} is an in nite chain of rhombs.
We can take each of these ribbons we have de ned and replace them with a straight line which is orthogonal to the set which has de ned them. By doing this we obtain a con guration of lines that is a superposition of k in nite families of parallel lines where k is the number of directions of edges in the tiling. This method can also be reversed so that we can create a tiling from a superposition of k in nite families of parallel lines.
De nition. 2.3.4 [1] A grid in En is an in nite family of parallel hyper-planes with xed inter-planar spacing d; a vector of length d orthogonal to the hyperplanes is called a grid vector.
This is then generalised in the following way
De nition. 2.3.5 [1] A multigrid (or k-grid) is a superposition of k grids. A set of k grid vectors ε1, ..., εn is the star of a multigrid.
This multigrid construction as described above can also be reversed. We note that the multigrid itself is a tiling of the plane. When de Bruijn considered these tilings, he called the tiles meshes in order to avoid confusion between tilings and multigrids. We can create the corresponding tiling from the multigrid by constructing the orthogonal dual of this
33 multigrid tiling. The orthogonal dual tiling is constructed as follows: We take the centre of each of the meshes and call this a vertex of the dual tiling and then we take the vertices of the meshes and replace each one of these with the centre of a tile in the dual tiling. But for now, let us focus on how multigrids are constructed. [1]
Step 1 : Choose a star of k unit vectors (k > n) Step 2 : Superimpose k grids, where each grid is orthogonal to one of our star vectors.
It is important that if more than n hyperplanes meet at any point, we shift su ciently many grids away from the origin through the direction of their star vectors so that at most n hyperplanes meet at any point. The reason for this is to avoid what we call `singular' cases. For example, if we were to take n = 2 and suppose we want the tiles of the dual tiling to be rhombs which have four sides. If we have a point where m lines meet on the multigrid, the corresponding tile will be a 2m − gon, which is a rhombs if and only if m = 2. Therefore the best way to navigate around this problem is to start by taking one hyperplane of each grid passing through the origin, and then shift the jth grid through
th some distance γj in the direction of the j star vector so that at most n hyperplanes meet at any point.
Step 3 : Using the grid star we proceed by constructing the prototiles of the tiling. The prototile corresponding to a particular vertex of the multigrid is the polygon spanned by the star vectors that meet there (Figure 2.11).
Step 4 : With the prototile copies and by using the multigrid as a 'blueprint` we con- struct the orthogonal dual of the multigrid.
−→ De nition. 2.3.6 [1] A shift vector for a k-grid is a k-tuple of real numbers γ = (γ1, ..., γk), −→ 0 ≤ γj < 1 for j = 1, ..., k. If γ is such that no more than n hyperplanes meet at any point, both it and the corresponding k-grid are said to be regular and all other −→γ 's and k-grids are called singular.
These are the vectors we considered in Step 2. It is important that 0 ≤ γj < 1 because if γj ∈ Z then the shift imposed by the vector would create a multigrid which coincides with the original.
This multigrid construction method also allows us to assign coordinates to the vertices
34 Figure 2.11: Constructing prototiles for a multigrid in the case where k=2
35 of the tiles. This will become very useful when we consider pentagrids in the near future.
In order to do this we need to rst of all take each grid and index the parallel hyperplanes with the numbers Z + 1/2. Therefore the equations of the hyperplanes are given by:
1 −→x · −→ε = l + , l ∈ , j = 1, ..., k j j 2 j Z
If we then take into account the shifts we will need to impose so that no more than n hyperplanes meet at any point we can write:
1 −→x · −→ε = l + + γ j j 2 j
We can de ne a slab as the area which two successive hyperplanes bound in the plane.
Every mesh of the multigrid is the intersection of k such slabs. This implies that we can label the meshes of our grid by k-tuples of integers (l1, ..., lk). Not all k-tuples of integers will correspond to a mesh because some slabs will not intersect at all. However if −→x is a point in the interior of a mesh, then it must satisfy the mesh condition:
Lemma. (The Mesh Condition): [1] There are integers lj, j = 1, ..., k such that the k- inequalities: 1 1 l − + γ < −→x · −→ε < l + + γ j 2 j j j 2 j have a simultaneous solution.
The k-tuples which do correspond to meshes are used to label the vertices of the dual tiling and the mesh condition tells us exactly which k-tuples these are.
2.3.4 Pentagrids
Pentagrids are a special case of multigrids. In fact they are the case when n = 2 and k = 5. This implies that they are the superposition of ve grids where each grid is orthogonal to one of the vectors of the grid star, which in this case point to the vertices of a regular pentagon (Figure 2.12). These vectors ε1, ..., ε5 de ne two rhombic prototiles, up to rotation. The rst of these have edges parallel to εj and εj+1 and the second have edges
36 parallel to εj and εj+2 when the subscripts are reduced modulo 5. The rhombs which they de ne are the T and t rhombs discussed before. However this tiling which is constructed from the pentagrid does not necessarily satisfy Penrose's rules just yet. de Bruijn was the one to show that one additional condition needs to be satis ed.
We saw earlier with multigrids that every mesh can be labeled with a k-tuple of integers.
Taking the inequality which we derived before and adapting it for the case when k=5, we obtain: −1 1 < l + γ − −→x · −→ε < , j = 1, ..., 5 2 j j j 2
If we then add the ve inequalities together we nd that:
5 5 5 X X 5 X 0 < l + γ + − −→x · ( ε ) < 5 j j 2 j j=1 j=1 j=1
The vectors εj point to the vertices of a regular pentagon which means that their sum is zero, so 5 5 X X 5 0 < l + γ + < 5 j j 2 j=1 j=1
This is satis ed by every pentagrid and pentagrid dual. But when: P5 −5 then we j=1 γj = 2 can simplify this further to:
5 5 X X 0 < lj < 5 ⇒ lj ∈ {1, 2, 3, 4} j=1 j=1
We will call this sum the index of the vertex that corresponds to a mesh.
De nition. 2.3.7 [1]The condition P5 1 is called the sum condition j=1 γj = 2 mod(1)
The following lemma is an important condition which we will use to prove that the pentagrid dual tilings are Penrose tilings:
Lemma. 2.3.8 [1] If a pentagrid satis es the sum condition, then the indices of the four vertices of a rhomb in the dual tiling can only be, in cyclic order, (1,2,3,2) or (2,3,4,3)
Proof. (Informal) As we pass from one mesh of the pentagrid to another by crossing a common edge, we pass from the nth slab of the jth grid to the (n + 1)st or (n − 1)st slab
37 Figure 2.12: The grid star of a pentagrid and a line from each grid corresponding to an orthogonal grid vector
of the same grid. This means that the only thing changing is our value of lj and it is increasing or decreasing by 1 each time we cross a boundary. We now look at a particular intersection point of the jth and kth grids and follow a small circle around this point. As we traverse around this point the kth and jth components of vector rst increase by one and then decrease by the same amount, which returns us to the starting vertex and starting index ⇒ we will cross four boundaries so each cycle of integers will be of four numbers and with the indices {1, 2, 3, 4} there are only two possible ways of doing this, namely (1,2,3,2) and (2,3,4,3). [1]
It is these conditions which lead to the formulation of the following theorem which de
Bruijn went on to prove:
Theorem. 2.3.9 [1] If the shift vector −→γ of a pentagrid is regular and satis es the sum condition then the dual tiling is a Penrose tiling.
I will not give a formal proof to this, but more of a sketch of the proof.
Let us assume that we are given a dual of a regular pentagrid that satis es the sum condition with indices attached to the vertices. There are three classes of edges in a pentagrid dual: Those joining vertices of index 1 and 2, those joining indices 2 and 3 and those joining indices 3 and 4. To begin with let's place single arrows on the edges
38 Figure 2.13: The three possibilities for when our double arrowed edges will point in opposite directions. [1] joining 1 and 2, having them pointing from 2 to 1. This choice in arrows implies that edges joining 3 and 4 must be decorated with a single arrow also, pointing from 3 to 4. In order for us to show that this is indeed a Penrose tiling it needs to be compliant with the
Penrose matching rules so we need to place double arrows on the edges joining indices 2 and 3. Since the single arrow edges directions have already been determined, so have the directions of the double arrows on each rhomb. The problem we face is showing that any two rhombs in our tiling which share a double arrow edge have the double arrows pointing in the same direction.
We now need to investigate under what conditions the arrows will point in opposite directions and then show that none of them are possible. There are three possibilities and they are shown in Figure 2.13.
Claim: Mirror image rhombs must alternate in a tiling of the plane.
Proof. The vectors εj, j = 1, 2 are the mirror images of the vectors ε5−j with respect to the line containing ε3 (Figure 2.12). Let l be any grid line of ε3 and consider the pattern of intersections of the lines 1 and 5 with l (Figure 2.14). Looking at the intersections of l with the lines of grid 1 correspond, in the dual, to t rhombs which are the mirror images of the rhombs corresponding to the intersection of the lines of grid 5 with l. Looking at the intersections of grids 2 and 4 in a similar way we see that they correspond to mirror image T rhombs. Each family of lines cuts l in a sequence of equispaced points (as the
39 Figure 2.14: The intersections of the lines of grids 1 and 5 with l correspond to the mirror image tiles in the dual tiling.[1] lines of the grid are equispaced). It follows that the intersection of the lines of the jth and
(5 − j)th grids with l must alternate with one another and thus in any ribbon of the tiling mirror image rhombs must also alternate.
This claim was something else which was rigorously proven by de Bruijn. The rst and last cases of Figure 2.13 are easily disproven by this claim and with a little more work we see that the second case also breaks the rules of this claim. This can be done by extending the con guration and we soon see that the claim is broken again.
Therefore we have disproven all of the possible cases where the arrows do not match which suggests that we have now shown that the dual tiling does satisfy the matching rules and is therefore a Penrose tiling.
40 Chapter 3
Projection Method for Constructing Tilings of the Plane
Previously when discussing projection methods we have talked about how to project a set of points, or a lattice, in n- dimensional space onto a d- dimensional subspace (where d < n). Now we want to nd a way of directly projecting tilings from higher dimensional spaces onto lower dimensional subspaces. In this chapter we will investigate the ways in which we can do this and give a few examples along the way. The chapter will be mainly focussed on constructing Penrose tilings from higher dimensional spaces, as it is these tilings which t in well with our crystallographic theory.
3.1 The Idea
First of all let us recall the canonical projection method which we discussed in Chapter 2.
There are three main ingredients we need to be able to carry out the projection method:
1. An n- dimensional lattice L, assumed to be integral.
2. A totally irrational d- dimensional subspace (where d < n)
3. A regular shift vector −→γ ∈ En, which was discussed at the end of Chapter 2.
The reason we need the shift vector is so that our totally irrational subspace does not contain a j- face of the Voronoi tessellation induced by L if 0 ≤ j < n − d. To put this
41 into context by considering the two-dimensional lattice as in Chapter 2, we need the shift vector to ensure that does not intersect any vertex of the Voronoi tessellation of L.
We will assume from now on that L is the standard lattice In. The Voronoi cells of
In are hypercubes which means that the acceptance domain is the orthogonal projection,
n−d into E , of an n - dimensional hypercube Qn. Going back to the two dimensional case we projected, onto the line the points of p I2 whose Voronoi cells were cut by . This gave us a tiling of the line, with lattice points as vertices of the tiles and the projections of the facet vectors of the Voronoi cells cut by were the faces of the tiles. What we want to nd now is the higher dimensional analogue of this. Constructing the vertices of the tiles is easy because that is simply the projection of the lattice points onto our subspace, but nding the faces of our tiles takes a little more intuition. The answer to this problem is provided in the multigrid construction, in which the tiles are obtained as duals of the multigrid. It can be shown that the n-grid is
n the intersection of the Voronoi tessellation of E , induced by In, with a d - dimensional subspace (we will show this later). Moreover the n - tuples of integers that satisfy the mesh condition are precisely the coordinates, in En, of the lattice points whose Voronoi cells are cut by . Therefore the multigrid construction has a natural interpretation in higher dimensional space and the projection method, for tilings, turns out to be a higher dimensional interpretation of dualization.
Before we move on we need to make it clear what a Delone tiling of En is. The Voronoi tessellation induced by Λ is a decomposition of space into tiles whose centers are points of the Delone tiling Λ. The Delone tiling is the orthogonally dual tiling to the Voronoi tiling, by which we mean
De nition. 3.1.1 [1] Two tilings are orthogonally dual if, for k > 0, the corresponding k and (n − k) faces are orthogonal.
Therefore the Delone tiling has the points of Λ as its vertices and the centers of their tiles are at the vertices of the Voronoi tiling. We have seen what it means for a point set to be the dual of another, but have yet to de ne what it means for a tiling to be a dual tiling.
42 De nition. 3.1.2 [1] Two tilings T and T ∗ of En are said to be dual if there is a one- to-one inclusion-reversing map between the k- dimensional faces of T and the (n − k) - dimensional faces of T ∗, k = 0, ..., n.
Therefore we can now see that the Delone tiling is the dual tiling (which happens to also be orthogonally dual) of the Voronoi tiling and vice versa.
Going back to the projection method, we now have to interpret what multigrid dual- ization means from a higher dimensional point of view. We know that the Delone tiling is orthogonally dual to the Voronoi tiling and that there is a one-to-one inclusion-reversing map (as described above). So when we cut a Voronoi tiling by , we select a subset of faces of the Voronoi tiling but at the same time we select the corresponding subset of faces of the Delone tiling. Our shift vector ensures that the dimensions of the faces of the Delone tiling that we select do not exceed dimension d. Now when we construct the tiling of what we are actually doing is projecting the selected faces of the Delone tiling onto . An example of this is shown in Figure 3.1.
p It is possible to show that projected tiling is the dual of the multigrid V (In) ∩ , where p p V (In) is the Voronoi tessellation induced by In.
3.2 Penrose Tilings as Projections
When considering Penrose tilings we are going to be looking at ve-dimensional space and therefore when we will be considering the pentagrid construction. It can be shown that −→ −→ the pentagrid star (Figure 2.12) of vectors 1 , ..., 5 is an orthogonal projection, onto a
−→ −→ 5 two-dimensional plane, of the vectors of an orthonormal basis e1 , ..., e5 of E . That is
−→ −→ ej −→ j , j = 1, ..., 5
Looking at the methods we used in the pentagrid construction it is easy to see how this can be interpreted in ve-dimensional space:
−→ 5 • The shift vector γ = (γ1, ..., γ5) can be interpreted as a translation in E .
43 Figure 3.1: A Voronoi tiling of the plane (solid lines) and a Delone tiling of the plane (dashed lines). The faces of the Voronoi tiling cut by (red) and their corresponding faces in the Delone tiling (black). To construct the tiling of we project these faces of the Delone tiling onto .
−→ • The vector l = (l1, ..., l5) which located a particular mesh in the pentagrid is a point
in the lattice I5.
The grid equations −→ −→ 1 represent families of parallel hyperplanes • x · j = lj + 2 + γj −→ orthogonal to the basis vector ej .
This last one means that the pentagrid hyperplanes partition E5 into ve-dimensional hyper-cubes, where each hyper-cube is the Voronoi cell of a lattice point in E5. We discussed in the last section that the multigrid construction is simply the intersection of the Voronoi tessellation of En with a d- dimensional subspace . Therefore the plane pentagrid which we constructed in Chapter 2 is the intersection of a plane and the Voronoi tessellation of I5 (Figure 3.2) −→ We know that k ∈ I5 is the location of a mesh in the pentagrid if and only if −→ intersects the interior of the translate of its Voronoi cell V ( k ), where we are translating by −→γ so that does not intersect any vertex, edge or two-face of the Voronoi tessellation.
44 Figure 3.2: The plane pentagrid is the intersection of the Voronoi tessellation of I5 and a plane . [1]
This is equivalent to requiring, in the canonical projection sense:
−→ Π⊥( k ) ∈ Π⊥(V (0) + −→γ )
This then allows us to construct Penrose tilings by canonical projection. However before we go on to discuss this, we will consider some of the properties related to the integer lattice I5.
3.2.1 I5 and the Unit Hypercube Q5
Q5 has ten four dimensional facets, with facet vectors
−→ −→ ±e1 = (±1, 0, 0, 0, 0), ..., ±e5 = (0, 0, 0, 0, ±1) and thirty two vertices of the form
5 X 1 1 α −→e , α ∈ {− , } j j j 2 2 j=1
45 −→ Figure 3.3: The unit hypercube Q5 projected onto a plane orthogonal to the vector w = (1, 1, 1, 1, 1) [1]
We de ne the body diagonal of Q5 as the line segment between the two points (0, 0, 0, 0, 0) and (1, 1, 1, 1, 1) which we can represent with the vector −→w = (1, 1, 1, 1, 1). Obviously trying to imagine how this unit hypercube looks is beyond our imagination, but what we can do is project it down to a plane and from there we can study its properties. Figure 3.3 −→ shows Q5 projected onto a plane orthogonal to the vector w . This projection identi es two vertices in the hypercube and these vertices are represented by the centre point of the diagram.
It can be shown that ve-fold rotation is a symmetry of Q5 which permutes the ve −→ basis vectors ej , j = 1, ...5 and because Q5 is invariant under ve-fold rotation, we can say that I5 is also invariant under the same rotation, as Q5 is the Voronoi cell for each point in the lattice. The rotation matrix of this rotation is given by
46 −→ Figure 3.4: Projections of the basis vectors ej onto the totally irrational subspaces and 0. [1]
0 0 0 0 1 1 0 0 0 0 A = 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0
We can easily compute the eigen values of A and we see that they are in fact the fth roots of unity 1, ζ, ζ2, ζ3, ζ4, where ζj = e2πij/5, j = 0, ..., 4. Note that ζ3 = ζ−2 and ζ4 = ζ−1.
This means that there is one real root and two pairs of complex conjugate pairs ζ, ζ−1 and ζ2, ζ−2. This implies that we can decompose E5 into two two-dimensional planes, which are both totally irrational and both orthogonal to one another, and the xed space
−→ −→ 5 ∆ generated by w [1]. The basis vectors ej of E project to the vertices of the regular pentagon on both of these totally irrational two-dimensional planes (as they did before), however on one of them, say , the angle between the basis vectors −→ and −−→ is 2π , whilst ej ej+1 5 in the other, say 0, the angle is 4π (Figure 3.4) 5 We can show that ⊥ = 0 ⊕ ∆. Note that ⊥ cannot be totally irrational, since ∆ contains multiples of −→w which will intersect other points in the lattice.
3.2.2 Projecting I5
The projection principle which we will use was described at the beginning of this Chapter: −→ We project, onto , the duals of the facets of the Voronoi tessellation V (I5) + γ that
47 are cut by . −→ only meets the faces of V (I5) + γ in dimensions 3, 4 and 5 which means the corre- sponding faces of the Delone tiling have dimensions 2, 1 and 0 respectively. In fact it can be shown that these two dimensional faces of the Delone tiling form a connected surface of square faces and the projections of these faces onto are what forms the rhombs in the rhomb Penrose tiling [1]. The set of lattice points which we project onto are decided by the canonical choice described in Chapter 2, that is those that project, under Π⊥, into the acceptance domain K = Π⊥(V (0) + −→γ ).
Consider the star of a vertex v of a Penrose tiling. By duality, this corresponds to the intersection of with a cluster of Voronoi cells in E5. Suppose that v = Π(x) for some
⊥ ⊥ −→ x ∈ I5. Then we know that Π (x) ∈ Π (V (0) + γ ). Due to the fact we are projecting down through the dimensions, when we project the Voronoi cell V (0)+−→γ onto ⊥ to create our acceptance domain K, a projected facet of this Voronoi cell will overlap with other facets of the same cell in K. Therefore Π⊥(x) will lie in the intersection of several projected −→ −→ −→ facets of V (0)+ γ . Let's name these facets f1, ...., fj and let f1, ..., fj be the corresponding facet vectors (where a facet vector, de ned in Chapter 2, are the vectors joining a point x to −→ −→ other lattice points whose Voronoi cell shares a facet with V (x)). Therefore f1−f1, ..., fj−fj
−→ ⊥ −→ ⊥ −→ ⊥ −→ are also facets of V (0) + γ , so Π (x − f1), ..., Π (x − fj ) ∈ Π (V (0) + γ ) which means −→ −→ that the plane must cut the Voronoi cells V (x − f1), ..., V (x − fk). Therefore the decomposition of the acceptance domain K by the projected facets of
V (0) determines the vertex con guration in the projected tiling. When we project V (0) to create the acceptance domain, we are projecting it onto a three dimensional subspace
⊥ 0 (which is the sum of a two dimensional subspace and ∆). The projection of Q5 onto a three dimensional space is a rhombic icosahedron (Figure 3.5). This has 22 vertices and
20 faces. The facets of Q5 (which are hypercubes Q4) are projected into K as rhombic dodecahedra (Figure 3.6). −→ −→ −→ −→ w is a lattice vector, so for all x ∈ I5 we have that x · w ∈ Z. Moreover the 22 vertices of K satisfy −→v · −→w ∈ {0, 1, 2, 3, 4, 5}. The points of X, the set of lattice points to be projected onto , are projected into four cross sections of K, namely the cross sections
48 Figure 3.5: The projection of Q5 onto a three dimensional space is a rhombic icosahedron.[13]
Figure 3.6: The projection of Q4onto a three dimensional space is a rhombic dodecahedra [14]
49 Figure 3.7: A Penrose tiling of the plane and the corresponding cross sections of the acceptance domain K (The black lines). The middle four cross sections represent levels {1, 2, 3, 4} respectively. [1] which are orthogonal to −→w at levels {1, 2, 3, 4}. If we recall Chapter 2, these are the indices of the rhombs in the pentagrid construction. This allows us to draw the conclusion that the index of a vertex is the level on which the preimage of the vertex lies in E5. Figure 3.7 shows a Penrose tiling corresponding to cross sections of K. The projected facets of Q5 have to overlap in K. The small polyhedron in the center of the diagram of K in Figure 3.7 represents the intersection of all ten dodecahedron.
Since the points Π(X) lie in four planar cross sections of K, these sections are the only features of K we need to study. However, to completely understand which regions of the cross sections represent which vertex stars, we have to consider the projected plane as the complex plane. Something initially described N.G de Bruijn in the early 1980's.
3.2.3 Projected Plane as Complex Plane
Recall from Chapter 2 (Figure 2.6) that in a rhomb tiling of the plane by T and t rhombs, there are eight possible vertex stars which we class as being `globally legal'. Looking at them as the vertex stars they are (ignoring the corresponding tiles) they look like those in
Figure 3.8. They have been given labels so we can distinguish between them.
If we were to consider all the possible directions the arrows from the vertices travel we
50 Figure 3.8: The eight dierent types of vertex star for rhomb Penrose tilings revisited [16]
nd that, if we consider the complex plane C, the directions are taken from the set
±1, ±ζ, ±ζ2, ±ζ3, ±ζ4 where . We also de ned the index of a vertex in Chapter 2 as the sum P . ζ = exp(2πi/5) j lj It wasn't explicitly stated before, but if our tiling is the result of a regular pentagrid, it is possible to label each vertex with the index 1,2,3,4. Moreover it can be shown that the index increases by one if we move in the direction of 1, ζ, ζ2, ζ3, ζ4 and decreases by one if we move in the direction of −1, −ζ, −ζ2, −ζ3, −ζ4 . This means that all vertices of the tiling are of the form
4 z0 + k0 + k1ζ + ... + k4ζ where z0 is xed, k0, ..., k4 ∈ Z and 0 < k0 + .... + k4 < 5. −→ 5 Consider the regular pentagrid de ned by the shift vector γ = (γ0, ...., γ4) ∈ R with the restriction that γ0 + ... + γ4 = 0. In the complex plane, this can be described by the
51 union of sets
−j z ∈ C : Re(zζ ) + γj ∈ Z where j = 0, ..., 4 [16]. For each z ∈ C we can also associate ve integers K0(z), ..., K4(z) where
−j Kj(z) = Re(zζ ) + γj where dxe is the least n ∈ Z such that n ≥ x. As discussed before we can attach rhombs to each intersection of two grid lines of the pentagrid and this forms a tiling of the plane. We can describe the vertices of this tiling in the following way. Consider the ve dimensional Euclidean space E5 and divide it into unit cubes in the standard way (That is, consider I5 and construct the Delone tiling). We −→ can then index each cube with ve integers k = (k0, ..., k4) and we can denote the interior of the cube as the set of points (x0, ..., x4) such that
(k0 − 1) < x0 < k0, ..., (k4 − 1) < x4 < k4
−→ We will call this the interior of the open unit cube for vector k . Now if we consider the two dimensional plane given by
4 X xj = 0 j=0
4 X 2j (xj − γj)Im(ζ ) = 0 j=0
4 X 2j (xj − γj)Re(ζ ) = 0 j=0 Theorem. 3.2.1 The vertices of a rhomb Penrose tiling produced by a regular pentagrid
4 are the points k0 + k1ζ + ... + k4ζ where (k0, ..., k4) runs through those elements of I5 whose open cube has a non empty intersection with the plane described above. [16]
52 This plane therefore describes the two dimensional totally irrational subspace of our
ve dimensional lattice. Therefore if the points (k0, ..., k4) are projected orthogonally onto the plane we get a gure that can be turned into the set of vertices of our rhomb Penrose tiling.
Considering the same rhomb tiling and the associated pentagrid, we ask the question whether there is a mesh in the pentagrid where
K0(z) = k0, ..., K4(z) = k4
with Kj(z) de ned as before. This is equivalent to asking if it is true that there exists z ∈ C such that for all j = 0, ..., 4 we have
−j kj − 1 < Re(zζ ) + γj < kj
−j Theorem. 3.2.2 The statement kj − 1 < Re(zζ ) + γj < kj on k0, ...., k4 is equivalent to P P 2j P P 2j ( kj, (kj − γj)ζ ) ∈ K, where K = ( λj, λjζ ) : 0 < λi < 1 i = 0, ..., 4 [16]
P 2j This means that kjζ lies in one of the four pentagrids K1, ..., K4 according to P kj = 1, 2, 3, 4. These Ki are cross sections of K and hence are the levels which we were P considering in the previous section. Hence the points of K with λj = r form a pentagon shaped Kr. As we have seen before K3 = −K2 and K4 = −K1 hence we only need to consider K1 and K2. These pentagons are shown in Figure 3.9.
4 The theorem above makes it easy for us to see whether a point k0 + k1ζ + ... + k4ζ is a vertex of the rhomb tiling of the plane. However we can also study the `neighbours' of a point, which we get by adding one of ±1, ±ζ, ..., ±ζ4. In this manner we can nd the type of vertex star which each section corresponds to. We set P 2j , then the j(kj − γj)ζ = θk result is as follows.
If P (the index equals 1) and if lies in the region marked by in the • j kj = 1 θk S 4 picture of K1 in Figure 3.9, then the point k0 + k1ζ + ... + k4ζ of the rhomb tiling has the type S from Figure 3.8.
If P (the index equals 1) and if lies in the region marked by in the • j kj = 1 θk Q
53 Figure 3.9: The pentagons K1 and K2
54 4 picture of K1 in Figure 3.9, then the point k0 + k1ζ + ... + k4ζ of the rhomb tiling has the type Q from Figure 3.8. There are ve orientations of Q representing the
ve dierent regions marked Q in K1.
If P (the index equals 1) and if lies in the region marked by in the • j kj = 1 θk K 4 picture of K1 in Figure 3.9, then the point k0 + k1ζ + ... + k4ζ of the rhomb tiling has the type K from Figure 3.8.
If P (the index equals 2) then the point 4 of the rhomb • j kj = 2 k0 + k1ζ + ... + k4ζ
tiling has the type D, J, S3,S4,S5 from Figure 3.8, depending on where θk lies in
the picture of K2 in Figure 3.9 (Note that in the gure, 3, 4 and 5 represent S3,S4 and S5 respectively).
For index values 3 and 4 we draw the same conclusions with θk = −θk. Associating these vertex stars to the tiles that t them, Figure 3.10 gives a summary of the decomposition of the pentagons by colour coding the section of the pentagon they account for.
3.2.4 Shift Vector −→γ
The shift vector −→γ plays a vital role in determining what the resulting rhomb tiling looks like. This is because a change in −→γ results in a change in our acceptance domain K and hence a change in which lattice points we are projecting. Assume we are given an initial
−→ −→ −→ 5 −→ −→ γ . If we think of it in terms of the orthonormal basis u1, ...., u5 of E , where u1, u2 span ,
−→ −→ 0 −→ −→ −→ u3, u4 span and u5 is parallel to w (as in 3.2.1). Therefore γ is the linear combination
−→ −→γ = α−→v + β v0 + $−→w
−→ where −→v ∈ and v0 ∈ 0. In order for the resulting tiling of the plane to be a Penrose tiling, we need to obey the sum condition, that is 1 . If we change the $ $ = 2 mod(1) component then we are simply translating the tiling since the acceptance domain is not associated to . However if we were to change the 0 component then we would be shifting and hence the set of lattice points of p which we select will be dierent. This results in K I5 a dierent Penrose tiling altogether. If we change so that 1 then the tiling is $ $ 6= 2 mod(1)
55 Figure 3.10: The decomposition of pentagons for indices 1 (a) and 2 (b) and their associated vertex atlases (colour coded)
56 Figure 3.11: The projected tiling when $ = 0 and the cross sections of K which we consider.[1] no longer a Penrose tiling. Consider the case when $ = 0 and look at the projected tiling (Figure 3.11). We can see clearly that there are con gurations which are not part of the
Penrose vertex star. The reason for this is one of the levels of K cuts the overlap region of ten dodecahedra, hence our tiling will contain a con guration of ten t rhombs, which are visible in the gure. These tilings for which 1 are called generalized Penrose $ 6= 2 mod(1) tilings. Nevertheless these tilings are generated by the canonical projection method and are therefore crystals.
57 Chapter 4
Diraction Geometry
Since 1912 the x-ray diraction pattern has been the principle image through which the structure of crystals and quasicrystals are studied. Diraction is the scattering of waves that takes place when the waves in question meet some form of obstacle. When a wave comes into contact with an obstacle, a number of things might happen to them: the waves may be de ected and spread out in dierent directions, they may come together and interfere with each other and during this interference they may combine producing a wave with a greater amplitude. What happens to the waves clearly depends on the object that they are initially coming into contact with. If su ciently many waves combine in phase, then we can use a number of signal receivers to determine the overall diraction pattern of the original source. When light is shone through a crystal, the waves combine in such a way that they produce so-called sharp bright spots on the screen. These are called Bragg peaks after the father and son team of William Lawrence Bragg and William
Henry Bragg, who carried out a great deal of research into the patterns this diracted light would produce [7]. X-rays are `light' waves on an extremely small scale. They have an extremely high frequency which happens to be exactly the right frequency for the atomic pattern of crystals to serve as a diraction grating. X-rays are not visible to the naked eye, therefore we need to use sophisticated signal receivers in order to view the diraction patterns created. When rst studied, diraction patterns were recorded on photographic plates which would react with the waves and over time produce a pattern depending on
58 Figure 4.1: The model we assume for far eld diraction. where the waves hit the plate with the highest intensity. However nowadays we tend to use more advanced methods of signal recovery such as computers which will measure the exact intensity at each point on the screen.
Diraction can be split into two distinct categories:
1. Near eld (or Fresnel) diraction - This is mathematically complicated so we will not
be focussing on this.
2. Far eld (Fraunhofer) diraction - This is easily modelled by Fourier transforms and
will be the most useful to us in the rest of the chapter. The diraction patterns that
we are used to seeing are those produced by far eld diraction.
4.1 Far Field Diraction Geometry
Figure 4.1 shows some of the assumptions we have to make when considering far eld diraction. The main assumption here is that the distance between the source and the mask and distance between mask and the source is arbitrarily `large'. The reason we do this is so that we can make the following assumptions which simplify the problem signi cantly:
• Light waves from the point source arrive at all points of the diraction mask in
59 Figure 4.2: A two-dimensional image of the aperture in the mask accompanied by the diraction pattern it produces. [8]
so-called parallel trains.
• The parallel trains of diracted light making the same angle with the mask meet at the same point on the screen.
• The light in question is monochromatic (i.e. all waves have the exact same wavelength
λ.
Now that we have our model in place we can begin to look at dierent scenarios in which the diraction mask contains dierent numbers of apertures in it's surface.
4.1.1 One Point Diraction
The rst case we consider is that when the diraction mask contains one aperture in it's surface with a set radius r. The way in which the light is diracted is shown in Figure 4.2.
What we begin to notice is that, as we decrease the value of r, the radius of the diraction disk increases. In fact these two radii are inversely proportional. As r −→ 0 the radius of the diraction disk will become larger and larger until, for r = 0, the entire screen would be illuminated with equal intensity. Let J(−→s ) be the value of the intensity
60 of the wave front in the direction of the (unit) vector −→s .Then we can say J(−→s ) has the same value in all directions, or in other words:
J(−→s ) = 1
This tells us that the value of J(−→s ) is independent of the aperture's position on the diraction mask. We know from studies carried out in physics that J(−→s ) is related to the amplitude function A(−→s ) in the following way:
J(−→s ) = |A(−→s )|2
A(−→s ) is not constant in all directions like the intensity function and does vary depending on the aperture's position. From this and the fact that |A(−→s )|2 = 1, we can gather that the amplitude function must be a complex valued function, so
A(−→s ) = exp(−2πif(−→s )) where f is some real valued function of −→s . We can conclude from this that A(−→s ) is a wave function. In order to work out the properties of this new function f(−→s ), we must consider the mask with two apertures.
4.1.2 Two Point Diraction
−→ Now we assume that we have two point apertures P1 and P2 with the vector d being the vector between these two points. The diracted waves will now interfere with each other as the waves emanating from P1 and P2 will overlap as they `spread out'. This means that J(−→s ) will no longer be constant and will instead vary with −→s in a way which depends on −→ d . One signi cant dierence comes from the lag between the two fronts which are in phase when they reach the mask. This is because when the in-phase waves reach the apertures and are diracted by the same angle, by assumption they will represent the same point on our diraction screen. However one of the wavefronts will have to travel a greater distance because of it's position above/below the other one. This creates a lag (Figure 4.3).
61 Figure 4.3: One front lagging behind the other [1]
Figure 4.4: Calculation of σ
This is something we can measure. The length of the lag, σ, is the scalar projection of −→ d onto −→s (Figure 3.4). −→s is a unit vector hence |−→s | = 1 and it follows that
−→ −→ −→ −→ σ = d · s = d · | s | cosθ = dcosθ −→ −→ where and is the angle between the two vectors and −→. d = d θ d s This helps us understand the following argument. Amplitudes are additive, hence:
−→ −→ −→ A( s ) = A0( s ) + Ad( s )
−→ −→ where A0( s ) and Ad( s ) represent the amplitudes emanating from each aperture in the
62 direction of the vector −→s . We know from before that |A(−→s )|2 = 1 so
−→ 2 −→ 2 |A0( s )| = |Ad( s )| = 1 and therefore we must have that
−→ −→ Ad( s ) = zdA0( s )
for some zd ∈ C with |zd| = 1. zd can be written in exponential form and since the translation of an aperture eects a phase shift in amplitude we can conclude that this −→ −→ −→ phase shift is the lag σ = d · s . This implies that multiplying A0( s ) by zd transforms −→ −→ A0( s ) in a way that varies inversely with d , so
−→ −→ −→ −→ Ad( s ) = exp(2πi d · s )A0( s ) which suggests that −→ −→ −→ −→ A( s ) = A0( s )(1 + exp(2πi d · s ))
Now because J(−→s ) = |A(−→s )|2 we have that: −→ −→ −→ −→ 2 −→ −→ −→ −→ J( s ) = A( s )A( s ) = |A0( s )| (1 + exp(2πi d · s ))(1 + exp(−2πi d · s )) −→ −→ = (1 + exp(2πi d · −→s ))(1 + exp(−2πi d · −→s )) −→ −→ −→ −→ = 1 + exp(2πi d · −→s ) + exp(−2πi d · −→s ) + exp(2πi d · −→s − 2πi d · −→s ) and by using Euler's formula −→ −→ −→ −→ = 1 + exp(0) + cos(2π d · −→s ) + isin(2π d · −→s ) + cos(−2π d · −→s ) + isin(−2π d · −→s ) −→ −→ −→ −→ = 2 + cos(2π d · −→s ) + isin(2π d · −→s ) + cos(2π d · −→s ) − isin(2π d · −→s ) −→ = 2 + 2cos(2π d · −→s ) and therefore
−→ J(−→s ) = 2(1 + cos(2π d · −→s ))
. −→ By looking at this formula we can now see that J(−→s ) attains in maxima when cos(2π d ·
63 −→ −→ −→ −→ s ) = 1, or in other words, when d · s ∈ Z. We can also see that J( s ) attains it's minima −→ when cos(2π d · −→s ) = −1. Thus the diraction pattern which is produced here will be a continuous in nite family of parallel lines that vary periodically from light to dark. It can also be shown that this period is proportional to 1/d . In fact any diraction pattern created by n apertures in the diraction mask is a superposition on families of these parallel lines [1].
4.1.3 N-Point Diraction
As you can imagine, it is possible to generalise the two-point diraction formulae to N- point diraction (i.e. the case when we have N apertures in our diraction mask). As with two-point diraction we can compute the intensity function for N apertures and the computation is very similar to the way we derived it before. −→ −→ −→ For N +1 point apertures at d0 = 0, d1, ..., dN we have that A( s ) = A0( s )+A1( s )+ −→ ... + AN ( s ) which is equivalent to
N −→ −→ X −→ A( s ) = A0( s )(1 + exp(2πidj · s )) j=1 and so because J(−→s ) = |A(−→s )|2 we have that: −→ −→ −→ −→ 2 PN PN −→ J( s ) = A( s )A( s ) = |A0( s )| ( j=0 k=0 exp(2πi(dj − dk) · s )) PN PN −→ = j=0 k=0 exp(2πi(dj − dk) · s )
N X X −→ = N + 1 + 2 cos(2π(dj − dk) · s ) j=1 k One thing that we do notice is that if we have an arrangement of N apertures which is su ciently `regular', then as we increase the number of regularly arranged apertures neither the number nor the spacing between the peaks of the value of J(−→s ) changes as we increase the value of N. The eect of increasing N is to makes the peaks in value of J(−→s ) higher and narrower and thus making the light parts of the diraction pattern sharper and brighter. As soon as we move away from this regular arrangement the graph of the value 64 Figure 4.5: The graphs of J(−→s ) for 6 equally spaced colinear points and for 6 colinear points with two spacings of relative sizes of J(−→s ) alters in a massive way. Figure 4.5 shows two cases of a graph for J(−→s ): One where the apertures are regularly arranged and one where they are not. The dierence is clear and points out the need for crystals to have such a regular structure about them. 4.2 Fourier Transforms and the Diraction Condition It is known that in diraction of the far- eld type, A(−→s ) is a Fourier transform of a function ρ(−→x ) describing the arrangement and nature of a set of apertures. ρ(−→x ) is often referred to as the density function. This implies that Fourier transforms play a vital role in our understanding of diraction theory. In this section we will cover what a Fourier transform does and how we can use it to describe diraction patterns created by a particular diraction mask. 65 4.2.1 The Fourier Transform All waveforms are the sum of simple sinusoids of dierent frequencies. What the Fourier transform does is take a given waveform and decomposes it into sinusoids. More speci cally it decomposes them into a sum of sinusoidal basis functions each of which is a complex exponential of a dierent frequency. De nition. 4.2.1 The Fourier Transform of a function f of n variables is the function fˆ de ned by the integral: F(f(−→x )) = fˆ(−→s ) = f(−→x )exp(−2πi−→x · −→s )d−→x ˆ Rn This tells us how much power f(−→x ) contains at frequency −→s . fˆ(−→s ) is often called the spectrum of f as we will see later on. We can also reverse the process. The inverse Fourier transform is given by: F −1(fˆ(−→s )) = f(−→x ) = fˆ(−→s )exp(2πi−→x · −→s )d−→s ˆ Rn We say that fˆ(−→s ) and f(−→x ) are Fourier Pairs, that is, they are equivalent to each other through a Fourier transform: fˆ(−→s ) ⇔F f(−→x ). Example. [9] To make it clear what this operation does, we will consider an example. Consider the box function shown in Figure 4.6. This function has an amplitude of A and extends across the time interval [−T/2,T/2]. We can now calculate the Fourier transform of this in the following way: ˆ ∞ F(f(t)) = f(s) = −∞ f(t)exp(−2πist)dt T/2 ´ = −T/2 Aexp(−2πist)dt ´ A h T/2 i A = −2πis exp(−2πist) |−T/2 = −2πis [exp(−πisT ) − exp(πisT ] h i AT exp(πisT )−exp(−πisT ) AT . This function has the plot shown in = πsT 2i = πsT sin(πsT ) Figure 4.7 Note that the integral does not converge and that the area under the graph of the function f(−→x ) is nite. For example, if we were to take f(−→x ) = 1 then the integral will 66 Figure 4.6: Box function Figure 4.7: Plot of the Fourier transform of the box function [9] 67 −→ 1, x ∈ C not exist for any −→s . However if we were to let f = where C is some 0, otherwise bounded, measurable set, then the transform will exist. We must also remember that the Fourier transformation F is an operator on a complex valued function f which associates f to it's Fourier transform fˆ. In other words: Ff = fˆ Before we go on to discuss some very important properties of the Fourier transformation, we need to discuss the convolution product of two functions. De nition. 4.2.2 The convolution product of two functions f and g, denoted by f ? g, is the integral of the product of the two functions after one of them has been reversed and shifted. In other words: ∞ (f ? g)(t) = f(τ)g(t − τ)dτ ˆ−∞ The convolution product is a kind of integral transform. Theorem. 4.2.3 The Fourier transformation F has the following properties for any com- plex valued functions f,g on En 1)F(f + g) = F(f) + F(g) 2) If f is translated, then it's Fourier transform is rotated and vice versa: Ff(−→x +−→c ) = exp(−2πi−→c · −→x )fˆ(−→x ) 3) The Fourier transform of a complex conjugate of f(−−→x ) is the complex conjugate of fˆ(−→s ): F(f(−−→x )) = f(−→s ) 4) The Fourier transform of the convolution product is the ordinary product of the Fourier transforms and vice versa: F(f ? g) = F(f)F(g) and F(fg) = F(f) ? F(g) These properties allow us to draw a connection between the various functions relating to the diraction patterns produced by a certain diraction grating. In fact, if we let g(−→x ) = f(−−→x ), then the convolution product f(−→x ) ? f(−−→x ) is called the autocorrelation function of f which we will denote by γ(−→x ). If we were to take the Fourier transform of 68 this autocorrelation function of f, then by 3) and 4) in Theorem 4.2.3 above, we will have that: 2 −→ −→ −→ −→ ˆ −→ ˆ −→ ˆ −→ F(f( x ) ? f(− x )) = F(f( x ))F(f(− x )) = f( s )f( s ) = f( s ) This allows us to form the following diagram: f(−→x ) −→ f(−→x ) ? f(−−→x ) l l 2 ˆ −→ ˆ −→ f( s ) −→ f( s ) where the double edged arrows represent the Fourier transformation from one function to the other. If we now let f(−→x ) = ρ(−→x ) (which we de ned at the beginning of this section, the density function) we can show that this diagram is equivalent to the following diagram linking the functions related to diraction patterns: ρ(−→x ) −→ γ(−→x ) l l A(−→s ) −→ J(−→s ) =γ ˆ(−→s ) This diagram is known as a Wiener Diagram named after Norbert Wiener who pointed out the usefulness of the autocorrelation function for deciphering x-ray diraction patterns [1] 4.2.2 Dirac Deltas and the Diraction Condition Now that we have a way of linking all of these diraction functions through various math- ematical operators, we will now turn our attention to trying to nd a way of describing ρ(−→x ), the density function, which ρ is the density of the N + 1 point apertures in our diraction mask. By the Wiener diagram, this is going to be the inverse Fourier transform of the amplitude function A(−→s ): N −→ −1 −→ X −→ −→ ρ( x ) = F (A0( s )(1 + exp(−2πixj · s ))) j=1 69 Which by Theorem 4.2.3 - 4) is equal to N −1 −→ −1 X −1 −→ −→ F (A0( s ) ? (F (1) + F (exp(−2πixj · s ))) j=1 If we split this into factors, the right hand factor is equivalent to: N −→ −→ −→ X −→ −→ −→ −→ 1exp(2πi x · s )d s + exp(−2πi(xj − x ) · s ) d s ˆ n ˆ n R j=1 R −→ −→ −→ −→ Now using the fact that Ad( s ) = exp(2πi d · s )A0( s ) we can nd all of the inverse transforms if we know the inverse transform of any one of them, say the constant function 1. We will now consider functions of one variable only (for simplicity). None of the inte- grals in the expression for ρ(−→x ) actually converge, but if we consider generalised functions, we can gain information from them. We de ne a generalised function as a continuous linear functional over a space of in nitely dierentiable functions such that all continuous func- tions have derivatives which are themselves generalised functions [15]. If we let f(s) ≡ 1 and replace f with a truncated version of itself, the characteristic function on the interval [−T,T ]. This function is zero outside the interval so its Fourier transform, ρT (x) is given by: T ρT (x) = −T 1exp(2πixs) ds T ´ = −T cos(2πxs) + isin(2πxs) ds ´ 1 i T = 2πx sin(2πxs) − 2πx cos(2πxs)|−T 1 i 1 i = 2πx sin(2πxT ) − 2πx cos(2πxsT ) − 2πx sin(−2πxT ) + 2πx cos(−2πxT ) 1 i 1 i = 2πx sin(2πxT ) − 2πx cos(2πxsT ) + 2πx sin(2πxT ) + 2πx cos(2πxT ) 1 = sin(2πxT ) πx Now the inverse Fourier transform of the function should be the limit: 1 ρT (x) = lim sin(2πxT ) T →∞ πx 70 If we plot the graph of ρT (x) for increasing values of T (Figure 4.8) we can see the graph tending to a particular shape of graph that is well known - The Dirac delta function. +∞ x = 0 De nition. 4.2.4 The Dirac delta function is de ned in the following way: δ(x) = 0 x 6= 0 and is also constrained to satisfy the identity: ∞ for su ciently well −∞ δ(x)g(x) dx = g(0) ´ behaved functions g. Hence we can say that: F −1(1) = δ. Moreover, if we were to speak in a more general sense we can say that: Proposition. 4.2.5 δ(x − d) and exp(−2πids) are Fourier pairs. We can still generalise further as the Dirac delta can always be generalised to spaces of arbitrary dimension. This means that if we were to look back to the equation A(−→s ) = −→ −→ A0( s ) + Ad( s ), we can conclude that the inverse Fourier transform of this equation is −→ δ(−→x ) + δ(−→x − d ) and speaking in even more general terms, the following generalised functions are Fourier pairs for any Delone set Λ: −→ X −→ −→ ρ( x ) = δ( x − dk), dk∈Λ −→ X −→ −→ ρˆ( s ) = exp(−2πidk · s ) dk∈Λ This was another subject which intrigued de Bruijn and it was he who coined the following term −→ De nition. 4.2.6 [1] A Dirac comb is any sum of the form P c(k)δ(−→x − d ) where dk∈Γ k c(k) ∈ C, Γ is the set of points at which the deltas are located (which need not be discrete) and the sum converges as a generalised function. The question we should be asking now is: Is the Wiener diagram (de ned before) valid for Dirac combs? The answer is somewhat vague, as it does hold for some cases but not all. The operations of convolution and multiplication are not de ned for all generalised functions so it is very dependent on the function which we're using. 71 Figure 4.8: The function ρT (x) for T = 2, 10, 50 [1] 72 So far we have discussed one of the roles that the Dirac delta plays in diraction theory - that is, it is a place marker for the points of a Delone set Λ. However it does play one more vital role - these deltas can also appear in the diraction patterns themselves! Let Λ be a −→ Delone set in En, whose points are denoted by d , k ∈ . Then ρ (−→x ) = P δ(−→x −d ) k Z Λ dk∈Λ k is a Dirac comb for this. We want to know whether the corresponding diraction pattern produces the so-called Bragg peaks which de ne a crystalline structure. This obviously will depends on the position of the points in our Delone set and will therefore depend on the behaviour of the Fourier transform of it's autocorrelation function γ. De nition. 4.2.7 [1] Let Λ be a Delone set in En with density ρ and autocorrelation γ. The spectrum of Λ is it's intensity function, de ned by the Fourier transform of it's autocorrelation function 2 −→ 1 X −→ −→ γˆ( s ) = lim exp(−2πidk · s ) T →∞ (2T )n dk∈ΛT n where ΛT = Λ ∩ [−T,T ] From inspection we can see that this will always be a positive measure and by Lebesgue we can decompose this into a discrete and a continuous component −→ X −→ −→ −→ −→ −→ γˆ( s ) = c( s )δ( s − s∗) +γ ˆc( s ), c( s ) ∈ R s∗∈S The set S is always countable. When it is countably in nite the rst term on the right forms a Dirac comb. We always have 0 ∈ S. If S = {0} then γˆ(−→s ) is said to be purely continuous and logically if γˆc = 0 then it is said to be purely discrete. When the spectrum is purely discrete, the spectrum is itself a Dirac comb. De nition. 4.2.8 [1] The diraction condition: Λ is a crystal if its spectrum has a count- ably in nite discrete component. The diraction condition could also be interpreted as a re ned de nition of what a crystal actually is. 73 So in other words our Delone set satis es the diraction condition if there is a countable in nity of deltas in its spectrum (i.e. the set S is countably in nite). If we have a Delone set and would like to decide whether it is a crystal or not, there are two problems which we face. The rst problem is the validity of the Wiener diagram on which all the calculations we have made prior to this have relied on and the second problem is determining the decomposition of the spectrum. If we were to assume that the diagram holds then there are powerful methods out there which allow us to determine whether the Λ is a crystal or not. We have to make this assumption, otherwise we can only conclude that every suitably chosen nite approximation to the Delone set is an approximation to a crystal. Therefore we will assume the diagram holds and carrying on searching for which Λ represent crystals. What we are seeking is the directions −→s∗ for which all of the expressions −→ −→ d · −→s∗ that appear in ρˆ(−→s ) = P exp(−2πid · −→s ) are identical modulo 1. Considering k dk∈Λ k −→ −→ these directions, letting zd = exp(−2πidk · s ) we have −→ X ρˆ( s ) = zd 1 dk∈Λ and therefore there will be bright spots at these −→s∗ in the diraction pattern. There is always one such −→s∗ for every Delone set and that is when s∗ = 0. This explains why there is a concentrated centre of maximal brightness at the origin. The remaining bright spots can be realised from the following statement: If Λ is a point lattice or a subset of a point lattice then every point s* of the dual lattice is also a point of maximal brightness. In fact p p∗ −→ if we have Λ = L then S = L and the continuous part of the spectrum, γˆc( s ) = 0. This gives us a relatively easy way of determining the locations of bright spots produced by a point lattice Delone set and it is thanks to the following formulation: Theorem. 4.2.9 [1] (Poisson Summation Formulation) Let L and L∗ be dual n-dimensional lattices. Then X −→ X −→ F( δ(−→x − l )) = δ(−→s − l∗) −→ −→ l ∈L l∗∈L∗ This formulation is one of the most useful tools in mathematical crystallography because 74 Figure 4.9: An example of the nite size eect from a subset of the square point lattice in the shape of a triangle [1] it says that the diraction pattern of a point lattice is in fact is the dual point lattice. However this is only true in the generalised framework when we have in nite sums of generalised functions (i.e. when the point set is in nite). When we have a nite point set there are regions of varying intensity and this is known as the ` nite size eect'. A good example of this comes from the point lattice in Figure 4.9. What we have in the gure is a subset of a square point lattice which is in the shape of a triangle. We can therefore think −→ −→ of the density function ρt( x ) as the product of the density function ρ( x ) of the square point lattice and the characteristic function χ of the triangle. Therefore: −→ −→ −→ ρˆt( s ) =ρ ˆ( s ) ? χˆ( s ) It will be the χˆ(−→s ) factor which causes the diraction pattern in Figure 4.9 to look like it does and not like the diraction pattern of a square point lattice. I will give one nal example of a non-periodic crystal which produces a diraction 75 Figure 4.10: Diraction pattern produced by a mask containing approximately 500 point apertures in a randomly occupied square of I2 [1] pattern with sharp bright spots. This is the case of in nite subsets of a point lattice. Let Lp be a point lattice in En and Λ be any in nite subset of L. The density associated to Λ has the form X −→ ρ(−→x ) = δ(−→x − l ) l∈Λ −→ −→ −→ The de nition of the dual lattice says that l · l∗ ∈ Z for every l∗ ∈ L∗ and so X −→ X ρˆ(−→s ) = exp(−2πi l · −→s ) = 1 l∈Λ l∈Λ for every −→s ∈ L∗ and because L∗ is countably in nite we have that Λ is indeed a crystal. So knowing this we can now consider a speci c example of a diraction mask which is most certainly non-periodic but produces a diraction pattern with sharp bright spots. If we let our diraction mask contain approximately 500 point apertures in a randomly occupied square region of the two dimensional integer lattice I2 we obtain a diraction pattern (Figure 4.10) which appears to have some form of order. The argument prior to this example provides the explanation for this phenomenon. 76 4.3 Topological Bragg Peaks - An Alternate Approach Diraction theory is not usually viewed as a topological theory and more statistical in nature. However a link has now been established and it is bene cial in many ways as for a lot of structures (including quasicrystals) their diraction patterns and Bragg peaks have a natural topological feel to them. This is a very new area of research for which many of the main results are still to be discovered. I will aim to give a feel for what topological Bragg peaks are all about. The lattices which we are going to be considering are called point patterns. They are similar to Delone sets in many ways and are in a sense more speci c n De nition. 4.3.1 [12] A point pattern Λ ⊂ R is a Delone set satisfying the following: 1. Λ has local nite complexity - up to translation, one nds only nitely many local con gurations of a given size 2. Λ is repetitive - local con gurations in repeat in Λ with bounded gaps. n If we de ne the ball B(0,R) = {y ∈ R : kxk < R} of radius R centered at the origin and let, for a point set Λ, Λ − x be the the point set shifted by some vector x, then we can re-write property 1 by saying the collection of R-patches {B(0,R) ∩ (Λ − x), x ∈ Λ} is nite. The conditions for a point set to be a point pattern certainly seem to be realistic for describing the atomic positions in a structure so for this section we will assume the point sets we describe are point patterns. Another type of point set which will be of use to us is what we call a Meyer set n De nition. 4.3.2 [12] A point set Λ ⊂ R is a Meyer set if it is relatively dense and the set of dierence vectors: ∆ = {x − y; x, y ∈ Λ} is uniformly discrete. These are often referred to as almost lattices. For example, the points of a lattice are a Meyer set, as are the vertices of any rhombic Penrose tiling. In fact, if we have a repetitive Meyer set, then it is a point pattern. n n Suppose that we have a point pattern Λ ⊂ R and a function f : R −→ C. We want to make clear what it means for f to depend only on the local con gurations in the 77 pattern. n De nition. 4.3.3 [12] A function f : R −→ C is called pattern equivariant for Λ if for all > 0 there exists R > 0 such that whenever the R- patches of x and y are the same then |f(x) − f(y)| < . When we say that the R- patches are the same, what we mean is B(0,R) ∩ (Λ − x) = B(0,R)∩(Λ−y), i.e. both have the same con guration in the ball of radius R once shifted to the origin. This de nition allows us to nally de ne what we mean by a topological Bragg peak −→ De nition. 4.3.4 [12] Let Btop denote the set of vectors k for which the plane wave −→ −→ −→ −→ −→ −→ −→ f−→( x ) = ei k · x = cos( k · x ) + i sin( k · x ) is pattern equivariant for Λ. k −→ −→ Thus a vector k ∈Btop if the phase of the plane wave at the point x is determined −→ −→ with more and more precision by the local con guration surrounding x . Any k ∈Btop therefore corresponds to a Bragg peak. However the Bragg peak we have identi ed may be an extinct one, that is, a Bragg peak whose intensity is zero. It is easy to show that Btop corresponds to the set of locations of topological Bragg peaks in our diraction patterns and Btop forms a group. This idea can be quite an abstract one, so to try and gain a better understanding of it we will look at a few examples. Consider the example where n = 2 and we are considering the integer lattice Λ = I2. −→ −→ −→ What we are looking for are the vectors k ∈ 2 for which the function f−→( x ) = cos( k · R k −→ −→ −→ −→ x ) + i sin( k · x ) is pattern equivariant, where x is a point in I2. Let's consider some −→ −→ −→ √ potential candidates for k . Suppose we choose the vectors k1 = (1, 0), k2 = ( 2, 0) and −→ −→ −→ k3 = (2π, 0). If we denote the vector x by x = (m, n) where m, n ∈ Z, then the dot products we return are −→ −→ k1 · x = (1, 0) · (m, n) = m −→ −→ √ √ k2 · x = ( 2, 0) · (m, n) = 2m 78 −→ −→ k3 · x = (2π, 0) · (m, n) = 2πm Now plugging these values into the function f we see that im e j = 1 √ f−→((m, n)) = 2im k e j = 2 j 2πim e j = 3 Now for a function to be pattern equivariant what we need is that for any two choices in our vector −→x = (m, n), the value of f will remain the same (or very close). From the three −→ values of k we have chosen it is clear to see that the only one which produces the same value −→ −→ 2πim for f regardless of the choice of x is k3 because e = cos(2πm)+i sin(2πm) = 1+0i = 1 −→ for all m ∈ Z. Therefore k3 = (2π, 0) is the location of a topological Bragg peak for I2. As −→ for the location of the rest of the topological Bragg peaks, note that k = (0, 2π) would work equally as well. Hence any integer combination of these two vectors would work as well. So the locations of the topological Bragg peaks of I2 have the form {(2πx, 2πy); x, y ∈ Z}. The work of Kellendonk [12] has brought to light many questions about the diraction patterns of materials. As mentioned before this is a relatively new area of research so there are no real experimental results as of yet, but work on this topic could result in yet another change in what we classify as a crystal. 79 Chapter 5 One Dimensional Crystals In this chapter we will look at the projection method we discussed in Chapter 2 & 3 in greater detail, that is, the example of the projection from two dimensional space onto a one dimensional totally irrational subspace of E2. The reason for studying this further is that, in practice, there are real materials (periodic or non-periodic) which can be pro tably studied as if they were one-dimensional. These sets are not as aesthetically pleasing as their higher dimensional counterparts, but they do reveal most of the underlying properties which make these crystals the way they are. 5.1 One-Dimensional Delone Sets A one dimensional Delone set is a set of co-linear points and can be identi ed by an in nite sequence of real numbers ..., x−2, x−1, x0, x1, x2, ... where the distances |xk − xk−1| are bounded. Obviously this sequence has to obey the properties of a Delone set, that is there exists two real numbers r0, R0 such that: 0 < r0 ≤ xk − xk−1 ≤ R0 80 From what we have learned in the previous chapter, if we want to discover if our Delone set is a crystal or not, we must give it a density function of the form X ρ(x) = δ(x − xk) xk∈Λ and then compute its Fourier transform X ρˆ(s) = exp(−2πixks) xk∈Λ The question that we are then faced with is whether it's intensity function J(s) =γ ˆ(s) satis es the diraction condition. It will obviously depend on each individual case whether it does or not. However in this section we are going to look at a few speci c examples and prove whether they do or not. One dimensional Delone sets whose points deviate from a point lattice by a periodic modulation are always crystals (that is their diraction spectra are purely discrete) [1]. In this family of sets there is restriction on the number of dierent step sizes, even though the step sizes themselves are bounded from above and below. We will restrict ourselves to the case where there are nitely many dierent step sizes. Then we can reinterpret our p Delone set as a projection of a path in the standard n-dimensional point lattice In. As we will see later on in this chapter, the geometry of this path will provide us with information about the behaviour of the Fourier transform of the density function. We will call these paths staircases from now on. Due to the niteness of the number of dierent steps, we have for all k ∈ Z xk+1 − xk ∈ {α1, ..., αn} where αj are positive real numbers. We can arbitrarily let x0 = 0 then all values of xm that proceed our initial value will have the form xm = m1α1 + ... + mnαn 81 −→ which is the dot product of something which we will call the population vector pm , which consists of the coe cients of the 0 , and the vector −→ . Note that Pn α s α = (α1, ..., αn) j=1 mj = m because we are working with an integer lattice, so movements between points can be n counted in the natural way. Let e1, ..., en be an orthonormal basis for our space E . The lattice generated by these basis vectors will be the standard n dimensional integer lattice. To construct the staircase we concatenate these vectors in a way which is decided by the sequence xk. It is clear from this that to obtain our original Delone set from this staircase, we simply project the nodes of the staircase onto the line generated by α. 5.2 Fibonacci Sequences The usual way of thinking about the Fibonacci numbers in action is if we consider a population consisting of A0s and B0s which obey rules which say at each generation B −→ A, A −→ A + B We can rewrite this in the form of a linear transform matrix 1 1 P = 1 0 −→ To see this. Consider the rst generation with initial population vector p0 = (0, 1) (repre- senting 0 A0s and 1 B). Then 1 1 Generation 1 (0, 1) = (1, 0) 1 0 1 1 Generation 2 (1, 0) = (1, 1) 1 0 1 1 Generation 3 (1, 1) = (2, 1) 1 0 . . . . 82 0 0 If we carry on in this manner we soon see that there are Fm A s and Fm−1 B s in the (m + 1)st generation. This can be deduced in the following way m+1 1 1 F F −→ m+1 m+1 m −−−→ p0P = (0, 1) = (0, 1) = (Fm,Fm−1) = pm+1 1 0 Fm Fm−1 Proposition. 5.2.1 The ratio Fm/Fm−1 approaches the golden number as m increases. √ 1+ 5 That is lim Fm/Fm−1 = = τ m→∞ 2 Proof. [1] The characteristic equation of P is given by λ2 − λ − 1 = 0 and its solutions are a) τ and b) − 1/τ 1 1 To nd the eigen vector −→ associated to the eigen value , we calculate −→ a) 1 τ 1 = 1 0 1 − τ 1 τ−→ ⇔ −→ = 0 and from this we nd that −→ = √ τ , √ 1 = 1 (τ, 1) 1 1 1 1+τ 2 1+τ 2 ν 1 −τ where ν2 = 1 + τ 2. b) In a similar manner to above we nd out that −→ = √−1 , √ τ = 1 (−1, τ) 2 1+τ 2 1+τ 2 ν These two eigen vectors form an orthonormal basis of E2 so we can write the population vectors and the matrix P in terms of them so 1 τ −→p = (0, 1) = −→ + −→ 0 ν 1 ν 2 and τ 1 −→p P = −→ − −→ 0 ν 1 ν 2 Therefore from this we can derive τ m (−τ −1)m−1 p−→ = −→p Pm = −→ − −→ m 0 ν 1 ν 2 as we see that −1 m−1 which means that −→ τ m −→ and therefore as m → ∞ (−τ ) → 0 pm − ν 1 → 0 0 0 −→ we increase m without bound we see that the ratio of A s and B s (pm) tends to the ratio 83 −→ of the components of 1 , which, from before, has ratio τ : 1. The golden number τ has many unique properties which will be useful to us later on. Of which are the following facts which can be easily proven by calculation 1. τ 2 = τ + 1 2. τ − τ −1 = 1 3. τ 3 − 2τ 2 = −1 We can rewrite the original growth rule in a multiplicative form, that is A −→ AB, B −→ A This substitution rule creates a sequence of A0s and B0s at each generation, where each generation is a concatenation of the previous two generations. If we carry on the substi- tution ad in nitum the in nite string which is created is nonperiodic. This is down to the previous proposition because an in nite string of symbols in which the ratio of the number of occurrences of some pair of symbols is irrational cannot be periodic [1]. This produces a string of symbols which is only in nite in one direction. To make it in nite in both directions, we begin by composing the string by grouping together symbols in the following way A0 = AB, B0 = A For example if we have a nite sequence ABAABABA then the grouped sequence will be ABAAB. We call this composed sequence the predecessor of the one we started with. De nition. 5.2.2 [1] A Fibonacci String F is a two-way in nite string of A0s and B0s that has predecessors of all levels with respect to the composition rule. Two way in nite strings are dierent to one way in nite strings in the sense that they are not translation invariant. Proposition. 5.2.3 Every Fibonacci string is repetitive 84 We can associate a Fibonacci string F with a Delone set by identifying A and B with line segments of two dierent lengths α and β. We choose the lengths to be the relative frequencies of A and B in the string so that α/β = τ De nition. 5.2.4 [1] A Fibonacci Sequence is a two way in nite sequence of points on the real line ..., −x2, −x1, x0, x1, x2, ... such that 1. xn+1 − xn ∈ c{1, τ} for some c > 0 2. The dierence sequence (xn+1 − xn) has predecessors of all levels under the compo- sition rule A0 = AB, B0 = A After we have composed the sequence, the size of the large steps is going to be α + β and the size of the small steps will be α. However the ratio will remain the same as before because α α + β α2 − αβ − β2 τ 2 − τ − 1 − = = = 0 β α αβ τ since |α| = τ, |β| = 1. Every Fibonacci sequence is therefore associated to an in nite staircase and the stair- case has an average slope of 1 [1]. τ 5.3 Projected One-Dimensional Crystals We will now discuss a method for producing one-dimensional Delone sets which are always crystals. This is the orthogonal projection method from Chapter 2 in the special case where we project from two-dimensional space onto a one-dimensional totally irrational subspace. Let p be the two dimensional integer point lattice whose Voronoi cell, as you may I2 recall, is a square. Let l be our totally irrational subspace which passes through the origin. From the previous section we will allow this line to have slope 1 . Let p so that the τ X ⊂ I2 points of are the points of p whose Voronoi cells are cut by . Due to the lattice points X I2 l 85 Figure 5.1: Staircase and subset (shaded area) of the integer lattice p formed by the X I2 line l of gradient 1/τ −→ being integer distances apart, we can assign a population vector pm = (m1, m2) to order st the points in X, where m1, m2 ∈ Z and m1 + m2 = m. Therefore the (m + 1) point of −−−→ −−−→ X is either pm+1 = (m1 + 1, m2) or pm+1 = (m1, m2 + 1). Therefore we can say that the points of form a staircase in p (Figure 5.1) X I2 As in Chapter 2, let Π be the orthogonal projector onto our subspace l, then Π(X) is a Delone set. It can be shown by trigonometry that the step lengths in this Delone set are of lengths τ 1 |Π(−→ )| = , |Π(−→ )| = 1 ν 2 ν We can nd the coordinates of the points in our Delone set explicitly. Let (u, v) be a point in our integer lattice. Then l cuts the Voronoi cell of (u, v) if and only if l cuts the principle diagonal of the Voronoi cell which lies on the line x + y = m. This implies that the points of intersection are where the line y = x/τ and x + y = m meet x mτ mτ and so ⇒x + τ = m ⇒xτ + x = mτ ⇒ x = 1+τ = τ 2 86 m m (x, y) = ( , ) τ τ 2 So therefore l will cut the cell of (u, v) if and only if 1 m 1 u − < < u + 2 τ 2 1 m 1 v − < < v + 2 τ 2 2 However m = u + v ⇒ v = m − u and so the second inequality becomes 1 m 1 1 m 1 1 m 1 m − u − 2 < τ 2 < m − u + 2 ⇒ −u − 2 < τ 2 − m < −u + 2 ⇒ u + 2 > m − τ 2 > u − 2 1 m 1 ⇒ u − 2 < m − τ 2 < u + 2 where m m(τ 2−1) mτ m , so the second inequality reduces down to the rst m − τ 2 = τ 2 = τ 2 = τ one, which we can rewrite as m = u τ where kxk is the nearest integer function, which assigns to the real number x the integer nearest to it. If we now look back at the staircase and the population vector we assigned to it, we can rewrite this vector as m m p−→ = ( , m − ) m τ τ because m . v = m − u = m − τ After we project these points onto l, we nd that the mth point of Π(X) has the coordinate m 1 m x = + m ν τν τ [1] Proposition. 5.3.1 In the Fibonacci case, the `acceptance domain' K is the interval τ τ [− 2ν , 2ν ] 87 Figure 5.2: Trigonometry behind the acceptance domain K. on the orthogonal complement of our line l, l⊥. Proof. We will follow the diagram (Figure 5.2) The points marked with a cross are the boundaries of the interval we want to nd. In order to nd them we will use the similar shaded triangles marked 1 and 2 which have the same shaded angle marked within them. First, let us consider triangle 1. We want to work out the length of the edge marked H. We calculate this by Pythagoras theorem to get: r r 1 1 r1 1 τ 2 + 1 ν H = ( )2 + ( )2 = + = = 2 2τ 4 4τ 2 4τ 2 2τ Now we can work out the shaded angle in the following way 0.5 τ cos θ = = ν/2τ ν We now use this in triangle 2 to nd the value A 88 τ A τ 2 cos θ = = ⇒ A = ν τ/2 2ν Where the value τ for the hypotenuse is calculated via the similar triangle which sits on 2 top of the Voronoi cell about the origin. Hence we have found our upper bound of the interval and by the symmetry of the lattice we can now say that τ 2 τ 2 ] K = [− 2ν , 2ν Proposition. 5.3.2 Λ = Π(X) is a Fibonacci sequence −−−→ −→ Proof. [1] What we need to show is that the dierence sequence Π(pm+1 − pm) has pre- decessors at all levels so that it obeys the properties of the de nition. Therefore what we need to show is that there is a rescaled copy of the integer lattice p that plays the same I2 role with respect to the predecessor of that p does with . Once we have found this Λ I2 Λ rescaled copy of p then the same argument can be applied to nd its predecessor and so I2 −→ −→ on. We claim that the lattice generated by τ 1 and τ 2 is τI2 and it is easy to check that it's Voronoi cell projects to τK on l⊥. Something else which we must also show is that the new construction does not introduce any new points, that is, all the points projected from are also projected from p. τI2 I2 A point of p has coordinates where p. It's projection onto is τI2 (τu, τv) (u, v) ∈ I2 l given by τ 2u+τv . We now have to show that this is a point in the Delone set . Since ν Π(X) τ 2u + τv = (τ + 1)u + v = τ(u + v) + u = (u + v, u) · (τ, 1) we have to show that ⊥ τ 2 τ 2 . However this follows easily from the fact νΠ (u + v, u) ∈ [− 2 , 2 ] that νΠ⊥(u, v) lies in the interval and 1 νΠ⊥(u + v, u) = −(u + v) + τu = (u − τv) τ The nal proof in this section is one which says that Fibonacci sequences are always crystals. Proposition. 5.3.3 The Fibonacci sequence is a crystal. 89 Proof. [1]There are many ways of proving this statement, but we will only give one. Our proof will make use of the point lattice dual. We want to study the sum X ρˆ(s) = exp(−2πixms) xm∈F The population vector we have discussed before can be written as a combination of two dierent bases. First of all as a linear combination of the orthonormal basis of p I2 −→ −→ −→ pm = m1 e1 + m2 e2 and secondly as a linear combination −→ −→ −→ pm = xm ε1 + ym ε2 −→ −→ −→ ⊥ where 1 is a unit vector on l, 2 is a unit vector orthogonal to 1 on l and ym is the −→ ⊥ projection of pm onto l . The point ym will always lie within our acceptance domain K −→ −→ −→ −→ because pm is a vector to points whose Voronoi cells are cut by l. Set w = s1 +t2 . Then −→ −→ −→ −→ −→ pm · w = xms |1 | + ymt |2 | = xms + ymt because i are unit vectors. This means that our expression for ρˆ(s) above becomes X −→ −→ ρˆ(s) = exp(−2πipm · w )exp(2πiymt) m∈Z If −→ is a vector in the dual lattice of p (which is a self dual, hence p p ) then w I2 I2 ∗ = I2 −→ −→ −→ −→ pm · w ∈ Z. Hence exp(−2πipm · w ) = 1 and therefore X ρˆ(s) = exp(2πiymt) m∈Z It can be shown that if the limit N 1 X lim exp(2πiymt) N→∞ N m=−N 90 is a non-zero constant then the diraction pattern it produces has a point of maximal brightness at s [1]. But this follows directly from the uniform distribution of the projection of p onto ⊥. Hence has a delta at every satisfying −→ −→ , where I2 l ρˆ(s) s pm · w = xms + ymt −→ p. Hence the Fibonacci sequence is a crystal. w ∈ I2 91 Concluding Remarks New results which change the way we look at crystals in a mathematical sense are appearing all of the time. As an example, the work of Kellendonk [12] on topological Bragg peaks was published less than one year ago. Therefore researching this eld has been very fruitful indeed as it is at the forefront of modern day mathematics and it is exciting to think that one day there may be future implications to some of the results I have proven or mentioned in this paper. It was only in the 1960's when the rst quasicrystalline structure was synthesised and since then many more man-made structures have been discovered. However it is thought that there may be naturally occurring substances or living organisms which have this aperiodic but ordered structure. For example it is believed that some viruses may have such a structure about them. They have a protein shell called a capsid which encloses the genetic material of the virus and it is this shell which is thought to have icosahedral structure about it in certain virus families. Understanding the capsid structures of these types of virus may lead to more eective vaccinations against deadly viruses, so an understanding of its quasicrystal structure could be vital. Quasicrystals made of an alloy of aluminium and manganese have also been shown to have very high electrical and thermal resistivity, almost equivalent to that of an insulator, which is very peculiar for a material which is made from 70% aluminium. This must therefore have something to do with it's structure. To summarise the potential uses of quasicrystals, I quote Olof Nilsson, manager of physical metallurgy for the R&D Centre of Sandvic Steel in Sweden, who said in a conference Three years ago I thought that quasicrystals were something that existed only in the academic world, but I have been forced to change my mind. As a matter of fact, we can't avoid quasicrystalline precipitates in our commercial materials, and furthermore, 92 we can make use of them.. I expect that, with future discoveries, the formal de nition of what constitutes a crystal will change once more and I will most de nitely keep myself updated with the breakthrough discoveries in this eld after what has been a fascinating year long project. 93 Bibliography [1] Marjorie Senechal Quasicrystals and Geometry , Cambridge University Press, 1995 [2] Wikipedia (2013). Bravais Lattice. Available: http://en.wikipedia.org/wiki/Bravais_lattice. [3] Gerhard Karlowatz. (2009). Advanced Monte Carlo Simulation for Semiconductor Devices. Available: http://www.iue.tuwien.ac.at/phd/karlowatz/img151.png. [4] Wikipedia. (2012). Checkerboard. http://en.wikipedia.org/wiki/File:Checkerboard_pattern.svg. [5] Wikipedia. (2014). Penrose Tiling. http://en.wikipedia.org/wiki/Penrose_tiling. [6] Wikipedia. (2007). Aperiodic Tiling. http://en.wikipedia.org/wiki/File:L_substitution_system.jpg [7] Wikipedia (2014). Bragg's Law. http://en.wikipedia.org/wiki/Bragg's_law [8] Crystallography. Scattering and Diraction (optical diraction diagrams). http://www.xtal.iqfr.csic.es/Cristalogra a/parte_05_3-en.html [9] The Fourier Transform (2009). The Fourier Transform of the box function. http://www.thefouriertransform.com/transform/fourier.php [10] Marjorie Senechal Crystalline Symmetries: An Informal Mathematical Introduction, Adam Hilger, 1990 [11] Tel Aviv University (2004). Lecture 8: Dual Lattices. http://www.cims.nyu.edu/~regev/teaching/lattices_fall_2004/ln/DualLattice.pdf [12] Johannes Kellendonk Topological Bragg Peaks And How They Characterise Point Sets. 1 October 2013. arXiv:1309.7632 94 [13] Wolfram Mathworld. Rhombic Icosahedron. http://mathworld.wolfram.com/RhombicIcosahedron.html [14] Wikipedia (2013). Rhombic Dodecahedron. http://en.wikipedia.org/wiki/Rhombic_dodecahedron [15] Wolfram Mathworld. Generalized functions. http://mathworld.wolfram.com/GeneralizedFunction.html [16] N.G. de Bruijn. Algebraic theory of Penrose's non-periodic tiling of the plane. 1981. alexandria.tue.nl/repository/freearticles/597566.pdf [17] Edita Pelantová, Zuzana Masáková. Quasicrystals: algebraic, combinatorial and geo- metrical aspects. 2003. http://arxiv.org/pdf/math-ph/0603065.pdf [18] Marjorie Senechal. Tilings, Diraction, and Quasicrystals. www.mathematica- journal.com/issue/v4i2/.../10-15_senechal42.mj.pdf [19] Simon Jazbec. The Properties and Applications of Quasicrystals. 2009. http://ma ja.fmf.uni-lj.si/seminar/ les/2009_2010/Quasicrystals.pdf [20] Diana Lutz. Putting Quasicrystals to Work. 1996. http://www.aip.org/tip/INPHFA/vol-2/iss-4/p26.pdf 95