Billiards in Nearly Isosceles Triangles

We prove that any sufficiently small perturbation of an isosceles triangle has a periodic billiard path. Our proof involves the analysis of certain infinite families of Fourier series that arise in connection with triangular billiards, and reveals some self-similarity phenomena in irrational triangular billiards. Our analysis illustrates the surprising fact that billiards on a triangle near a Veech triangle is extremely complicated even though Billiards on a Veech triangle is very well understood.


Introduction
Very little is known about irrational polygonal billiards. For instance, it is unknown if every triangular shaped table has a periodic billiard path. See [MT02] and [Tab95] for surveys about billiards, mainly rational. It has long been known that all acute and right triangles have periodic billiard paths; and the papers [VGS92], [HH00], [Hoo06], [Sch06a], [Sch06b] exhibit some infinite families of periodic billiard paths for some obtuse triangles. In particular, in [Sch06a], [Sch06b] it is shown that a triangle has a periodic billiard path provided that all its angles are at most 100 degrees.
Every isosceles triangle has a periodic billiard path of combinatorial length 4. This path is perpendicular to two sides of the triangle and looks like the letter V when the triangle is drawn with its apex pointing up. Unfortunately, this path disappears as soon as we perturb the triangle so that it is no longer isosceles. One might wonder if there is more to the story for such perturbations. Here is the main result of this paper.
Theorem 1.1 Any sufficiently small perturbation of an isosceles triangle has a periodic billiard path.

Overview of the Proof
The parameter space of (obtuse) triangles is an open triangle ∆ ⊂ R 2 , where the point (x, y) corresponds to the obtuse triangle whose small angles are x and y radians. To each infinite periodic word W , with digits in the set {1, 2, 3}, we assign the region O(W ) ⊂ ∆ as follows: A point belongs to O(W ) if W describes the combinatorics of a periodic billiard path in the corresponding triangle. By this we mean that we label the sides of the triangle 1, 2, and 3, and then read off W as the sequence of successive edges encountered by the billiard path. We call O(W ) an orbit tile and W a combinatorial type.
The open line segment {x = y} ⊂ ∆ parametrizes the obtuse isosceles triangles. We prove Theorem 1.1 by covering a neighborhood of this segment by orbit tiles. This innocentsounding approach gives rise to an extremely intricate covering involving infinitely many combinatorial types. As we go along, we will try to explain why the complexity seems necessary.
We define the Veech points V n = (π/2n, π/2n); n = 3, 4, 5... (1.1) These points are special because they correspond to triangles which have Veech's famous lattice property [Vee89]. First of all, we will prove the following result. Statement 2 of Theorem 1.3 takes care of all the Veech points except V 2 k for k = 3, 4, 5... Statement 1 suggests that perhaps the remaining points will be trouble. Indeed, computer evidence strongly supports the following conjecture.
Conjecture 1.4 For k = 3, 4, 5... no neighborhood of V 2 k has a finite covering by orbit tiles.
Remark: In [Sch06a] it is proved that no neighborhood of (π/3, π/6) has a finite covering by orbit tiles. Thus, the covering constructed in [Sch06a] and [Sch06b] for the "100 degree result"mentioned above is necessarily infinite, partly because of the "trouble spot" at the point corresponding to the (30, 60, 90)-triangle. In the same way, Conjecture 1.4 states that there are an infinite number of "trouble spots" at various Veech points.
Though we only need to deal with the points V 2 k to finish the proof of Theorem 1.1, we find it useful to treat the points V n in a uniform way. We decompose neighborhoods of the Veech points into quadrants. Let B(ǫ) denote the ball of radius ǫ, centered at the origin. Let B ±,± (ǫ) denote the intersection of B ǫ with the open (±, ±) quadrant. Now define N ±,± (n, ǫ) = V n + B ±,± (ǫ).
Theorem 1.5 For each n ≥ 4 there are words A n , B n , and C n , and some ǫ n > 0 such that • N −+ (n, ǫ n ) ⊂ O(B n ) • N +− (n, ǫ n ) ⊂ O(C n ).
The last statement is present to take care of the boundaries of the quadrants. See Figures 1.1  and 1.3 below. It is worth remarking that the words A n are part of a larger family discovered by Halbeisen and Hungerbühler [HH00]. Theorem 1.5 focuses our attention on the regions N ++ (n, ǫ). Theorem 1.5 and Conjecture 1.4 together imply that these regions do not have finite covers by orbit tiles, at least when n is a power of 2. We deal with all values of n at once, by introducing a doubly infinite family {W nk } of words. Here n = 4, 5, 6... and k = 0, 1, 2.... Figure 1.3 below shows some of the corresponding orbit tiles for n = 4. Theorem 1.6 For each n ≥ 3, there is some ǫ = ǫ n such that Theorems 1.2, 1.5, and 1.6 take care of neighborhoods of all points except V 2 and V 3 . The example we work out in the next chapter shows that V 3 ∈ O(W ) for a certain word W of length 22. See Corollary 2.4. Alternatively, Theorem 1.3 handles a neighborhood of V 3 . Finally, in [Sch06b] we proved that a neighborhood of V 2 , the point corresponding to the right-angled isosceles triangle, is contained in the union of 9 orbit tiles. This completes the proof of Theorem 1.1.

Some Pictures of the Tiles
We discovered all the results in this paper using our computer program, McBilliards, a welldocumented Java-based program which is publicly available. 1 The reader can see great pictures of our tiles using McBilliards. Here we reproduce a few of these pictures.  .1 shows a picture the first few orbit tiles in the A series. The horizontal grid lines have the form x 2 = π/n for n = 4, 5, 6... and the vertical grid lines have the form x 1 = π/n for n = 4, 5, 6. The right-angled tips of these tiles are the Veech points.
The tiles in the A series are also part of the tiles of we use to prove Theorem 1.2. Theorem 1.2 uses a double-infinite family {Y n,m } of words with m ∈ N and n ∈ {2, 3, 4...}. We have A n = Y n,1 . For n fixed, the tiles {Y n,m } live between the two consecutive Veech points V n and V n+1 . Figure 1.2 shows some of these tiles.
The convergence is such that any compact subset Q ′ n ⊂ Q n is contained in S n (O nk ) for k sufficiently large in comparison to n.
Proof of Theorem 1.6: Fixing n, Theorem 1.7 implies that there are constants 0 < ǫ 1 < ǫ 2 such that, for k sufficiently large, O(W nk ) contains the set V n + Λ k , where Λ k is the convex hull of (ǫ 1 /k 2 , 0); (ǫ 2 /k 2 , 0); (0, ǫ 1 /k 2 ); (0, ǫ 2 /k 2 ) for j large. But the union of the sets V n + Λ k covers N(n, ǫ) for some ǫ > 0. These sets "bunch up" as k → ∞. Compare Figure 1.3. So, Theorem 1.7 proves Theorem 1.6. ♠ Our technique for proving Theorem 1.7 involves looking at the Fourier transforms of the analytic functions which define the edges of the orbit tiles of interest to us. The Fourier transforms of these functions are functions defined on Z 2 . It turns out that the supports of these Fourier transforms grow "linearly" with the parameter k in a way we make precise in §6. To deal with with the situation as k → ∞ we prove the Quadratic Rescaling Theorem, a result which describes the asymptotic limits of a family of functions which vary with the prescribed growth. One of the main technical innovations in the paper is a combinatorial method for understanding such growing families of Fourier series. Our technique seems to be more general than the application we give here, but so far this is the main application.

Paper Outline
In §2 we will give background information about triangular billiards, and in particular discuss how one computes the orbit tile O(W ) based on the combinatorics of the word W . All of the constructions in §2 are programmed into McBilliards. The interested reader can see these constructions in action when using the program.
In section 3, we prove theorem 1.2.
In §4 we will prove Theorem 1.5. §5-8 are devoted to the proof of Theorem 1.7. In §5 we will introduce our words W nk , and prove some preliminary results about the orbit tiles O nk := O(W nk ). In particular, we will isolate a region R nk ⊂ ∆ with the property that (independent of n and k) a certain 16 functions define O nk ∩ R nk .
In §6 we will prove the Quadratic Rescaling Theorem, a result that is designed to analyze infinite families of defining functions, such as the 16 families we isolate in §5. In §7 and §8 we will use the Quadratic Rescaling Theorem to finish the proof of Theorem 1.7.
In §9, which is logically independent from the rest of the paper, we prove theorem 1.3. This classifies the Veech triangles V n which lie in the interior of an orbit tile.

Billiard Paths and Defining Functions 2.1 Unfoldings
The unfolding of a word W with respect to a triangle T , which we denote by U(W, T ), is the union of triangles obtained by reflecting T out according to the digits of W . This classic constructed is discussed in detail in [Sch06a] and in e.g. [Tab95]. We will persistently abuse our notation in the following sense: A point X in parameter space represents a triangle T = T X . We will often write U(W, X) in place of U(W, T ).
There is a sequence of vertices which runs across the top of U(W, T ). We call these the top vertices and label them a 1 , a 2 , ... from left to right. There is a sequence of vertices which runs across the bottom of U(W, T ) and we label these b 1 , b 2 , ... from left to right.   The first side of U has been highlighted in both examples. W represents a periodic billiard path in T iff the first and last sides of U(W, T ) are parallel and the interior of U(W, T ) contains a line segment L, called a centerline, such that L intersects the first and last sides at corresponding points. In both examples above, the first and last sides are parallel. However, the centerline only exists for Figure 2.1. In particular, Figure 2.1 shows that the given word describes a periodic billiard path for the triangle corresponding to V 3 . As in Figures 2.1 and 2.2 we always rotate the picture so that the first and last sides are related by a horizontal translation. We call this horizontal translation the holonomy.

Stability and Hexpaths
A word W is called stable if the first and last sides of U(W, T ) are parallel for any triangle T . This implies that O(W ) is an open set. In this section we will explain a combinatorial criterion for stability. The proof is well known, and we omit it. See [Sch06a] for details. Let H 0 be the outer hexagon shown in Figure 2.3. The shape if H 0 is a bit strange, but the inscribed hexagon has vertices on the integer lattice Z 2 as shown. Also, H is well related to a square of side-length 2, as shown on the right hand side of Figure 2.3. The sides of H 0 are divided into 3 types, according to their label. Let H denote the tiling of R 2 by translates of H 0 . By H we really mean the union of edges of the tiling. By construction, the midpoints of edges in H lie in Z 2 .
Given the word W , we can draw a path in H by following the edges as determined by the word: we move along the dth family when we encounter the digit d.
This condition in the Hexpath Lemma is equivalent to the better known condition, which appears as lemma 3.3.1 in [Tab95]. We have restricted this lemma to our context. Lemma 2.2 A word W is stable iff the number of times each letter ℓ = 1, 2, 3 appears in an odd position in W equals the number of times ℓ appears in an even position.
This condition can easily be verified for the example we have been considering. In addition it happens for squares of words of odd length. Corollary 2.3 (Odd squares are stable) If W is a word of odd length, then W 2 is stable. Now we can take care of the loose end from the introduction.

Corollary 2.4 V 3 is contained in the interior of an orbit tile.
Proof: Figure 2.1 shows that the given word describes a periodic billiard path for the triangle corresponding to V 3 . Figure 2.4 below shows that the hexpath corresponding to this word is closed. Hence, the corresponding billiard path is stable. ♠

The Squarepath
It turns out that the hexpath H(W ) contains precisely the same information as a certain rectilinear path, which we call the squarepath. Each vertex of the hexpath has a unique type 3 edge emanating from it. The squarepath is obtained by connecting the midpoints of these type-3 edges together, in order. We denote the squarepath by Q(W ). We can also define similar paths based on the edges of type 1 or 2. These paths are somewhat more complicated, though they will be of theoretical importance for us. In practice, however, we will always try to work with the type 3 edges.
If we mark off points on the squarepath at integer steps (starting with a vertex) the resulting points are naturally in bijection with the type 3 edges of the unfolding. In the next section we will elaborate on this bijection.  It is possible to reconstruct H(W ) from Q(Q), when Q(W ) is a closed loop. When Q(W ) is embedded, this loop bounds a finite union of squares. We simply replace each square by the associated hexagon. Then H(W ) is the boundary of the union of hexagons. In general, H(W ) is the union of all the edges of H which intersect Q(W ). There is a natural ordering to these edges, and so the union of all these edges naturally has the structure of a closed loop.
It turns out that there is a simple algorithm for deducing the combinatorics of the unfolding from the squarepath. Say that a k-dart is a union of k isosceles triangles, arranged around a common vertex, in the pattern shown in Figure 2.6 for k = 2, 3, 4.. A k-dart is just an unfolding with respect to either the word (13) k−1 1 or the word (23) k−1 2. Figure 2.6: k-darts for k = 1, 2, 3 We say that the 3-spine of the dart is the union of the two outermost long edges. We have highlighted the spines of our darts in Figure 2.6.
The relation of U(W, * ) to Q(W ) is as follows: • The maximal darts of the unfolding are in bijection with the edges of the square path.
(The maximal k-darts correspond to edges of length 2k.) The maximal darts are glued together along their 3-spines.
• Two consecutive maximal darts lie on opposite sides of their common 3-edge iff Q(W ) makes a northwest or southeast turn at the vertex corresponding to this 3-edge.
To make this work precisely, we need to take the infinite periodic continuation of U, or else identify the first and last sides of U to make an annulus. As it is, the reader needs to take special care in figuring out how the rightmost maximal dart fits together with the leftmost one. We have included a copy of Figure 2.2, except with the spines of the maximal darts drawn in black. See Figure 2.7. We have taken a lot of trouble to describe the squarepath and its relation to the hexpath and the unfolding because we plan to specify all our words in terms of their squarepaths. Using the square path gives a very simple description of the word, and lets the reader best see the patterns which arise in our families.

Edge Labellings
We label each edge of H by the coordinates of its midpoint. This labelling is canonical, once we decide which point of Z 2 gets labeled (0, 0). The McBilliards convention is to assign the label (0, 0) to the edge of H(W ) corresponding to the last digit of W . This edge is the leftmost edge of the unfolding U(W, * ). In Figure 2.8 we have labeled the origin and several nearby points. We identify e with its label. Our labelling has a geometric significance. Let X = (x 1 , x 2 ) be a parameter point and let T X be the corresponding triangle. Let e 1 and e 2 be two edges of the unfolding U(W, T X ). Let θ(e 1 , e 2 ) be the counterclockwise angle through which we must rotate e 1 so as to produce an edge parallel to e 2 . We take θ mod π, so that the orientations of e 1 and e 2 are irrelevant. Then, as is easily established by induction: θ(e 1 , e 2 ) = X · ( e 2 − e 1 ). (2.1)

Defining Functions
We frequently write for notational convenience. Given two points p, q ∈ R 2 we write p ↑ q; p q; p ↓ q iff the y coordinate respectively is greater than, equal, or less than the y coordinate of q. Suppose that p and q are two vertices of our unfolding. In this section we will give the formula for a function F = F p,q which has the property that F = 0 iff p q. These defining functions are computed purely from the word W . The orbit tile O(W ) can be described as the region where the defining functions corresponding to the (a i , b j ) pairs are all positive. The edges of O(W ) is defined in terms of the 0-level sets of the defining functions.
For any d ∈ {1, 2, 3} there is an infinite, periodic polygonal path made from type-d edges of the infinite periodic continuation of U(W, T ). The image of this path in U(W, T ) is what we call the d-spine. We have already encountered the 3-spine: It is the union of the 3-spines of the maximal darts of U(W, * ). Let e 0 , ..., e m denote the list of edges, ordered from left to right, which appear in the d-spine. We say that the vertices p and q are d-connected if there is a polygonal path of type-d edges connecting p to q. In this case, let f 0 , ..., f n denote these edges, ordered from left to right. We order p and q so that p is the left endpoint of the d-path and q is the right endpoint. The 3 thick grey edges in Figure 2.9 show the 3-path connecting p = b 4 to q = a 6 .
We define We will explain the global sign in front of P below. The reason for the general alternation of the signs is explained in [Sch06a]. Our functions have the following geometric interpretation: If we normalize so that the d edges have length 1 and rotate U(W, T ) so that the first edge is horizontal, then ±P (X) is the vector pointing from p to q and Q(X) is the translation vector. Therefore, For the above example the sign in front of P turns out to be a (+). (See below.) We therefore have Here is what we call the function tableau for P .
(+) 4 −1 6 −1 6 −3 When we reconstruct the function from its tableau, we use the convention that the signs of the terms alternate. The (+) of (−) indicates the global sign in front of P . When d = 3, we can represent both P and Q in terms of the squarepath. First of all, the list of vertices of Q is precisely the function tableau for Q. The situation for P is more involved: The edges of U(W, * ) are in canonical bijection with the edges which emanate from the vertices of the hexpath. Say that a 3-edge of Q is a starter if it corresponds to an edge of U(W, * ) which is incident to p. Say that a 3-edge of H is a finisher if it corresponds to an edge of U(W, * ) which is incident to q. Let P denote the shortest sub-path of Q d (W ) whose initial endpoint is the midpoint of a starter and whose final endpoint is the midpoint of a finisher. Then the function tableau for P is just the list of coordinates of the vertices of P . Figure 2.10 shows the paths corresponding to P and Q. The origin is marked with a grey dot.
We can interpret the path Q as a function from Z 2 to Z, as follows. We alternately color the vertices encountered by Q black and white, starting with white. Q assigns the value x 1 − x 2 to X ∈ Z 2 if x 1 white vertices of Q coincide with X and if x 2 black vertices of Q coincide with X. We make the same definition for P , except that we have to take care whether or not to color the first vertex encountered by P black or white. (See below.) With this interpretation, Q is the Fourier series of Q. (2.5) The same goes for P and P .
The Global Sign: This discussion supposes that F > 0 if q ↑ p. (As above, p is on the left.) We also suppose that the initial vertex of P is also a vertex of Q. In this case, the sign in front of P is (−1) u , where u is the number of vertices of Q (starting with the first one) which lie before the first vertex of P . That is, the initial vertex of P should get the same color whether it is considered a vertex of P or a vertex of Q. For example, we can see from Figure 2.10 that u = 2 and so the sign is a (+). This rule has a simple geometric proof: When p and q are the first and last vertices of the 3-spine of U, then P = Q and so the sign definitely should be a (+). If we move q along the 3-spine, the sign does not change, by "continuity": Moving either vertex by 1 "click" should produce a nearby value for P . However, moving the p vertex changes the global sign, given the form of Equation 2.3. In general the first vertex of P need not be a vertex of Q. This irritating situation does not arise in this paper. McBilliards has a general algorithm which correctly determines the sign in every possible case.  Recall that the unfolding U can be written as a union of maximal darts. We say that a vertex of U is inferior if it is an inferior vertex of one of the maximal darts. Let δ(W ) denote the largest k such that U(W, * ) contains a k-dart. Here is the main result of this section:

The Dart Lemma
.
Suppose also that all the top superior vertices of U(W, X) lie above all the bottom superior vertices of U(W, X). Then X ∈ O(W ).
Remark: There is a more restrictive angle condition that almost immediately guarantees that the maximal darts are acute. This condition is given by This is precisely the condition we will use in the proof of Theorem 1.2. Our condition is weaker than this and does not, in itself, guarantee that the maximal darts are acute. However, our weaker condition combines with the second hypothesis of the Dart Lemma to establish the acuteness. See the very end of our proof. We mention this because the discrepancy is likely to otherwise cause confusion.
We will prove the Dart Lemma in several stages. We say that the base of a dart is the vertex which is common to all the triangles. The base is denoted by a big black dot in Figure  2.11. We say that the centerline of the dart is the ray of bilateral symmetry, emanating from the base. The centerline is indicated by a ray in Figure 2.11. We say that the dart points up if the ray points upward, and points down if the ray points downward. Let D V denote the union of outermost edges of D which are not the longest edges. This set is highlighted in Figure 2.11. We say that D is acute if D V makes an acute angle towards the centerline of the dart. Figure 2.11 shows an acute dart. Proof: We will prove this statement for the top vertices. The proof for the bottom vertices is the same. Let v be an inferior top vertex of U. Then there is some maximal dart D such that v is an inferior vertex of D. Each edge of D, except possibly the edges on the 3-spine, is an edge of reflection of U. Thus, the inferior vertices of D all have the opposite type (top or bottom) from the base. Likewise for the superior vertices of D. Hence, the inferior vertices and the superior vertices of D all have the same type. Since one inferior vertex of D is a top vertex, the base of D is a bottom vertex. Since U is controlled, D points up. Lemma 2.6 now implies that v is higher than one of the superior vertices v ′ of D. As we already mentioned, v ′ is also a top vertex of U. Hence, we have found another top vertex, v ′ , which is lower than v. ♠ To finish the proof of the Dart Lemma, we just have to establish that U = U(W, X) is a controlled unfolding. Let D be a maximal dart of U. Assume without loss of generality that the basepoint of D is a bottom vertex. Each edge of D V is an edge of reflection of U. Hence the endpoints of D V are top vertices. All these vertices are superior vertices. Hence, the endpoints of D V lie above the basepoint of D V . Our restriction on X guarantees that the angle of D V is at most 2π. Hence D V must actually make an acute angle, since both its edges point up. The centerline lies between these two up-pointing edges. Hence D itself is up-pointing. Since D is arbitrary, we see that U is controlled. This completes the proof of the Dart Lemma.

Pseudo-Parallel Families
Suppose that e is an edge of the unfolding U(W, * ). When U(W, X) is rotated so that it has horizontal holonomy, the line containing e is parallel to the complex number E( e · X)U Q (X). (2.6) We say that the edges {e 0 , ..., e n } form a pseudo-parallel family relative to the point X 0 if the dot product e j · X 0 is independent of j. In this case, the edges e 0 , ..., e n are all parallel in U(W, X 0 ). We assume that these edges have negative slope in U(W, X 0 ). The points e 0 , ..., e n must lie on a line segment in R 2 . In our examples in this paper, the line in question always has slope −1 because X 0 lies on the isosceles line. We order our edges so that e 0 , ..., e n appear in order on the line.
Let R ′ (e j ) denote the region in parameter space such that e j has negative slope in the unfolding. Let R(e j ) denote the path connected component of R ′ (e j ) which contains X 0 . Proof: We think of {X t } as a path in R(e 0 ) ∩ R(e n ) which connects X 0 to some other point X 1 . Let S 1 denote the unit complex numbers and let E : R → S 1 be the universal covering map. For each object z ∈ S 1 we let z denote the lift to R, so that E( z) = z. In particular, we define Let I t ⊂ R be the interval whose endpoints are E 0 (t) and E n (t). By convexity E j (t) ⊂ I t for all t. The edges e 0 (t) and e n (t) have negative slope for all t. Hence I t has length less than π/2 for any t ∈ [0, 1]. Hence E t (j) lies in the arc I t , which has length less than π/2. If we rotate S 1 so that U(t) = 1 then the endpoints of I t , namely E 0 (t) and E j (t), are both contained in the same negative quadrant of R 2 . (Either (−+) or (+−).) Hence I t is contained in one of the negative quadrants. Hence E j (t) is also contained in one of these quadrants. That is, e j has negative slope for any parameter value t. ♠ 3 Proof of Theorem 1.2 Our proof of Theorem 1.2 is based on the 2 parameter family Y m,n of odd-length words.
Y n,m = 1(W n ) m 32; W n = (31) n−1 (32) n−1 (3.1) Figure 1.4 of the introduction shows the orbit tiles for some of the words Y n,m . The family W n , which we call the unstable family, describes unstable periodic billiard paths in certain isosceles triangles of interest. The square words Y 2 m,n are stable by Corollary 2.3. Theorem 1.2 is therefore a consequence of the following result.
Theorem 3.1 For every integer n ≥ 2 and real number x so that π 2n+2 < x < π 2n , there is a periodic billiard path in T x with combinatorial type Y n,m for some m ∈ N .
Here T x denote the obtuse isosceles triangle corresponding to the point (x, x) in the plane. The two small angles of T x have measure x-radians.

The Unstable Family
Proposition 3.2 W n describes a periodic billiard path in T x for all x < π 2n−2 .
Proof: The unfolding for the word W n consists of two maximal n−1 darts. Given our bounds on x, we satisfy the hypotheses given in the remark immediately following the statement of the Dart Lemma. Thus, it suffices to consider the superior vertices of the unfolding. With this choice, the unfolding is horizontal as desired. (That is, A D.) Figure 3.1: An unfolding for the word W 5 . One period is shown, which begins at edge AE and ends at the parallel edge DH.
By symmetry, each point has a partner-point at the same height:

A D B C E H F G
Thus, it is sufficient to concentrate on the central rhombus, CDF E. Given a vector v ∈ R 2 , we use d(v) to denote the angle of the vector made with the horizontal. It is sufficient to check that the four vectors That is, for v equal each of those four vectors, we must have 0 < d(v) < π. We compute In all cases, we have 0 < d(v) < π for 0 < x < π 2n−2 . ♠

The stable family Y n,m
The word Y n,m has an additional special symmetry. If you write Y n,m in reverse and swap the letters 1 and 2, you get Y n,m back. Given a word W , let W denote W written in reverse with 1 swapped with 2. There is some word W = W m,n such that Remark: It is a consequence of work in [Hoo07] that every stable periodic billiard path in an isosceles triangle has an combinatorial type W with the symmetry W = W . This fact, however, is not necessary for our proof here.
We now record some special properties of words having the form given by the right hand side of Equation 3.2.
Proposition 3.3 Let Y be a word of the form Y = 1W 3 W 2, and T be an obtuse isosceles triangle. Consider the unfolding U(Y 2 , T ) chosen so that the translation bringing the first edge to the last is horizontal. Then the long edge (edge 3) of the first triangle in the unfolding is horizontal.
Proof: Consider the bi-infinite repeating word Y . This word has some symmetry, which is revealed by expanding the word out.
Reflection in the vertical line above swaps the letters 1 and 2 while preserving 3. This is precisely how the reflective symmetry of the isosceles triangles permutes the labeling of the sides. Thus, this symmetry extends to the bi-infinite unfolding U(Y , T ). The direction of the holonomy of U(Y 2 , T ) must be the eigenvector corresponding to eigenvalue −1 of the reflective symmetry of U(Y , T ). But this reflection is just the reflective symmetry of the first triangle in the unfolding. So these two directions are parallel. ♠ We will use the following principle for detecting our billiard path. Recall that side 3 denotes the long side of an isosceles triangle.
Proposition 3.4 Suppose that a billiard path in an obtuse isosceles triangle starts out parallel to side 3, and has initial combinatorial type 1W 3, where the final 3 corresponds to an edge which the path hits at the midpoint, M. Then the billiard path is closed and has combinatorial type 1W 3 W 2.
See Figure 3.2 for a case when this proposition applies. Proof: The trajectory t 1 described in the proposition lies within the unfolding of the initial word 1W 3 and then hits M. The unfolding for the word 1W 3 W 2 has 180 degree rotational symmetry φ around the point M. Thus, the longer trajectory t 2 = t 1 ∪ φ(t 1 ) lies within the unfolding of 1W 3 W 2. Now consider the unfolding of the even length word (1W 3 W 2) 2 . This unfolding has vertical reflective symmetry ρ which swaps the two halves of the word. (It is vertical assuming the trajectory is horizontal.) The trajectory t 3 = t 2 ∪ ρ(t 2 ) lies within the unfolding of 1W 3 W 2. ♠ We will break the proof of Theorem 3.1 into two cases. The first case is easiest.
Lemma 3.5 For each x satisfying π 2n+1 ≤ x < π 2n there is a periodic billiard path in T x with combinatorial type Y n,1 .
Proof: Given the triangle T x , unfold the triangle according to the square of the word Y n,1 as in Figure 3.2. Let M 1 be the midpoint we must hit. This is the first midpoint of a long side which is the fixed point of a 180 degree rotational symmetry of the the bi-infinite unfolding, U(Y n,1 , T x ).
We coordinatize the unfolding so that M 1 is given coordinates (0, 0). We will show that all the top vertices have positive y-coordinate, and all the bottom vertices have negative ycoordinate. Regardless of n, the Dart Lemma tells us that most of the vertices are irrelevant. It is enough to prove this statement for those vertices, who are given names in Figure 3.2. We have named four vertices A, B, C and D. The other vertices are either images of these under the rotational symmetry about M 1 (denoted by * ′ ), images under reflection in the vertical line through D ′ (denoted by * r ), or images under the composition. So, it is enough to show that the statement is true for the vertices A, B, C and D.
The points A and B have the same y-coordinate by Proposition 3.3. M 1 lies below them, because angle ABM 1 = 2nx < π. Also angle ABC = (2n + 1)x ≥ π, so C has y-coordinate greater than or equal to the y-coordinates of A and B. Finally, M 1 lies closer to B then D. Furthermore, the vector − −− → BM 1 is closer to horizontal than the vector − − → DB. ( DBA = x = M 1 BC, but the horizontal direction lies strictly between the directions of − −− → BM 1 and − − → BC.) Thus the y-coordinate of D must be negative. ♠

Remark:
The words Y n,1 are the same as the words A n−1 , which appear in Figure 1.1 and play a prominent role in Theorem 1.5.
The second case is more complicated. While we could give a constructive proof, as above, we find that a non-constructive proof clarifies the situation. To illustrate this case, we consider the word Y 4,2 and the triangle T in Figure 3.3. The unfolding U(Y 4,2 , T ) depicted in this Figure contains a horizontal segment joining the first triangle to the midpoint of the long side. This segment hits the sequence of sides 1(31) 4 (32) 4 3. Let W = (31) 4 (32) 4 . By Proposition 3.4, there is a periodic billiard path in T with combinatorial type Y 4,2 = 1W 3 W 2. The significant point is that by Proposition 3.4, we only need to consider the unfolding for an initial subword. Lemma 3.6 For each x with π 2n+2 < x < π 2n+1 , there is a periodic billiard path in T x with combinatorial type Y n,m for some m ∈ N .
Proof: Consider the unfolding of T x according to the infinite word 1(W n ) ∞ . See Figure 3.4. We normalize the unfolding so that the initial long side of T x is horizontal.
We will show that there is some index m such that M m lies below all preceding top vertices and above all preceding bottom vertices. Here the points M 0 , M 1 , ... are the midpoints of some of the long segments. See  Many of the vertices in Figure 3.4 are given names. The unlabelled vertices are inferior, and may be ignored by the Dart Lemma. Understanding this unfolding is made much easier by the fact that W n is the combinatorial type of a periodic billiard path in T x . See Proposition 3.2. We consider the vector This vector points in the direction (n + 1)x − π 2 (measured relative to the horizontal in polar coordinates). In particular, v has positive y-coordinate, since x > π 2n+2 . To compute this direction, note that − −− → S 1 M 1 points in the direction 2nx and − −− → S 1 M 2 points in direction 2x. We always normalize so that the long side has length 1. This gives us the formula v = 1 2 (cos 2x, sin 2x) − 1 2 (cos 2nx, sin 2nx). Now we will eliminate some top vertices. The vector It follows that A 1 has the least y-coordinate of any vertex in the unfolding 1(W n ) ∞ . Now we will eliminate some bottom vertices.
It follows that the bottom vertex with greatest y-coordinate in the unfolding up to the appearance of M m is either D or B m . (We ignore the fact that when m is even B m appears later in the unfolding than M m .) Letting P y denote the y-coordinate of an arbitrary point P , we want to show that for some m. When m is even, So, for any choice of m, the first inequality holds. The first endpoint M 0 lies below D since Here g(x) is independent of i. We compute The formula for f follows from the fact that the length of the short side is 1 2 cos x , and − − → P D and − − → P A 1 point in directions π + x and π + (2n + 1)x respectively. A sufficient criterion for the second equation is that g(x) < f (x). That this is true follows from some trigonometry. First we reduce f (x) and g(x) to more convenient forms.

The A Family
Here we introduce the words {A n } for n ≥ 2. These words already appear in [HH00], and their analysis is quite easy. A n is the square of a word of odd length. Listing out the first few words explicitly and then writing the general pattern, we have:    If X is a parameter point sufficiently close to V n then the lowest top vertices of U(A n , X) remain a 1 and a 2 and the highest bottom vertices remain b 2n and b 2n+1 . When X ∈ N −− (n, ǫ) the 3-spine for U(A n , X) is no longer a straight line segment, but rather makes a zig-zag. Both obtuse angles in the unfolding are slightly smaller, and this causes the 3 spine to make an acute angle in the directions of the centerlines of the maximal darts, as shown in  From this geometric picture we see easily that the a 1 and a 2 lie above b 2n and b 2n+1 for points in N −− (n, ǫ n ).
As an alternate argument we note that, since A n is an odd square, the unfolding U(A n , * ) has glide-reflection symmetry. Thus, if a i is the lowest top vertex then b 2n+i−1 is the highest bottom vertex. Thus, it suffices to show that a 1 ↑ b 2n and a 2 ↑ b 2n+1 . We compute the defining function F for (a 1 , b 2n ) and find that F (x 1 , x 2 ) = −4 sin 2 (nx) sin(2nx). (4.2) For (x 1 , x 2 ) near V n = (π/2n, π/2n), the above expression is negative iff x 2 < π/2n. That is, a 1 ↑ b 2n iff x 2 < π/2n and x 2 is sufficiently close to π/2n. The calculation for the pair (a 2 , b 2n+1 ) yields the same result, but with x 1 and x 2 interchanged. This takes care of the first statement of Theorem 1.5.

The B Family
We will show that By symmetry, Our argument will show that the two segments bounding N −− (n, ǫ n ) (except for V n itself) are respectively contained in O(B n ) and O(C n ). This takes care of the fourth statement of Theorem 1.5. The word B n has length 40n − 60. This word is determined by its squarepath Q n := Q(B n ), which we now describe. We will draw Q 4 and Q 5 , with the understanding that Q n+1 is obtained from Q n by lengthening each edge by 2 units. The small grey squares in Figure  4.6 have edgelength 2. We have drawn some of the edges in grey to help the reader parse the loops. These loops are homeomorphic to figure 8 curves. The grey dot indicates the origin.  Equation 4.5 shows three lists. The list L j is the list of jth coordinates of the successive vertices of the squarepath. The list L computes either L 1 + L 2 mod 4n or L 1 + L 2 + 2n mod 4n depending on the parity of the vertex. (4.6) We will often write ω = ω n when the dependence on n is clear. The holonomy of U(B n , V n ) is obtained as the alternating sum of the vertices of Q. Since we are evaluating this sum at V n , each vertex contributes some power of ω to the sum. The list L above tells us which power. In deriving this list, we used the relation ω a+2n = −ω a . Note that L is independent of n. We have: This agrees with the McBilliards Calculations. An easy calculus argument shows that Q(V n ) lies between 1 and ω on the unit circle.

Reducing to Six Vertices
Any point X sufficiently near V n satisfies the hypothesis of the Dart Lemma with respect to B n . Hence, we just have to show that all the top superior vertices of U(B n , X) lie above all the bottom superior vertices of U(B n , X) for X ∈ N −+ (n, ǫ) when ǫ is sufficiently small. In this section we will reduce this problem to checking 6 superior vertices.
Since U n decomposes into 20 maximal darts, there are at most 80 superior vertices, independent of n. Each maximal dart has 2 (O)uter superior vertices and 2 (I)nner superior vertices. The outer superior vertices lie on the 3-spine and the inner superior vertices do not.
The superior vertices in each maximal dart are naturally ordered from left to right. There are two superior vertices on the (L)eft and two on the (R)ight. We denote the 4 superior vertices of the Kth maximal dart (perhaps redundantly) by (K, L, O); (K, L, I); (K, R, I); (K, R, O). (4.8) Sometimes we will decorate our notation with an asterisk to indicate whether it is a top vertex ( ) * or a bottom vertex ( ) * . A leader is either a lowest top vertex or a highest bottom vertex. Here is the main result of this section: Moreover, these points all have the same height.
Any top superior vertex a 2 which is not on our list should lie above b 1 = (1, L, I) * . We will symbolically compute for such pairs and show that the imaginary part of this function is positive. Hence . Finally we will show that the P (V n ) is a real multiple of Q(V n ) when P is defined relative to the pair (b 1 , c) and c is on the list given in Lemma 4.1. The key to our calculations is a slick procedure for computing these points.
In computing our points we will slightly modify the method described in §2, so as to use the 3-spine as much as possible. Unfortunately, it is not possible to directly connect all the points of interest to us by a 3-path. The work-around we explain below works for every point except for a 1 = (1, L, O) * , which we easily observe to lie above the points listed in Lemma 4.1. (The edge connecting (1, L, O) to (1, L, I) has negative slope.) For the rest of the superior vertices we do the following: • Connect (1, L, I) * to (1, L, O) * using the common edge e 0 .
• Connect (I, L, O) * to a point p ′ on the 3-spine which is adjacent to p, using the fewest number of edges e 1 , ..., e s from the 3-spine.
• Connect p ′ to p by the edge e s+1 which is incident to both vertices.
Once p ′ is determined, there are either 2 or 3 choices for p. To use an analogy, the 3-spine is like the highway and the other edges we use are like the off and on ramps. Our first step is to get onto the highway using e 0 . Then we drive along the highway using e 1 , ..., e s . At this point we can either take one of the off-ramps and stop the car or else go one more mile and stop the car, depending on our final destination.  We normalize so that the type 3 (long) edges of our triangles have length 1. It follows from the Law of Sines that the short sides have length (4.9) The vector that points from (1, L, I) * to p is (4.10) Referring to Figure 4.8,, the number δ is −1 (bottom) or 0 (middle) or 1 (top) depending on which of the three choices we make for e s+1 . Here L refers to the labeling in Equation 4.5.
To demonstrate Equation 4.10, we note that the three paths suggested by Figure 4.8 lead to the three sums Let's concentrate on the right sum. We will use the notation x → y to denote that Im(x) and Im(y) are positive multiples of each other. Our middle expression simplifies to Recalling our formula for the holonomy, we compute that Hence Im(F (V n )) > 0. The point of U n corresponding to the third sum above is (3, R, I) * , and this shows that (3, R, I) * ↑ (1, L, I) * . It turns out that our sums always lead to the general expression  In listing the results of our calculations it suffices to list vector (a 1 , a 2 , a 3 ). For each δ ∈ {−1, 0, 1} and each β ∈ {1, ..., 20} we compute P (δ, s)Q and extract the coefficient vector (a 1 , a 2 , a 3 ).
All the expressions on the first two tables correspond to bottom vertices. An inspection of  (4.17)

The End of the Proof
Let H ij be the defining function which measures the height of α i minus the height of β j , when the unfolding is normalized to that the long edges have length 1. Below we will prove Lemma 4.2 For every relevant pair of indices, we have There is equality in the second equation iff (i, j) = (1, 1).
Note that H ij (V n ) = 0. Lemma 4.2 therefore says that H ij (X) > 0 provided that (i, j) = (1, 1) and X ∈ N −+ (n, ǫ n ) for sufficiently small ǫ n . If we knew that H 11 vanished identically on the line x 1 = π/2n then we could conclude the same result for (i, j) = (1, 1). Before proving Lemma 4.2 we will take care of the exceptional pair of indices. Proof: We will draw the picture for the case n = 4, but the phenomenon we describe is completely general. Let P and Q and F be the defining functions associated to our two points. Figure 4.9 shows the paths Q and P . Note that P covers half the vertices of Q and the vertices in the complement Q − P are, as a subset of Z 2 , isometric to the vertices of P . The isometry is given by the translation (x 1 , x 2 ) → (x 1 + 2n, x 2 ). Taking care to get the sign right, we see that Q = P + P ′ , where there is a bijection between the terms of P and the terms of P ′ , having the form When x 1 = π/2n we see that corresponding terms take on the same value. Hence Q = 2P when x = π/2n. But this means that F ( π 2n , x 2 ) = 0. ♠

Variation of Edgelength
In proving Lemma 4.2 we will connect various vertices of the unfolding together by the same sorts of paths we used in the proof of Lemma 4.1. These paths mainly involve the long edges, which all have unit length, but sometimes they involve a short edge as well. Even though we are evaluating the derivatives of our defining functions at V n , we still need to understand how these short edges vary in length for points nearby V n . In this section, we will deal with this issue.
Suppose T is a triangle with small angles x 1 and x 2 , normalized so that the long side of T has unit length. Let l j denote the length of the side of T which is opposite the x j angle. Of course, l j depends on the parameters x 1 and x 2 . When x 1 = x 2 we have l 1 = l 2 . When x 1 = x 2 = π/2n we have λ := l 1 = l 2 = sin(x 1 ) sin(x 1 + x 2 ) = sin(π/2n) sin(π/n) = 1 2 cos(π/2n) = 1 ω + ω −1 . (4.18) As usual, ω = E(π/2n). Here λ is as in Equation 4.10. Our calculations below require the quantities: (4.19)

Proof of Lemma 4.2
We define F α,j = height(α j ) − height(a 1 ); F β,j = height(β j ) − height(a 1 ). (4.20) here a 1 = (1, L, O) * . Again, we measure these heights when the U n is normalized so that the long edges are unit length. We have the obvious equation (4.21) We will deduce Lemma 4.2 from our computations of F α,i and F β,j . In our proof of Lemma 4.1 we constructed a path from b 1 = (1, L, I) to and given point p. The first edge of this path joined b 1 to a 1 . So, the path we use to connect a 1 to p is just the same one we used above, except with the first edge chopped off. In describing our paths, we let Y k denote the path made from the first k edges of the 3-spine. We let e ± k denote the short edge such that e ± k = e k + (±1, 0). Here e k is the kth edge of the 3-spine. (The correspondence e → e is discussed in detail in §2.) The three edges e − k , e k , e + k correspond to 3 consecutive horizontal dots in Figure 4.8 With this notation, we have: • The path connecting a 1 to α 1 is Y 13 .
• The path connecting a 1 to α 2 is Y 5 ∪ e + 6 . The short edge has type 2.
• The path connecting a 1 to α 3 is Y 9 ∪ e + 10 . The short edge has type 2.
• The path connecting a 1 to β 1 is Y 3 .
• The path connecting a 1 to β 2 is e − 1 . This (short) edge has type 1.
• The path connecting a 1 to β 3 is Y 15 ∪ e − 16 . This edge has type 2.
We will leave the details of our calculation to Mathematica, but here we outline the main points. In each case we F = P Q, so that F = Im( F ). By the product rule we have (4.22) We evaluate P and Q using Equation 4.10 (without the first term). We evaluate ∂ j P and ∂ j Q essentially by differentiating Equation 4.10 (without the first term.) We now explain how the differentiation works. Our calculations use the lists from Equation 4.5.
Let R be one of the expressions we want to differentiate. If Y k appears in the definition of the path associated to R then we see a contribution of in the expression for ∂ j Y (V n ). In conjugating (for the case R = Q) we simply reverse the signs of the list of numbers in L. For instance 2i ω 3 [(−4 − 4ω 2 − 7ω 4 − 3ω 6 )n + (10 + 14ω 2 + 16ω 4 + 8ω 6 )]. (4.23) If we see e ± k in our expression, and this edge has type 1, then we see a contribution of in the expression for ∂ 1 R(V n ) and a contribution of in the expressions for ∂ 2 R(V n ). If e ±1 k has type 2, then we see the same contributions, but with λ 1 and λ 2 switched.
These are the ingredients for our calculations. We let H ij = F α,i − F β,j . When we compute these quantities using the expressions above, we find that the result always has the form Here f is some polynomial in ω, ω −1 and n which is linear in n. (This polynomial, and the exponent a, both depend on the indices i, j, k.) The first case occurs 5 times and the second case occurs 13 times.
Since we only care about the sign of the imaginary part of ∂ i H jk we clear denominators by multiplying the second form by the positive real expression (4.27) Call the resulting expression L ijk . When ∂ i H jk has the first form we simply set L ijk = ∂ i H jk . In all cases we find that This time c j has the form 5 The Words for Theorem 1.7 Theorem 1.7 is the most delicate of our existence results. Here we introduce the necessary words. In later chapters, we will analyze these words, as we did for the proofs of Theorems 1.2 and 1.5.

The Squarepaths
Theorem 1.7 involves the words {W nk } for n = 3, 4, 5... and k = 0, 1, 2.... In this chapter we introduce these words and consider the corresponding unfoldings. It turns out that W nk has length 24n + 30k 2 − 68k − 20. The shortest word, W 30 , has length 52. Rather than present W nk as a long string of digits, we will draw the square path Q nk := Q 3 (W nk ). The path Q nk is not embedded, but is the union of two embedded halves. Reflection about a diagonal line swaps these two halves. We will draw one half of Q nk , we well as the diagonal line. Q nk is based an in (n − 1) × n grid of squares, which we call an n-stamp. Each square in the stamp has edge-length 2, as in Figure 2.3.    The path Q n+1,k is obtained by increasing the length of each edge of Q nk by 2 units.

The Unfoldings
We will see that U nk consists of 4 "strips", attached along 4 "hinges".  When we change the point relative to which we unfold, the strips do not change much and the hinges open and close, so to speak. For points in the orbit tiles, the hinges adjust so that the whole unfolding is practically a straight line. In Figure 4.4, we have chosen a point that is far from the relevant tile. Our remaining pictures show the unfoldings for points actually in the relevant orbit tile.
The strips are essentially composed of units that we call blocks. The left hand side of Figure 5.5 shows what we call a block . In general, a k-block is defined to be k blocks lined up in sequence. The right hand side of Figure 5.5 shows a 2-block. The triangles in a k-block all have the same shape, and this shape depends on the point in parameter space of interest to us. If we glue the opposite sides of a block together we get a space which is naturally the union of two 2-darts. The strips are essentially composed of blocks. Once we describe U 30 , we will describe U nk as a modification which amounts to changing the combinatorial structure of each strip.  Note that U(W 30 , V 3 ) has 12 long edges which are all parallel and nearly vertical. We have highlighted 4 of these edges. The unfolding U 3k is obtained by cutting U 30 open along each of the 4 highlighted edges and inserting a k-block. Figure 5.7 shows U(W 31 , V 3 ). The pattern continues in the obvious way. In describing our surgery, we have used the geometry of U(W 30 , V 3 ) to highlight 4 particular edges along which we cut. However, this surgery has a combinatorial meaning for any parameter. We obtain U n,k from U 3k by replacing each maximal m-dart with a maximal m ′ -dart, where m ′ = m + (n − 3).
This fits exactly with our description of Q nk as being obtained from Q 3k by lengthening each edge by 2(n − 3) units.  We end this section with a computation. The formulas in the next result will be useful when we make explicit computations in §6 and §7.
Lemma 5.1 Let n be fixed and let e k be the leftmost edge of U(W nk , V n ). As k → ∞, the slope of e k converges to 0.
Proof: We normalize so that e k has unit length. If we trim off the portions of the darts from the set U := U(W nk , V n ) we see that the resulting set is the union of 4 parallel annuli attached along 4 edges. We compute by elementary trigonometry and induction that the 4 strips have total length Ψ 1 + Ψ # k; Ψ 1 = 12(1 + cos( π n )); Ψ # = 8(1 + cos( π n )).
(5.1) Each annulus has width Ψ 2 = sin( π n ) (5.2) If we rotate U so that the first edge is horizontal then the holonomy has coordinates The line determined by this complex number converges to a horizontal line as k → ∞. Thus, if we rotate U so that the holonomy is horizontal, then the slope of the first edge tends to 0 as k → ∞. ♠

Remark:
In §7 we will give a more explicit and combinatorial derivation of the formulas in Lemma 5.1.

The Pivot Region
Here we isolate 4 basic features of the unfolding U(W nk , V n ).
• There is a family of 12 + 8k parallel and nearly vertical edges. We call these edges quasi-vertical , or QV for short.
• There is a family of 24 + 16k parallel and nearly horizontal edges. We call these edges quasi-horizontal , or QH for short.
• Each QV edge is flanked by two QH edges, in the sense that reflection in this QV edge swaps the two QH edges flanking it.
• There are exactly 4 QH edges which connect top to bottom vertices. We call these edges the hinges.
These facts are all established inductively. They hold true for the parameter (3, 0), and then we check easily that they remain true when we perform one of the surgeries described above.
We have distinguished the above edges just for the unfoldings attached to specific parameters. However, we extend our definitions of QH and QV, using continuity, to the unfoldings attached to any point of parameter space. Of course, for points remote to the regions of interest to us, the QH edges need not be close to horizontal and the QV edges need not be close to vertical. Moreover, these edges need not be parallel to each other at other parameters. To talk precisely about the situation we make the following definition: Definition: Let A n denote the region (x 1 , x 2 ) ⊂ ∆ such that The points V n and V n−1 are two opposite corners of the little square defined by these conditions. Let R ′ nk ⊂ A n denote the set of points such that all the QH edges have negative slope. Let R nk denote the path connected component of R ′ nk which contains V n . We call R nk the pivot region.
Lemma 5.2 For any point in R nk the QV edges all have positive slope.
Proof: As we pointed out above, each QV edge V is flanked by two QH edges H 1 and H 2 . That is, reflection in V swaps H 1 and H 2 . This is a property that holds for all parameters: It is a combinatorial symmetry. Now, let X ∈ R nk be some point. We consider what happens as we vary the parameter continuously from V n to X, staying inside R nk . If the slope of V changes from positive to negative then V must be either vertical or horizontal at some point. But then it is impossible for H 1 and H 2 to both have negative slope at this point and still flank V . This is a contradiction. ♠ Referring to §2.6, we note that a vertex of U nk is superior if and only if it is incident to a QH edge. This fact is seen by inspection for U 30 , and then is unchanged by any of the surgeries we perform. We call a superior vertex a pivot if it is incident to one of the pivot edges. There are 4 top pivots and 4 botton pivots.
Lemma 5.3 Let X ∈ R nk be any point. If all the top pivots lie above all the bottom pivots of U(W nk , X) then X ∈ O(W nk ).
Proof: Let v be a top superior vertex of U nk which is not a pivot. Inspecting our unfoldings, we see that v has one of two properties: • v is the left vertex of a QH edge which is not a hinge.
• v lies to the left of another superior vertex v ′ , and reflection in a QV edge swaps v and v ′ .
This property is easily seen, by inspection, for U 30 , and our surgery operations do not destroy this property. In either of the above cases, our conditions on the slopes of the QV and QH edges forces v to lie above v ′ . Since this works for all superior vertices which are not pivots, we see that only a pivot can be the lowest top superior vertex. Likewise, only a pivot can be a bottom superior vertices. Hence, by hypothesis, all the top superior vertices lie above all the bottom superior vertices. Hence X ∈ O(W nk ) by the Dart Lemma. ♠

The Structure of the Quasi-Horizontal Edges
Say that a QH point in Z 2 is a point e which corresponds to a QH edge. Here we describe the pattern of QH points associated to our unfoldings. We use the McBilliards labeling convention that the leftmost edge e 0 of U nk , which happens to be a QH edge, corresponds to (0, 0) ∈ Z 2 . If e is any other QH edge, then e is parallel to e 0 at the point V n = (π/2n, π/2n). But the angle between e and e 0 is given by e · (π/2n, π/2n). This angle must be an integer multiple of π. Hence e 1 + e 2 ≡ 0 mod 2n. (5.4) The map (x, y) → x + y maps the hexpath H nk to a subset of Z having diameter less than (4 + 1 2 )n. Hence Equation 5.4 forces the QH points to lie along at most 4 lines of slope −1 in Z 2 . Hence the QH edges fall into at most 4 pseudo-parallel families, in the sense of §2.7.
After some trial and error we figured out how to draw the QH points. We find that these points fall into exactly 4 pseudo-parallel families, and we compute the extreme points of these families as follows: Setting Actually, we don't care so much about these formulas. The main feature of the QH points we use is that the coordinates of the northwest extreme endpoints−the first ones listed in each line above−are independent of k. This property, together with symmetry, will tell us everything we want to know about these edges. 6 The Quadratic Rescaling Theorem

Overview
We are interested in studying infinite sequences {W nk } of words introduced in the last chapter. We will hold n fixed and set W k = W nk . We wish to understand the asymptotic shape of the orbit tiles O(W k ) as k → ∞. Each O(W k ) is a piecewise analytic polygon, whose sides are given as the 0-level sets of analytic functions. It turns out that O(W k ) has a uniformly bounded number of sides, independent of both n and k, and we will be able to make sense of the notion of a side of O(W k ) which is independent of k. This allows us to group the various defining functions involved into families. Our analysis is done one function-family at a time.
As we saw in §2, the function F k has the special form F k (X) = Im(P k (X − X 0 )Q k (X − X 0 )), (6.1) (The only difference between the set-up here and in §2 is that we take special care to translate F so that (0, 0), rather than X 0 , is the main point of interest to us in the domain.) Here X 0 is the point in parameter space to which the orbit tiles converge. Summarizing the discussion in §2, P k is the development image of a certain saddle connection associated to W k , and Q k is the holonomy of the unfolding. We want to place conditions on {P k } and {Q k } so that the rescaled functions {G k } converges to a linear map (whose formula we can compute explicitly). Here The conditions we place on {P k } and {Q k } have to do with the growth patterns of the supports of their Fourier transforms. A few glances at the figures in §5 should be enough to convince the reader that the conditions we discuss are satisfied, in particular, by the functions associated to the words introduced in §5. We will state the Quadratic Rescaling Theorem in the next section and then spend the rest of the chapter proving it. As in §2 we frequently let E(x) = exp(ix).

The Main Result
All our constructions are based on a translation T : Z 2 → Z 2 and a point X 0 ∈ πQ 2 . More specifically, we have There is a natural homomorphism associated to X 0 : φ(x 1 , x 2 ) = p 1 q 2 x 1 + p 2 q 1 x 2 G.C.D.(q 1 , q 2 ) ∈ Z/N; N = q 1 q 2 /G.C.D.(q 1 , q 2 ). (6.4) Here G.C.D. stands for greatest common divisor . We require that T is compatible with X 0 in the sense that φ(M 1 , M 2 ) = 0.
Going back to the general case, we let {R k } stand for either the sequence {P k } or {Q k }. According to the theory developed in §2, we can write (6.6) Here R k : Z 2 → Z is the Fourier transform of R k . We say that R k has linear growth if there is some map R # : Z 2 → Z such that Here T is the translation above. We call R # the growth generator for {R k }. We require that both R # and R 0 are supported on finitely many points of Z 2 . Intuitively, the support of R k grows linearly along the fibers of the homomorphism φ. As the notation suggestions, define Supposing that both { P k } and { Q k } have linear growth, we define Here ∂ j is the partial derivative with respect to x j . The rest of this chapter is devoted to proving: Theorem 6.1 (Quadratic Rescaling) Suppose that { P k } and { Q k } have linear growth with respect to T , and the quantities P # (X 0 ) and Q # (X 0 ) and δ are all real. Then {G k } converges in the C ∞ -topology to G, whose equation is given by

Remarks:
The C ∞ convergence means that each partial derivative of G k converges, uniformly on compact subsets, to the corresponding partial derivative of G.

Quadratic Growth Conditions
Let A denote the set of all globally defined and analytic complex-valued functions on R 2 . Given a multi-index I = (i 1 , i 2 ) we define (6.10) Given an infinite sequence {F k } ∈ A we can write out the power series expansions F k (X) = I C k,I X I (6.11) We say that {F k } forms a quadratic growth family if {F k (0, 0)} is a constant sequence we have the following finite limits for some ǫ > 0.
Note that ǫ only enters into the third equation.
Recall that {G k } is the rescaled version of {F k }, as in Equation 6.2.
Lemma 6.2 (Convergence) Suppose that {F k } is a quadratic growth family. Then {G k } converges in the C ∞ topology to the linear function G, whose formula is given by Proof: From the chain rule, we get the following series expansion: Here L k (x) is a linear function whose coefficients vanish as k → ∞, and R k is everything else. It suffices to to show that R k and all its derivatives tend to 0 uniformly on compact subsets. Let ∂ stand for some partial derivative and let Ω be some big constant. Suppose X = (x 1 , x 2 ) is such that |x j | ≤ Ω for j = 1, 2. There is some constant N, depending on ∂, such that For k > ΩN N 1/ǫ the term in braces is less than 1. Hence By hypothesis, this last sum tends to 0 as k → ∞. ♠

A Fact about the Fourier Transform
Now we begin to use the information about the sequences {P k } and {Q k } to establish the conditions on {F k } discussed in the Convergence Lemma above. Let X 0 be as in Equation 6.3 and let φ be the associated homomorphism, given in Equation 6.4. In particular, the value N is given by Equation 6.4. Consider a function of the form (6.14) Choosing any residue class k ∈ Z/N we define the modular transform: In all cases of interest to us, the sum in Equation 6.14 is a finite sum. This sum defines a function R φ : Z/N → C.
Lemma 6.3 (Modular Transform) With the notation as above, we have We can write R(X 0 ) = N j=1 R j , where Summing over j we get the result. ♠

Growth Formulas
Let X 0 and T be as in Equation 6.3. In particular, recall that T represents translation by the vector (M 1 , M 2 ) ∈ Z 2 . Let V denote the set of sequences of the form {R k } which have (M 1 , M 2 )-linear growth. We want to be clear that each individual element of V is a sequence of functions. V is a vector space. The vector space laws on V are given by componentwise scaling and addition. That is We say that an element {R k } of V 0 is simple if its growth generator R # is the indicator function for a single lattice point. That is, there is some integral point A such that R # (X) = 1 iff X = A and R # (X) = 0 otherwise. The simple elements of V 0 form a basis for V 0 .
Lemma 6.4 Let {R k } be any element of V and let I = (i 1 , i 2 ) be any multi-index. Then Thus, it suffices to prove this lemma for elements of V 0 . Given the scaling and additivity properties of Equation 6.17, it suffices to establish Equation 6.17 for the simple elements of V 0 . Suppose R # is the indicator function for (a 1 , a 2 ) ∈ Z 2 . Then (a 1 + jM 1 , b 1 + jM 2 ), (6.18) and otherwise this function vanishes. Let β = φ(a 1 , a 2 ) ∈ Z/N. Note that β = R # (X 0 ) for simple elements. All the points in Equation 6.18 lie in the same fiber of φ, namely φ −1 (β). From the Modular Transform Lemma we have This completes the proof. ♠ One useful special case of Lemma 6.4, stated more precisely, is: Remark: We can relate Equation 6.20 to Lemma 5.1 as follows.
In our examples we have X 0 = V n , the Veech Point. If R is the holonomy function (the Q-function) associated to either the 1-spine or 2-spine of the unfolding U nk , then R 0 (X 0 ) = Ψ 1 +i4Ψ 2 and R # (X 0 ) = Ψ # .
Lemma 6.5 Let {R k } be any element of V. Then for j = 1, 2, Hence Equation 6.21 holds, with zero constant term, for the simple elements. Both sides of Equation 6.17 (with zero constant term) can be interpreted as homomorphisms from V into C. Hence, Equation 6.21 holds, with zero constant term, for all elements of V 0 . Finally we note that ∂ j S k (X 0 ) and ∂ j R k (X 0 ) differ by a constant. ♠

Consequences of the Growth Formulas
Now we assume that {P k } and {Q k } and {F k } are all as in the Quadratic Rescaling Theorem.
Proof: For easy reference we repeat Equations 6.20 and 6.21. For ease of notation we write ρ = ρ(X 0 ), understanding that all our functions ρ are evaluated at X 0 unless we explicitly indicate otherwise. We will also set M = M j and ∂ = ∂ j .
Using these equations and the identity we just expand everything out and cancel many terms in pairs. We find that Here Now we get to the nontrivial quantities. In our calculations we use the fact that P # and Q # are both real at X 0 .
This completes the proof. ♠ Now we turn to the task of establishing Equation 6.12. Let D be a differential operator of order α. We have where trig stands for either the sine or the cosine function, depending on the parity of α. Equation 6.24 gives us max j (|J D,kj |) < M α+1 k α . Given that the sine and cosine functions lie between −1 and 1, and that there are at most Mk 2 terms in the sum for DF k , we have Let (n, k) stand for "n choose k". If I = (i 1 , i 2 ) is a multi-index, with α = |I|, then we have Summing over all multi-indices of weight α we get

This is Equation 6.22. ♠
Here is an improvement on Equation 6.22, at least when |I| = 2.
Lemma 6.9 There is some constant M such that Proof: To simplify our notation, an expression like D I P k shall stand for D I P k (X 0 ). The functions P k and Q k are always evaluated at X 0 . We will consider D 2,0 F k (0), the other partial derivatives of interest having a similar analysis. By the chain rule we have D 2,0 F k (0) = From Lemma 6.17 and our assumptions, there are constants a, b ∈ R such that The term C has the same treatment as A. All in all, |D 2,0 F | = O(k 3 ). ♠ Lemma 6.10 Equation 6.12 holds for ǫ = 1/4.
Proof: Without loss of generality, we may take k > 2 + M 100 . Given Equations 6.22 and 6.27 we have This last expression tends to 0 as k → ∞. ♠ The Quadratic Rescaling Theorem is an immediate consequence of the Convergence Lemma, Lemma 6.6, Lemma 6.7, and Lemma 6.10.

The Main Result
We defined the pivot region R nk in §5.3. (We will recall the definition in the next section.) In this section we compute the asymptotic shape of this region when n is fixed and k → ∞. Let T nk denote the dilation which maps V n to (0, 0) and dilates by a factor of k 2 . Define ). (7.1) Lemma 7.1 (Pivot) For any n, the set T nk (R nk ) converges to the infinite strip Σ n defined by the inequalities |x − y| < C n .
Remarks: (i) When we restrict to any bounded region of the plane, the convergence we have in mind is the same discussed in Theorem 1.7.
(ii) The Pivot Lemma is sharp. As we will see in the next chapter, the limit turns out to have vertices on both components of ∂Σ n ., The rest of the chapter is devoted to proving the Pivot Lemma.

Reducing to Defining Functions
Suppose for the moment that e(t) is a continuously varying segment in R 2 for t ∈ [0, 1].
Suppose that e(0) has negative slope. Let f (t) denote the height of the left endpoint of e(t) minus the height of the right endpoint of e(t). Note that f (0) > 0. We would like to make the following statements, which we call the slope statements: 1. e(t) has negative slope ∀t ∈ [0, 1] if f (t) > 0 ∀t ∈ [0, 1].
2. If f (t) < 0 for some parameter t then e(t) has positive slope at t.
We would like the slope statements to be true because we would like to define the pivot region in terms of the defining functions associated to the endpoints of the QH edges. Unfortunately, the slope statements are not necessarily true. The problem is that e(t) could become vertical at some point. However, the slope statements are true if e(t) has finite slope for all t ∈ [0, 1]. Most of this section is devoted to dealing with this irritating hitch in the slope statements.
Once we have the kink worked out, we will proceed to define the pivot region in terms of defining functions.
Recall that R nk is the path component of R ′ nk which contains V n ; and R ′ nk is the subset of A n consisting of points where all the QH edges of U nk have negative slope. (We defined A n in §5.3; this set is about to be replaced so we will not bother to recall the definition here.) Let A nk ⊂ ∆ denote the subset consisting of points which are within k −3/2 of P n . For k sufficiently large, A nk is a subset of A n , the set defined in §5.3. Notice that T nk (A nk ) is a disk of radius k 1/2 . Hence lim k→∞ T nk (R ′ nk ) = lim k→∞ T nk (R ′ nk ∩ A nk ). For this reason, we will always work within A nk when we analyze R ′ nk and R nk . For the next several results we choose two points X 0 , X 1 ∈ A nk . We might as well take X 0 = V n . Let U j = U(W nk , X j ) for j = 0, 1.
Lemma 7.2 Let e 0 and e 1 be corresponding edges of U 0 and U 1 , which are edges of the jth triangle from the left. If U 0 and U 1 are both rotated so that the leftmost edge is horizontal then the angle between e 1 and e 2 is at most O(jk −3/2 ).

Proof:
The point e ∈ Z 2 corresponding to e 0 and e 1 has norm O(j). Also |X 0 − X 1 | = O(k −3/2 ) by hypothesis. But the angle between our two edges is | e · (X 0 − X 1 )|, a quantity which is O(jk −3/2 ). ♠ Corollary 7.3 If U 0 and U 1 are both rotated so that the leftmost edges are horizontal, then the angle between the holonomy of U 0 and the holonomy o U 1 is O(k −1/2 ).
Proof: Here we will use the fact that X 0 = V n . Let L 0 denote the line which joins up the endpoints of the 3-spine S 0 of U 0 . Given the structure of U 0 discussed in §5, we see that L has length which is linear in k. The holonomy of U 0 maps the left endpoint of L 0 to the right endpoint of L 0 . Let S 1 be the 3-spine of U 1 . If the left endpoints of corresponding jth edges of L 0 and L 1 are matched up, then the right endpoints differ by at most O(jk −3/2 ). Hence, by vector addition, we see that the right endpoints of S 0 and S 1 differ by at most Cnk j=1 O(jk −3/2 ) = O(k 1/2 ), assuming that the left endpoints have been matched up. Our notation in the last estimate is a bit informal. The constant C n is present in the sum to indicate that there are at most C n k edges in U nk . Hence L 1 also has length which is linear in k. It now follows from basic trigonometry that the angle between L 0 and L 1 is O(k −1/2 ). ♠ Corollary 7.4 If X ∈ A nk and k is sufficiently large then none of the QH edges in U(W nk , X) is vertical.
Proof: If U 0 and U 1 are both rotated so as to have horizontal holonomy then the angle between corresponding edges of U 0 and U 1 is at most O(k −1/2 ). This is an immediate consequence of the previous two results, and the fact that there are O(k) triangles in U 0 and U 1 . This lemma now follows from Lemma 5.1. The idea is that the QH edges are nearly horizontal for one point in A nk , and then that cannot rotate much as we move around in A nk . ♠ Now we can proceed with the analysis of the region R nk by means of defining functions. For each QH edge e, let F nk,e denote the defining function which measures the height of the right endpoint of e minus the height of the left endpoint of e. (We normalize so that e has length 1, as in §2.) In particular, let e 1 , ..., e 8 be the Qh edges corresponding to the extreme QH points, as discussed in §5.4. Compare Figure 5.10. Let R nk ⊂ A nk denote those points X such that F nk,a (X) < 0 for a = 1, ..., 8. Here is the main result of this section.
Lemma 7.5 If lim k→∞ T nk ( R nk ) = Σ n , then the Pivot Lemma is true.
Proof: As above, we use the convention that our defining functions measure the height of the left vertex minus the height of the right vertex. Let X ∈ R nk . The Convex Hull Lemma says that F nk,e (X) > 0 for all QH edges e. Given Corollary 7.4 we now know that all the QH edges have negative slope. Hence R nk ⊂ R ′ nk for k sufficiently large. If T nk ( R nk ) converges to Σ n then the connected component U k of T nk ( R nk ) containing (0, 0) also converges to Σ n . Since R nk ⊂ R ′ nk , we see that U k is a connected subset of R ′ nk which contains V n . Hence U k ⊂ T nk (R nk ). In summary, some subset U k of T nk (R nk ) converges to Σ n .
Suppose we could find a sequence {X k } of points such that X k ∈ R nk but T nk (X k ) converges to some point of R 2 − Σ n . Then some defining function F nk,a would be negative at X k . But then, by the second slope statement and Corollary 7.4, some QH edge would have positive slope at X k . This is a contradiction. Hence T nk (R nk ) itself converges to Σ n . ♠

Bilateral Symmetry
Now we will focus our attention on the 8 extreme QH points, the ones listed in §5.4. The important fact for us is that the coordinates of the 4 northwest extreme points are independent of k and the 4 southwest extreme points are, in a geometric sense, symmetrically located with respect to the northwest extreme points. Each pseudo-parallel family is bounded by a northwest point and a southeast point. We call two such extreme points partners.
Lemma 7.6 Suppose that F 1 and F 2 are the defining functions associated to a pair of partner extreme points. Then F 1 (x, y) = F 2 (y, x).

Proof:
The hexpath H nk has bilateral symmetry across a diagonal line. If we reflect H nk in its line of symmetry and trace it backwards we get the same path. Correspondingly, if we rotate U nk by 180 degrees and trace it backwards we get a cyclic permutation of U nk , except with the edges of type 1 and 2 reversed. From this symmetry we see that partner extreme points correspond to edges of different types. We label so that the extreme point corresponding to F j has type j.
Consider the effect of changing the origin in Z 2 . This amounts to adding some vector V 0 to all the lattice points. If we compute our defining functions with the new labeling at X we simply multiply both P and Q by the same quantity E(X · V 0 ). Hence F is unchanged. For the duration of the lemma, we change the origin so that it lies on the line of bilateral symmetry for the hexpath H nk . In this case we have P 1 (x, y) = P 2 (y, x) by symmetry. Reflection in the main diagonal interchanges the two sets Q 1 and Q 2 . Hence Q 1 (x, y) = Q 2 (y, x). From Equation 6.1 we see that F 1 (x, y) = F 2 (y, x). ♠

The Computation
According to the symmetry above, we just have to analyze the defining functions associated to the 4 northwest extreme points. Let {P k } and {Q k } and {F k } be the functions associated to one of these points. We use the convention that F k measures the height of the right vertex minus the height of the left vertex. Referring to the setup for the Q.R.T., we have the basic constants The fundamental translation T moves points 2n − 2 units south and east. Inspecting the figures in §5.1, we see that the hexpath H nk , interpreted as a function from Z 2 into Z has T -linear growth. The same therefore is true of the path Q, which is derived from H nk as discussed in § §2.4. By Lemma 5.1 we have Q # (V n ) = Ψ # ∈ R. (We will give a second derivation of this fact in §8, based on the combinatorics of the 3-spine of the unfolding.) Note that P k is the indicator function for a single point whose coordinates do not change with P . Hence P # is the 0-function. The value of P k (V n ) is independent of both k and the choice of northwest extreme point. For the point (0, 0) we can see that P k (V n ) = ±1. Our convention of the position of the left vertex minus the position of the right vertex leads to P k (V n ) = −1. All in all, we have .6 we see that According to the Q.R.T, the family of functions {G k } converges in the smooth topology to the function This calculation agrees, for small n, with the automatic computations done by McBilliards.
We now see that G is one of the two linear functions defining Σ n . The other function comes from symmetry, as we have discussed above. This completes the proof of the Pivot Lemma.

Overview
We continue using the notation from previous chapters. In particular, T nk is the dilation which maps V n to 0 and expands by k 2 . We are interested in understanding the limits, as k → ∞, of the sets T nk (O(W nk )).
Let a 1 , a 2 , a 3 , a 4 be the top pivot vertices of U nk , labeled from left to right. Likewise let b 1 , b 2 , b 3 , b 4 be the bottom pivot vertices of U nk , labeled from left to right. For any pair (p, q) of vertices amongst these, let F k [p, q] be the corresponding defining function. We are suppressing n from our notation. If we want to evaluate our function at a point X, we write it as F k [p, q](X). In computing these functions we will use the following sign conventions Assuming that {F k [p, q]} satisfies the G[p, q] denote the rescaled limit of the sequence {F k [p, q]}. Below we will prove satisfies the hypotheses of the Q.R.T. for j = 2, 3, 4.
We will compute the scaling limits of these functions explicitly below. The work in §7 shows that each of the function families {F k [a i , b i ]} satisfies the hypotheses (and hence the conclusion) of the Q.R.T. Any other function F k [p, q] can be written as a linear combination of the functions F k [a i , b i ] and the functions from Lemma 8.1. (We will find these linear combinations explicitly below.) Hence, all our function families satisfy the conclusions of the Q.R.T.
Let Ω n ⊂ R 2 denote the convex set on which all the functions G[a i , b j ] are positive. By the Q.R.T. Ω n is a convex polygon−possibly empty or infinite−with at most 16 sides. By the same symmetry as discussed in §7, we know that Ω n is symmetric with respect to reflection in the line x 1 = x 2 . We also know that Ω n ⊂ Σ n , because the defining functions for Σ n , namely G[a i , b i ], are by definition positive on Ω n .
Lemma 8.2 Assume that Ω n is bounded. Then T nk (O(W nk )) converges to Ω n , in the sense mentioned in the introduction.
Proof: Let U be any open set whose closure is contained in the interior of Ω n . Let X be any point of T −1 nk (U). By the Pivot Lemma, X ∈ R nk for k sufficiently large. Lemma 5.3 now says that the lowest top vertex of U(W nk , X) is one of the a i and the highest bottom vertex is one of the b j . Once k is sufficiently large, the functions G k [a i , b j ] will all be positive on U. Hence F k [a i , b j ](X) > 0 provided that k is sufficiently large. But now we know that all the top vertices of U(W nk , X) lie above all the bottom vertex. Hence X ∈ O(W ). Hence U ⊂ T nk (O(W )). This is Property 1 of our definition of convergence.
Suppose that we can find points X k ∈ O(W nk ) such that T nk (X k ) converges to a point in R 2 which is not contained in the closure of Ω n . Then, for k sufficiently large, at least one of the functions F k [a i , b j ] is negative on X k . But then some bottom vertex of U(W nk , X k ) lies above some top vertex. Hence X k ∈ O(W nk ). This is a contradiction. This contradiction, establishes Property 2 of our convergence. ♠ During our proof of Lemma 8.1 we will gather enough information to compute all the functions defining Ω exactly. It is then a simple matter to check that these functions cut out precisely the region advertised in Theorem 1.7.

Asymptotic Limit Calculations
Here will prove Lemma 8.1 and compute the rescaled limits for the relevant families of functions. For each of the 3 function families of interest to us, the corresponding vertices can be connected to each other using part of the 3-spine. Figure 8.1 shows by example the general pattern for the paths P and Q. The example corresponds to U 41 , but every picture has the same combinatorial structure. In our figures in this chapter, the grids have edgelength 1 rather than 2, as in §5. We make the change because we want to point out integer coordinates of various points in the squarepath, and some of these coordinates might be odd. The whole path represents Q. The white dot denotes the origin. Each of the three black paths corresponds to a different one of our function families. Tracing the path around clockwise, starting at the origin, we encounter the paths in the same order that the function families are listed. From this picture we see clearly that both P and Q have T linear growth, where T is as in §7. It remains to compute all the quantities relevant to the Q.R.T., for each of the three families. We will do this in a step by step fashion. To keep our pictures concrete, we will draw the case n = 4, though the general case is extremely similar.

The Holonomy Calculation
Here we compute the quantities associated to the family {Q k }. Just by scaling we get We explain the constant λ n in §8.3 below. Unfortunately, the geometric method used in the proof of Lemma 5.1 does not readily shed light on the derivatives of Q # . So, here we will use the combinatorial method explained in §2. At any rate, our calculations here serve as a second proof of Lemma 5.1. we arrive at the following tableaux for Q 0 and Q # .
We determined the sign for the second tableaux by trial and error: We expect Q # (X 0 ) to be positive real rather than negative real because our unfoldings grow in the positive direction. To help show the pattern we have staggered the entries in the tableau for Q # to indicate their corresponding lines in the tableau for Q 0 .
(8.4) as in §7. We could use the Modular Transform Lemma from §6 to evaluate the above functions and their derivatives at X 0 . However, we will do the calculations symbolically in Mathematica. From Equation 8.3 we get: Q 0 (X 0 ) = 12c + 4is; In particular, using the double angle formulas we see that Equation 8.1 is indeed true.

The First Function Family
Here we compute the quantities associated to {P k }, for P k [a 1 , b 2 ]. Figure 8.3 shows P 0 and P # . The bottom two dots represent P # . Setting M = 2n−2 we read off the following function tableaux for P and P # . Remarks: (i) The tableau for P 0 consists of the first 5 lines of the tableau for Q 0 , and the tableau for P # consists of the first 2 lines of the tableau for Q # . In this sense, the tableaux for Q 0 and Q # form a kind of master list.
We compute symbolically from Equation 8.6 that Next, we compute that: Equations 8.2.1 and 8.7 tells us, in particular, that Q # (X 0 ) and P # (X 0 ) are both real. (We could also deduce this geometrically. Combining Equations 8.2.1 and 8.7 we compute that This is the same value we got for the family considered in §7 in connection with the Pivot Lemma. We don't have an explanation for this agreement. Next, we compute (8.10) All the relevant quantities are real, and so the associated family {F k } satisfies the hypotheses of the Quadratic Rescaling Theorem. Using the formulas above we compute Im(δ 1 ) = Im det −2c −12i(n − 1)c 8c 8i(4n − 5)c = 16(2n − 1)c 2 .

The Second Function Family
Here we compute the quantities associated to {P k }, for P k [b 2 , b 3 ]. Comparing Figures 8.1 and 8.2 we find that the tableau for P 0 consists of lines 6, 7, 8 of the tableau for Q 0 and the tableau for P # consists of lines 3, 4 of the tableau for Q # . Here are these tableaux: From these tableaux we compute that P 0 (X 0 ) = 3cis; P # (X 0 ) = 2c; (8.14) Using these equations, and the ones above for Q, we compute that F (0, 0) = 0; δ = 0; ∆ 1 = Im(δ 1 ) = 16c 2 ; δ 2 = Im(δ 2 ) = −16c 2 . (8.15) There {F k } satisfies the hypotheses of the Q.R.T. and the rescaled limit is Remark: For small values of n this agrees with the calculations made by McBilliards. For this example, it took many tries before we got the sign right. The problem is that the constant term vanishes, making it much trickier to deduce the correct sign without making an error.

The Third Function Family
Here we compute the quantities associated to {P k }, for P k [b 3 , a 4 ]. Comparing Figures 8.1 and 8.2 we find that the tableau for P 0 consists of line 9 of the tableau for Q 0 and the tableau for P # consists of lines 5, 6 of the tableau for Q # . Here are these tableaux: From these tableaux we compute that P 0 (X 0 ) = c + is; P # (X 0 ) = 2c; Using these equations, and the ones above for Q, we compute that F (0, 0) = 8cs; δ = −16c 2 ; δ 1 = (16 + 32n)c 2 ; δ 2 = 16c 2 ; Therefore {F k } satisfies the hypotheses of the Q.R.T, and the rescaled limit is For small values of n this agrees with the calculations made by McBilliards.

The Fudge Factor
Suppose that p 1 , p 2 , p 3 are three vertices on our unfolding. Let F ij be the defining function which computes (up to scale) the height difference between p i and p j . Here we refer to the function defined in §2. We would like to say that F 13 = F 12 + F 23 but there is a catch. The functions might not all be computed with respect to the same spine of the unfolding, and thus the function values might represent differences in heights measured at different scales. This explains the fudge factor λ n in Equation 8.1. In this section we will address this issue systematically. Let θ d (X) denote the sine of the angle of the triangle T X which is opposite the dth edge. Supposing that F has been computed in terms of the d-spine, we define Then F measures the height difference between the relevant points when the edge of type d is scaled to have length sin(θ d ). the exponent 2 appears in the definition because the functions P and Q both scale linearly with the edge length, and F is the imaginary part of their product. It follows from the Law of Sines that a triangle may be scaled, with a single scale, so that its type d edge has length sin(θ d ). Therefore, the functions F is computed at the same scale, independent of which spine it uses. Thus we really do have F 13 = F 12 + F 23 . (8.22) Our modification works well with the Q.R.T. The set of functions that satisfy the conclusion of the Q.R.T. with respect to the same point X 0 forms an algebra. That is, they can be added, scaled, and multiplied together. Letting A k (X) = sin(θ d (X)), a function which is actually independent of the parameter k. We see easily that the sequence {A k } satisfies the conclusions of the Q.R.T. and has rescaled limit function sin d (X 0 ). Therefore, if {F k } satisfies the conclusions of the Q.R.T. and has rescaled limit G, then { F k } also satisfies the conclusions of the Q.R.T. and has rescaled limit In the examples of interest to us, we have the following conversions: If G is based on the 3-spine then If G is based on either the 1-spine or the 2-spine then Working with the G limits instead of the G limits, we can add and subtract with impunity.

The Shapes of the Tiles
In this section we compute the region Ω n , using the explicit formulas for the functions defining Ω n . Our first task is to write down all these functions. To make our notation simpler, we will understand that our functions are always evaluated at the point (x 1 , x 2 ). We will also use the notation  In the language of §7.3, the first and third hinges of the unfolding U nk correspond to northwest extreme points, and the second and fourth hinges correspond to southeast extreme points. Combining Lemma 7.6 and Equation 7.8 we see that Again, these are the functions which define the strip Σ n from the Pivot Lemma. Summarizing the calculations we made in this chapter, we have (8.29) To explain the rules we use to compute the rest of the defining functions, we simplify our notation. We let [a i b j ] stand for G[a i , b j ]. Using our sign conventions above (and checking the signs against the output from McBilliards) we find that Since we are only interested in pairs of the form [a i b j ] we further compress our notation and write We computed all the above quantities symbolically and noticed a lot of symmetry. To help express this symmetry we write [i 1 j 1 ] ∼ [i 2 j 2 ] if the map (x 1 , x 2 ) → (x 2 , x 1 ) conjugates the one function to the other. We compute the following relations: Hence Ω n is defined by these 4 alone. Now that we are done adding these functions together, we can replace them by positive multiples and still define the same region in the plane. Setting σ = sec(π/n) = 1 2c 2 − 1 (8.30) we compute [13] ∝   1 −2(n − 1)(c/s) −2(n + 1)(c/s)   ; [24] ∝   −1 2(c/s)(n + 1 + σ) 2(n − 1)(c/s)(1 + 2σ)   . (8.31) Now we dilate the plane by ζ n := 2(n − 1)(c/s).
The region ζ n Ω n is the region cut out by the defining functions obtained from the ones above (and their symmetric conjugates) by dividing all the second and third coordinate entries by ζ n . That is, ζ n Ω n is defined by the functions: we now list the vertices from Theorem 1.7, modified so that their first coordinate is padded with a 1.  We claim that ζ n Ω is the convex hull of the vertices from Theorem 1.7. To see this, it suffices to show that the matrix of dot products between the vectors in Equation 8.32 and the vectors in Equation 8.34 is non-negative, and has two zeros in each row and column. Here is the matrix  This completes our verification that ζ n Ω n is the convex hull of the above mentioned vertices. We got Ω n as the limit of rescaling by a quadratic family {T nk } of dilations. If we had used the quadratic family {ζ n T nk } instead, we would get ζ n Ω right on the nose. This is the family {S nk } we use for the proof of Theorem 1.7. This completes our proof of Theorem 1.7.
As a final remark, we calculate that both functions [13] and [31] vanish at a common vertex of Ω n . Hence the function [11], one of the defining functions for Σ n , also vanishes at this vertex. Hence Ω n has a vertex on ∂Σ n . By symmetry, Ω n intersects both components of ∂Σ n in a vertex. This justifies our comment, in §7, that the Pivot Lemma is sharp.

Stability questions
In this section, we prove 1.3. Recall that V n for n ≥ 3 is the obtuse isosceles triangle with two angles of measure π/(2n). We will prove that if n is a power of two, then V n has no stable periodic billiard paths. For n not a power of two, we will construct a stable periodic billiard path.

A Homological Condition for Stability
Given a triangle T , let DT denote its double with the vertices removed. (The double of a polygon can be thought of a pillowcase for a pillow made in the shape of the polygon.) See Figure 9.1. There is a natural folding map f : DT → T which sends each of the two triangles making up DT isometrically to T . (This map folds in the sense that is 2-1 except on the edges of the triangle, where it is 1-1.) Ifp is a closed geodesic on DT , then f (p) is a periodic billiard path. Conversely, if p is a periodic billiard path in T (which hits an even number of sides in a period), then there is a closed geodesic liftp with f (p) = p. The liftp is unique up to the single non-trivial automorphism of the folding map f , which preserves the labeling of edges and swaps the two triangles.
Lemma 9.1 A periodic billiard path p in a triangle T is stable if and only if its liftp to the double DT is null homologous (equivalent to zero in H 1 (DT, Z)).
Remark: Because we removed the vertices of the triangles from DT , DT is topologically a 3-punctured sphere. Thus, H 1 (DT, Z) ∼ = Z 2 Proof: This is formally a restatement of Lemma 2.2. The total sign counted for the edge labeled 1 in Lemma 2.2 is equivalent to computing the algebraic sign of the intersection of p with the lift of the edge labeled 1 to DT . Having zero algebraic intersection number with each of the lifts of an edge to DT is equivalent to being null homologous. ♠

Translation surfaces and Veech's lattice property
We will need to understand some of the implications of work of Veech [Vee89]. For this, We introduce some of the ideas appearing in the study of translation surfaces. See [MT02] for a more thorough introduction.
A translation surface is a union of polygonal subsets of the plane with edges glued together pairwise by translation. There is a natural translation surface associated to every triangle T . Let G = r 1 , r 2 , r 3 denote the subgroup of Isom(R 2 ) generated by the reflections in the sides of the triangle T . The translation surface S(T ) is the disjoint union of the triangles g(T ) with g ∈ G with some identifications. Two triangles g 1 (T ) and g 2 (T ) are identified by translation if g 1 • g −1 2 is a translation. Also, we identify two triangle g 1 (T ) and g 2 (T ) along the i-th edge by translation if g 1 •g −1 2 can be written as a composition of r i and a translation. The resulting surface S(T ) is the smallest translation surface cover of DT . The covering map is the map which sends each triangle in S(T ) isometrically to the triangle in DT with the same orientation. In this section, we follow the convention that the vertices of the triangles making up S(T ) are removed. (These points are really cone points, we only remove them to make our topological notation simpler.) Since a translation surface is built from polygonal subsets of the plane glued together by translations, the surface inherits a notion of the direction of a unit tangent vector. This notion of direction is just the measure of angle compared to a horizontal vector. This notion of direction is a fibration from the unit tangent bundle T 1 S(T ) to the circle R/2πZ.
There is a natural action of the affine group SL(2, R) on the space of translation surfaces. Suppose A ∈ SL(2, R) and S is a translation surface, then we will define A(S). Suppose S is the disjoint union of the polygonal subsets of the plane P i with i ∈ Λ with edges identified pairwise by translation. Let A(S) be the disjoint union of the polygons A(P i ) with A acting linearly on the plane with the same edge identifications. The new edge identifications are also by translation. This is possible because A sends parallel lines to parallel lines and preserves the ratio's of lengths of pairs of parallel segments.
The Veech group Γ(S) is the subgroup of elements A ∈ SL(2, R) such that there is a direction preserving isometry ϕ A : A(S) → S. We abuse notation by using A to denote the natural map from S → A(S) given by the restriction of the action of A on the plane to the polygonal subsets of the plane making up S. The map ϕ A • A : S → S is known as an affine automorphism of S. The set of such maps forms a group known as the affine automorphism group of S. The Veech group Γ(S) is always discrete. We say S has the lattice property if Γ(S) is a lattice in SL(2, R). We will utilize the following consequence of Veech's work. If a translation surface has the lattice property, then the collection of closed geodesics on S can be well understood.
A direction θ ∈ R/2πZ is called a completely periodic direction for a translation surface S if every bi-infinite geodesic in this direction is closed.
Theorem 9.2 (Veech Dichotomy) Suppose S has the lattice property. Let θ ∈ R/2πZ be a direction. Then either θ is a completely periodic direction for S or the geodesic flow in the direction θ is uniquely ergodic. Moreover, θ is completely periodic if and only if there is a parabolic A ∈ Γ(S) for which θ is an eigendirection. Veech showed that S(V n ) has the lattice property. We will now carefully describe S(V n ), so that we can explicitly use Veech's property. See Figure 9.2 for visual guidance. To build S(V n ), start with a copy of V n oriented in the plane so that longest side lies on the x-axis. Cut this triangle in two along the axis of symmetry. Reflecting each half repeated along the edges it shares with V n yields two regular 2n-gons. The halves can then be reassembled by gluing according to appropriate translations. This amounts to gluing each edge of the left regular 2n-gon to the opposite side of the right 2ngon by translation. We remove the center point of each 2n-gon and also the vertices of the 2n-gon, since these points correspond to vertices of lifts of our triangle V n . (This removal of vertices will make our homological notation simpler.) We enjoy the following consequence.
Lemma 9.4 (Enumeration) If p 1 is a closed geodesic on S(V n ) then there is a closed geodesic p 0 in the directions 0 or π 2n and an affine automorphism ϕ D • D which maps p 1 onto p 0 .
Proof: The direction of p 1 is not uniquely ergodic. Thus by Veech dichotomy, this direction must be an eigendirection for a parabolic C ∈ Γ S(V n ) . Because of the structure of the group, there must be an D ∈ Γ S(V n ) and an integer k = 0 such that either D • C • D −1 = ±A k or D • C • D −1 = ±B k . Then D maps the direction of p 1 onto either the direction 0 or π 2n . It follows that the affine automorphism corresponding to D sends the geodesic p 1 to a geodesic which travels in either the direction 0 or π 2n . This image is our p 0 . ♠

Generators for the Affine Automorphism Group
In this subsection, we combine the idea of Lemma 9.1 with Veech's lattice property to yield necessary and sufficient conditions for stability of a periodic billiard path in V n constructed via the affine automorphism group of S(V n ). Topologically, DV n is a 3-punctured sphere. We choose generators α 1 and α −1 for H 1 (DV n , Z) ∼ = Z 2 . See Figure 9.1. We choose a basis for the homology group H 1 (S(V n ), Z) ∼ = Z 2n+1 . Our basis is the homology classes of the collection of curves B = {β 1 , β −1 , γ 1−n , . . . , γ n−1 } depicted in Figures 9.2 and 9.3. The homology classes γ k with k even all contain horizontal geodesics. We order them so that the portions of these geodesics in the left polygon increase in y coordinate as the index k increases. The homology class γ 0 is chosen so that a geodesic in this class travels below the centers of the two 2n-gons. The homology classes γ k with k odd all contain geodesics which travel in the direction of angle π 2n . With the correct choice of indices, we have that the algebraic intersection number γ 2i ∩ γ 2i+1 = 1, but no other pairs of curves intersect.
Using φ, we get elements of the cohomology group H 1 (S(V n ), Z), given by the coefficents of α 1 and α −1 . Denote these elements by φ * 1 and φ * −1 . We think of these as maps of the form H 1 (S(V n ), Z) → Z. They satisfy φ(x) = φ * 1 (x)α + φ * −1 (x)α −1 . We have φ * 1 = 2nβ * 1 + γ * 1−n + 2γ * 2−n + . . . + nγ * 0 − (n − 1)γ * 1 − (n − 2)γ * 2 − . . . − γ * n−1 φ * −1 = 2nβ * −1 − γ * 1−n − 2γ * 2−n − . . . − (n − 1)γ * −1 + nγ * 0 + (n − 1)γ * 1 + . . . + γ * n−1 . (9.1) We now describe the action of the affine automorphism group on S(V n ). For the formulas that follow, we follow the convention γ −n = γ n = 0 and γ * −n = γ * n = 0. The affine automorphism group of S(V n ) is generated by the following elements. (From the point of view of matrices the action on the dual basis H 1 (S(V n ), Z) is the inverse transpose of the action on H 1 (S(V n ), Z).) • The involution which swaps the two regular 2n-gons. The actions on homology and cohomology are given by • The right Dehn twists in the odd cylinders. The actions on homology and cohomology are given by if k is odd γ * k if k is even (9.5) • The right Dehn twists in the even cylinders. The actions on homology and cohomology are given by if k is odd and k ∈ {−1, 1} (9.6) if k is even (9.7) The work above gives us an method to prove the existence or non-existence of a stable periodic billiard path. By the Enumeration Lemma, every closed geodesic in S(V n ) is the image of one of the geodesics in one of the homology classes γ 1−n , . . . , γ n−1 under an affine automorphism. Let w ∈ σ, τ o , τ e denote the action of this affine automorphism on homology, and let w * be the equivalent element of σ * , τ * o , τ * e acting on cohomology. Thus, given the homology class x of a closed geodesic, there is a w ∈ σ, τ o , τ e such that x = w(γ k ). For x to be stable, we must have φ(x) = 0 ∈ H 1 (DT, Z). This is equivalent to saying that φ * i (x) = 0 for i ∈ {−1, 1}. Then φ * i (x) = φ * i w(γ k ) = w * (φ * i ) (γ k ) for both i ∈ {−1, 1}. That is, when w * (φ * i ) is written in the basis B * , the coefficient of γ * k must be zero. We summarize these conclusions in the lemma below.
Lemma 9.5 (Stability) Suppose p is a periodic billiard path in V n . Then there is a lift to a closed geodesicp in S(V n ). Let x be the homology class ofp. Then x = w(γ k ) for w an action of an affine automorphism on homology and for some k. It follows that p is stable if and only if when w * (φ * i ) is written in the basis B * , the coefficient of γ * k is zero.

Instability
Suppose that n = 2 m for an integer m ≥ 2. We will use the Stability Lemma to show that V n has no stable periodic billiard paths. We will show that the condition for stability given in the Stability Lemma cannot even hold modulo 2n = 2 m+1 . The following holds for all n, though we will only use it for powers of two.
Proposition 9.6 Let w * ∈ σ * , τ * o , τ * e . We will show that there exist odd integers r, s such that for some (irrelevant) coefficients b 1 and b −1 and the remaining coefficients given by c(i) = r(i + n) if i is odd s(i + n) if i is even.
Proof of part 1 of Theorem 1.3: We assume n = 2 m . For all γ * i we have 1 ≤ i+n ≤ 2n−1, so i + n ≡ 0 mod (2 m+1 ). Since an odd number r can not divide 2 m+1 , we must have that r(i + n) ≡ 0 mod (2 m+1 ) for any odd r and any 1 ≤ i + n ≤ 2n − 1. Therefore, the Stability Lemma implies that there can be no stable periodic billiard paths in V n . ♠ Proof of Proposition 9.6: The proof is by induction on the group element w * . The statement is true for the identity element with r = s = 1. See Equation 9.1. Now suppose that the statement is true for the group element w * 0 for the odd numbers r 0 and s 0 . Let w * 1 = σ * • w 0 . Then by Equation 9.3 the statement is true for w * 1 with the odd numbers r 1 = −r 0 and s 1 = −s 0 . Let w * 2 = (τ * o ) ±1 w 0 . By Equation 9.5, the statement is true for the odd numbers r 2 = r 0 ∓ 2s 0 and s 2 = s 0 . Let w * 3 = (τ * e ) ±1 w 0 . By Equation 9.7, the statement is true for the odd numbers r 3 = r 0 and s 3 = ∓2r 0 + s 0 . ♠

Existence of Stable Trajectories
Suppose that n is not a power of two. We will show the second part of Theorem 1.3 that V n has a stable periodic billiard path. This follows directly from the two propositions below. The first deals with the case when n is odd and the second when n is even. . We must show that for each i = ±1 when w * (φ * i ) is written in the basis B * , the coefficient of γ * −2 a is zero. Then the Stability Lemma implies the proposition.