AN EXPLICIT CONSTANT FOR SULLIVAN’S CONVEX HULL THEOREM

CHRISTOPHER J. BISHOP

Abstract. For any simply connected domain Ω in the plane, let S be the boundary of the convex hull of ∂Ω which faces Ω. We show that there is always a 7.82quasiconformal map from S to Ω which extends to the identity on ∂S = ∂Ω.

Date: September 27, 2000. 1991 Mathematics Subject Classification. Primary: 30C62 . Key words and phrases. Brennan conjecture, Sullivan’s theorem, quasiconformal mappings,

convex hulls, hyperbolic manifolds, locally Lipschitz . The author is partially supported by NSF Grant DMS 9800924.

AN EXPLICIT CONSTANT FOR SULLIVAN’S THEOREM 1

1. Introduction

Let Ω ⊊ ℝ^{2} be a simply connected domain and let C(∂Ω) ⊂ ℍ^{3} = ℝ_{+}^{3} be the hyperbolic convex hull of ∂Ω (this is the hyperbolic convex hull of the set of all hyperbolic geodesics with endpoints in ∂Ω). Let S be the boundary component of C(∂Ω) which separates Ω from C(∂Ω). Equivalently, let Ω̃ ⊂ ℍ^{3} be the union of all hemispheres whose bases are contained in Ω. Then S = ∂Ω̃ ∩ ℍ^{3}. Let ρ_{S} denote the intrinsic path metric on S (using hyperbolic arclength) and let ρ denote the usual hyperbolic metric on the unit disk 𝔻, the upper half space ℍ^{3} or Ω. It is an observation of Thurston that there is an isometry ι from the metric space (S, ρ_{S}) to (𝔻, ρ) (see [7] [4], [5]). Moreover, there is a K_{0}-biLipschitz map σ from (Ω, ρ) to (S, ρ_{S}) which extends continuously to the identity on ∂Ω (and K_{0} is independent of Ω). Consequently, there is a universal K < ∞ so that σ is K-quasiconformal. The map σ is conformally natural in the sense that if Ω is invariant under a group of Möbius transformations then σ commutes with the group action. This appeared in Sullivan’s paper [6] in the case when Ω is simply connected and invariant under a convex co-compact quasi-Fuchsian group. Epstein and Marden prove in [4] the more general statement above. Their proof gives biLipschitz constant K_{0} ≈ 88.2 and quasiconformal constant K ≈ 82.6. The purpose of this paper is to prove

Theorem 1.1. For any simply connected domain Ω, there is a 7.82-quasiconformal map σ of Ω to S which extends to the identity on ∂Ω = ∂S. This map is also biLipschitz from the hyperbolic metric on Ω to the path metric on S. There is a 13.3-biLipschitz map from Ω with its hyperbolic metric to S with its path metric.

Strictly speaking, this result is not an improvement of the Epstein-Marden estimates since our construction is not conformally natural. It follows from our proof that there is also a conformally natural map (with a possibly larger, but uniformly bounded constant). This is because if Φ : 𝔻 → Ω is conformal, and G is a group of Möbius transformations acting on Ω, then ι ∘ σ ∘ Φ is quasiconformal from the disk to itself and hence has quasisymmetric boundary values h : 𝕋 → 𝕋 (with bounds depending only on the quasiconformal constant of σ). Moreover, this boundary map h conjugates the Fuchsian group G_{1} = Φ^{−1}∘G∘Φ to the Fuchsian group G_{2} = ι∘G∘ι^{−1}. Hence by the Douady-Earle extension theorem [3], there is a conformally natural extension h̃ of h to the disk with quasiconformal constant depending only on h (and

2 CHRISTOPHER J. BISHOP

hence only on the quasiconformal constant of σ). Thus Φ^{−1} ∘ h̃ ∘ ι^{−1} : Ω → S is the desired conformally natural, quasiconformal map. It is not clear whether the best constant for Sullivan’s theorem is the same if we also require the map to be conformally natural.

This paper is a companion to [2]. That paper gives a short proof of Sullivan’s theorem with an unspecified constant K < ∞ and discusses the connection of Sullivan’s theorem to properties of conformal mappings. In particular, it shows that if Sullivan’s theorem holds for a domain Ω with a constant K, then there is a K-quasiconformal map F of Ω to the unit disk which has bounded derivative. Moreover, Corollary 1.7 of [2] and Theorem 1.1 imply

Corollary 1.2. Any conformal map f : 𝔻 → Ω can be written as f = g ∘ h where h is a 7.82-quasiconformal self-map of the disk and |g′| is bounded away from zero.

I also prove in [2] that is we could prove this with K = 2, the well known Brennan conjecture would follow. This provides some extra motivation for finding the best constant in Sullivan’s theorem. See [2] for further results, remarks and questions regarding Sullivan’s theorem and its applications.

I would like to thank John Garnett, Al Marden and Michel Zinsmeister for reading an earlier version of this paper and for their numerous helpful comments and suggestions. The reader familiar with Epstein and Marden’s paper [4] will see the obvious debt this paper owes to it; although many of the details vary, the construction we give here is motivated by their ideas.

2. The construction

Epstein and Marden’s proof of Sullivan’s theorem in [4] comes in two steps: first showing the nearest point retraction from Ω to the ϵ-distance surface S_{ϵ} of S is biLipschitz and then averaging the nearest point retraction map from S_{ϵ} to S over certain intervals in a foliation of S_{ϵ}. Our construction is motivated by the second step, although we shall apply it to construct a map from S directly to Ω.

2.1. Overview of the construction. Before beginning the details, we will describe the overall plan of the construction.

AN EXPLICIT CONSTANT FOR SULLIVAN’S THEOREM 3

1. The first step is to reduce the case of a general Ω to the special case of “finitely bent” by showing that every simply connected domain has finitely bent approximations and that if we prove the estimate with uniform constants for these approximations, we can pass to the limit for Ω.

2. Next we assume Ω is finitely bent. This means that the corresponding convex hull surface S is a finite union of geodesic triangular faces {F_{j}}. Moreover, Ω itself can be written as a finite union of “triangles” {T_{j}} (ideal hyperbolic triangles for certain disks D_{j} inside Ω) with vertices on ∂Ω, “interior crescents” {C_{j}} (regions bounded by two intersecting circles, each the boundary of a triangle) and “boundary crescents” {B_{j}} (bounded by two orthogonal circles, one the boundary of a triangle, the other a boundary arc of Ω). We will define a metric ρ̃ on each piece which is comparable to the usual hyperbolic metric ρ on Ω.

3. We divide each triangle T_{j} into a central triangle and three “cusp regions” (to be described below). We define a subregion Ω_{2} ⊂ Ω be removing all the central triangles for the T_{j}’s. This region has a natural foliation E as in the paper of Epstein and Marden [4]. We also define Ω_{0} = ∪_{j}T_{j} ∪ B_{j} and a region Ω_{1} = Ω_{0} ∩ Ω_{2} = Ω_{2} \ ∪_{j}C_{j}. The desired map F : S → Ω will be a composition of maps

S ⟶^{f_{0}} Ω_{0} ⟶^{f_{1}} Ω_{1} ⟶^{f_{2}} Ω_{2} ⟶^{f_{3}} Ω ⟶^{f_{4}} Ω.

4. The map f_{0} : S → Ω_{0} is simply the inverse of the nearest point retraction from faces on S to the triangles T_{j} and boundary crescents B_{j}. It is isometric on each face of S, so has quasiconformal constant K_{0} = 1.

5. To describe the map f_{1} : Ω_{0} → Ω_{1}, we need to decompose each triangle into pieces in two different ways. For simplicity, assume our triangle is in the upper half-plane with vertices at 0, 1 and ∞, i.e., T = \{z = x+iy : 0 < x < 1, |z− \frac{1}{2}| > \frac{1}{2}. First we cut T into three isomorphic pieces (see Figure 1),

U_{1} = {z : 0 < x < 1, |z| > 1, |z − 1| > 1},

U_{2} = \{z : 0 < x < \frac{1}{2}, |z − \frac{1}{2}| > \frac{1}{2}, |z| < 1\},

U_{3} = \{z : \frac{1}{2} < x < 1, |z − \frac{1}{2}| > \frac{1}{2}, |z − 1| < 1\}.

The three domains meet at the “center” of the triangle u = \frac{1}{2} +i\frac{\sqrt{3}}{2}. The second

Figure 1. Subdividing the triangles

decomposition of T is into four pieces: three isometric “cusps”

W_{1} = {z : 0 < x < 1, |y| > 1, },

W_{2} = \{z : 0 < x < \frac{1}{2}, |z − \frac{1}{2}i| < \frac{1}{2}, |z − \frac{1}{2}| ≥ \frac{1}{2}\},

W_{3} = \{z : \frac{1}{2} < x < 1, |z − 1 − \frac{1}{2}i| < \frac{1}{2}, |z − \frac{1}{2}| > \frac{1}{2}\}

and a “central triangle”

X = T \ (W_{1} ∪ W_{2} ∪ W_{3}) = \{z : y < 1, |z − \frac{1}{2}i| > \frac{1}{2}, |z − 1 − \frac{1}{2}i| > \frac{1}{2}\}.

See Figure 1. The map f_{1} : T → T is a quasiconformal map from each U_{i} to W_{i}, i = 1, 2, 3 which is the identity on ∂T. Let g : U_{1} → W_{1} be the unique conformal map which takes ∞ → ∞, i → i and 1 + i → 1 + i and let h be the map h(x, y) = (x, g^{−1}(yi)). Then we define f_{1} = h ∘ g on U_{1} and define it symmetrically on U_{2} and U_{3}. Then f is the identity on ∂T and its quasiconstant on each U_{i}, i = 1, 2, 3, is the same as for h, which we will show is K_{1} = max_{1≥t} |g′(it)| (Lemma 2.7). The map g can be written explicitly in terms of the modular function λ and using this we will show that

K_{1} = max_{t≥1} |g′(it)| = |g′(i)| ≈ 1.20656

6. The map f_{2} : Ω_{1} → Ω_{2} is defined by sliding points along leafs of the foliation E. We will define this more precisely later, but roughly the idea is to fix a parameter 0 < s ≤ \frac{1}{2} and look at the interval I in a leaf of E so that I ∩ Ω_{2} has length 2s and is “centered” at x. The map f_{2} slides x along the leaf by an amount which depends on width and position of the crescents crossed by I. The

AN EXPLICIT CONSTANT FOR SULLIVAN’S THEOREM 5

quasiconformal constant of this map will be bounded by an explicit function K_{2}(s). For example, we will compute K_{2}(1/2) = 6.93808.

7. The map f_{3} : Ω_{2} → Ω is the identity on the crescents and the inverse of f_{1} on the triangles. Thus its quasi-constant is also K_{1} ≈ 1.20655.

8. The final map f_{4} : (Ω, ρ̃) → (Ω, ρ) is the identity map. The metric ρ̃ is chosen so this map is conformal, and hence can be ignored for computing the quasiconformal dilatation of F, but it needs to be considered to compute the biLipschitz constant of F.

9. The map F constructed in this way has quasiconformal constant K = K_{1}^{2}K_{2} ≈ 10.1002. We can do a little better using two observations. First, the places where the maps f_{1} and f_{2} attain their maximum dilatations are different and thus the constant for the composition is a little better than just the product K_{1}K_{2}. Second, the map f_{3} is conformal on the crescents, but K_{1}-quasiconformal on the triangles. Thus we could modify it to make it K_{1}-quasiconformal on the crescents as well, without having to pay any extra “price”. Using these ideas, we get K ≈ 7.8127.

We now start describing each step of the construction in more detail.

2.2. Passing to a limit. We will only do the construction explicitly for certain “finitely bent” cases. To pass to the limit we use the following easy lemma (which is left to the reader).

Lemma 2.1. Suppose Ω ⊂ ℂ is simply connected, 0 ∈ Ω and Ω ≠ ℂ. Suppose 0 ∈ Ω_{1} ⊂ Ω_{2} ⊂ ⋯ ⊂ Ω = ∪_{n}Ω_{n} and for each n there is an onto map f_{n} : Ω_{n} → 𝔻 which is K-quasiconformal and hyperbolically M-biLipschitz and such that f_{n}(0) = 0. Then there is a f : Ω → 𝔻 which is also K-quasiconformal and hyperbolically MbiLipschitz.

It is also convenient to record the following known result about quasiconformal mappings (e.g., Theorem 35.1 [8]). Using this, it is enough to estimate the dilatation of our map on the faces of S and check that it is continuous across the boundaries of these pieces.

Lemma 2.2. Suppose 𝔻 = ∪Ω_{n} where {Ω_{n}} is a finite collection of regions with disjoint interiors and rectifiable boundaries. If f : 𝔻 → Ω is a homeomorphism which is K-quasiconformal on the interior of each Ω_{n} then f is a K-quasiconformal mapping on the whole disk.

2.3. Approximation by triangulated polygons. Given a simply connected Ω we will describe how to approximate it by “finitely bent” regions. A different method is described in [4]. Our first step is to approximate Ω from within by a polygonal region P. Fix some small ϵ > 0. The vertices of P should be a finite set E = {z_{k}}_{0}^{N} with z_{0} = z_{N} which satisfy

1. dist(z_{k}, z_{k+1}) < ϵdist(z_{k}, ∂Ω)

2. The angle formed by any three consecutive points is within ϵ of π. (We can easily create such a polygon by first approximating Ω by a smooth region and then taking vertices on the boundary of this region, very close to each other in comparison to the distance to ∂Ω). Next we wish to triangulate P, i.e., write P = ∪_{j}T_{j} where {T_{j}} is a union of Euclidean triangles which satisfy

1. they have pairwise disjoint interiors,

2. every vertex of T_{j} is an element of E,

3. the open Euclidean disk D_{k} whose boundary contains the three vertices of T_{j}

does not intersect E. The last condition is not obvious, so we will show that such a triangulation is always possible.

Figure 2. Triangulation with disks

First, for each pair of consecutive points z_{k}, z_{k+1} let I_{k} = [z_{k}, z_{k+1}]. If I_{k} is not already the side of a previously chosen triangle then we choose a triangle with I_{k} as

AN EXPLICIT CONSTANT FOR SULLIVAN’S THEOREM 7

one side as follows. Consider the family of circles containing these two points and centered on the perpendicular bisector L of the segment I_{k}. When the center of the circle is the midpoint of I_{k} then clearly z_{k} and z_{k+1} are the only points of E on the circle. Move the center of the circle continuously along L into P until the first time the circle meets three (or more) points of E. If it meets just three points of E then the corresponding triangle satisfies all our conditions and is put into our triangulation. If it meets n ≥ 4 or more points of E, the resulting convex polygon is easily divided into n − 2 triangles with the desired properties and each is put in our triangulation.

To see that two triangles arising in this way must have disjoint interiors, consider two such triangles T_{1} and T_{2} and let D_{1} and D_{2} be the corresponding disks determined by the vertices. If the disks are disjoint, then so are the triangles so there is nothing to do. If D_{1} = D_{2} then T_{1} and T_{2} had to be formed by splitting a convex n-gon, n > 3, into triangles, and hence are disjoint by choice. Finally suppose D_{1} and D_{2} intersect but are not identical. Let L be the Euclidean line determined by the two points of ∂D_{1} ∩ ∂D_{2}. Since E does not intersect these disks the vertices of T_{1} must lie on ∂D_{1} \ D_{2} and the vertices of T_{2} must lie on ∂D_{2} \ D_{1} and hence the triangles lie on opposite sides of L. Thus the two triangles can share at most an edge or vertex in common.

Thus the triangles we have constructed satisfy all the desired conditions, except that their union may not be all of P. To complete the triangulation of P take a triangle T from our collection and an edge I of T which borders an “uncovered” part of P. Let L be the perpendicular bisector of I and let c ∈ L be the center of the disk D corresponding to T. Deform D by moving its center c along L in the direction towards I and keeping the endpoints of I on its boundary. Continue until the boundary of disk hits three or more points of E. Clearly, the new boundary points lie in the half-plane determined by L which does not contain c, so the convex polygon determined by these points has I as one side and covers some points not previously covered by the triangulation. Triangulate the resulting convex polygon if necessary and add the triangles to our collection. The proof of disjointness works just as before, so all the desired properties hold.

Now repeat this construction until all of P is covered (since there are only a bounded number of triangles in any triangulation of P and we we add at least one new triangle each time, the process must terminate in a finite number of steps). We will let T denote the collection of Euclidean triangles we have constructed.

2.4. Approximation by finitely bent regions. We now wish to modify the polygon P by replacing its straight line edges by circular arcs, in such a way that the resulting region has a finitely bent hyperbolic convex hull, i.e., the boundary component S of the convex hull of ∂Ω facing Ω will consist of a finite number of geodesic faces joined along infinite geodesics.

Suppose {I_{k}} is an enumeration of the boundary segments of P and let T_{k} be the unique triangle in our triangulation which has I_{k} as an edge and let D_{k} be the corresponding disk (determined by having the three vertices of T_{k} on its boundary). Let J_{k} be the smaller arc of ∂D_{k} whose endpoints are the same as the endpoints of I_{k}. If ϵ in our choice of E is close to zero, then the arcs J_{k} are very close to I_{k} (they make angle ≤ ϵ with I_{k} at the endpoints) and hence they are pairwise non-intersecting and their union is a Jordan curve bounding a simply connected region Ω′.

From this point on, we will only work with Ω′ and never the original Ω, so for convenience we will assume that Ω = Ω′.

We claim that the corresponding convex hull boundary S is finitely bent, i.e., is a finite union of geodesic triangles.

First we show that every D_{k} is in Ω. Suppose D_{1} is a disk which is not. Then ∂D_{1} crosses ∂Ω. Since D_{1} does not contain any point of E, ∂D_{1} must cross some arc J_{k} ⊂ ∂Ω twice. Thus three vertices defining D_{1} must lie in D_{k}, which is impossible since D_{k} does not contain any points of E. The contradiction implies D_{1} ⊂ Ω, as desired.

Let W ⊂ ℍ^{3} be the union of Euclidean hemispheres with bases {D_{k}}. By the previous paragraph each disk is in Ω so that W is contained in a complementary component of the convex hull of ∂Ω. We claim that it is the whole component, i.e., ∂W ∩ ℍ^{3} = S.

To prove this, it is enough to show that given any disk D ⊂ Ω, the hemisphere with base D is contained in W. Let x ∈ D and choose a T_{k} ∈ T such that x ∈ T_{k}. Then the interiors of D and D_{k} intersect since they both contain x. If D = D_{k}, there is nothing to do. Otherwise, ∂D ∩ ∂D_{k} consists of two points. Let L be the line through these two points and let H be the halfplane bounded by L such that

AN EXPLICIT CONSTANT FOR SULLIVAN’S THEOREM 9

D_{k} ∩ H ⊂ D ∩ H. The triangle T_{k} must lie in H (otherwise it would have a vertex in ∂D_{k} \ H ⊂ D contrary to the assumption that D ⊂ Ω and hence contains no points of E). Hence x ∈ H and hence above x, the hemisphere for D lies under the hemisphere for D_{k}, as desired. This completes the proof that S is finitely bent.

2.5. The nearest point retraction from Ω to S. The nearest point retraction from Ω to S is now quite simple to describe. Given a point x ∈ Ω one takes a horoball in ℍ^{3} based at x and expand it until it first hits S. Since S is the boundary of a convex set, there is a unique point of S hit and this is the “nearest point” to Ω.

The surface S consists of a finite number of geodesic triangles, one corresponding to each element of our triangulation. These are called the faces of S and will be denoted by {F_{k}}. The face F_{k} is the part of S which projects vertically to T_{k}. On each of these faces we define a map F_{k} → D_{k} ⊂ Ω by following the unique geodesic which is orthogonal to F_{k} at a point x in the direction towards D_{k}. The image of F_{k}, is an triangle S_{k} in the hyperbolic geometry of D_{k}. More precisely, if T_{k} is a boundary triangle then S_{k} is bounded by J_{k} ⊂ ∂D_{k} on one side and the other two sides are circles orthogonal to ∂D_{k} which connect the endpoints of J_{k} to the third vertex of T_{k}. If T_{k} is an interior triangle, then ∂S_{k} consists of three circular arcs orthogonal to ∂D_{k} and connecting the vertices of T_{k}. It is easy to verify that the nearest point retraction map takes S_{k} back to F_{k} and hence different S_{k}’s are disjoint (since they map to different sets). Moreover, the retraction map is isometric from the hyperbolic of D_{k} restricted to S_{k} to the hyperbolic metric of ℍ^{3} restricted to F_{k}. Note that in general we do not have S_{k} ⊂ T_{k}; this happens iff T_{k} contains the center of the disk D_{k}.

Figure 3. Geodesic face projects to triangles.

10 CHRISTOPHER J. BISHOP

The union of of the S_{k}’s may not be all of Ω (in fact, it won’t be unless Ω is a disk). If T_{j} and T_{k} are triangles which share and edge I then the corresponding “hyperbolic” triangles both have vertices at the endpoints of I, but the corresponding edges I_{j}, I_{k} need not be the same. The region C_{jk} between these edges we will call an “interior crescent” and to each such crescent we will associate a number 0 ≤ α ≤ π which is the angle between I_{j} and I_{k} at either common endpoint. Under the nearest point retraction onto S this crescent maps to the geodesic G_{jk} in S which is the common boundary of the adjacent faces F_{j} and F_{k}. The number α is also equal to the exterior angle the two faces make along their common boundary and is called the bending angle (or bending measure) of this geodesic.

Figure 4. Hyperbolic triangles and crescents (shaded).

2.6. Foliating the crescents. The retraction map collapses each crescent onto a geodesic and the preimage of a single point is a circular arc in the crescent which is orthogonal to both boundary curves. This gives a foliation of the crescent. There is an equally obvious orthogonal foliation by circular arcs connecting the two endpoints of the crescent. Following Epstein and Marden we call the first foliation E and the second G. See Figure 5.

If we map the crescent by a Möbius transformation sending the endpoints to 0 and ∞, the crescent becomes a wedge of angle α bounded by straight lines through the origin and the foliation is just the union of circular arcs centered at the origin. By taking logarithms we can map the wedge to an infinite horizontal strip of width α and the foliation E is mapped to vertical line segments and G to horizontal lines.

Figure 5. The foliations in crescents

We define a metric ρ̃ on the original crescent by mapping the Euclidean metric on the strip back to the crescent. We will prove a little later that this metric is comparable (with absolute constants) to the hyperbolic metric on Ω restricted to the crescent. See Lemma 2.3.

2.7. Foliating the triangles. Next we would like to extend these foliations into the triangles (or more precisely, into the triangles minus their central triangles).

If S is a boundary triangle with boundary edge I then divide it into two pieces by the geodesic in D joining the endpoints of I. One piece is a crescent with angle α = π/2 (which we will call a boundary crescent) and the other is an ideal triangle in D. On the boundary crescent take the foliation E as before by mapping to a infinite horizontal strip and taking inverse images of vertical line segments. The foliation G is again by circular arcs connecting the two endpoints.We define the metric ρ̃ on boundary crescents by restricting the hyperbolic metric on D.

Thus it only remains to describe the foliation in an ideal triangle T. We will do this by cutting the triangle T into three cusp pieces W_{1}, W_{2}, W_{3} and a central triangle X as described earlier. See Figure 6. Each W_{j} can be conformally mapped in a unique way to W_{1} = {z = x + iy : 0 < x < 1, y > 1} and in W_{1} we define the foliation E by horizontal lines and the foliation G by vertical lines. We define a metric ρ̃ on W_{j} by pulling back the hyperbolic metric of the upper half plane restricted to W (this is the same as restricting the hyperbolic metric on the disk D corresponding to T to W_{j}).

Figure 6. The foliations in triangles (E is solid, G is dashed)

2.8. The metric ρ̃ is comparable to ρ. In this section we show the metric ρ̃ we have defined on Ω is comparable to the usual hyperbolic metric ρ on Ω. First we review a few basic facts about the hyperbolic metric.

Suppose Ω_{1} ⊂ Ω_{2} and ρ_{Ω_{1}} and ρ_{Ω_{2}} are the corresponding hyperbolic metrics. Then dρ_{Ω_{2}} ≤ dρ_{Ω_{1}}. A well know consequence of the Koebe \frac{1}{4}-theorem that if D = D(z_{0}, r) is a disk and Ω is a simply connected domain containing D and such that ∂Ω ∩ ∂D ≠ ∅, then

dρ_{Ω} ≤ dρ_{D} ≤ 4ρ_{Ω}. (1)

If Ω = H is a half-plane then we have dρ_{H}(z_{0}) = \frac{1}{2}dρ_{D}(z_{0}) (just map D → H by a Möbius transformation and note the derivative has absolute value 2 at z_{0}). Thus if Ω is simply connected and contains a half-plane H and x ∈ ∂H ∩ ∂Ω, then

dρ_{Ω}(z) ≤ dρ_{H}(z) ≤ 2dρ_{Ω}(z), (2)

for any point z ∈ H which is on the line perpendicular to ∂H at x.

Lemma 2.3. Let ρ be the hyperbolic metric on Ω and let ρ̃ be the metric on Ω defined in the previous two sections. Then on interior crescents we have

\frac{1}{2} ≤ \frac{dρ}{dρ̃} ≤ 1,

AN EXPLICIT CONSTANT FOR SULLIVAN’S THEOREM 13

on triangles we have

\frac{\sqrt{3}}{4} ≤ \frac{dρ}{dρ̃} ≤ 1,

and boundary crescents we have

\frac{1}{\sqrt{2}} ≤ \frac{dρ}{dρ̃} ≤ 1.

Proof. First consider interior crescents. By construction, each interior crescent C has sides which are orthogonal to the boundaries of the disks D_{1}, D_{1} ⊂ Ω and C ⊂ D_{1} ∪ D_{1}. Suppose the crescent has angle α. Then if we map the endpoints of the crescent to 0, ∞ so that the crescent is mapped to V_{1} = {z : | arg(z)| < α/2}, the domain D_{1} ∪ D_{2} is mapped to V_{2} = {z : | arg(z)| < (α + π)/2} and Ω is mapped to a domain Ω^{∗} containing this wedge.

The metric ρ̃ on V_{1} is given by dρ̃ = \frac{1}{|z|}|dz|. If z_{0} ∈ V_{1}, let H = \{z : | arg(z) − arg(z_{0})| < π/2}. This is a half-plane and V_{1} ⊂ H ⊂ V_{2} and dρ_{H}(z_{0}) = dρ̃(z_{0}). By (2), dρ_{Ω^{∗}} ≤ dρ_{H} ≤ 2dρ_{Ω^{∗}}, which proves the desired inequality.

Next, consider a triangle T and corresponding disk D ⊂ Ω. Since D ⊂ Ω we have ρ̃ ≥ ρ. To prove the other inequality, map D to the upper half plane ℍ by a Möbius transfomation so T is mapped to the triangle with vertices 0, 1, ∞ and Ω is mapped to some domain Ω^{∗} containing the upper half plane. By symmetry, we need only consider points in U = {z = x + iy : 0 < x < 1/2, |z| > 1}. Fix a point z = x + iy ∈ U and let D_{1} = D(z, y) and D_{2} = D(z, dist(z, ∂Ω^{∗})) = D(z, r_{2}). Since 0 ∈ ∂Ω^{∗} we know y ≤ r_{2} ≤ \sqrt{x^{2} + y^{2}}. Thus

dρ̃(z) = dρ_{ℍ}(z) = \frac{1}{2}dρ_{D_{1}}(z) = \frac{1}{2} \frac{r_{2}}{y} dρ_{D_{2}}(z) ≤ 2\frac{r_{2}}{y} dρ_{Ω^{∗}}(z) = 2\frac{r_{2}}{y} dρ(z).

Since x ≤ 1/2 and y ≥ \sqrt{3}/2, this gives dρ̃(z) ≤ \frac{4}{\sqrt{3}}dρ, as desired.

Finally we consider boundary crescents. As in the case of triangles assume we have a boundary crescent B ⊂ D ⊂ Ω. Just as above ρ̃ ≥ ρ because of the containment D ⊂ Ω. Again map D to the upper half plane ℍ sending B to the half disk with base [0, 1] ⊂ ∂Ω^{∗}. For z = x + iy in this half disk, we have dist(z, Ω) = y and so dρ̃ = dρ_{ℍ} ≤ 2dρ_{Ω^{∗}} = 2dρ, as desired.

The following is now immediate

Corollary 2.4. The identity map (Ω, ρ̃) → (Ω, ρ) is 4/\sqrt{3}-biLipschitz.

14 CHRISTOPHER J. BISHOP

2.9. Mapping U to W. In this section we define a (1.20656)-quasiconformal map from U_{1} to W_{1} which is the identity on the boundary. If one is willing to take a larger constant, then this section may be skipped. In Section 2.14 we will describe an much easier map which gives the constant 1.76759.

As noted in the beginning of the section, the map f_{1} is defined using the conformal map g : U_{1} → W_{1}. It slightly easier to consider the map g : U → W where

U = {z : 0 < x < 1/2, |z| > 1},

W = {z : 0 < x < 1/2, |y| > 1, },

and g is uniquely determined by the requirement that g(i) = i, g(∞) = ∞ and g(u) = \frac{1}{2} + i (where u = \frac{1}{2} + i\sqrt{3}/2). If we can construct this map, then the map from U_{1} to W_{1} can be obtained by reflecting over the line x = \frac{1}{2}. In this section we gather together some facts we will need about the map g.

Fortunately, the map g can be written as a composition of maps with explicit formulas. The first element of this composition is the well known modular function λ(z), which gives a conformal map from the region T = {z = x + iy : 0 < x < 1, |z − \frac{1}{2}| > 1/2\} onto the upper half-plane. This function can be explicitly written as λ(z) = λ_{1}(z)/λ_{2}(z), where

λ_{1}(z) = ∑_{n=−∞}^{∞}(\frac{1}{cos^{2}(π(n − \frac{1}{2})z)} − \frac{1}{sin^{2}(π(n − \frac{1}{2})z)}),

λ_{2}(z) = ∑_{n=−∞}^{∞}(\frac{1}{cos^{2}(πnz)} − \frac{1}{sin^{2}(π(n − \frac{1}{2})z)}).

The fact that λ is given by this formula and maps T to the upper half-plane is proven in Section 7.3.5 of Ahlfors’ book [1].

Since our region U is strictly contained in the upper half-plane, these series representations converge exponentially fast on U, so the truncations will give good approximations. In fact, if we let

λ_{1,N}(z) = ∑_{n=−N}^{N}(\frac{1}{cos^{2}(π(n − \frac{1}{2})z)} − \frac{1}{sin^{2}(π(n − \frac{1}{2})z)}),

λ_{2,N}(z) = ∑_{n=−N}^{N}(\frac{1}{cos^{2}(πnz)} − \frac{1}{sin^{2}(π(n − \frac{1}{2})z)}),

then

AN EXPLICIT CONSTANT FOR SULLIVAN’S THEOREM 15

Lemma 2.5. Suppose z = x + iy. Then for all x and all y ≥ \frac{1}{2}, and j = 1, 2,

|λ_{j}(z) − λ_{j,N}(z)| ≤ 200e^{−πN}. (3)

For y ≥ 1, j = 1, 2,

|\frac{d}{dy} λ_{j}(z) − \frac{d}{dy} λ_{j,N}(z)| ≤ 400e^{−πN}. (4)

Proof. Since cos(iy) = cosh(y) and sin(iy) = sinh(y), we have

| cos(z)|, | sin(z)| ≥ \frac{1}{4}e^{y}, y ≥ \frac{1}{2}

Thus each of the terms

cos^{2}(π(n − \frac{1}{2})iy), sin^{2}(π(n − \frac{1}{2})iy), cos^{2}(πniy), sin^{2}(π(n − \frac{1}{2})iy)

is bounded below by (4e)^{−2}e^{2πny}. Thus for y ≥ 1/2,

|λ_{j}(z) − λ_{j,N}(z)| ≤ 16e^{2} ∑_{n=N+1}^{∞} e^{−πn} = 16e^{2} \frac{e^{−πN}}{1 − e^{−π}} ≤ 200e^{−πN},

as desired. The second estimate follows from the first by the usual Cauchy estimate for the derivative of an analytic function.

Thus if we use N = 10, the lemma says the truncations will be accurate to within 400e^{−10π} ≤ 10^{−11}. Using Mathematica to compute the truncated series we get

λ_{1}(i) ≈ 6.87519, λ′_{1}(i) ≈ 22.9211i,
λ_{2}(i) ≈ 13.7504, λ′_{2}(i) ≈ 13.7504i.

This gives

|λ′(i)| = |\frac{λ′_{1}(i)λ_{2}(i) − λ_{1}(i)λ′_{2}(i)}{λ_{2}(i)^{2}}| ≈ 1.09422.

As noted above, λ maps the region T to the upper half-plane with λ(0) = 1, λ(1) = ∞ and λ(∞) = 0 (see Theorem 7, page 281 of [1]). Using symmetry it is easy to see that λ(i) = 1/2 and λ(u) = u and the region U ⊂ T is mapped univalently onto V = \{z = x+iy : 0 < x < \frac{1}{2}, y > 0, |z −1| < 1\}. This region, in turn, is mapped to the lower half-plane by the function

J(z) = \frac{4}{27} \frac{(\frac{3}{4} + (\frac{1}{2} − z)^{2})^{3}}{z^{2}(1 − z)^{2}} .

See page 282 of [1]. (We are abusing notation; usually the composition J ∘ λ is called the “J-function”.) Moreover, J(0) = ∞, J(\frac{1}{2}) = 1 and J(u) = 0.

16 CHRISTOPHER J. BISHOP

Next note that z → i + \frac{1}{2π} cos^{−1}(2z − 1) is a univalent map of the lower half-plane to the half-strip W taking ∞ → ∞, 0 → i and 1 → 1 + i. Thus, we see that the desired map is

g(z) = i + \frac{1}{2π} cos^{−1}(2J(λ(z)) − 1).

Figure 7. The map g as a composition

Using the chain rule and the series expansions

J(z) = 1 + 12(z − \frac{1}{2})^{2} + O(|z − \frac{1}{2}|^{4}),

cos(z) = 1 − \frac{1}{2}z^{2} + O(|z|^{4}),

it is easy to show that

|g′(i)| = |λ′(i)|\frac{2\sqrt{3}}{π}.

Combined with our earlier estimate for |λ′(i)| this gives |g′(i)| ≈ 1.20655. The other fact that we will need about g is Lemma 2.6. For all y ≥ 1, |g′(iy)| is decreasing in y (in particular, |g′(iy)| ≤ |g′(i)| for y ≥ 1).

Proof. Let U′ be a polygonal approximation to U in which the circular arc connecting i and u is replaced by a polygon which is convex down and for which the segment containing i is horizontal. Then by the Schwarz-Christoffel formula, the conformal mapping F_{1} from the upper-half plane to W which sends 0 → i, 1 → 1 + i and ∞ → ∞ has derivative

F′_{1}(z) = z^{−1/2}(z − 1)^{−1/2},

and the conformal mapping F_{2} from the upper half-plane to U′ sending 0 → i, 1 → u and ∞ → ∞ has derivative

F′_{2}(z) = z^{−1/2}(z − 1)^{−1/2−∑_{j} α_{j}} ∏_{j}(z − x_{j})^{α_{j}}

for some choice of {x_{j}} ∈ (0, 1) and {α_{j}} > 0.

The conformal map h from U′ to W sending ∞ → ∞, i → i and u → 1 + i can be written as h = F_{1} ∘ F_{2}^{−1} and so its derivative at iy is F′_{1}/F′_{2} at t = F_{2}^{−1}(iy) < 0. Since 0 < x_{j} < 1, and

\frac{F′_{1}(t)}{F′_{2}(t)} = ∏_{j}(\frac{t − 1}{t − x_{j}})^{α_{j}},

we see that this is an increasing function on (−∞, 0] since each (t − 1)/(t − x_{j}) is increasing there. Since the map y → F_{2}^{−1}(iy) is clearly decreasing, we see that |h′(iy)| is decreasing in y. The conformal map g : U → W can be written as a uniform limit of maps h as above (by reflection the maps can be extended to an open neighborhood of {z = x + iy : y > 1}) and hence g′ is a limit of the derivatives. A limit of non-decreasing functions is non-decreasing, so the lemma is proven.

Lemma 2.7. Let

U_{1} = {z : 0 < x < 1, |z| > 1, |z − 1| > 1},
W_{1} = {z : 0 < x < 1, |y| > 1, }.

Then there is a K_{1} quasiconformal map from U_{1} to W_{1} which is the identity on ∂U_{1} ∩ ∂W_{1} = {z = x + iy : x ∈ {0, 1}, y ≥ 1}. One can take K_{1} = 1.20656.

18 CHRISTOPHER J. BISHOP

Proof. Let g : U → W be the conformal map described above and by reflection extend it to a conformal map from U_{1} to W_{1}. Define h : W_{1} → W_{1} by h(x, y) = h(x, |g^{−1}(iy)|). It is very easy to see that a map of the form (x, y) → (x, f(y)) has quasiconformal dilatation max(|f′(y)|, |f′(y)|^{−1}|) at the point (x, y) (for a more general case, see Lemma 2.8 below).

Thus h ∘ g maps U_{1} to W_{1}, is the identity on ∂U_{1} ∩ ∂W_{1} and has quasiconformal constant

K_{1} = max_{y≥1} |g^{−1}(iy)′|^{−1} = |g′(i)|,

by Lemma 2.6. The value of K_{1} then follows from our earlier computation.

2.10. Computing the dilatation. In this section we record a fairly straightforward computation for the quasiconformal dilatation of certain maps which we will need later. Given a mapping G defined on a domain U and a point z ∈ U we define

L_{1}(G, z) = lim inf_{w→z} \frac{|G(z) − G(w)|}{|z − w|},

L_{2}(G, z) = lim sup_{w→z} \frac{|G(z) − G(w)|}{|z − w|},

Lemma 2.8. Suppose G : (x, y) → (x, g(x, y)) is differentiable. Then

\frac{1}{2}(\sqrt{Y} − \sqrt{X}) = L_{1}(G, z) ≤ L_{2}(G, z) = \frac{1}{2}(\sqrt{Y} + \sqrt{X}), (5)

where

X = 1 + (g_{x})^{2} + (g_{y})^{2} − 2|g_{y}|, Y = 1 + (g_{x})^{2} + (g_{y})^{2} + 2|g_{y}|.

Thus the quasiconformal dilatation of G is

lim sup_{r→0} \frac{max_{|z−w|=r} |G(z) − G(w)|}{min_{|z−w|=r} |G(z) − G(w)|} = \frac{(\sqrt{Y} + \sqrt{X})}{(\sqrt{Y} − \sqrt{X})}.

Proof. The tangent map for G is (x, y) → (x, xg_{x} + yg_{y}) = (u, v). Thus the preimage of the circle u^{2} + v^{2} = 1 is the ellipse given by x^{2} + (xg_{x} + yg_{y})^{2} = 1. If r_{1}, r_{2} are lengths of the semiminor and semimajor axes respectively of this ellipse then (5) is equivalent to the claim that

\frac{1}{r_{1}} = \frac{1}{2}(\sqrt{Y} + \sqrt{X}), \frac{1}{r_{2}} = \frac{1}{2}(\sqrt{Y} − \sqrt{X}).

To prove these, rewrite the equation for the ellipse as (1 + (g_{x})^{2})x^{2} + 2g_{x}g_{y}xy + (g_{y})^{2}y^{2} = Ax^{2} + Bxy + Cy^{2} = 1.

It is well known from the theory of conic sections that

r_{1}r_{2} = \frac{2}{\sqrt{4AC − B^{2}}}

\frac{r_{1}^{2} + r_{2}^{2}}{r_{1}^{2}r_{2}^{2}} = \frac{1}{r_{1}^{2}} + \frac{1}{r_{2}^{2}} = A + C

(since A + C and 4AC − B^{2} are invariant under rotations of the ellipse, and these equations hold when B = 0, i.e., the axes are parallel to the coordinate axes). Thus

r_{1}^{2} + r_{2}^{2} = 4\frac{A + C}{4AC − B^{2}}
\frac{1}{r_{2}^{2}} \frac{4}{4AC − B^{2}} + r_{2}^{2} = 4\frac{A + C}{4AC − B^{2}}
\frac{1}{r_{2}^{2}} + \frac{1}{4}(4AC − B^{2})r_{2}^{2} = A + C
\frac{1}{u} + αu = β.

Since r_{1}^{2} and r_{2}^{2} play symmetric roles, they must be the two solutions of this equation. Using the quadratic formula to solve it, we get

r_{1}^{2}, r_{2}^{2} = \frac{β ± \sqrt{β^{2} − 4α}}{2α}
= 2\frac{(A + C) ± \sqrt{(A + C)^{2} − (4AC − B^{2})}}{4AC − B^{2}}.

Setting

X = A + C − \sqrt{AC − B^{2}}
= 1 + (g_{x})^{2} + (g_{y})^{2} − \sqrt{4(1 + (g_{x})^{2})(g_{y})^{2} − 4(g_{x}g_{y})^{2}}
= 1 + (g_{x})^{2} + (g_{y})^{2} − 2|g_{y}|,

and similarly

Y = A + C + \sqrt{AC − B^{2}} = 1 + (g_{x})^{2} + (g_{y})^{2} + 2|g_{y}|,

we get

r_{2}^{2} = \frac{\frac{1}{2}(X + Y) + \sqrt{XY}}{\frac{1}{2} \frac{1}{4}(Y − X)^{2}}
= 4\frac{Y + 2\sqrt{XY} + X}{(Y − X)^{2}}
= 4\frac{(\sqrt{Y} + \sqrt{X})^{2}}{(\sqrt{Y} − \sqrt{X})^{2}(\sqrt{Y} + \sqrt{X})^{2}}
= \frac{4}{(\sqrt{Y} − \sqrt{X})^{2}} ,

which clearly gives the desired result for r_{2}. The calculation for r_{1} is similar.

First note that for a map of the form (x, y) → (x, y + g(x, y)) with g_{y} ≥ 0, the formulas for X and Y become X = (g_{x})^{2} + (g_{y})^{2}, Y = (g_{x})^{2} + (2 + g_{y})^{2}. (6) Note that \sqrt{Y} + \sqrt{X} clearly increases if either |g_{x}| or g_{y} increases. Thus to find an upper bound for \sqrt{Y} + \sqrt{X} it suffices to find upper bounds for |g_{x}| and g_{y}.

On the other hand, by differentiating \sqrt{x^{2} + (2 + y)^{2}} − \sqrt{x^{2} + y^{2}}, it is easy to check that \sqrt{Y} − \sqrt{X} decreases if |g_{x}| increases or g_{y} decreases. Thus to find a lower bound for \sqrt{Y} −\sqrt{X} we need an upper bound for |g_{x}| and a lower bound for g_{y} (which will usually be 0).

Finally, we claim that the dilatation

K = \frac{\sqrt{Y} + \sqrt{X}}{\sqrt{Y} − \sqrt{X}} ,

increases if either |g_{x}| or g_{y} increases. To see this, consider

μ = \frac{K − 1}{K + 1} = \sqrt{Y} /\sqrt{X} = \sqrt{g_{x}^{2} + (2 + g_{y})^{2}}/\sqrt{g_{x}^{2} + g_{y}^{2}}.

This is clearly increasing in both |g_{x}| and g_{y} (just view it as |g_{y} + ig_{x}|/|2 + g_{y} + ig_{x}|). Since the map μ → K is increasing, this proves the claim. Thus to find an upper bound on K it suffices to find upper bounds on |g_{x}| and g_{y}.

2.11. The averaging map. In this section we define the map f_{2} : Ω_{1} → Ω_{2}. Recall that every point z ∈ Ω_{1} ⊂ Ω_{2} is on a leaf of the foliation E and the map f_{2} will be defined by “sliding” the point a certain distance along the leaf (i.e., f_{2} will preserve leaves of E).

AN EXPLICIT CONSTANT FOR SULLIVAN’S THEOREM 21

Given an interval I on a leaf of the foliation E we will measure its length in two ways. One is just the length of I with respect to the metric ρ̃. The other is the ρ̃ length of I ∩ Ω_{2}, i.e., we only measure the length of I in the triangles and boundary crescents and ignore the length contributed by interior crescents. This length will also be called the S-length because it is the length of R(I) on S (recall R is the nearest point retraction from Ω to S).

Given a point z ∈ Ω_{1}, choose an orientation for the leaf L which contains it. For a real number s with |s| ≤ 1/2, let I(z, s) be the closed interval in L of S-length |s| with one endpoint z and other endpoint in the positive direction if s > 0 and the negative direction if s < 0. Let L(z, s) be the signed length of I(z, s) in E and let b(z, s) = L(z, s) − s. Another way to define b(z, s) is to travel along the leaf of E containing z in the direction sign(s) until we have traversed distance s in Ω_{1} (so we don’t count the distance crossing crescents). Then b(s, z) is sign(s) times the sum of the widths of the crescents we have crossed. (If the stopping point is on the boundary of a crescent we count that crescent in the sum.) We shall prove later (Lemma 2.9) that for s < 1, b(s, z) is bounded independently of z and Ω; thus in some sense a leaf of E must spend a fixed fraction of its length outside interior crescents.

For z ∈ Ω_{1} and 0 < s ≤ \frac{1}{2}, define

B(z, s) = \frac{1}{2s} ∫_{−s}^{s} b(z, s)ds,

and we define f_{2}(z) to be the point in Ω_{2} which is on the same leaf L as z and (signed) distance B(z, s) from it. In other words, we define f_{2}(z) by sliding z (signed) distance B(z, s) along L. If L is regular this define a single point. If L is singular, there may be two points distance B(z, s) from z, but these two points then must be on the boundary of a central triangle and must correspond to each other via the map f_{1}.

If z_{1}, z_{2} ∈ ∂Ω_{1} are points on the same leaf of E which are separated by a single interior crescent of width α, we claim that f_{2}(z_{1}) = f_{2}(z_{2}). To see this, suppose z_{2} is in the positive direction from z_{1} and note that the crescent contributes +α/2 to B(z_{1}, s) and contributes −α/2 to B(z_{2}, s). Since z_{1} and z_{2} are α apart, this proves the claim. Thus the composition f_{2}∘f_{1}∘f_{0} is continuous across the bending geodesics in S. Combined with the previous paragraph, this shows f_{3} ∘ f_{2} ∘ f_{1} ∘ f_{0} is continuous from S to Ω.

22 CHRISTOPHER J. BISHOP

Now we have to show the map f_{2} is piecewise quasiconformal. Since our map f_{2} preserves leaves of E it is locally of the form f : (x, y) → (x, y+g(x, y)), so by Lemma 2.8, the quasiconformal dilatation is

K = \frac{\sqrt{(g_{x})^{2} + (2 + g_{y})^{2}} + \sqrt{(g_{x})^{2} + (g_{y})^{2}}}{\sqrt{(g_{x})^{2} + (2 + g_{y})^{2}} − \sqrt{(g_{x})^{2} + (g_{y})^{2}}} .

We will write this a little more compactly as k(g_{x}, g_{y}), where

k(u, v) = \frac{\sqrt{u^{2} + (2 + v)^{2}} + \sqrt{u^{2} + v^{2}}}{\sqrt{u^{2} + (2 + v)^{2}} − \sqrt{u^{2} + v^{2}}} . (7)

We will give upper bounds for g_{y} and |g_{x}| in terms of the “bending function” B_{1} which is defined as follows. For any z ∈ Ω on a leaf L of E and 0 < s ≤ 1 let I be the interval in L centered (with respect to S-length) at z of S-length 2s and let B_{1}(z, s, Ω) = L(z, s). Then

B_{1}(s) = sup_{Ω} sup_{z∈Ω} B_{1}(z, s, Ω).

In other words, for any finitely bent Ω and any interval I of length 2s in the foliation R(E) ⊂ S on the corresponding convex hull surface, the total bending along I is less than B_{1}(s). Of course, it is not obvious that there is a finite value of B_{1}(s) with this property, but we will prove that there is in the next section.

Now for the upper bounds on |g_{y}| and |g_{x}|. Since we only need bounds that hold almost everywhere, we may assume z in the interior of triangle. First, by the fundamental theorem of calculus

g_{y}(z) = \frac{1}{2s}[b(z, s) − b(z, −s)] ≤ \frac{1}{2s}B_{1}(s). (8)

Also note that since b(z, s) ≥ 0 and b(z, −s) ≤ 0 we have that g_{y}(z) ≥ 0 for all z. Since b is a step function

g(z) = ∑_{j} α_{j}^{+}(s − d_{j}^{+}(z)) − ∑_{k} α_{k}^{−}(s − d_{k}^{−}(z))

where {α_{j}^{+}} are the crescent widths hit by I in the positive direction and {α_{k}^{−}} are in the negative direction. The function d_{j}^{±}(x, y) is the S-distance from the crescent C_{j} to the G-leaf containing z. Then d_{j}^{±}(z) = w_{0}^{±}(z) + ∑_{n=1}^{j} w_{n}^{±}(z) where w_{n}^{±}(z) is the length of the interval connecting C_{n} to C_{n+1} on the E-leaf containing z and w_{0}^{±}(z) is the length of the interval connecting z to C_{1} .

AN EXPLICIT CONSTANT FOR SULLIVAN’S THEOREM 23

It is easy to compute that |\frac{d}{dx}w_{j}^{±}(z)| = w_{j}^{±}(z). To do this, map the triangle to the ideal triangle 0, 1, ∞ in the upper half plane so that the leaf of E in question is a horizontal line. Then at (Euclidean) height y, the (hyperbolic) length of the leaf is w = 1/y and its rate of change is \frac{dw}{dy} = −1/y^{2}. Since hyperbolic distance in the vertical direction is h = log y, we see from the chain rule that \frac{dw}{dh} = \frac{dw}{dy}/\frac{dh}{dw} = (−1/y^{2})/(1/y) = −1/y = −w; in other words, the rate of change of the width has absolute value equal to width, as desired.

Since d_{j}^{±}(z) ≤ s this implies

|g_{x}(z)| ≤ \frac{1}{2s}s(∑_{j} α_{j} + ∑_{k} α_{k}) ≤ \frac{1}{2}B_{1}(s). (9)

Thus we get the estimate

(10)  K_{2}(s) ≤ k(\frac{1}{2}B_{1}(s), \frac{1}{2s}B_{1}(s)),

where k is as in (7).

2.12. A bound on bending. In this section, the E and G foliations will always refer to the foliations on the convex hull surface S corresponding to E and G in Ω via the retraction map R. As before, let B_{1}(s) be the maximum sum of angles crossed by any S-length 2s segment of the E foliation.

Lemma 2.9. With B_{1} defined as above, and 0 ≤ s < 1,

B_{1}(s) ≤ B_{2}(s) ≡ \frac{π + log \frac{1+s}{1−s}}{\sqrt{1 − s^{2}}} .

Proof. The proof follows the extremely clever idea of the proof of Proposition 2.14.2 of Epstein and Marden’s paper [4], although some of the details vary. The basic idea it to take a foliation path γ_{1} of S-length 2s and center x ∈ S and approximate it by another path γ_{2} and then compare the angles crossed by the two paths. It is also interesting to note that the proof of this lemma in the only place in this proof of Sullivan’s theorem which is really three dimensional: we will use in an essential way that S is the boundary of a convex region in hyperbolic space and I do not know how to derive the same estimates from merely two dimensional arguments (although this may be possible).

Without loss of generality we assume x is on a face of S and not a bending geodesic so that γ_{1} is differentiable at x. Let v_{0} be a tangent vector to S at x which is

24 CHRISTOPHER J. BISHOP

perpendicular to γ_{1} at x. The new path is γ_{2} = P ∩ S where P is the unique hyperbolic plane in ℍ^{3} which contains x and which is orthogonal to the vector v_{0} at the point x. We can estimate the angles crossed by γ_{2} using the following lemma from [4].

Lemma 2.10 (Lemma A.6.2, [4]). Let Q be a noncompact convex polygon with a finite number of geodesic sides in the hyperbolic plane. Let A be an arc of length s in ∂Q. Then the sum of the exterior angles of Q met by A (including the endpoints) is at most π + s.

The elegant proof of this given in [4] is attributed to Francis Bonahon. Briefly, it consists of replacing (if necessary) Q by the convex hull of A and a vertex p of Q at ∞. Then connect p to the finite vertices of Q, thus writing Q as a union of triangles. Each such has one finite length side, and this length is an upper bound for the area of the triangle. The Gauss-Bonet theorem implies the sum of the exterior angles is less than 2π + s and subtracting the exterior angle π at p gives the result.

Since S is the boundary of a convex set (namely the convex hull of ∂Ω) and is finitely bent, γ_{2} = P ∩ S is the boundary of a convex polygon Q in P. Moreover, this polygon is noncompact because otherwise the base of P (i.e. the corresponding disk in the boundary of hyperbolic space) would lie inside Ω and P would have to be tangent to S wherever they intersected, contrary to our choice of P. Thus Q is compact and the lemma applies to γ_{2}.

So we now have to turn our estimate for the bending along γ_{2} into an estimate for bending along γ_{1}. To do this we will not use the hyperbolic arclength parameterization of γ_{2}, but instead will use the parameterization given by the leaves of the foliation G, i.e., parameterize γ_{1} by arclength and parameterize γ_{2} so that γ_{1}(t) and γ_{2}(t) lie on the same leaf of G. We want to estimate two quantities: first the distance D(t) between γ_{1}(t) and γ_{2}(t) (since leaves of G are geodesics in S this is the same as the distance along the leaf containing both points) and second the angle A(t) between γ_{2}(t) and the foliation E. We claim that both are bounded by the quantities we obtain when γ_{1} is the horizontal line segment {(t, 1) : 0 ≤ t < 1} in ℍ and γ_{2} is the circular arc \{(t, \sqrt{1 − t^{2}}) : 0 ≤ t < 1\} (note that the parameterizations are

AN EXPLICIT CONSTANT FOR SULLIVAN’S THEOREM 25

related as above if G consists of vertical lines). More concretely, we claim

D(t) ≤ D_{0}(t) ≡ log \frac{1}{\sqrt{1 − t^{2}}} , (11)

A(t) ≤ A_{0}(t) ≡ arcsin(t). (12)

Figure 8. The definition of D_{0} and A_{0}

The angle A(t) is a little difficult to deal with directly since it may be discontinuous at bending geodesics (since the tangents to both γ_{2} and E have discontinuities there). To get around this problem, we again follow Epstein and Marden by bounding A(t) by another angle which does vary continuously.

For two tangent vectors v, w in ℍ^{3} with the same base point, we let ang(v, w) denote the angle between these vectors. Given a point x ∈ S and a tangent vector v at x we let ang(v, E) denote the minimal angle between v and the two possible tangent vectors to E at x. Similarly for ang(v, G).

Let v(t) be the unit tangent vector of ℍ^{3} at γ_{2}(t) which is orthogonal to P. Since leaves of G are parallel to the bending geodesics, the tangent direction to G (ignoring signs) is continuous along γ_{2}, even when we cross a bending geodesic. Since v(t) is obviously continuous along γ_{2}, we see that A_{2}(t) = ang(v(t), G) is continuous.

By construction v(t) is always perpendicular to γ′_{2}(t) (since this vector is tangent to P) and γ′_{2} lies in the plane generated by E and G at γ_{2}(t). We claim that

ang(γ′_{2}(t), E) ≤ ang(v(t), G),

i.e., A(t) ≤ A_{2}(t). To prove this, fix G, E and γ′_{2} and rotate v(t) around the axis parallel to γ′_{2}. It is easy to see that the smallest angle with G occurs when v(t) is the plane containing G, E and γ′_{2} and that this angle equals the angle between E and

26 CHRISTOPHER J. BISHOP

γ′_{2}. Thus for v(t) in its original position, the angle with G must have been larger, proving the claim. See Figure 9. Thus we only need to prove (12) with A_{2} in place of A.

Figure 9. Proof that ang(γ′_{2}(t), E) ≤ ang(v(t), G)

To prove (11) and (12) we will enumerate the triangles of S hit by the paths and consider the intersections γ̃_{1} = γ_{1} ∩ T and γ̃_{2} = γ_{2} ∩ T. Fix one such triangle T. Then γ̃_{1} connects two edges of T and using a Möbius transformation we may place the common vertex at ∞. Thus we may assume T lies in a vertical plane and that γ̃_{1} corresponds to a horizontal line segment at height 1 (so G consists of vertical lines and E of horizontal lines in T). See Figure 10.

Figure 10. The angle between γ_{2} and G

AN EXPLICIT CONSTANT FOR SULLIVAN’S THEOREM 27

The plane P then corresponds to some Euclidean hemisphere and v is the normal vector to this hemisphere. Moreover, γ̃_{2} is the circular arc P ∩ T. It is clear that the angle v makes with the vertical and height of γ̃_{2} both change most rapidity if γ̃_{2} is radial (i.e., its vertical projection onto the base of P is a radius of that disk). Thus we need only consider that case and the desired estimates become the following claims about circles in the plane.

Lemma 2.11. Suppose 0 ≤ a < b < 1 and let L_{1} = {(a, y) : 0 ≤ y < ∞} and L_{2} = {(b, y) : 0 ≤ y < ∞}. Suppose x is a point on the real axis and y is a point on L_{1} such that

\sqrt{1 − a^{2}} ≤ y ≤ 1/\sqrt{1 − a^{2}} (13)

and

(14) θ_{a} = arcsin(|x − a|/|x − y|) ≤ arcsin(a).

Let z ∈ L_{2} be such that |x − z| = |x − y| ≡ r. Then

\sqrt{1 − b^{2}} ≤ z ≤ 1/\sqrt{1 − b^{2}} (15)

(16) |θ_{b}| = | arcsin(|x − b|/r)| ≤ arcsin(b).

Figure 11. The proof of Lemma 2.11

28 CHRISTOPHER J. BISHOP

Proof. This is elementary geometry and calculus and breaks into several cases. First we prove the left side of (15). Let C_{0} be the circle of radius 1 around 0 and let C_{1} be the circle of radius r around x. Since

r = |x − y| = \sqrt{y^{2} + x^{2} + 2ax + a^{2}} ≥ |x − 1|,

the circle C_{1} contains the point 1 in its interior disk. Thus C_{1}’s intersection with L_{1} and L_{3} = {(1, y) : y ≥ 0} are above C_{0}’s intersections with these lines and hence C_{1}’s intersection with L_{2} is above C_{0}’s intersection with this line, as desired (otherwise C_{1} and C_{0} would have to intersect at least 4 times which is impossible).

Next we prove the right side of (15). The arc of C_{1} between L_{0} and L_{1} is convex down whereas the upper bound is convex up. By assumption the circle has smaller derivative at t = a, and hence remains below the upper bound on the whole interval [a, b].

Now we prove (16). First suppose x ≤ 0. Then r ≥ 1. Consider θ_{t} = arcsin((|x| + t)/r) as a function on [a, b]. At t = a it is less than arcsin(t) by assumption and since r > 1 is has smaller derivative than arcsin(t) and hence it remains smaller on the whole interval, and hence is smaller at t = b, as desired. Next suppose 0 ≤ x ≤ b. Since |x − b| ≤ b and z ≥ \sqrt{1 − b^{2}} we get

θ_{b} = arctan(|x − b|/z) ≤ arctan(b/\sqrt{1 − b^{2}}) = arcsin(b),

as desired. Finally if x ≥ b then (16) is obvious since |x − t|/r as t goes from a to b angle decreases from a to b.

Conditions (13) and (15) are equivalent to (11) and conditions (14) and (16) are equivalent to (12). Thus the lemma says that if (11) and (12) hold when γ_{2} enters T they still hold when it leaves (and at all points in γ_{2} ∩ T). Thus by induction we have now proven the desired estimates (11) and (12).

Finally we must estimate the exterior angle of γ_{1} when it crosses a bending geodesic L in terms of the exterior angle for γ_{2} when it crosses L. Since γ_{1} is perpendicular to L, its exterior angle at L is the same as the exterior angle made by the two faces F_{1} and F_{2} of S which meet along the geodesic L. Let α denote this exterior angle. Let β be the exterior angle made by γ_{2} when it crosses L. Choose t so that γ_{2}(t) ∈ L.

AN EXPLICIT CONSTANT FOR SULLIVAN’S THEOREM 29

We claim that

α ≤ \frac{β}{cos(A_{0}(t))}. (17)

To prove this we need the following simple lemma about planes in Euclidean 3-space.

Lemma 2.12. Suppose two half-planes H_{1}, H_{2} meet along a line L with exterior angle α. Suppose 0 ∈ L and v is a unit vector such that ang(v, L) ≤ θ. Let P be the plane perpendicular to v. Let v_{i} be the unit vector in the direction of the ray P ∩ H_{i} for i = 1, 2. Let β = π − ang(v_{1}, v_{2}) be the exterior angle made by these vectors. Then α ≤ β/ cos(θ).

Proof. We may as well assume L is the z-axis, and

H_{1} = {(x, y, z) : x > 0, y = sin(α/2)x},
H_{2} = {(x, y, z) : x < 0, y = sin(α/2)|x|}.

Also, because cos(θ) is decreasing for 0 ≤ θ ≤ π/2 it suffices to prove the estimate when ang(v, L) = θ, and without loss of generality we may assume the z-coordinate of v is positive.

We want to show β ≥ cos(θ)α. Think of β as a function of v. We claim that β will be minimized when v lies in the xz-plane. This is the same as saying ang(v_{1}, v_{2}) will be maximized when v is in the xz-plane. Let P_{1} = {(x, y, z) : y = 1} and let z_{i} be the z-coordinate of the point P ∩ H_{i} ∩ P_{1} for i = 1, 2. It is easy to see that |z_{1} − z_{2}| is maximized when v is in the xz-plane (since the line P ∩ P_{1} has steepest slope in this case) and that z_{1} = −z_{2} in this case. It is also easy to see that for a given displacement |z_{1} −z_{2}| the angle between v_{1} and v_{2} is maximized when z_{1} = −z_{2}. This implies the claim that β is minimized when v is in the xz-plane. See Figure 12.

Thus we may also assume the y-coordinate of v is 0 and its x-coordinate is negative. It is now straight forward to compute β in terms of α. See Figure 13. Take points as in the figure with coordinates

A = (cot(θ), 0, 0), B = (cot(θ), 0, 1),
C = (cot(θ), cot(θ) tan(α/2), 0), D = (cot(θ), cot(θ) tan(α/2), 1).

The the line [0, D] lies in H_{1} ∩ P, hence it is parallel to the vector v_{1}. The segment [OB] also lies in P. Thus the angle DOB will be half the exterior angle formed by

30 CHRISTOPHER J. BISHOP

Figure 12. β is minimized when v is in the xz-plane Figure 13. Computing relation between α, β, θ

v_{1} and v_{2}, i.e. is β/2. Thus |BD| = tan(β/2)|OB| = tan(β/2)/ sin(θ), |AC| = tan(α/2)|OA| = tan(α/2)/ tan(θ).

Since |AC| = |BD|, we get

tan(α/2) = tan(β/2)/ cos(θ).

Since arctan(x) is increasing and convex down on x ≥ 0, We have

\frac{arctan(cx)}{cx} ≤ \frac{arctan(x)}{x},

if c ≥ 1, or equivalently arctan(cx) ≤ c arctan(x). Thus

α = 2 arctan(\frac{1}{cos(θ)} tan(β/2)) ≤ \frac{2}{cos(θ)} arctan(tan(β/2)) = \frac{β}{cos(θ)},

as desired.

The lemma implies (17) by taking v = v(t) and H_{i} to be the tangent plane to F_{i} at γ_{2}(t) for i = 1, 2. We have now proved that if γ_{2} crosses a bending geodesic L at parameter t with exterior angle β, then the angle between the corresponding faces is at most α = β/ cos(A_{2}(t)).

Since |A_{2}(t)| ≤ A_{0}(t) = arcsin t, we have cos(A_{2}(t)) ≥ \sqrt{1 − t^{2}}, and hence the sum of exterior angles crossed by γ_{1} in the interval [−s, s] is at most

∑_{j} \frac{β_{j}}{\sqrt{1 − t_{j}^{2}}} ≤ \frac{1}{\sqrt{1 − s^{2}}} ∑_{j} β_{j}

where we assume γ_{2} has exterior angle β_{j} at parameter value t_{j} in the interval [−s, s]. The hyperbolic length of the path γ_{2} is bounded by the length of the arc of |z| = 1 from x = −s to x = s which is easily computed to be

∫_{−s}^{s} \frac{1}{1 − t^{2}} dt = log \frac{1 + s}{1 − s}.

Thus the amount of bending in γ_{2} is at most π + log \frac{1+s}{1−s} and this proves Lemma 2.9.

The estimate is certainly not sharp for s near 1. For such values γ_{2} has hyperbolic length greater than π and in this case, not all the bending can be concentrated at the two endpoints. Moreover, for s > 1 one can deduce a finite bound for B_{1}(s) by cutting into subintervals of length < 1 and adding the estimates for these subintervals.

2.13. A quasiconformal constant for Sullivan’s theorem. The desired map S → Ω is a composition of maps F = f_{4} ∘ f_{3} ∘ f_{2} ∘ f_{1} ∘ f_{0}. The map f_{0} is piecewise conformal and the maps f_{1} and f_{3} are piecewise quasiconformal with constant K_{1} = 1.20656. The map f_{2} is piecewise quasiconformal with constant K_{2} given by (10). Combining this with Lemma 2.9 we get

K_{2}(s) = k(\frac{1}{2}B_{2}(s), \frac{1}{2s}B_{2}(s)),

32 CHRISTOPHER J. BISHOP

and setting s = \frac{1}{2}, we can use Mathematica to compute K_{2}(1/2) = 6.93808. Thus the map F has quasi-constant at most

K = K_{1}K_{2}K_{1} = 10.1002

There are two observations we can use to improve this a bit. The first is that the maps f_{1} and f_{2} take on their maximum dilatations in different places and so the dilatation for the composition should be strictly less than K_{1}K_{2}. More precisely, the map f_{1} has its largest dilatation close to the central triangles and the dilatation tends to 1 as we go out the cusps. On the other hand, the map f_{2} has a better estimate near the central triangles because for such z’s the corresponding E-leaf must cross the triangle and hence has large length not in a crescent. For example, if I is an Einterval of S-length 2s < 1 centered at z ∈ T and z is less than hyperbolic distance r = − log 2s from a central triangle then I \ T has at most one component and this component has S-length at most s/2. Thus the bending along I is at most B_{2}(s/2) instead of B_{2}(s). This improves our estimate of the dilatation of f_{2} at such points to

K_{3}(s) = k(\frac{1}{2}B_{2}(\frac{s}{2}), \frac{1}{2s}B_{2}(\frac{s}{2})).

where k is given by (7).

If z = x + iy ∈ W_{1} is more than hyperbolic distance r from a central triangle then y ≥ \frac{1}{2s} and we estimate the dilatation of f_{2} using B_{2}(s) just as before, but for f_{1} we have the improved estimate

K_{1}(z) = |g′(ig^{−1}(\frac{1}{2s}))|,

at such points (since we know |g′| is decreasing along the positive imaginary axis). Since |g′(iy)| ≤ |g′(i)| ≤ K_{1} = 1.20656 we know that

|g^{−1}(\frac{1}{2s})| ≥ h(s) = 1 + (\frac{1}{2s} − 1)/K_{1}.

Thus

K_{1}(z) ≤ |g′(h(s)i)|,

which we can estimate from our explicit formula for g. Thus for any given 0 < s ≤ \frac{1}{2}, F will be quasiconformal with constant at most

K_{4}(s) = K_{1} max(K_{2}(s)|g′(h(s)i)|, K_{3}(s)K_{1}).

Sampling different values of s leads to taking s = .39 which gives K_{4} = 9.18558.

AN EXPLICIT CONSTANT FOR SULLIVAN’S THEOREM 33

The second idea for improving our estimate concerns the map f_{3} : Ω_{2} → Ω. On triangles this map is just f_{1}^{−1} and it is the identity on the crescents. Thus it is K_{1} = 1.20656 quasiconformal on the triangles, but conformal on the crescents. We can take advantage of this by redefining the metric on the crescents in Ω_{2} by shrinking distances along the E-foliation by a factor of K_{1} and then defining f_{3} on the crescents to be the map which expands distances along the E-foliation by a factor of K_{1}. Thus f_{3} is still a K_{1}-quasiconformal map. However, since the crescents in Ω_{2} are “thinner” now, our estimate of B_{2}(s) is reduced by a factor of K_{1}. Therefore in our estimate of the dilatation of f_{2} we may replace each occurance of B_{2}(s) by B_{2}(s)/K_{1}. Thus for any 0 < s ≤ 1/2, F has dilatation at most

K_{5}(s) = K_{1} max(K_{7}(s)|g′(h(s)i)|, K_{6}(s)K_{1}),

where

K_{6}(s) = k(\frac{1}{2K_{1}} B_{2}(\frac{s}{2}), \frac{1}{2sK_{1}} B_{2}(\frac{s}{2})).

and

K_{7}(s) = k(\frac{1}{2K_{1}} B_{2}(s), \frac{1}{2sK_{1}} B_{2}(s)).

Searching values of s leads to s = .39 which gives K_{5}(.39) = 7.8127.

2.14. A biLipschitz constant for Sullivan’s theorem. In this section we will note the map constructed above is biLipschitz from the path metric on S to the hyperbolic metric on Ω, although we will not explicitly compute the constant. We will compute the constant for another map in which f_{1} is replaced by a simpler map.

As we shall see below, the maps f_{0}, f_{2} and f_{4} are all bilipschitz. However, the f_{1} : U_{1} → W_{1} we constructed is not; it has zero derivative at the point u = \frac{1}{2} + i\frac{\sqrt{3}}{2}. The mapping f_{1} is biLipschitz outside a neighborhood of this point (with constant depending on the neighborhood). If the map f_{2} is chosen with a parameter s < \frac{1}{2}, then f_{3} ∘ f_{2} ∘ f_{1} is the identity on a neighborhood of the center of each triangle. Since f_{1} (and hence f_{3}) are biLipschitz outside this neighborhood, we see that our map F is biLipschitz with a constant depending on s (with a little more work one can show it is also biLipschitz for s = 1/2).

In order to give a map with an explicit constant we will replace f_{1} and f_{3} by other maps obtained by repeating the construction of f_{1}, but instead of the map g : U → W

34 CHRISTOPHER J. BISHOP

defined earlier, we use the map g_{0} : U → W given by

(x, y) → (x, y + 1 − \sqrt{1 − x^{2}}).

Then by using (6) and the bounds g_{y} = 0, |g_{x}| ≤ 1/\sqrt{3}, we get g_{0} is quasiconformal with constant

K_{g_{0}} = k(\frac{1}{\sqrt{3}}, 0) ≈ 1.76759.

If we let

k_{1}(x, y) = \frac{1}{2}(\sqrt{x^{2} + (y + 2)^{2}} − \sqrt{x^{2} + y^{2}}),

k_{2}(x, y) = \frac{1}{2}(\sqrt{x^{2} + (y + 2)^{2}} + \sqrt{x^{2} + y^{2}}),

then

L_{1}(f_{1}) = k_{1}(\frac{1}{\sqrt{3}}, 0)(1 − \sqrt{3/4}) ≈ .651388 ≈ \frac{1}{1.53518}

L_{2}(f_{1}) = k_{2}(\frac{1}{\sqrt{3}}, 0) ≈ 1.32951 ≈ \frac{1}{.752158}.

The extra factor of (1 − \sqrt{3/4}) occurs in L_{1} because k_{1} only computes L_{1} with respect to the Euclidean metric. When a point at height y is shifted upwards by t the hyperbolic metric decreases by a factor of \frac{y}{y+t}. For g_{0} the worse case occurs at the point u where y = 1 − \sqrt{3/4} and y + t = 1.

Using Lemma 2.8 and our bounds for f_{2} gives for s = 1/2,

L_{1}(f_{2}) = k_{1}(B_{2}(s)/2, B_{2}(s)/2s) ≈ .356553,

and

L_{2}(f_{2}) = k_{2}(B_{2}(s)/2, B_{2}(s)/2s) ≈ 6.39594.

Since f_{3} is the identity on interior and boundary crescents and the inverse of f_{1} on triangles, we get

L_{1}(f_{3}) = \frac{1}{L_{2}(f_{1})} ≈ .752158,

L_{2}(f_{3}) = \frac{1}{L_{1}(f_{1})} ≈ 1.53518.

Finally, by Lemma 2.3

L_{1}(f_{4}) = \frac{\sqrt{3}}{4} ≈ .433013, L_{2}(f_{4}) = 1.

and hence the whole composition satisfies

L_{1}(f) = L_{1}(f_{1})L_{1}(f_{2})L_{1}(f_{3})L_{1}(f_{4}) ≈ .0756439 ≈ \frac{1}{13.2198},

L_{1}(f) = L_{2}(f_{1})L_{2}(f_{2})L_{2}(f_{3})L_{2}(f_{4}) ≈ 13.0544.

As in the previous section, we could probably reduce the constants by a more careful analysis of where the worse case occurs in each step of the composition, but we will not do this here.

2.15. Computer code. For the reader’s convenience we give Mathematica code for evaluating the functions described in the last two sections.

lambda1[z_]:= Pi^2 Sum[ Cos[Pi(n-.5)z]^{-2}
                          - Sin[Pi(n-.5)z]^{-2},{n,-10,10}];
lambda2[z_]:= Pi^2 Sum[ Cos[Pi(n)z]^{-2}
                           -  Sin[Pi(n-.5)z]^{-2},{n,-10,10}];
lambda[z_]:= lambda1[z]/lambda2[z];
dlambda[z_]=D[lambda[z],z];
J[z_]:= (4/27)(1-z+z^2)^3/(z^2(1-z)^2);
K1= N[Abs[ dlambda[I] 2 Sqrt[3] /Pi]];   (* = 1.2065... *)
g[z_]:= ArcCos[ 2 J[lambda[z]]-1]/(2Pi);
dg[z_]=Abs[D[g[z],z]];
L1[x_,y_]:= (1/2)(Sqrt[x^2 + (2+y)^2] -Sqrt[x^2+y^2]);
L2[x_,y_]:= (1/2)(Sqrt[x^2 + (2+y)^2] +Sqrt[x^2+y^2]);
k[x_,y_]:= L2[x,y]/L1[x,y];
B2[t_]:= (Pi+Log[(1+t)/(1-t)])/Sqrt[1-t^2];
K2[s_]:= k[B2[s]/2,B2[s]/(2s)];
K3[s_]:= k[B2[s/2]/2,B2[s/2]/(2s)];
h[s_]:= 1+ ((2s)^{-1} -1)/1.20656;
K4[s_]:= K1 Max[ K2[s] dg[ h[s] I ], K3[s]  K1];
K6[s_]:= k[B2[s/2]/(2 K1),B2[s/2]/(2s K1)];
K7[s_]:= k[B2[s]/(2 K1),B2[s]/(2s K1)];
K5[s_]:= K1 Max[ K7[s] dg[ h[s] I],  K6[s]  K1];
Kg0=k[1/Sqrt[3],0];
L1f1=L1[1/Sqrt[3],0]Sqrt[3]/2;
L2f1=L2[1/Sqrt[3],0];
L1f2[s_]:=L1[B2[s]/2,0];
L2f2[s_]:=L2[B2[s]/2,B2[s]/(2s)];
L1F= L1f1 L1f2[.5] (L2f1)^{-1}  Sqrt[3]/4;
L2F= L2f1 L2f2[.5] (L1f1)^{-1};

References

[1] L.V. Ahlfors. Complex Analysis. McGraw Hill, 1979.

[2] C.J. Bishop. Quasiconformal Lipschitz maps, Sullivan’s convex hull theorem and Brennan’s conjecture. 1999. Preprint.

[3] A. Douady and C. J. Earle. Conformally natural extension of homeomorphisms of the circle. Acta Math., 157(1-2):23–48, 1986.

[4] 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), pages 113–253. Cambridge Univ. Press, Cambridge, 1987.

[5] C. Rourke. Convex ruled surfaces. In Analytical and geometric aspects of hyperbolic space (Coventry/Durham, 1984), pages 255–272. Cambridge Univ. Press, Cambridge, 1987.

[6] D. Sullivan. Travaux de Thurston sur les groupes quasi-fuchsiens et les variétés hyperboliques de dimension 3 fibrées sur S^{1}. In Bourbaki Seminar, Vol. 1979/80, pages 196–214. Springer, Berlin, 1981.

[7] W.P. Thurston. The geometry and topology of 3-manifolds. Geometry Center, University of Minnesota, 1979.

[8] Jussi Väisälä. Lectures on n-dimensional quasiconformal mappings. Springer-Verlag, Berlin, 1971. Lecture Notes in Mathematics, Vol. 229.

C.J. Bishop, Mathematics Department, SUNY at Stony Brook, Stony Brook, NY 11794-3651

  E-mail address: bishop@math.sunysb.edu