On the Penrose and Taylor-Socolar Hexagonal Tilings
OON THE PENROSE AND TAYLOR-SOCOLAR HEXAGONALTILINGS
JEONG-YUP LEE , ROBERT V. MOODY Abstract.
We study the intimate relationship between the Penrose and the Taylor-Socolar tilings, within both the context of double hexagon tiles and the algebraiccontext of hierarchical inverse sequences of triangular lattices. This unified approachproduces both types of tilings together, clarifies their relationship, and offers straight-forward proofs of their basic properties.
Keywords: Penrose tiling, Taylor-Socolar tiling, double hexagon tiling, nested triangu-larizations, inverse sequences of hierarchical lattices1.
Introduction
From the very beginning, aperiodic tilings have played a significant role in unravellingthe mysteries of aperiodic crystals. Knowing what is mathematically possible has oftenturned out to be a crucial element in conceiving what might be physically realizable.In this paper we discuss two remarkable aperiodic tilings of the plane that are builtout of one of the most basic of all crystallographic structures: the standard periodichexagonal lattice.Also from the very beginning, there arose the question of what might be the mini-mum number of different prototiles necessary for a system of tiles and correspondingmatching rules that permit, and only permit, aperiodic tilings. The very first aperi-odic tilings involved thousands of prototiles. The famous aperiodic tilings of the planelike the rhombic Penrose and the Ammann-Beenker tilings are each based on just twoprototiles, the allowable motions being translations and rotations. This of course im-mediately raises the question of whether aperiodic tilings based on just one prototileare possible.Taylor[7] and Taylor and Socolar[6] introduced a planar aperiodic tiling which can bebuilt from a single hexagonal prototile allowing translations, rotations, and reflections .The tiling is based on the familiar hexagonal tiling of the plane, but if one distinguishesthe prototile in its direct and reflected forms, then the matching rules allow onlyaperiodic tilings to appear. Their work revived interest in much earlier work of R.Penrose. In [4], he had introduced an aperiodic tiling, also based on marked hexagonaltiles, but additionally involving two other types of thin edge tiles and small corner tiles,which he called a (1 + (cid:15) + (cid:15) )-tiling. However, already in this paper he had introduced a r X i v : . [ m a t h . M G ] J a n JEONG-YUP LEE , ROBERT V. MOODY arrowed double hexagon tiles as an alternative way to represent the tiling, see Fig. 1.Later, in his online notes [5] he expanded upon the double tile theme, and pointed outthe essential matching rules that make them work.In both the Taylor-Socolar and Penrose cases, the ‘matching rules’ somewhat stretchthe original notion of matching rules, so there can remain some controversy aboutwhether these are strictly aperiodic monotiles. However, that is not an issue here.These tilings are interesting, and puzzling too, for although the Taylor-Socolar hexag-onal tilings, henceforth called Taylor-Socolar tilings or T-S tilings, and the Penrosehexagonal tilings (Penrose tilings ) seem deeply related, that relationship is somewhatobscure. The two tilings are not mutually locally derivable (MLD) [2, 1] in the technicalsense, but are in mutually derivable in a rather different sense that we shall explain.In [3] we put forward a development of the T-S tilings based on the underlying hi-erarchical system of nested equilateral triangles that are so prominent in both the T-Stilings and the Penrose tilings. The aperiodicity of the tilings comes from this hier-archical structure, and indeed these tilings seem to have been invented with preciselythis feature in mind. The structure of nested triangles has an algebraic interpretationas an inverse system of finite groups, arising from the standard triangular lattice andits natural triangular sublattices, and is closely related to the 2-adic integers. In [3]we made this algebraic interpretation the basis out of which we constructed the T-Stilings. In fact, as long as the nested system of triangles is generic, meaning that itis free of singularities (like points which are simultaneously the vertices of triangles ofunbounded size), then there is a unique T-S tiling belonging to it. We shall see thatthe same type of mathematics applies to the Penrose tilings, and not surprisingly thetwo inverse systems are deeply connected.A more detached look at the double hexagon tilings reveals that they actually incor-porate both types of tilings simultaneously. This association is not entirely new [2, 1],but in this paper it is the double hexagon tilings that are taken as the fundamentalobjects, and they serve as the parents of the two individual types (Penrose and T-S) ofhexagonal tilings. Thus we may think of the two tilings as siblings of each other. Alge-braically a double hexagon tiling corresponds to a matched pair of inverse sequences,and with these we can see how the algebra and geometry fit seamlessly to elucidateeach other. The paper offers a unified treatment of the two tilings along with proofsof the implied hierarchical structuring and the aperiodicity. Since Penrose tilings based on five-fold symmetry are so much a part of the aperiodic culture, weshould emphasize that the Penrose tilings of this paper are based on hexagons and have nothing todo with the rhombic or kite/dart Penrose tilings.
N THE PENROSE AND TAYLOR-SOCOLAR HEXAGONAL TILINGS 3 Double hexagon tiles and their tilings
An arrowed hexagon is a regular hexagon in which each side has been given a di-rection, indicated by an arrowhead. An arrowed hexagon is called well-arrowed if,up to rotation, the arrows form the pattern shown on the right side of Fig. 1. In factall three hexagons in this figure are well-arrowed. The structure of the well-arrowedhexagon gives it a well-defined orientation in the plane, namely that provided by thetwo parallel arrows facing in the same direction.
Figure 1.
A well-arrowed hexagon is shown on the right. It has one pair of opposite sideswhose arrows face in the same direction, thus providing an orientation for the tile. A doublehexagon tile is on the left, the key feature being that the orientations of the inner and outerarrowed hexagons are at right-angles. When three double hexagon tiles meet at a vertex, thegray parts around the vertex form corner hexagons, see Fig. 2. The assumption of legality allows completion of the arrowing on the corner hexagons to well-arrowed hexagons, asindicated on the right.
If we start at any edge of a well-arrowed hexagon and then look every alternate edgeas we go around it, we notice that arrows on the three edges always have mixed typeof clockwise and counter-clockwise. We notice also the useful fact that if we have ahexagon and three alternate edges have been arrowed so as to be of mixed type, thenthere is a unique way to complete the arrowing to make it into a well-arrowed hexagon.A hexagonal tiling of the plane with well-arrowed hexagons is called well-matched ifthe hexagons meet edge-to-edge and the arrows of these coinciding edges also coincide– that is, they point in the same direction. We are only interested in well-matchedtilings of arrowed hexagons.A double hexagon tile (or double hex tile) consists of a pair of well-arrowed hexagons,one within the other, as shown on the left side of Fig. 1. The inner hexagon is centeredwithin the outer one with its orientation at right-angles to the orientation of the outerone. Its size is chosen so as to make the outer hexagon 3 times the area of the innerhexagon (so there is a linear scaling factor of √ corner hexagons . JEONG-YUP LEE , ROBERT V. MOODY Suppose that we have a hexagonal tiling of the plane with double hexagons. Bythis we mean that we are using the outer hexagons as the tiles. Suppose that fromthe perspective of the arrowing on the outer hexagons this hexagonal tiling is well-matched. Three outer hexagons meet at every common vertex v , and the three edgesof the corresponding inner hexagons that are closest to v form three edges of a cornerhexagon H centered on v . With the terminology introduced above we can ask whetheror not these three arrows are mixed. If they are mixed then we can extend the arrowingto make H well-arrowed.Suppose that all the corner hexagons of the tiling can be well-arrowed in this way.Collectively the inner hexagons together with the corner hexagons form another hexag-onal tiling of the plane, if we ignore the question of their arrows matching. However,there does arise the question of whether or not all the common edges of adjacentsmall hexagons actually do have matching arrows, that is, whether or not this newtiling is well-matched. The double hexagon tiling is called legal if they do. Thus adouble hexagon tiling is legal if its outer hexagons are well-matched, all of its cornerhexagons can be completed to well-arrowed hexagons, and the consequent well-arrowingof the small hexagons completes to a small hexagon tiling of the plane which is well-matched. In this situation we have two well-matched hexagonal tilings, one using thelarge hexagons and the other using the small ones inner and corner hexagons. Fig. 2shows a patch of a legal double hexagon tiling. Figure 2.
A legal patch of double hexagon tiles. Where three double hexagons meet ata vertex, a corner hexagon is created. Note the mixed arrowing of these small grey cornerhexagons.
The double hexagon tiles that we are discussing are also called
Penrose hexagonaltiles , and a legal tiling is a
Penrose hexagonal tiling . We shall use both names
N THE PENROSE AND TAYLOR-SOCOLAR HEXAGONAL TILINGS 5 in this paper, because within the context of understanding the intimate relationshipbetween Penrose hexagonal tilings (based on the large hexagons) and Taylor-Socolarhexagonal tilings (based on the small hexagons), it is convenient at times to simplythink in terms of legal double hexagon tilings.3.
Decorations and triangles
There are other decorations of well-arrowed hexagons and double hexagon tiles thatare equivalent representations of the arrowing but help to make the underlying geom-etry of the tilings more transparent. The first of these is the marking of well-arrowedhexagons shown in Fig. 3, which replaces the arrows of a well-arrowed hexagon with ablack stripe and two black corner markings. Initially we will use this representation ofthe arrowing with the small hexagons, later for the outer hexagons.Notice that when two well-arrowed hexagons are attached along some edge so that thecorresponding arrows match (i.e. they are well-matched), the black markings line up,either stripe to stripe, stripe to corner, or corner to corner, to create an extended blackpath. In fact, that the stripes and corner markings match to form extended paths, isexactly the same as arrow matching. In the resulting paths the corner markings indeedserve as corners at which the direction of the path changes. If we have a well-matchedtiling of well-arrowed tiles, we will have also a set of paths. It is easy to see that if,purple in following a path, it turns right or left at a corner, then at its next corner itwill turn in the same sense (again right or again left) and so the resulting paths willbe equilateral triangles (the corners create 60 ◦ angles).The only way this can fail is if there are paths that extend infinitely in some directionalong some straight line. Such a tiling is called a singular (or non-generic ) tiling.Later on we will examine the similar paths created by the stripes and corner markingson the large hexagon tiles, and the same issue of singularity will arise.The generic situation is that of non-singular (or generic ) tilings, that is, all ofthe paths form triangles. In this paper, in order to keep all the essential ideas clear,we shall always assume non-singularity, though at this point we need it only with thesmall hexagons and their markings. The resulting triangles come in various sizes andarrangements, and this is something we address in the next section.The second decoration is one that we make to double hexagon tiles, replacing theouter arrowing by colored short diagonals (short diagonals are the ones that pass atright-angles between opposite edges, as opposed to long diagonals that pass from vertexto opposite vertex). This is explained in Fig. 4.If we begin with a legal double hexagon tiling then we know that we end up with twowell-matched hexagonal tilings: one of large hexagons and one of small hexagons. Sincewe are assuming non-singularity, the well-matching of the small hexagons leads to acollection of triangles on the plane – equilateral triangles created by the stripes on the JEONG-YUP LEE , ROBERT V. MOODY Figure 3.
Well-arrowed hexagonal tiles can be converted into hexagonal tiles with stripes.These decorations fit together to make triangulations of the plane.
Figure 4.
Arrows on the edges of the outer hexagon which are oriented in the counter-clockwise direction are represented by short red half-diameters. For arrows in the clockwiseorientation we use short blue half-diameters. If we include the striping decoration of theinner hexagons as well (Fig. 3), we arrive at the fully decorated double hexagons shownon the right-hand side of the figure. Evidently the decorated tiles carry information fullyequivalent to the arrowing. Notice that proper matching of the edges of outer hexagonsis equivalent to a change of color as the short diagonal passes through the common edge.Also notice that if one holds the black stripe horizontally, then as one moves along a fullblue diagonal from right to left, the diagonal passes through the black stripe from above tobelow. It is the other way around for red stripes. We use this observation to make the colordetermination of the short diagonals in Fig. 10. small hexagons. Each small hexagon has an inner part of some edge of a triangle acrossit and the corners of two other triangles, one on each side of that edge, so altogethereach small hexagon is involved with three triangles.The very smallest triangles (called level 0 triangles) are those composed by puttingthree corner markings together around a common vertex of three small hexagons. Everystripe in a hexagon obviously belongs to a triangle larger than these smallest ones.Indeed there are triangles of ever increasing sizes, without limit. It is this result thatwe will establish in the next section.
N THE PENROSE AND TAYLOR-SOCOLAR HEXAGONAL TILINGS 7 Nesting and hierarchy
Let us continue with a non-singular legal double hexagon tiling D , in which we havecompleted its small hexagons to a well-matched hexagonal tiling and then resolvedeverything into triangles by decorating each of the small hexagons.A triangle is nested in another one if it appears as in Fig. 5, where the smallertriangle is nested in the larger. In this section we will prove that except for the verysmallest triangles (the ones made from three corners) every triangle has another onenested inside it. From this, we will see that every triangle has inside it a sequence oftriangles nested within each other, diminishing in size to the smallest size triangles.We refer to this phenomenon by saying that all triangles are nested within . Figure 5.
The green (smaller) triangle is nested in the larger (black) one. Notice the twoways of drawing this. A patch of triangles that arise from our tiling presents the innertriangles as being totally in the interior of the outer ones, as shown on the left-hand sidehere. We often will wish to allow the inner triangles to stretch to meet the boundaries ofthe outer triangles, as shown on the right hand side of the figure. This creates three newtriangles called bf corner triangles, so that the outer triangle is now decomposed as fourequal sized triangles.
There is more to this. Let us stretch out, or expand, the triangles created by thedecorations of the small hexagons (inner hexagons and corner hexagons) so that thecorners meet the edges of the triangle surrounding them as illustrated in Fig. 5. Indoing this each nested triangle produces three neighbors that fill out the whole trianglethat it lies in. In fact there is a nesting that involves one triangle sitting inside anotherof exactly twice the linear size, so that the larger triangle is decomposed into fourequal-sized equilateral triangles of which the nested triangle is one. The three trianglesthat emerge as neighbors of the nested one are called corner triangles (not to beconfused with the smallest triangles that we formed out of three corner markings).We will speak of the patterns of triangles (expanded or not) which are formed bythe decorations of the small hexagons as arrangements of triangles and derive theirnesting properties as we proceed. Notice that without the implications derived from
JEONG-YUP LEE , ROBERT V. MOODY the decorations of the outer hexagons, it is possible to get an arrangement of triangleslike the one shown in Fig. 6, which is visibly periodic.In all we shall see that triangles that are nested within appear on ever increasingscales, so there is a hierarchical structure. We shall call such an arrangement of trianglesa nested triangulation . Figure 6.
Periodic arrangements of triangles are possible if only decorations of the smallhexagons are used.
When we refer to the sizes of triangles in one of arrangements of triangles, we willalways refer to the side lengths of the stretched out versions. Lengths are normalizedso that the smallest triangles will be of side length equal to 1. We will see that withthis normalization all lengths are powers of 2.In the sequel we will commonly use both versions of the triangles and nested trianglesthat emerge out of our discussion–the original ones that come from the decoratedhexagons, and the stretched out ones that give us the arrangements of triangles. Oncewe have proved that all triangles are nested within, it is trivial to convert from onepicture to the other.In the stretched out version, two triangles are said to make an opposite pair ifthey are of the same size and share a common edge, see Fig. 7. Notice that there is nospecification of how each of the two opposing triangles fits into the overall arrangementof triangles.
Proposition 4.1.
Let D be a non-singular legal double hexagon tiling. Complete itssmall hexagons to a well-matched hexagonal tiling and let T be the resulting arrange-ment of expanded triangles formed from the decorations of the small hexagons. Then N THE PENROSE AND TAYLOR-SOCOLAR HEXAGONAL TILINGS 9
AA’ BB’ CC’
Figure 7.
On the left side the black triangles form an opposite pair of triangles. So too dothe pairs of smaller triangles labelled
A, A (cid:48) and
B, B (cid:48) . On the right-hand side the two blacktriangles do not form an opposite pair, but the two purple triangles
C, C (cid:48) do. (i) all triangles occur in opposite pairs; (ii) the side lengths of the triangles are all of the form k for some k = 0 , , , . . . ( k iscalled the level of the triangle );(iii) every triangle is nested within.The proof of Prop. 4.1 is by induction on the size of triangles. The smallest triangleshave side length 1 = 2 (level 0). There is no nesting within to take place. Thestretched triangle pattern created by these triangles is in itself a genuine triangularlattice of the plane and, in particular, every triangle edge borders an opposite pair oftriangles.We now assume that the three statements of Prop. 4.1 have been proved up to somelevel k .First we check that there must be triangles of size larger than 2 k . Fig. 8 shows why.It shows a triangle of level 2 k and uses matching triangles to see that there must belarger ones.We now take any triangle T of the next size, say m , that is larger than 2 k . We seeimmediately that m = 2 k +1 and it is internally nested, in the right way, Fig. 9. Thiscompletes the induction steps for parts (ii) and (iii).We now come to the proof of part (i). It is useful to prepare this by looking at thesituation pictured in Fig. 10. What this shows is how the coloring of the tile decorationsis related to the matching of opposite triangles. The color rules show that as a colordiagonal crosses a triangle edge at right-angles it changes color. When it crosses anedge that is not at right angles to it then it does not change color, but, as we have , ROBERT V. MOODY v vv Figure 8. v is a vertex of a triangle (shown in black) of level k . At v there is a smallhexagon, and its stripe allows for only two things to happen: either one of the edges of thetriangle extends through v , in which case it is an edge from a larger triangle, or there is anedge passing through v that is parallel to the opposite edge of the triangle. In the latter casewe use (i) to place down the two opposite pairs of triangles of adjoining triangles, shown ingreen (the adjacent edges are actually coincident edges of course). Then we see that the edgethrough v must actually be an edge that includes both the top edges of the green triangles:thus again a larger triangle. noted in the caption to Fig. 4, its color is related to the way in which it crosses theedge. This figure is the basis for our proof of matching triangles.Continuing to the proof of (i) we start with a triangle T of level k + 1 and showthat it must be matched by a triangle of the same level on each of its sides. Let S be the largest equilateral triangle nested in it. Now, any equilateral triangle of anylevel 2 r in our arrangement of triangles has exactly one vertex at the center of a largehexagon of the double hexagon tiling. This has to do with the lattice structure inducedby the arrangement of triangles coming from double hexagon tiling, and though prettyself-evident in the figures, is explained algebraically in §
5. In Fig. 11 we have madesuch a choice, indicating it by the small yellow circle at v . We shall use this coloringconvention to mark other vertices that are centers of the large hexagons. We shall startby showing that there must be matching triangles to T on the two sides of T on which v does not lie.The triangle S creates a partition of T into itself and three surrounding triangles,and we know that each of these must have an opposite match. We show these matchingtriangles along the lower edge of T solid edge. The shape of the small hexagon at theirintersection w must be of the type shown in the Fig. 11. What does the small hexagon N THE PENROSE AND TAYLOR-SOCOLAR HEXAGONAL TILINGS 11 T Figure 9.
We are given a triangle T of minimum size larger than 2 k . From one of itsvertices (we have taken it as the top one here) we fit in the largest sub-triangle possible (atthe very least, there is always a triangle of level 0 that can be fitted in). Its lower side isindicated in green. It must turn inwards at the sides of T and complete to the oppositegreen triangle. By the induction assumption its other two sides also complete to oppositepairs, and this leads to the new black triangle with the green triangle nested in it. Since westarted from a maximal sized sub-triangle, this larger black triangle must in fact the entiretyof our original triangle T . This shows that T has edge length 2 k +1 . The visible nesting andthe induction hypotheses show that the new triangle nested within. at v (cid:48) look like? The caption to Fig. 11 shows that neither of the two possibilities shownthere is possible. Thus the remaining possibility, which is that of a matched trianglefor T , must occur. This is illustrated on the left side of Fig. 12. This same argumentcan be applied to the other side of T which does not contain the point v .There remains the task of proving that the side that contains v also matches T toa triangle of the same size. Let us suppose that on this side the matching fails. Theargument we have just used tells us that in this case on this side we will see a triangle X which aligns its corner at the point v . The point x is the center of a double hexagon,just like v was, so it follows by what we have proved that the triangle Y shown existsand matches it. Again it has a point y which is a double hexagon center and so Y produces the matching triangle Z . But this is clearly a contradiction since Z overlapsbut does not coincide with the triangle T . This contradiction shows that there is amatching opposing triangle along the edge of T containing v .With this we conclude part (i) of Prop. 4.1, and so the entire proposition. (cid:3) Since there are triangles of every level, it is impossible that there are any translationalsymmetries.
Proposition 4.2.
Every non-singular legal double hexagon tiling is aperiodic. (cid:3)
Looking at Fig. 13 we see: , ROBERT V. MOODY vv’ vv’ Figure 10.
In the center of the figure we see the large pair of triangles making a diamondshape between the two extreme vertices v and v (cid:48) , which are assumed to be vertices arisingfrom centers of the large hexagons of the double hexagon tiling. The diamond is made up ofan opposite pair of fully internally nested triangles. The triangles are all in their stretchedform, but the line thicknesses indicate the nesting relationships. On the left side we see thecorresponding arrangement of double hexagons that surround the small hexagons between v and v (cid:48) . The arrows must match, but their common orientation in the horizontal directionis irrelevant here. There is a color change as we cross each triangle edge at right angles.The key point is what happens at the ends of the diamond as the color line crosses edges(indicated by the thickest lines) which are not at right angles to it. The main edge at v isshown by the heavy black line. The rules for coloring hexagons show that the color stripeis fully red here, see Fig. 4. The main edge at v has to be matched with its partner at v (cid:48) ,where color strip changes to fully blue. Notice the correct color change at the arrow. Thescenario shown at the right side of the Figure, where the main edge at v (cid:48) is shown in purple,cannot occur because of the color change violation at the brown arrow. Proposition 4.3.
In any small hexagonal tile the triangles that arise from its twoopposite corner markings are of the same size. (cid:3)
For future reference we also note (see Fig. 5):
Proposition 4.4.
In the nested triangulation created by the standard edge and cornermarkings of arrowed hexagonal tiles, every stripe forms the central part of an edge ofa some triangle.
N THE PENROSE AND TAYLOR-SOCOLAR HEXAGONAL TILINGS 13
Figure 11.
The vertex v is the unique vertex of triangle S which is the center of a doublehexagon tile. The two lower corner triangles formed in T , with common vertex w , haveopposite triangles shown in purple. The question is, what happens at the vertex v (cid:48) ? Neitherof the two possibilities shown here can occur. On the left, the triangle with vertices v (cid:48) and w would be left unclosed at w . On the right we are in the situation shown on the right sideof Fig. 10, which we know violates the color change property. The algebra of nested triangulations
If we start with a non-singular legal double hexagon tiling then we obtain a tilingof the plane with the small hexagons. The centers of these hexagons form a triangularlattice of the plane composed of level 0 (side length 1) equilateral triangles, as we haveseen. For definiteness we now specify this lattice as a set of points in R , namely theset of points Q = Z a + Z a ⊂ R , where a = (1 , a = ( − , √ ), Fig. 14. Joiningnearest neighbors of Q produces the triangular lattice of level 0 triangles, indicated bythe thin lines in Fig. 15.We know that there are also triangles of level 1 (side length 2). They are matchedacross each of their edges, and so there is a second triangular lattice of the plane byequilateral triangles. This meshes precisely with the first, in the sense that each level 1triangle is composed of four level 0 triangles. The vertices of the level 1 triangles forma coset q + 2 Q of Q .We may repeat this process, now looking at triangles of level 2, whose vertices lie ona coset q + q + 4 Q (where q ∈ Q and q ∈ Q ). Continuing this way we led to viewour nested triangulations in terms of ever refined cosets from the sequence Q ⊃ Q ⊃ Q ⊃ Q ⊃ · · · . , ROBERT V. MOODY Figure 12.
The left-hand side shows the matching that has to take place on the lowerside of T . The right-hand side shows what happens if there is not an opposite match to T on the side containing v . This corresponds to the right-hand side of Fig. 11. Chasingaround the pairs of opposite matching triangles yields X, Y, Z and the latter is clearly totallymis-matched with T . Figure 13.
The two triangles whose corners make up the pair of corner markings on awell-arrowed tile are always of the same size. Here the pair of triangles is shown in black.The matching on opposite sides of triangles leads to the red and green matched triangles.Since these too must match, we see that all four triangles are of the same size. Note thatthere is no presumption here about how these triangles lie in larger triangles.N THE PENROSE AND TAYLOR-SOCOLAR HEXAGONAL TILINGS 15
Figure 14.
The basis vectors for the lattices Q and P . The directions of edges in the Q -triangulations are ± a , ± a , ± ( a + a ), which are called a-directions, and those of P -triangulations are ± w , ± w , ± ( w − w ), which are called w-directions. Figure 15.
A nested triangulation of the plane. Triangulations of increasing scales (four areshown here) coexist in one underlying triangulation. Each increase in scale can be createdin four different ways by choosing one vertex of the previous scale as a reference point.
Thus the double hexagon tiling leads to the sequence q = ( q , q + q , q + q + q , . . . ) , where each q k ∈ k − Q . We refer to such a sequence as a Q -nested triangulation T . Indeed we see that any such sequence corresponds to a sequence of triangularlattices with each level nested within the next, that is to say every triangle of one levelappears as a corner triangle or as a central triangle within a triangle of one level higher.Specifically, up to rotation a typical triangle of level k has vertices x, x + 2 k a , x + 2 k a ,all of them lying in one coset q + · · · + q k + 2 k Q , where we assume k ≥
1. Themid-points of its edges are x + 2 k − a , x + 2 k − a , x + 2 k − a + 2 k − a , which form thevertices of a triangle of level k − q + q + · · · + q k − + 2 k − Q , hence thenesting. , ROBERT V. MOODY The sequence q can be interpreted as an Q -adic element of the inverse sequence(5.1) Q : Q/Q ← Q/ Q ← Q/ Q ← Q/ Q ← · · · . This sets up a bijective correspondence between Q -adic numbers and nested triangu-lations, and we will write T = T ( q ) when we wish to make the connection explicit. Inthe sequel Q will be written as Q , the first in a series of such inverse limits.The condition of non-singularity has algebraic consequences. The Q -nested triangu-lation is singular if some of the paths created by the stripes of the small hexagons donot close, but rather extend indefinitely. Since the directions of the stripes are all inthe a -directions of the lattice Q (see Fig. 14) this is equivalent to saying that there isan infinite path of edges consecutive edges in some direction a ∈ {± a , ± a , ± ( a + a ) } and this in turn implies that q lies in x + Z a for some x ∈ Q . Here Z is the 2-adicintegers. We need to avoid q having this form. See [3] for more details.In order to interpret the double hexagon tiles in this algebraic setting, we need, alongwith Q , its Z -dual P , relative to the standard dot product on R . Thus P = Z w + Z w where w = a + a and w = a + a , Fig. 14. We note that P ⊃ Q ⊃ P ⊃ Q ⊃ P ⊃ · · · , all the steps being of index 3. In fact each of the lattices in this chain is a scaled androtated version of the one before it, and in particular a scaled and rotated version ofthe original triangular lattice Q . They are all triangular lattices. We do not use P directly in what follows, but rather 3 P , since we wish to keep everything inside theinitial lattice Q .There are two clear differences between the triangular lattices arising from Q and3 P . The first is that the basic triangles in 3 P have side length √
3, so factors of √ Q - and 3 P -nested triangulations. The second is that the directions ofthe sides in all the Q -nested triangulations (at all scales) are {± a , ± a , ± ( a + a ) } ,which we are referring to as a -directions, and those for 3 P -nested triangulations are {± w , ± w , ± w − w ) } , which we have called w -directions. These two sets ofdirections are interchanged by 90 ◦ rotations.In a legal double hexagon tiling the large hexagons share one third of the vertices ofthe small hexagons, and their centers form some coset c +3 P of Q mod 3 P . This bringsus to a second inverse sequence of groups based on 3 P, P, P, . . . and correspondinggroup Q which is related to Q as shown in the commutative diagram (5.2). All themappings here are the natural homomorphisms that arise from factoring out largersubgroups. N THE PENROSE AND TAYLOR-SOCOLAR HEXAGONAL TILINGS 17 Q : Q/Q ←− Q/ Q ←− Q/ Q ←− Q/ Q ← · · ·↑ ↑ ↑ ↑ ↑ Q : Q/ P ←− Q/ P ←− Q/ P ←− Q/ P ←− · · · (5.2)Given the choice of q ∈ Q and c ∈ Q/ P , there is a unique element r = ( r , r + r , r + r + r , . . . ) , where r k ∈ k − P for each k , which maps onto q and has r ≡ c mod 3 P . Thisfollows from the more general fact: Lemma 5.1.
For all k, l ∈ N , k l − P ∩ k − l Q = 3 k l P and k l P ∩ k l − Q = 3 k l Q .
Proof: Dividing out common powers of 2 and 3, we are reduced to proving that 3 P ∩ Q = 6 P and 2 P ∩ Q = 2 Q , respectively, both of which are trivial to verify. (cid:3) So, given a non-singular legal double hexagon tiling, we arrive not only at an element q ∈ Q but also an element r ∈ Q . This element picks out one coset r k +2 k P for each k = 0 , , , . . . and so should determine a family T + ( r ) of nested triangulations, justthe same way as q did. To guarantee that we really do have a non-singular 3 P -nestedtriangulation, that is to avoid infinite lines, we have to make an assumption similarto the non-singularity assumption that we have already seen. This time the triangleedges are in w -directions, so we must assume that r does not lie in x + Z w for any w ∈ { w , w , w − w } .Thus the joint conditions equivalent to non-singularity are that for all x ∈ Q ,(i) q does not in x + Z a for any a ∈ {± a , ± a , ± ( a + a ) } ;(ii) r does not lie in x + Z w for any w ∈ { w , w , w − w } .These are the same conditions as appeared in [3], though we did not use double hexagontilings there. We will pursue the detailed study of the singular double hexagon tilingsin another paper.Returning to our discussion, the situation is this. We are given a generic doublehexagon tiling whose small hexagons are centered on Q and whose large hexagons arecentered on c + 3 P . We thus have two nested triangulations T ( q ) and T + ( r ), the firstbeing determined by the markings on the small hexagon tiles and the second simplyby the algebra of the commutative diagram (5.2). Since the large hexagon tiles can begiven their own stripe and corner markings based on their arrowing in just the sameway as we did for the smaller hexagons, it is natural to ask whether or not this newnested triangulation based on r ∈ Q is the one that these markings create. In fact this , ROBERT V. MOODY z z + 2 k w z k w z + 2 k a z k a Figure 16.
How triangle sides of triangles from T ( q ) and T + ( r ) right-bisect each other. is indeed the case. Here is the argument. As a matter of nomenclature, we will use thesame concept of level for the new triangulation T + ( r ) as we did before. The smallesttriangles are said to be of level 0 with side lengths equal to √
3, and subsequent sizeshave levels 1 , , . . . of side lengths 2 √ , √ , . . . .Take any edge e of a level k + 1 triangle T + from the triangulation T + ( r ) and let z be its midpoint. The two ends of e have the form z ± k w for w in one of the w-directions of the lattice, see the red line segment in Fig. 16. Since both end points lie in r + · · · + r k +1 + 2 k +1 P , we see that z ∈ r + r + · · · + r k + 2 k P ⊂ r + · · · + r k + 2 k Q = q + · · · q k + 2 k Q , and r k +1 ≡ k w mod 2 k +1 P . Let a be in the a-direction at right-angles to w and consider the two points z ± k a . These are two points of a level k + 1triangle of T ( q ) and z is its midpoint.To see this explicitly we take the case where w = w and a = a = 2 w − w . Then2 k w − k a = 2 k (3 w − w + w ) = 2 k +1 (2 w − w ) ≡ k +1 Q .
This shows that z + 2 k a ∈ r + · · · + r k + 2 k w + 2 k +1 Q = r + · · · + r k +1 + 2 k +1 Q = q + · · · + q k +1 + 2 k +1 Q .
See Fig. 16. This proves:
Proposition 5.2.
At their midpoints, the edges of level k + 1 triangles of T + ( r ) bothright-bisect and are right-bisected by edges of level k + 1 of T ( q ) . These midpoints are centers of double hexagon tiles and, as in all double hexagontiles, the stripes of the inner and outer hexagons are at right-angles. This appliesto triangles of all levels k = 0 , , , . . . . Since by Prop. 4.4 every stripe of a smallhexagon lies at the middle of some edge of some triangle, we conclude that the nestedtriangulation T + ( r ) is directly related to the stripes on the large hexagon tiles, namelythe path created by these stripes form the triangles of this triangulation. Thus the N THE PENROSE AND TAYLOR-SOCOLAR HEXAGONAL TILINGS 19 vv’ v”u u’z
Figure 17. v, v (cid:48) are the ends of an edge e of a triangle T + from the nesting determinedby T + ( r ). At its midpoint z we see the edge u, u (cid:48) of a triangle T from T ( q ). The blacktriangles all come from the triangulation of T ( q ), the largest ones being of level 3. The edge e is maximal, in the sense that it is not part of an edge of some larger triangle from T + ( r ).Thus at its ends, the stripes of the large hexagons at v and v (cid:48) are in the directions of theother sides of T + . The inner hexagons along vv (cid:48) are centered at double hexagon centers andtheir stripes are all oriented in the same direction, namely perpendicular to the vv (cid:48) . At theleft we have separated out the outer hexagons that overlay the small hexagons along vv (cid:48) .We see their matching arrows and how their stripes align to form the edge vv (cid:48) (in green).The colors (red/blue) of the short diameters of these large hexagons are determined by (ordetermine, whichever way one wants to put it) the color rule that we see in Fig. 10, thoughnote that the stripe of the large hexagons is perpendicular to that of the small ones, so theright/left crossing rule is opposite! The fact that the stripe orientation changes at the enddictates that the edge vv (cid:48) is an interior edge of a larger triangle. The shift indicated by theorientation of the arrows matches this. triangular tiling produced by the outer hexagons is indeed the one produced by thenested triangulation of T + ( r ). Fig. 17 illustrates what is going on here and also showsthat the edge shifting (involved in truly nesting the triangles) is also properly indicatedby the outer hexagon tiles.A direct consequence of Prop. 5.2 is that generic (resp. singular) 3 P -nested trian-gulations give rise to generic (resp. singular) Q -nested triangulations: , ROBERT V. MOODY Figure 18.
Starting from the pair ( q , r ), with r ≡ c mod 3 P , hexagons centered at thelattice points of Q are shown, with those centered on a coset c + 3 P indicated in yellow. Thenesting of the triangulation T ( q ) arising from q is indicated. Proposition 5.3.
Let r ∈ Q and let q be its image in Q . Then T ( q ) is generic ifand only if T + ( r ) is generic. (cid:3) From nested triangulations to double hexagon tilings
At this point it is rather clear that given any triangulation T ( q ) and any choiceof one coset c + 3 P leading to T + ( r ) ought to lead to a legal double hexagon tiling.Here are the details. We will assume that both triangulations are non-singular. Thisguarantees that there are no infinite edges, we get proper triangulations, and they arenested. This nesting can be geometrically manifested by laterally shifting the edges asindicated by the nesting. The Voronoi cells of the lattices (nearest neighbor cells) arehexagons centered respectively on the points of T ( q ) and T + ( r ). We know that everyhexagon from T ( q ) has a triangle edge passing through it and this edge will be shiftedlaterally in nesting. This is shown in Fig. 18. The hexagon is made into a well-arrowedhexagon by placing the pair of parallel arrows in the direction of the shift. The smallhexagons now make a well-arrowed and well-matched hexagon tiling.Now we do the same thing with the triangulation T + ( r ), leading to the new hexag-onal tiling, again with arrows indicating edge shifting in the nesting, see Fig. 19. Thisis the second well-arrowed and well-matched hexagonal tiling. The outcome is that wehave a tiling of double hexagon tiles which is non-singular and legal, see Fig. 2.7. Penrose tilings, Taylor-Socolar tilings, and beyond
By definition a Penrose tiling is precisely a legal double hexagon tiling. Taylor-Socolar tilings (T-S tilings) are usually defined by the T-S tiles shown in Fig. 20 andthey are assembled as regular hexagonal tilings, but under the matching rules
RT1 the black lines must join continuously when tiles abut;
RT2 the ends of the diameters of two hexagonal tiles that are separated by an edgeof another tile must be of opposite colors.
N THE PENROSE AND TAYLOR-SOCOLAR HEXAGONAL TILINGS 21
Figure 19.
Continuing from Fig. 18, from r ∈ Q we obtain a nested triangulation T + ( r ),part of which is shown here. Figure 20.
The two prototiles for the Taylor-Socolar tilings
In Fig. 10 we have seen that the diagonals of the inner hexagons of a double hexagontiling can be colored, and if we restrict this coloring to the actual physical area of theinner hexagons then we have the colorings of Fig. 20. The matching of the outer arrowsof the double hexagon tiling amounts to the color rule
RT2 , so we have in this way onethird of a T-S tiling. If the double hexagon tiling is legal then we know that this partialhexagonal tiling of inner hexagons along with the corner hexagons completes to a newproperly arrowed hexagonal tiling together with a corresponding nested triangulation.If we assume that the nested triangulation is non-singular, which is generically thecase, then this tiling-triangulation corresponds to a unique T-S tiling. That is, theone-third tiling we have completes uniquely to a T-S tiling. The proof of this is givenin [3]—each non-singular nested triangulation corresponds to a unique T-S tiling andvice-versa.Thus every non-singular Penrose tiling produces inside it a non-singular T-S tilingmade out of its inner and corner hexagons. Now let us go in the other direction. Ifwe begin with a non-singular T-S tiling then it produces a nested triangulation out ofthe stripes on each hexagon. Relative to a fixed coordinate system, this triangulationcorresponds to an element q in the Q -adic completion Q . In order to obtain a doublehexagon tiling from this we need to select which hexagons will be the inner hexagonsand which the corner hexagons for the new tiling. This amounts to choosing one coset , ROBERT V. MOODY from the Q/ P . Choose one, say, c + 3 P . Then there is a unique r ∈ Q that maps q under the natural mapping of Q to Q and for which r ≡ c mod 3 P . This producesthe centers and the nested triangulation that determines a legal double hexagon tiling,as we have pointed out in § q once we know r .A noteworthy observation comes by comparing Fig. 20 and Fig. 4: it shows that thedistinction between the parity (that is, the difference between the two types of smallhexagon tiles (respectively large hexagon tiles) is the same for the T-S tiling and thePenrose tiling. Thus the parity distribution of a Penrose tiling is the same as the paritydistribution of one coset modulo 3 P of the T-S tiles.Although it is shown in [2] that the two tiling spaces generated by T-S tilings andPenrose tilings define distinct MLD (mutual local derivabillity) classes, it is clear bynow that the two types of tilings are intimately related, and indeed, modulo the choiceof a coset, there is a mutual derivability. We can summarize some key points as follows: Theorem 7.1. (i)
Taylor-Socolar tilings and Penrose tilings are aperiodic. (ii)
Given a non-singular Taylor-Socolar tiling on Q , one can build, in a canonicalway, three different non-singular Penrose tilings, one for each of the three cosets c + 3 P of Q mod 3 P . At any point of c + 3 P one knows exactly what type ofPenrose tile should be put in that position, and this uses only local informationof the T-S tiling. (iii) Given a non-singular Penrose tiling on some coset c + 3 P , there is a uniquenested triangulation on Q formed by the decoration of inner hexagons and cornerhexagons of Penrose tiles. This nested triangulation gives a unique non-singularT-S tiling. Note that unlike the situation in (ii) , this construction, is not local. The process of producing double hexagon tiles from a pair ( q , r ) suggests that wemight do it again, choosing a coset d + 3 Q with d ≡ c mod 3 P and then determining s ∈ Q . This triple ( q , r , s ) leads to triple-hexagon tiles and a triple-hexagon tiling.The rules for admissibility follow the same principles as we have used above. Thelargest hexagonal tiles have middle sized hexagonal tiles at their centers, and createmiddle sized corner tiles around them. The requirement is well-arrowing throughout.This yields a well-arrowed hexagonal tiling of middle sized tiles. In the same way, createfrom this a hexagonal tiling of small tiles, and where again we require well-arrowingthroughout. N THE PENROSE AND TAYLOR-SOCOLAR HEXAGONAL TILINGS 23
Figure 21.
The four types of triple (or 3-tiered) hexagons
These triple tiles come in four types and produce a new type of hexagonal tiling,Fig. 21. There is no reason to stop there. This new hierarchical situation is illustratedin the commutative diagram (7.1). Q : Q/Q ←− Q/ Q ←− Q/ Q ←− Q/ Q ← · · ·↑ ↑ ↑ ↑ ↑ Q : Q/ P ←− Q/ P ←− Q/ P ←− Q/ P ←− · · ·↑ ↑ ↑ ↑ ↑ (7.1) Q : Q/ Q ←− Q/ Q ←− Q/ Q ←− Q/ Q ←− · · ·↑ ↑ ↑ ↑ ↑ Q : Q/ P ←− Q/ P ←− Q/ P ←− Q/ P ←− · · ·↑ ↑ ↑ ↑ ↑ ... ... ... ... ...More generally there are ‘ n -tuple hexagons’ or to have a better sounding name, n - tiered hexagons , each of which consists of n hexagons stacked within each other,which tile the plane according to the n -th line of (7.1). There are two choices forthe orientation of a hexagon at each stage of layering, but after taking into account , ROBERT V. MOODY rotations, this gives 2 n − types of these tiered tiles. Non-singularity (resp. singularity)is a common property to all levels. 8. Outlook
The purpose of this paper has been to clarify the unity that exists between the Taylor-Socolar tilings and the Penrose hexagonal tilings—a unity that can be expressed bothgeometrically and algebraically in terms of double hexagon tiles. Each non-singularlegal double hexagon tiling encompasses both a Penrose tiling and a T-S tiling, and thispairing can be interpreted algebraically in terms of (5.2). Each of the two hexagonaltilings leads to a nested triangulation, and these two are bound together by the simplerule that triangle edges of each right-bisect edges of the other.There are two issues that arise here that we have not discussed, but plan to pursuein a future work. The first is the nature of singularities in these tilings from boththe geometric and algebraic perspectives, and their detailed manifestation in the cor-responding tiling hulls. The second is the study of the n -tiered hexagonal tilings. Thealgebraic setting which uses the first n rows of the commutative diagram (7.1) sug-gests that the n -tiered hexagons lead to aperiodic tilings in which there are potentially2 n − types of tiles. Thus there is a hierarchy of aperiodic hexagonal tilings, and theircorresponding tiling hulls, about which we know very little. Acknowledgment
The first author would like to acknowledge that this work was supported by a Na-tional Research Foundation of Korea (NRF) Grant funded by the Korean Government(MSIP) (No. 2014004168) and by research fund of Catholic Kwandong University(CKURF-201604560001). She is also grateful for the support by the Korea Institute for AdvancedStudy (KIAS) .
References [1] Baake, M.; Grimm, U.
Aperiodic order, Vol. 1: A mathematical invitation , CambridgeU. Press, 2013.[2] Baake, M.; G¨ahler, F.; Grimm, U.
Hexagonal inflation tilings and planar monotiles ,Symmetry , 581.[3] Lee, J.-Y.; Moody, R. V. Taylor-Socolar hexagonal tilings , Symmetry , , 1-46; doi:10.3390/sym5010001 .[4] Penrose, R. Remarks on a tiling: Details of a (1+ (cid:15) + (cid:15) )- aperiodic set , In The mathemat-ics of long-range aperiodic order ; Moody, R.V. Ed.; NATO ASI Series C: 489, KluwerAcad. Publ.: Dordrecht, Netherlands,1997, pp 467–497.[5] Penrose, R.
Twistor Newsletter , Reprinted in
Roger Penrose: Collected Works , Volume6: 1997-2003, Oxford University Press: New York, USA, 2010. http://people.maths.ox.ac.uk/lmason/Tn/ , see specifically:
N THE PENROSE AND TAYLOR-SOCOLAR HEXAGONAL TILINGS 25 http://people.maths.ox.ac.uk/lmason/Tn/41/TN41-08.pdf , http://people.maths.ox.ac.uk/lmason/Tn/42/TN42-09.pdf , http://people.maths.ox.ac.uk/lmason/Tn/43/TN43-11.pdf , (1996, 1997).[6] Socolar, J.; Taylor, J. An aperiodic hexagonal tile , Journal of Combinatorial Theory,Series A , 2207–2231.[7] Taylor, J.
Aperiodicity of a functional monotile , Preprint (2010),, Preprint (2010),