Taylor–Socolar Hexagonal Tilings As Model Sets
Abstract.
The Taylor–Socolar tilings [19, 20] are regular hexagonal tilings of the plane but are distinguished in being comprised of hexagons of two colors in an aperiodic way. We place the Taylor–Socolar tilings into an algebraic setting which allows one to see them directly as model sets and to understand the corresponding tiling hull along with its generic and singular parts.
Although the tilings were originally obtained by matching rules and by substitution, our approach sets the tilings into the framework of a cut and project scheme and studies how the tilings relate to the corresponding internal space. The centers of the entire set of tiles of one tiling form a lattice in the plane. If denotes the set of all Taylor–Socolar tilings with centers on then forms a natural hull under the standard local topology of hulls and is a dynamical system for the action of . The adic completion of is a natural factor of and the natural mapping is bijective except at a dense set of points of measure in . We show that consists of three LI classes under translation. Two of these LI classes are very small, namely countable orbits in . The other is a minimal dynamical system which maps surjectively to and which is variously , , and at the singular points.
We further develop the formula of [19] that determines the parity of the tiles of a tiling in terms of the coordinates of its tile centers. Finally we show that the hull of the parity tilings can be identified with the hull ; more precisely the two hulls are mutually locally derivable.
1. Introduction
This paper concerns the aperiodic hexagonal monotilings created by Joan Taylor. We learned about these tilings from the unpublished (but available online) paper of Joan Taylor [20], the extended paper of Socolar and Taylor [19], and a talk given by Uwe Grimm at the KIAS conference on aperiodic order in September, 2010 [4]. These tilings are in essence regular hexagonal tilings of the plane, but there are two forms of marking on the hexagonal tile (or if one prefers, the two sides of the tile are marked differently). We refer to this difference as parity (and eventually distinguish the two sides as being sides and ), 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 Fig. 1 and Fig. 8.
The two Taylor–Socolar tiles are shown in Fig. 2, the main features being the black lines, one of which is a stripe across the tile, and the three colored diameters, one of which is split in color^{1}^{1}1Note that the two tiles here are not mirror images of each other, unless one switches color during the reflection. In [19] there is an alternative description of the tiles in which the diagonals have flags at their ends, and in this formulation the two tiles are mirror images of each other.. The difference in the two tiles is only in which side of the colorsplit diameter the stripe crosses. In the figure the tiles are colored white and gray to distinguish them, but it is the crossingcolor of the black stripe that is the important distinguishing feature.
Taylor–Socolar tilings can be defined by following simple matching rules (R1, R2) and can also be constructed by substitution (the scaling factor being ). In this paper it is the matching rules that are of importance.

the black lines must join continuously when tiles abut;

the ends of the diameters of two hexagonal tiles that are separated by an edge of another tile must be of opposite colors, Fig. 3.
The paper [19] emphasizes the tilings from the point of view of matching rules, whereas [20] emphasizes substitution (and the halfhex approach). There is a slight mismatch between the two approaches, see [4], which we will discuss later.
If one looks at part of a tiling with the full markings of the tiles made visible, then one is immediately struck by how the black line markings of the tiles assemble to form nested equilateral triangles, see Fig. 8. Although these triangles are slightly shrunken (which ultimately is important), we see that basically the vertices of the triangles are tied to the centers of the hexagons, and the triangle sidelengths are in suitable units. This triangle pattern is highly reminiscent of the square patterns that underlie the famous Robinson tilings [17, 10] which also appear in sizes which scale up by factors of . These tilings are limitperiodic tilings and can be described by model sets whose internal spaces are adic spaces. The TaylorSocolar tilings are also limitperiodic and it seems natural to associate some sort of adic spaces with them and to give a modelset interpretation of the picture.
One purpose of this paper is to do this, and it has the natural consequence that the tilings are pure point diffractive. It is convenient to base the entire study on a fixed standard hexagonal tiling of the coordinate plane . The centers of the hexagonal tiles can then be interpreted as a lattice in the plane (with one center at ). The internal space of the cut and project scheme that we shall construct is based on a adic completion of the group consisting of all translation vectors between the centers of the hexagons. We shall show that there is a precise onetoone correspondence between triangulations and elements of . But the triangulation is not the whole story.
The set of all Taylor–Socolar tilings associated with a fixed standard hexagonal tiling of the plane form a tiling hull . This hull is a dynamical system (with group ) and carries the standard topology of tiling hulls. Each tiling has an associated triangulation, but the mapping so formed, while generically , is not globally . What lies behind this is the question of backing up from the triangulations to the actual tilings themselves. The question is how are the tile markings deduced from the triangulations so as to satisfy the rules 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 offcenteredness 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?
In this paper we explain this and give a complete description of the hull and the mapping , Theorem 6.9. There are numerous places at which is singular (not bijective); in fact the set of singular points in is dense. Two special classes of singular points are those corresponding to the central hexagon triangulations (CHT)(see Fig. 17) and the infinite concurrent line tilings (iCwL) (see Fig. 18). In both cases there is fold rotational symmetry of the triangulation and in both cases the mapping is manytoone. These two types of tilings play a significant role in [19].
The hull has a minimal invariant component of full measure and this is a single LI class. There are two additional orbits, whose origins are the iCwL 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 , 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 [19] (with some minor modifications in notation). We use this to give parity formulas for all the tilings of .
A couple of comments about earlier work on aperiodic hexagonal tilings are appropriate here. D. Frettlöh [8] discusses the halfhex tilings (created out of a simple substitution rule) and proves that natural point sets associated with these can be expressed as model sets. Halfhexes don’t play an explicit role in this paper, though the hull of the halfhex tilings is a natural factor of lying between and [8, 9, 11]. They were important to Taylor’s descriptions of her tilings and are implicitly embedded in them.
In [15], Roger Penrose gives a fine introduction to aperiodic tilings and then goes on to create a class of aperiodic hexagonal tilings, which he calls tilings in which there are three types of tiles that assemble by matching rules. The main tiles are hexagonal, with keyed edges. The other two are a linearlike tile with an arbitrarily small width ( tiles) which fit along the hexagon edges, and some very tiny tiles () which fit at the corners of the tiles.
Clearly his objective was to create a single tile that only tiles aperiodically, although that was not achieved in [15]. Subsequently, however, Penrose did find a solution to the problem that uses a single hexagonal tile with matching rules for the edges and corners, [16]. This has only recently become more widely known after Joan Taylor’s work started to circulate.
One can quibble about whether or not Taylor’s tiling stretches the concept of matching rules since the second rule relates nonadjacent tiles and also in her tiling there are two tiles, though (at least in the right markings) they are mirror images of each other. However, the tilings of Penrose and Taylor tilings are a fascinating pair. Extensive computational work of F. Gähler indicates that the two tilings are quite distinct from one another, though they both have as a factor and apparently both have the same dynamical zeta functions [3].
There is an algorithmic computation for determining that certain classes of substitution tilings have purepoint spectrum. It has been used to confirm that the Taylor–Socolar substitution tilings have pure point spectrum or, equivalently, are regular model sets [2].
2. The triangulation
In principle the tilings that we are interested in are not connected to the points of lattices and their cosets in , but are only point sets that arise in Euclidean space as the vertices and centers of tilings. However, our objective here to realize tiling vertices in an algebraic context and for that we need to fix an origin and a coordinate system so as to reduce the language to that of . Let be the triangular lattice in defined by
where and . Then where and is a lattice containing as sublattice of index , see Fig. 4. For future reference we note that and .
Joining the points of that lie at distance from one another creates a triangular tiling. Inside each of the unit triangles so formed there lies a point of , and indeed consists of three cosets: itself, the centroids of the “up” triangles (those with a vertex above a horizontal edge), and the “down” triangles (those with a vertex below a horizontal edge), see Fig. 5. What we aim to do is to create a hexagonal tiling of . When this tiling is complete, the points of will be the centers of the hexagonal tiles and the points of immediately surrounding the points of will make up the vertices of the tiles.^{2}^{2}2 Nearest neighbours in are distance apart and the short diameters of the hexagons are of length while the edges of the hexagons are of length . The main diagonals of the hexagons are of length in the directions of . One notes that each of these vectors of is also of length .
Each of the hexagonal tiles will be marked by colored diagonals and a black stripe, see Fig. 2. These markings divide the tiles into two basic types, and it is describing the pattern made from these two types in modelset theoretical terms that is a primary objective of this paper (see Fig. 8). The other objective is to describe the dynamical hull that encompasses all the tilings that belong to the Taylor–Socolar tiling family.
We let the coset of up (respectively down) points be denoted by and respectively:
Remark 2.1.
There are three cosets of in . In our construction of the triangle patterns we have taken the point of view that itself will be used for triangle vertices and the other two cosets for triangle centroids. However, we could use any of the three cosets as the triangle vertices and arrive at a similar situation. This amounts to a translation of the plane by or . We come back to this point in §9.
We now wish to retriangularize the plane still using points of as vertices, but this time making triangles of side length equal to using as vertices a coset of in . There are four cosets of in and they lead to four different ways to make the triangularization. Fig. 6 shows the four types of triangles of side length . The lattices generated by the points of any one of these triangles is a coset of and together they make up all four cosets of in .
Choose one of these cosets, call it , where , and thereby triangulate the plane with triangles of side length . The centroids of the new triangles are a subset of the original set of centroids and, in fact, together with the vertices they form the coset . This is explained in the Fig. 7, which also explains the important fact that the new centroids, namely those of the new edgelength triangles of , make up two cosets of in depending on the orientation of the new triangles, and these orientations are opposite to those that these points originally had. Thus we obtain (which is in !), (which is in ), and the coset decomposition
with and .
We now repeat this whole process. There are four cosets of in and we select one of them, say , with , and this gives us a new triangulation with triangles of side length . Their centroids in form cosets and , and we have the decomposition
Continuing this way we obtain with , and sets with and for all , and the partition
(1) 
We now carry out the entire construction based on an arbitrary infinite sequence
where for all . This results in a pattern of overlapping triangulations based on triangles of edge lengths (these are referred to as being triangles of levels ). In §4 we shall make our tiling out of this pattern. But certain features of the entire pattern are clear:

all points involved as vertices of triangles are in ;

all triangle centroids are in ;

there is no translational symmetry.
The last of these is due to the fact that there are triangles of all scales, and no translation can respect all of these scales simultaneously.
A point is said to have an orientation (up or down) if there is a positive integer such that for all , . Every element of is in or for , and some for other values of as well. For the elements which have an orientation there is a largest for which this is true and this gives its final orientation. If has an orientation, we shall say that the level of its orientation is if its orientation stabilizes at . If it does not stabilize we shall say that is not oriented. We shall see below (Prop. 3.3) what it means for a point not to have an orientation.^{3}^{3}3We shall introduce levels for a number of objects that appear in this paper: points, lines, edges, triangles.
3. The adic completion
In this section we create and study a completion of under the adic topology. The adic topology is the uniform topology based on the metric on defined by if and when are in different cosets of . This metric is translation invariant. is the completion of in this topology and is the closure of in , which is also the completion of in the adic topology.
may be viewed as the set of sequences
where for all and .
is a group under componentwise addition and is the subgroup of all such sequences with all components in . There is the obvious coset decomposition
so has index in . We note that and are compact topological groups.
We have via
We often identify as a subgroup of via the embedding .
Note that the construction of expanding triangles of §2 depends on the choice of the element , where . Then we can obtain the compatible sequence
and thus we can identify each possible construction with an element of . Let denote the pattern of triangles arising from .
Let denote the unique Haar measure on for which . The key feature of is that for all . We note that is countable and has measure , and that and .
Remark 3.1.
We should note a subtle point here. In one can divide by . In fact, for all , exists since , and
Thus we can find an element of corresponding to and similarly corresponding to . However, our view is that and is the adic completion of this, with each of the three cosets leading to a different coset of in . Thus but and we conclude that has torsion.
Two examples of this are important in what follows. Define and for . and similarly based on . Their limits are denoted by respectively. They lie in .
Lemma 3.2.
For all ,
Similarly for , interchanging the indices .
In particular and . Furthermore, .
Proof.
From the definitions, and . This gives the case of the Lemma. Now proceeding by induction,
as required. Similarly
Taking the limits and using the formula for multiplication by at the beginning of Remark 3.1, we find that and similarly with the indices interchanged. ∎
Consider what happens if there is a point which does not have orientation. This means that there is an infinite sequence with . Then from (2), for each . This means or .
Proposition 3.3.
has at most one point without orientation. A point without orientation can occur if and only if or . These two families are countable and disjoint.
Proof.
If does not have an orientation then either and , which gives one of the cases; or , which gives the other. Conversely, in either case we have points without orientation. Since in one case and in the other case , we see that the two families are disjoint. ∎
Remark 3.4.
We do not need to go into the exact description of the orientations of triangles, but confine ourselves to a few remarks here. For any fixed , define the sequence of sets and , , inductively by and
and similarly for . In other words we put together into all the points which are oriented upwards at step , and likewise all that are oriented downwards at step .
Since and have measure we see that the sets change by less and less as increases. Furthermore it is clear that for all .
Proposition 3.5.
For all the sets and are clopen and disjoint. They each have measure . ∎
For each we define , and similarly for .
Proposition 3.6.
where is disjoint from , and
is the union of an open set and The same goes for . In particular and are the closures of their interiors. Both and are sets of measure .∎
4. The Tiles
Let us assume that we have carried out a triangulation as described in §2. We now have an overlaid pattern of equilateral triangles of side lengths . Each of these triangles has vertices in and its centroid in . The points of the two cosets of different from (shown as blue points in Fig. 4) form the vertexes of a tiling of hexagons made from the triangulation, see Fig. 9. This tiling, with the tiles suitably marked, is the tiling that we wish to understand. Our objective is to give each hexagon of the tiling markings in the form of a black stripe and three colored diagonals as shown in Fig. 2.
Apart from the lines of the triangulation (which give rise to short diagonals of the hexagons of the tiling) we also have the lines on which the long diagonals of the hexagons lie and which carry the color. To distinguish these sets of lines we call the triangulation lines alines (since they are in the directions ) and the other set of lines wlines (since they are in the directions ). We also call the wlines coloring lines, since they are the ones carrying the colors red and blue. The wlines pass through the centroids of the triangles of the triangulation. We say that a wline has level if there are centroids of level triangles on it, but none of any higher level. We shall discuss the possibility of wlines that do not have a level in this sense below. Note that every point of is the centroid of some triangle, some of several, or even many!
There are two steps required to produce the markings on the tiles. One is to shift triangle edges off center so as to produce the appropriate stripes on the tiles. We refer to this step as edge shifting. The second is to appropriately color the main diagonals of each tile. This we refer to as coloring. The two steps can be made in either order. However, each of the two steps requires certain generic aspects of the triangulation to be respected in order to be carried out to completion. We first discuss the nature of these generic conditions and then finish this section by showing how edge shifting is carried out.
We need to understand the structure of the various lines (formed from the edges of the various sized triangles) that pass through each hexagon. Let us say that an element of is of level if it is a vertex of a triangle of edge length but is not a vertex of any longer edge length. Similarly an edge of a triangle is of level if it is of length , and an line (made up of edges) is of level if the longest edges making it up are of length . All lines of all levels are made from the original set of lines arising from the original triangulation by triangles of edge length , so a line of level has edges of lengths on it.
The word ‘level’ occurs in a variety of senses in the paper. These are summarized in Table 1.
of a triangle  §2 if the side length is , 

where a side length is the length of and  
of orientation of  §2 at which stops switching between and 
of a line  §4 max. of centroids of level triangles on it 
of a point of  §4 max. for which it is a vertex of a triangle of level 
of a triangle edge  §4 for which it is an edge of a level triangle 
of an line  §4 max. for edges on this line 
There are two types of generic assumptions that we need to consider.
Definition 4.1.
A triangulation (or the value of associated with it) in which every wline has a finite level is called generic. This means that for every wline there is a finite bound on the levels of the centroids (points of ) 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 wlines of higher level cut through that ball. See Fig. 18 for an example that shows failure of the genericw condition.
A triangulation (or the value of associated with it) is said to be generic if every aline has a finite level. This means for every aline 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 Fig. 16 and Fig. 18.
A tiling is said to be generic if it is both genericw and generica. All other tilings (or elements ) are called singular. One case of the failure of genericw is discussed in Prop. 3.3 above. The only way for one of our generic conditions to fail is that there are alines or wlines of infinite level. This situation is discussed in §6.
Every element of has a hexagon around it and three lines passing through it in the directions . These lines pass through pairs of opposite edges of the hexagon at rightangles to those edges. We shall call these lines 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 if they occur at level (and no higher). The original triangulation has level . One should note that a line may occur as part of the edges of many levels of triangles, but under the assumption of generica 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.
In looking at the construction of level triangles out of the original triangulation of level triangles, we note immediately that every point of has at least one line of level through it (though by the time the triangularization is complete this line may have risen to higher level), see Fig. 6. The vertices of the level triangles have three lines of level through them, and the rest (the midpoints of the sides of the level triangles) have just one of level and the other two of level . Thus at this stage of the construction each hexagon has either one short diameter from a level line or it has short diameters all of level .
This is the point to remember: at each stage of determining the higher level triangles, we find that the hexagon around each element of is of one of two kinds: it either has three short diameters of which two have equal level and the third a higher level, or three short diameters all of the same level . The latter only occurs when the element of is a vertex of a triangle of level . Since we are in the generica case, there is no element of which is a vertex of triangles of unbounded scales, and the second condition cannot hold indefinitely. Once an element of is not a vertex at some level then it never becomes a vertex at any other higher level (all vertices of triangles at each level are formed from vertices of triangles at the previous level).
We conclude ultimately that in the generica cases every hexagon has three short diameters of which two are of one level and one of a higher level. See Fig. 9.
Lemma 4.2.
For satisfying generica each hexagonal tile of has three short diameters of which exactly one has the largest level and the other two equal but lesser levels.
We now describe edge shifting. Fix any with . This is going to be the distance by which lines are shifted. It is fixed throughout, but it exact value plays no role in the discussion. Take a tiling based on .
Now consider any edge that has level but does not occur as part of an edge of higher level. This edge occurs as an edge inside some triangle of level , and this allows us to distinguish two sides of that edge. The side of the edge on which the centroid of lies is called the 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 ) by the distance . Note that the shifting distance is independent of . This shifted edge then becomes the black stripe on the hexagonal tiles through which this edge cuts, see Fig. 11. Fig. 10 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 Fig. 11.
In the case that satisfies generica, the edges of every line of the triangulation are of bounded length. Thus every edge undergoes a shift by the prescription above. Thus,
Proposition 4.3.
If satisfies the condition generica then there is a uniquely determined edge shifting on it. ∎
5. Color
So far we have constructed a triangulation from our choice of , and shown how edges can be shifted to produce the corresponding hexagonal tiling with the tiles suitably marked by black stripes. We wish now to show how the (long) diagonals of the hexagons are to be colored. This amounts to producing a color (red, blue, or redblue) for each of the long diagonals of each hexagon of the tiling. The only requirement is that the overall coloring obey the rule that is used to make Taylor–Socolar tilings.
As we mentioned above, coloring is made independently of shifting in the sense that the two processes can be done in either order. In fact, in this argument we shall suppose that the stripes have not been shifted, so they still run through the centroids of the tiles.
We shall show that for in the generic case there is exactly one allowable coloring.
Assume that we have a generic tiling (this means both and generic). Now consider any hexagon of the tiling. We note from Lemma 4.2 that it has three short diameters, one of which is uniquely of highest level, and it is this last short diameter which determines (after shifting) the black stripe for this hexagon. We will refer to this short diameter as the stripe, even though in this discussion it has not been shifted. The other two colored (long) diameters are a red one which lies at clockwise of the stripe and a blue one which lies counterclockwise of the stripe. The redblue diameter cuts the stripe at rightangles, but which way around it is (redblue or bluered) is not determined yet.
Consider Fig. 12 in which we see two complete level triangles overlaid on the basic level triangles. Tiles of the hexagonal tiling are shown on points of with the hexagons at the vertices of the level triangles shown in green. These latter are points of . At each point of there are three edge lines running through it. But notice that at the midpoints of the sides of the level triangles (white hexagons), the edge belonging to the level triangle has higher level than the other two. This is the edge that will become the stripe for the hexagon at that point. This stripe forces the red and blue diameters for this hexagon.
The idea behind coloring is based on extrapolating this argument to wlines passing through midpoints of higher level triangles. Consider Fig. 13. The point is the midpoint of an edge of a triangle of level . Drawing the line towards the centroid of the top left corner triangle of level we see first of all that the edge of the level triangle through is the highest level edge through and hence the coloring along the line starts off red, as shown. Now the rule R2 forces the next part of the coloring to be blue and we come to the hexagon center . This has three edges through it, but the one that our line crosses at rightangles 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 .
And so it goes, until we reach the point . Here meets the midpoint of the edge of another level triangle. This edge produces the stripe for the hexagon at , but it is not at rightangles to , so there is no color change on at . Since is the midpoint of this level triangle, the same argument that we used at shows that the coloring should start off blue, as indeed we have seen it does. At this point one can see by the glide reflection symmetry along that the entire line will ultimately be colored so as to fully respect the rule 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 .
One can see a similar line coloring of the line passing through and . This time the point is the midpoint of an edge of a level triangle and is the centroid of one of the level corner triangles of this level triangle. The pair produces another example, with this time the first color out of being blue.
Finally, we show part of a potential line coloring starting at towards . We say ‘potential’ because from the figure we do not know how the level triangles lie. If is a midpoint of an edge of a level triangle, then the indicated line is colored as shown. If is not a midpoint then this line is not yet colorable.
We can thus continue in this way indefinitely. The important question is, does every tile get fully colored in the process? Using condition genericw, the answer is yes. To see this note that each element of of has three coloring lines through it. It will suffice to prove that the process described above will color these three coloring lines.
Now assuming the condition genericw we know that has an orientation. This means that it is the centroid of some triangle of level in the triangulation, and it is not the centroid of any higher level triangle. The triangle then sits as one of the corner triangles in a triangle of level . Up to orientation, the situation is that shown in Fig. 14. The colors of the two hexagons shown are then determined because the edges of produces stripes on them. Thus the two corresponding coloring lines that pass through are indeed colored. Thus the colorings of these two coloring lines through , the ones that pass through the midpoints of two sides of are forced.
What about the third line through (shown as the dotted line in Fig. 14)? We wish to see this as a wline through a midpoint of an edge, just as we saw the other two lines. We look at the centroid of , since the coloring line , which is through and , is the same as the line through and and the centroid is of higher level than and also has an orientation. We can repeat the process we just went through with with instead, to get a new triangle of which is the centroid, and a triangle in which sits as one of its corners ( is the centroid of but it may be the centroid of higher level triangles as well).
If this still fails to pick up the line then it must be that still passes through a vertex of (as opposed to through the midpoint of one of its edges) and the line passes through the centroid of . However, is of higher level still than that of . The upshot of this is that if we never reach a forced coloring of (so that it remains forever uncolored in our coloring process) then we have on the line centroids of triangles of unbounded levels. This violates condition genericw. Thus in the generic situation the coloring does reach every coloring line and the coloring is complete in the limit.
This completes the argument that there is one and only one coloring for each generic triangularization.
If one is presented with a triangularization and wishes to put in the colors, then one sees that the coloring becomes known in stages, looking at the triangles (equivalently cosets) of ever increasing levels. Figure 15 shows the amount of color information that can be gleaned at level .
Proposition 5.1.
Any generic tiling is uniquely colorable. ∎
We note that in the generic situation, the shifting and coloring are determined locally. That is, if one wishes to create the marked tiles for a finite patch of a generic tiling, one need only examine the tiling in a finite neighbourhood of that patch. That is because the shifting and coloring depend only on knowing levels of lines, and what the levels of various points on them are. Because of the generic conditions, these levels are all bounded in any finite patch and one needs only to look a finite distance out from the patch in order to pick up all the appropriate centroids and triangle edges to decide on the coloring and shifting within the patch. Of course the radii of the patches are not uniformly bounded across the entire tiling.
Here we offer a different proof of a result that appears in [19]:
Proposition 5.2.
In any generic tiling and at any point which is a hexagon vertex, the colors of the three concurrent diagonals of the three hexagons that surround are not all the same (where they meet at ).
Proof: The point is the centroid of come corner triangle of one of the triangles of the triangulation. Fig. 14 shows how the coloring is forced along two of the medians of the corner triangle and that they force opposite colorings at . See also Fig. 27. ∎
5.1. Completeness
We have now shown how to work from a triangularization to a tiling satisfying the matching rules R1,R2. Does this procedure produce all possible tilings satisfying these rules? The answer is yes, and this is already implicit in [19]. 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 edgeshifting of a triangulation 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 .
In the generic cases, the coloring that we impose on this triangulation is precisely that forced by R2. When we discuss the nongeneric cases in §6.2, we shall see that for nongeneric 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 must be among those that we construct from and so we see that our procedure does create all possible tilings conforming to the matching rules.
When we determine the structure of the hull in §6.4 we shall also see that it is comprised of a minimal hull and two highly exceptional countable families of tilings. The former contains triangulations of all types and, as the terminology indicates, the orbit closure of any one of its tilings contains all the others in the minimal hull, so in a sense if you have one then you have them all. (The two exceptional families of tilings appear in the rule based development of the tilings, but do not appear in the inflation rule description.)
6. The Hull
6.1. Introducing the hull
Let denote the set of all Taylor–Socolar hexagonal tilings whose hexagons are centered on the points of and whose vertices are the points of . The group (with the discrete topology) acts on by translations. We let be the set of all translations by of the elements of . We call and the hulls of the Taylor–Socolar tiling system. We give and 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 one can do away with ‘the small shifts’ part. See [13] for the topology.
In fact it is easy to see that, although we have produced it out of , is just the standard hull that one would expect from the set of all Taylor–Socolar tilings when they have not been anchored onto the points of . Thus is compact, and since is a closed subset of it, it too is compact.
The translation actions of on and on are continuous. We note that the hulls and are invariant under sixfold rotation and under complete interchange of the two tile types. Our task is to provide some understanding of and . Here we shall stick primarily to since the corresponding results for are easily inferred. We let denote the set of all the generic tilings in .
Each element produces a triangularization of the plane, and the triangularizations are parameterized precisely by elements in . In particular there is an element corresponding to , and we have a surjective mapping
(3)  
Proposition 6.1.
The mapping is continuous (with respect to the local topology on and the adic topology on ). Furthermore is on .
Proof: Any is determined by its congruence classes modulo , which are represented equally well by the congruence classes of modulo . These congruence classes are the sets of vertices of the triangles of increasing sizes, starting with those of level . Now any patch of tiles containing a ball , , will determine part of the triangulation with triangles of all levels for some , and we have as . The larger the patch the more congruence classes we know, and this is the continuity statement.
In the case of a generic tiling , already determines the entire markings of the tiles and hence determines . Thus is on .∎
Below we shall see that with respect to the Haar measure on the set of singular (i.e. nongeneric) is of measure . A consequence of this is [5], Thm. 6:
Corollary 6.2.
is uniquely ergodic and the elements of are regular model sets. ∎
We shall make the model sets rather explicit in §7.
6.2. Exceptional cases
We now consider what happens in the case of nongeneric tilings. To be nongeneric a tiling must violate either generica or genericw. We consider these two situations in turn.
6.2.1. Violation of generica
In the case of violation of generica, there is an aline of infinite level (that is, it does not have a level as we have defined it). Let be the adic completion of .
Proposition 6.3.
A tiling , where , has an aline of infinite level if and only if for some and some . Furthermore when this happens the points of lying on the infinitelevelline are those of the set .
Proof: All lines of the triangulation are in the directions and all lines of the triangulation contain edges of all levels from up to the level of the line itself. Thus the points of on any line of the triangulation are always a set of the form where and .
Suppose that we have a line of infinite level and its intersection with is contained in . The line has elements where is a vertex of a triangle of level . This means that . We conclude that . Furthermore . This is true for all if we define . Writing with and , we have
where . Thus . This proves the only if part of the Lemma.
Going in the reverse direction, if then this is a prescription for a line of points in that have vertices of all levels. Then the line is of infinite level. ∎
Proposition 6.4.
If a tiling has an infinite aline then it is in and has either precisely one infinite aline or three infinite alines which are concurrent. The latter case occurs if and only if , where .
Proof: Let have an infinite aline . We already know that and for some , , and some from Lemma 6.3. If it has a second (different) infinite line then similarly where , , and . Certainly for otherwise the two lines are parallel and this leads to overlapping triangles of arbitrarily large size, which cannot happen. But we have . Since are linearly independent over and we see that and are actually in . Then . We will indicate this by writing for .
In this case, since we find that is a vertex of a level triangle, for all . Since this is true for all , is a point through which infinite level lines in all three directions pass. Thus the existence of two infinite lines implies the existence of three concurrent lines.
In the other direction, if then as we have just seen there will be three concurrent infinite lines passing through it. ∎
The case of a tiling with three concurrent infinite alines in Proposition 6.4 is called a central hexagon tiling (CHT tiling) in [19]. We also refer to them as iCaL 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 aline are designated as iaL tilings.
6.2.2. Violation of genericw
The case of violation of genericw is somewhat similar, though it takes more care. One aspect of this is to avoid problems of torsion in , which we shall do by staying inside where this problem does not occur. Thus in the discussion below the quantity , where , is of course in , but when we see it with coefficients from we shall understand it as being in (as opposed to being in ). Another problem is that the violation of genericw is not totally disjoint from the violation of generica, as we shall see.
Proposition 6.5.
has a wline of infinite level if and only if for some and some . Furthermore when this happens the points of lying on the infinitelevelline are those of the set .
Proof: All wlines deriving from the triangulation are necessarily in the directions . Of course , and iff . It really makes no difference which of the six choices is, but for convenience in presentation we shall take herewith so that . This is in the vertical direction in the plane.
All wlines contain centroids of levels up to the level of the line itself. Furthermore if a wline contains a centroid of level then it also contains one of the vertices of the corresponding triangle and so also at least one point of of level . It follows that for any wline in the direction there is an so that the set of points of on is the set .
Suppose that we have a wline of infinite level and its intersection with is . Then the line has elements where is a vertex of a triangle of level . This means that . We conclude that . Furthermore . This is true for all if we define . Writing with and , we have
Thus . This proves the only if part of the Lemma.
Going in the reverse direction, if then this is a prescription for a line of points in that have vertices of all levels. The corresponding wline has centroids of unbounded levels, so the line is a wline of infinite level. ∎
Proposition 6.6.
If a tiling has an infinite wline then and it has either precisely one infinite wline or three infinite wlines which are concurrent. The latter case occurs if and only if the point of concurrency is either a point of infinite level (discussed in Prop. 6.4) or a nonorientable point (discussed in Prop. 3.3).
Proof: Let have an infinite wline . We again take this to be in the direction of . Then for some , .
Suppose that it has a second (different) infinite line . Then similarly where , , and . As above, we note that because if , the two lines are parallel and each of the two lines contains vertices of arbitrarily large levels. But the parallel lines through vertices and centroids of level are spaced at a distance of