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OPTIMAL ANGLE BOUNDS FOR QUADRILATERAL MESHES CHRISTOPHER J. BISHOP

Abstract. We show that any simple planar n-gon can be meshed in linear time by O(n) quadrilaterals with all new angles bounded between 60 and 120 degrees.

1991 Mathematics Subject Classification. Primary: 30C62 Secondary: Key words and phrases. Quadrilateral meshes, Riemann mapping, thick/thin decomposition, linear time. The author is partially supported by NSF Grant DMS 04-05578. 1

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1. Introduction In this note we answer a question of Bern and Eppstein by proving: Theorem 1.1. Any simply connected planar domain Ω whose boundary is a simple n-gon has a quadrilateral mesh with O(n) pieces so that all angles are between 60◦ and 120◦ , except that original angles of the polygon with angle < 60◦ remain. The mesh can be constructed in time O(n). The theorem is sharp in the sense that no shorter interval of angles suffices for all polygons: using Euler’s formula, Bern and Eppstein proved (Theorem 5 of [4]) that any quadrilateral mesh of a polygon with all angles ≥ 120◦ must contain an angle ≥ 1200 . On the other hand, any boundary angle θ > 120◦ must be subdivided by the mesh in Theorem 1.1 and hence there must be a new angle ≤ θ/2 in the mesh. Thus taking polygons with an angle θ ց 120◦ shows 60◦ is the optimal lower bound. It is perhaps best to think of Theorem 1.1 as an existence result. Although we give a linear time algorithm for finding the mesh, the constant is large and the construction depends on other linear algorithms, such Chazelle’s linear time triangulation of polygons, which have not been implemented (as far as I know). The three main tools in the proof of Theorem 1.1 are conformal maps, thick/thin decompositions of polygons and hyperbolic tesselations. We will decompose Ω into O(n) “thick” and “thin” parts. The thin parts have simple shapes and we can easily construct an explicit mesh in each of them. The thick parts are more complicated, but we can use a conformal map to transfer a mesh from the upper half-plane, H, to the thick parts of Ω with small distortion. The mesh on H is produced using a finite piece of an infinite tesselation of H by hyperbolic right pentagons. Of these ideas, the thick/thin decomposition is the most novel, so we will take a few moments to describe it a little further. A standard technique in the theory of hyperbolic manifolds is to partition the manifold into its thick and thin parts (based on the size of the injectivity radius; see Section 2.4). Thin parts often cause technical difficulties, but this is partly compensated for by the fact that there are only a few possible types of thin parts and each has a well understood shape. Thus we can think of the manifold as consisting

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of “interesting” thick parts attached to some annoying, but explicitly described, thin parts. The manifold is considered especially nice if no thin parts occur. The thin parts of a polygon are subdomains of the interior, Ω, which are generalized triangles and quadrilaterals, with two straight edges (which are sub-segments of two sides of the polygon) and either one or two circular arc edges (which are cross cuts of the polygon). See Figure 1. In Section 5 we will give a definition of thin parts in terms of extremal distance,a conformally invariant way to measure the distance between disjoint arcs on the boundary of a Jordan domain. It will turn out that for each thin part, there is a conformal map of Ω to a thin rectangle R = (0, 1ǫ ) × (0, 1) or a half infinite strip (0, ∞) × (0, 1) with the straight sides of the thin part mapping to the horizontal edges of R. The first case is called a hyperbolic ǫ-thin part and the second is called a parabolic thin part, in analogy to the two types of thin parts of a Riemann surface (see Section 2.4).

Figure 1. Parabolic and hyperbolic thin parts are triangles and quadrilaterals respectively.

Figure 2. A polygon with one hyperbolic thin part (darker) and six parabolic thin parts.

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Figure 3. A polygon with five hyperbolic thin parts (and 42 unmarked parabolic thin parts). This figure is not to scale. In practice, we would choose the ǫ in the definition of a hyperbolic ǫ-thin part to be quite small, which gives thin parts which are much “wider” than those shown here. The parabolic thin parts occur exactly at the vertices of Ω. At such points, two edges touch and hence have extremal distance 0 in Ω. A hyperbolic ǫ-thin part corresponds to a pair of non-adjacent sides whose extremal distance in Ω is ≃ ǫ. A polygon with no hyperbolic ǫ-thin parts will be called “ǫ-thick”. Although thin parts have been defined using extremal distance and conformal maps they can be identified in linear time using the medial axis of the domain (at least if we allow a bounded ambiguity in the size of ǫ), as described in [10] and [8]. We can always add more vertices to the boundary of a polygon to make it thick, although there is no bound on how many are needed (consider a 1 × n rectangle). Also note that adding vertices can convert a thick polygon to a non-thick one, i.e., adding two vertices close to the middle of one side of a unit square creates a new hyperbolic thin part. Once we have removed the thin parts, the remaining thick components of Ω are bounded by pieces of the original boundary and the circular arc crosscuts adjacent to the thin parts. Moreover, every angle is close to 90◦ . For each such component we will construct a subdomain of H and a conformal map from this subdomain to the thick component. This conformal map has small distortion in a precise sense, and a fine enough mesh on H can be transfered to the thick part of Ω with only a small distortion of angles. The domain in H is easily meshed with angles between 60◦ and 120◦ , so transferring it by the conformal map gives a mesh with angles between 60 − ǫ and 120 + ǫ (with ǫ small if the mesh is fine enough). To remove the “±ǫ”, we will construct our mesh so that angles near 60◦ and 120◦ are only used at a few points. In a neighborhood of these points we use a conformal linear map to transfer

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the mesh (and hence have no distortion of the angles). Outside these neighborhoods, the angles are bounded uniformly away from 60◦ and 120◦ and hence transferring the mesh by a conformal map keeps the angles strictly between 60◦ and 120◦ . Since the thin parts have simple shapes, we will can give explicit meshes for them. Moreover, along the circular cross cuts, we can arrange for the vertices to be in almost any arrangement we want (with just a restriction on the minimum number used). It is therefore a simple matter to arrange for the meshes of the thin and thick parts to agree along the common boundaries and so give a mesh of the polygon. It will be clear from our construction that only the hyperbolic thin parts require long narrow pieces in the mesh. We say that a quadrilateral has “bounded geometry” if all its angles are in a compact subinterval of (0, 180) and if all four sides are of comparable length. If Ω is ǫ-thick, then our proof will show it can be meshed with O(n/ǫ) quadrilaterals of bounded geometry (still with angles between 60◦ and 120◦ ). Moreover, if {Hk }N k=1 is an enumeration of the hyperbolic thin parts and if Hk is P −1 ǫk -thin, then the polygon can be meshed with O(n + N k=1 ǫk ) quadrilaterals with bounded geometry, and this is sharp. Because of the lower bound on new angles in Theorem 1.1, angles < 120◦ on the polygon cannot be further subdivided, since this would create a new angle of size < 60◦ . By construction, vertices with angle between 120◦ and 2400 will be subdivided into two equal angles, and vertices with angle between 240◦ and 360◦ will divided into three equal angles. Next we give a more detailed sketch of the proof of Theorem 1.1 and describe the structure of the rest of the paper. Definitions and details will be given later. Given a simple polygon bounding a simply connected domain Ω, let f : H → Ω be conformal and let E ∪ {∞} be the preimages of the vertices. We form a sawtooth region W ⊂ H with these points as vertices and compute the medial axis of W (Section 5). Using the medial axis, we divide W into “thick” and “thin” parts and define a corresponding thick and thin decomposition of Ω. On each thin part of Ω we define an explicit quadrilateral mesh with all new angles between 60◦ and 120◦ (Section 7). We fix a component of the thick part of W and cover it by a finite number of isometric hyperbolic pentagons, so that the union is hyperbolically convex. The remainder of hyperbolic space is covered by a finite union of half-disks, circular arc

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triangles and circular arc quadrilaterals. The half-disks correspond to where the thin parts will be attached and do not have to be meshed. The quadrilaterals can be explicitly meshed by circular arc regions with all angles equal to 90◦ . The pentagons can be explicitly meshed by circular arc regions with angles between 72◦ and 108◦ (Section 3). The triangular regions can be explicitly meshed by circular arc regions with angles between 60◦ and 120◦ (Section 4). The meshings on different regions are chosen so that along any common boundaries, the choice of vertices agree (i.e., the meshes are consistent). We then transfer the mesh of hyperbolic space (minus the half-disks) to Ω (minus the thin parts) by mapping the vertices forward by the conformal map and connecting adjacent vertices by line segments (Section 6). If the mesh elements are small enough, the procedure of mapping forward and replacing the edges by line segments can only change angles by a small amount. For the pentagons and quadrilaterals this is fine since the angles used lie in [72, 108] and a small perturbation is still in [60, 120]. However, for the triangles some more care is needed. First, we will show that replacing all the circular arc edges by line segments in the mesh of the triangular pieces in H keeps all angles between 60◦ and 120◦ (this depends on the particular construction of the circular arc mesh). Next we show that angles near 60◦ and 120◦ are only used in a small neighborhood of one point in each triangle, called the center. When we map to Ω we use the conformal map on most of the triangular region, but use a linear approximation on a neighborhood of the center. This gives a mesh with angles between 60◦ and 120◦ . Once the details have been verified, this shows there is a quadrilateral mesh with the given angle bounds. It is easy to verify that most of the steps used are linear in the number of vertices. The non-trivial steps are the construction of the medial axis of W and of the computation of the conformal map. The medial axis of any polygon can be computed in linear time by a result of [12], although our region W is a monotone histogram, so an even easier linear time method is available (based on the convex case given in [1]). Approximation of conformal maps in linear time is described in [10] and will be discussed in greater detail in Section 6. The rest of the paper in organized as follows: Section 2: Background on M¨obius transformations and hyperbolic geometry. Section 3: Meshing hyperbolic right pentagons and quadrilaterals.

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Section 4: Meshing hyperbolic right triangles Section 5: The thick and thin parts of a sawtooth domain. Section 6: Conformal maps and transferring a mesh from H to Ω Section 7: Meshing the thin parts of a polygon. Section 8: Further remarks and questions. I would like to thank Marshall Bern for asking me the question which lead to Theorem 1.1 and pointing out his paper [4] with David Eppstein. Also thanks to Joe Mitchell for many helpful conversations on computational geometry. This paper is part of a series ([6], [7], [9], [10]) which exploits the close connection between the medial axis of a planar domain, the geometry of its hyperbolic convex hull in H3+ and the conformal map of the domain to the disk. This was originally motivated by a result of Dennis Sullivan [27] about boundaries of hyperbolic 3-manifolds and its generalization by David Epstein (only one “p” this time) and Al Marden [15]. Many thanks to those authors for the inspiration and insights they have provided.

¨ bius transformations and hyperbolic geometry 2. Mo 2.1. M¨ obius transformations. A linear fractional (or M¨obius) transformation is a map of the form z → (az + b)(/cz + d). This is a 1-1, onto, holomorphic map of the Riemann sphere S 2 = C ∪ {∞} to itself. Such maps form a group under composition and are well known to map circles to circles (if we count straight lines as circles which pass through ∞). M¨obius transforms are conformal, so they preserve angles. The non-identity M¨obius transformations are divided into three classes. Parabolic transformations have a single fixed point on S 2 and are conjugate to the translation map z → z + 1. Elliptic maps have two fixed points and are conjugate to the rotation z → λz for some |λ| = 1. The loxodromic transformations also have two fixed points and are conjugate to z → λz for some |λ| < 1. If, in addition, λ is real, then the map is called hyperbolic. Given two sets of three distinct points {z1 , z2 , z3 } and {w1 , w2 , w3 } there is a unique M¨obius transformation that sends wk → zk for k = 1, 2, 3. This map is given by the formula w1 − ζw3 w2 − w1 (z − z1 )(z2 − z3 ) τ (z) = , ζ= . 1−ζ w2 − w3 (z − z3 )(z2 − z1 )

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A M¨obius transformation sends the unit disk 1-1, onto itself iff it is if the form z→λ

z−a , 1 − a¯z

for some a ∈ D and |λ| = 1. Any loxodromic transformation of this form must actually be hyperbolic. 2.2. The hyperbolic metric. The hyperbolic metric on the unit disk is given by Z 2|dz| , ρ(z, w) = inf 2 γ 1 − |z| where the infimum is over all rectifiable arcs connecting z and w in D. This is a metric of constant negative curvature. In some sources, the “2” is omitted; this changes the magnitude of the curvature of the metric, but most properties are the same. We have chosen this version to be consistent with the trigonometric formulas found in [3] (see below). Geodesics for this metric are circular arcs which are perpendicular to the boundary (including diameters). The hyperbolic metric is well known to be invariant under M¨obius transformations of the disk, so it is enough to compute it when one point has been normalized to be 0 and the other rotated to be on the positive axis. In this case ρ(0, x) = log

1+x , 1−x

and for z ∈ D with ρ = ρ(0, z), eρ − 1 . |z| = ρ e +1 It is also convenient to consider the isometric model of the upper half-space, H. In this case the hyperbolic metric is given by ρ(z, w) = inf

Z

γ

|dz| , y

where z = x + iy, but geodesics are still circular arcs perpendicular to the boundary. In this paper, we will generally consider the upper half-space model, only because certain pictures are easier to draw in this case. Although we will not explicitly use them in this paper, the computation of our mesh depends on the hyperbolic trigonometric laws (e.g., these were used in the computation of many of the figures). Assume we have a triangle with geodesic sides

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of lengths a, b, c (possibly infinite) and opposite angles denoted α, β, γ. See Figure 4. Then there is a Law of Sines and two Laws of Cosines: sinh a sinh b sinh c = = sin α sin β sin γ cosh c = cosh a cosh b − sinh a sinh b cos γ, cosh c =

cos α cos β + cos γ sin α sin β

γ a

b

β

α c

Figure 4. Labeling of a triangle for Sine and Cosine laws. The first two correspond to the usual laws in Euclidean geometry, but the third is particular to hyperbolic geometry. Note that it implies that in hyperbolic geometry the side lengths of a triangle are determined by its angles. Any proper, simply connected plane domain has a hyperbolic metric defined by mapping back to the disk by a conformal map (this is well defined since conformal self-maps of the disk are hyperbolic isometries). It is often more convenient to deal with the more geometrically defined quasi-hyperbolic metric Z |dz| ρ(z, w) = inf , γ dist(z, ∂Ω) where the infimum is over all rectifiable curves connecting z and w in Ω. This is well known to be boundedly equivalent to the hyperbolic metric (e.g., (4.15) of [17]). It will also be convenient to consider another variant of the hyperbolic metric adapted to polygonal domains. Suppose E ⊂ R and for points z, w ∈ H define Z |dz| ρE (z, w) = inf , γ dist(z, E \ E(z)) where E(z) a point of E closest to z. In other words, ρE is defined by dividing by the distance to the second closest point of E. This metric is important to us because far from the boundary it looks like the hyperbolic metric, but near the boundary it

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remains bounded (the meaning of near and far depend on E) and we will show that if E has n points, then H has a certain quadrilateral mesh with O(n) pieces, all of which are small with respect to this metric (this could not be true for the hyperbolic metric since boundary pieces of the mesh would have infinite diameter). Moreover, if f is a conformal map of H to a polygon so that f (E) contains all the vertices, then f distorts angles of this mesh very little (depending on the ρE -diameter of the pieces), outside a uniform ρE -neighborhood of E. This will be explained further in Section 6.

Figure 5. The graph of the distance to the second nearest point. Above the solid curve ρE is comparable to the hyperbolic metric and below the curve it looks like the reciprocal of the height of the curve. We think of this as the hyperbolic metric adapted to stop growing once we are close to a single point of E or an interval of its complement. 2.3. Extremal distance. This section is not strictly needed for the proof of Theorem 1.1, but to clarify the meaning of the thick/thin decomposition, it is useful to include a few remarks about conformal modulus and extremal distance. Suppose Γ is a family of locally rectifiable paths in a planar domain Ω and ρ is a non-negative Borel function on Ω. We say ρ is admissible for Γ if Z ℓ(Γ) = inf ρds ≥ 1, γ∈Γ

and define the modulus of Γ as Mod(Γ) = inf Ω

γ

Z

ρ2 dxdy, M

where the infimum is over all admissible ρ for Γ. The reciprocal of the modulus is called the extremal length of the path family. These are important conformal invariants which are fundamental to the theory of conformal and quasiconformal mappings, and whose basic properties are discussed in many sources such as Ahlfors’

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book [2]. Basics examples to keep in mind include the path family connecting opposite sides of an a × b rectangle (if sides of length a are connected the modulus is a/b) and the family connecting the inner and outer circles of an annulus (the extremal length is log(R/r) if R > r are the radii of the circles). If Ω is a Jordan domain and I, J ⊂ ∂Ω are disjoint arcs, then the extremal distance from I to J in Ω is defined as the extremal length of the path family connecting I and J inside Ω. Roughly speaking, if the distance between two subarcs of ∂Q is much smaller than the minimum of their diameters, then the extremal distance will be small. The main case of interest to us is when Ω = H, I = [ 1r , r] and J = −I. In this case the extremal distance between I and J in Ω is

1 π

log 1r + O(1).

Computing the exact extremal distance between two sides of a general polygon is difficult in practice, (as hard as computing the conformal map from Ω to H), but for polygons the extremal distance can be estimated to within a multiplicative factor of 1 + ǫ in time O(n) (with constant depending on ǫ) by results in [10]. Moreover, there is a relatively simple way to compute extremal distance to within a multiplicative factor of 8 given the medial axis of Ω, using the so called “iota map”, as described in [8], [10]. 2.4. Thin parts of Riemann surfaces. It may also be helpful to recall what the thin part of a Riemann surface is, since the thick/thin decomposition of polygons is inspired by this case. If we have have a discrete group G of M¨obius transformations acting on the disk and G contains no elliptic elements, then the quotient space R = D/G is a Riemann surface and every Riemann surface occurs in this way, except for a short, well understood list of exceptions (the sphere, the plane, the punctured plane and genus 1 tori). The group G acts as isometries of the hyperbolic metric on D and so R has a well defined hyperbolic metric as well. Closed, homotopically non-trivial loops on R correspond to elements of the covering group G. Such a loop either has a shortest homotopic representative (i.e., a closed geodesic on R), or can be homotoped to loops which are arbitrarily short (i.e., is a loop surrounding a puncture on the surface). In the first case the loop is represented by a hyperbolic element of G and in the second case it is represented by a parabolic element. The ǫ-thin part of R consists of the points through which there is a non-trivial closed loop of length ≤ ǫ (this

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Figure 6. A surface with a hyperbolic thin part (darker) and three parabolic thin parts (lighter). says the injectivity radius at x is ≤ ǫ/2). The Margulis lemma says that there is an ǫM > 0 so that for ǫ < ǫM , the ǫ-thin part of R consists of connected components of two types: hyperbolic components which are an annular neighborhoods of a closed geodesics of length ≤ ǫ and parabolic components which consist of a neighborhood of a puncture on the surface. See Figure 6. Hyperbolic thin parts are compact and have two boundary components. Parabolic thins parts are non-compact and have a single boundary component in the surface. A similar definition of thin part makes sense for hyperbolic manifolds of any dimension. The presence of thin components can often cause technical difficulties (especially in the higher dimensional cases), but because they have a very specific structure, these difficulties can often be dealt with using explicit constructions and estimates. In this paper, we will use an analogous decomposition of a polygon into thick and thin parts and will similarly use explicit constructions on the thin parts. 2.5. Right circular triangles. A circular triangle is a collection of three vertices z1 , z2 , z3 connected by disjoint circular arcs (or line segments). We say it is a right circular triangle if all three interior angles are 90◦ and the interior is on the left as we traverse the points in the given order. For example, if the vertices are 0, 1, i we can take the edges to be the line segments [i, 0], [0, 1] and the arc of the unit circle in the first quadrant. See the left side of Figure 7. Any M¨obius image of a right circular triangle is another one and since we can map any three points to any three points, we see that there is a least one such triangle for each oriented triple of distinct points. Moreover, there is only one. Consider a second triangle T2 for the same points in the same order. It has an edge e which

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Figure 7. Two right circular triangles corresponding to same triple of points, but in opposite orientations. is not an edge of the first triangle T1 and hence is either inside or outside the first triangle. If inside, then the next edge must be outside T1 (since the interior angle for both triangles is 90◦ ). By the same reasoning, the third edge of T2 is inside T1 . But the first and third edges can’t both be inside T1 since then the angle would be strictly less than 90◦ . Thus there is exactly one triangle associated to any three points in a given orientation. Note that there is a second right circular triangle with the same three vertices, but in the opposite orientation. See the right side of Figure 7. We will also need the following observation. If C is the circle which passes through all three vertices of a right circular triangle, then it makes an angle of 45◦ with each of the edges of the triangle. This is obvious when the vertices are three corners of a square and the general case is a M¨obius image of this. See Figure 7. 2.6. Right circular quadrilaterals. We will also need to consider right circular quadrilaterals. We want to note that every such quadrilateral has two orthogonal foliations by circular arcs connecting opposite sides. Moreover, except for one case, we can take a M¨obius image of the quadrilateral so that one foliation consists of segments on rays through the origin and the other consists of arcs of circles centered at the origin. To see this, take two opposite sides. Each lies on a circle and these circles either intersect in 0, 1 or 2 points or are the same circle. In the first case we can conjugate by a M¨obius transformation so both disks are centered at 0. Then the two other sides must map to radial segments and the foliations are as claimed. Secondly, if the circles intersect in two points, we can assume these points are 0 and ∞ so the circles are both lines passing through 0 and again the foliations are radial rays and

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circles centered at 0. Thirdly, if the opposite sides belong to the same circle, we can conjugate it to be the real line, with the two sides being arcs symmetric with respect to the origin. Then the other two sides must be circular arcs centered at 0 and the two foliations are as before. The exceptional case is if the two circles intersect in one point. Then we can conjugate this point to infinity and the intersecting sides to two parallel lines. The other two sides must map to perpendicular segments and the region is foliated by perpendicular straight lines.

Figure 8. Any right circular quadrilateral is M¨obius equivalent to one of these cases and hence has an orthogonal foliations by circular arcs. A special case of a right circular quadrilateral occurs when one side is an interval on R and the other three sides are hyperbolic geodesics in H (two half-infinite and one finite). We call such a region a hyperbolic Carleson quadrilateral because of the similarity to Carleson squares (which are Euclidean squares in H with one side on R). See Figure 9. Note that a hyperbolic Carleson quadrilateral is determined up to hyperbolic isometry by the hyperbolic length of the one finite geodesic side. A hyperbolic Carleson triangle will refer to a right circular triangle in H with one edge on R. Any two such are clearly hyperbolically isometric.

Figure 9. Hyperbolic Carleson quadrilateral and triangle.

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2.7. Hyperbolic right n-gons. A proper hyperbolic polygon is a bounded region in hyperbolic space bounded by a finite number geodesic segments. The polygon is “right” if every interior angle is 90◦ . There are no proper hyperbolic right triangles or quadrilaterals, but there are hyperbolic right n-gons for every n ≥ 5 (the Carleson quadrilaterals and triangles in the previous subsection were not proper polygons since they are not bounded in hyperbolic space). See Figure 10. Each of these n-gons can be extended to a tesselation of hyperbolic space by successive reflection across the edges. In Figure 11 we show the first two generations of such a tesselation for pentagons (n = 5). Let Tn denote the collection of n-gons in the tesselation. In the complete tesselation, each edge of an n-gon lies on some hyperbolic geodesic. Each of these geodesics divides hyperbolic space into two hyperbolic half-planes and we will let Hn denote the collection of hyperbolic half-planes thus generated.

Figure 10. Examples of hyperbolic right n-gons for n = 5, 6, 7, 10.

Figure 11. On the left is a hyperbolic right pentagon and all 10 of the touching pentagons. The rest of the disk is covered by Carleson quadrilaterals and triangles. On the right is the same figure drawn in the upper half-plane model. Because the point −1 ∈ ∂D has been mapped to ∞, one of the Carleson regions is now unbounded, and adjacent ones appear very large.

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3. Meshing quadrilaterals and pentagons In this section we show how to mesh hyperbolic right pentagons and hyperbolic Carleson quadrilaterals using circular arc meshes. These cases are “easy” in the sense that we will only use angles that are strictly within [60, 120]. 3.1. Carleson quadrilaterals. We simply have to take a finite set of leaves from the orthogonal foliations discussed in Section 2.6. More specifically, the “top” edge of a Carleson quadrilateral Q is a finite segment of an infinite hyperbolic geodesic γ that has two endpoints a, b on the boundary of hyperbolic space. Thus Q has a foliation by arcs which lie on circles through a and b and an orthogonal foliation by circular arcs which are hyperbolic geodesics perpendicular to the top edge. Using a finite set of each type of foliation leaf we can mesh Q with hyperbolic quadrilaterals. See Figure 12.

Figure 12. A quadrilateral mesh of a Carleson quadrilateral. “Horizontal” edges lie on circles which pass through the same two points on the boundary. “Vertical” edges are hyperbolic geodesics perpendicular to the top edge. 3.2. Hyperbolic right pentagons. Connect the center c of the pentagon by hyperbolic geodesics to the (hyperbolic) center of each edge. This divides the pentagon into five quadrilaterals each of which has 3 right angles and an angle of 72◦ at the center. Consider one of these quadrilaterals Q with sides S1 , S2 , S3 , S4 where S1 , S2 each connects the center of the pentagon to midpoints of adjacent sides. Then S3 , S4 are each half of a side of the pentagon adjacent at a vertex v, with S3 opposite S1 and S4 opposite S2 . See Figure 13.

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S4

y 90

90

v b

Q’

Q S1

a S3 90

φ

ex

90

x

90

ey 72

c

S2

Figure 13. Definitions used in the mesh of a hyperbolic right pentagon. Each pentagon is divided into five quadrilaterals as shown.

Place a point x along S3 and let ex be the geodesic segment from S1 to S3 which meets S3 at x and makes a 90◦ angle with S3 . Similarly define an segment fy which joints y ∈ S4 to S2 . We claim that the segments cross at an angle (labeled φ in Figure 13) which is between 72◦ and 90◦ . The two segments ex , fy divide Q into four quadrilaterals, one of which contains the vertex v. This subquadrilateral, Q′ is a Lambert quadrilateral, i.e., bounded by four hyperbolic geodesic segments and having 3 right angles. The one non-right angle, φ, is a function of the hyperbolic lengths of the two opposite sides (in this case a function of a = ρ(x, v) and b = ρ(y, v)), cos(φ) = sinh a sinh b. See Theorem 7.17.1 of [3]. Clearly, φ decreases as either a or b increase. For a and b close to zero we have φ ≈ 90◦ and when a, b take their maximum value (a = b is the hyperbolic length of S3 ) we get Q′ = Q and φ = 72◦ . Thus φ takes values between 72◦ and 90◦ , as claimed. To define a mesh of Q, take a finite set of points {xk } ⊂ S3 and {yk } ⊂ S4 and take the union of segments exk , fyk . This divides Q into hyperbolic quadrilaterals with geodesic boundaries and angles between 72◦ and 108◦ . Doing this for each of the five quadrilaterals which make up the hyperbolic right pentagon gives a mesh of the pentagon. See Figure 14.

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Figure 14. On the left is a quadrilateral mesh of a hyperbolic right pentagon. The edges of the mesh are hyperbolic geodesic segments and the mesh has angles between 72◦ and 108◦ . On the right is the mesh of 11 hyperbolic right pentagons (one containing the origin and its 10 neighbors), each meshed as above. Because vertices are evenly spaced (in the hyperbolic metric) along edges of the pentagons, meshing of adjacent pentagons match up.

4. Meshing Carleson triangles Using a procedure similar to the case of pentagons, it is fairly easy to mesh Carleson triangles with small hyperbolic quadrilaterals whose angles lie between 60◦ and 120◦ . Taking a conformal image of this mesh would then give a mesh with angles bounded above by 120◦ + ǫ and below by 60◦ − ǫ. If we were willing to prove Theorem 1.1 with this weaker bound, then this section could be considerably shortened. However, to get the stronger bounds of 60 and 120 we will have to work a little harder, constructing a mesh with a particular structure and pushing the mesh to Ω by a combination of conformal and linear maps. The basic idea is to take a Carleson triangle in H and subdivide it into a smaller right circular triangle (this is called the “inner” triangle and is a Euclidean similarity of a fixed shape, independent of the shape of the outer triangle) and surround it by nine right circular arc quadrilaterals as shown in Figure 15. The inner triangle is symmetric under Euclidean rotations of 120◦ and by reflection through a vertical line. Moreover, it is divided into three quadrilaterals by connecting the center of the triangle to the midpoint of each edge by a straight line. The edges of the surrounding

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quadrilaterals connect the vertices of the inner triangle and the midpoints of its edges to points on the outer triangle.

Figure 15. The outer triangle is a Carleson triangle in the upper half-plane. Its interior is divided into an inner triangle and nine surrounding right circular quadrilaterals. The quadrilaterals are meshed as described in Section 3.1 and the inner triangle has an easy mesh similar to the mesh for pentagons in Section 3.2. The intricate part is to place the inner triangle so corresponding segments on the “left” and “right” sides of the outer triangle have equal hyperbolic length. We will describe how to do this later in this section. 4.1. Meshing a circular right triangle. Before worrying about the details of placing the inner triangle, we start with the “obvious” mesh of it. Such a triangle is a circular arc polygon with three sides and three 90◦ angles. As noted before, given three ordered points in the plane there is exact one such triangle, and any two are related by the M¨obius transformations that sends one set of three vertices to the other. Our construction is M¨obius invariant and in our figures, we consider the case when the vertices are 0, 1, i (see e.g., Figures 16 and 17.) We define three foliations on this triangle T . For each vertex v, reflect v through the circular arc on the opposite side to define a point v ∗ and foliate T by arcs which lie on circles passing through both v and v ∗ . For the specific right triangle defined above, one family consists of radial arcs (circles passing through v = 0 and v ∗ = ∞), another of arcs of circles passing through ±1 and the last consists of circles passing through ±i. Note that each foliation leaf passes through one of the vertices of T and

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is perpendicular to the opposite side. See Figure 16. There is a unique point p in the triangle which can be connected to each of the three sides by a foliation leaf and so that the three leaves meet at p making an angle of 120◦ . This is called the center of the triangle.

Figure 16. Three foliations of a Carleson triangle. Each foliation passes through an associated vertex and is perpendicular to the opposite side. The center of the triangle is the point that can be connected to all three sides by foliation paths meeting with three 120◦ angles. The three arcs connecting the center to the sides divide T into three quadrilaterals. Restrict each foliation to the two quadrilaterals which are not adjacent to the vertex it passes through. This gives two foliations on each quadrilateral. See Figure 17. Taking a finite set of leaves for each foliation gives a quadrilateral mesh of the right circular triangle.

Figure 17. The three foliations, each restricted to two of the three quadrilaterals and their union. Angles near the center are close to 120◦ and 60◦ , near the edges they are close to 90◦ . However, replacing arcs by segments in this picture gives an angle > 120◦ at the center. 4.2. A special case: the right circular equilateral triangle. When we replace the circular arc edges in the mesh by straight line segments, the angles change slightly. For example, in the rightmost picture in Figure 17, this replacement would decrease

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the central angle facing the origin, but increase the other two. In particular, replacing arcs by segments gives two angles larger than 120◦ . There is one case, however, where this does not happen. Consider the three vertices of a Euclidean equilateral triangle. The corresponding right circular arc triangle is also invariant under Euclidean rotations of 120◦ and the center of the triangle defined above agrees with the usual center. In particular, the three edges which connect the center to the sides are all Euclidean straight lines. See Figure 18.

Figure 18. For a right circular triangle with vertices on a Euclidean equilateral triangle, the edges connecting the center to the sides are straight. In this case, replacing the circular arcs of the mesh by straight lines gives all angles between 60◦ and 120◦ . For this particular triangle, we claim that replacing the circular arc edges of the quadrilateral mesh by straight line segments gives quadrilaterals whose angles lie between 60◦ and 120◦ . To prove this, consider a point z in one of the three quadrilaterals and the two foliation paths γ1 , γ2 which connect it to the two opposite vertices, v1 , v2 respectively. Let L1 , L2 be the lines through the center c and the points v1 , v2 . If we think of the arc γ1 as a graph over the line L1 it is monotonically increasing as we move away from v1 and remains increasing so as long as we stay inside the triangle (since γ1 is perpendicular the the opposite side of the triangle, the point of greatest distance from L1 occurs outside the triangle). Thus if we translate L1 to pass through the point z, we see that γ1 stays on one side of this new line up to z and on the other side beyond z. Thus any chord of γ1 in the triangle with one endpoint at z also stays on the same side of the line as the corresponding arc of γ1 . Similar for γ2 and L2 . See the right side of Figure 19. Thus if we mesh the triangle using points on the foliation paths the sides of the quadrilaterals are chords of paths like γ1 and γ2 and at each vertex there will be two

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v2 γ2

L2

c

z

L1

v1

γ

γ2 1

γ

1

Figure 19. If we choose any point z of the equilateral triangle then chords of the foliation paths with endpoint z form angles which are bounded between 60◦ and 120◦ . angles less than 120◦ and two greater than 60◦ (which are the angles formed by L1 and L2 ). Thus meshing the triangles by foliation paths and then replacing arcs of the foliations by segments gives a quadrilateral mesh with all angles between 60◦ and 120◦ (strictly between except for the three 120◦ angles at the center). 4.3. Meshing the outer triangle. We can now mesh the outer triangle by simply using foliation leaves for the inner triangle and the surrounding quadrilaterals. This gives a foliation of the whole Carleson triangle which has only one singular point at the center of the inner triangle. The foliation paths break up into six families which are illustrated in Figure 20. Taking any finite subset of leaves (as long as it includes the leaves passing thought the vertices and midpoints of the inner triangle) gives a quadrilateral mesh of the outer triangle. See Figure 21. 4.4. Placing the inner triangle. In order that our mesh of the outer triangle be consistent with meshes on adjacent pieces, we want the foliation leaves to end at specified points on the boundary. It is convenient to assume these points are determined by their hyperbolic distance to the top vertex and the same set of distances are used for both the left and right sides (we don’t care about the bottom, since there are never adjacent pieces along that edge). To satisfy this condition, any foliation leaf that connects the left and right edges must have endpoints that are equidistant from the top vertex. This places a constraint on where we can place the inner triangle. We want the inner triangle to be placed inside the outer triangle so that these conditions hold:

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Figure 20. The foliation paths corresponding to different segments on the left and right sides of the outer triangle.

Figure 21. The full mesh of a Carleson triangle. by circular quadrilaterals with all angles between 60◦ and 120◦ . If we replace the circular arcs by straight line segments, then the angles remain between 60◦ and 120◦ . Outside the inner triangle the angles are as close to 90◦ as we wish.

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(1) The surrounding quadrilaterals are right circular quadrilaterals. (2) The top vertex of the inner triangle is connected to points on the boundary of the outer triangle that are hyperbolically equidistant from the top vertex of the outer triangle (3) The inner triangle is as small as we wish compared to the outer triangle. (4) The two sides of the inner triangle adjacent to its top vertex are hyperbolic geodesics, i.e., they extend to be perpendicular to the real line. The tricky part is to position the top vertex of the inner triangle so that conditions (1), (2) and (3) hold. We spend the rest of this section showing that the set of points which satisfy (1) and (2) form a straight line which we can easily compute. We start be reviewing some simple facts about circles. 4.5. Parabolic families of circles. Suppose F denotes the set of horizontal lines in the plane. See the left side of Figure 22. Any image F ′ of F under a M¨obius transformation τ is either a set of lines in a fixed direction (if τ fixes ∞ and is thus a linear map) or is a single straight line L and the family of circles tangent to L at the point p = τ (∞). We call this a parabolic family of circles, since it consists of the orbits of a 1-parameter group of parabolic M¨obius transformations. Suppose that p = 0. See the center of Figure 22. If we superimpose F and F ′ then every point, other than p, is the intersection of a curve from each family and we can define an angle between these curves at this point. Both families are invariant under the dilations z → λz for λ > 0 and so the angles remain unchanged under dilations. Thus if we ask for the set on which the angle between the two families is a certain value, the answer is a radial line from 0 to ∞. See right side of Figure 22. Since angles are preserved by M¨obius transformations we deduce that given any two parabolic families of circles, the set where the families intersect at a given angle is a circular arc (or straight line) connecting the centers of the two families. 4.6. Suitable points form a line. A crescent is a region bounded by two circular arcs which meet at distinct points. Any such region is M¨obius equivalent to a sector {z : 0 < arg(z) < θ}. There is a natural identification between the two sides of a crescent by applying the unique elliptic M¨obius transformation that fixes the two vertices and rotates one side into the other.

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Figure 22. On the left is the horizontal family with base point at ∞, i.e., the family of horizontal lines. In the center is a parabolic family, centered at 0. Since both families are invariant under dilations z → λz, λ > 0, the angle of intersection is constant on radial lines. This means the slope of a parabolic family is constant on rays through the base point (see figure on right). Suppose T is a right circular triangle with one edge I on the real line and let e1 , e2 denote the other two edges of T and assume they meet at a vertex w in the upper half-plane. The bisector of the angle of T at w makes an angle θ between 0◦ and 45◦ with the vertical (so θ = 0 means the bisector is vertical and T is symmetric with respect to it; θ = 45◦ means one of e1 or e2 is a vertical line segment). See the left side of Figure 23. w

w

L e2

e1

x1

γ

e1

I a v

γ

b

a

γ2

1

v I

x2

e2 b

Figure 23. The definitions used in the proof of Lemma 4.1. Let L be the set of points, v, such that there there are circular arcs γ1 , γ2 from v to points x1 ∈ e1 , x2 ∈ e2 so that (1) these arcs are are perpendicular to e1 , e2 at these points, (2) both γ1 , γ2 make angle 45◦ with the vertical at v, and (3) the points x1 , x2 are equidistant from w in the hyperbolic metric of H. See the right side of Figure 23. Lemma 4.1. L is a Euclidean straight line segment that starts at w and makes angle θ/2 with the vertical.

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Proof. If we reflect the circular arc triangle T through the real axis we get a crescent with vertices at w and its conjugate w¯ and with angle 90◦ at both vertices. There is an elliptic transformation R that fixes both these vertices and maps one side of the crescent to the other. Since R preserves the upper half-plane it must be an isometry of the hyperbolic metric and therefore ρ(w, x) = ρ(w, R(x)). There is also a 1-parameter family of hyperbolic transformations Tt that fixes each of vertices and preserves the crescent. Suppose I is the base of T on R and a, b are its endpoints. Let γ be the circular arc in the lower half-plane from a to b which forms a crescent of interior angle 45◦ with I. From our previous remarks about right circular arc triangles, γ is an arc of the circle which passes through all three vertices of T . See Figure 23. Because of our previous remarks we know that γ is exactly the set of points where the horizontal parabolic families with base points a and b meet at angle 90◦ , i.e., at each point z of this arc there are a pair of circular arcs (from the two parabolic families) that connect z to each of a and b, have horizontal tangents at a and b (and hence are perpendicular to e1 and e2 ) and meet at angle 90◦ at z. Thus the bottom of this arc (where it has a horizontal tangent), is a point of L because the two arcs both make angle 45◦ with the vertical. Now apply the transformations Tt to the arc γ to bring it closer to w. Each image is still a circular arc connecting a point at of e1 to a point bt of e2 and these points are interchanged by R, so they are equidistant from w. Since M¨obius transformations preserve angles, each point x of γt is connected to two arcs to at and bt which are perpendicular to the sides of T and meet with angle 90◦ at x. At the place where γt has a horizontal tangent these arcs each meet the vertical at angle 45◦ . See Figure 24. Thus the set L consists of the points where each γt has a horizontal tangent. But each γt is an arc of a circle passing through w (since γ was and Tt fixes w). Thus each γt is an arc from a fixed parabolic family of circles. Hence the set where they make a fixed angle with the horizontal (namely zero) is a straight line. Determining the angle of this line is a simple calculation. See Figure 25.

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Figure 24. The arc γ is part of a circle which passes through all three vertices of the Carleson triangle. When we apply the 1-parameter hyperbolic family to this circle we get the family of circles with a common tangent as shown. This is part of a parabolic family, so the points with horizontal tangent form a straight line. τ θ

θ

ψ ψ

Figure 25. Suppose a circle is tangent to a line that makes angle θ with the horizontal. Label angles τ, ψ as shown. Then τ = 90 − ψ and θ = 180 − 2ψ, which implies τ = θ/2. Thus the line from the tangent point to the bottom of the circle makes angle θ/2 with the vertical, as claimed. We now know that the possible locations for the top vertex of the inner triangle form a straight line in the triangle and that this line meets the real axis at a point bounded away from the endpoints of the base of the triangle. Thus by choosing a point close enough to the real axis we can make the inner triangle as small as we wish, and by rescaling it correctly we can make its two “upper” sides geodesic segments. There is a compact 1-parameter family of Carleson triangles that we have to make this choice for, but we can choose the inner triangle so that the resulting side lengths on outer triangle are always the same. This is because when we choose a curve γt

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whose lowest point is near the real axis, the same is true for all the M¨obius images of it taken from our compact family. Thus we can choose the same value of t for every triangle (and hence the endpoints of γt on the outer boundary are the same distance from the top vertex). This completes our description of how to mesh a Carleson triangle. 5. Thick and thin parts of sawtooth domains We have discussed how to mesh certain special regions in hyperbolic space, but so far there has been no use of the particular polygon given to us in Theorem 1.1. In this section, we will use the special regions described above to construct a mesh of hyperbolic space which is adapted according to a given finite set on the boundary. In the next section, this finite set will be taken to be the preimages of the vertices of our polygon under a conformal map and we will see how to transfer the mesh to the polygon. This will complete the description of the mesh on the thick parts of the polygon. Then we will describe the mesh on the thin parts and how to join the thick and thin parts. Given a set of n points S ⊂ R, we define the associated “sawtooth” domain as the finite union W = ∪s∈S Cs where Cs is a vertical cone of angle 90◦ and with vertex at s, i.e., Cs = {x + iy : |x − s| < y}. See Figure 26. Clearly, this is an (unbounded) polygon with 2n sides. The particular choice of angle in the sawtooth domain is unimportant, and sometimes it is convenient to consider sawtooth domains W (θ) with angle other than 90◦ , i.e., each Cs is replaced by a cone of angle θ. If θ > 90◦ then W (θ) is a neighborhood of W = W (90◦ ) and consists of all points that are within a constant hyperbolic distance of W (the constant depending on θ).

Figure 26. The sawtooth domain W associated to the set S. The dashed lines indicate a hyperbolic neighborhood of the form W (θ), θ > 90◦ .

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The medial axis of a planar domain Ω is the set of centers of all open disks D ⊂ Ω such that ∂D ∩ ∂Ω has at least two points. We can compute the medial axis of W in time O(n) by a result of Chin, Snoeyink and Wang [12]. The full strength of their result is not needed; since W is a monotone histogram, the medial axis can be computed (as described in [12]) using a modification of the linear time method for convex polygons given in [1]. The medial axis of a polygon with n sides is a tree with O(n) edges. The interior of each edge consists of centers of disks which hit the boundary in exactly two points. Thus there are three types of possible edges: point-point bisectors, point-edge bisectors and edge-edge bisectors. An edge-edge bisector for W must be a vertical line segment. Enumerate all the edge-edge bisectors and compute their hyperbolic length (which is log y/x if the endpoints are at Euclidean heights x < y above R).

Figure 27. The dotted line is the medial axis of the sawtooth domain. There are five edge-edge bisectors: four parabolic (including the unbounded one) and one hyperbolic. The curved sections are parabolic arcs and correspond to point-edge bisectors. This figure is only an illustration of the approximate shape; not an exact computation. Fix some A > 1 and consider all the edge-edge bisectors that have hyperbolic length ≥ 3A. Such an edge may either have finite or infinite hyperbolic length and we refer to these cases as hyperbolic and parabolic, respectively (these correspond to hyperbolic and parabolic thin parts of surfaces which are bounded and unbounded respectively). For any such edge e, let e′ ⊂ e be the points which are more than hyperbolic distance A from the endpoints of e. This is called the central part of e and e′′ = e \ e′ is called the outer part. For each edge e, the outer part e′′ consists

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of either one or two segments (in the parabolic and hyperbolic cases respectively), each of hyperbolic length = A. Each central part e′ has one or two endpoints in W (in the same cases) and always has hyperbolic length ≥ A. Let Sˆ be the vertical projection of all the central axes onto the real line; this consists of S (which is the projection of the parabolic axes, except the one associated to ∞) and the projection of the hyperbolic axes.

e’’ e’’

x

x e’

e’

e’’

Figure 28. Examples of long edge-edge bisectors and the corresponding thin parts (parabolic on left and hyperbolic on right). We have increased the slope of ∂W to improve visibility. For each endpoint x of a central part e′ , draw a horizontal line segment in W through x with its endpoints on ∂W . This segment divides W into exactly two components. Let H denote the collection of all such horizontal segments (there are O(n) of them since there are O(n) edges of the medial axis of W ). This collection of segments divides W into a finite number of components. The components which contain a central segment e′ of a edge-edge bisector e will be called the thin parts of W (parabolic or hyperbolic depending on the type of edge it contains) and the remaining components are called the thick parts of W . Note that hyperbolic thin parts are trapezoids and parabolic thin parts are triangles one vertex on R at the corresponding point of S (except for one unbounded parabolic thin part which is a truncated cone; we can think of this as a triangle with a vertex at ∞). See Figures 28 and 29.

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thin

thick

thin

thin

thick thin

thin

Figure 29. The division of W into thick and thin regions. This is only a cartoon. For large values of A, the thick regions (gray) would extend further into each “tooth” and the thin regions (white) would be much wider. For example, if A = 10 then different components of the thick part should occur at Euclidean scales at least e−30 apart (if drawn to scale we could only show one component).

Consider a thick component W0 ⊂ W . If ∂W0 has k sides, we want to find a set of O(k) hyperbolic half-planes in H5 whose intersection K satisfies W0 ⊂ K ⊂ W0 (θ) for some absolute θ to be determined below. It was proven in [10] that if W0 has k edges, then it has hyperbolic area O(k), and so does W0 (θ) (with constant depending on θ). Thus K will be a hyperbolically convex region that is the union of O(k) right pentagons from our tesselation. Recall that each right pentagon in our tesselation is the image of the “center” pentagon in the disk as pictured on the rightmost side of Figure 10. This image is by a M¨obius transformation that takes the disk to the half-plane and maps some point of the unit circle to ∞. It is clear from Figure 10 that each point of the circle in contained on the boundary of at most two hyperbolic half-planes that bound the pentagon and that the point is bounded uniformly away from both endpoints of at least one of them (a point can’t be arbitrarily close to two distinct points). The image of this hyperbolic half-plane is a Euclidean half-disk in H whose Euclidean diameter is comparable to the Euclidean diameter of the pentagon we started with. See Figure 30. Suppose we have a tesselation pentagon which hits ∂W0 (θ) and consider the associated hyperbolic half-plane, as above (in Figure 30 the dashed line indicates part

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Figure 30. Each right pentagon is associated to a half-disk of comparable size that contains it. of ∂W0 (θ)). This half-disk contains the pentagon (hence the part of ∂W0 (θ) that it covers). If θ is large enough, then the half-disk does not hit W0 , since the hyperbolic distance from W0 to ∂W0 (θ) is large and the pentagon is a bounded hyperbolic distance from the Euclidean “top” of the half-plane (since otherwise the Euclidean sizes of the pentagon and half-plane would not be comparable). Thus we can cover ∂W0 (θ) by half-planes by first covering it with pentagons from the tesselation: given one that hits ∂W0 (θ) we can check its neighbors in bounded time and will cover the whole boundary in time O(k). Finally, using a similar argument, applied to a pentagon that hits the thin part above W0 , we can find a half-disk that contains W0 (i.e. the complement is a hyperbolic half-plane that misses W0 ). Thus in linear time we can find the desired half-planes. See Figure 31.

Figure 31. W0 is contained in an intersection of hyperbolic halfplanes which is a union of O(k) pentagons. The dashed line indicates part of ∂W0 (θ).

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Next, we have to enumerate the right pentagons whose union is K, but this is easy. For example, in [10] we showed how to cover a thick component like W0 by O(k) Whitney boxes in linear time. This also shows W has hyperbolic area O(k) since each Whitney box has area O(1). See Figure 32. These boxes have an obvious tree structure and for each box it only takes bounded time to find and test all the pentagons in T5 that hit it, assuming we know the pentagons which hit the parent box. Thus we can enumerate all the pentagons in K in linear time.

Figure 32. It is already known how to list the Whitney boxes covering a thick component and only takes bounded time per box to find the pentagons in K which hit the that box. Once we have the pentagons that make up K, We divide H \ K into hyperbolic half-planes, Carleson quadrilaterals and Carleson triangles. One half-plane is the unbounded one discussed above. The others are in 1-to-1 correspondence with the thin parts of the original sawtooth domain W that are adjacent to the component W0 . Each of these corresponds to some central edge of the medial axis of W . For each such edge, we project it vertically onto the real line and choose a half-plane from our covering of W0 (θ) which contains this point and whose endpoints are bounded away from the point (relative to the Euclidean diameter of the hyperbolic half-plane). See Figure 33. To see that there is such a half-plane, suppose x ∈ R and let z be the point on ∂W0 (θ) directly above x and let H be the hyperbolic half-plane chosen to cover z. Then the Euclidean diameter of H is comparable to the height of z which implies dist(x, I c ) > ǫ|I| for some absolute ǫ > 0, where I is the base of H. b of H is a union When we remove these half-planes from H\K, the remaining part Ω

of O(k) Carleson quadrilaterals and triangles. Since we already know how to mesh pentagons, quadrilaterals and triangles (compatibly across any common boundaries) b i.e., we can mesh a region that looks like Figure 34. Also note we can now mesh Ω,

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Figure 33. For each thin part adjacent to W0 we choose a hyperbolic half-plane that covers the projection of the corresponding central axis. b omits a ρS -neighborhood of fixed size around each point of S and that on Ω, b that Ω ρS is comparable to the quasihyperbolic metric on the complement of S.

Figure 34. When we remove hyperbolic half-planes that cover the ˆ of hyperbolic space looks like adjacent thin parts, the remainder, Ω, this. Any two different components of the boundary are at least hyperbolic distance 2A apart. Thus the figure is only qualitative: in a true drawing the smaller components would be at most e−2A times the size of the largest component. The final step of our adapted mesh involves a “collar” around each of the attached thin parts. We have meshed up to the edge of a half-disk H which contains a thin part. This thin part has an associated axis e and this axis projects to a point x on the boundary. The point x is bounded away from the endpoints of the base I of H, but we would like to have it be the center of I. So replace H by a smaller half-disk H ′ which is centered at x and mesh the region H \ H ′ as illustrated in Figure 35. The idea is simply to take the obvious mesh by radial segments and concentric circles, which works when H and H ′ are concentric and map it by the inverse of the M¨obius transformation which fixes the endpoint of I and centers H ′ .

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Since M¨obius transformations are conformal, this mesh will have all angles close to 90◦ if the quadrilaterals have small enough Euclidean size compared to H and H ′ .

Figure 35. Meshing the region between two non-concentric circles by taking a M¨obius image of the obvious meshing between concentric ˆ is circles. By this procedure we can make sure our meshed region, Ω, bounded by half-circles which are centered at points of Sˆ (i.e., points of S or projections of hyperbolic central axes).

6. Conformal mapping The Riemann mapping theorem says that given any simply connected planar domain Ω (other than the whole plane) there is a 1-1, onto, holomorphic map of the disk onto Ω. Moreover, we may map 0 to any point of Ω and specify the argument of the derivative at 0. Such a mapping is conformal, i.e., it preserves angles locally. More importantly, a conformal mapping is close to linear on small balls with estimates that depend on the ball but not on the mapping. Koebe’s estimate (e.g. Cor. 4.4 of [17]) says that if f is a conformal map between simply connected domains Ω1 and Ω2 and z ∈ Ω1 then dist(z, ∂Ω1 ) 1 ′ |f (z)| ≤ ≤ 4|f ′ (z)|. 4 dist(f (z), ∂Ω2 ) If Ω1 = D we can replace the right hand “4” by a “1”. The closely related distortion theorem states (Equation (4.17) of [17]) that if f is conformal on the unit disk, then 1 − |z| |f ′ (z)| 1 + |z| ≤ ≤ , (1 + |z|)2 |f ′ (0)| (1 − |z|)3 This says that on small balls a conformal map is uniformly close to a linear map. More precisely, if f is conformal on a ball B(w, r) then (6.1)

|f (z) − L(z)| ≤ O(ǫ2 |f ′ (z)|),

for |z − w| ≤ ǫr, where L(z) = f (w) + (z − w)f ′ (w) is a Euclidean similarity.

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We are considering maps of the half-plane to a polygon and in this case the map f extends conformally by Schwartz reflection across each boundary arc not containing a prevertex (= preimage of a vertex) of the polygon. Thus (6.1) applies to the extension, and r is the Euclidean distance to the nearest prevertex. Hence the map f is close to linear as long as the distances involved are small compared to the distance to the nearest prevertex (which we noted earlier was comparable to the ρS distance b Thus as long as our quadrilaterals have angles bounded uniformly away from on Ω).

60◦ and 120◦ and we take them small enough, the image mesh will also have angles between 60◦ and 120◦ .

The only place in our mesh where angles near 60◦ or 120◦ are used are near the centers of the inner triangles in the meshing of Carleson triangles. Moreover, the inner triangles could be chosen to be as small as we wish compared to the outer triangle (and hence small compared to the distance to the nearest prevertex point). Therefore we can use a single linear approximation on the inner triangle and use the conformal map on the remainder of the outer triangle. Near the boundary of the inner triangle, all angles used are near 90◦ and because the conformal map is a close approximation to the linear map there, the images of the quadrilaterals adjacent to the boundary of the inner triangle are almost preserved. Thus in all cases, the image mesh has angles between 60◦ and 120◦ . The next step is to consider what the mesh looks like on the half-circle boundary b We claim that in the image these boundary comcomponents of the filled region Ω. ponents still look like circular arcs, but possibly with angle measure other than 180◦ .

Moreover, the images of the vertex points have approximately the same distribution as before, up to a linear change of angle. Finally, the errors are small with estimates that only depend on size of A in the definition of thin part. We claim that if A is large enough then the conformal map is close to a power b ∩ H, where, as before, x is the function z → a + b(z − x)α on each component of ∂ Ω center of the base I of H (and by our collar construction at the end of the previous

section this is the vertical projection of the associated medial axis edge onto the real line). For parabolic thin parts this is easy since the thin part is mapped into a neighborhood of a vertex in Ω which looks like a cone. The conformal map onto Ω can

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CHRISTOPHER J. BISHOP

then be compared to the conformal map onto the cone (which is a power function) using the following result (which is Lemma 24 of [10]). See Figure 36.

Figure 36. In a parabolic thin part the map looks like a power map. A power functions sends circular arcs around the prevertex to circular arcs around the vertex and rescales the argument linearly.

Lemma 6.1. Suppose R is the rectangle [0, L] × [0, 1] and Ω is a simply connected domain containing R which is formed by replacing the two vertical sides of R by curves. Let f : R → Ω be the conformal map such that f (c) = c and f ′ (c) > 0, where c=

L +i 21 2

denotes the center of R. Then |f (z)−z| ≤ O(e−L/2π ) for |z −c| < 2. More,

generally, the conclusion remains true if R is replaced by any generalized quadrilateral Ω normalized so that d = dist(z, ∂Ω) = 1 and the replacements are performed on two sides whose extremal distance from Dz = D(z, 21 .) is at least L. This says that for z in the central section of a thin part of the sawtooth domain Ω, the conformal map in a neighborhood of z only depends on f (z), f ′ (z) and the shape of the thin part in the image, at least up to an error which tends to zero exponentially as A → ∞. This should be clearer in Figure 37. The left side shows a parabolic thin part which is shaded gray. The region between the dashed arcs is the outer section and close to the vertex is the central part. The lemma says that, except for a very small error, the conformal map into the central part does not depend on what Ω looks like outside the outer part. In particular, if we were to replace the part of Ω outside the thin part by an infinite cone that extended the sides of the thin part, then the conformal map to the cone is a good approximation to the map to Ω (inside the central section of the thin part). The conformal map from H to a cone is just a power function, as claimed earlier. A function of the form z → z α maps circles centered at 0 to other circles centered at 0 and arguments are changed linearly by the factor α. This proves our other claims.

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Figure 37. On the left is a parabolic thin part and on the right is a hyperbolic thin part. At points which are far (more than hyperbolic distance A) from the ǫ0 -thick part the conformal map is close to a power function in the sense of Lemma 6.1.

The right side of Figure 37 shows the analogous situation for hyperbolic thin parts. As before, the map inside the central section of the thin part does not depend on (up to a small error) what Ω looks like outside the whole thin part. The only difference from the parabolic case is that when we try to simplify Ω by changing it outside the thin part, there are a few more possibilities to consider than just infinite cones. Define a collection E (for “elementary”) of conformal maps of the upper half-plane to regions bounded by at most three straight edges (segments, rays or lines) with at most one bounded segment used. As special cases we take the functions f (z) = az +b (one boundary line), f (z) = az α + b (two boundary rays) and f (z) = a log z + b (two parallel boundary lines). When there are three sides there must be two rays and a finite segment. These are the possible regions we use to simplify Ω. The argument above shows that the map for Ω can be approximated by the map onto one of these regions. Next we have to observe that each region has a map which is also approximately a power function (at least if A is large).

Figure 38. These are the special domains corresponding to E.

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The Schwarz-Christoffel formula gives a formula for the Riemann map of the upper half-plane onto a polygonal region Ω: if the interior angles of P are απ = {α1 π, . . . , αn π} and the preimages of the vertices are {z1 , . . . , zn , then Z n−1 Y f (z) = A + C (w − zk )αk −1 dw. k=1

See e.g., [14], [21], [28]. The formula was discovered independently by Christoffel in 1867 [13] and Schwarz in 1869 [24], [25]. For other references and a brief history see Section 1.2 of [14]. The problem with the formula is that we know the α’s but not the z’s, so all we have done is reduce a 2n-parameter problem to finding n unknown prevertices. However, if there are only three vertices, then we can place them anywhere we want by a M¨obius transformation. For example, to map H to a polygonal region with one vertex at ∞, we can place the prevertices at ∞ and ±ǫ so the derivative of the conformal map is f ′ (z) = (z − ǫ)α1 −1 (z + ǫ)α2 −1 . For points z with |z| ≥ 1, this satisfies f ′ (z) = z α1 +α2 −2 + O(ǫ/|z|2 ). In particular, f is close to the power map onto the cone corresponding to ǫ = 0, i.e., radial segments in {|z| > 1} are mapped to curves that are almost radial, in the sense that the angle they make with radial lines is O(ǫ/|z|2 ). So for hyperbolic thin parts, the mapping must approximate a mapping onto an elementary region, and the mapping for the elementary region approximates a power function, in the sense that radial lines map close to radial lines. The approximation is as close as we wish if A is sufficiently large. This completes the argument for the hyperbolic thin parts. We have now proven the existence of the desired quadrilateral mesh on the thick part of Ω. However, this is not quite a linear time algorithm for computing it, since we have used evaluations of conformal maps without an estimate of the work involved. We address this now. The exact conformal map onto a general polygon probably can’t be computed in finite time, but we can compute an approximate map onto a simple n-gon in time O(n) with a constant depending only on the desired accuracy.

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In [10] I show that a (1 + ǫ)-quasiconformal map from H to Ω can be computed and evaluated at n points in time O(n) where the constant depends only on ǫ. I will refer to [10] for the definition and relevant properties of quasiconformal mappings, and only point out here that if f : H → Ω is conformal and g : H → Ω is the (1 + ǫ)-quasiconformal approximation constructed in [10], then ρE (f −1 (g(z)), z) = O(ǫ), for all z ∈ H. In particular, if we have a Euclidean quadrilateral Q in our mesh, then the g images of the vertices agree with the f images of points that are very close in the ρE metric. Suppose these points define a quadrilateral Q′ . Then the angles of Q′ are within O(ǫ) of those of Q and so the f image of these points form a quadrilateral in Ω with angles close to Q. But this is the same as the g image of the original points, so we see that using g to transfer the mesh vertices also gives quadrilaterals with angles that are as close as we wish to angles used on H. The fast Riemann mapping theorem given in [10] implies: Theorem 6.2. Suppose we are given a simply connected region Ω bounded by an n-gon and any ǫ > 0. We can compute a set S ⊂ R, the corresponding quadrilateral mesh of H, and a map g on vertices of the mesh which extends to a (1 + ǫ)quasiconformal map of the disk to Ω taking the set S to the vertices of ∂Ω. The vertices of a quadrilateral with ρE -diameter ≤ ǫ and at least ρE distance ǫ0 away from S have images which form a quadrilateral whose angles are within O(ǫ) of the original. The total work is O(n) where the constant may depend on ǫ, but is no worse than O(log 1ǫ log log 1ǫ ). 7. Meshing the thin parts We have now completed the most difficult portion of the proof of Theorem 1.1, which is meshing of the thick parts using conformal maps and hyperbolic geometry. What remains is to mesh the thin parts so that the vertices used match the mesh of the thick parts along the common boundaries. All of this can be done by elementary, explicit constructions which we now consider, case by case. 7.1. Parabolic thin parts. If a vertex of our original polygon has angle between 0◦ and 120◦ we mesh a neighborhood of it as illustrated in Figure 39. The labeled

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angles satisfy 0 < θ ≤ 120 1 1 60 ≤ θ1 = 180 − 60 − θ = 120 − θ ≤ 120 2 2 1 1 1 60 ≤ θ2 = (180 − θ) = 90 − θ ≤ 90 2 2 4

θ2

θ1

120 120

120

θ

Figure 39. This is the mesh in a conical neighborhood of a vertex of angle between 0 and 120. For a vertex with angle between 120 and 240 we divide the angle into two equal subangles and mesh each smaller angle as above. For a vertex with angle between 240 and 360 we divide into three equal angles and mesh each as above. In every case we have a quadralateral mesh of a polygonal neighborhood of the vertex whose vertices lie either on the original edges of the polygon or on a circular arc concentric with the vertex. 7.2. Attaching a parabolic thin part to the thick part. Parabolic thin parts are cones with tip at a vertex of the polygon. This cone can be divided into a smaller cone and a number of sectors using circular arcs centered at the tip of the cone. See Figure 40. By making the parabolic part smaller, we can separate the thick and thin parts by a sector and this sector may be as “wide” as we wish (i.e., the extremal distance between the two circular arc edges may be as large as we wish). On each circular arc edge we have a number of vertices from the meshings on the thick and thin parts respectively. However, the number of straight segments inscribed on each edge need

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Figure 40. A cone and a sector. not be the same (based on our construction we would expect far more vertices on the edge of the thick part than on the edge of the thin part). For example, Figure 41 shows a such a sector with N1 segments on one arc and N2 on the other. Let N = max(N2 , N1 ). We subdivide the sector into two further sectors with a circular arc concentric with the other two and place N + 1 equally spaced vertices on it. To join the meshes on the thick and thin parts we have to mesh each of these subsectors with only the given vertices on the circular arcs (but adding any vertices we want on the straight segments).

N2

N N1

Figure 41. Because the construction in the previous subsection subdivides angles > 120◦ and then subdivides the corresponding outer arcs, we may assume our sectors have angle ≤ 60◦ .

7.3. Increasing the number of vertices on an arc. There are basically two cases to consider, N1 < N2 and N2 < N1 , which require slightly different constructions. If N1 < N2 the basic building block will be the mesh illustrated in Figure 42, which meshes a sector with one segment on the inner arc and two on the outer.

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CHRISTOPHER J. BISHOP

θ3

θ1 θ

θ2

120

60

60

120

90 θ 5 90 θ4

Figure 42. This shows how we can cut a sector into two equal angles using a quadrilateral mesh with angles between 60◦ and 120◦ . We assume the angle of the original sector is ≤ 60◦ . Every angle is the mesh is equal to one of the labeled angles or the complement of one of them, and it is elementary to check that the labeled angles satisfy: 0 ≤ θ ≤ 60 1 1 90 ≤ θ1 = 180 − (180 − θ) = 90 + θ ≤ 120 2 2 60 ≤ θ2 = 360 − 60 − 2θ1 = 120 − θ ≤ 120 1 75 ≤ θ3 = 90 − θ ≤ 90 4 1 60 ≤ θ4 = 90 − θ ≤ 90 2 1 90 ≤ θ5 = 360 − 120 − 60 − θ4 = 90 + θ ≤ 120 2 By repeatedly using the same construction we can increase the number of vertices on the outer arc by a factor of two at each stage. If N2 ≤ 2k N1 (this must hold for some uniformly bounded k since N1 , N2 are both ≥ 1 and uniformly bounded), then we form k generations of sectors as illustrated on the left side of Figure 43, where the angle of the sectors goes down by a factor of two at each stage. In the first k − 1 generations we put a copy of the mesh from Figure 42, suitable reflected so that adjacent copies of the same generation have matching vertices along the common radial edges. In the last generation we similarly place copies of the mesh into enough sectors to attain the desired number of vertices on the outer arc. In the remaining

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sectors of the kth generation we simply put chords of the appropriate circles to join the radial vertices of the constructed mesh to the sides of the original sector. See Figure 43.

Figure 43. If N1 < N2 we subdivide the sector repeatedly as shown on the left and insert copies of the mesh from Figure 42 to get the desired number of vertices on the outer arc. On the right is an example with N1 = 1, N2 = 6 and k = 3. If N1 > N2 then we follow a similar procedure, but this seems to require a slightly more complicated construction than in the N1 < N2 case, as illustrated in Figure 44. This case can only arise when attaching a hyperbolic thin part to a thick part. In this situation, we can assume the number of vertices on the circular arc boundaries is large (by choosing the mesh in hyperbolic space to have lots of vertices on the edges of the unmeshed half-spaces) and hence assume that the angle measure of each sector that needs to be split is small. Here the angles satisfy 0 < θ ≤ 20 70 ≤ θ1 = 180 − 90 − θ ≤ 90 1 1 80 ≤ θ2 = (180 − θ) = 90 − θ ≤ 90 2 2 1 1 1 85 ≤ θ3 = (180 − θ) = 90 − θ ≤ 90 2 2 4 3 90 ≤ θ4 ≤ θ5 ≤ θ6 ≤ θ7 = 90 + θ ≤ 105 4 1 θ9 − 90 = (90 − θ2 ) + (90 − θ8 ) = θ + (90 − θ8 ) 2 1 It easy to see that |90−θ8 | → 2 θ as the distance between the points A and B increases. Thus if the sector is long enough we have 70 ≤ θ8 ≤ 110 and 60 ≤ θ9 ≤ 120, as desired.

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θ3 θ2 θ1

C B θ7

90

θ6

120

θ5

120

θ4

90

A

θ9

90

θ8

60

60

90

Figure 44. This is the analog to Figure 42 when N1 > N2 . The outer arc has two segments and the inner arc has four. We also need to assume a smaller range for θ.

Figure 45. This is the analog of Figure 43 when N1 > N2 . Inside each of these sectors we place a copy of the mesh in Figure 44 (except for possibly some sectors in the innermost ring where we only place chords to connect vertices to the edge of the big sector.)

7.4. Meshing between arcs with the same number of vertices. We now have to mesh a sector that has the same number of vertices on each of the circular arc edges, but which are not necessarily arranged with the same spacing. This is easy to do once we notice how easy the corresponding problem on a rectangle is. If we have N points on each length 1 side of a 1 × r rectangle then connecting corresponding points gives quadrilaterals with all angles uniformly near 90◦ if r is large. For sectors, we simply have to “exponentiate” this construction.

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Suppose we have a sector with N vertices on each of the two bounding circles. We map the sector to a rectangle by a logarithm, so that the concentric circular arcs go to two vertical sides of height ≤ 2π and the two radial edges go to two horizontal sides of length r = log(r2 /r1 ). Thus we are given a rectangle R and N points on each of N the vertical sides, say {xk }N k=1 and {yk }k=1 in increasing order. Then draw a straight

line from xk to yk for each k = 1, 2, . . . , N and add vertical lines any way you wish. This gives a quadrilateral mesh of the rectangle with all angles within arcsin(π/r) of π 2

= 90◦ . Take the image of this mesh under the exponential map and we get the

vertices of a quadrilateral mesh of the sector that has all angles close to 90◦ , if r2 /r1 is large and the maximum angular spacing between the given vertices is small. See Figure 46.

Figure 46. Meshing the region between two concentric circles with non-aligned vertices by taking exponential map image of obvious meshing between vertical sides of a rectangle. 7.5. Hyperbolic thin parts. When connecting a parabolic thin part to a thick part, we could assume the connecting region is part of a sector, i.e., the straight edges meet at a point and the circular arc edges are concentric at this point. Moreover, the inner edge (i.e., the circular arc edge closer to the intersection point) may be assumed to have fewer vertices since at most seven are used for the outer edge of a parabolic thin part, and we may choose any larger number we want for the edge corresponding to the thick part. A hyperbolic thin part may also be a sector, but more generally is a region bounded by two straight line segments and two circular arcs. The lines need not intersect (or may intersect far from the thin part) but they must both hit a disk which is small compared to diameter of the thin part (depending on the size of A in the definition of

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thin part). For example, Figure 47 shows a hyperbolic thin part where this happens. This thin part is not a sector, but we can write it as the union of a sector and a thin quadrilateral.

Figure 47. Hyperbolic thin parts need not be sectors, but they become sectors after removing a thin quadrilateral along one edge.

The hyperbolic thin parts are then easy to mesh. We just have to mesh piece of the part that looks like a sector (any vertices that occur on the side of the sector shared by the narrow quadrilateral are simply connected to the opposite side of the quadrilateral by a perpendicular segment). Moreover, using the sector meshes we described above for connecting a parabolic thin part to a thick part, we can reduce to the case when both circular arc edges have a equal number of equidistributed vertices. We then simply use radial segments to mesh the sector into a bounded number of narrow quadrilaterals. It is clear that all the angles are close to 90◦ if the number of points is large and the size of A in the definition of thin part is large. Note that this is the only step where we need long, narrow quadrilaterals in the mesh. Thus if there are no hyperbolic thin parts, not only are all the (new) angles in the mesh bounded between 60◦ and 120◦ , all the quadrilaterals have bounded geometry. Within a hyperbolic ǫ-thin part O(1) quadrilaterals of modulus ǫ are used. Each can be broken up into O(1/ǫ) quadrilaterals of bounded geometry. Thus if {Hk } enumerates the hyperbolic thin parts and Hk is ǫk -thin, the whole polygon can be P meshed by O(n + ǫ−1 k ) quadrilaterals of bounded geometry and this is optimal.

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8. Concluding remarks 8.1. Non-simple polygons. Theorem 1.1 also holds for non-simple polygonal boundaries, if we use a certain definition of what a quadrilateral mesh of a non-simple polygon is. We have to treat original edges of the original domain differently than the interior edges we introduce. Boundary edges must be treated as having two distinct sides and new vertices we add while meshing must belong to one side or the other, but not necessarily both. For example, the mesh on the left in Figure 48 would be considered a valid quadrilateral mesh. The shaded region looks like it has five vertices, but the one labeled b was introduced because of a mesh edge “below” the boundary slit and is not counted as a vertex of the shaded region above the boundary.

a b

Figure 48. The result is true for non-simple boundaries, if well agree that the shaded region in the left figure is a quadrilateral. Otherwise, a single vertex can generate a path of mesh edges giving rise to n new vertices. Adding n such points requires n2 new vertices in the mesh. If we do not adopt the “two-sided” interpretation of boundary edges, then there need not be an O(n) mesh. See the right side of Figure 48. At the vertex in the center of the bottom edge there is an angle of 180◦ which must be subdivided and so there must be an edge of the mesh which goes “up” making at least angle 60◦ with the bottom edge and creating a new vertex on the line above. If this vertex is considered as a vertex on the top side of the edge, then there is an angle of 180◦ there which must be subdivided, so there must be a mesh edge going up from this point. Continuing, we get a path that stays in a 60◦ cone above the starting point (the shaded region on right side of Figure 48) and hits every horizontal line. By taking n horizontal lines and n points on the bottom line (spaced far enough apart) we see that at least n2 vertices may be needed.

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8.2. Multiply connected domains. If we agree to the “one-sided” interpretation of boundary edges discussed above, then it seems reasonable that any region bounded by n points and segments has a O(n) size quadrilateral mesh with all new angles between 60◦ and 120◦ , although the methods of this paper are not quite enough to establish this. To extend the proof to multiply connected domains will require a few more steps. First, a new type of thin part needs to be introduced. This will be an annulus which surrounds isolated boundary points, or boundary components which are small compared to the distance to the next component or clusters of small boundary components. Once these thin parts have been removed, what remains is a union of uniformly perfect domains, i.e., there are no isolated boundary points in a quantitative sense. Uniformly perfect domains include many common examples from analysis and dynamics, and share several properties with simply connected domains. There is a large literature dealing with them including [11], [19], [22], [23], [26]. In [18] Mar´ıa Jos´e Gonz´alez showed that any uniformly perfect domain has a set of cross cuts which are chord arc curves and so that the complement of these cross cuts is a simply connected subdomain that lifts to a fundamental domain of the covering group whose boundary is a chord-arc curve in D. We need to reprove Gonzales’ theorem and show that if the domain has polygonal boundary with n sides then her cross cuts can also be taken to be polygonal with at most O(n) sides. We then “thicken” the crosscuts and apply the construction in this paper to the remaining simply connected domain and then show that the mesh can be extended across the thickened cross cuts. However, it is not clear that the linear time bound can be maintained. If we do not take the one-sided interpretation, is there always a O(n2 ) mesh? 8.3. Mumford-Bers compactness. One of the most important results about the thick/thin decomposition of Riemann surfaces is Mumford-Bers compactness [20], [5], which basically says that the set of surfaces with a given genus and no ǫ-hyperbolic thin parts form a compact family. A similar statement can be formulated for polygons: given any sequence of n-gons we can form a subsequence in which the thick parts are 1-1 correspondence and have linear renormalizations which converge to thick polygons (with total number of sides O(n)). Thus a sequence of n-gons can only degenerate because of the degeneration of thin parts. Is this useful?

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49

Later in this section we will define the dome of a domain Ω as a surface in the upper half-3-space which meets the plane exactly along the boundary of Ω. For a polygon we can reflect this surface to the lower half-space, and glue the two copies together along the interiors of the edges of the polygon. This gives a Riemann surface which is homeomorphic to the sphere with n punctures. The usual thin parts of this surface are in 1-to-1 correspondence with the thin parts of the polygon (with the types also matching), although the size of the ǫ can vary by at most a bounded multiplicative factor. Does this Riemann surface (or its associated covering group) tell us anything useful about the polygon? 8.4. Gap-Crescent decompositions of finite unions of disks. The proof in this paper depends upon a precise form of the Riemann mapping theorem which gives a linear bound on the work needed to approximate a conformal mapping to a polygon. In practice, this approximation may be too complicated to actually compute, but there are some ideas in the proof of the fast mapping theorem which may be more directly related to meshing. These ideas involve the “normal crescent decomposition”, of a finite union of disks, so we start with a few words about approximating polygons by unions of disks. As described in [10], polygons can be approximated by taking a finite union of medial axis disks. If there are no hyperbolic thin parts, then an n-gon can be nicely approximated by a union of O(n) disks in the sense that there is an “close to conformal” quasiconformal map from one to the other. For example, consider the polygon in Figure 49. The approximating domain can be quasiconformally mapped to the original polygon by a simple map that “straightens” the boundary and is close to conformal on the interior, so that a mesh of the approximation can be transfered to the original with small distortion of angles (this works away from the vertices; near the vertices will require special constructions, as in this paper). Also as is described in [10], any simply connected domain Ω which is a finite union of disks has a “normal crescent” decomposition as a finite union of crescents (a region bounded by two circular arcs) and ideal hyperbolic triangles (a region bounded by three pairwise tangent circles). Connected unions of the triangles are called gaps. In the lower left of Figure 49, crescents are shaded and the gaps are white. A few more examples are shown in Figures 53 and 54. This decomposition is closely associated

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Figure 49. A simple polygon, an approximation by a finite union of medial axis disks, the corresponding normal crescent decomposition and dome. to a surface in the upper half space called the dome of the domain. This is the upper envelope of the union of all hemispheres whose bases on R2 are subdisks of the domain (in fact, it is the union of those whose bases are medial axis disks). When the base domain is a finite union of disks, the dome is a finite union of hyperbolic geodesic faces (i.e., they lie on Euclidean hemispheres) joined along hyperbolic geodesics. The lower right picture in Figure 49 shows the dome of the finite union of disks above it. Different faces are shaded in alternating colors to make them easier to see. There is a natural map from the base domain to the dome which is the boundary extension of the nearest point projection from hyperbolic space onto the dome. The preimages of the faces are the gaps and the preimages of the joining geodesics are the normal crescents. For more details and illustrations, see [10]. Viewed as a surface in hyperbolic space, the dome is the isometric to the hyperbolic unit disk and the isometry is simple to explicitly compute. If one takes the isometry from the disk to the dome and then projects the dome down to the base domain (in a certain non-obvious way), one can get an approximation of the Riemann map from the disk to the domain. This was the result of Sullivan, Epstein and Marden referred

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to at the end of the introduction. The fact that this map can be computed in linear time and provides a uniformly good approximation to the Riemann map (independent of the geometry of the domain) is the inspiration for the more complicated linear time conformal mapping algorithm given in [10]. Is the rough approximation, by itself, sufficient to transfer meshes from the disk to a domain and retain good properties? 8.5. The E and G foliations. The crescents have an obvious pair of orthogonal foliations: circular arcs through the pair of vertices and circular arcs perpendicular to the boundary arcs. See Figure 50. There is an analogous foliation of ideal triangles, as shown in Figure 51, where the vertices of the triangle are taken to be 0, 1, ∞ (and can be transfered to any other ideal triangle by a M¨obius transformation). Thus we can foliate any finite union of disks by two simple orthogonal foliations by circular arcs, except for a small triangle in the center of each ideal triangle. See Figure 52 for an example for a simple union of disks. Can we use these foliations to find a “nice” mesh of Ω? For more details about the foliations see [9] and [16].

Figure 50. The E and G foliations of a crescent.

Figure 51. The E and G foliations of an ideal triangle. Note that there is a central triangle which is not foliated.

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Figure 52. The E and G foliations for the finitely bent domain Ω0 (plotted separately and together) 8.6. Continuation methods. The normal crescent decomposition is also useful because we can define a family of domains Ωt by multiplying the angle of each crescent by a factor t ∈ [0, 1] and moving the gaps by appropriate M¨obius transformations. When t = 1 we have the original domain. For t = 0 we get the disk (i.e., when we collapse the crescents to zero angle, the gaps come together to form a disk). The important thing about this path of domains is that it is continuous in a quasiconformal sense, i.e., for 0 ≤ s < t ≤ 1 there is an explicit map from ∂Ωs to ∂Ωt which has a quasiconformal extension to the interiors with QC-constant 1+O(|s−t|). This means that the map is close to conformal, i.e., perturbs angles by at most O(|s−t|). Suppose that we have a problem that (1) we can solve for points on a circle and (2) given a solution for a domain we can find a solution on small quasiconformal perturbations. Then we can find a solution on Ω using O(1) intermediate domains between D and Ω. This is exactly the strategy used to prove the fast Riemann mapping theorem in [10]. Are there meshing problems that can be attacked in this way? A few examples of angle scaling families are given in Figures 53 to 54. References [1] A. Aggarwal, L. J. Guibas, J. Saxe, and P. W. Shor. A linear-time algorithm for computing the Vorono˘ı diagram of a convex polygon. Discrete Comput. Geom., 4(6):591–604, 1989. [2] L. V. Ahlfors. Lectures on quasiconformal mappings. The Wadsworth & Brooks/Cole Mathematics Series. Wadsworth & Brooks/Cole Advanced Books & Software, Monterey, CA, 1987. With the assistance of Clifford J. Earle, Jr., Reprint of the 1966 original. [3] A.F. Beardon. The geometry of discrete groups. Springer-Verlag, New York, 1983.

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Figure 53. The one parameter family connecting the disk to an approximation of the square. In each picture the angles have been multiplied by t = 0, .2, .4, .6, .8, 1

Figure 54. Another angle scaling family corresponding to a nonsimple polygon.

53

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[4] M. Bern and D. Eppstein. Quadrilateral meshing by circle packing. Internat. J. Comput. Geom. Appl., 10(4):347–360, 2000. Selected papers from the Sixth International Meshing Roundtable, Part II (Park City, UT, 1997). [5] L. Bers. A remark on Mumford’s compactness theorem. Israel J. Math., 12:400–407, 1972. [6] C.J. Bishop. Divergence groups have the Bowen property. Ann. of Math. (2), 154(1):205–217, 2001. [7] C.J. Bishop. Quasiconformal Lipschitz maps, Sullivan’s convex hull theorem and Brennan’s conjecture. Ark. Mat., 40(1):1–26, 2002. [8] C.J. Bishop. A fast QC-mapping theorem for polygons. 2003. Preprint. [9] C.J. Bishop. An explicit constant for Sullivan’s convex hull theorem. In In the tradition of Ahlfors and Bers, III, volume 355 of Contemp. Math., pages 41–69. Amer. Math. Soc., Providence, RI, 2004. [10] C.J. Bishop. Conformal mapping in linear time. 2006. Preprint. [11] Richard D. Canary. The Poincar´e metric and a conformal version of a theorem of Thurston. Duke Math. J., 64(2):349–359, 1991. [12] F. Chin, J. Snoeyink, and C. A. Wang. Finding the medial axis of a simple polygon in linear time. Discrete Comput. Geom., 21(3):405–420, 1999. [13] E.B. Christoffle. Sul problema della tempurature stazonaire e la rappresetazione di una data superficie. Ann. Mat. Pura Appl. Serie II, pages 89–103, 1867. [14] T.A. Driscoll and L.N. Trefethen. Schwarz-Christoffel mapping, volume 8 of Cambridge Monographs on Applied and Computational Mathematics. Cambridge University Press, Cambridge, 2002. [15] D.B.A. Epstein and A. Marden. Convex hulls in hyperbolic space, a theorem of Sullivan, and measured pleated surfaces. In Analytical and geometric aspects of hyperbolic space (Coventry/Durham, 1984), volume 111 of London Math. Soc. Lecture Note Ser., pages 113–253. Cambridge Univ. Press, Cambridge, 1987. [16] D.B.A. Epstein and A. Marden. Convex hulls in hyperbolic spaces, a theorem of Sullivan and measured pleated surfaces. In Analaytical and geometric aspects of hyperbolic spaces, London Math. Soc. Lecture Notes Series 111, pages 113–253. Cambridge University Press, 1987. [17] J.B. Garnett and D.E. Marshall. Harmonic measure, volume 2 of New Mathematical Monographs. Cambridge University Press, Cambridge, 2005. [18] M. J. Gonz´alez. Uniformly perfect sets, Green’s function, and fundamental domains. Rev. Mat. Iberoamericana, 8(2):239–269, 1992. [19] A. Hinkkanen. Julia sets of polynomials are uniformly perfect. Bull. London Math. Soc., 26(2):153–159, 1994. [20] D. Mumford. A remark on Mahler’s compactness theorem. Proc. Amer. Math. Soc., 28:289–294, 1971. [21] Z. Nehari. Conformal mapping. Dover Publications Inc., New York, 1975. Reprinting of the 1952 edition. [22] Ch. Pommerenke. Uniformly perfect sets and the Poincar´e metric. Arch. Math. (Basel), 32(2):192–199, 1979. [23] Ch. Pommerenke. On uniformly perfect sets and Fuchsian groups. Analysis, 4(3-4):299–321, 1984. [24] H.A. Schwarz. Gesammelte Mathematische Abhandlungen. Springer, Berlin, 1890. [25] H.A. Schwarz. Confome abbildung der oberfl¨ache eines tetraeders auf die oberfl¨ache einer kugel. J. Reine Ange. Math., pages 121–136, 1969. Also in collected works, [24], pp. 84-101.

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[26] Toshiyuki Sugawa. Uniformly perfect sets: analytic and geometric aspects [translation of S¯ ugaku 53 (2001), no. 4, 387–402; mr1869018]. Sugaku Expositions, 16(2):225–242, 2003. Sugaku Expositions. [27] D. Sullivan. Travaux de Thurston sur les groupes quasi-fuchsiens et les vari´et´es hyperboliques de dimension 3 fibr´ees sur S 1 . In Bourbaki Seminar, Vol. 1979/80, pages 196–214. Springer, Berlin, 1981. [28] L. N. Trefethen and T.A. Driscoll. Schwarz-Christoffel mapping in the computer era. In Proceedings of the International Congress of Mathematicians, Vol. III (Berlin, 1998), number Extra Vol. III, pages 533–542 (electronic), 1998. C.J. Bishop, Mathematics Department, SUNY at Stony Brook, Stony Brook, NY 11794-3651 E-mail address: [email protected]

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