# Taylor–Socolar Hexagonal Tilings as Model Sets

^{1}

^{2}

^{*}

## Abstract

**:**

**MSC**52C23

## 1. Introduction

**parity**(and eventually distinguish the two sides as being sides 0 and 1), and in terms of parity the tilings are aperiodic. In fact the parity patterns of tiles created in this way are fascinating in their apparent complexity, see Figure 1 and Figure 2.

**Figure 1.**A section of a Taylor–Socolar tiling showing the complex patterning arising from the two sides of the hexagonal tile, here indicated in white and gray. Notice that there are islands (Taylor and Socolar call the llamas) both of white and gray tiles.

**Figure 2.**The figure shows a pattern of triangles emerging from the construction indicated in Section 2, manifesting the rule

**R1**. The underlying hexagonal tiling is indicated in light and dark shades, which indicate the parity of the hexagons. The underlying diagonal shading on the hexagons manifests the rules

**R2**.

**Figure 3.**The two basic hexagonal tiles. One is a white tile and the other a light gray. These are colored with red and blue diameters. The rotational position of the tiles is immaterial. Note how the tiles are identical as far as the red diagonal and blue diagonal are concerned. The distinction is in which color of the red-blue diagonal cuts the black stripe.

**R1**,

**R2**and can also be constructed by substitution (the scaling factor being 2). In this paper it is the matching rules that are of importance.

**R1**- the black lines must join continuously when tiles abut;
**R2**- the ends of the diameters of two hexagonal tiles that are separated by an edge of another tile must be of opposite colors, Figure 4.

**Figure 4.**Rule

**R2**: Two hexagon tiles separated by the edge of another hexagon tile. Note that the diameter colors of the two hexagons are opposite at the two ends of the separating edge. It makes no difference whether or not the diameters color-split—the diameters must have different colors where they abut the separating edge.

**R1**,

**R2**? There are two aspects to this. The triangulations themselves are based on hexagon centers, whereas in an actual tiling the triangles are shrunken away from vertices. This shrinking moves the triangle edges and is responsible for the off-centeredness of the black stripe on each hexagon tile. How is this shrinking (or edge shifting, as we call it) carried out? The second feature is the coloring of the diagonals of the hexagons. What freedom for coloring exists, given that the coloring rule

**R2**must hold?

**CHT**) (see Figure 5) and the infinite concurrent w-line tilings (

**iCw-L**) (see Figure 6). In both cases there is 3-fold rotational symmetry of the triangulation and in both cases the mapping ξ is many-to-one. These two types of tilings play a significant role in [2].

**Figure 5.**The (central part of a) central hexagon (

**CHT**) tiling. Full (edge-shifted) triangles of levels $0,1,2$ are shown. At the outside edges one can see the beginnings of triangles of level 3. The rays from the central hexagon in the six a-directions will have infinite a-lines in them. However the edge shifting rules cannot be applied to them because they are of infinite level—they are not composed of edges of finite triangles. In the end a full tiling is obtained by placing a fully decorated tile into the empty central hexagon. There are 12 ways to do this, and each way then determines the rest of the tiling completely. These tilings violate both forms of generic condition.

**Figure 6.**The

**iCw-L**tilings. The triangulation is generic-a but not generic-w. Of course the partial tiling shown is perfectly consistent with generic tilings—in fact all Taylor–Socolar tilings contain this type of patch of tiles. However, if the pattern established in the picture is maintained at all scales, then indeed the result is a not a generic tiling since it fails generic-w.

**iCw-L**triangulations, and although they perfectly obey the matching rules they are not in the same LI class as all the other tilings. On the other hand the

**CHT**tilings (those lying over the

**CHT**triangulations) are in the main LI class and, because of the particular simplicity of the unique one whose center is $(0,0)$, the question of describing the parity (which tiles are facing up and which are facing down) becomes particularly easy. Here we reproduce the parity formula for this

**CHT**tiling as given in [2] (with some minor modifications in notation). We use this to give parity formulas for all the tilings of ${X}_{Q}$.

## 2. The Triangulation

**Figure 7.**The figure shows the standard triangular lattice Q (the red points) and the larger lattice P (red and blue points) in which Q lies with index 3. The points of Q may be viewed as the vertices of a triangularization of the plane by equilateral triangles of side length 1. The blue points are the centres of these triangles. The color here has nothing to do with the coloring of the diagonals of the tiles—it only distinguishes the two cosets.

**Figure 8.**The generators ${a}_{1},{a}_{2}$ of Q and the generators ${w}_{1},{w}_{2}$ of P, showing how the cosets of Q in P split into the points of Q and the centroids (centers) of the up and down triangles. Around the point $0\in Q$ we see the hexagonal tile centered on 0 with vertices in $P\backslash Q$.

**Remark**

**2.1**

**Figure 9.**Four superimposed triangles, each indicated by its circled bottom lefthand vertex. The vertices of each triangle generate a different coset of Q modulo $2Q$.

**Figure 10.**The figure shows how the centroids (indicated with solid blue dots) of the new side-length-2 triangles (indicated with solid red dots) are obtained as vectors from ${q}_{1}+2P$. The two $2Q$-cosets of ${q}_{1}+2P$ which are not ${q}_{1}+2Q$ itself indicate the centroids of the new up and down triangles. Notice that the orientations of the new triangles, and hence the orientations associated with the new centroids, are opposite to the orientations associated with these points when they were viewed as centroids of the original triangulation. This explains why ${S}_{2}^{\uparrow}\subset {S}_{1}^{\downarrow}$ and ${S}_{2}^{\downarrow}\subset {S}_{1}^{\uparrow}$ .

- all points involved as vertices of triangles are in Q;
- all triangle centroids are in $P\backslash Q$;
- there is no translational symmetry.

**have an orientation**(up or down) if there is a positive integer k such that for all ${k}^{\prime}>k$, $x\notin {S}_{{k}^{\prime}}^{\uparrow}\cup {S}_{{k}^{\prime}}^{\downarrow}$. Every element of $P\backslash Q$ is in ${S}_{k}^{\uparrow}$ or ${S}_{k}^{\downarrow}$ for $k=1$, and some for other values of k as well. For the elements x which have an orientation there is a largest such k for which this is true. We call this k the

**level of its orientation**. If there is no such k we shall say that x is

**not oriented**. We shall see below (Proposition 3.3) what it means for a point not to have an orientation [14].

## 3. The Q-Adic Completion

**Remark**

**3.1**

**Lemma**

**3.2**

**Proof:**

**Proposition**

**3.3**

**Proof:**

**Remark**

**3.4**

**Proposition**

**3.5**

**Proposition**

**3.6**

## 4. The Tiles

**Figure 11.**A partial triangulation of the plane overlaid on the basic lattice of hexagons which will make up the tiles. The levels of the triangles are indicated by increasing thickness. One can clearly see triangles of levels $0,1,2,3,4$ and one can also see how triangle edges of level k ultimately become edges passing through the interior of triangles of level $k+1$. This will be used to make the shifting of edges later on.

**lines**(since they are in the directions ${a}_{1},{a}_{2},{a}_{1}+{a}_{2}$) and the other set of lines w-

**lines**(since they are in the directions ${w}_{1},{w}_{2},{w}_{2}-{w}_{1}$). We also call the w-lines

**coloring lines**, since they are the ones carrying the colors red and blue. The w-lines pass through the centroids of the triangles of the triangulation. We say that a w-line has

**level**k if there are centroids of level k triangles on it, but none of any higher level. We shall discuss the possibility of w-lines that do not have a level in this sense below. Note that every point of $P\backslash Q$ is the centroid of some triangle, some of several, or even many!

**level**k if it is a vertex of a triangle of edge length ${2}^{k}$ but is not a vertex of any longer edge length. Similarly an edge of a triangle is of

**level**k if it is of length ${2}^{k}$, and an a-line (made up of edges) is of

**level**k if the longest edges making it up are of length ${2}^{k}$. All lines of all levels are made from the original set of lines arising from the original triangulation by triangles of edge length 1, so a line of level k has edges of lengths $1,2,\cdots ,{2}^{k}$ on it.

**Table 1.**Uses of the word “level” k and section number where it is defined. If there is no such k the level is infinite.

of a triangle | Section 2 k if the side length is ${2}^{k}$, |

where a side length $1={2}^{0}$ is the length of ${a}_{1}$ and ${a}_{2}$ | |

of orientation of $x\in P$ | Section 2 k at which x stops switching between ${S}_{k}^{\uparrow}$ and ${S}_{k}^{\downarrow}$ |

of a w-line | Section 4 max. k of centroids of level k triangles on it |

of a point of Q | Section 4 max. k for which it is a vertex of a triangle of level k |

of a triangle edge | Section 4 k for which it is an edge of a level k triangle |

of an a-line | Section 4 max. k for k-edges on this line |

**Definition**

**4.1**

**generic**-w. This means that for every w-line there is a finite bound on the levels of the centroids (points of $P\backslash Q$) that lie on that line. In this case for any ball of any radius anywhere in the plane, there is a level beyond which no w-lines of higher level cut through that ball. See Figure 6 for an example that shows failure of the generic-w condition.

**generic**-a if every a-line has a finite level. This means for every a-line there is a finite bound on the levels of edges that lie in that line. In this case for any ball of any radius anywhere in the plane, there is a level beyond which no lines of the triangulation of higher level cut through that ball. See Figure 12 and Figure 6.

**generic**if it is both generic-w and generic-a. All other tilings (or elements $\mathbf{q}\in Q$) are called

**singular**. One case of the failure of generic-w is discussed in Proposition 3.3 above. The only way for one of our generic conditions to fail is that there are a-lines or w-lines of infinite level. This situation is discussed in Section 6.

**short diameters**. These short diameters arise out of the edges of the triangles of the triangulations that we have created. Each triangle edge is part of a line which is a union of edges, all of the same level. As we have pointed out, the line (and its edges) have level k if they occur at level k (and no higher). The original triangulation has level 0. One should note that a line may occur as part of the edges of many levels of triangles, but under the assumption of generic-a there will be a highest level of triangles utilizing a given line, and it is this highest level that gives the line its level and determines the corresponding edges.

**Figure 12.**This shows a sketch of how a tiling with an infinite a-line (the horizontal line) can be constructed so that it is generic-w. Here $\mathbf{q}=z{a}_{1}$, where z is the 2-adic integer $(1,1,5,5,21,21,85,85,\dots )$ (the $\phantom{\rule{3.33333pt}{0ex}}mod\phantom{\rule{0.277778em}{0ex}}2,4,8,16,32,64,128,256,\dots $ values). Some triangles of edge lengths $2,8,32$ are shown. There are triangles of arbitrary large side lengths on the horizontal line, but the triangulation does not admit a second infinite a-line and cannot admit an infinite w-line since $\mathbf{q}$ is of the wrong form.

**Lemma**

**4.2**

**inner**side of the edge, and the other side its

**outer**side. This edge (but not the entire line) is shifted inwards (i.e., towards the centroid of T) by the distance ϵ. Note that the shifting distance ϵ is independent of k. This shifted edge then becomes the black stripe on the hexagonal tiles through which this edge cuts, see Figure 13. Figure 14 shows how edge shifting works. At the end of shifting, each hexagon has on it a pattern made by the shifted triangle edges that looks like the one shown in Figure 13.

**Figure 14.**The figure shows how edge shifting is done. Part of a triangulation is shown in thin black lines. The shifted edges are shown in thicker gray lines. The extended black lines indicate that the largest (level 2) triangle sits in the top right corner of a level 3 triangle, which is not shown in full. Note how the edges of the level 2 triangle shift.

**Proposition**

**4.3**

## 5. Color

**R2**forces the next part of the coloring to be blue and we come to the hexagon center e. This has three edges through it, but the one that our w-line crosses at right-angles has the highest level, and so will produce the stripe for the corresponding hexagon. The color must switch at the stripe, and so we see the next red segment as we come to d.

**R2**. For a full example where one can see the translational symmetry take over, the reader can fill in the coloring on the gray line through y.

**Figure 15.**The figure shows some hexagonal tiles, each centered on a point of Q. The point q is assumed to be in the coset ${q}_{1}+2Q$ and the gray hexagons are those in the picture whose centers are in this coset. The white hexagons are centered at points from all three of the remaining cosets of Q relative to $2Q$. These are the midpoints of the edges of the level 1 triangles. Notice in each, the red and blue diagonals clockwise and counterclockwise of the direction of the black stripes. At the bottom we see the three vectors ${a}_{1},{a}_{2},{a}_{1}+{a}_{2}$. The centers of the white hexagons are, reading left to right and bottom to top, ${q}_{1}+{a}_{1},{q}_{1}+3{a}_{1};{q}_{1}+{a}_{1}+{a}_{2},{q}_{1}+2{a}_{1}+{a}_{2},{q}_{1}+3{a}_{1}+{a}_{2};{q}_{1}+3{a}_{1}+2{a}_{2}$. The picture manifests the rule

**R2**and shows that elements of the same coset carry the same orientation of diameters. Note that from the rotational symmetry of the process and the fact that the hexagons centered on ${q}_{1}+{a}_{1}$ and ${q}_{1}+3{a}_{1}$ have identically aligned diagonals, we can infer that this property is retained across each of the cosets ${q}_{1}+{a}_{1}+2Q,{q}_{1}+{a}_{2}+2Q,{q}_{1}+{a}_{1}+{a}_{2}+2Q$.

**Figure 16.**Coloring of lines. Colors are forced on w-lines as they pass through the midpoint of a triangle directed towards the centroid of one of its corner triangles.

**Figure 17.**The figure shows how the centroid of the triangle T, which is in the top corner of the main triangle, is on two coloring lines. The third coloring line through p is the dotted line through v. This passes through the centroid ${p}^{\prime}$ of ${T}^{\prime}$.

**Figure 18.**This figure shows how the coloring appears if one determines the coloring by the information in increasing coset levels. This figure corresponds to the process at $k=2$. The triangle vertices and their corresponding hexagons are indicated at levels $0,1,2$ and the corresponding partial coloring is noted.

**Proposition**

**5.1**

**Proposition**

**5.2**

**Proof:**

**Figure 19.**In the figure the small circle indicates $(0,0)$. A vertical color line is shown, which meets the ${a}_{3}$ axis in the point $p=(4,-4,0)$. The points on this axis are all of the form $(u,-u,0)$, $u\in \mathbb{Z}$. In the

**CHT**tiling set-up, $D\left(u\right)$ indicates that the point is a vertex of a level ${log}_{2}D\left(u\right)$ triangle, in this case $D\left(4\right)=4$, so we are on a level 2 triangle (${2}^{2}=4$). We already saw that the color starts with a full blue diagonal at $(4,-4,0)$. Moving down the line to the next point decreases ${x}_{1}$ by 2 and increases ${x}_{2},{x}_{3}$ by 1. We note that ${x}_{3}-{x}_{2}$ remains constant, and $D({x}_{3}-{x}_{2})=4$. At the second step we cross, at right-angles into another triangle of level 2, and the color proceeds without interruption. At the 4th step we are at a vertex that is the midpoint of the side of a triangle of level 3 and looking up our vertical line we can see that it is passing into a corner triangle of level 4—namely where we just came from, and we see a forced full red diagonal. In the first three steps the diameters are all red-blue (top to bottom), whereas after the red-red diagonal the next three steps are blue-red diameters, so there is a switch that affects parity. We can ignore the points with full diameters (they get sorted out in a different w-direction). We note that $\lfloor \frac{{x}_{3}}{D({x}_{3}-{x}_{2})}\rfloor $ maintains the value 0 on steps $1,2,3$ and maintains the value 1 on steps $5,6,7$, showing that the formula notices the change of diameters correctly. If we continue $\lfloor \frac{{x}_{3}}{D({x}_{3}-{x}_{2})}\rfloor =2$ on the next three step sequence, but modulo 2 this is the same as 0.

#### 5.1. Completeness

**R1**,

**R2**. Does this procedure produce all possible tilings satisfying these rules? The answer is yes, and this is already implicit in [2]. We refer the reader to the paper for details, but the point is that in creating a tiling following the rules a triangle pattern emerges from the stripes of the hexagons. This triangularization can be viewed as the edge-shifting of a triangulation $\mathcal{T}$ conforming to our edge shifting rule. Thus we know that working with all triangulations, as we do, we are bound to be able to produce the same shrunken triangle pattern as appears in T.

**R2**. When we discuss the non-generic cases in Section 6.2, we shall see that for non-generic triangulations there are actually choices for the colorings of some lines, but these choices exhaust the possibilities allowed by the rule

**R2**. Thus the tiling T must be among those that we construct from $\mathcal{T}$ and so we see that our procedure does create all possible tilings conforming to the matching rules.

## 6. The Hull

#### 6.1. Introducing the Hull

**hulls**of the Taylor–Socolar tiling system. We give ${X}_{Q}$ and X the usual local topologies—two tilings are close if they agree on a large ball around the origin allowing small shifts. In the case of ${X}_{Q}$ one can do away with “the small shifts” part. See [15] for the topology.

**Proposition**

**6.1**

**Proof:**

**Corollary**

**6.2**

#### 6.2. Exceptional Cases

#### 6.2.1. Violation of Generic-a

**Proposition**

**6.3**

**Proof:**

**Proposition**

**6.4**

**Proof:**

**CHT**tiling) in [2] (see Figure 5). We also refer to them as

**iCa-L**tilings. Edge shifting is not defined along these three lines, and we shall see that we have the freedom to shift them arbitrarily to produce legal tilings. The tilings in which there is one infinite a-line are designated as

**ia-L**tilings.

#### 6.2.2. Violation of Generic-w

**Proposition**

**6.5**

**Proof:**

**Proposition**

**6.6**

**Proof:**

**CHT**tiling.

**CHT**tilings.

Type | Single | Three concurrent |
---|---|---|

infinite a-line | $\mathbf{q}\in Q+\overline{{\mathbb{Z}}_{2}}a$ | $\mathbf{q}\in Q$ CHT |

infinite w-line | $\mathbf{q}\in Q+\overline{{\mathbb{Z}}_{2}}w$ | $\mathbf{q}\in Q$ CHT |

$\mathbf{q}\in -{\mathbf{s}}_{\mathbf{2}}-2{\mathbf{s}}_{\mathbf{1}}+Q$ or $\mathbf{q}\in -{\mathbf{s}}_{\mathbf{1}}-2{\mathbf{s}}_{\mathbf{2}}+Q$ |

**Lemma**

**6.7**

**Lemma**

**6.8**

**CHT**tiling.

**Proof:**

#### 6.3. Coloring for the **iCw-L** Tilings

**iCw-L**tilings (infinite concurrent w-line tilings). We also refer to the underlying triangulations with the same terminology. See Figure 6.

**iCw-L**tiling can be made in an arbitrary way without violating the tiling conditions

**R1**,

**R2**[2]. Of the 8 possible colorings the two truly symmetric ones (the ones that give an overall 3-fold rotational symmetry—including color symmetry—to the actual tiling) are exceptional in the sense that no other tilings in the Taylor–Socolar system have a point $p\in P\backslash Q$ (i.e., a tile vertex) with the property that the three hexagon diagonals emanating from it are all of the same color (see Proposition 5.2). These exceptional symmetric

**iCw-L**tilings are called

**SiCw-L**tilings. In [2] these tilings are described as having a “defect” at this point, and indeed they are not LI to any other tilings except other

**SiCw-L**tilings.

**iCw-L**triangulation. In the other 6 colorings there are at each hexagon vertex two diameters of the same color and one of the opposite color, and we shall soon prove that they all occur in ${X}_{Q}$.

**iw-L**tilings.

#### 6.4. The Structure of the Hull

**Theorem**

**6.9**

**SiCw-L**tilings with three blue-red (blue first) diameters emanating from some hexagon vertex q, which form a single Q-orbit in ${X}_{Q}$, and ${X}_{Q}^{r}$ is the companion orbit with red-blue diameters. Both of these orbits are dense in ${X}_{Q}$.

**iCw-L**tilings that are not color symmetric. Restricted to the minimal hull ${X}_{Q}^{\u2020}$, the mapping ξ defined in (3) is:

- (i)
- $1:1$ on ${X}_{Q}^{gen}$;
- (ii)
- $6:1$ at
**iCw-L**points except**SiCw-L**points; - (iii)
- $12:1$ at
**CHT**points; - (iv)
- $2:1$ at all other non-generic points.

**Remark**

**6.10**

**CHT**tilings (or

**iCa-L**tilings) described above. Both ${X}_{Q}^{gen}$ and ${X}_{Q}^{\u2020}$ are of full measure in ${X}_{Q}$.

**Proof:**

**iCw-L**cases, where $x:=\mathbf{q}+{w}_{2}+{\mathbf{s}}_{\mathbf{1}}+2{\mathbf{s}}_{\mathbf{2}}\in {w}_{2}+Q$ or $x:=\mathbf{q}+{w}_{1}+{\mathbf{s}}_{\mathbf{2}}+2{\mathbf{s}}_{\mathbf{1}}\in {w}_{1}+Q$. In these cases x is a non-orientable point and there exists a nested sequence of triangles of all levels centered on x. This sequence begins either with an up triangle of level 0 or a down triangle of level 0. In either case everything about the triangulation is known and the entire tiling is determined except for the coloring of the three w-lines through x. In fact all of the 8 potential colorings of these three lines are realizable as tilings, as we shall soon see.

**iCw-L**triangulations we have the

**SiCw-L**tilings in which the colors of the diagonals of the three hexagons of which x is a vertex start off the same—all red or all blue. This arrangement at a hexagon vertex never arises in a generic tilings, and it is for this reason that these tilings produce different LI classes than the one that the generic tilings lie in: one “red” LI class and one “blue” LI class. As pointed out in [2] these

**SiCw-L**tilings have the amazing property that they are completely determined once the three hexagons around x have been decided (It is also pointed out in [3] that the

**SiCw-L**tilings do not arise in the substitution tiling process originally put forward in Taylor’s paper. However, they do arise as legal tilings from the matching rule perspective, though they could be trivially removed by adding in a third rule to forbid them. A similar situation has been shown to occur with the Robinson tilings for which there is a matching rule and also a substitution scheme that result in a hull and its minimal component [5]. As pointed out in [2], this is different from tilings like the Penrose rhombic tiling where the matching rules determine the minimal hull). The form of the points x with no orientation shows that there are just two Q orbits of them, one for each of the two non-trivial cosets of Q in P.

**iCw-L**tiling. Since each element of the sequence has a unique coloring and coloring in generic tiles is locally determined by local conditions, there must be a subsequence of these tilings that converges to one of some particular coloring. This must produce a coloring of diameters with two diameters of one color and one of the other color since we are using only generic tilings in the sequence. Now the rotational three-fold symmetry and the color symmetry of ${X}_{Q}$ shows that all 6 possibilities for the coloring will exist. This also shows that all these tilings are in the orbit closure of ${X}_{Q}^{gen}$.

**CHT/iCa-L**triangulations have the form $\mathcal{T}\left(q\right)$ where $q\in Q$. They have three concurrent a-lines and three concurrent w-lines at q and leave the central tile completely undetermined. This tile can be placed in any way we wish, and this fixes the entire tiling. There are a total of 12 ways to place this missing tile (6 for each parity), whence ξ is $12:1$ over q.

**iCw-L**and

**CHT/iCa-L**tilings, the remaining singular values of

**q**correspond to the

**ia-L**and

**iw-L**triangulations where there is either a single infinite level a-line or a single infinite level w-line, Section 6.3. Fortunately these two things cannot happen at the same time, see Lemma 6.8. That means that there is only one line open to question and there is only one line on which either the shift or color is not determined. In the

**ia-L**case there is an a-line for which edge shifting is un-defined and we wish to show that all the two potentially available shifts lead to valid tilings. Likewise in the

**iw-L**case there is a w-line to which no color can be assigned, and we wish to prove that both coloring options are viable.

**CHT**tiling Λ centered at 0. If one forms a sequence $\{{q}_{1}+\cdots +{q}_{k}+\Lambda \}$ of

**CHT**tilings and if $\{{q}_{1}+\cdots +{q}_{k}\}$ converges to a point of on the line $\overline{{\mathbb{Z}}_{2}}{a}_{1}$ that is not in Q then the point of

**CHT**concurrence has vanished and one is left only with the x-axis as an single infinite level a-line, and it will have the shifting induced by the original shifting along the x-axis in Λ (which can be either of the two potential possibilities). Of course one can do this in any of the a directions. A similar type of procedure works to produce all of the

**iw-L**tilings. This concludes the proof of the theorem. ☐

## 7. Tilings as Model Sets

#### 7.1. The Cut and Project Scheme

**cut and project scheme**.

**model set**defined by the

**window**W. Most often we wish to have the additional condition that the boundary $\partial W:=\overline{W}\backslash {W}^{\circ}$ of W has Haar measure 0 in $\overline{P}$. In this case we call $\u22cf\left(W\right)$ a

**regular model set**.

#### 7.2. Parity in Terms of Model Sets: The Generic Case

**Theorem**

**7.1**

**Proof:**

**Figure 20.**The tiles associated with points of ${V}_{1},{V}_{2},{V}_{3}$ are indicated by increasingly dark shades.

**Figure 21.**Ω and a line through a vertex x in the direction w meeting the opposite edge at $3\phantom{\rule{0.166667em}{0ex}}{2}^{k-1}w$.

**Figure 22.**Showing ${x}_{0}$ as a midpoint of an edge f of a triangle of level ${2}^{k+1}$ and the direction of w from it.

**Figure 23.**Showing a level k edge inside a triangle T of level $k+1$ and its corresponding shift (dotted line). Note how edge shifting for edges of level k repeats modulo ${2}^{k+1}$.

#### 7.3. Parity in Terms of Model Sets: The Non-Generic Case

**iw-L, iCa-L**, etc.).

**SiCw-L**tilings. Let Λ be such a tiling, which we may assume to be associated with $\xi (\Lambda )=0\in Q$. Comparing the

**SiCw-L**tiling Λ with an

**iCw-L**tiling Γ for which $\xi (\Gamma )=0$, we notice that the only difference between Λ and Γ is on the lines through 0 in the w-directions where $w\in \Omega $. The total index is introduced in [19]. Notice that it is enough to compute that the total index of the set of all points off these w-lines is 1 (see cite[LM1]). Because the set of points off the lines of w-directions is the disjoint union of cosets ${V}_{k}$ (we have seen this earlier), we only need to show that the total index of ${\cup}_{1}^{\infty}{V}_{k}$ is 1, i.e.,

## 8. A Formula for the Parity

**CHT**tilings centered at $(0,0)$. These correspond to the triangulation for $\mathbf{q}=0$. The parity formula for a tile is based on the coordinates of the center of the hexagonal tile. Due to the non-uniqueness of the

**CHT**tilings along the 6-rays at angles $2\pi k/6$ emanating from the origin, the basic formula is valid only off these rays. Later we show how to adapt this formula to arbitrary $\mathbf{q}$.

**triple coordinates**.

#### 8.1. **CHT** Formula

**CHT**tiling [2]. The

**CHT**tiling has the advantage that all the shifting due to the choice of the triangulation is taken out of the way, and this makes it easier to see what is going on. Our notation and use of coordinates is different from that in [2], but the argument is essentially the same. Figure 25 shows how the

**CHT**triangulation looks around its center $(0,0)$. The formula for parity is made of two parts each of which corresponds to one the two features which combine to make parity: edge shifting and the color.

**CHT**triangulation, different from the ${a}_{1}$-axis. This line meets the ${a}_{3}$-axis at a point $(n{2}^{k},-n{2}^{k},0)$ for some non-negative integer k and some odd integer n. This point is the apex of a level k triangle and is the midpoint of an edge from a triangle of level $k+1$ (though the ${a}_{3}$-axis itself is of infinite level here). As such we see that the horizontal edge to the right from $(n{2}^{k},-n{2}^{k},0)$ is shifted downwards. As the edge passes into the next level $k+1$ triangle we see that the shift is upwards. This down-up pattern extends indefinitely both to the right and to the left. In Figure 26 $n=1$ and k is unspecified, but the underlying idea does not depend on the value of n. We now note that the points along the horizontal edge rightwards from $({2}^{k},-{2}^{k},0)$ are $({2}^{k},-{2}^{k}-1,1),({2}^{k},-{2}^{k}-2,2),\dots $, or $x=({2}^{k},-{2}^{k}-{x}_{3},{x}_{3})$ in general. Now we note that

**Figure 25.**Part of the triangulation corresponding to $\mathbf{q}$ centered at 0 (indicated by the dot). Triangles of scales $1,2,4$ are shown, as well as part of a triangle of scale 8.

**CHT**tiling as indicated in Figure 27. The figure indicates how the color must be on the ${a}_{3}$-axis as we proceed in the vertical direction.

**Figure 26.**The horizontal line through the point $({2}^{k},-{2}^{k},0)$ is seen as passing through midpoints of consecutive ${2}^{k+1}$ triangles. The corresponding edges along this line shift downwards and upwards alternately. The points $({2}^{k},-{2}^{k}-m,m)$ with $m\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}1,2,\dots ,{2}^{k}-1$ are on a shift-down edge. The next set for $m\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}{2}^{k}\phantom{\rule{3.33333pt}{0ex}}+\phantom{\rule{3.33333pt}{0ex}}1,\dots ,{2}^{k}\phantom{\rule{3.33333pt}{0ex}}+\phantom{\rule{3.33333pt}{0ex}}{2}^{k}\phantom{\rule{3.33333pt}{0ex}}-\phantom{\rule{3.33333pt}{0ex}}1$ are on a shift-up edge. The next set, $m\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}2.{2}^{k}+1,\dots 2.{2}^{k}+{2}^{k}-1$ are on a shift-down edge, etc. The formula $\lfloor \frac{{x}_{3}}{D({x}_{3}+{x}_{2})}\rfloor =\lfloor m/\left({2}^{k}\right)\rfloor $ accounts for this precisely, varying between 0 and 1 according to down and up.

**Theorem**

**8.1**

**CHT**tilings centered at $(0,0)$, the parity of a hexagonal tile centered on $x=({x}_{1},{x}_{2},{x}_{3})$ is

**Proof:**

**Remark**

**8.2**

**Remark**

**8.3**

**CHT**triangulation centered at $(0,0)$ the hexagon diameters along the three axes defined by ${a}_{1},{a}_{2},{a}_{3}$ have no shift forced upon them and can be shifted independently either way to get legal tilings. These are the hexagons centered on the points excluded by the condition ${x}_{j+1}+{x}_{j+2}\ne 0$. Similarly the three w-lines through the origin have no coloring pattern forced upon them and can independently take either. The centers of the hexagons that lie on these lines are excluded by the condition ${x}_{j+1}-{x}_{j+2}\ne 0$. In the

**CHT**tiling, the central tile can be taken to be either of the two hexagons and in any of its six orientations. Having chosen one of these 12 options for the central tile the rest of the missing information for tiles is automatically completed. The parity function $\mathbb{P}$ can be then extended to a function ${\mathbb{P}}^{e}$ so as to take the appropriate parity values on the 6 lines that we have just described.

**Figure 27.**Vertical color lines start with a full blue diameter or full red diameter as shown. The green triangle indicates why the second blue line segment up from the origin is in fact blue. The vertical line is centered at the mid-point of an edge of a level 4 triangle and passes into one of the level 2 corner triangles of this level 4 triangle. The discussion on color shows that it must be blue. All the other blue line segments are explained in the same way. We cannot assign color at the origin itself since the origin is not the mid-point of any edge.

#### 8.2. Parity for Other Tilings

**CHT**triangulation now centered at c, then the formulae above become

**CHT**triangulation, its limit $\mathcal{T}\left(\mathbf{q}\right)$ need not be. In fact we know that the translation orbit of any of the

**CHT**tilings centered at $(0,0)$ is dense in the minimal hull, and so we can compute a parity function ${\mathbb{P}}_{\mathbf{q}}$ of any tiling of the minimal hull in this way. In the case of generic $\mathbf{q}$ this results in a complete description of the parity of the tiling. In the case that there is convergence of either a-lines or w-lines (so one is not in a generic case) one can still start with one of the extension functions ${\mathbb{P}}^{e}$ and arrive at a complete parity description ${\mathbb{P}}_{\mathbf{q}}^{e}$ of any of the possible tilings associated to $\mathcal{T}\left(\mathbf{q}\right)$.

**Corollary**

**8.4**

## 9. The Hull of Parity Tilings

#### 9.1. Scaling

**Figure 28.**This shows how the scaled up hexagon inherits the colors of the smaller hexagons on which they are centered. The coloring is carried out by the same procedure that was used to color the original sized hexagons, though now the diameter sizes have doubled and only the lines of level at least 1 are used.

**CHT**tilings,

**iCw-L**tilings, and even

**SiCw-L**tilings all transform into tilings of the same type.

#### 9.2. From Parity to Taylor–Socolar Tilings

**Figure 29.**Patches of (parity) tiles in which the central tile has 5 of its surrounding tiles of the same color.

**The vertices of 1-level triangles**: We wish to show that around each hexagon centered on a point of ${q}_{1}+2Q$ there are at least two different pairs of tiles with mismatched colors amongst its six surrounding tiles (and hence the no five-hexagons rule is true). This is explained in the text below and in Figure 30.**Figure 30.**Examples are shown that demonstrate how we can see that around each hexagon centered on a level 1 vertex of ${q}_{1}+2Q$ there are at least two different pairs of tiles with mismatched colors amongst its six surrounding tiles. The two cases correspond to a level 1 vertex at the mid-point of a level 2 triangle and a level 1 vertex a non-midpoint of a higher level triangle. As explained in the text, one pair is found in a uniform way in both cases. The other pair is found by one method in the first case and another method in the other.There are two different situations for a hexagon H centered on the vertex v of a 1-level triangle. One is the case that v is at the midpoint of a 2-level triangle. This is indicated on the left side of Figure 30. The other is the case that v is on the edge of a n-level triangle where $n\ge 3$. This is indicated on the right side of Figure 30. In both cases, we note that the two tiles on opposite sides of H which share the long edge of a triangle of level $\ge 2$ have different colors. The reason is the following. Apart from the red-blue diameters, the long red and blue diameters and black stripes are same for the both the Taylor–Socolar tile and the reflected Taylor–Socolar tile. However the red and blue diameters of the middle tiles of 2-level triangles determine different red-blue diameters for the two tiles, and so they have different parity.Now we wish to find another pair of tiles with opposite colors for each of the cases. Let us look at the first case (see the left side of Figure 30). Consider the corner of the level 3 triangle T that is defined by H, and consider the two edges of T that bound this corner. The red and blue diameters of the tiles at the mid-points of these two edges of T determine different red-blue diameters for two neighbouring tiles in the surrounding tiles of H. This again results in different parities.Finally consider the second case (see the right side of Figure 30). Notice that the long black stripe of H is the part of the long edge of 3 or higher level triangle. If this long edge is from a level 4 or higher triangle, we consider just the part of it that is the level 3 triangle T whose edge coincides with the stripe of H. The red and blue diameters of the tile centered on the mid-point of this edge of T determine different red-blue diameters for two of the ring of tiles around H, as shown.**The vertices of level $\ge 2$ triangles**: Next we look at the six tiles surrounding the corner tiles of level $\ge 2$ triangles in a Taylor–Socolar tiling. Notice that the pattern of the colored diameters of six surrounding tiles is same for every corner tile of a level 2 or higher triangle, Figure 31. So what determines the basic patches of 7 tiles around the corner tiles of level 2 or higher triangles is the pattern of black stripes on it. Furthermore, there is a one-to-one correspondence between these basic patches of 7 tiles and the basic patches of 7 tiles of white and gray colors. This is shown in Figure 32. The key point is that this means that the basic patches of 7 tiles around the vertices of the level 2 or higher triangles already determine the coloring.**Seeing how the other cosets of Q mod $2Q$ violate the “two tiles of each color in the ring of any basic patch of 7 tiles” rule.**From a given parity tiling, there are four choices in determining the 1-level triangles. One of these is the coset ${q}_{1}+2Q$, and we know that the 7 tile patches around each of these points satisfy the “no five-hexagon” rule. However the three choices, corresponding to the other three cosets of $Q\phantom{\rule{3.33333pt}{0ex}}mod\phantom{\rule{0.277778em}{0ex}}2Q$ all have some 7 tile patches that violate the rule. We can see violations to the rule for each of the other three cosets in Figure 33, which is a small piece from the lower right corner of Figure 1. Since any parity tilings in the hull are repetitive, we observe the patches frequently over the parity tiling. Furthermore, since there is only one way that is allowed to determine 1-level triangles, it does not depend on where one starts to find the 1-level triangles. They will all match in the end.

**Figure 32.**The edge and color patterns and corresponding parity patterns that can occur in the hexagons surrounding the corners of triangles of level $\ge 2$.

**Figure 33.**Here we see three violations to the five-hexagons rule, the rings of hexagons being indicated by the circles. The shaded triangles show how the cosets determined by the center look, and show that the violating hexagons are from three different cosets. The hexagonal pattern comes from the lower right corner of Figure 1.

**Corollary**

**9.1**

## 10. Concluding Remarks

## Acknowledgments

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Lee, J.-Y.; Moody, R.V.
Taylor–Socolar Hexagonal Tilings as Model Sets. *Symmetry* **2013**, *5*, 1-46.
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Lee J-Y, Moody RV.
Taylor–Socolar Hexagonal Tilings as Model Sets. *Symmetry*. 2013; 5(1):1-46.
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Lee, Jeong-Yup, and Robert V. Moody.
2013. "Taylor–Socolar Hexagonal Tilings as Model Sets" *Symmetry* 5, no. 1: 1-46.
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