Ammann Tiling

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Table of contents

1 Constructing the Tiling 2 Iterating Tilings 2.1 Theorem 1 2.2 Proof 2.3 Theorem 2 2.4 Proof 2.5 Theorem 3 2.6 Proof 2.7 Construction Algorithm 2.8 Theorem 4 2.9 Proof of (a) 2.10 Proof of (b) 2.11 Proof of (c) 2.12 Proof of (d) 3 Forming an Index Sequence 3.1 Process 3.2 Why didn't we model this construction on the Penrose case? 3.3 Forming the Groupoid 4 Differences Between Penrose and Ammann Tilings 5 What's Left to Do? 6 Works Cited

Constructing the Tiling

An Ammann tiling is constructed from a Penrose tiling by rhombs (we label these "type 1" and "type 2" for convenience; type 1 is shown above on the far left, and type 2 is to its right ). Note that in the original Penrose tiling, the edges of the rhombs are all of equal length. The Penrose tiles must be assembled so that sides with matching markings are adjacent (see diagram above).

The construction of the Ammann tiling is determined entirely by the choice of point Q. The rest of the divisions of the rhombs are also shown in the diagram above such that lines that are the same length by construction are displayed in the same color. This construction is clearly unique once Q is chosen, because the triangle in the lower left of the type1 rhomb above is congruent to the triangle on the lower left of the type 2 rhomb next to it. Similarly, the triangle on the upper right of the type 1 rhomb is congruent to the triangle on the lower right of the type 2 rhomb. The dark purple line is constructed by joining the vertices of the two triangles that lie within the left rhomb.

In order to ensure a non-periodic tiling of the plane, Q must be chosen so that the distances from Q to the three vertices it is connected to and the distance between the two points in the interior of the type 1 rhomb are all different (Grunbaum/Shephard, 548). More simply, the segments shown above in pink, orange, light purple, and dark purple are all of different lengths.

As displayed above, the construction of the Ammann tiles necessarily causes a = c = e = f = g = n, b = h = i = o, j = k = l = m, and d = p. The diagram at the very bottom of this page shows part of an Ammann Tiling without edge coloring.

Iterating Tilings

Just as with a Penrose tiling, it is possible to create an index sequence for the Ammann tiling shown at right. The first image to the right shows the original tiling, T_n, and the third image shows the new tiling, T_{n + 1}. (Check the third image yourself to see that T_{n + 1} is composed of two pentagons and a hexagon.) In order to clarify the process of going from T_n to T_{n + 1}, we label the pentagon in T_n with three sides of equal length type A(n), the hexagon type B(n), and the other pentagon type C(n).

Theorem 1

Given an Ammann tiling, T_n, the six coronas illustrated below are the only possible coronas of an A(n) tile (the only possible arrangements of tiles of T_n around A(n)) for any .

Proof

Let T_n be an Ammann Tiling. There is necessarily a Penrose tiling by rhombs that corresponds to T_n. The vertex star atlas of a Penrose tiling by rhombs contains the eight distinct vertex stars shown below (Senechal, 177).

As can be seen by inspection, a type 1 rhomb of the Penrose tiling is needed to form a type A tile in the corresponding Ammann tiling. So, to look at all possible coronas of A, we must look at all possible coronas of a type 1 rhomb. In forming these, I have positioned the central rhomb with the same orientation in each case. Based on the vertex star atlas, there are six possible configurations: (Image)

Now, here are the same images again, with only the relevant lines colored:

(Image)

As you can see, the six possible coronas of a type 1 rhomb yield exactly six possible coronas of the corresponding type A tile (the ones we wanted). Therefore, these six coronas are the only possible coronas of A.

Theorem 2

The set of coronas of A is a reduced corona atlas.

Proof

We will use proof by contradiction. Assume that the set of coronas of A does not cover the tiling. Then, there is at least one corona of a B or C tile that does not contain any A tiles. Consider the corresponding Penrose tiling. As stated above, a type 1 rhomb necessarily results in a type A tile. So, the corresponding Penrose patch does not have any type 1 rhombs. Note that for an Ammann patch of n tiles, the corresponding Penrose patch contains at least n tiles. This means that there exists a patch in the Ammann tiling of at least six tiles, none of which is type A. Furthermore, the corresponding Penrose patch contains at least six type 2 (skinny) rhombs and no type 1 rhombs. But consider the Penrose vertex star atlas (Senechal, 177):

The vertex star atlas shown above determines every Penrose tiling (Senechal, 177). Note that in each vertex star there is at least one type 1 rhomb. So, we have reached a contradiction. Therefore, this set of four coronas of type A tiles covers the tiling, and is thus a reduced corona atlas.

Theorem 3

A, B, and C are an aperiodic protoset.

Proof

From the Penrose vertex atlas, we get the following eight Ammann vertex stars:

Let the vertices of the three tiles be numbered as shown below.

Using the angle numbering shown above, we find from the first eight vertex stars:

But since the construction of the Ammann tiles added three vertices, we need to consider the vertex stars of these. The neighborhoods of A determined in Theorem 1 yield six more Ammann vertex stars:

From these six vertex stars, we have:

Combining these sets, we have fourteen vertex stars that appear in an Ammann tiling that has been derived from a Penrose tiling by rhombs. These thirteen necessarily appear in the Ammann vertex star atlas.

So, the angle relations that result from construction are:

With these angle relations and the known edge congruences, there are ten more possible vertex stars, which are shown below. (We leave as an exercise the proof that there are no other locally legal Ammann vertex stars.) The central vertex of each vertex star is labeled with the numbers of the vertices that meet there.

We prove that none of these ten vertex stars can be extended to an infinite tiling:

(7,13,9) and (7,11,9) are not globally legal because there is no vertex with angle 2Θ that is connected to two purple edges.

If two instances of vertex 6 meet, then vertices 7 and 11 meet. But if vertices 7 and 11 meet, then vertex 9 meets there as well, because 7 + 11 = 10Θ − 9. However, as stated above, (7,11,9) is not globally legal, so a vertex star where two instances of vertex 6 meet is not globally legal.

Therefore, (6,6,6,6,6), (6,6,6,6,5), (6,6,6,5,5), (6,6,5,5,5), (6,5,6,5,6), and (2,14,6,6) are not globally legal.

Now, we focus our attention to vertex (2,14,6,5). We have already shown that there are only six possible coronas of an A tile. But the configuration of the vertex star of (2,14,6,5) does not fit with any of these. Specifically, two A tiles may not be placed so that vertices 5 and 2 are incident.

So, (2,14,6,5) and (2,14,5,5) are not globally legal. Therefore, none of these ten vertex stars is globally legal, and Ammann vertex star atlas contains only the aforementioned fourteen vertex stars.

The Penrose vertex star atlas of eight vertex stars completely determines all Penrose tilings (Senechal, 177). So, the set of fourteen Ammann vertex stars that are derived from a Penrose tiling completely determines all Ammann tilings that can be derived from Penrose tilings. But the set of vertex stars of an Ammann tiling that has been derived from a Penrose tiling is identical to the vertex star atlas of an arbitrary Ammann tiling. Therefore, an arbitrary Ammann tiling can be derived from a corresponding Penrose tiling. Since Penrose tilings are non-periodic, so too are Ammann tilings. Therefore, A, B, and C form an aperiodic protoset.

Construction Algorithm

By connecting vertices within the six coronas of A_n as shown below, we create a new set of three tiles. We label the new tiling T_{n + 1} and label the new tiles A(n+1), B(n+1), and C(n+1). (Eventually, we want to show that tiling T_{n + 1} is an Ammann tiling with protoset {A(n + 1),B(n + 1),C(n + 1)}.)

To clarify the labeling of the diagram:

The edges of type A tiles are labeled with letters 'a' through 'e', type B tiles are labeled with letters 'f' through 'k', and type C tiles are labeled with letters 'l' through 'p'. The "main edge" is shown below in bold in each case where the tiles are shown separately. The main edge always carries the label of the earliest alphabetical letter. More explicitly, type A tiles have main edge labeled 'a', type B tiles have main edge labeled 'f', and type C tiles have main edge labeled 'l'. In the new tiling, the labeling proceeds alphabetically in the opposite direction.

The diagram below is a key for how to proceed from T_n to T_{n + 1}.

In the diagram, both (1) and (2) form tiles of type C(n + 1), (3) forms a tile of type A(n + 1), and (4) forms a tile of type B(n + 1).

A proof will be given shortly that the corona atlas of T_{n + 1} is the same form as the corona atlas of T_n, but for now, you can see that this is a plausibly claim by looking at figure (1'), which shows the tiles of T_{n + 1} assembled as a corona of an A(n + 1) tile.

Theorem 4

Let T_n be an Ammann tiling. Then, under the algorithm mentioned above, T_{n + 1} is also an Ammann tiling. We will do this by proving the following:

(a) Edge congruences are conserved from T_n to T_{n + 1}.

(b) T_{n + 1} has the same corona atlas (its tiles fit together in the same way) as T_n.

(c) We can construct a tiling of 2 rhombs from the vertices of T_{n + 1}.

(d) This tiling is a Penrose tiling.

Proof of (a)

Since T_n is an Ammann tiling, a = c = e = f = g = n, b = h = i = o, j = k = l = m, and d = p. By construction, as shown in the picture below, A = C = E = F = G = N, B = H = I = O, K = M in tiling T_{n + 1}.

Since in T_n, a = e = g, we have J = K = M = L. Also, it can be easily shown that only an A tile would fit between the C and B tiles on the right in figures (1) and (2), so D = P.

Therefore, (keeping in mind that the edges of the new tiling are labeled in the opposite direction) we get the edge congruences shown below: A = C = E = F = G = N, B = H = I = O, J = K = M = L, D = P.

So, the algorithm preserves congruence of edges between T_n and T_{n + 1}.

Proof of (b)

Clearly, the angles of T_n are not congruent to the angles of T_{n + 1}. We want to show that the restrictions imposed by our algorithm for constructing T_{n + 1} are the same as the restrictions present in T_n

In proving Theorem 2, we came up with the Ammann vertex atlas:

And we arrived at the following angle relations for T_n:

Now, we show that the construction algorithm gives us exactly the same restrictions on T_{n + 1}.

In the picture below, we label the interior angles by the corresponding vertex labels.

From this picture, we get immediately: * 12 = 5 = 2θ,

* 5 = * 6 = 12 = 2θ,

* 8 = 10 = 4θ,

* 10 = 5 + 6 = 4θ,

* 11 = * 13, and

* 1 = * 7

Look at the vertex circled in orange to see: .

Look at the vertex circled in pink to see: .

Look at the vertex circled in purple to see: .

Now, look at vertex * 2. The green arc marks angle * 2 and the blue arc marks angle * 14. Also, the purple arc marks angle 10 = 4θ, and the brown arc marks angle 12 = 2θ.

So, .

The angle restrictions are:

These are exactly the same angle relations present in T_n. Since angle relations are conserved from T_n to T_{n + 1} and length congruences are conserved as well (by (b)), T_{n + 1} has the same corona atlas (and reduced corona atlases) as T_n.

Proof of (c)

Since T_n is an Ammann tiling, it has a unique corresponding Penrose tiling. The first picture below shows the coronas of of A(n) with the corresponding Penrose tiles drawn in. The picture directly below it shows the tiles of T_n with numbered vertices and black lines drawn where they are cut by possible corresponding Penrose tiles. The multiple labels at a vertex indicate the two or three vertices that coincide there.

Now, in order to make a tiling by rhombs that corresponds to T_{n + 1}, we note the label numbers of the vertices that are connected in T_n and connect the corresponding vertices in T_{n + 1}. Below the three tiles of T_{n + 1} are drawn with these cuts and are superimposed upon the four Amman coronas.

From this image, we see that the segments and (shown in light blue) are of equal length by construction. Similarly, and (shown in green) are the same length.

Below is corona 4 redrawn with the blue line, the green line, and some length labels. Note that both rhombs have all sides of length Φ.

The green line is clearly of length Φ + 1. We determine the length of the blue line by using the law of cosines (c² = a² + b² − 2abcosC):

X = Φ(Φ)

X = Φ² = Φ + 1

So, the blue and green lines are the same length, giving us: . Since we have already shown that the T_{n + 1} tiles assemble themselves in exactly the same patterns as the Tn tiles, we cut the new tiling along these lines and have produced a tiling by rhombi.

Proof of (d)

Note that rhombs have opposite angles equal. Let the small angle of a 1rhomb (the rhomb corresponding the the Penrose "fat rhomb") be x, and let the large angle be y. Let the small angle of a 2rhomb be z, and let the large angle be w.

Since we know that the T_{n + 1} tiles fit together in the same patterns as the T_n tiles, we extend corona (5') as shown below. Since five identical 1rhombs are need to complete the cycle of 2π with angle x in the center, . So, . Therefore, 1rhombs are Penrose type 1 rhombs.

Since, T_{n + 1} and T_n have the same corona atlas, the configuration shown below is valid. Here, the large angle of the 2rhomb supplements the sum of two large angles of 1rhombs. , and the last vertex is: . Therefore, 2rhombs are Penrose type 2 rhombs.

Since 1rhombs are Penrose type 1 rhombs, and 2rhombs are Penrose type 2 rhombs, the tiling from part (c) is a Penrose tiling. Therefore, T_{n + 1} is an Ammann Tiling.

Forming an Index Sequence

Process

Now, to create an index sequence, pick a point, P that lies within an A(1) tile of T₁. Consider which of the four possible coronas surround this A(1) tile. Record the corresponding number (1, 2, 3, or 4). Now, iterate the tiling twice, yielding T₃. P will lie within an A(3) tile of T₃ (proof to come). Consider the corona of this A(3) tile. Record the number of this corona, 1, 2, 3, or 4. Continue this process ad infinitum. I claim that given a point, P, this index sequence is unique to the tiling, and a tiling is uniquely determined by such an index sequence.

Why didn't we model this construction on the Penrose case?

In the process of iterating a Penrose tiling (by triangles) edges are deleted, but never added. In the process described above for iterating an Ammann tiling, each iteration adds and deletes edges. Identical Penrose index sequences are constructed whenever two points P₁ and P₂ lie within the same tile of the original tiling. In the Ammann case, something very different happens: when the tiling is iterated, a new edge may divide a tile, so although P₁ and P₂ start out in the same tile of T₁, they may not lie in the same tile of T₂. Whenever this is the case, the index sequences of T₁ corresponding to P₁ and P₂ are not identical (they differ in at least one digit), and it is extremely difficult to determine whether , where x_n is the index sequence corresponding to P₁ and y_n is the index sequence corresponding to P₂. But since the index sequences track the same tiling, we need them to agree eventually. The process mentioned above clearly fulfills this requirement.

Forming the Groupoid

Let X_m be the set of all possible infinite sequences of 1, 2, 3, and 4. Just as in the Penrose case, we define an equivalence relation, R_m on this set:

From this equivalence relation, we follow the procedure we used for Penrose Tilings to construct a Groupoid.

Differences Between Penrose and Ammann Tilings

• The process for iterating an Ammann tiling creates new edges, while the process for iterating a Penrose tiling does not. (Neither process creates new vertices.)

• Ammann tiles have no adjacency rules for assembly (though they will only fit together in certain configurations). Penrose tiles have marked vertices and edges to indicate proper assembly.

• No matter what the size of the tiling, Penrose tiles are always similar to the original two tiles (though they may each have a different scaling factor). Ammann tiles preserve all of the edge congruences, but only some of the angle congruences as the tiling is iterated. (Perhaps some of the iterations involve similar tiles.)

• It is not known whether the Ammann tiling is useful in studying quasicrystals.

What's Left to Do?

• What is the relationship between the tiles in T_n and T_{n + 1}?

• Are T_n tiles similar to T_{n + 2} tiles?

• Is there a bijection between X_p, the set of all sequences that represent Penrose tilings, and X_m, the set of sequences that represent Ammann tilings?

Works Cited

The construction of the Ammann tiling was taken from Tilings and Patterns by Grunbaum/Shephard.

The possible configurations of the Penrose tiling by rhombs were taken from Quasicrystals and Geometry by Marjorie Senechal.

The rest of this page is original work by Vivian Olsiewski Healey with guidance from Terry Bisson and BJ Kahng.

Related pages: Tilings, Penrose Tilings