An aperiodic monotile that forces nonperiodicity through dendrites
AAN APERIODIC MONOTILE THAT FORCES NONPERIODICITYTHROUGH DENDRITES
MICHAEL MAMPUSTI AND MICHAEL F. WHITTAKER
Abstract.
We introduce a new type of aperiodic hexagonal monotile; a prototile that admitsinfinitely many tilings of the plane, but any such tiling lacks any translational symmetry.Adding a copy of our monotile to a patch of tiles must satisfy two rules that apply only toadjacent tiles. The first is inspired by the Socolar–Taylor monotile, but can be realised byshape alone. The second is a dendrite rule; a direct isometry of our monotile can be addedto any patch of tiles provided that a tree on the monotile connects continuously with a treeon one of its neighbouring tiles. This condition forces tilings to grow along dendrites, whichultimately results in nonperiodic tilings. Our dendrite rule initiates a new method to producetilings of the plane. Introduction
Almost 60 years ago, Hao Wang posed the Domino Problem [16]: is there an algorithmthat determines whether a given set of square prototiles, with specified matching rules, cantile the plane? Robert Berger [4] proved the Domino Problem is undecidable by producingan aperiodic set of 20,426 prototiles, a collection of prototiles that tile the plane but onlynonperiodically (lacks any translational periodicity). This remarkable discovery began thesearch for other (not necessarily square) aperiodic prototile sets. In the 1970s, there weretwo stunning results giving examples of very small aperiodic prototile sets. The first was byRaphael Robinson who found a set of six square prototiles [10]. The second was by RogerPenrose who reduced this number to two [5, 9]. Penrose’s discovery led to the planar einstein(one-stone) problem: is there a single aperiodic prototile?In a crowning achievement of tiling theory, the existence of an aperiodic monotile was re-solved almost a decade ago by Joshua Socolar and Joan Taylor [12, 15]. Several candidates hadbeen put forth prior to their monotile, but the experts immediately recognised the importanceof Socolar and Taylor’s discovery [1, 2, 3, 7, 8]. The Socolar–Taylor monotile is a hexagonal tilewith two local rules that enforce aperiodicity. The first rule forces tiles to arrange themselvesinto collections of triangles, and the second rule ensures these triangles are nested, therebyforcing the resulting tiling to be nonperiodic. One limitation of the Socolar–Taylor monotile isthat the second local rule applies to pairs of non-adjacent tiles, so aperiodicity is not enforcedby adjacencies. Another limitation is that reflected copies of the monotile are required to tilethe plane. The search for an aperiodic monotile with local rules that only apply to adjacenttiles or does not require reflections has been a driving force of research in tiling theory sinceSocolar and Taylor’s amazing discovery.
Mathematics Subject Classification.
Primary: 52C23; Secondary: 37E25; 05B45.
Key words and phrases. aperiodic tilings; dendrites; fractal; monotile; nonperiodic.This research was partially supported by EPSRC grant EP/R013691/1, ARC Discovery ProjectDP150101595, and the Australian Government Research Training Program Scholarship. a r X i v : . [ m a t h . M G ] M a y MICHAEL MAMPUSTI AND MICHAEL F. WHITTAKER −−− + Figure 1.
Representations of our monotile: the R1-curves appear as blackcurves or edge decorations, and the R2-tree as red trees or magnetic dipoles.In this paper, we put forward a new type of aperiodic monotile that does not require areflection, and has rules that only apply to adjacent tiles. We start with a hexagonal tilesatisfying Socolar and Taylor’s first local rule, and add a rule that only allows finite patchesof tiles to connect along a dendrite. Our second rule is motivated by the proposed growth ofcertain quasicrystals [6, 17]. Interestingly, a consequence of not requiring a reflected copy ofour monotile is that the first rule can be enforced by shape alone, while this does not hold forthe Socolar–Taylor monotile [13, p.22]. Although we are comparing our monotile with Socolarand Taylor’s construction, these two monotiles are different in character. The Socolar–Taylormonotile is defined using local rules, whilst our monotile is defined by pairing a local rule anda dendritic rule. Indeed, our dendrite rule is local in the sense of building tilings, but is notlocal as a rule on tilings. So the monotile we present here is not an einstein in the technicalsense, but rather a variant that requires tilings to be constructed starting from a seed tile.Three representations of our monotile appear in Figure 1, and each of these representationsmust satisfy the local rule R1 and the dendrite rule R2 outlined below.Before introducing our monotile, we briefly define the terminology used in the paper. A tiling is a covering of the plane by closed topological discs, called tiles, that only intersecton their boundaries. A patch is a finite connected collection of tiles that only intersect ontheir boundary. The building blocks of a tiling are the prototiles : a finite set of tiles withthe property that every tile is a direct isometry (an orientation preserving isometry) of aprototile. If the prototile set consists of a single tile, or a single tile and its reflection, we callit a monotile . A tiling is said to be nonperiodic if it lacks any translational periodicity, and aset of prototiles is called aperiodic if it can only form nonperiodic tilings.Our monotile has two distinct features: a disconnected set of three curves that meet tileedges off-centre, and a connected tree that meets itself whenever two edges with a tree inter-sect. Once a single tile has been placed, a direct isometry of our monotile can be added tothe plane provided the resulting collection of tiles is a patch, and R1 : the black off-centre lines and curves must be continuous across tiles (c.f. [12, R1 ]) and R2 : the new tile’s red tree continuously connects with at least one tree of an adjacent tile.We note that R1 can be realised by shape alone, represented as puzzle like edge contours, while R2 can be represented by magnetic dipole-dipole coupling. These representations appear inFigure 1.In what follows, we will refer to the off-centre decorations that determine R1 as R1-curves.These combine to form R1-triangles. Similarly, we will refer to the decorations that determine R2 as R2-trees. We say that a tiling T satisfies rule R if every patch in T is contained in apatch that can be constructed following rule R . Therefore, a tiling T satisfies R2 if and onlyif the union of R2-trees in T is connected. Two legal patches satisfying R1 and R2 appear DENDRITIC MONOTILE 3 in Figure 2. From this point forward we will use the representation from the left hand side ofFigure 1, since we use R1-triangles heavily in the arguments that follow.
Figure 2.
Two sets of equivalent legal patches. The patches on the top can beconstructed directly using R1 and R2 , while the patches on the bottom cannot,but are subsets of legal patchesThe goal of the paper is to prove the following theorem. Theorem A.
The monotile in Figure 1 is aperiodic. That is, there is a tiling of the planeusing only direct isometries of the monotile that satisfies R1 and R2 , and any such tiling T is nonperiodic. The proof of Theorem A is essentially the contents of the rest of this paper. Since the proofis quite involved, we now provide a brief sketch. In Section 2, we identify two classes, C and C , of tilings satisfying R1 and R2 . We prove that both classes are non-empty and onlycontain nonperiodic tilings. To prove C contains only nonperiodic tilings, we use a cleverconstruction of Socolar and Taylor to build tilings satisfying R1 , but not necessarily R2 . Wethen build a tiling in C by recursively constructing a spiral fixed point of R2-trees about theorigin. We prove that C is non-empty and that every tiling is nonperiodic by building allpossible tilings in the class. The key to this construction is Lemma 2.4, which shows that nolegal tiling can contain an infinite R1-triangle (two R1-rays connected by an R1-corner). In MICHAEL MAMPUSTI AND MICHAEL F. WHITTAKER
Section 3, we show that R2 rules out two patterns of R2-trees, which we call R2-cycles andR2-anticycles. These patterns are exactly those that are formed between R1-triangles whenthey are arranged into a periodic lattice, as illustrated in Figure 16. A nice consequence ofruling these patterns out is that the union of R2-trees in any tiling is always a connected tree . We are left to show that every tiling satisfying R1 and R2 must be in C or C . This isachieved by proving that the absence of R2-cycles and R2-anticycles implies that every tilingfits into one of these two classes.After proving Theorem A we discuss the continuous hull of tilings arising from our monotile.We show that all tilings in the continuous hull satisfy R1 and a weak version of R2 .Our use of dendrites in constructing the rules for our monotile was motivated by growth inquasicrystals. Several papers hypothesise that dendritic structures in molecules, particularlyin soft-matter quasicrystals, are the mechanism that force nonperiodicity, see [6, 17]. In [18],magnetic dipole-dipole coupling of quasicrystals is mentioned, which can also be modelled byour monotile as indicated in Figure 1. According to the recent survey paper of Steurer [14],one of the most pressing questions in quasicrystal theory is understanding how they form, andwhen they grow periodically and quasiperiodically. The dendrite rules in this paper show thatdendritic growth can lead to aperiodic tile sets. Acknowledgements.
We are grateful to Michael Baake, Franz G¨ahler, Chaim Goodman-Strauss,Jamie Walton, and Stuart White for their helpful comments and mathematical insights.2.
Classes of tilings arising from the monotile
In this section, we consider two classes of tilings satisfing R1 and R2 . We show that eachclass is non-empty and only contains nonperiodic tilings. In the following section, we showthat these classes exhaust the possible tilings that can be constructed from our monotile.We begin with a closer look at the R1-triangles. Notice that the small R1-curves only occurat angle π/
3, so any R1-triangle must be equilateral. We also observe that the R1-curves canonly give rise to nested R1-triangles, an infinite R1-triangle (two infinite R1-rays connectedby an R1-curve) or a bi-infinite R1-line. Let us denote the length of a finite R1-triangle bythe number of tiles comprising the straight R1-line segment of any given side, as in Figure 3.Length 0 Length 1 Length 3 Length 7
Figure 3.
R1-triangles have length 2 n − n = 0 , , , . . . . DENDRITIC MONOTILE 5
The following lemma is easily deduced from R1 and the geometry of the R1-curves. Lemma 2.1.
Any nested R1-triangle has length n − for some n = 0 , , , . . . . Definition 2.2.
Let C denote the collection of tilings whose prototile set is the monotile fromFigure 1 satisfying R1 and R2 . Consider the subcollections of C defined by the properties: C : if the corners of a pair of R1-triangles meet at a common tile in T , then these R1-triangles have the same length; C : T contains a bi-infinite R1-line.We first consider the collection C . Let us introduce a convention that will be used in theproof of the following proposition. Define R θ to be the rotation operator that rotates a patchcounterclockwise around the origin by θ . Proposition 2.3.
The collection C is non-empty and only contains nonperiodic tilings. zeroth R1-grid first R1-grid n th R1-grid
Figure 4.
The interlaced honeycombs of R1-triangles in the Socolar–Taylor construction
Proof.
To see that C contains only nonperiodic tilings, we appeal to a construction of Socolarand Taylor [12, Theorem 1], which we now summarise. Tilings satisfying R1 are producedby adding markings to tiles that form successively larger hexagonal grids of interlaced R1-triangles, see Figure 4 for a pictorial representation of their construction. To force thesehoneycomb lattices of length 2 n − C , this condition is one of ourhypotheses. The honeycomb lattices of R1-triangles have no largest translational periodicityconstant, so that all of the infinite tilings produced must be nonperiodic.We are left to show that C is non-empty. To construct a tiling T ∈ C , we recursivelydefine patches P n with three key properties:(1) P n satisfies R1 and R2 ,(2) the patch P n is a strict subset of P n +1 , and(3) as n increases, the patch around the origin in P n increases exponentially.The union of patches P n is a tiling of the plane, see (2.2) below.To define the recursive algorithm, we carefully look at how patches grow with respect to R2 . The first few steps of our algorithm appear in Figure 5, which should help decipher therecursive definition below. Essentially, the patch P n is constructed from P n − by gluing P n − MICHAEL MAMPUSTI AND MICHAEL F. WHITTAKER P P P P P Figure 5.
Constructing an infinite tiling by piecing together R2-treesand three direct isometries of P n − together by a single connecting tile in the centre of thenew patch P n . These central tiles have centre at a point x n (explicitly described below), andeach such x n is marked by a dot in Figure 5, which helps to see P n in P n +1 .Let P be our monotile in exactly the orientation appearing in Figure 1, placed with itscentre on the origin. Using polar coordinates ( r, θ ), the recursive formula for patch P n isdefined by points x := (0 ,
0) and x n := n (cid:88) i =1 (cid:0) i − , iπ/ (cid:1) , along with patches P n := R nπ ( P + x n ) (cid:71) P n − (cid:71) R π ( P n − − x n − ) + x n + (cid:0) n − , nπ/ − π/ (cid:1) (2.1) (cid:71) R π ( P n − − x n − ) + x n + (cid:0) n − , nπ/ (cid:1)(cid:71) R π ( P n − − x n − ) + x n + (cid:0) n − , nπ/ π/ (cid:1) . Note that the points x n appear at each corner of the superimposed spiral in Figure 6. DENDRITIC MONOTILE 7
Figure 6.
Part of the patch P with the spiral branch of the R2-tree highlightedThe method we have used to build P n ensures that both R1 and R2 are satisfied. Moreover,the patches P n overlap where they intersect, and are space filling in a spiral pattern thatsuccessively connects the points x n around the origin, see the Figure 6. Thus, the union(2.2) T := (cid:91) n =0 P n is a tiling of the plane satisfying both R1 and R2 .To finish the proof, we show that T satisfies the defining condition of C . Observe that eachpatch P n has two R2-tree straight segments of length 2 n − x n at angles nπ and ( n +3) π . The recursive definition extending P n into P n +1 ensuresthat the straight segments of the R2-tree arms terminate at length 2 n −
1, which implies thatthe lengths of the R1-triangles (realised in P n +3 ) along those arms also have length 2 n − T is in C . (cid:3) We now consider the collection C , but first a lemma. Lemma 2.4.
Any tiling satisfying R1 and containing an infinite R1-triangle (two R1-raysconnected by an R1-corner) does not satisfy R2 . MICHAEL MAMPUSTI AND MICHAEL F. WHITTAKER
A B
Figure 7.
The pictorial argument for Lemma 2.4
Proof.
Suppose T is a tiling satisfying R1 and containing an infinite R1-triangle. Let A and B be the respective connected components of the union of R2-trees associated with each sideof the infinite R1-triangle, see Figure 7. We will argue that A and B cannot be connected.Notice that the branches of A and B into the interior of the infinite R1-triangle never reachthe opposite side of the R1-triangle, and are disjoint. At the infinite R1-triangle corner, oneof the trees extends into the corner tile, while the other does not. Let us assume A does notextend. The only remaining connection possible between A and B is along R1-branches of A growing outside the infinite R1-triangle. Every such R2-branch extends along the side ofan R1-triangle, and terminates at the corner of the R1-triangle, if the R1-triangle is finite,or extends infinitely if the R1-triangle is infinite. However, as noted above, R2-branches intothe interior of an R1-triangle never reach the opposite side of the R1-triangle and are disjointfrom any R2-branches extending from the opposite side. It follows that A never meets B , andso T cannot satisfy R2 . (cid:3) Proposition 2.5.
The collection C is non-empty and only contains nonperiodic tilings.Proof. We will construct a tiling in C , starting with a bi-infinite string of tiles forming anR1-line. Note that the union of R2-trees along this string is connected. In order to simplifythe argument, we fix the orientation of this string of tiles, as in Figure 8, and will refer tothe top, bottom, left and right as per the orientation depicted. Our construction will produceevery possible tiling in C up to direct isometry. DENDRITIC MONOTILE 9
Figure 8.
The bi-infinite string of tiles forming an R1-line in Proposition 2.5.We begin by adding tiles above the R1-line. Lemma 2.4 implies that we can never addan infinite R1-triangle with a corner meeting the R1-line. So every R1-triangle meeting theR1-line must have length 2 n − n ∈ N by Lemma 2.1. Figure 9.
Adding tiles to attach a length 3 R1-triangle to the R1-line
Figure 10.
Adding the forced tiles between the R1-triangle and the R1-lineSuppose we add an R1-triangle of length 2 n − n = 2. We note that the union of R2-trees is no longerconnected, but this will be rectified shortly. Between the length 2 n − n tiles to the leftor right of the bottom corner of the 2 n − n +1 − n +2 − n +1 − n + j tiles along the R1-line from its successorcarries on ad infinitum. Lemma 2.4 implies that tilings in C cannot contain infinite triangles,so we must change the direction of our choice an infinite number of times so that every tileon the R1-line contains the corner of some finite length R1-triangle. Moreover, these trianglesare forced to occur periodically. That is, placing a length 2 m − l on the R1-line yields a length 2 m − l + k m +1 for Figure 11.
A length 7 R1-triangle is forced to one side of a length 3 R1-triangleall k ∈ Z , and at least one R1-triangle of length 2 m − m +1 . This construction leads to a half-plane of tiles that satisfies R1 . Figure 12.
To the right of an R1-triangle, the R2-trees of the shaded tiles areconnected to the R2-trees along the infinite stringWe now argue that the half-plane of tiles constructed above also satisfies R2 . Indeed, theunion of R2-trees along the bi-infinite string of tiles is connected. Given a length 2 m − R1 and R2 .An analogous argument implies that all tiles in the lower half-plane satisfy R1 and R2 .Since the upper and lower half-planes intersect along the bi-infinite R1-line, the resultingtiling is in C . Since there are arbitrarily large R1-triangles arranged in interlaced periodic DENDRITIC MONOTILE 11 patterns whose corners meet the R1-line, the resulting tiling is nonperiodic, giving the desiredresult. (cid:3)
We note that the classes C and C have non-trivial intersection. Indeed, if the R1-triangleson opposite sides of the bi-infinite R1-line of a tiling in C have the same length, then it isalso in C . 3. Proof of Theorem A
We have shown in Propositions 2.3 and 2.5, that the classes C and C from Definition2.2 are non-empty and contain only nonperiodic tilings. In this section, we will prove that C = C ∪ C , which will prove Theorem A. The key is to prove that any tiling in C that doesnot belong to C must have an infinite R1-line through it, so it belongs to C . n = 2 n = 3 n = 4 Figure 13.
Monotiles with dashed lines represent legal patches of length 2 n − n ∈ N , amonotile with dashed lines represents a patch of tiles depicted in Figure 13, where the maindiameter of R2-trees has length 2 n −
1. We note that these are the patches P n that appearedwhen we constructed a tiling in Proposition 2.3. Moreover, notice that these patches fittogether in the manner depicted in Figure 14.The fundamental tool of this section is to prove that R2 rules out three R1-triangles meetingcorners to sides in the cyclic fashion appearing in Figure 15, where the solid lines have lengthone, and the dotted lines have length 2 n − n ∈ { , , , . . . } . Due to the behaviour ofthe R2-trees in the centre of the cyclic R1-triangles, we will refer to these configurations as R2-cycles and
R2-anticycles , respectively. We note that Lemma 2.1 implies that R2-cyclesand R2-anticycles only occur with side length 2 n − R2 . However, a growth rule that disallows these periodic lattices wasthe key to the results of this paper. n = 3 n = 2 Figure 14.
For n = 2 and n = 3, the patches in Figure 13 fit together as abovean R2-cycle of length 2 n − n − Figure 15.
The illegal patches appearing in Lemmas 3.1 and 3.3
Lemma 3.1.
Suppose T is a tiling in C , then T does not contain an R2-cycle.Proof. We begin with a patch containing a R2-cycle of length 2 n −
1, and show that any tilingthat extends the patch fails to satisfy R2 . Fix n ∈ { , , , . . . } , and suppose we start withthe patch on the left-hand side of Figure 15.Since R2 implies that the R2-tree must be infinite, at least one branch of the R2-treeleaving the central R2-cycle must be infinite. We will refer to the R1-triangle associated withthe infinite tree as triangle A . Lemma 2.4 implies that triangle A cannot be infinite. Wewill show that triangle A cannot be finite either. Lemma 2.1 implies that triangle A musthave length 2 n + m +1 − m ∈ { , , . . . } . Since the union of R2-trees terminates atthe R1-corner of A , there must be an infinite branch leaving the main tree. All R2-branchestowards the interior of A are finite, so any infinite branch must turn away from triangle A .A straightforward, but geometrically technical, induction proves that if triangle A has length DENDRITIC MONOTILE 13
Figure 16.
A periodic lattice of R1-triangles
A B m = 1 An R1-trianglewith length2 n (2 − (2 − − m = k = 1),which is impossibleby Lemma 2.1 Figure 17.
The case for n ∈ N and m = 1 in the induction from Lemma 3.12 m + n +1 −
1, then the next R1-triangle clockwise in the R2-cycle of length (2 n −
1) (labelled B in Figures 17 and 18) forces an R1-triangle of length 2 n (2 m +1 − (2 k − − m = 1 m = 2 A B 5 13 9
Figure 18.
The case for n = 1 and m = 1 , k ∈ { , . . . , m } . Lemma 2.1 implies that such triangles cannot exist in a tiling, giving us thedesired contradiction. So T cannot belong to C . (cid:3) Remark . A comment on the omitted geometric induction argument from the proof ofLemma 3.1 is in order. Figure 17 shows the argument for arbitrary n ∈ N and m = 1. Figure18 shows the geometric argument for n = 1 and m = 1 ,
2. Using Figure 18, the generalargument for m = 1 , n −
1, as depicted in Figures13 and 14.
Lemma 3.3.
Suppose T is a tiling in C , then T does not contain an R2-anticycle.Proof. We begin with a patch containing a R2-anticycle of length 2 n −
1, and show that anytiling that extends the patch does not belong to C . Fix n ∈ { , , , . . . } , and suppose we startwith the patch on the right-hand side of Figure 15.Rule R2 implies that all three branches of the R2-anticycle must be infinite and must allbe connected. Let us concentrate on just one of these branches. Lemma 2.4 implies that theR1-triangle associated with this branch cannot be infinite, and then Lemma 2.1 implies itmust have length 2 n + m − m ∈ { , , . . . } . However, if this R1-triangle has length2 n + m −
1, then an R2-cycle of length 2 n − T is not in C . See Figure 19 for apictorial representation, where x is the location of the R2-cycle in the case m = 2. (cid:3) We are now able to tackle the proof of Theorem A. The reader is encouraged to considerFigures 20 and 21 while reading through the proof.
DENDRITIC MONOTILE 15 x R2-anticycle
Figure 19.
The pictorial idea of the proof of Lemma 3.3, for n ∈ N and m = 2CentreLeft R1-anticycle Right R1-cycleCentreLeft R1-anticycle Right R1-cycleCentreLeft R1-anticycle Right R1-cycle Figure 20.
Three possibilities for m = 0 and n > Proof of Theorem A.
Propositions 2.3 and 2.5 prove that C and C are both non-empty andonly contain nonperiodic tilings. We will prove that C = C ∪ C , thereby proving the result. CentreLeft R1-anticycle Right R1-cycle
Figure 21.
One possibility for m = 1 and n > C and C satisfy R1 and R2 so that C ∪ C ⊆ C . We are left to prove thereverse inclusion. Suppose T is in C . If all pairs of R1-triangles in T that meet at a commontile have the same length, then T is in C . If there exists a pair of R1-triangles which meetat a common tile, but do not have the same length, we claim that T is in C , which wouldimply that C ⊆ C ∪ C . To do this, we must show that such a tiling T has a bi-infinite R1-linethrough it.Suppose that T contains a tile where a pair of R1-triangles meet that have lengths 2 m − n − m (cid:54) = n . Since these two R1-triangles meet at their respective R1-corners ona common tile, they are separated by the R1-line segment in this tile. We will argue thatthis line segment must extend indefinitely in both directions. For the sake of contradiction,suppose the extended R1-line segment has an R1-corner, which must occur at length 2 m + j along the straight R1-line segment from the corner of the 2 m − j ∈ { , , . . . } .At any such R1-corner, there is either an R2-cycle or an R2-anticycle of length (2 k −
1) for k ∈ { , . . . , min { m, n }} . Since Lemma 3.1 and Lemma 3.3 imply that R2-cycles and R2-anticycles cannot exist in T , the R1-line segment must extend infinitely in both directions.Thus, T ∈ C so that C ⊆ C ∪ C , as required. (cid:3) The continuous hull of our aperiodic monotile
We conclude the paper with a brief discussion of the tiling space, or continuous hull, ofthe tilings in the class C . Recall that the continuous hull of a collection of tilings Λ is thecompletion of { T + R d : T ∈ Λ } in the tiling metric, typically denoted by Ω. Under mildassumptions, the continuous hull is a compact topological space endowed with a continuous R d action, making (Ω , R d ) a dynamical system. For further details see [11, Section 1.2]. Weare interested in the continuous hull of C = C ∪ C . Theorem 4.1.
All tilings in the continuous hull of C are nonperiodic and satisfy R1 . More-over, in any such tiling, there are at most three connected components of R2-trees, and eachcomponent crosses an infinite number of tiles.Proof. We begin by considering the class C . As described in Proposition 2.5, a tiling is in C ifit contains a bi-infinite R1-line with intertwined 2 n -periodic patterns of R1-triangles of length2 n − C . Note that on each side of the bi-infinite R1-line, tilings in C look locally like DENDRITIC MONOTILE 17 tilings in C , so the completion of C contains tilings in C . However, we will handle the tilingsin C later, so we ignore these elements of the completion for now. Figure 22.
A patch that extends to a typical tiling in C , note that R1-trianglesthat meet at a tile on the bi-infinite R1-line need not have the same length.We now consider tilings in the completion of C that contain a bi-infinite R1-line. Since wecan have arbitrarily large R1-triangles on either side of the bi-infinite R1-line, the completionof C contains tilings with an infinite R1-triangle whose corner meets the R1-line. Suchan infinite R1-triangle meeting a bi-infinite R1-line forces a half plane of R1-triangles withexactly two connected R2-components as shown in the proof of Lemma 2.4, and both of thesecomponents cross an infinite number of tiles. Since the behaviour of R1-triangles above andbelow the bi-infinite R1-line are independent, there are tilings with infinite R1-triangles onone or both sides of the bi-infinite R1-line in the completion. Thus, tilings in the completionof C that contain a bi-infinite R1-line have at most three connected R2-components.We now consider the continuous hull of C . As described in Proposition 2.3, such tilingshave successively larger hexagonal grids of interlaced R1-triangles. For any n ∈ N , a tilingof interlaced R1-triangles decomposes into patches P n connected by single tiles, where P n isdefined in Proposition 2.3 (also see Figure 14). The image on the left-hand side of Figure 23shows all possible arrangements of P connected by a single tile. Up to direct isometry, thereare exactly two configurations. These are shown for P in Figure 24. For n ∈ N , let us callthese two patches R n and S n . Placing the origin at the centre of each patch, as shown inFigure 24, we see that R n ⊂ R n +1 and S n ⊂ S n +1 . For example, the reader can compare thepatches in Figure 23, where n = 2, with the patches in Figure 24, where n = 3. Therefore, R := ∞ (cid:91) n =0 R n and S := ∞ (cid:91) n =0 S n are tilings satisfying R1 . Thus, in addition to tilings in C , the continuous hull of C containsdirect isometries of the tilings R and S . As n tends to infinity, the patches P n converge to apartial tiling, which can be scaled down at each step so as to be depicted as the fractal on theright-hand side of Figure 23. The fractal nature of this partial tiling implies that, up to directisometry, R and S are the only two additional elements in the completion of C . Notice thatthe tilings R and S have exactly three connected components of R2-trees and each componentcrosses an infinite number of tiles, as desired. (cid:3) Figure 23.
All possible patches of P n connected by a single tile in C arerepresented on the left, and a representation of the partial tiling P with infinitelysmall tiles appears on the right • • Figure 24.
Cauchy sequences of the above configurations of patches P n , withthe central dots on the origin, converge to tilings R and S in the continuoushull of C . Each of these tilings has three connected components of R2-treesThe astute reader will notice that the hull of C contains all possible tilings satisfying R1 with the condition that either C or C holds. That is, the dendrite rule R2 ensures that thereare no periodic tilings, but does not otherwise factor into the final description of the hull. DENDRITIC MONOTILE 19
Finally, we comment on the differences between tilings coming from our rules and Socolar-Taylor tilings. In the Socolar-taylor tilings, their rule R2 is designed to ensure that all theR1-patterns in their tilings fall into the class C . Since their rules are local matching rules,their hull is automatically complete. On the other hand, none of the R1-patterns in C \ C are possible in the Socolar-Taylor tilings, and hence we have two genuinely different classes oftilings that are not MLD to one another. References [1] S. Akiyama and J-Y Lee,
The computation of overlap coincidence in Taylor–Socolar substitution tiling ,Osaka J. Math. (2014), 597–609.[2] M. Baake, F. G¨ahler and U. Grimm, Hexagonal inflation tilings and planar monotiles , Symmetry (2012),581–602.[3] M. Baake and U. Grimm, Aperiodic Order. Volume 1: A Mathematical Invitation , Cambridge UniversityPress, Cambridge, 2013.[4] R. Berger,
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E-mail address : [email protected] Michael F. Whittaker, School of Mathematics and Statistics, University of Glasgow,University Place, Glasgow Q12 8QQ, United Kingdom
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