Outer Billiards and the Pinwheel Map

In this paper we establish a kind of bijection between the orbits of a polygonal outer billiards system and the orbits of a related (and simpler to analyze) system called the pinwheel map. One consequence of the result is that the outer billiards system has unbounded orbits if and only if the pinwheel map has unbounded orbits. As the pinwheel map is much easier to analyze directly, we think that this bijection will be helpful in attacking some of the main questions about polyonal outer billiards.

introduced outer billiards in the late 1950s and J. Moser [M1] popularized the system in the 1970s as a toy model for celestial mechanics. Outer billiards is a discrete self-map of R 2 − P , where P is a bounded convex planar set as in Figure 1.1 below. Given x 1 ∈ R 2 − P , one defines x 2 so that the segment x 1 x 2 is tangent to P at its midpoint and P lies to the right of the ray − − → x 1 x 2 . The map x 1 → x 2 is called the outer billiards map. The map is almost everywhere defined and invertible. The purpose of this paper is to establish an equivalence between polygonal outer billiards and an auxilliary map which we call the pinwheel map. We call our main result the Pinwheel Theorem. We worked out this equivalence for a restricted version of outer billiards on kites 1 in [S2,Pinwheel Lemma].
A straightforward version of the Pinwheel Theorem, which works for all points sufficiently far from the polygon, appears in almost every paper on polygonal outer billiards. (See §2.2 of this paper for a description and proof.) Here we mention three papers specifically. In Vivaldi-Shaidenko [VS], Kolodziej [Ko], and Gutkin-Simanyi [GS], it is proved (each with different methods) that outer billiards on a quasirational polygon has all orbits bounded. (See §1.2 for a definition.) For other work on outer billiards, see [M2], [D] (bounded orbits for sufficiently smooth convex domains), [G] (bounded orbits for trapezoids), [S1] (unbounded orbits for the Penrose kite), [T2] (aperiodic orbits for the regular pentagon), [DF] (unbounded orbits for the half-disk).
In contrast to the "far away Pinwheel Theorem", which has an almost instantaneous proof, the full result is much more subtle. The full result makes a statement about all outer billiards orbits, and not just those sufficiently far from the polygon. This stronger statement allows us to give a kind of bijection between the unbounded orbits of the pinwheel map and the unbounded outer billiards orbit -a bijection that is not deducible just from the "far away" result. See §1.3 for precise statements. In [S2] we use this bijection to show that outer billiards has unbounded orbits relative to any irrational kite.
Another motivation for studying the pinwheel map is that it has a geometrically appealing acceleration -i.e., speed up of the time parameter. This acceleration, in turn, has a higher dimensional compactification as a polytope exchange map. The one case we worked out, [S2,Master Picture Theorem], had a very rich structure and was quite decisive for our theory of outer billiards on kites. The general case promises to be equally rich, and we hope that it will be useful in studying the fundamental questions about general polygonal billiards -e.g., the existence of unbounded orbits.

The Main Result
We will prove our result for convex polygons that have no parallel sides. We call such polygons nice. There is probably a similar result that works for any convex polygon, but allowing parallel sides introduces annoying complications. Let P be a nice n-gon. We orient the edges of P so that they go clockwise. To each edge e of P , one associates a pair (Σ, V ), where Σ is an infinite strip in the plane and V is a vector that points from one edge of Σ to the other. Let v be the head vertex of e. Since P is a nice polygon, there is a unique vertex w of P that is as far as possible from the line L containing e. Let V = 2(w − v). Let L ′ be the line parallel to L such that w is equidistant from L and L ′ . We let Σ be the strip bounded by L and L ′ . This is a construction that comes up often in polygonal outer billiards. We call (Σ, V ) a pinwheel pair and we call Σ a pinwheel strip.
The n pinwheel strips Σ 1 , ..., Σ n are cyclically ordered, according to their slopes. Our convention is that the strips rotate counterclockwise as we move forward through the indices.
Remark: The n-gon P is quasirational if and only if it may be scaled so that the n parallelograms Σ j ∩ Σ j+1 all have integer areas. Here indices are taken cyclically.
Given the pair (Σ, V ), we define a map µ on R 2 − ∂Σ as follows.
The map µ moves points "one step" closer to lying in Σ, if they don't already lie in Σ. Note that µ is not defined on the boundary ∂Σ.
In other words, we try to move p by the (k + 1)st strip map. If the point doesn't move, we increment the index and give the next strip map a chance to move the point.
Let ψ denote the second iterate of the outer billiards map. We define ψ to be the identity inside the polygon P . Let π : R 2 n → R 2 be the projection map. A section is a map ι : R 2 → R 2 n such that π • ι is the identity.
This relation holds on all points for which ψ is well defined.
Far from the origin, we have k(p) = 1 unless ψ(p) lies in a pinwheel strip. In this case k(p) = 2 because one extra iterate of ψ * is required to shift the index. As we mentioned above, the Pinwheel Theorem is well known, and very easy to prove for points far from the polygon. (See Lemma 2.1.) The new information given by the Pinwheel Theorem is that correspondence extends in some way to the whole plane. When k(p) > 2 it means that there is a funny cancellation that happens in order to make the two systems line up. The fact that k(p) ≤ 3n puts a bound on the complexity of this cancellation. The bound we get on k(p) probably isn't sharp but it has the right order of dependence on n.

Corollaries
Just from knowledge of the relation far from the origin, one can conclude nothing about how the unbounded orbits of one system compare to the unbounded orbits of the other. It is easy to construct two maps of the plane that agree outside a compact set, such that one of the maps has unbounded orbits and the other one doesn't.
The Pinwheel Theorem adds the information that rules out such wierd pathologies. We say that an unbounded orbit of ψ * is natural if it lies in ι(R 2 ) sufficiently far from the origin. In the next result, the word orbit means both the forwards and backwards orbit. This makes sense because both ψ and ψ * are invertible. We state our result for the forward direction of the orbit, though a similar statement holds for the backwards direction, and for both directions at the same time.
Corollary 1.2 Relative to any nice polygon, there is a canonical bijection between the forward unbounded ψ-orbits and the natural forward unbounded ψ * -orbits. The bijection sending the orbit O to the orbit O * is such that O = π(O * ) outside a compact subset.
One reason why one might want to study ψ * in place of ψ is that ψ * has an appealing acceleration. Define Topologically, X is the disjoint union of n strips. X agrees with ι(R 2 ) outside a compact set. We let ψ : X → X be the first return map of ψ * to X. Geometrically, we start with a point in Σ 1 and iterate µ 2 until we land in Σ 2 , then iterate µ 3 until we land in Σ 3 , etc. Once again, we state the result in terms of the forward direction just for convenience.

Corollary 1.3
There is a canonical bijection between the forward unbounded orbits of ψ and the forward unbounded orbits of ψ.
One can accelerate somewhat further. The map Ψ = ( ψ) n preserves each individual strip in X. The action on each strip is conjugate to the action on any of the other ones outside of a compact set. Thus, we can pick on of the strips, say Σ 1 , and consider the map Ψ : Σ 1 → Σ 1 . We call Ψ the pinwheel return map.
At the same time, we can consider the first return map of ψ to Σ 1 . We call this map Ψ. Again, it is well known that Ψ = Ψ outside a compact set, but this is not enough information to produce a correspondence between the unbounded orbits of the two systems. Our next corollary fills in this information. Once again, we have picked out the forward direction just for convenience.
Corollary 1.4 There is a canonical bijection between the forward unbounded orbits Ψ and the forward unbounded orbits of Ψ.
This last corollary is pretty close to a direct generalization of the Pinwheel Lemma we proved for kites in [S2]. In [S2] we actually proved that Ψ = Ψ if Σ 1 was properly chosen from amongst the 4 possible strips. However, this stronger result does not seem to be true in general.

Outline of the Paper
In §2 we reduce the Pinwheel Theorem to two auxilliary results, Theorem 2.3 and Theorem 2.4. Roughly, Theorem 2.4 describes an infinite family of "local substitution rules" that allows one to convert between between the pinwheel map and the polygonal outer billiards map. Theorem 2.3 guarantees that all these local rules patch together globally.
In §3 and §4 we establish some geometric and combinatorial facts that we will need for the proofs of Theorems 2.3 and 2.4. In §5 we prove Theorem 2.3, and in §6-7 we prove Theorem 2.4. Ultimately, the argument boils down to robust general properties like convexity and induction, but only after we find the combinatorial structure of what is going on.
In §8 we prove the theorem, due to [VS], [K], and [GS], that all outer billiards orbits are bounded relative to a quasi-rational polygon. Our proof only uses the trivial part of the Pinwheel Theorem, namely the part that works for points sufficiently far from the polygon. There is nothing new in our proof, but we think that we have boiled the matter down to its essence. We include this proof because we already have the notation set up, and because we like the result so much.
I would like to thank John Smillie and Sergei Tabachnikov for helpful conversations about outer billiards and the pinwheel map. This work is part of an ongoing conversation John Smillie about the pinwheel map. In particular, the formulation we give of the Pinwheel Theorem emerged in conversations with Smillie.
I would also like to thank the Clay Mathematics Institute for their support during my sabbatical this year.

The Forward Partition
Let P be a nice n-gon. Let ψ be the square outer billiards map. We have already mentioned that ψ is a piecewise translation. For almost every point p ∈ R 2 − P , there is a pair of vertices (v p , w p ) of P such that The dependence of V p on p is locally constant, and the regions where the map p → V p is constant are convex and polygonal. Since there are only finitely many pairs of vertices of P , we have a partition of R 2 −P into convex polygonal sets. We call this partition the forward partition associated to P . Here we summarize some of the results we establish in the next chapter.
• P has 2n unbounded regions, which we provisionally call R i (±) for i = 1, ..., n. The region R i (±) is labelled by the pair of vertices (v, w) such that 2(w − v) = ±V i . Compare Figure 1.2.
• Sufficiently far from the origin, the regions R i (+) and R i (−) share a ray with Σ i . A sufficiently large circle centered at the origin, oriented counterclockwise, encounters R 1 , Σ 1 , R 2 , Σ 2 , etc. The ordering of the plusses and minuses depends on the geometry of P . See Figure 2.3 below.

Far From the Origin
Let µ 1 , ..., µ n be the strip maps defined in §1.2. Let X be as in Equation 1. Given a compact set K, let If K is sufficiently large, the map π : X K → X K is a bijection (and a local isometry.) The point is that the pinwheel strips are all disjoint far from the origin, thanks to the fact that P is a nice polygon.
Proof: Figure   Proof: We will take k = 1 and set p 1 = p. Provided that p 1 ∈ Σ 1 starts sufficiently far away from ∆, the map ψ simply adds V 1 to p 1 until the resulting orbit reaches a point p 2 ∈ Σ 2 . The map ψ * does exactly the same thing, and after the same number steps, we arrive at the point (p 2 , 2). And so on. ♠

The Pinwheel Identities
We have already associated n vectors to our nice n-gon P , namely V 1 , ..., V n . Each V k defines a line segment S k that joins the two ends of S k . Referring to Figure 1.2, the spoke S k joins the vertices v and w. There are n spokes associated to P and we write these as S 1 , .., S n . This cyclic ordering is compatible with the ordering that comes from the slopes of the spokes.  We say that an oriented, connected polygonal path γ is admissible if the following holds.
• γ consists of an odd number of spokes of P .
• The ordering on the spokes of γ is compatible with the cyclic order.
• Let γ ′ be the polygonal path in R ∪ ∞ obtained by connecting the slopes of the spokes of γ. Then γ ′ is a proper subset of R ∪ ∞.
The last condition means that γ does not wrap all the way around P . We use the notation a → b to name admissible paths. The first spoke is a and the last one is b. We often take the indices mod n and use b + n in place of b in case b < a. Thus, 7 → 10 is another name for the the path in Figure 2.5.
We prove the following result in §4.2 Lemma 2.2 There is a bijection between tiles in the forward partition and admissible paths. The tile corresponding to a → b is labelled by the vertex pair (v, w), where v is the first vertex of a → b and w is the last one.
We let T (a → b) denote the tile in the forward partition corresponding to the path a → b. Let ψ * denote the pinwheel map, defined in §1.2. Here are our two main technical results.
In §5 and §6-7 we prove Theorems 2.3 and 2.4 respectively. Combining these two results, we have Corollary 2.5 Let T (a → b) and T (c → d) be two tiles such that there is a point p such that p ∈ T (a → b) and ψ(p) ∈ T (c → d). Then there is some integer k ∈ {0, ..., 3n} such that (ψ * ) k (p, a − 1) = (ψ(p), c − 1). Now we deduce the Pinwheel Theorem from Corollary 2.5. The map ψ is defined precisely on the interiors of the tiles in the forward partition. For p in the interior of the tile T (a → b), we define ι(p) = (p, a − 1).
As usual, we take the indices mod n. Let p ∈ R 2 be a point with an unbounded ψ-orbit O. Let p * = ι(p). Let O * be the ψ * orbit of p * . By the Pinwheel Theorem, there is an infinite sequence t 1 , t 2 , ... such that This clearly shows that O is unbounded if and only if O * is unbounded. The analysis in Lemma 2.1 shows that Since π is a bijection outside of K * , there is only one orbit O * such that π(O * − K) = O − K. Hence, the assignment O → O * , which first seems to depend on the choice of p, is well defined independent of the choice. If O 1 and O 2 are different unbounded orbits, they differ outside of K. Hence O * 1 and O * 2 differ as well. This shows that the assignment O → O * is injective. Finally, let O * be some unbounded natural orbit. We just choose some p ∈ O * − K * and let p = π(p * ). Then the argument above shows that O * is the image of O under our correspondence. Hence, our correspondence is a bijection.

Proof of Corollary 1.3
In light of Corollary 1.2, we just have to construct a bijection between the set of forward unbounded natural orbits of ψ * and the set of forward unbounded orbits of ψ. But ι(R 2 ) and X agree outside a compact set. So, suppose that O * is a forward unbounded natural orbit. The set O = O * ∩ X is a forward unbounded orbit of X. The nature this construction makes it clear that the correspondence O * → O is a bijection.

Proof of Corollary 1.4
For each orbit O of ψ, the intersection O ∩ Σ 1 is the corresponding orbit of Ψ. In light of the analysis in Lemma 2.1 this gives a bijection between the set of unbounded orbits of ψ and the set of unbounded orbits of Ψ. Similarly, there is a canonical bijection between the set of unbounded orbits of ψ and the set of unbounded orbits of Ψ. Finally, Corollary 1.3 gives us a canonical bijection between the unbounded orbits of ψ and the set of unbounded orbits of ψ. Composing all these bijections gives us the desired result.

Basic Definitions
Throughout the chapter, P is a nice n-gon. We say that a minimal strip is a strip Σ that contains the interior of P in its interior and is not a proper subset of any other such strip. (The pinwheel strips are twice as fat as certain minimal strips.) We call an ordered pair (v, w) of vertices of P a maximal pair if there is a minimal strip Σ such that v and w lie in distinct components of ∂Σ. The spokes we defined in §2.3 are precisely the line segments joining vertices of maximal pairs. As in the previous chapter, we let S k denote the spoke corresponding to the vector V k .
Lemma 3.1 Let S 1 and S 2 be two consecutive spokes. Then S 1 and S 2 share a common vertex. Moreover, there is a minimal strip that contains all three vertices of ∂S 1 ∪ ∂S 2 .
Proof: For each θ ∈ R ∪ ∞, there is a unique minimal strip Σ θ having slope θ. For all but n value of θ, the strip Σ θ contains 2 vertices of P on its boundary. For such value, there is a unique spoke S θ selected by Σ θ . The assignment θ → S θ is locally constant and changes only at one of the n special values. At these special values, Σ θ contains 3 vertices of P in its boundary, and these are precisely the endpoints of two consecutive spokes. Our lemma is a restatment of this geometric picture. ♠ Recall that e 1 , ..., e n are the edges of P , ordered according to slope. The correspondence e j → S j is what we call the clockwise correspondence. Figure  2.4 shows an example.
Lemma 3.2 The clockwise correspondence is a bijection between edges and spokes.
Proof: By the Pidgeonhole Principle, it suffices to prove that the correspondence e → P e is an injection. So, for the sake of contradiction, suppose that there are two edges e and f such that P e = P f = P . Let a, b, c, d be the edges of e and f , ordered as in Figure 3.1. The vertex b is farther from cd than is a and the vertex c is farther from ab than is d. This is a geometrically impossible situation. ♠

Smooth Approximation
We say that an oval is a smooth and convex simple closed curve of strictly positive curvature. The condition of strict positive curvature guarantees that the tangent lines to the curve vary strictly monotonically in small neighborhoods. This monotonicity is all we really need.
Minimal strips and spokes are defined for ovals just as we defined them for nice polygons. The spokes are easier to understand for ovals than for polygons. Here are two properties of spokes associated to ovals.
1. Any oval has a unique spoke of a given slope.

Any two distinct spokes cross in their interior.
These properties follow from the monotonicity of the tangent lines and continuity. We omit the easy proofs.
We are really only interested in the spokes of a nice polygon, but one efficient way to understand certain things about them is through smooth approximations. We can find a sequence {P n } of ovals that converges to P in the Hausdorff topology. More precisely, we mean that P n is contained in the (1/n)-tubular neighborhood of P , and vice versa. We omit the straightforward proof of this fact. Lemma 3.3 For any spoke S of P there is a sequence {S n } that converges to S. Here S n is a spoke of P n .
Proof: There is a spoke S n of P n that has the same slope as S. Evidently S n converges to S. ♠

Admissible Paths and Vertex Pairs
Here we give a quick illustration of the utility of smoothly approximating nice polygons by ovals. (We will give somewhat more substantial applications in later sections of this chapter.) Lemma 3.4 Any two spokes of a nice polygon P intersect.
Proof: The corresponding result for ovals is fairly obvious. We mentioned it above. The corresponding result for nice polygons follows by continuity. ♠ We say that a pair (v, x) of vertices of P is admissible if there is some vertex w such that (v, w) is a maximal pair and The clockwise path joining v to w contains x. Lemma 3.5 Suppose that (v, w) and (w, v) are both admissible pairs. Then (v, w) is a maximal pair.
Proof: If (v, w) is an admissible pair that is not maximal, then there are maximal pairs (v, w ′ ) and (w, v ′ ) that relate to (v, w) and (w, v) as described in the definition of admissible pairs. By construction, these two maximal pairs do not intersect. This contradicts. Lemma 3.4. ♠ The next lemma refers to admissible paths, which we defined in §2.3. Proof: Each maximal pair of vertices is clearly admissible. These correspond to single spokes. Conversely, each admissible path of length 1 is an oriented spoke and hence corresponds to a maximal pair of vertices. If we start with an admissible path a → b and minimally lengthen it to the new admissible path a → b ′ , the new endpoint is a vertex adjacent to, and counterclockwise from, the old endpoint. At the same time, the admissible vertices are obtained from the maximal vertices by moving the endpoints counterclockwise. Our result follows from these facts and from induction. ♠

Interpolation Properties of Spokes
Suppose that S a and S b are two spokes of P . The indices correspond to the ordering on the spokes that comes from their slopes. We take indices mod n and arrange that a < b. Let S a (1) and S a (2) be the two vertices of S a . Let S b (1) and S b (2) be the two vertices of S b . We choose these labels so the vertices S a (1); are counterclockwise cyclically ordered, as in We distinguish 2 arcs of P . The first one, V (a, b, 1), connects S a (1) to S b (1) while avoiding S a (2) and S b (2). We define V (a, b, 2) the same way, with the roles of (1) and (2) interchanged. These two paths are highlighted in grey in Figure 3.3. The way we have assigned spokes to edges in the previous section is not symmetric with respect to order reversal. This subtle asymmetry makes itself felt in the "lopsided" condition a ≤ j < b in the next result.
Here e j is the edge associated to the spoke S j .
Proof: The same kind of limiting argument as in Lemma 3.4 shows that the endpoints of S j lie in V (a, b, 1) ∪ V (a, b, 2) for each relevant j. This lemma fails only if an endpoint of S j is also an endpoint of V (a, b, 1) ∪ V (a, b, 2) and the edge e j sticks out the side. As j increases to b, the spoke S j turns counterclockwise, but its endpoints cannot leave V (a, b, 1) ∪ V (a, b, 2). One of the endpoints is already stuck at the right endpoint of V (a, b, 1) and so this endpoint cannot move at all. The only possibility is that the spokes S j+1 , ..., S b all share the endpoint S b (1). In particular, we may assume that j = b − 1. In this case, the two unequal endpoints of S j and S b are the endpoints of an edge, say f . Since S j and S b are consecutive, it follows from Lemma 3.1 that there is a minimal strip whose boundary contains ∂S j ∪ ∂S b . One of the sides of this strip must contain f . But then, the clockwise correspondence assigns S j to f . On the other hand, the clockwise correspondence assigns S j to e j and e j = f . This contradicts Lemma 3.2, which says that the clockwise correspondence is a bijection. ♠ Let p 1 , p 2 , p 3 ∈ R 2 − P be a portion of an outer billiards orbit. We say that the triple (p 2 , S a , S b ) is harmonious if the ray − − → p 1 p 2 is tangent to P at an endpoint of S a and the ray − − → p 2 p 3 is tangent to P at an endpoint of S b . Figure  3.5 shows the situation. Figure 3.5 also shows a distinguished arc A(p) having endpoints S a (1) and S b (2). We say that this A(p) is the arc subtended p. The arc A(p) consists of those points q ∈ P such that the line segment joining q to p only intersects P at q. Say that two closed arcs of P are almost disjoint if they share an endpoint.
Proof: All the same definitions make sense for ovals, and the result there is obvious. The result now follows for nice polygons by continuity. ♠

The Pinwheel Orientation
We define a canonical orientation on the spokes of a nice n-gon P . Each spoke S j corresponds to a unique pinwheel strip Σ j in the sense that the vector V j associated to Σ j points from one endpoint of S j to the other. We orient S j by saying that oriented(S j ) = V j , We say that a spoke S j is special if the three spokes S j−1 , S j , S j+1 share a common vertex, and otherwise ordinary. For instance, in Figure 2.2 the spokes S 1 and S 3 are special and the rest are ordinary.
Lemma 3.9 Let γ be an admissible path of length at least 3. The orientation on the spokes of γ induced by the orientation on γ coincides with the pinwheel orientation on all edges but the last one. On the last spoke, the two orientations agree if and only if the spoke is ordinary.
Proof: Consider a subpath of γ that contains two spokes, say, S 1 and S b . In all cases, it follows from Lemma 3.1 that there exists a minimal strip Σ ′ such that ∂Σ ′ contains ∂S 1 ∪ ∂S 2 . In particular, ∂Σ ′ contains the edge e shown in Figure 3.6. But then the strip Σ that is twice as wide as Σ ′ is a pinwheel strip containing e in its boundary. Comparing Figures 1.2 and 3.6, we see that the orientation given to S 1 by γ coincides with the pinwheel orientation on S 1 .
It remains to describe what happens for the last spoke of γ. The argument we just gave never uses the third property satisfied by an admissible path. In case the last spoke of γ is ordinary, we can prolong γ by two more spokes in such a way that the longer path γ ′ satisfies the first two properties for an admissible path. The previous argument then shows that the pinwheel orientation and the γ-induced orientations coincide the third-to-last spoke of γ ′ , which is the last spoke of γ. Suppose that the last spoke of γ is special. Just for the sake of argument, let's suppose that the last spoke of γ is labelled S 2 . Figure 3.7 shows the picture of the last two spokes of γ when S 2 is special. In this case e ∈ Σ 2 . Comparing Figures 1.2 and 3.7 we see this time that the orientation on S 2 given by γ is −V j , as claimed. ♠ We also record the following result.
Lemma 3.10 The edges e 0 and e 1 are adjacent if and only if S 1 is a special spoke.
Proof: If S 1 is a special spoke, then the spokes S 0 , S 1 , S 2 all share a common point. The other three points are the vertices of the arc e 0 ∪ e 1 . Hence e 0 and e 1 are adjacent. Conversely, if e 0 and e 1 are adjacent, then it follows from Lemma 3.1 that e 0 is the edge connecting two points of ∂S 0 ∪ ∂S 1 and e 1 is the edge connecting two points of S 1 ∪ S 2 . But then S 0 , S 1 , S 2 share a common vertex. Hence S 1 is special. ♠

The Combinatorics of the Partition
We say that a forward wall is a line segment or ray that lies in the closure of 2 tiles of the forward partition. We say that a forward wall W is primary if the outer billiards map is not defined on interior points of W . We say that W is secondary if W is not primary. For a point in the interior of a secondary wall, the outer billiards map is defined but the next iterate is not defined. Put another way, the secondary walls are obtained by pulling back the union of the primary walls under the outer billiards map. Figure 4.1 shows pattern for a pentagon. The primary walls divide R 2 − P into n distinct primary cones.  Proof: Let p 0 be the endpoint of a secondary wall w 0 . Let p 1 be the image of p 0 under the outer billiards map. If p 0 is in the interior of a cone then p 1 must lie on an interior point of a primary wall w 1 . Pulling back a neighborhood of p 1 in w 1 , we see that w 0 contains a 2-sided neighborhood of p 0 . Hence p 0 is not an endpoint of w 0 . The second statement follows from the fact that the square-outer billiards map is clearly defined in a neighborhood of the vertex of a cone. ♠ Lemma 4.2 No two secondary walls intersect.
Proof: Suppose that p 0 ∈ w 1 ∩ w 2 is a point of intersection between two secondary walls. We can approximate p 0 by a sequence {p 0 (k, n)} of points in w k for k = 1, 2. We can arrange that both sequences lie in the interior of the same cone. Hence, the outer billiards map is defined on both sequences. Let p 1 (k, n) denote the image of p 0 (k, n) under the outer billiards map. Since the outer billiards map is an isometry on the interior of each cone, the two sequences {p 1 (k, n)} converge to the same point q 1 . By construction, q 1 lies on two different primary walls. But the primary walls only intersect at the vertices of P . Hence q 1 is a vertex of P . Given the definition of the outer billiards map, p 0 must be the same vertex of P . This is ruled out by the previous lemma. ♠ According to the lemmas above, the secondary walls cut across the cones like crosshatching, as shown schematically in Figure 4.2. Every bounded tile is either a triangle or a quadrilateral. The triangular bounded tiles contain a neighborhood of a cone apex. The quadrilateral bounded tiles have one pair of opposite sides that are primary walls and another pair of opposite sides that are secondary walls. As we justify in the proof of Lemma 4.3 below, each secondary wall is parallel to an edge of P , and the pattern is as indicated by the labels in

Proof of Lemma 2.2
Just for this section we say that the pair (v, w) of vertices is a label if it labels some nonempty tile in the forward partition.
Lemma 4.3 Every maximal pair of vertices is a label.
Proof: Let (v, w) be a maximal pair, and let Σ be the corresponding pinwheel strip. We rotate so that Σ is horizontal, as in Figures 4.3 and 4.4 below. If p ∈ R 2 lies just below L and far to the right, then w v p It is a consequence of Equations 5 and 6 that there is an unbounded region labelled by (w, v) and an unbounded region labelled by (v, w). ♠ Now we show that the set of labels coincides with the set of admissible pairs. Let's first show that such any admissible pair (v, x) is a label. First choose a point p 0 ∈ R 2 −P in the unbounded tile labelled by (v, w). Consider the outer billiards orbit p 0 , p 1 , p 2 , with p 2 = ψ(p 0 ) as usual. The situation is shown in Figure 4 Let p 1 (t) denote a point that varies along the line segment connecting p 1 to v. Let p 0 (t) and p 2 (t) be the backwards and forwards images of p 1 (t) under the outer billiards map. As p 1 (t) varies from p 1 to v, the moving tangent line from p 1 (t) to P encounters every vertex along the oriented path from v to w. Hence, there is some value of t for which p 0 (t) lies in a region labelled by (v, x). Hence (v, x) is a label. This argument, incidentally, justifies the comments about the secondary walls mentioned at the end of the last section.
To show that all the labels are admissible, we observe that we can reverse the process we just described, starting with a point p 0 in the region labelled by (v, x) and then moving outward along the ray pointing from v to p 0 until we arrive at a point that lies in a region labelled by the maximal pair (v, w). This shows that (v, x) is admissible.
Combining the result here with Lemma 3.6, we establish Lemma 2.2.

Some Geometric Estimates
The bounded tiles in a single cone C all have a common boundary with one of the strips Σ, but are disjoint from that strip. If we rotate so that Σ is horizontal, as in Figure 4.6, then the shaded region T beneath Σ is the union of the bounded tiles in C.  Proof: The boundary wall w of T opposite the cone apex v 0 is the largest bounded secondary wall in C. The secondary wall w is parallel to a line L which extends one of the sides of P . The fact that w is bounded gives control on the slope of L, and the basic geometric fact is that the point p 1 necessarily lies in the lower half of Σ. On the other hand, p 0 and p 1 are equidistant from the lower boundary of Σ. Here p 1 is the image of p 0 under the outer billiards map. But this shows that the distance from p 0 to the lower boundary of Σ is less than half the width of Σ. Since p 0 is the extreme point of T , the same statement can be made for any point of T . ♠ Let e 1 , ..., e n be the edges of P . Let ∂ 1 Σ j be the component of ∂Σ j that contains e j . Let ∂ 2 Σ j be the other component.
Lemma 4.5 If e 0 is adjacent to e 1 then T (1 → b) and P lie on the same side of ∂ 1 Σ 0 . Otherwise, T (1 → b) and P lie on the same side of ∂ 2 Σ 0 .
Proof: If e 0 and e 1 are adjacent, then e 0 = f 0 in Figure 4.7. The lines extending e 0 and e 1 contain the boundary of the primary cone that contains T (a → b). This is the shaded region in Figure 4.7. In this case, the result is obvious. In the second case, e 0 = g 0 , and we want to rule out the possibility that T (a → b) intersects the black cone. But one checks easily that points lie in the black cone lie in tiles labelled by vertex pairs (x, y) where y lies in the interior of the clockwise arc connecting the two endpoints of the spoke S a . This is a contradiction. ♠ Recall that the vectors V 1 , ..., V n are associated to the pinwheel strips Σ 1 , ..., Σ n . In our proof of Theorem 2.4 we will be interested in sets such as Corollary 4.6 If e 0 is adjacent to e 1 then the line ∂ 2 Σ 0 separates P from T (1 → b; 1). Otherwise, the line ∂ 1 Σ 0 separates P from T (1 → b; 1).
Proof: Suppose that e 0 and e 1 are adjacent. Since e 0 and e 1 are adjacent, the vector V 1 joins a a point on ∂ 1 Σ 0 to a point on the centerline of Σ 0 . Hence The first case of this lemma now follows from Lemma 4.5. The point is that adding V 1 ejects T (1 → b) outside of Σ 0 , and onto the correct side.
Suppose that e 0 and e 1 are not adjacent. This time V 1 joins a point on the centerline of Σ 0 to a point on ∂ 2 Σ 0 . Hence ∂ 2 Σ 0 + 2V 1 = ∂ 1 Σ 0 . The rest of the proof is the same in this case. ♠ Recall from §3.9 that a spoke S a is special if the spokes S a−1 , S a , S a+1 share a common vertex. Otherwise, S a is ordinary. Combining Lemma 4.6 with Lemma 3.10 and relabelling, we have the following result.
Corollary 4.7 If S a is a special spoke, then the line ∂ 2 Σ a−1 separates P from T (a → b; a). Otherwise, the line ∂ 1 Σ a−1 separates P from T (a → b; a).
The last result we prove is not needed anywhere in this paper. We include it because we think gives a nice piece of extra information about the geometry of the partition.

Lemma 4.8 A partition tile T is unbounded if and only if ψ(T ) ∩ T = ∅.
Proof: If T is unbounded, then T contains an infinite cone. Being a uniformly bounded piecewise translation, ψ cannot map an infinite cone off itself. It remains to show that ψ(T ) ∩ T = ∅ implies that T is unbounded. Let (v, w) be the pair of vertices labelling T . It suffices to prove that (v, w) is a maximal pair. Let p 0 , p 1 , p 2 ... be the forwards outer billiards orbit of a point p 0 ∈ T , Suppose p 2 = ψ(p 2 ) ∈ T . By construction, P ⊂ Q, where Q is the quadrilateral with vertices v, w, p 2 , and p 0 p 1 ∩ p 3 p 4 . Since v and w are opposite vertices of Q, there is an infinite strip that contains v and w on its boundary and containes the rest of Q in its interior. This strip picks out (v, w) as a maximal pair. ♠ 5 Proof of Theorem 2.3

Reformulation of the Result
We can define the backwards partition for P just as we defined the forwards partition. We just use the inverse map ψ −1 . The map ψ sets up a bijection between the tiles in the forward partition and the tiles in the backward partition. A tile in the forward partition labelled by a pair of vertices (v, w) corresponds to a tile in the backward partition labelled by a pair of vertices (w, v).
We change our notation so that T + (a → b) = T (a → b), and T − (a → b) denotes a tile in the backwards partition corresponding to the "backwards admissible path" The backwards admissible paths have the same definition as the forwards admissible paths, except that the spokes are traced out in reverse cyclic order. In short, if we reverse a forwards admissible path, we get a backwards admissible path. From this structure, we have We define If c = b there is nothing to prove. If necessary, we add n to c so that b < c. In this chapter, we will prove the following result.
Theorem 2.3 is a quick corollary.
Proof of Theorem 2.3: Our point q in Theorem 2.3 lies

A Result about Strips
Let P be a nice polygon. Let Σ the pinwheel strip associated to an edge e of P . We rotate and scale so that Σ is bounded by the lines y = 0 and y = 1, as in Figure 5.2. Let v be the vertex of P farthest from the bottom edge e of P . We translate so that v = (0, 1/2). Let T be the triangle formed by lines extending the two edges of P incident to v and the top boundary component of Σ. Let {p i } be an outer billiards orbit, with p i+2 = ψ(p i ). Let q i ∈ P be the midpoint of the segment joining p i to p i+1 .
Lemma 5.1 Suppose that q 1 , q 2 ∈ P L and the clockwise arc from q 2 to q 3 does not contain e. Then p 2 ∈ Σ L .
Proof: Let R be the ray that starts at v and moves up and to the left, extending the left edge of T . The conditions q 1 , q 2 ∈ P L force p 2 to lie in the region Σ ′ L bounded by P L , by R, and by the negative x-axis. Σ L is exactly the portion of Σ ′ L that lies below the line y = 1. If p 2 ∈ Σ ′ L − Σ L then q 2 = v and p 3 lies below the line y = 0. But then the clockwise arc from q 2 to q 3 contains e. This contradiction shows that p 2 ∈ Σ L . ♠ Lemma 5.2 Suppose that q 1 , q 2 ∈ P R and the counterclockwise arc from q 1 to q 0 does not contain e. Then p 2 ∈ Σ R .

The End of the Proof
We label the vertices of P so that b < c. Let Our goal is to show that p ∈ Σ j . The sets of interest so us, such as P L and P R , all depend on an edge e j , and we sometimes write P L (e j ) and P R (e j ) to denote this dependence. We will establish the stronger result that We set p 2 = p and we let {p k } be the outer billiards orbit of p. We let q k be the midpoint of the line segment joining p k to p k+1 , as in the previous section. The next lemma refers to a definition made in §3.4.

Lemma 5.3
The triple (p 2 , S b , S c ) is a harmonious triple.
Proof: The point p 2 lies in some tile T − (b, a). This means that b is the starting point of an admissible path whose first spoke is S b . Since (b, a) labels the tile T − (b, a) in the backwards partition, the line connecting p 1 to p 2 is tangent to P at an endpojnt of S b . In short q 1 is an endpoint of S a . Similarly, p 2 is in some tile T + (c, d). The same argument shows that q 2 is an endpoint of S b . ♠ We will treat the case when the spokes S b and S c do not share a vertex, though this is mainly for the sake of drawing one picture rather than two. When S b and S c share a vertex, the proof is essentially identical.
Let X = S b ∩ S c . In the case we are considering, X is a point interior to both S b and S c . We rotate so that S b has positive slope and S c has negative slope.
Let Σ = Σ j for some relevant index j. Let e j be the edge of P contained in ∂Σ j . We can further rotate P , if necessary, so that e = e j lies below the intersection S b ∩ S c . Figure 5.3 shows the situation.
Referring to Lemma 3.7, we let V (b, c, 1) denote the vertex set on the bottom of the picture and V (b, c, 2) the vertex set on the top. We will treat the case when p 2 lies to the left of X. This case relies on Lemma 5.1. The case when p 2 lies to the right of X has practically the same proof, and relies on Lemma 5.2 instead.
Let H b and H c respectively denote the right halfplanes bounded by S b and S c . Figure 5.3 shows a circle in place of a nice polygon. This is our attempt to draw a somewhat schematic picture of what is going on.
1. Combining Lemma 5.3 and Lemma 3.8, we see that q 1 = S b (1) and q 2 = S c (2) both lie to the left of X.
2. By Lemma 3.7, we have e j ⊂ closure(H b − H c ).
3. By Lemma 3.7 again, v j ⊂ H c . Here v j is the vertex farthest from e j .

By
Steps 3 and 4, we have A(p 2 ) ⊂ P L (e j ). See §3.4 for a defintion of A(P 2 ). Hence q 1 , q 2 ∈ P L (e j ).
6. Combining Steps 2 and 5, we conclude that the clockwise arc connecting q 2 to q 3 does not contain e j .

Combining
Steps 4 and 6 and Lemma 5.1, we see that p 2 ⊂ Σ L (e j ), as desired. This completes the proof. The unbounded regions correspond to the admissible paths of length 1. When a = b, the result is a tautology. So, we assume that a = b. The admissible path in question has length at least 3. We are going to prove Theorem 2.4 by induction. As previously, we replace b by b + n if necessary so that a < b.
The admissible path a → b does not necessarily involve all the spokes between a and b. If S i is not involved in the path a → b, let W i = 0. Otherwise, let W (a, b, i) be the vector that points from the first endpoint of S i to the last one. By Lemma 3.9, we have W i = V i for i < b. For i = b we have W i = ±V i , where the (+) option is taken if and only if S b is an ordinary spoke. The vector W i depends (mildly) on a → b, but we suppress this from our notation.
Proof: Let (v, w) be the admissible pair of vertices associated to T (a → b).
Recall that a and b respectively name the first and last spoke of the admissible path associated to T (a → b) whereas v and w respectively name the first endpoint of the path and the last endpoint of the path. The path a → b simply traces out the involved spokes. By definition At the same time ψ(p) − p = 2(w − v). Putting these two equations together gives the lemma. ♠ Lemma 6.1 establishes an identity between certain multiples of the vectors involved in the relevant strip maps. This is a good start. What connects the result in Lemma 6.1 to the pinwheel map is the claim that the multiples involved are precisely the ones that arise in the relevant strip maps. This amounts to showing that certain translates of the tile T (a → b) lie in the right place with respect to the relevant strips. Here is the main construction in our proof. Define The sets T (a → b; k) for i = a, ..., b are translates of T (a → b).
Recall that µ b is the local strip map with index b. In this chapter we will prove In the next chapter, we will prove Lemma 6.3 T (a → b; k) ⊂ Σ k for all k = a, ..., (b − 1).

Remarks:
(i) Lemma 6.3 is the much more interesting of the two results. It turns out that Lemma 6.2 is just a disguised version of Corollary 4.7.
(ii) It would be easier to say simply that T (a → b; k) ⊂ Σ k for all k = a, ..., b, but this is not true. Lemma 6.2 makes the strongest statement we can make.
Proof of Theorem 2.4: We relabel so that a = 1. Let Let p = p 0 ∈ T 0 be an arbitrary point. Define This is just a pointwise version of Equation 13. By Lemma 6.1, we have Choose any k = 0, 1, ..., b − 2 and consider the pair (p k , k). There are two cases to consider. Suppose first that the index k is involved in the path 1 → b. Then by Lemma 6.3. Therefore Hence Now suppose that the index k is not involved in the path 1 → b. Then p k+1 = p k and this common point lies in Σ k+1 . Hence In either case, we see that for some exponent e = e k . Applying this argument for as long as we can, we see that for some exponent e. Finally, by Lemma 6.2, we have Hence (p b , b − 1) is in the forward ψ * -orbit of (p 0 , 0). Combining this information with Equation 16, we see that there is some positive k < 2n such that This completes the proof. ♠ The rest of this chapter is devoted to the proof of Lemma 6.2. The material in the next section will also be used in §7.

The Conjugate Polygon
As in the proof of Theorem 2.3, we find it useful here to consider both the forward and backward partition of P . Since our labelling conventions have been geared towards the forward partition, we find it useful to set things up in a way that doesn't require us to deal directly with the backward partition.
Let R denote reflection in the x-axis. For any subset A ⊂ R 2 , let A = R(A). One could equally well think of R as complex conjugation.
A basic and easy fact is that R maps the backward partition of P to the forward partition of P . For that reason, we will consider the forward partitions of P and P at the same time. We rotate so that one edge of P is horizontal and lies in the x-axis. We call this edge e 0 . The associated pinwheel strip is Σ 0 and the associated spoke is S 0 . We also arrange that P lies in the upper half plane, so that e 0 is the bottom edge. The labellings of the remaining objects of P are then forced by our earlier conventions. To emphasize the dependence on P , we write e 0 (P ), etc. We let e 0 (P ) be the horizontal edge of P .
The cyclic ordering forces two sets of equations.
e k (P ) = e −k (P ); Σ k (P ) = Σ −k (P ); (25) This second set of equations is more subtle. Figure 6.3 illustrates why S 0 (P ) = S 1 (P ). There are several possible pictures, depending on the geometry of P , and we have picked one of the possiblities. The other possibilities have the same outcome. The rest of Equation 26 is then forced by the cyclic ordering. We can define the sets T (a → b; k) relative to P just as well as for P . We tack a P or P on the end of our notation to indicate which polygon we mean.
Proof: We've already remarked that ψ sets up a bijection between the tiles in the forward partition of P to the tiles in the backward partition. Briefly using the notation from the previous chapter, we have The path b → a is the same as the path a → b but it is given the opposite orientation. We call this the reversal property. We will use it below. The composition R • ψ carries the tiles in the forward partition of P to the tiles in the forward partition of P . Combining Equations 26 and 27, we see that Note that ψ(T (a → b; P )) = T (a → b; b; P ).
Combining Equations 29 and 28, we have By the reversal property and Equation 26, we have To consider the case k = b in detail, we have Applying the same argument, inductively, to each of the vectors W b−1 (P ), W b−2 (P ), etc., we establish the lemma. ♠

Proof of Lemma 6.2
We are going to apply Corollary 4.7 to P . There are two cases for us to consider, depending on whether or not the spoke S b is ordinary. We'll first consider the case when S b is ordinary. It is convenient to set By Lemma 6.4, By definition and Lemma 3.9, By Corollary 4.7, the line ∂ 1 Σ α−1 (P ) separates T (α → β; P ) + V α (P ) from P . Applying the reflection R and using Equation 26, we see that the line separates T (a, b; b − 1; P ) from P . But then for any p ∈ T (1 → b; b − 1; P ). Since S b is an ordinary spoke, V b = W b by Lemma 3.9. Now we know that µ b (p) = p + W b , as desired. When S b is special, the proof is the same except for some sign changes. This time, the line We say that an abstract admissible path is a finite tree τ with the following structure. First, there is a distinguished maximal path γ in τ having odd length at least 3. Every other edge of τ is incident to γ. We call the edges of τ − γ special . We call the edges of γ ordinary, except perhaps for the first and last edge of γ. The first and last edges of γ can be either special or ordinary. We draw γ as a zig-zag with lines of alternating negative and positive slope. We insist that every edge of τ intersects x-axis. We orient γ from left to right. We draw the ordinary edges with thick lines and the special edges with thin lines. Figure 7.1 shows an example. A dotted line represents the x-axis. γ is the path 13678. Here, the first edge of γ is ordinary and the last edge is special. The initial vertex is the left endpoint of the γ. The final edge if the last edge of γ. Linear Order: There is a natural linear ordering on the edges of τ , induced from the order in which they intersect the x-axis, goind from left to right. The numerical labels in Figure 7.1 indicate the ordering. We see that τ ′ ⊂ τ is a prefix of τ of τ ′ and τ share the same initial set of edges and if τ ′ is an abstract admissible path in its own right.

Flags and Sites:
To each edge e of τ , except the last one, we assign a vertex v e . If e is an edge of γ, then v e is the leading vertex of e. If e is not an edge of γ, then v e is the vertex of γ incident to e. We say that a flag is a pair (e, v e ). For instance, the flags in Figure 7.1 are (1, a); (2, a); (3, b); (4, b); (5, b); (6, c); (7, d).
We say that a site is a pair (τ, f ), where f is a flag of τ .
Natural Involution: There is a natural involution R on the set of abstract admissible paths: Simply rotate the path about the origin by 180 degrees and you get another one. We call this map R. The map R carries flags of τ to flags of R(τ ) in a slightly nontrivial way. We first create reverse flags of τ by interchanging the notion of left and right, and then we apply R to these reverse flags to get ordinary flags of R(τ ). In Figure 7.1, the reverse flags are Reduction: We say that the site (τ ′ , f ) is a direct reduction of (τ, f ) if τ ′ is a prefix of τ . The flag f is the same in both cases. We say that (τ ′ , f ′ ) is an indirect reduction of (τ, f ) if (τ ′ , f ′ ) is a direct reduction of R(τ, f ). We say that one site (τ 2 , f 2 ) a reduction of another site (τ 1 , f 1 ) if (τ 2 , f 2 ) is either a direct or an indirect reduction of (τ 1 , f 1 ). In this case, we write (τ 1 , f 1 ) → (τ 2 , f 2 ).

Hereditary Properties:
Let C be a collection of sites. We say that C is hereditary if it has the following properties.
• C is closed under the natural involution.
• C is closed under reduction.
Say that a site (τ, f ) is prototypical if f is the first flag of τ . Let Ω be a map from {0, 1}. We say that Ω is hereditary if Ω has the following properties.
• Ω evaluates to 1 on all prototypical sites in C.

The Reduction Lemma
In this section we will prove the following result.
Lemma 7.1 (Reduction) Suppose that C is a hereditary collection of sites and Ω is a heriditary function on C. Then Ω ≡ 1 on C.
Proof: It suffices to prove that, through the two operations of R and reduction, every site can be transformed into a prototypical site. It henceforth goes without saying that all sites belong to C.
Let (τ, f ) be a site. Let γ be the maximal path of τ . Let f = (e, v). Either v lies in the left half of γ or the left half. (There are an even number of vertices.) Applying R if necessary, we can assume that v lies in the left half of τ . If γ has length 5 we let τ ′ denote the subtree of τ obtained by deleting the last two vertices of γ and all incident edges. Then τ ′ is a prefix of τ and (τ, f ) → (τ ′ , f ). We just have to worry about the case when γ has length 3. Let (τ 1 , f 1 ) = (τ, f ) and let (τ 2 , f 2 ) = R(τ, f ). Also, let e k be the edge of f k for k = 1, 2. If e 1 is not the first edge of τ 1 then (τ 2 , f 2 ) has the following two properties.
1. e 2 is neither of the last two edges of τ 2 .
2. At least 3 edges of τ 2 are incident to the right vertex of γ. Figure 7.2 shows a typical situation. The thick grey lines represent e 1 and e 2 . The upshot is that after applying R, we can assume that f = (e, v), where e is neither of the last two edges of τ . We let τ ′ be the prefix obtained by cutting off these last two edges. The second property mentioned above guarantees that τ ′ is a prefix of τ . Again we have (τ, f ) → (τ ′ , f ).
In summary, the process above only stops when we reach a prototypical site. ♠

The Pinwheel Collection
Let P be a nice polygon. Each admissible path associated to P gives rise to an abstract admissible path. This path encodes the way the spokes in the path (and the skipped spokes) meet at their endpoints. Figure 7.3 shows an example. The unusual labelling on the abstract admissible path is present only to underscore the correspondence. We call the abstract admissible path produced in this way the abstraction of the admissible path. The maximal path in the abstract admissible path corresponds to the actual admissible path. We let C be the class of all sites (τ, f ), where τ is the abstraction of an admissible path for some nice polygon.
Lemma 7.2 C is a heriditary collection.
Proof: The fact that C is closed under reduction comes from the fact that we have built in the basic properties of admissible paths into our definition of abstract admissible paths. If τ is the abstraction of a → b, then any prefix τ ′ is the abstraction of some admissible path The analysis in §6.2 shows that C is closed under the natural involution. The basic reason, as we have discussed already, is that ψ carries the forward partition to the backward partition. ψ maps the forward tile associated to the path a → b to the backward tile associated to the path b → a. The abstraction of b → a is exactly the image of the abstraction of a → b under the natural involution. ♠

The Binary Function
Now we are going to define a function Ω : C → {0, 1}. We will first consider the situation for a given nice polygon P , and then we will take into account all nice polygons at the same time.
Let (τ, f ) be a site in C. This means that there is a nice polygon P and an admissible path a → b such that τ is the abstraction of a → b. Moreover, f is just one of the sites. The edges of τ are naturally in correspondence with the strips Σ a , ..., Σ b . Moreover, there is a natural correspondence between the sets T (a → b; a), ..., T (a → b, b − 1) and the sites of τ . The correspondence is set up in such a way that each site (τ, f ) corresponds to a pair All of this depends on P . We define Ω ′ (τ, f ; P ) = 1 if Lemma 6.3 is true for the pair in Equation 36 Finally, we define Ω(τ, f ) = 1 if and only if Ω ′ (τ, f ; P ) = 1 for every instance in which the site (τ, f ) arises. Lemma 6.3 is equivalent to the statement that Ω ≡ 1 on C. Accordingly, we will prove Lemma 6.3 by showing that Ω is a heriditary function and then invoking the Hereditary Lemma.

Property 1
The prototypical sites correspond to the case k = a in Equation 36. Cyclically relabelling, we take a = 1. The following lemma implies that Ω = 1 on all prototypical sites.
Proof: We rotate so that Σ 1 is horizontal, and T (1 → b) lies beneath the lower boundary of Σ 1 . Let LΣ 1 and UΣ 1 denote the lower and upper boundaries of Σ 1 respectively. T (1 → b) is contained in the shaded region T shown in Figure 4.3. By Lemma 4.4, every point of T (1 → b) is closer to LΣ 1 than (half) the distance between LΣ 1 and RΣ 1 . We have the following 2 properties.
It follows from these two properties that T (1 → b) + W 1 ⊂ Σ 1 . ♠ Let a → B denote the maximal admissible path that starts with S a . Then a → B corresponds to the triangular tile T (a → B) that v as a vertex. Define Note the similarity between this equation and Equation 13. In the next section we will prove the following result. Assume this result for now.
Corollary 7.6 Suppose c > b is that that a → c is an admissible path. Then Ω(a → b; k) implies Ω(a → c, k).
Proof: The admissible paths a → b and a → c agree except for possibly the last edge of a → b. Hence the sets T (a → b; k); T (a → c; k); p k are respectively translates, by the same vector X, of the sets Let T (a → b) denote the convex hull of T (a → b) and p 0 . Let T (a → b; k) denote the convex hull of T (a → b; k) and p k . First of all, by the analysis done in connection with Figure 4.2. Translating the whole picture by X, we have The second equality follows from the convexity of Σ k . ♠ Corollary 7.6 is just a restatement of Property 3.

Proof of Lemma 7.5
Now we take care of the final piece of business. We relabel so that a = 1. Let v 0 = v = p 0 and let {v k } be the successive vertices of our admissible path, where we only advance the point if the spoke is actually involved in the path. Put another way, the vertex v k is incident to the spokes S k and S k+1 . Figure 7.4 shows an example. The strip Σ k contains the three points of ∂S k ∪ ∂S k+1 and v k = S k ∩ S k+1 . Compare Lemma 3.1. From this structure, we see that v k lies on the centerline of the strip Σ k for all k. The fixed point of H, namely v 0 , belongs to every strip. In particular, v 0 ∈ Σ k . This situation forces p k = H(v k ) ∈ Σ k . ♠ 8 Quasirational Polygons

Notation
Let P be a quasirational n-gon. We can scale P so that the quantities are all integers. Let D be the least common multiple of {A 1 , ..., A n }. Define The quantities D 1 , ..., D n are all integers. Let e j be the edge of P contained in the boundary of Σ j . As usual, we orient e j so that, taken together, all these edges go clockwise around P . Let W j be the vector parallel to e j that spans Σ j+1 .
The vector V j+1 associated to the strip Σ j+1 points from one corner of Σ j ∩ Σ j+1 to the opposite corner. The point is that the spoke associated to Σ j+1 joins the head point of e j+1 to the tail point of e j .
We draw a picture in the case j = 1. The whole construction is affinely invariant, so to draw pictures we normalize so that Σ 1 is horizontal and Σ 2 is vertical, and both have the same width. Up to dihedral symmetry, there are two possible local pictures. These are shown in Figure 8.1. Which case occurs depends on whether e j and e j+1 are adjacent. We will carry out the construction in case e j and e j+1 are not adjacent. The other case has essentially the same treatment.
As in Equation 1, define Let P j = P × {j}. The polygons P 1 , ..., P n are just copies of P . Let v j be the vertex of P j that lies on the centerline of Σ j × {j}. Let Q j be the polygon obtained by rotating P j by 180 degrees about v j . Figure 8.2 shows a picture of these polygons for j = 1, 2. Technically, these polygons do not live in the plane, but rather in R 2 n . However, we have drawn their projections into the plane. The polygons P j all have the same image in the plane, so we set P = P 1 = P 2 in the picture.

The Main Argument
We define R m j = P m j ∪ Q m j ; P m j = P j + (mW j , j); Q m j = Q j + (mW j , j). (43) We think of these as open polygons.
Lemma 8.1 Let m be any integer. Then Moreover, suppose p ∈ Σ j × {j} lies between R j and R mD j j . Then ψ(p) lies between R j+1 and R mD j+1 j+1 . Proof: We first consider the strips Σ 1 and Σ 2 in detail. We normalize by an affine transformation so that the picture looks as in Figures 8.1 and 8.2. Our two strips (as normalized) are tiled by squares -i.e., translates of Σ 1 ∩ Σ 2 . The strip map associated to (Σ 2 , V 2 ) maps the squares in Σ 1 to the squares in Σ 2 , translating along the diagonal lines. This is shown in Figure 8 • The dark polygon at bottom right is the projection to the plane of Q M 1 . • The light polygon at top left is the projection to the plane of ψ(P M 1 ). • The dark polygon at top left is the projection to the plane of ψ(Q M 1 ). Note that ψ(P 1 ) = P 1 and ψ(Q 1 ) = Q 2 . From the picture, we can see that ψ maps any point in Σ 1 lying between R 1 and R M 1 to a point in Σ 2 lying between ψ(R 1 ) and ψ(R M 1 ). Now, Σ 2 is also tiled by parallelograms which are translates of the parallelogram Σ 2 ∩ Σ 3 . Figure 8.4 shows the picture. Let h denote the length of the vertical edge of Σ 2 ∩ Σ 3 . We have From this we see that provided that M ′ is also an integer. Now we use the fact that P is quasirational. We set This gives us M ′ = mD 2 ∈ Z.
Observing that the same argument can be carried out for any choice of j and not just j = 1, we see that we have completed the proof. ♠ Lemma 8.1 immediately implies that any point in Σ j between R j and R mD j j has a forwards bounded orbit. Hence, all orbits of ψ are bounded. By Corollary 1.3, all forward outer billiards orbits are bounded. By symmetry, all backward outer billiards orbits are also bounded. This completes the proof.
Remarks: (i) The outer billiards orbits O m corresponding to R mD j j , at least for m large, are known as necklace orbits in [GS]. These orbits consist of a "necklace" of polygons winding once around P , and touching vertex to vertex as shown in Figure 8.3. (ii) We have stated the proof in such a way that it appears to require the full force of the Pinwheel Theorem. However, this is not the case. Outside of a large compact region, there is an obvious correspondence between the pinwheel maps and the outer billiards maps. For m large, our argument shows that the region of the plane bounded by the orbit O m is invariant under the outer billiards map. This gives an exhaustion of R 2 by bounded invariant sets.