Koebe conjecture and the Weyl problem for convex surfaces in hyperbolic 3-space
KKOEBE CONJECTURE AND THE WEYL PROBLEM FOR CONVEX SURFACESIN HYPERBOLIC 3-SPACE
FENG LUO AND TIANQI WUA
BSTRACT . We prove that the Koebe circle domain conjecture is equivalent to the Weyl typeproblem that every complete hyperbolic surface of genus zero is isometric to the boundary ofthe hyperbolic convex hull of the complement of a circle domain. It provides a new way toapproach the Koebe’s conjecture using convex geometry. Combining our result with the workof He-Schramm on the Koebe conjecture, one establishes that every simply connected non-compact polyhedral surface is discrete conformal to the complex plane or the open unit disk.The main tool we use is Schramm’s transboundary extremal lengths. C ONTENTS
1. Introduction 21.1. The main result 21.2. History, motivations and general form of Weyl problem in H X in a circle 52.3. Outline of the proof of Theorem 1.2 (a) 62.4. Outline of the proof to Theorem 1.2 (b) 63. Hyperbolic geometry and convex surfaces 73.1. The Klein and Poincar´e models of hyperbolic 3-space 73.2. Euclidean area of hyperbolic convex surfaces 83.3. Estimates for the nearest point projection 104. Hausdorff convergence and the Poincar´e metrics 124.1. Alexandrov’s work on convex surfaces 134.2. The Poincar´e metrics and their convergence 134.3. A generalized form of Arzela-Ascoli theorem 165. Proof of part (a) of Theorem 1.2 assuming equicontinuity 175.1. The Hausdorff limit set Y of the sequence { Y ( n ) } φ n to the Riemann sphere 185.3. Finish the proof of part (a) of Theorem 1.2 216. Proof of part (b) of Theorem 1.2 assuming equicontinuity 217. Transboundary extremal lengths and a duality theorem 237.1. Transboundary extremal lengths 237.2. An example of transboundary extremal length and a duality theorem 247.3. Transboundary extremal length estimates on annuli 25 Mathematics Subject Classification.
Key words and phrases. circle domains, convex hull, convex surfaces, hyperbolic and conformal geometries. a r X i v : . [ m a t h . G T ] F e b FENG LUO AND TIANQI WU
8. Proof of the equicontinuity Theorem 5.1 278.1. Extremal length estimate I: lim n EL (Γ n ) = 0 lim inf n EL (Γ (cid:48) n ) > lim inf n EL (Γ (cid:48) n ) > lim EL (Γ n ) = 0 NTRODUCTION
The main result.
Given a compact set X in the Riemann sphere ˆ C whose complementis connected and X contains at least three points, there are two natural hyperbolic metricsassociated to X . The first is the Poincar´e hyperbolic metric on the complement of X and thesecond is the hyperbolic metric on the boundary of the convex hull of X in the hyperbolic3-space H with ∂ H = ˆ C . The latter was first investigated in details by W. Thurston in1970’s. There are close relationships between the two hyperbolic metrics (see [34], [5], [11]and others). The paper explores these two metrics, the Koebe circle domain conjecture and theWeyl isometric embedding problem.In 1908, P. Koebe [18] made the circle domain conjecture that any connected open set, i.e.,a domain, in the Riemann sphere is conformally diffeomorphic to a new domain whose bound-ary components are either round circles or points, i.e., a circle domain . The Riemann mappingtheorem is a special case of the conjecture. In our investigation [21] of discrete conformalgeometry of polyhedral surfaces, we discovered close relationships among the Koebe conjec-ture, the Weyl problem on convex surfaces, and discrete uniformization. The main result ofthe paper shows that the Koebe conjecture is equivalent to a version of the Weyl problem fornon-compact convex surfaces in the hyperbolic 3-space. Together with the important workof He-Schramm on the Koebe conjecture, this implies the existence of isometric embeddingof any genus zero hyperbolic surface of countably many ends into the hyperbolic 3-space asthe boundary of the convex hull of the complement of a circle domain in ∂ H . The latterimplies the existence part of the discrete uniformization conjecture for non-connected simplyconnected polyhedral surfaces.Let S be the unit 2-sphere, ˆ C = C ∪ {∞} be the Riemann sphere, ( H , d P ) or H P be thePoincar´e ball model of the hyperbolic 3-space whose boundary is S , and C H ( Y ) be the convexhull of a closed set Y in H ∪ S in the hyperbolic 3-space. A circle type closed set Y ⊂ S or ˆ C is a closed set whose complement is a circle domain, i.e., each connected component of Y is either a closed round disk or a point. It is well known by the work of W. Thurston ([36], seealso [11]) that if Y is a closed set in S containing more than two points, then the boundary ofthe convex hull ∂C H ( Y ) ⊂ H , in the intrinsic path metric, is a complete hyperbolic surface.As a convention in this paper, if C H ( Y ) is 2-dimensional, we use ∂C H ( Y ) to denote the metricdouble of C H ( Y ) across its boundary, i.e., ∂C H ( Y ) := C H ( Y ) ∪ id ∂ C H ( Y ) . In particular, ∂C H ( Y ) is a complete hyperbolic surface without boundary. IRCLE DOMAIN 3
Conjecture 1. ([21], see also [23]) Every genus zero complete hyperbolic surface is isometricto ∂C H ( Y ) for a circle type closed set Y in S .The main result of the paper shows, Theorem 1.1.
The circle domain conjecture of Koebe is equivalent to Conjecture 1.
This theorem provides a new way to approach the Koebe conjecture using convex geometry.By the uniformization theorem, every connected open set U ⊂ S whose boundary containsmore than two points admits a unique complete hyperbolic metric d U conformal to the complexstructure on U . The metric d U is usually called the Poincar´e metric of the domain U . On theother hand, Koebe proved that any genus zero Riemann surface is conformally diffeomorphicto a domain in the Riemann sphere. It follows that every genus zero complete hyperbolicsurface is isometric to a Poincar´e metric ( U, d U ) where U ⊂ S with | ∂U | ≥ . This showsthat Theorem 1.1 is equivalent to the following result. Theorem 1.2. (a) For any circle domain U ⊂ ˆ C whose boundary contains at least three points,there exists a circle type closed set Y ⊂ S such that the Poincar´e metric ( U, d U ) is isometricto ∂C H ( Y ) .(b) For any circle type closed set Y ⊂ S with | Y | ≥ , there exists a circle domain U ⊂ ˆ C such that ∂C H ( Y ) is isometric to the Poincar´e metric ( U, d U ) . Using He-Schramm’s theorem [16] that Koebe conjecture holds for domains with countablymany boundary components, we obtain,
Corollary 1.3.
Every genus zero complete hyperbolic surface with countably many topologicalends is isometric to ∂C H ( Y ) for a circle type closed set Y in S . The relationship between the Koebe conjecture and Conjecture 1 was discovered duringour investigation of the discrete uniformization conjecture for polyhedral surfaces. Indeed,Conjecture 1 can be considered as a generalized version of the existence part of discrete uni-formization conjecture. The above corollary implies that every non-compact simply connectedpolyhedral surface is discrete conformal to the complex plane C or the unit disk D . See [13],[14], [21] and [23].The main tool we use to show Theorem 1.2 is Schramm’s transboundary extremal lengths.1.2. History, motivations and general form of Weyl problem in H . Organization of thepaper Koebe conjecture is known to be true for connected open set U ⊂ ˆ C which has finitelymany boundary components ([18]). The best work done to date is by He-Schramm [16] wherethey proved the conjecture for U with countably many boundary components. Conjecture 1 isknown to be true for finite area hyperbolic surfaces and hyperbolic surfaces of finite topologicaltypes whose ends are funnels by the works of Rivin [30] and Schlenker [31] respectively. F.Fillastre [12] proved that Conjecture 1 holds for many symmetric domains with countablymany boundary components.Conjecture 1 is a Weyl type problem for convex surfaces. It is well known that a smoothconvex surface in the Euclidean space (respectively the hyperbolic space) has Gaussian cur-vature at least zero (respectively − ). Weyl’s problem asks the converse. Namely, whetherany Riemannian metric of positive curvature on the 2-sphere is isometric to the boundary ofa convex body in the 3-space. The problem was solved affirmatively by Levy, Nirenberg andAlexandrov. The natural generalization of Weyl’s problem to the hyperbolic 3-space states that FENG LUO AND TIANQI WU every complete path metric of curvature at least − on a genus zero surface is isometric to theboundary of a closed convex set in the hyperbolic 3-space. This was established by Alexandrovin [1]. Using the work of Alexandrov, we are able to show [22] that the converse holds, i.e.,every complete hyperbolic surface of genus zero is isometric to ∂C H ( Y ) for some closed set Y ⊂ ∂ H . Take a simply connected domain U ⊂ C with U (cid:54) = C and let Y = ˆ C − U beits complement. Then in the upper-half space model of the hyperbolic 3-space, the boundarysurface ∂C H ( Y ) is homeomorphic to U and hence is simply connected. Thurston’s theoremsays that there exists an isometry Φ from ∂C H ( Y ) to ∂C H ( D c ) where D is the unit disk. On theother hand, the Riemann mapping theorem says that there exists a conformal diffeomorphism φ from U to D . Thus Thurston’s isometry Φ can be considered as a geometric realization ofthe Riemann mapping φ .The Koebe conjecture and Conjecture 1 are the corresponding Riemann mapping-Thurston’sisometry picture for non-simply connected domains. It is also conjectured by [21] that if X and Y are two circle type closed sets in the Riemann sphere such that ∂C H ( X ) is isometricto ∂C H ( Y ) , then X and Y differ by a Moebius transformation. Though it is known that theuniqueness part of the Koebe circle domain conjecture is false, it is possible that the conjecturemay still be true in view of Pogorelov’s rigidity theorem [28] (see also Theorem 1 in [9]). Inany case, the following conjecture of us has strong supporting evidences. Conjecture 2.
Suppose X and Y are two circle type closed sets in S with countably manyconnected components such that ∂C H ( X ) is isometric to ∂C H ( Y ) . Then X and Y differ by aM¨obius transformation.The strongest version of the Weyl problem in hyperbolic 3-space is the following. Conjecture 3.
Suppose ( X, d ) is a planar surface with a complete path metric whose curvatureis at least − . Then there exists a complete convex surface Y in H isometric to ( X, d ) suchthat each end of Y is either a circle or a point in the sphere at infinity of H . Furthermore, theconvex surface Y is unique up to isometry of H .1.3. Organization of the paper and acknowledgement.
The paper is organized as follows.An outline of the proof of the main theorem is in §
2. In §
3, we proved an area estimate theoremfor convex surfaces in the hyperbolic 3-space and a few results on the nearest point projectionmaps. In §
4, we establish some results relating to Hausdorff convergence and the convergenceof the Poincar´e metrics. Theorem 1.2 is proved in § § §
7, 8, and 9. In the Appendix, we recall the work ofReshetnyak [29] on the complex structure on non-smooth convex surfaces which is used in thepaper.We thank Michael Freedman and Francis Bonahon for stimulating discussions. Part of thework was carried out while the first author was visiting CMSA at Harvard. We thank S.T.Yau for the invitation. The work is supported in part by NSF 1760527, NSF 1737876, NSF1811878 and NSF 1405106.2. A
PROOF OF A SPECIAL CASE OF T HEOREM
AND OUTLINE OF THE PROOF OF T HEOREM X lies in a circle. The proof is easy and has to be dealt with separately. Then we outlinethe main steps of the proof of Theorem 1.2 since the proof takes the rest of the seven sections. IRCLE DOMAIN 5
The strategy of proving Theorem 1.2 is to approximate an arbitrary circle domain by circledomains of finite topology. The key result which enables us to show that the limiting domain isstill a circle domain is the equicontinuity of the family of approximation conformal maps, i.e.,Theorems 5.1 and 6.1. To prove the equicontinuity, we use Schramm’s transboundary extremallength [32] and a duality Theorem 7.4 of transboundary extremal lengths on annuli. The latterenables us to estimate the modules of annuli in possibly non-smooth convex surfaces.2.1.
Notations and conventions.
We use S , C , ˆ C , D , H P and H K to denote the standard 2-sphere, the complex plane, the Riemann sphere, the open unit disk, the Poincar´e model of thehyperbolic 3-space and the Klein model of the hyperbolic 3-space respectively. The sphericaland the Euclidean metrics on S , ˆ C and C are denoted by d S and d E respectively. The Poincar´emetric on a domain U ⊂ ˆ C with | ∂U | ≥ is denoted by d U . The interior of a surface S isdenoted by int ( S ) . The convex hulls of a set Y in H P and H K are denoted by C H ( Y ) and C K ( Y ) . The ball of radius r centered at p in a metric space ( Z, d ) is B r ( p, d ) and the diameterof X ⊂ Z is diam d ( X ) . If X is a path connected subset of a metric space ( Z, d ) , then theinduced path metric on X is denoted by d X . The closure of a set X ⊂ R is denoted by X .2.2. Theorem 1.2 in the special case of X in a circle. A circle domain U = ˆ C − X is specialif the boundary ∂U is contained in a circle, i.e., X lies in a circle. This is the same as thehyperbolic convex hull of X in H is 2-dimensional. Theorem 1.2 for these domains is aneasy consequence of the Riemann mapping theorem and Caratheodory’s extension theorem ofRiemann mapping. Theorem 2.1.
Suppose Y is a compact subset of the circle S ⊂ ˆ C such that each connectedcomponent of Y is a single point and | Y | ≥ . Then there exist two closed sets X , X ⊂ S whose connected components are points such that(a) ˆ C − Y is conformal to ∂C H ( X ) , and(b) ∂C H ( Y ) is conformal to ˆ C − X .Proof. To see (a), let d U be the Poincar´e metric on U = ˆ C − Y . Since U is invariant under theorientation reversing conformal involution τ ( z ) = z , by the uniqueness of the Poincar´e metric,we see that τ is an isometric involution of ( U, d U ) . In particular, the fixed point set S ∩ U of τ is a union of complete geodesics. This implies that U = {| z | ≤ } − Y is a simply connectedhyperbolic surface with geodesic boundary in the metric d U . By the monodromy theorem,there exists an isometry ψ from ( U , d U | ) onto a closed convex domain D in the hyperbolicplane H which will be viewed as the totally geodesic plane in H with ∂ H = S . Since U has geodesic boundary, the convex domain D is the convex hull C H ( X ) of a closed set X ⊂ S . Now both int ( U ) and int ( D ) are Jordan domains and ψ is a conformal map betweenthen. Therefore, by Caratheodory extension theorem, ψ extends to be a homeomorphism Φ between their closures which are U ∪ Y and D ∪ X . This extension homeomorphism Φ sends Y to X . In particular, each component of X is a point. By the Schwarz reflectionprinciple, Φ can be naturally extended to a conformal homeomorphism between ( U, d U ) and ∂C H ( X ) := C H ( X ) ∪ id ∂ C H ( X ) .To see part (b), since Y ⊂ S , the hyperbolic convex hull C H ( Y ) ⊂ H is a topologicaldisk contained in the hyperbolic plane H ⊂ H whose boundary is S . Then by the Riemannmapping theorem, there exists a conformal diffeomorphism φ from int ( C H ( Y )) to the unit disk D = { z ∈ C (cid:12)(cid:12) | z | < } . By Carath´eodory extension theorem, φ extends to a homeomorphism Φ from the closure C H ( Y ) in R to the closed disk D . Let X = Φ( Y ) whose components FENG LUO AND TIANQI WU are all points. By the Schwarz reflection principle, Φ can be naturally extended to a conformalhomeomorphism between ∂C H ( Y ) and ˆ C − X . (cid:3) Outline of the proof of Theorem 1.2 (a).
Now suppose U = ˆ C − X is a circle domainin ˆ C such that X contains at least three points. Let d U be the Poincar´e metric on U . We willfind a circle type closed set Y ⊂ S such that ( U, d U ) is isometric to ∂C K ( Y ) ⊂ ( H , d K ) .Here ( H , d K ) is the Klein model of the hyperbolic space and C K ( Y ) is the convex hull in theKlein model.Produce a sequence of circle domains U n = ˆ C − X ( n ) with ∂U n consisting of finitely manycircles such that { X ( n ) } converges to X in Hausdorff distance. More precisely, using a Moe-bius transformation, we may assume X ⊂ { z ∈ C | ≤ | z | ≤ } . Let W , ..., W i , ... bea sequence of distinct connected components of X such that ∪ ∞ i =1 W i is dense in X . Since X is a circle type closed set, each disk connected component of X is in the sequence. Let X ( n ) = ∪ ni =1 W ( n ) i where W ( n ) i = W i if W i is a disk and W ( n ) i = { z ∈ C | d E ( z, W i ) ≤ l n } if W i consists of a single point. We make l n small such that l n decreases to and W ( n ) i ∩ W ( n ) j = ∅ for i (cid:54) = j . This shows that X ( n ) is a circle type closed set having finitely many connected com-ponents and X ( n ) converges to X in the Hausdorff distance in C .For each X ( n ) , by Schlenker’s work [31], we construct a circle type closed set Y ( n ) ⊂ S such that there exists an isometry φ n : ( ˆ C − X ( n ) , d n ) → ∂C K ( Y ( n ) ) where d n is the Poincar´emetric on U n . By composing with Moebius transformations, we may assume that (0 , , ∈ ∂C K ( Y ( n ) ) and φ n (0) = (0 , , ∈ ∂C K ( Y ( n ) ) . By taking a subsequence if necessary, wemay assume that Y ( n ) converges in Hausdorff distance to a closed set Y ⊂ S . We will show:(1) The sequence { φ n } contains a subsequence converging uniformly on compact subsetsto a continuous map φ : U → ∂C K ( Y ) . This is achieved by showing { φ n : ( U n , d S ) → ( ∂C K ( Y ( n ) ) , d E ) } is an equicontinuous family. The later is proved in § § φ : ( U, d U ) → ∂C K ( Y ) is an isometry. This follows from Alexandrov’sconvergence Theorem 4.1 and convergence of Poincar´e metrics (Theorem 4.2) in § Y is of circle type. Since the Hausdorff limit of a sequence of rounddisks is a round disk or a point, we will prove in § Y is the Hausdorfflimit of a sequence of components of Y ( n ) ’s. This is proved using the results obtained in step(1).2.4. Outline of the proof to Theorem 1.2 (b).
Part (b) of Theorem 1.2 states that for anycircle type closed set Y ⊂ S with | Y | ≥ , there exists a circle domain U = ˆ C − X withPoincar´e metric d U such that Σ := ∂C H ( Y ) is isometric to ( U, d U ) . The strategy of the proofis the same as that for part (a) of Theorem 1.2. The only technical complication is due to theestimate of modules of rings in non-smooth convex surfaces.By Theorem 2.1, we may assume that the set Y is not contained in any circle, i.e., C H ( Y ) is3-dimensional. By composing with a Moebius transformation, we may assume that (0 , , ∈ Σ . Since Y is a circle type closed set, there exists a sequence { Y n } of components of Y suchthat ∪ ∞ i =1 Y i is dense in Y and (0 , , ∈ C H ( Y ∪ ... ∪ Y ) . Denote Y ( n ) = ∪ ni =1 Y i and Σ n = ∂C H ( Y ( n ) ) . By construction (0 , , ∈ ∂C H ( Y ( n ) ) for n ≥ and the sequence { Y ( n ) } converges in Hausdorff distance to Y in S . IRCLE DOMAIN 7
Now each Σ n is a genus zero Riemann surface of finite type. By Koebe’s theorem fordomains of finite topology, there exists a circle domain U n = ˆ C − X ( n ) and a conformaldiffeomorphism φ n : Σ n → U n . Using Moebius transformations, we normalize U n such that ∈ U n , φ n (0 , ,
0) = 0 and the open unit disk D is contained in U n such that ∂ D ∩ ∂U n (cid:54) = ∅ . Bytaking a subsequence if necessary, we may assume that X ( n ) converges in Hausdorff distanceto a compact set X in the spherical metric d S . We will show:(1) The sequence { φ n } contains a subsequence converging uniformly on compact subsetsto a continuous map φ : ∂C H ( Y ) → U := ˆ C − X . This is achieved by showing { φ n :( ∂C H ( Y ( n ) ) , d E ) → ( U n , d S ) } is equicontinuous. The later is proved in § φ : ∂C H ( Y ) → ( U, d U ) is an isometry. This is a consequence of Alexan-drov’s convergence theorem and convergence of Poincar´e metrics theorem in § X is of circle type. We will prove in § X isthe Hausdorff limit of a sequence of components of X ( n ) ’s. This is proved using the resultsobtained in step (1).3. H YPERBOLIC GEOMETRY AND CONVEX SURFACES
We prove several results on hyperbolic geometry to be used in the paper. One of them is,
Theorem 3.1.
Suppose X is a convex set in the Poincar´e model H P of the hyperbolic 3-space.Then the Euclidean area of the convex surface ∂X is at most π . We believe the constant π can be improved to π which is optimal. It is easy to see that π cannot be improved.As a convention, for a Riemannian metric d on M , we also use d to denote the distance on M induced by the Riemannian metric d .3.1. The Klein and Poincar´e models of hyperbolic 3-space.
Let B = { ( x , x , x ) ∈ R | (cid:112) x + x + x < } be the open unit ball in the 3-space. The Poincar´e model ( H , d P ) orsimply H P of the hyperbolic space is B equipped with the hyperbolic metric d P x = 4 (cid:80) i =1 dx i (1 − | x | ) . It is a complete metric of constant sectional curvature − and is conformal to the Euclideanmetric (cid:80) i =1 dx i . By definition, d P x ≥ (cid:80) i =1 dx i . Let d P ( x, y ) and d E ( x, y ) = | x − y | bethe distances associated to the Poincar´e and the Euclidean metrics respectively. Then(1) d P ( x, y ) ≥ | x − y | . The Klein model of the hyperbolic 3-space ( H , d K ) or simply H K is the unit ball B equipped with the Riemannian metric d K x = (cid:80) i =1 dx i − | x | + ( (cid:80) i =1 x i dx i ) (1 − | x | ) . Since d K x ≥ d E x , we have(2) d K ( x, y ) ≥ | x − y | . The geodesics and totally geodesic planes in the Klein model are the intersections of Eu-clidean lines and Euclidean planes with B . Therefore, convex surfaces and convex sets FENG LUO AND TIANQI WU in ( H K , d K ) are the same (as point sets) as those in B in Euclidean geometry. The map Ψ( x ) = x | x | : B → B is an isometry from the Poincar´e model H P onto the Klein model H K . See [26]. Furthermore, we have Proposition 3.2.
For all x, y ∈ R , | Ψ( x ) − Ψ( y ) | ≤ | x − y | . Proof.
We have | Ψ( x ) − Ψ( y ) | = 2(1 + | x | )(1 + | y | ) | (1 + | y | ) x − (1 + | x | ) y |≤ | x − y | (1 + | x | )(1 + | y | ) + 2(1 + | x | )(1 + | y | ) | x | y | − y | x | || . Now using | x | y | − y | x | | = | x | | y | | x − y | , we see that | Ψ( x ) − Ψ( y ) | ≤ | x − y | (1 + | x | )(1 + | y | ) + 2 | x − y || x || y | (1 + | x | )(1 + | y | )= 2 | x − y | (1 + | x || y | )(1 + | x | )(1 + | y | ) ≤ | x − y | (1 + | x || y | )1 + 2 | x || y |≤ | x − y | . (cid:3) Euclidean area of hyperbolic convex surfaces.
We begin with the following,
Lemma 3.3.
Suppose f ( z ) is the inversion about a circle C and B is a circle which intersects C at π angle. Then | f (cid:48) ( z ) | ≤ for all z ∈ B .Proof. By composing with a rotation and a translation, we may assume that B is a circlecentered at the origin and C is a circle of radius r centered at p . See Figure 1. Let s bethe point in B closest to the center p . By the assumption, the angle ∠ qp is π and the angle θ = ∠ sqp is in the interval [ π , π ] . The inversion about the circle C is given by f ( z ) = r z − p + p .Hence | f (cid:48) ( z ) | = r | z − p | . If z ∈ B , then | f (cid:48) ( z ) | ≤ ( r | s − p | ) = ( sin( φ )sin( θ ) ) ≤ ( θ ) ≤ where wehave used the Sine law for the triangle ∆ pqs , φ = ∠ qsp and θ ∈ [ π , π ] . pq s B C . . . r p /4 F IGURE
1. Lipschitz property of inversion (cid:3)
IRCLE DOMAIN 9
Recall that A denotes the closure of a set A in the Euclidean space R n . Given a non-emptyclosed convex set X in the n-dimensional hyperbolic space H nP , considered as the Poincar´e ballmodel, the nearest point projection π : H n → X which sends H n onto X is defined as follows(see [11]). If p ∈ H n , then π ( p ) ∈ X is the point in X closest to p . If p ∈ ∂ B n , let L be thelargest horoball centered at p such that int ( L ) ∩ X = ∅ . Then π ( p ) = L ∩ X . The convexity of X shows that L ∩ X consists of one point. It is well known that π is continuous. Furthermore, Lemma 3.4.
Suppose π : H nP → X is the nearest point projection.(a) If p / ∈ X , then the hyperbolic codimension-1 plane W p through π ( p ) perpendicular tothe geodesic from p to π ( p ) is a supporting plane for the convex set X .(b) If X is a complete geodesic in the Poincar´e model of the hyperbolic plane H P and A is aconnected component of ∂ H P − ∂X , then the nearest point projection π | A coincides with theinversion about the circle C which contains ∂X and bisects the angles formed by X and A .Proof. The proof of part (a) can be found in [11]. To see part (b), we use the upper-half-planemodel { z ∈ C | im ( z ) > } of H and take X = { iy | y ∈ R > } and A = { x | x ∈ R > } . Thenearest point projection onto X when restricted to A is π ( x ) = ix which is the same as thereflection about the line C = { xe πi/ | x ∈ R } on A . Clearly the line C bisects the angle formedby A and X . (cid:3) To prove Theorem 3.1, we may assume the hyperbolic convex set X is closed in H P . Let d E∂X be the induced path metric on the convex surface ∂X from the Euclidean metric d E . Bythe work of Alexandrov [1], the metric space ( S, d
E∂X ) is a surface of bounded curvature (in thesense of Alexandrov. See § ( ∂X, d E∂X ) and can be estimated using the path metric. This shows that Theorem 3.1 is a consequence ofthe following property of the nearest point projection π . Theorem 3.5.
Let X be a closed convex set in the Poincar´e model H P of the hyperbolic 3-space and π be the nearest point projection to X . Then for all p, q ∈ ∂ H = S , (3) d E∂X ( π ( p ) , π ( q )) ≤ d S ( p, q ) . We remark that the above theorem also holds for high dimensional hyperbolic spaces H n . Proof.
By the definition of induced path metrics on ∂X (see for instance [10]), it suffices toprove that for any p, q ∈ S ,(4) d E ( π ( p ) , π ( q )) ≤ d S ( p, q ) . Now if π ( p ) = π ( q ) , then the result holds trivially. If π ( p ) (cid:54) = π ( q ) , let Y be the completehyperbolic geodesic joining π ( p ) to π ( q ) and W p and W q be the codimension-1 hyperbolicplanes perpendicular to Y at π ( p ) and π ( q ) respectively. Note that W p ∩ W q = ∅ . By Lemma3.4, the convex set X lies in the region in H bounded by W p and W q . Let U p and U q be thedisjoint spherical caps in S bounded by ∂W p and ∂W q such that p ∈ U p and q ∈ U q .Consider the Euclidean plane P through the origin O and containing the geodesic Y . Then P is perpendicular to both W p and W q . Therefore, there exist p (cid:48) ∈ P ∩ ∂W p and q (cid:48) ∈ P ∩ W q such that(5) d S ( U p , U q ) = d S ( p (cid:48) , q (cid:48) ) ≤ d S ( p, q ) . We will show that d E ( π ( p ) , π ( q )) ≤ d S ( p (cid:48) , q (cid:48) ) which implies the inequality (4). Y . p q p (p) p ( q ) . .. W p W q P p ( q ) p (p) P . . p' q 'p'q 'Y Y . . F IGURE
2. Lipschitz property of rearest point projectionThe inequality d E ( π ( p ) , π ( q )) ≤ d S ( p (cid:48) , q (cid:48) ) can be proved by focusing on the plane P since d S ( p (cid:48) , q (cid:48) ) = d ∂P ( p (cid:48) , q (cid:48) ) where d ∂P is the induced path metric on ∂P from the Euclidean 3-space.In the disk model D = P ∩ H of the hyperbolic plane H . Take the geodesic Y to be theconvex set and let π Y be the nearest point project from D to Y . Then by definition π Y ( p (cid:48) ) = π ( p ) and π Y ( q (cid:48) ) = π ( q ) . Let M be the component of ∂D − ∂Y which contains p (cid:48) . Then due to(5), we see that q (cid:48) ∈ M . By Lemma 3.4(b), the map π Y | M is the same as the inversion f on thecircle C which contains ∂Y and bisects the angles formed by Y and M . The angle between C and ∂D could be either π/ or π/ . In the case of π/ , by Lemma 3.3, the inequality d E ( π ( p ) , π ( q )) = d E ( π Y ( p (cid:48) ) , π Y ( q (cid:48) )) ≤ d S ( p (cid:48) , q (cid:48) ) follows. In the case of π/ , M lies outsideof the circle C . So | f (cid:48) ( z ) | ≤ for any z ∈ M and d E ( π ( p ) , π ( q )) = d E ( π Y ( p (cid:48) ) , π Y ( q (cid:48) )) ≤ d S ( p (cid:48) , q (cid:48) ) (cid:3) Here is how to see that the best constant in Theorem 3.1 cannot be smaller than π . Take acircle domain U ⊂ S and consider the Euclidean area of ∂C H ( U ) . The surface ∂C H ( U ) is adisjoint union of totally geodesic planes. Now each totally geodesic plane P in H correspondsto a spherical cap P ∗ . The ratio of the Euclidean areas of P and P ∗ can be arbitrary close to as the Euclidean diameters of P and P ∗ tend to zero. This shows there is a sequence of circledomains U n such that the Euclidean area of ∂C H ( U n ) approach π .3.3. Estimates for the nearest point projection.
Let us recall the Hausdorff distance. Sup-pose A is a subset of a metric space ( Z, d ) and r > . The r -neighborhood of A , denoted by N r ( A, d ) , is the open set { z ∈ Z | d ( z, A ) < r } . If A, B are two closed subsets of ( Z, d ) , thentheir Hausdorff distance d h ( A, B ) is inf { r | A ⊂ N r ( B, d ) and B ⊂ N r ( A, d ) } . A sequenceof closed subsets { X n } in ( Z, d ) is said to converge in Hausdorff distance to a close set X if lim n d h ( X n , X ) = 0 . For a compact metric space ( Z, d ) , the set of all closed subsets in theHausdorff distance is compact. See [10]. Lemma 3.6.
For any (cid:15) > , there exists δ > such that for any closed (hyperbolic) convex sets D, D (cid:48) ⊂ H P with the Hausdorff distance induced from the Euclidean metric d Eh ( D, D (cid:48) ) < δ ,then d E ( π D ( x ) , π D (cid:48) ( x )) < (cid:15) for all x ∈ ∂ H . IRCLE DOMAIN 11 x n q n q n 'p n p n '(a) (b)0 q p x .. .. p (x) . J L F IGURE
3. Hausdorff distance estimate (a), nearest projection (b)
Proof.
Suppose otherwise there exist a constant (cid:15) > , two sequences of closed convexsets D n , D (cid:48) n ⊂ H and a sequence of points x n ∈ ∂ H such that the Hausdorff distance d Eh ( D n , D (cid:48) n ) ≤ n and | q n − q (cid:48) n | ≥ (cid:15) where q n ∈ D n and q (cid:48) n ∈ D (cid:48) n are the nearest points in D n and D (cid:48) n to x n . By taking subsequences if necessary, we may assume that x n → p ∈ ∂ H , q n → q and q (cid:48) n → q (cid:48) for q (cid:54) = q (cid:48) ∈ H .Since the Hausdorff distance between D n and D (cid:48) n is at most n and q n ∈ D n , q (cid:48) n ∈ D (cid:48) n ,we find p n ∈ D n (resp. p (cid:48) n ∈ D (cid:48) n ) such that | p n − q (cid:48) n | ≤ n (resp. | p (cid:48) n − q n | ≤ n ). Since q n is the nearest point in D n to x n , the hyperbolic inner angle θ n of ∠ x n q n p n is at least π/ .Similarly, θ (cid:48) n = ∠ x n q (cid:48) n p (cid:48) n ≥ π/ . See Figure 3(a). By the construction p n → q (cid:48) and p (cid:48) n → q in H . If q, q (cid:48) , x are pairwise distinct, taking the limit as n tends to infinity and using thecontinuity of the angles, we see the hyperbolic triangle ∆ xqq (cid:48) has inner angles , θ, θ (cid:48) where θ = lim n θ n ≥ π/ , θ (cid:48) = lim n θ (cid:48) n ≥ π/ . But this contradicts the Gauss-Bonnet theorem. If q or q (cid:48) is equal to x , say q (cid:48) = x and q (cid:54) = q (cid:48) , consider the angle ∠ q (cid:48) qx which is the limit of theangles ∠ p n q n x n and therefore is at least π/ . However, q (cid:54) = q (cid:48) = x , the limiting angle ∠ q (cid:48) qx is . This again is a contradiction. (cid:3) Lemma 3.7.
Assume X is a closed convex set in H P containing the origin O and a point q ∈ ∂ H . Let π be the nearest point projection from H P to X . Then for any x ∈ ∂ H , d S ( x, q ) ≤ d E ( π ( x ) , q ) ≤ d S ( x, q ) . Proof.
For simplicity we denote d S ( x, q ) = θ ∈ (0 , π ) . Assume L is the largest horoballcentered at x such that int ( L ) ∩ X = ∅ . Let r and p be the Euclidean radius and center ofthe horoball L . See Figure 3(b). Since the interior of L does not intersect the closed interval [ q, O ] ⊂ X , we have r ≤ / and if θ < π/ , then the Euclidean distance from p to the linethrough q, O is at least r , i.e.,(6) sin θ ≥ r − r . Since π ( x ) ∈ ∂X , we have the following bound on d E ( q, π ( x )) = | q − π ( x ) | . d E ( p, q ) − r = d E ( q, p ) − d E ( p, π ( x )) ≤ d E ( q, π ( x )) ≤ d E ( q, p ) + d E ( p, π ( x )) = d E ( q, p ) + r. By the Cosine Law on the triangle (cid:52) qOpd E ( p, q ) = (cid:112) − r ) − − r ) cos θ. Denote f ( t ) = (cid:112) − t ) − − t ) cos θ for t ∈ [0 , . Then we have | f ( t ) | ≥ | cos θ − (1 − t ) | and f (cid:48) ( t ) = cos θ − (1 − t ) f ( t ) ∈ [ − , . This shows that f ( t ) + t is an increasing function of t and f ( t ) − t is a decreasing function of t . Note that d E ( p, q ) = f ( r ) .Now if θ ≥ π/ , using r ≤ / , we have d E ( p, q ) − r = f ( r ) − r ≥ f (1 / − / (cid:112) / − cos θ − / ≥ (cid:112) / − / ≥ θ ,d E ( p, q ) + r = f ( r ) + r ≤ f (1 /
2) + 1 / (cid:112) / − cos θ + 1 / ≤ (cid:112) / / ≤ θ. If θ < π/ , by (6), we have r ≤ sin θ/ (1 + sin θ ) and d E ( p, q ) − r = f ( r ) − r ≥ f ( sin θ (1 + sin θ ) ) − sin θ (1 + sin θ )= (cid:112) (1 + sin θ ) + 1 − θ ) cos θ − sin θ θ ≥
12 ( (cid:113) (1 + sin θ − cos θ ) + sin θ − sin θ ) ≥
12 ( (cid:112) sin θ + sin θ − sin θ )= √ −
12 sin θ ≥ θ . Similarly, d E ( p, q ) + r ≤ (cid:112) θ ) − θ ) cos θ + sin θ θ ≤ ( (cid:113) (1 + sin θ − cos θ ) + sin θ + sin θ ) ≤ ( (cid:113) (2 sin θ ) + sin θ + sin θ )=( √ θ ≤ θ. (cid:3)
4. H
AUSDORFF CONVERGENCE AND THE P OINCAR ´ E METRICS
We begin with a recall of Alexandrov convergence which is used extensively on convergenceof convex surfaces in the hyperbolic and Euclidean spaces. A sequence { X n } of closed subsetsin a metric space ( Z, d ) is Alexandrov convergent to a closed subset X if (i) for any p ∈ X ,there exists a sequence { p n } with p n ∈ X n such that lim n p n = p and (ii) for any convergentsequence { p n i } with p n i ∈ X n i and lim i p n i = p , p ∈ X . Alexandrov convergence is indepen-dent of the choice of the distance d . Clearly if { X n } converges to X in the Hausdorff distance, IRCLE DOMAIN 13 then { X n } Alexandrov converges to X . In general, the converse is not true. But if the space ( Z, d ) is a compact, then Alexandrov convergence is equivalent to the Hausdorff convergence.In particular, for a compact space ( Z, d ) , the Hausdorff convergence of compact subsets isindependent of the choice of metrics. See [10] and [1] for details.4.1. Alexandrov’s work on convex surfaces.
By a complete convex surface S in E or H we mean the boundary of a closed convex set of dimension at least . The convergence theoremof Alexandrov which will be used extensively is, Theorem 4.1 (Alexandrov) . Suppose { S n } is a sequence of complete connected convex sur-faces in the Euclidean or hyperbolic 3-space Alexandrov converging to a complete connectedconvex surface S . If x, y ∈ S and x n , y n ∈ S n such that lim n x n = x and lim n y n = y , then lim n d S n ( x n , y n ) = d S ( x, y ) , where d Σ is the induced path metric on the convex surface Σ . The proof of this theorem for the Euclidean case is in Chapter 3 of [1]. The hyperbolic casewas stated in section 3 of Chapter 12 of [1].4.2.
The Poincar´e metrics and their convergence. If X is a closed set in the Riemann sphere ˆ C which contains at least three points, then none of the connected component of ˆ C − X isconformal to the complex plane C or the punctured plane C − { } . Therefore, by the uni-formization theorem, each connected component of ˆ C − X carries the Poincar´e metric. The Poincar´e metric on ˆ C − X is defined to be the Riemannian metric whose restriction to eachconnected component is the Poincar´e metric.The main result in this section is the following. Theorem 4.2.
Suppose { X n } is a sequence of compact sets in the Riemann sphere convergingto a compact set X in the Hausdorff distance such that X contains at least three points and X (cid:54) = ˆ C . Let d n = a n ( z ) | dz | and d U = a ( z ) | dz | be the Poincar´e metrics on U n = ˆ C − X n and U = ˆ C − X respectively. Then a n ( z ) converges uniformly on compact subsets of U to a ( z ) .Furthermore, if p, q are two points in a connected component of U and p n , q n are two points ina connected component of U n such that lim n p n = p and lim n q n = q , then (7) lim n d n ( p n , q n ) = d U ( p, q ) . Note that by Hausdorff convergence, for any compact set K ⊂ U , K ⊂ U n for n large.Therefore a n is defined on K for large n . Proof.
It suffices to prove the theorem for the case that all X n contain a fixed set of threepoints. Indeed, let { u, v, w } ⊂ X with u, v, w pairwise distinct and consider sequences u n , v n , w n ∈ X n such that lim n u n = u, lim n v n = v and lim n w n = w . There exists aMoebius transformation ψ n sending u n , v n , w n to u, v, w respectively. By the construction, ψ n converges uniformly to the identity map in the spherical metric d S on ˆ C . This implies that ψ n ( X n ) converges in Hausdorff distance to X and { u, v, w } ⊂ ψ n ( X n ) . Now suppose thetheorem has been proved for the sequence ψ n ( X n ) . Using the fact that ψ n induces an isometrybetween Poincar´e metrics on the open sets ˆ C − X n and ˆ C − ψ n ( X n ) , we see Theorem 4.2 holdsfor the general case.Now using a Moebius transformation, we may assume that X n contains { , , ∞} and con-verges to X in Hausdorff metric. The strategy of the proof is as follows. First we show that for any compact set K in U ,the family of functions { a n | K } contains a uniformly convergent subsequence. Next we showthat the limit b ( z ) of any convergent subsequence of { a n } (in the topology of uniform conver-gence in compact sets) produces a hyperbolic metric b ( z ) | dz | on U . Finally, we show that thehyperbolic metric b ( z ) | dz | is complete. Therefore, by the uniqueness of the Poincar´e metric, b ( z ) | dz | is the Poincar´e metric d U = a ( z ) | dz | . Since all limits of convergent subsequences arethe same, it follows that the sequence { a n ( z ) } converges to a ( z ) .To show that { a n | K } contains a convergence subsequence in the l ∞ -norm, express the openset U as a union of open Euclidean round disks B j such that the Euclidean closure B j is stillin U . Then each compact set K in U is contained in a finite union of these closed disks B j . Byusing Cantor’s diagonal argument, it suffices to prove the statement for K to be a compact ball { z || z − p | ≤ r } in U . We will use the following well known consequence of the Schwarz-Picklemma whose proof is omitted. Lemma 4.3. (Schwarz-Pick) Suppose A and B are two open sets in ˆ C − { q , q , q } suchthat A ⊂ B and d A = a A ( z ) | dz | and d B = b B ( z ) | dz | are the Poincar´e metrics on A and B respectively. Then d A ≥ d B , i.e., a A ( z ) ≥ b B ( z ) . Let d , be the Poincare metric on C − { , } . Since U n ⊂ C − { , } , by Lemma 4.3,(8) d n ≥ d , . Lemma 4.4. If K = { z || z − p | ≤ r } is in U n for all n large, then sequence { a n | K } contains aconvergent subsequence in the l ∞ -norm.Proof. Let D be an open disk centered at p containing K such that D ⊂ U n for n large.Consider the incomplete simply connected hyperbolic surface ( D, d n | D ) . There exists an ori-entation preserving isometric immersion f n : ( D, d n | D ) → ( D , d D ) such that f n ( p ) = 0 . Inparticular, f n : D → D are holomorphic maps bounded by 1. Therefore { f n } forms a normalfamily and contains a subsequence which converges uniformly on compact sets to an analyticfunction h in D . For simplicity, we assume that the subsequence is { f n } . We claim that | f (cid:48) n ( p ) | is bounded away from . Indeed, consider the standard tangent vector v = ∂∂x at p . By (8), thelength of v in d n is at least the length δ of v in d , . It follows that the length of f (cid:48) n ( p ) in thePoincare metric on D is at least δ . This shows that h (cid:48) ( p ) (cid:54) = 0 and therefore h is not a constant.Furthermore, the same argument shows h (cid:48) ( z ) (cid:54) = 0 for all z ∈ D . Since the Poincar´e metricon D is | dw | −| w | , it follows that a n ( z ) = 2 | f (cid:48) n ( z ) | −| f n ( z ) | . By the uniform convergence of the analyticfunctions f n , we conclude that a n converges uniformly to b ( z ) = 2 | h (cid:48) ( z ) | −| h ( z ) | on K . Since h isanalytic and h (cid:48) ( z ) (cid:54) = 0 , hence b ( z ) | dz | is a hyperbolic metric. (cid:3) Lemma 4.5.
Let d ∞ be the limiting conformal hyperbolic metric on U from a convergentsubsequence of d n . Then d ∞ is a complete metric on each component of U . Assume that Lemma 4.5 holds. By the uniqueness of the Poincar´e metric, we have d ∞ = d U .It follows that d n converges uniformly on compact sets to d U .To prove Lemma 4.5, take a connected component W of U and a Cauchy sequence x n in ( W, d ∞ | W ) . Since U ⊂ C − { , } , by (8), we have for all n, md ∞ ( x n , x m ) ≥ d , ( x n , x m ) . IRCLE DOMAIN 15
Therefore, { x n } is a Cauchy sequence in C − { , } in the d , metric. In particular, there is apoint p ∈ C − { , } so that lim n x n = p . We claim that p ∈ U .Assuming the claim and using the fact that the topology determined by d ∞ and the Euclideanmetric d E on U are the same, we see that p is in W and x n converges to p in the d ∞ metric.To see the claim, suppose otherwise that p ∈ X . Then U ⊂ C − { , , p } . By definition ofHausdorff convergence, there exists p n ∈ X n such that lim n p n = p and p n (cid:54) = 0 , , ∞ . Let d (cid:48) n be the Poincar´e metric on C − { , , p n } . By Lemma 4.3 and C − X n ⊂ C − { , , p n } , we have d n ≥ d (cid:48) n . In particular, for any indices i, j , d n ( x i , x j ) ≥ d (cid:48) n ( x i , x j ) . Let d , ,p be the Poincar´emetric on C − { , , p } . Then due to lim p n = p , the Poincar´e metrics d (cid:48) n converge uniformlyon compact subsets to d , ,p . As a consequence, we have lim n d (cid:48) n ( x i , x j ) = d , ,p ( x i , x j ) . Since lim n d n ( x i , x j ) = d ∞ ( x i , x j ) , we obtain d ∞ ( x i , x j ) ≥ d , ,p ( x i , x j ) . Hence { x n } is a Cauchy sequence in the d , ,p metric on C −{ , , p } . Since, d , ,p is a completemetric space, it follows that there is q (cid:54) = p in C − { , } such that lim n x n = q in C . Thiscontradicts the assumption that lim n x n = p in C .Now we prove (7) by showing d U ( p, q ) ≥ lim sup n d n ( p n , q n ) and d U ( p, q ) ≤ lim inf n d n ( p n , q n ) . Let γ be the shortest geodesic joining p to q in d U metric. By Hausdorff convergence, thereexists r > such that the r -neighborhood N r ( γ, d U ) is in U n for n large. Now join p n to p and q n to q in N r ( γ, d U ) by Euclidean line segments α n and β n . By uniform conver-gence of Poincar´e metrics (as Riemannian metrics) and lim n p n = p , lim n q n = q , we have lim n l d n ( α n ) = lim n l d n ( β n ) = 0 and lim n l d n ( γ ) = l d U ( γ ) . Here we have used l d ( c ) to denotethe length of a path c in the metric d . Therefore(9) d U ( p, q ) = l d U ( γ ) = lim n l d n ( γ ) + lim n l d n ( α n ) + lim n l d n ( β n )= lim n l d n ( α n γβ n ) ≥ lim sup n d n ( p n , q n ) . Here α n γβ n is the product of paths. Next to see d U ( p, q ) ≤ lim inf n d n ( p n , q n ) , by (9), wecan choose R > large such that d n ( p n , q n ) ≤ R/ and p n , q n ∈ B R (0 , d U ) . Consider thecompact set W = B R (0 , d U ) = { x ∈ U | d U ( x, ≤ R } . Note that p ∈ B R (0 , d U ) and d U ( p, ∂W ) ≥ R . Since W is compact, W ⊂ U n for n large. We claim that d n ( p n , ∂W ) ≥ R and d n ( q n , ∂W ) ≥ R for n large. If otherwise, after taking a subsequence if necessary, say l n = d n ( p n , ∂W ) < R for n large. Let δ n : [0 , l n ] → ( U n , d n ) be a shortest geodesic from p n to ∂W parameterized by the arc length. Since d n converges uniformly on compact sets(as Riemannian metrics) and l n is bounded, by Arzela-Ascoli theorem, we find a subsequence δ n i converging uniformly on an interval [0 , l ] to a Lipschitz map δ : [0 , l ] → ( U, d U ) where δ (0) = p = lim n p n and δ ( l ) ∈ ∂W . By the continuity of distance function and the Fatou’slemma lim n d U ( p n , ∂W ) = d U ( p, ∂W ) ≤ l d U ( δ ) ≤ lim inf l d ni ( δ n i ) ≤ R. This contradicts d U ( p, ∂W ) ≥ R . This shows, using d n ( p n , q n ) ≤ R/ , that the shortestgeodesic γ n in ( U n , d n ) joining p n to q n is in the compact set W . By Arzela-Ascoli theorem, wefind a subsequence γ n i converging uniformly to a Lipschitz path γ ∗ joining p to q . Therefore, by Fatou’s lemma again,(10) d U ( p, q ) ≤ l d U ( γ ∗ ) ≤ lim inf i l d ni ( γ n i ) . Since replacing { γ n } by any subsequence we still have (10), we conclude d Ω ( p, q ) ≤ lim inf n l d n ( p n , q n ) . (cid:3) A generalized form of Arzela-Ascoli theorem.
For our application, we need a slightlymore general form of Arzela-Ascoli theorem. Recall that a family of maps f n : ( X n , d n ) → ( Y n , d (cid:48) n ) between metrics spaces is called equicontinuous if for any (cid:15) > , there exists δ > such that if x n , y n ∈ X n with d n ( x n , y n ) < δ , then d (cid:48) n ( f n ( x n ) , f n ( y n )) < (cid:15) . Theorem 4.6.
Suppose ( Z, d ) and ( Y, d (cid:48) ) are compact metric spaces and { f n : X n → Y } is an equicontinuous family where X n ⊂ Z are compact. Let X ⊂ Z be a compact subsetsuch that for any x ∈ X there exists a sequence x n ∈ X n converging to x . Then there existsa subsequence { f n i } , converging uniformly to some continuous function f : X → Y , i.e., forany (cid:15) > there exist δ > and N > such that for any i ≥ N , x n i ∈ X n i and x ∈ X with d ( x n i , x ) < δ , we have d (cid:48) ( f n i ( x n i ) , f ( x )) < (cid:15) .Proof. Since Z is compact, we can find a countable dense subset A ⊂ X . Then for any (cid:15) > there exists a finite subset A (cid:15) ⊂ A such that X ⊂ ∪ a ∈ A (cid:15) B (cid:15) ( a, d ) . For any a ∈ A , by the assumption, there are a n ∈ X n such that lim n a n = a . By the standarddiagonal method, we find a subsequence of { f n } which, for simplicity, we may assume is { f n } itself, such that f n ( a n ) converge for all a ∈ A . Define f ( a ) = lim n →∞ f n ( a n ) . We first claim that f : A → Y is equicontinuous. Indeed, for any (cid:15) > there exists δ > such that if x, y ∈ X n with d ( x, y ) < δ , then d (cid:48) ( f n ( x ) , f n ( y )) < (cid:15) . Now if a, a (cid:48) ∈ A with d ( a, a (cid:48) ) < δ/ , then d ( a n , a (cid:48) n ) < δ for n sufficiently large, and d (cid:48) ( f ( a ) , f ( a (cid:48) )) = lim n →∞ d (cid:48) ( f n ( a n ) , f n ( a (cid:48) n )) ≤ (cid:15). Since f is uniformly continuous on A , we can extend f to a uniformly continuous function,still denoted by f , to X . Now to see the uniform convergence of f n to f , take any (cid:15) > . Thereexists δ > such that(1) for any n , x, y ∈ X n with d ( x, y ) ≤ δ , d (cid:48) ( f n ( x ) , f n ( y )) < (cid:15)/ ; and(2) for any x, y ∈ X with d ( x, y ) ≤ δ , d (cid:48) ( f ( x ) , f ( y )) < (cid:15)/ .Find N = N ( (cid:15) ) such that for any n ≥ N and any a ∈ A δ/ , d ( a n , a ) < δ/ and d (cid:48) ( f n ( a n ) , f ( a )) < (cid:15)/ . This is possible since A δ/ is a finite set. Then for any x ∈ X ,find an a ∈ A δ/ such that d ( x, a ) < δ/ . If n ≥ N and x n ∈ X n with d ( x, x n ) < δ/ , find a n such that d ( a n , a ) < δ/ . Then d ( x n , a n ) ≤ d ( x n , x ) + d ( x, a ) + d ( a, a n ) ≤ δ/ δ/ δ/ δ IRCLE DOMAIN 17 and d (cid:48) ( f n ( x n ) , f ( x )) ≤ d (cid:48) ( f n ( x n ) , f n ( a n )) + d (cid:48) ( f n ( a n ) , f ( a )) + d (cid:48) ( f ( a ) , f ( x )) ≤ (cid:15)/ (cid:15)/ (cid:15)/ (cid:15). (cid:3)
5. P
ROOF OF PART ( A ) OF T HEOREM
ASSUMING EQUICONTINUITY
In this section, we prove Theorem 1.2(a) assuming Theorem 5.1. Suppose U = ˆ C − X is acircle domain such that | X | ≥ and d U is the Poincar´e metric on U . We will find a circle typeclosed set Y ⊂ S such that ( U, d U ) is isometric to ∂C K ( Y ) ⊂ ( H , d K ) .Using a Moebius transformation, we may assume that X ⊂ { z ∈ C | < | z | < } . Let W , ..., W i , ... be a sequence of distinct connected components of X such that ∪ ∞ i =1 W i is densein X . Each disk component of X is in the sequence. Let X ( n ) = ∪ ni =1 W ( n ) i where W ( n ) i = W i if W i is a disk and W ( n ) i = { z ∈ C | d E ( z, W i ) ≤ l n } if W i consists of a single point. We make l n small such that l n decreases to and W ( n ) i ∩ W ( n ) j = ∅ for i (cid:54) = j and W ( n ) i ⊂ { z ∈ C | < | z | < } . This shows that X ( n ) is a circle type closed set having finitely many connectedcomponents and X ( n ) converges to X in the Hausdorff distance in ˆ C . Let d n be the Poincar´emetric on U n = ˆ C − X ( n ) . By Schlenker’s work [31], there exists a circle type closed set Y ( n ) ⊂ S with (0 , , ∈ ∂C K ( Y ( n ) ) and an isometry φ n : ( U n , d n ) → ∂C K ( Y ( n ) ) such that φ n (0) = (0 , , . By taking a subsequence if necessary, we may assume that { Y ( n ) } convergesin Hausdorff distance to a compact set Y .The main technical theorem which will be proved in § Theorem 5.1.
The sequence { φ n : ( U n , d S ) → ( ∂C K ( Y ( n ) ) , d E ) } is equicontinuous. In the rest of the section, we will prove part (a) of Theorem 1.2 assuming Theorem 5.1.5.1.
The Hausdorff limit set Y of the sequence { Y ( n ) } .Lemma 5.2. (a) There exists a positive lower bound on the injectivity radii of ( U n , d n ) at .(b) The closed set Y contains at least three points and Y (cid:54) = S .(c) There exists a positive lower bound on the Euclidean diameters of φ n ( B / (0 , d E )) .Proof. To see part (a), take three points p , p , p , which are in X ( n ) for all n large, and let d W be the Poincar´e metric on W = ˆ C − { p , p , p } . Since U n ⊂ W , the Schwarz-Pick’s lemmashows that d n ≥ d W and B r (0 , d n ) ⊂ B r (0 , d W ) . Let r be an injectivity radius of d W at ,i.e., B r (0 , d W ) is isometric to the radius r ball in the hyperbolic plane H . Then r is also aninjectivity radius for d n at . Indeed, if otherwise, there exists a homotopically non-trivial loop δ at whose length in d n is less than r . Then d n ≥ d W implies the length of δ in d W is lessthan r . This contradicts the choice of r .To see part (b), since Y ( n ) converges to Y and (0 , , ∈ ∂C K ( Y ( n ) ) , we see that (0 , , ∈ ∂C K ( Y ) and hence Y contains at least two points and Y (cid:54) = S . Now if Y contains only twopoints, say Y = { p , p } , then there exists a sequence of positive numbers r n → such that Y ( n ) ⊂ B r n ( p ) ∪ B r n ( p ) in S where B r ( p ) is the ball of radius r centered at p in the sphericalmetric. Therefore, there exists a sequence of homotopically non-trivial loops γ n ⊂ ∂C K ( Y ( n ) ) through (0 , , whose hyperbolic lengths tend to . Using the isometry φ n , we see that thehomotopically non-trivial loops γ (cid:48) n = φ − n ( γ n ) in ( U n , d n ) pass through such that their lengths in the Poincar´e metrics d n tends to zero, i.e., l d n ( γ (cid:48) n ) → . But this contradicts the part (a) thatthe injectivity radii of d n at are bounded away from .To see part (c), choose a small radius r > such that B r (0 , d W ) ⊂ B / (0 , d E ) . Then B r (0 , d n ) ⊂ B r (0 , d W ) ⊂ B / (0 , d E ) . Let Z n be the surface φ n ( B r (0 , d n )) which is con-tained in φ n ( B / (0 , d E )) . Part (c) follows by showing that lim inf n diam d E ( Z n ) > . We will prove a stronger result that the Euclidean area of Z n is bounded away from . Indeed,since Z n lies in a (Euclidean) convex surface ∂C K ( Y ( n ) ) , one sees that πdiam d E ( Z n ) ≥ Area d E ( Z n ) . Here we have used the fact that if a closed convex surface A contains a closedconvex surface B in its interior, then area d E ( A ) ≥ area d E ( B ) . Now by the construction, Z n in ∂C K ( Y ( n ) ) with the induced path metric from d K is isometric to the standard hyperbolicdisk of radius r in H . In particular, the hyperbolic area of Z n is π sinh ( r / . On the otherhand, since φ n is an isometry, we see that Z n is contained in the compact subset B r (0 , d K ) of H K . On the compact set B r (0 , d K ) , there exists a constant C > such that the hyperbolicmetric d K ≤ Cd E . It follows that area d E ( Z n ) ≥ C − area d K ( Z n ) = 4 C − π sinh ( r / andthe result follows. (cid:3) Extension of φ n to the Riemann sphere. Let Σ n = ∂C K ( Y ( n ) ) . By Theorem 5.1, eachmap φ n is uniformly continuous and hence can be extended continuously to a map, still denotedby φ n , from ( U n , d S ) to (Σ n , d E ) . Here ¯ Z is the closure of a set Z in R or ˆ C . Furthermore,the family of the extended maps { φ n : ( U n , d S ) → (Σ n , d E ) } is again equicontinuous.Our next goal is to extend φ n continuously to a map from ˆ C = U n ∪ X ( n ) to Σ n ∪ Y ( n ) such that the extended family is still equicontinuous with respect to the spherical metric on ˆ C and the Euclidean metric. In the spherical metric d S z = | dz | | z | on the Riemann sphere ˆ C , allEuclidean disks and half spaces are spherical closed balls. We extend each homeomorphism φ n to a continuous map from ( ˆ C , d S ) to Σ n ∪ Y ( n ) by coning from the centers of disks. Moreprecisely, let D r = { z ∈ C || z | ≤ r } and S r = ∂D r be the disk of radius r and its boundary in C . Given any homeomorphism f : S r → S R , its Euclidean central extension F : D r → D R isthe homeomorphism defined by the formula(11) F ( z ) = | z | r f ( rz | z | ) . For a round disk W = B r ( p, d S ) in the 2-sphere S ⊂ R of radius r ≤ π/ , let ˆ W be the Eu-clidean disk ˆ W ⊂ R such that ∂ ˆ W = ∂W . The projection ρ : ˆ W → W from − p sends point x ∈ ˆ W to the intersection of the ray from − p to x with S . It is a bi-Lipschitz homeomorphismwhose the bi- Lipschitz constant is at most π . We extend a homeomorphism f from the bound-ary of a spherical disk to the boundary of a spherical disk by the formula ˆ f = ρ ◦ F ◦ ρ − where F is the central extension to the Euclidean disk and ρ and ρ are the bi-Lipschitz homeomor-phisms produced above. For simplicity, we still call ˆ f the central extension of f with respectto the spherical metrics. Take a disk component Z of X ( n ) . Then φ n ( ∂Z ) is the boundary ofa disk component Z (cid:48) of Y ( n ) . Both Z and Z (cid:48) have spherical radii at most π/ by the normal-ization condition that X ( n ) ⊂ { z ∈ C | < | z | < } and (0 , , ∈ ∂C K ( Y ( n ) ) . Extending φ n to Z by the spherical central extension produces a homeomorphism, still denote it by φ n ,which is now defined on ˆ C with image in Σ n ∪ Y ( n ) . Proposition 5.7 below shows that the IRCLE DOMAIN 19 family of extended continuous maps { φ n : ( ˆ C , d S ) → (Σ n ∪ Y ( n ) , d E ) } is equicontinuous. It isproved in two steps. In the first step, we show that spherical central extensions of functions inan equicontinuous family of maps between circles form an equicontinuous family. Due to thebi-Lipschitz property of projections ρ ’s, it suffices to show that the Euclidean central exten-sions of members of an equicontinuous family of maps between circles is still equicontinuous.This is in Proposition 5.3. In the second step, we show that the extended maps { φ n } on ( ˆ C , d S ) is equicontinuous.Suppose g : ( X, d ) → ( Y, d (cid:48) ) is a map between two length metric spaces. Its modu-lus of continuity function is ω ( g, δ ) = sup { d (cid:48) ( g ( x ) , g ( x )) | d ( x , x ) ≤ δ } . By definition, d (cid:48) ( g ( x ) , g ( y )) ≤ ω ( g, d ( x, y )) and if X is path connected, ω ( g, N δ ) ≤ N ω ( g, δ ) for N ∈ Z > .If δ > δ , then ω ( g, δ ) ≥ ω ( g, δ ) . Also, if g : X → Y is a homeomorphism between twocompact spaces X and Y , then(12) ω ( g, diam d ( X )) = diam d (cid:48) ( Y ) . We consider ∂D r as the length metric space with induced path metric d ∗ from d E and thestandard Euclidean metrics on the balls D r and D R in the following proposition. Proposition 5.3.
Suppose F : D r → D R is the central extension of a continuous map f : ∂D r → ∂D R given by (11). Then ω ( F, δ ) ≤ ω ( f, πδ ) . Proof.
The proof is based on several lemmas.
Lemma 5.4.
Given r , r ≥ , | r e √− θ − r e √− θ | ≥ r | e √− θ − e √− θ | . Proof.
We divide the proof into two cases. In the first case cos( θ − θ ) ≤ . Then | r e √− θ − r e √− θ | = | r − r e √− θ − θ ) | ≥ Re ( r − r e √− θ − θ ) ) = r − r cos( θ − θ ) ≥ r ≥ r | e √− θ − e √− θ | . Here Re ( z ) is the real part of a complex number z . In the second case that cos( θ − θ ) ≥ . Then | sin ( θ − θ ) | ≥ | e √− θ − e √− θ | . Therefore, | r e √− θ − r e √− θ | = | r e √− θ − θ ) − r | ≥ | Im ( r e √− θ − θ ) − r ) | = r | sin( θ − θ ) | ≥ r | e √− θ − e √− θ | . Here Im ( z ) is the imaginary part of a complex number z . (cid:3) Lemma 5.5. If x ∈ (0 , is a real number, then xω ( g, δ ) ≤ ω ( g, xδ ) . Proof.
It follows from the definition and triangle inequality that if k ∈ Z > is a natural number,then(13) ω ( g, δk ) ≥ k ω ( g, δ ) . Therefore, for x ∈ (0 , , ω ( g, xδ ) ≥ ω ( g, δ [ x ]+1 ) ≥ x ]+1 ω ( g, δ ) ≥ /x ω ( g, δ ) = x ω ( g, δ ) . (cid:3) Now we prove Proposition 5.3. Assume that | r e √− θ − r e √− θ | ≤ δ . This implies | r − r | ≤ δ since | r − r | ≤ | r e √− θ − r e √− θ | . Also, by Lemma 5.4, r | e √− θ − e √− θ | ≤ δ . Then by Lemmas 5.5, 5.4, (13) and (12), we have | F ( r e √− θ ) − F ( r e √− θ ) |≤ | F ( r e √− θ ) − F ( r e √− θ ) | + | F ( r e √− θ ) − F ( r e √− θ ) | = r r | f ( re √− θ ) − f ( re √− θ ) | + | r − r | r | f ( re √− θ ) | ≤ r r ω ( f, d ∗ ( re √− θ , re √− θ )) + | r − r | r R ≤ r r ω ( f, d ∗ ( re √− θ , re √− θ )) + | r − r | r ω ( f, r ) ≤ ω ( f, d ∗ ( r e √− θ , r e √− θ )) + 2 ω ( f, | r − r | ) ≤ ω ( f, πδ ) + 2 ω ( f, δ ) ≤ ω ( f, πδ ) . (cid:3) As a consequence, we have
Corollary 5.6. (a) If { f n : S r n → S R n : n ∈ N } is an equicontinuous family of maps betweencircles, then the central extensions { F n : D r n → D R n | n ∈ Z > } is equicontinuous.(b) If { f n : ∂W n → ∂W ∗ n : n ∈ N } is an equicontinuous family of maps between boundariesof spherical balls W n and W ∗ n of radii at most π/ , then the spherical central extensions { F n : W n → W ∗ n } is equicontinuous. Proposition 5.7. (a) The family of extended homeomorphisms { φ n : ( ˆ C , d S ) → (Σ n ∪ Y ( n ) , d E ) } is equicontinuous.(b) The limit function φ of a convergent subsequence { φ n i } is a surjective map from ˆ C to ∂C K ( Y ) ∪ Y such that φ ( X ) = Y and φ | U is an isometry from ( U, d U ) to ∂C K ( Y ) . Inparticular, ∂C K ( Y ) is connected.Proof. To see (a), by the normalization conditions that (0 , , ∈ ∂C K ( Y ( n ) ) and X ( n ) ⊂{ z ∈ C : 1 < | z | < } , each component of Y ( n ) and X ( n ) has radius at most π/ in d S . ThusCorollary 5.6(b) applies. Take any (cid:15) > , by Theorem 5.1 and Corollary 5.6, there exists δ > such that if d S ( x, y ) ≤ δ and either (i) x, y ∈ ˆ C − X ( n ) or (ii) x, y are both in a connectedcomponent of X ( n ) , then | f n ( x ) − f n ( y ) | ≤ (cid:15) . It remains to prove the cases where the pair x, y with d S ( x, y ) ≤ δ satisfy that (1) one of them is in X ( n ) and the other is in ˆ C − X ( n ) , or (2) x, y are in different connected components of X ( n ) . Consider a shortest geodesic γ joining x to y in ( ˆ C , d S ) . In the first case (1), we may assume that x ∈ X ( n ) and y ∈ ˆ C − X ( n ) . Let z bean intersection point of γ ∩ ∂X ( n ) such that x and z are in the same component of X ( n ) . Then d S ( x, z ) ≤ δ and d S ( z, y ) ≤ δ . Therefore, | f n ( x ) − f n ( y ) | ≤ | f n ( x ) − f n ( z ) | + | f n ( z ) − f n ( y ) | ≤ (cid:15) . In the second case (2) that x, y are in different components of X ( n ) , then the geodesicsegment γ contains a point z / ∈ X ( n ) with d S ( x, z ) ≤ δ and d S ( z, y ) ≤ δ . Then by the casejust proved, | f n ( x ) − f n ( y ) | ≤ | f n ( x ) − f n ( z ) | + | f n ( z ) − f n ( y ) | ≤ (cid:15) . In all cases, we haveestablished the equicontinuity.To see part (b), by the generalized Arzela-Ascoli theorem (Theorem 4.6) and part (a), wecan find a subsequence which, for simplicity we assume is { φ n } , converges uniformly to a map φ : ˆ C → ∂C K ( Y ) ∪ Y . Then Proposition ?? implies that φ ( ˆ C ) = ∂C K ( Y ) ∪ Y and φ ( X ) = Y .For any x ∈ U , by Theorem 4.2 we have lim n d n (0 , x ) → d U (0 , x ) . Since φ n is an isometryfrom ( U n , d n ) to ∂C K ( Y ( n ) ) , d K ((0 , , , φ n ( x )) = d n (0 , x ) ≤ sup m d m (0 , x ) < ∞ , and φ ( x ) = lim n φ n ( x ) ∈ H K ∩ φ ( ˆ C ) = ∂C K ( Y ) , i.e., φ ( U ) ⊂ ∂C K ( Y ) . Using φ ( U ∪ X ) = ∂C K ( Y ) ∪ Y , φ ( X ) = Y and φ ( U ) ⊂ ∂C K ( Y ) , we see that φ ( U ) = ∂C K ( Y ) . Further, by IRCLE DOMAIN 21 the work of Alexandrov and convergence of Poincar´e metrics (Theorems 4.1 and 4.2), φ | U isan isometry from ( U, d U ) to ∂C K ( Y ) . (cid:3) Finish the proof of part (a) of Theorem 1.2.
Now take a connected component Y k of Y . To show it is a round disk or a point, we use Proposition 5.7 to find Y k = φ ( X k ) for someconnected component X k of X . Indeed since φ is onto, there exists a connected component,say X k of X , which is mapped by φ into Y k . Since φ | U is a homeomorphism from U to ∂C K ( Y ) and φ | U induces bijection on spaces of ends (see [27]), we have φ ( X k ) = Y k . Since X is a circle domain, there exists a sequence { X ( n ) k n } of components of X ( n ) converging in theHausdorff distance to X k . By the uniform convergence of φ n to φ and Lemma ?? , φ n ( X ( n ) k n ) converges in Hausdorff distance to φ ( X k ) = Y k . But each φ n ( X ( n ) k n ) is a component of Y ( n ) which is a round disk or a point. Therefore Y k is a round disk or a point.6. P ROOF OF PART ( B ) OF T HEOREM
ASSUMING EQUICONTINUITY
Recall Theorem 1.2 (b) states that for any circle type close set Y ⊂ S with | Y | ≥ ,there exists a circle domain U = ˆ C − X with Poincar´e metric d U such that the boundary ofthe hyperbolic convex hull Σ = ∂C H ( Y ) is isometric to ( U, d U ) . The basic strategy of theproof is the same as that in Theorem 1.2 (a). In this section, we prove the part (b) using anequicontinuity property which will be established in § ( H , d P ) of the hyperbolic 3-space in the rest of thesection unless mentioned otherwise. By Theorem 2.1, we may assume that the set Y is notcontained in any circle, i.e., C H ( Y ) is 3-dimensional. Composing with a Moebius transfor-mation of ∂ H , we may assume that (0 , , ∈ Σ = ∂C H ( Y ) . By Caratheodory’ theorem onconvex hull, there exist four components Y , .., Y of Y such that (0 , , ∈ C H ( ∪ i =1 Y i ) . Since Y is a circle type closed set, there exists a sequence { Y , ..., Y n , ... } of components of Y suchthat ∪ ∞ i =1 Y i is dense in Y . Note that the density implies each disk component of Y is in the se-quence. Define Y ( n ) = ∪ ni =1 Y i and Σ n = ∂C H ( Y ( n ) ) . By construction (0 , , ∈ ∂C H ( Y ( n ) ) for n ≥ and the sequence { Y ( n ) } converges in the Hausdorff distance to Y in H .Since Σ n is a genus zero Riemann surface of finite topological type, by Koebe’s circledomain theorem, there exist a circle domain U n = ˆ C − X ( n ) and a conformal diffeomor-phism φ n : Σ n → U n . Using Moebius transformations, we normalize U n such that ∈ U n , φ n (0 , ,
0) = 0 and the open unit disk D is a maximum disk contained in U n , i.e., X ( n ) ⊂ ˆ C − D and X ( n ) ∩ ∂ D (cid:54) = ∅ . By taking a subsequence if necessary, we may assume that { X ( n ) } con-verges in the Hausdorff distance to a compact set X in ˆ C such that X ⊂ ˆ C − D and X ∩ ∂ D (cid:54) = ∅ . Lemma 6.1.
The hyperbolic injectivity radii of the surfaces (Σ n , d P Σ n ) at (0,0,0) are boundedaway from zero.Proof. Since the isometry Ψ( x ) = x | x | from the Poincar´e metric d P to the Klein metric d K is bi-Lipschitz when restricted to any compact set in H , it suffices to show that the injectivityradii of the surface S n := ∂C K ( Y ( n ) ) at (0 , , in the metric d KS n are bounded away from zero.Here C K ( Y ( n ) ) is the convex hull of Y ( n ) in the Klein model. In particular C K ( Y ( n ) ) and S n are Euclidean convex body and Euclidean convex surfaces. By the assumption that Y is not ina circle, the convex set C K ( Y ) is 3-dimensional and contains a Euclidean ball. It follows thatthere exists a Euclidean ball B which is contained in C K ( Y ( n ) ) for all large n . This implies that the Euclidean injectivity radii of ∂C K ( Y ( n ) ) in d ES n at (0 , , are bounded away fromzero. Since d K ≥ d E and B r ( p, d K ) ⊂ B r ( p, d E ) , the injectivity radii of ( S n , d KS n ) are boundedaway from . (cid:3) Lemma 6.2.
The closed set X contains at least three points, i.e., | X | ≥ . Furthermore, forany r > , there exists r (cid:48) > such that the spherical ball B r (cid:48) (0 , d S ) are contained in B r (0 , d n ) in U n for all n where d n = a n ( z ) | dz | is the Poincar´e metric on U n Proof. If X contains at most 2 points, say X ⊂ { a, b } ⊂ { z ∈ ˆ C : 1 < | z | < } . Then for any (cid:15) > , X ( n ) ⊂ B (cid:15) ( a, d S ) ∪ B (cid:15) ( b, d S ) for sufficiently large n . We claim that(14) lim n a n (0) = 0 . Assume f is a M¨obius transformation on ˆ C such that f ( a ) = 0 , f ( b ) = ∞ , f (0) = 1 , and b n ( z ) | dz | is the Poincar´e metric on f ( U n ) = ˆ C − f ( X ( n ) ) . Then a n (0) = b n (1) | f (cid:48) (0) | . Wewill show that lim n b n (1) = 0 . Consider a sequence of rings A R n = { z | R − n < | z | < R n } with lim n R n = ∞ such that A R n ∩ f ( X ( n ) ) = ∅ . Let the Poincar´e metric on A R = { z ∈ C || R − < | z | < R } be c R ( z ) | dz | . Then by the Schwarz-Picks lemma c R n ( z ) ≥ b n ( z ) . It is well known([4], page 49) that c R (1) = π R .
It follows | b n (1) | ≤ π R n and hence lim n a n (0) = lim n b n (1) | f (cid:48) (0) | = 0 .On the other hand, since (Σ n , d P Σ ) and ( U n , d n ) are isometric, by Lemma 6.1, there ex-ists a hyperbolic disk of radius r > isometrically embedded in U n and centered at .Let f n be an orientation preserving isometry from the ball B r (0 , d P ) in the Poincar´e diskto B r (0 , d n ) in ( U n , d n ) . Since f n is an isometry, we have | dz | −| z | = a n ( f n ( z )) | f (cid:48) n ( z ) | . Thisshows | f (cid:48) n (0) | = 2 /a n (0) and by (14), | f (cid:48) n (0) | → ∞ . Recall Koebe’s quarter theorem saysthat if g : B r (0 , d E ) → C is an injective analytic map, then its image g ( B r (0 , d E )) containsthe Euclidean ball of radius | g (cid:48) (0) | r centered at g (0) . Applying it to the injective analytic maps f n defined on B r (0 , d P ) , we see that f n ( B r (0 , d P )) contains the Euclidean disk of radius 2centered at for n large. This contradicts the assumption that ∂ D (cid:54)⊂ U n and shows | X | ≥ .Finally to see the second part of the Lemma, by the Schwarz-Pick lemma applied to D ⊂ U n ,we have | dz | −| z | ≥ a n ( z ) | dz | and in particular a n (0) ≤ . Therefore, | f n (0) | ≥ . Then Koebequarter Theorem implies that f n ( B r (0 , d P )) contains B r (cid:48) (0 , d E ) for some r (cid:48) independent of n .Since d E and d S are bi-Lipschitz equivalent when restricted to D , the result follows. (cid:3) Now we prove Theorem 1.2(b). The key result used in the proof is the following equiconti-nuity theorem to be proved in § Theorem 6.3.
The sequence { φ n : ( ∂C H ( Y ( n ) ) , d E ) → ( U n , d S ) } is equicontinuous. Assuming the above theorem, each map φ n : Σ n → U n extends to a continuous map, stilldenoted by φ n : Σ n → U n between their closures in R and ˆ C . Furthermore, the extendedfamily { φ n : (Σ n , d E ) → ( U n , d S ) } is still equicontinuous. Now each boundary component of Σ n and U n is a round circle or a point. Use central extension to extend φ n to be a continuousmap, still denoted by φ n , from Σ n ∪ Y ( n ) to U n ∪ X ( n ) = ˆ C . By the normalization condi-tion that (0 , , ∈ ∂C H ( Y ( n ) ) and X ( n ) ⊂ ˆ C − D , each component of Y ( n ) and X ( n ) has IRCLE DOMAIN 23 spherical radius at most π/ for n > . Using Corollary 5.6 and Proposition 5.7, the extendedfamily { φ n : (Σ n ∪ Y ( n ) , d E ) → ( ˆ C , d S ) } is equicontinuous. By the generalized Arzela-AscoliTheorem 4.6, and by taking a subsequence if necessary, we may assume that φ n convergesuniformly to a continuous map φ : ∂C H ( Y ) ∪ Y → ˆ C . Since φ n ( Y ( n ) ) = X ( n ) , by Lemma ?? , φ ( Y ) = X and φ is onto.By Lemma 6.2, Alexandrov’s convergence theorem 4.1 and convergence theorem of Poincar´emetrics (Theorem 4.2), we see that φ | ∂C H ( Y ) is an isometric embedding of ∂C H ( Y ) into a com-ponent Ω (cid:48) of ˆ C − Y . In particular, φ ( ∂C H ( Y )) ⊂ ˆ C − X . Together with φ ( Y ) = X and that φ is onto, we see that φ ( ∂C H ( Y )) = ˆ C − X . Therefore, φ is an isometry from ∂C H ( Y ) to ˆ C − X .Now we claim that X is a circle type closed set. Indeed, by the same argument as in § X k of X is of the form φ ( Y k ) for some component Y k of Y . By Proposition ?? and that X is of circle type, each φ ( Y k ) is the Hausdorff limit of a sequence of components φ n ( Y ( n ) k n ) = X ( n ) k n of X ( n ) . Therefore the result follows.7. T RANSBOUNDARY EXTREMAL LENGTHS AND A DUALITY THEOREM
The transboundary extremal length introduced by O. Schramm ([32]) is a very useful con-formal invariant and has been used in many works (see [7], [8] and others).Suppose Σ is a Riemann surface homeomorphic to an annulus and F and F ∗ are two familiesof curves in Σ such that F consists of closed curves separating two boundary components of Σ and F ∗ consisting of paths joining different boundary components of Σ . Then a well knownduality theorem states that the extremal lengths satisfy EL ( F ) EL ( F ∗ ) = 1 . The goal in thissection is to show that the duality theorem still holds for transboundary extremal lengths. Thelatter result (Theorem 7.4) is the key tool for estimating the modules of rings on non-smoothconvex surfaces.7.1. Transboundary extremal lengths.
We call a pair ( S, X ) admissible if (1) S is a surface,(2) X is a compact subset of S such that X has finitely many connected components, and(3) S − X is a Riemann surface. Given an admissible pair ( S, X ) , let ∼ be the equivalentrelationship on S such that x ∼ y if and only if x = y , or x, y are in the same componentof X . Denote S X = S/ ∼ the quotient space with the quotient topology. If X , ..., X n arecomponents of X , we use [ X i ] denote the image of X i in S X and let [ X ] = { [ X ] , ..., [ X n ] } .Then S X is the disjoint union of S − X and [ X ] . Each point [ X i ] is an end of the surface S − X .Given an admissible pair ( S, X ) and a conformal Riemannian line element g on S − X , an extended metric m on ( S, X ) is a pair ( ρg, µ ) such that ρ : S − X → R ≥ is a Borel measurablefunction and µ : [ X ] → R ≥ . The area of the extended metric m is defined to be A ( m ) = (cid:90) S − X ρ dA g + (cid:88) a ∈ [ X ] µ ( a ) , where dA g is the area form of the Riemannian metric associated to g . By a curve in S X wemean a continuous map γ from an interval to S X . We often identify a curve with its image in S X . The length of γ in the extended metric m is defined to be l m ( γ ) = (cid:90) ( S − X ) ∩ γ ρds + (cid:88) a ∈ [ X ] ∩ γ µ ( a ) , where ds is the line element associated to g . If Γ is a family of curves in S X , its length in theextended metric m is defined to be l m (Γ) = inf { l m ( γ ) : γ ∈ Γ } . Schramm’s transboundary extremal length [32] of Γ is(15) EL (Γ) = sup m l m (Γ) A ( m ) where the supremum is over all finite area extended metrics m .We will drop the adjective “transboundary” when we refer to extremal lengths below. Someof the basic properties of extremal lengths follow from the definition (see [3] for a proof). Lemma 7.1. (a) Suppose Γ and Γ are two families of curves in S X such that for any γ ∈ Γ ,there exists γ ∈ Γ satisfying γ ⊂ γ . Then EL (Γ ) ≥ EL (Γ ) . In particular, if Γ ⊂ Γ ,then EL (Γ ) ≥ EL (Γ ) .(b) Suppose Γ and Γ are two families of curves in S X such that they are supported in twodisjoint Borel measurable subsets A , A of S X , i.e., for any γ ∈ Γ , γ ⊂ A and for any γ ∈ Γ , γ ⊂ A . Then EL (Γ ∪ Γ ) − ≥ EL (Γ ) − + EL (Γ ) − . One of the key properties of extremal lengths is the following conformal invariance.
Lemma 7.2 (Schramm [32] Lemma 1.1) . Suppose φ : S X → ( S (cid:48) ) X (cid:48) is a homeomorphism suchthat φ | S − X is a conformal diffeomorphism between S − X and S (cid:48) − X (cid:48) . Then for any curvefamily Γ in S X , EL (Γ) = EL ( φ (Γ)) , where φ (Γ) = { φ ( γ ) : γ ∈ Γ } . An example of transboundary extremal length and a duality theorem. A flat cylinder is a Riemannian surface isometric to S = S × (0 , h ) equipped with the product metric g = dx + dy where ( e √− x , y ) are points in S . A square in S is a compact subset of the form I × I where I and I are two closed intervals of the same length. We consider a point as aclosed interval. If X is a disjoint union of finitely many squares in S , then ( S, X ) is called a square cylinder pair . The following lemma is known to Schramm [33]. Lemma 7.3.
Suppose ( S, X ) is a square cylinder pair. Let Γ ∗ be the family of curves in S X joining the two boundary components of S and Γ be the family of all simple loops in S X separating the two boundary components of S . Then EL (Γ) EL (Γ ∗ ) = 1 . Proof.
We will show that EL (Γ) = πh and EL (Γ ∗ ) = h π . Since the computations are similar,we only compute EL (Γ ∗ ) . Let the components of X be X , ..., X n of edge lengths h , ..., h n with h i ≥ . Let the coordinate in S be ( e √− x , y ) . Construct an extended metric m =( (cid:112) dx + dy , µ ) on ( S, X ) such that µ ([ X j ]) = h j . Then the area A ( m ) of m is πh . Forany curve γ ∈ Γ ∗ , we have l m ( γ ) ≥ h by definition. Therefore, l m (Γ ∗ ) ≥ h . This shows, EL (Γ ∗ ) ≥ l m (Γ ∗ ) /A ( m ) ≥ h π . To see that EL (Γ ∗ ) ≤ h π , take any extended metric m =( ρ (cid:112) dx + dy , µ ) and for each e √− t ∈ S , let γ t be the line segment { e √− t } × (0 , h ) in S . IRCLE DOMAIN 25
Then l m (Γ ∗ ) ≤ l m ( γ t ) = (cid:90) ( S − X ) ∩ γ t ρ ( e √− t , y ) dy + (cid:88) j :[ X j ] ∈ γ t h j . This shows, πl m (Γ ∗ ) ≤ (cid:90) π l m ( γ t ) dt = (cid:90) S − X ρ ( e √− t , y ) dydt + (cid:88) j h j . By Cauchy inequality we have π l m (Γ ∗ ) ≤ ( (cid:90) S − X ρ ( e √− t , y ) dydt + (cid:88) j h j )( (cid:90) S − X dydt + (cid:88) j h j ) . = A ( m )(2 πh ) . This shows l m (Γ ∗ ) /A ( m ) ≤ h π and the result follows. (cid:3) The main tool which enables us to estimate the module of rings on convex surfaces is thefollowing theorem. A version of it for quadrilaterals was proved by Schramm [33] (Theorem10.1). Recall that an open ring is a Riemann surface conformal to a flat cylinder.
Theorem 7.4.
Suppose R is an open ring, and X is a compact subset of R with finitely manycomponents. Let Γ ∗ be the family of curves in R X joining different boundary components of R and Γ be the family of all simple loops in R X separating the two boundary components of R .Then the transboundary extremal lengths satisfy (16) EL (Γ) EL (Γ ∗ ) = 1 . Proof.
By Lemmas 7.2 and 7.3, it suffices to prove that there exists a square cylinder pair ( R , X ) and a homeomorphism f from R X to R X such that f | : R − X → R − X isconformal. The later result was established by Jenkins (the corollary of Theorem 2 in [17]). (cid:3) Transboundary extremal length estimates on annuli.
The following result is essen-tially a special case of a result of Schramm on transboundary extremal length of curves inplanar co-fat domains.
Proposition 7.5.
Suppose S is an annulus in C − { } containing R r = { r < | z | < r } and W ⊂ S is a disjoint union of finitely many round closed disks and points. Let Γ be the familyof all loops in S W separating the two boundary components of S . Then EL (Γ) ≤ . Proof.
Let Γ ∗ be the family of all paths in S W joining two boundary components of S . Thenby Theorem 7.4, we have EL (Γ) − = EL (Γ ∗ ) . Thus it suffices to show EL (Γ ∗ ) = 1 / .This is a direct consequence of Schramm’s Theorem 6.1 in [32]. (cid:3) We also have a counterpart of Proposition 7.5 for non-smooth convex surfaces. Given acompact set Z ⊂ ∂ H , by the work of Reshetnyak [29], the surfaces ∂C K ( Z ) and ∂C H ( Z ) are surfaces of bounded curvature and therefore naturally Riemann surfaces whose conformalstructures are induced by the path metrics d KS and d PS . If Z is a finite disjoint union of closedround disks and points, then the surface ∂C H ( Z ) ∪ Z is an isometric gluing of two surfaces ofbounded curvature along round circles. It is again a surface of bounded curvature by the workof Alexandrov-Volkov. Therefore Reshetnyak’s work implies that it is naturally a Riemann surface. See §
10 for details. The extremal lengths of curves on these surfaces will be computedusing these complex structures.Fix q ∈ S , let D r = B r ( q, d S ) be the disk of radius r centered at q in S , C r = ∂D r and E r = { z ∈ S | d S ( z, q ) ≥ r } be the complement of int ( D r ) . The disk D r is convex if andonly if r ≤ π/ . Proposition 7.6.
Let r ∈ (0 , / , V ⊂ D r and V (cid:48) ⊂ E r be two compact connected sets,and W ⊂ S be a finite disjoint union of closed convex disks and points such that W ∩ ( V ∪ V (cid:48) ) is a union of components of W . Suppose the nearest point projection π : H P → C H ( W ) satisfies (17) d E ( q, π ( x )) ≤ r for all x ∈ D r and d E ( q, π ( x )) ≥ r for all x ∈ E r .Then,(a) π ( V ) , π ( V (cid:48) ) are disjoint and ∂C H ( W ) − π ( V ) − π ( V (cid:48) ) contains a unique annulus com-ponent S whose boundary intersects both π ( V ) and π ( V (cid:48) ) ,(b) W (cid:48) = W ∩ S is a union of components of W ,(c) for the family Γ of all loops in S W (cid:48) separating the two boundary components of S , EL (Γ) ≤ π. g p ( C r ) g c r (a) (b) W is a union of 's S p ( c )E D r VV' g F IGURE
4. Transboundary extremal length estimates on convex surfaces in H Proof.
To see part (a), it suffices to show that π ( D r ) and π ( E r ) are disjoint. This followsfrom (17) that π ( D r ) ⊂ B r ( q, d E ) and π ( E r ) ∩ B r ( q, d E ) = ∅ . The second statementof part (a) follows from the fact that ∂C H ( W ) ∪ W is topologically a 2-sphere and π ( V ) and π ( V (cid:48) ) are disjoint compact connected sets in ∂C H ( W ) ∪ W .Part (b) follows by showing that each component W j of W is either contained in or disjointfrom π ( V ) ∪ π ( V (cid:48) ) = π ( V ∪ V (cid:48) ) . Indeed if W j ∩ π ( V ∪ V (cid:48) ) contains a point x , then x =( π | ∂ H ) − ( x ) ∈ V ∪ V (cid:48) . By our assumption on W , W j ⊂ V ∪ V (cid:48) and thus W j = π ( W j ) ⊂ π ( V ∪ V (cid:48) ) .To see (c), let Γ ∗ be the family of paths in S W (cid:48) joining two boundary components of S . Thenby Theorem 7.1, we have EL (Γ) − = EL (Γ ∗ ) . It suffices to prove EL (Γ ∗ ) ≥ π . To this IRCLE DOMAIN 27 end, consider the extended metric m = ( ρd ES , λ ) where ρ ( z ) = 1 for z ∈ π ( D r ) ∩ S − W and is zero otherwise, and for each component W i of W , λ ([ W i ]) = diam ( W i ∩ D r ) is thespherical diameter of W i ∩ D r in ( S , d S ) . Let µ be the spherical measure on S . We have(18) diam ( W i ∩ D r ) ≤ µ ( W i ∩ D r ) . Indeed, if W i ⊂ D r or W i ∩ D r = ∅ , then diam ( W i ∩ D r ) ≤ diam ( W i ) and theresult follows from the well known estimate that (2 r ) ≤ µ ( B r ( x, d S )) . If W i is not inside D r and W i intersects D r , then W i ∩ D r contains a spherical disk of radius r . Then diam ( W i ∩ D r ) ≤ diam ( D r ) ≤ µ ( W i ∩ D r ) and (18) holds again.The area A ( m ) = (cid:82) π ( D r ) ∩ S − W dµ (cid:48) + (cid:80) ni =1 ( λ ([ W i ])) ≤ µ (cid:48) ( π ( D r ) − W )+ (cid:80) ni =1 ( λ ([ W i ])) where µ (cid:48) is the surface area measure on ∂C H ( W ) induced by the Euclidean metric d E . By (4), π is 2-Lipschitz, π ( D r ) − W = π ( D r − W ) and (18), we have A ( m ) ≤ µ ( D r − W ) + 2 n (cid:88) i =1 µ ( W i ∩ D r ) ≤ µ ( D r ) ≤ · πr = 360000 πr . Here we have used the fact that µ ( B r ( q, d S )) ≤ πr . For each path γ in Γ ∗ joining the twoboundary components of S W (cid:48) , let ˜ γ be the path on S obtained by gluing to γ ∩ ( S − W ) theshortest geodesic path β i in each component W i of W (cid:48) such that [ W i ] ∈ γ ∩ [ W (cid:48) ] . Note thatthere may be component W j of W (cid:48) outside of D r . Since V ⊂ D r and V (cid:48) ⊂ E r , the path ˜ γ contains a subarc γ ∗ joining a point p ∈ π ( C r ) to p ∈ π ( C r ) such that γ ∗ ⊂ π ( D r ) .By the assumption (17),(19) d E ( p , q ) ≥ r and d E ( p , q ) ≤ r. See figure 4(b). Clearly l m ( γ ) ≥ (cid:82) γ ∗ − W ds + (cid:80) W i ∩ ˜ γ ∩ D r (cid:54) = ∅ λ ([ W i ]) . The later is at least thespherical length l d S ( γ ∗ ) of γ ∗ since each W i ∩ D r is spherically convex and the sphericallength of β i ∩ γ ∗ is at most the spherical diameter λ ([ W i ]) of W i ∩ D r . Now the spheri-cal length l d S ( γ ∗ ) is at least the Euclidean length l d E ( γ ∗ ) of γ ∗ which is at least d E ( p , q ) − d E ( p , q ) . By (19), we see d E ( p , q ) − d E ( p , q ) ≥ r − r = 2 r . Therefore, we have l m ( γ ) ≥ r for all γ ∈ Γ ∗ and l m (Γ ∗ ) ≥ r. This implies EL (Γ ∗ ) ≥ l m (Γ ∗ ) A ( m ) ≥ r πr = 190000 π . (cid:3)
8. P
ROOF OF THE EQUICONTINUITY T HEOREM
Theorem.
The sequence { φ n : ( U n , d S ) → ( ∂C K ( Y ( n ) ) , d E ) } is equicontinuous. We prove it by contradiction. Suppose otherwise that there exists (cid:15) > and x n , x (cid:48) n ∈ U n such that d S ( x n , x (cid:48) n ) → and(20) d E ( φ n ( x n ) , φ n ( x (cid:48) n )) ≥ (cid:15) . By taking subsequences if necessary, we may assume that x n , x (cid:48) n → p ∈ ˆ C . Let Σ n = ∂C K ( Y ( n ) ) Lemma 8.1.
The limit point p is in the Hausdorff limit X of the sequence X ( n ) = C − U n . Proof.
Suppose otherwise that p / ∈ X , i.e., p ∈ U . Then by the Hausdorff convergence, thereexists r > and n such that B r ( p ) = { z ∈ C || z − p | < r } is in U n for all n ≥ n . Let d B and d n be the Poincare metrics on B r ( p ) and U n respectively. Then the Schwartz-Pick’s lemmashows d n ( x, y ) ≤ d B ( x, y ) for all x, y ∈ B r ( p ) . But we also have d B ( x n , x (cid:48) n ) → . Therefore d n ( x n , x (cid:48) n ) → . By (2) that d K ( x, y ) ≥ d E ( x, y ) , we have d n ( x n , x (cid:48) n ) = d K Σ n ( φ n ( x n ) , φ n ( x (cid:48) n )) ≥ d E Σ n ( φ n ( x n ) , φ n ( x (cid:48) n ))) ≥ d E ( φ n ( x n ) , φ n ( x (cid:48) n ))) ≥ (cid:15) . This is contradictory to d n ( x n , x (cid:48) n ) → . (cid:3) By Lemma 8.1 and x n ∈ U n , we see that p ∈ X ∩ ∂U . Let X ∗ be the connected componentof X which contains p . Due to the normalization condition on U n , the disk B / (0) = { z ∈ C || z | < / } is contained in U and U n for all n .Let ˜Σ n be ∂C K ( Y ( n ) ) ∪ Y ( n ) which is a topological 2-sphere. By the work of Reshetnyak[29], the surface ˜Σ n and Σ n are naturally Riemann surfaces. The conformal map φ n : U n → Σ n implies that the two admissible pairs ( ˆ C , X ( n ) ) and ( ˜Σ n , Y ( n ) ) are conformally equivalent.Indeed, the homeomorphism φ n induces a bijection between the ends of U n and Σ n . Theends of U n and Σ n are naturally identified with [ X ( n ) ] and [ Y ( n ) ] respectively. Therefore,each component X i of X ( n ) corresponds to a unique component Y i of Y ( n ) under φ n . Wedefine φ n ( X i ) = Y i and this extension produces the conformal equivalence between the pairs ( ˆ C , X ( n ) ) and ( ˜Σ n , Y ( n ) ) .Construct two families of paths Γ n and Γ (cid:48) n as follows.If X ∗ is a single point, then Γ n is defined to be the set of simple loops in ˆ C X ( n ) separat-ing { x n , x (cid:48) n } and B / (0) and Γ (cid:48) n is defined to be the set of simple loops in ˜Σ Y ( n ) n separating { φ n ( x n ) , φ n ( x (cid:48) n ) } and φ n ( B / (0)) .If X ∗ is a round disk, then by the construction of X ( n ) , X ∗ is a connected component of X ( n ) for n sufficiently large. Therefore p ∈ ∂X ( n ) for n large. We define Γ n to be the set of simplearcs γ in ˆ C X ( n ) − { [ X ∗ ] } such that γ ∪ { [ X ∗ ] } is a simple loop separating { x n , x (cid:48) n } and B / (0) in C X ( n ) , and Γ (cid:48) n to be the set of simple arcs in ˜Σ Y ( n ) n − { [ φ n ( X ∗ )] } such that γ ∪ { [ φ n ( X ∗ )] } separates { φ n ( x n ) , φ n ( x (cid:48) n ) } and φ n ( B / (0)) in ˜Σ Y ( n ) n .The conformal invariance of extremal length implies that EL (Γ n ) = EL (Γ (cid:48) n ) . We willderive a contradiction by showing that lim inf n EL (Γ (cid:48) n ) > and lim EL (Γ n ) = 0 .8.1. Extremal length estimate I: lim n EL (Γ n ) = 0 . Let D r = B r ( p, d E ) be the Euclideanball radius r centered at p in C , C r = ∂D r , and E r = { z ∈ ˆ C || z − p | ≥ r } . By thenormalization condition that X ( n ) ∩ D = ∅ , B / (0) ⊂ E / . Lemma 8.2.
For any r > , there exist r (cid:48) < r/ and N such that for n > N , no component of X ( n ) − X ∗ intersects both C r and C r (cid:48) .Proof. Let us prove for the case that X ∗ is a round disk. The same argument also worksfor the case of a single point. Suppose otherwise, there exists a sequence of components Z n of X ( k n ) − X ∗ such that Z n intersects both C r and C /n where lim n k n = ∞ . By taking asubsequence if necessary, we may assume that Z n converges in Hausdorff distance to a disk Z (cid:48) which intersects C r and contains p . Since the sequence { X ( n ) } converges in Hausdorff distanceto X , there exists a disk component X j of X such that Z (cid:48) ⊂ X j . But p ∈ X ∗ and p ∈ X j .Therefore X j = X ∗ . This shows that Z (cid:48) ⊂ X ∗ . However, since Z n and X ∗ are differentcomponents of X ( k n ) , we see the distance from the center of Z n to X ∗ is bounded away fromzero. This shows that the center of Z (cid:48) is outside of X ∗ and contradicts Z (cid:48) ⊂ X ∗ . (cid:3) IRCLE DOMAIN 29
By Lemma 8.2, we construct a sequence of disjoint circles C R i inside D / and a sequenceof integers N i increasing to infinity such that(1) R i +1 < R i / ,(2) if n ≥ N i , each connected component of X ( n ) − X ∗ intersects at most one circle of C R , C R , ...., C R i ,(3) if n ≥ N i , { x n , x (cid:48) n } ⊂ D R i and,(4) if X ∗ is a disk, X ∗ ∩ C R (cid:54) = ∅ .For any (cid:15) > , let j = (cid:100) /(cid:15) (cid:101) + 1 . We will show that if n ≥ N j , then EL (Γ n ) < (cid:15) . This isachieved by finding j disjoint open subsets A , ..., A j of ˆ C X ( n ) such that for each i = 1 , ..., j ,the curve family G i = { γ ∈ Γ n : γ ⊂ A i } satisfies(21) EL ( G i ) ≤ . Assuming (21), then by Lemma 7.1(a) and (b), EL (Γ n ) ≤ EL ( ∪ ji =1 G i ) ≤ ( j (cid:88) i =1 EL ( G i ) − ) − ≤ j ≤ (cid:15). This shows lim n EL (Γ n ) = 0 .Now to establish (21), fix n ≥ N j and construct A i ’s as follows. Define V i = D R i ∪ ( ∪{ X ( n ) k : X ( n ) k ∩ D R i (cid:54) = ∅ and X ( n ) k (cid:54) = X ∗ } ) which is a compact topological disk and V (cid:48) i = E R i − ∪ ( ∪{ X ( n ) : X ( n ) k ∩ E R i − (cid:54) = ∅ and X ( n ) k (cid:54) = X ∗ } ) . By the construction, V i ∩ V (cid:48) i = ∅ , { x n , x (cid:48) n } ⊂ V i and B / (0) ⊂ V (cid:48) i . Thus A (cid:48) i := ˆ C − V i − V (cid:48) i is a topological annulus containing the ring { R i − < | z | < R i − } where R i − > R i − .Furthermore, A (cid:48) i separates { x n , x (cid:48) n } from B / (0) . Let W i = ( X ( n ) − X ∗ ) ∩ A (cid:48) i . It is a compactF IGURE
5. Construction of rings A i ’s subset of A (cid:48) i and is a finite disjoint union of disks and points.If X ∗ is a single point set, we define A i = ( A (cid:48) i ) W i ⊂ ˆ C X ( n ) . Then G i = { γ ∈ Γ n : γ n ⊂ A i } is the family of simple loops γ in A i that separate the two boundaries of A i . By Proposition7.5, we see that inequality (21) that EL ( G i ) ≤ holds and the result follows. AA JJLLMMNNPP QQ VV WWZZAA CC DD II JJ NN F IGURE
6. Construction of A (cid:48)(cid:48) i in the case X ∗ is a disk AA JJLLMMNNPP QQ TTVV WWZZAA CC DD II JJ NN F IGURE
7. Construction of A i in the case X ∗ is a diskIf X ∗ is a disk, consider A (cid:48)(cid:48) i = int ( A (cid:48) i ∪ X ∗ ) which is a topological annulus containing thering { R i − < | z | < R i − } . Furthermore, A (cid:48)(cid:48) i separates { x n , x (cid:48) n } and B / (0) . Let A i = IRCLE DOMAIN 31 ( A (cid:48)(cid:48) i ) W i and define ˜ G i to be the family of simple loops in A i separating two boundaries of A (cid:48)(cid:48) i . Then by Proposition 7.5, EL ( ˜ G i ) ≤ . It remains to prove EL ( G i ) ≤ EL ( ˜ G i ) . Thisfollows from Lemma 7.1 (a) by proving that any curve ˜ γ ∈ ˜ G i contains a curve γ in G i . Indeed, ∂X ∗ ∩ A i consists of two disjoint arcs α and β such that each of them joins the two boundarycomponents of A (cid:48)(cid:48) i . Since ˜ γ separates different boundary components of A (cid:48)(cid:48) i , it contains an arc γ such that (1) γ ∩ X ∗ = ∅ and (2) the end points of γ are in α and β . By definition, γ is in G i := { δ ∈ Γ n : δ ⊂ A i } .8.2. Extremal length estimate II: lim inf n EL (Γ (cid:48) n ) > . For a convex surface S in the Kleinmodel ( H , d K ) , the induced path metrics d KS and d (cid:48) S on S are conformal where d (cid:48) ( x, y ) = | Ψ − ( x ) − Ψ − ( y ) | with Ψ( x ) = x | x | being an isometry between H K and H P . This is due tothe conformal equivalence of d P and d E on H . By Proposition 3.2 and Theorem 3.1, d (cid:48) ( x, y ) ≥ | x − y | and the area of a convex surface S in d (cid:48) metric is at most π . Define an extendedmetric m n on ˜Σ Y ( n ) n to be the pair ( d (cid:48) Σ n , ν n ) where d (cid:48) Σ is the induced path metric from d (cid:48) on Σ n and ν n on a connected component of Y ( n ) is the spherical diameter of the component. Hencethe area of m n is uniformly bounded from above by π since the square of the diameter of aspherical ball is at most twice of its area. Since EL (Γ (cid:48) n ) ≥ l m n (Γ (cid:48) n ) /A ( m n ) ≥ π l m n (Γ (cid:48) n ) ,it suffices to show that lim inf n l m n (Γ (cid:48) n ) > . We prove lim inf n l m n (Γ (cid:48) n ) > by replacing themetric d (cid:48) in m n by the Euclidean metric d E . Let m (cid:48) n = ( d E | Σ n , ν n ) be the extended metric on ˜Σ Y ( n ) n . Then due to d (cid:48) ( x, y ) ≥ | x − y | , lim inf n l m n (Γ (cid:48) n ) ≥ lim inf n l m (cid:48) n (Γ (cid:48) n ) and the resultfollows by showing lim inf n l m (cid:48) n (Γ (cid:48) n ) > .Note that by the assumption that (0 , , ∈ ∂C K ( Y ( n ) ) , each component Y (cid:48) of Y ( n ) isa spherical ball of radius at most π/ and hence is convex. Now suppose otherwise that lim inf n l m (cid:48) n (Γ (cid:48) n ) = 0 . After taking a subsequence, we may assume that there exists a sequenceof simple loops γ n ∈ Γ (cid:48) n such that l m (cid:48) n ( γ n ) → . For each γ n , construct a new path ˜ γ n obtainedby gluing to γ n − [ Y ( n ) ] the shortest spherical geodesic segment in each component Y ( n ) i of Y ( n ) for which [ Y ( n ) i ] ∈ γ n ∩ [ Y ( n ) ] . By the construction of m (cid:48) n , we have l m (cid:48) n ( γ n ) ≥ l E (˜ γ n ) where l E ( β ) is the Euclidean length of a path β . It follows that lim n l E (˜ γ n ) = 0 .If Y ∗ is a single point set, by construction, ˜ γ n is a simple loop in ˜Σ n separating two compactsets A n = φ n ( B / (0)) and B n = { φ n ( x n ) , φ n ( x (cid:48) n ) } . By Lemma 5.2 (c) and (20), both Eu-clidean diameters of A n and B n are bounded away from . After taking a subsequence, we mayassume that A n and B n converge in Hausdorff distances to two compact sets A and B of posi-tive Euclidean diameter and ˜ γ n converges uniformly, as Lipschitz maps, to ˜ γ in ∂C K ( Y ) ∪ Y .By a standard fact on path metric convergence, l E (˜ γ ) ≤ lim inf n l E (˜ γ n ) . Therefore l E (˜ γ ) = 0 ,i.e. ˜ γ is a single point. Consider the Euclidean convex hull Z of the set A ∪ B . Due to diam E ( A ) > and diam E ( B ) > , Z has to be 1-dimensional since otherwise any simpleloop α in ∂Z separating A from B has positive length. This shows that A ∪ B lies a Euclideanline segment. Therefore, we see that the convex set C K ( Y ) ∪ Y is one dimensional. Thiscontradicts Lemma 5.2 that Y contains at least three points.If Y ∗ is a disk, then γ n is a simple arc ending at [ Y ∗ ] . Construct a new loop γ ∗ n by gluing to γ n − [ Y ∗ ] the shortest geodesic segment β n in S . Since Y ∗ is convex, β n ⊂ Y ∗ . By the construc-tion γ ∗ n separates φ n ( B / (0)) from { φ n ( x n ) , φ n ( x (cid:48) n ) } in ˜Σ n . We claim that lim l E ( γ ∗ n ) = 0 andtherefore reduce this case to the case just proved above. To this end, let β (cid:48) n be the Euclideanline segment having the same end points as β n . Since β n has length at most diam d S ( Y ∗ ) ≤ π ,we have l E ( β (cid:48) n ) ≥ π l E ( β n ) . On the other hand, l E ( β (cid:48) n ) ≤ l E (˜ γ n ) since they have the same end points. It follows that l E ( γ ∗ n ) = l E (˜ γ n ) + l E ( β n ) ≤ (1 + π l E (˜ γ n ) . Therefore lim n l E ( γ ∗ n ) = 0 .9. P ROOF OF THE EQUICONTINUITY T HEOREM
Theorem.
The family { φ n : ( ∂C H ( Y ( n ) ) , d E ) → ( U n , d S ) } is equicontinuous. We prove it by deriving a contradiction. Suppose otherwise, there exist (cid:15) > and twosequences x n , x (cid:48) n ∈ Σ n := ∂C H ( Y ( n ) ) such that d E ( x n , x (cid:48) n ) ≤ n and(22) d S ( ϕ n ( x n ) , ϕ n ( x (cid:48) n )) > (cid:15). By taking a subsequence if necessary, we may assume { x n } converges to some point q ∈ Σ ∪ Y where Σ = ∂C H ( Y ) . We claim that q ∈ Y . If otherwise q ∈ Σ , by Lemma 9.1below that there is a constant C > independent of n such that the Poincar´e metric d n on U n = ˆ C − X ( n ) satisfies d n ≥ C d S , we have d P Σ n ( x n , x (cid:48) n ) = d n ( ϕ n ( x n ) , ϕ n ( x (cid:48) n )) ≥ C d S ( ϕ n ( x n ) , ϕ n ( x (cid:48) n )) > C (cid:15) . But lim n x n = lim n x (cid:48) n = q ∈ Σ , by Alexandrov’s convergenceTheorem 4.1, lim n d P Σ n ( x n , x (cid:48) n ) = d P Σ ( q, q ) = 0 . This contradicts that d P Σ n ( x n , x (cid:48) n ) ≥ C (cid:15). Lemma 9.1.
For the above sequence { d n } , there exists a constant C > independent of n such that d n ( ϕ n ( x n ) , ϕ n ( x (cid:48) n )) ≥ C d S ( ϕ n ( x n ) , ϕ n ( x (cid:48) n )) . Proof.
Since the sequence { X ( n ) } Hausdorff converges to X and | X | ≥ (Lemma 6.3), wecan choose a 3-point set { w , w , w } ⊂ X . Let W = ˆ C − { w , w , w } and d W = a ( z ) | dz | be the Poincar´e metric on W . Note that a ( z ) tends to infinity as z approaches ∂W since each w i is a cusp. Therefore, there exists a constant C W > such that d W ≥ C W d S on ˆ C . Nowtake three points u ( n ) i ∈ X ( n ) , i = 1 , , , such that lim n u ( n ) i = w i . We claim that there existsa constant C > independent of n such that(23) d V n ≥ C d S , where d V n is the Poincar´e metric on V n = ˆ C − { u ( n )1 , u ( n )2 , u ( n )3 } . This follows from d W ≥ C W d S and the convergence of d V n to d W uniformly on compact subsets of W . Using theSchwarz-Pick’s lemma that d n ≥ d V n , we see d n ( φ n ( x n ) , φ n ( x (cid:48) n )) ≥ d V n ( φ n ( x n ) , φ n ( x (cid:48) n )) ≥ C d S ( φ n ( x n ) , φ n ( x (cid:48) n )) . (cid:3) Let r ∈ (0 , / be the radius produced in Lemma 6.1 such that K n = { x ∈ Σ n : d P Σ n ( x, (0 , , ≤ r } is an embedded disk in Σ n for all n . By Lemma 6.2 the image φ n ( K n ) contains a spherical ball B r (0 , d S ) for some radius r > independent of n . Let Y ∗ be thecomponent of Y which contains q and let ˜Σ n = Σ n ∪ Y ( n ) which is a topological 2-sphere.By the same argument as in §
8, we see that φ n induces a conformal equivalence between theadmissible pairs ( ˜Σ n , Y ( n ) ) and ( ˆ C , X ( n ) ) and a bijection between components of Y ( n ) and X ( n ) . If Y i is a component of Y ( n ) , we use φ n ( Y i ) to denote the corresponding component of X ( n ) .Construct two families of paths Γ n and Γ (cid:48) n as follows. IRCLE DOMAIN 33 If Y ∗ is a single point, then Γ n is defined to be the set of simple loops in ˜Σ Y ( n ) n separat-ing { x n , x (cid:48) n } and K n , and Γ (cid:48) n is defined to be the set of simple loops in ˆ C X ( n ) separating { φ n ( x n ) , φ n ( x (cid:48) n ) } and φ n ( K n ) .If Y ∗ is a round disk, then by the construction of Y ( n ) , Y ∗ is a connected component of Y ( n ) for n sufficiently large. Therefore p ∈ ∂Y ( n ) for n large. We define Γ n to be the set of simplearcs γ in ˜Σ Y ( n ) n − { [ Y ∗ ] } such that γ ∪ { [ Y ∗ ] } is a simple loop separating { x n , x (cid:48) n } and K n in ˜Σ Y ( n ) n , and Γ (cid:48) n to be the set of simple arcs in ˆ C X ( n ) − { [ φ n ( Y ∗ )] } such that γ ∪ { [ φ n ( Y ∗ )] } separates { φ n ( x n ) , φ n ( x (cid:48) n ) } and φ n ( K n ) in ˆ C X ( n ) .Since φ n induces a conformal equivalence between the admissible pairs ( ˜Σ n , Y ( n ) ) and ( ˆ C , X ( n ) ) , the conformal invariance of extremal length implies that EL (Γ n ) = EL (Γ (cid:48) n ) . Wewill derive a contradiction by showing that lim inf n EL (Γ (cid:48) n ) > and lim EL (Γ n ) = 0 .9.1. Extremal length estimate I: lim inf n EL (Γ (cid:48) n ) > . Consider the extended metric m =( d S | U n , µ ) on ˆ C X ( n ) such that µ ([ X ( n ) i ]) = diam d S ( X ( n ) i ) .By definition, EL (Γ (cid:48) n ) ≥ l m (Γ (cid:48) n ) A ( m ) . Now A ( m ) ≤ Area d S ( U n ) + (cid:88) i diam d S ( X ( n ) i ) ≤ Area d S ( U n ) + (cid:88) i Area d S ( X ( n ) i ) = 4 Area d S ( ˆ C ) = 16 π. It remains to show l m (Γ (cid:48) n ) > .Note that by the normalization that D ⊂ U n , each component X ( n ) i of X ( n ) is sphericallyconvex. For any γ ∈ Γ (cid:48) n , construct the new loop γ ∗ ⊂ ˆ C obtained by gluing to γ ∩ ( ˆ C − X ( n ) ) a spherical geodesic segment of length at most λ ([ X ( n ) i ]) inside the X ( n ) i for each intersectionpoint [ X ( n ) i ] of [ X ( n ) ] ∩ γ . Then by the construction of extended metric m , l m ( γ ) ≥ l d S ( γ ∗ ) .Note that φ n ( K n ) contains a spherical ball B r (0 , d S ) and hence its spherical diameter isbounded below by r . Also, d S ( φ n ( x n ) , φ n ( y n )) > (cid:15) by (22).If X ∗ is a single point, then γ ∗ separates { φ n ( x n ) , φ n ( y n ) } from the ball φ n ( K n ) in ˆ C .Hence, there exists δ > independent of γ and n such that l d S ( γ ∗ ) ≥ δ. Indeed, consider thetwo Jordan domains Q and Q bounded by γ ∗ in ˆ C . One of them contains a spherical ballof radius r and the other contains two points of spherical distance (cid:15) apart. Therefore, thespherical diameters of Q and Q are at least δ > for some δ independent of γ and n . Nowthe length of γ ∗ is at least min( diam d S ( Q ) , diam d S ( Q ) , π/ . Therefore, the result follows.If X ∗ is a round disk, then by the construction the set γ ∗ ∪ X ∗ separates { φ n ( x n ) , φ n ( y n ) } and φ n ( K n ) in ˆ C . Let q (cid:48) , q (cid:48)(cid:48) be the end points of γ ∗ in X ∗ and β be the shortest spherical geodesicfrom q (cid:48) to q (cid:48)(cid:48) . Since X ∗ ∩ D = ∅ , we see that β is contained in X ∗ . The loop γ ∗ β separates { ϕ n ( x n ) , ϕ n ( y n ) } and φ n ( K n ) in ˆ C . Since β is a shortest spherical geodesic, l m ( γ ∗ ) ≥ l m ( β ) .It follows that l m ( γ ∗ ) ≥ l m ( γ ∗ β ) ≥ δ . Extremal length estimate II: lim EL (Γ n ) = 0 . The methods of proofs are similar tothose in section 8.1. The complication is due to the non-smooth convex surfaces. Recall D r = B r ( q, d S ) , C r = ∂D r and E r = S − D r . Lemma 9.2.
For any r > , there exists r (cid:48) < r/ such that no component of Y − Y ∗ intersects both C r and C r (cid:48) .Proof. By a simple area estimates, there exist finitely many components Z , Z , ..., Z n of Y − Y ∗ such that Z i intersects both C r and C r/ . Take r (cid:48) > to be any positive number such that r (cid:48) < r/ and r (cid:48) < min ≤ i ≤ n d S ( Y ∗ , Z i ) . (cid:3) We will prove that for any (cid:15) > , EL (Γ n ) < (cid:15) for sufficiently large n . By Lemma 7.1(b) and the same argument as in subsection 8.1, it suffices to prove that for any (cid:15) > , if n issufficiently large, we can find j = (cid:100) π/(cid:15) (cid:101) disjoint open subsets A , ..., A j of ˜Σ Y ( n ) n suchthat for any i = 1 , ..., j , the curve family G i = { γ ∈ Γ n : γ ⊂ A i } satisfies(24) EL ( G i ) ≤ π. By the Lemma 9.2 and Lemma 3.6, we can take a sequence of disjoint circles C R , ..., C R j inside D / , and N = N (cid:15) such that(1) R i +1 < R i / ,(2) each connected component of Y − Y ∗ intersects at most one circle of C R , ..., C R j ,(3) for n ≥ N , d E ( q, x n ) < R j / , d E ( q, x (cid:48) n ) < R j / , and(4) for n ≥ N , d E ( π ( x ) , π n ( x )) < R j / for all x ∈ S where π = π C H ( Y ) and π n = π C H ( Y ( n ) ) .Given ≤ i ≤ j , let V i = D R i ∪ ( ∪{ Y ( n ) k : Y ( n ) k ∩ D R i ) (cid:54) = ∅ , Y ( n ) k (cid:54) = Y ∗ } . By the construction, V i ⊂ D R i − . Let V (cid:48) i = E R i − ∪ ( ∪{ Y ( n ) k : Y ( n ) k ∩ E R i − ) (cid:54) = ∅ , Y ( n ) k (cid:54) = Y ∗ } . By the construction, V (cid:48) i ⊂ E R i − ⊂ E R i − . By Lemma 3.7 that d E ( π n ( x ) , q ) ≤ d S ( x, q ) ≤ R i − for any x ∈ D R i − , we have(25) d E ( π n ( x ) , q ) ≤ d E ( π n ( x ) , q ) + R j ≤ R i − + R j ≤ R i − . Similarly by Lemma 3.7 that R j − ≤ d E ( π n ( x ) , q ) for any x ∈ E R i − , we have(26) d E ( π n ( x ) , q ) ≥ d E ( π n )( x ) , q ) − R j ≥ R i − − R j ≥ R i − . This shows that π ( D R i − ) ∩ π ( E R i − ) and π ( V i ) ∩ π ( V (cid:48) i ) are empty sets. Then by Proposition7.6 for V = V i , V (cid:48) = V i with respect to D R i and E R i − , we see if n > N , then(a) there exists a unique annulus component A (cid:48) i in ˜Σ n − π n ( V i ) − π n ( V (cid:48) i ) such that theboundary components of A (cid:48) i intersects both π n ( V i ) and π n ( V (cid:48) i ) ,(b) W (cid:48) := ( Y ( n ) − Y ∗ ) ∩ A (cid:48) i is a union of components of Y ( n ) ,(c) for the family ˜ G i of all loops in ( A (cid:48) i ) W (cid:48) separating the two boundary components of A (cid:48) i , EL ( ˜ G i ) ≤ π .By the construction, we see that { x n , x (cid:48) n } ⊂ π n ( D R j ) and K n ⊂ π n ( E R ) . So ˜ G i is the setof simple loops in ( A (cid:48) i ) W (cid:48) separating { x n , x (cid:48) n } and K n . IRCLE DOMAIN 35
Now if Y ∗ is a single point, we take A i = ( A (cid:48) i ) W (cid:48) and G i = ˜ G i . By Proposition 7.6, theinequality (24) holds. Furthermore, since A (cid:48) i ⊂ π n ( D R i − ) − π n ( D R i ) are pairwise disjoint, A i are pairwise disjoint and the result follows.If Y ∗ is a disk, let A (cid:48)(cid:48) i = A (cid:48) i − Y ∗ and A i = ( A (cid:48)(cid:48) i ) W i and define G i to be the family of simplearcs γ in A i such that γ ∪ { Y ∗ } separates { x n , x (cid:48) n } and K n . By Proposition 7.6, EL ( ˜ G i ) ≤ π . Furthermore, since A (cid:48)(cid:48) i ⊂ π n ( D R i − ) − π n ( D R i ) are pairwise disjoint, A i are pairwisedisjoint. It remains to prove EL ( G i ) ≤ EL ( ˜ G i ) . This follows from Lemma 7.1 (c) by provingthat any curve ˜ γ ∈ ˜ G i contains a curve γ in G i . Indeed, ∂Y ∗ ∩ A (cid:48)(cid:48) i consists of two disjoint arcs α and β such that each of them joins the two boundary components of A (cid:48) i . Since ˜ γ separatesdifferent boundary components of A (cid:48) i , it contains an arc γ such that (1) γ ∩ Y ∗ = ∅ and (2) theend points of γ are in α and β . By definition, γ is in G i := { δ ∈ Γ n : δ ⊂ A i } .10. A PPENDIX : C
ONFORMAL STRUCTURE ON NON - SMOOTH SURFACES OF BOUNDEDCURVATURE
In this appendix, using the work of Reshetnyak [29], we will justify the computations of ex-tremal lengths of curve families in conformal structures on non-smooth surfaces like ∂C H ( Y ) and ∂C H ( Y ) ∪ Y . We begin with a brief recall of Alexandrov’s theory of surfaces of boundedcurvature and then discuss the associated conformal structure.10.1. Surfaces of bounded curvature and Alexandrov-Zalgaller gluing theorem.
Suppose ( S, d ) is a surface with a path metric d . This means that the distance d ( x, y ) between two pointsis equal to the infimum of lengths of all paths from x to y measured in d . Geodesics in ( S, d ) are locally distance minimizing curves. Suppose β and γ are two geodesics from a point O = β (0) = γ (0) parameterized by arc lengths. Then the (upper) angle α between them at O is defined to be α = lim sup s → ,t → arccos( d (0 , β ( s )) + d (0 , γ ( t )) − d ( β ( s ) , γ ( t ))2 d (0 , β ( s )) d (0 , γ ( t )) ) . For a geodesic triangle T , the (upper) excess of the T is δ ( T ) = θ + θ + θ − π where θ i arethe angles of the triangle T at three vertices. The surface ( S, d ) is called of bounded curvature if for every point p ∈ S , there exists a neighborhood U of p and a constant M ( U ) < ∞ suchthat for any collection of pairwise disjoint geodesic triangles T , T , ..., T n in U , we have n (cid:88) i =1 δ ( T i ) ≤ M ( U ) . All smooth Riemannian surfaces, polyhedral surfaces, convex surfaces in the Euclidean andhyperbolic spaces are surfaces of bounded curvature. For a smooth Riemannian surface ( S, g ) with Gaussian curvature K , the area form dA g , the curvature form KdA g and the total geodesiccurvature of a smooth curve are well defined. In a surface of bounded curvature, Alexandrovredefined these notations using the distance d and they become the area measure, the curvaturemeasure w and the total geodesic curvature. We say a sequence of metrics d n on a space X converges uniformly to a metric d on X if d n ( x, y ) → d ( x, y ) uniformly on X × X . One of thekey theorem on surfaces of bounded curvature is the following approximation theorem. SeeTheorem 6.2.1 in [29] Theorem 10.1 (Alexandrov-Zalgaller) . Suppose ( S, d ) is a surface of bounded curvature and p ∈ S . Then there exist a disk neighborhood U of p , a constant C ( U ) > and a sequence ofpolyhedral metrics d n on U such that d n converges uniformly to ( U, d | U ) and (27) | w d n | ( U ) + | κ d n | ( ∂U ) ≤ C, where w d n is the curvature measure and κ d n is the total geodesic curvature of a path. Con-versely, if ( S, d n ) is a sequence of polyhedral surfaces converging uniformly to a path metric ( S, d ) such that (27) holds on S for some constant C , then ( S, d ) is a surface of boundedcurvature. Surfaces of bounded curvature are flexible for gluing. The following useful theorem appearsas Theorem 8.3.1 in [29] or Theorem 6 in [2]. We state a simpler version of it which is sufficientfor our situation.
Theorem 10.2 (Alexandrov-Zalgaller) . ) Suppose ( S , d ) and ( S , d ) are two compact sur-faces of bounded curvatures whose boundary curves have bounded variation of turn. Let f : ∂S → ∂S be an isometry. Then the space obtained by gluing S and S along theirboundary via f is a surface of bounded curvature. As a corollary, if Z is a closed set in S , then ( ∂C H ( Z ) , d P∂C H ( Z ) ) and ( ∂C H ( Z ) , d E∂C H ( Z ) ) are surface of bounded curvature and if Y is a finite disjoint union of round disks in S , thenthe surface ∂C H ( Y ) ∪ Y in the induced path metric from the Euclidean distance d E is a surfaceof bounded curvature.10.2. Conformal structures on surfaces of bounded curvature.
The construction of con-formal charts for surfaces of bounded curvature by Reshetnyak goes as follows. Suppose U isan open disk in the plane and w is a signed Borel measure on U . Then the function ln λ ( z ) = π (cid:82) U | z − ζ | w ( dζ ) + h ( z ) is the difference of two subharmonic functions on U where h is a har-monic function. Since the Hausdorff dimension of the set of points where ln λ ( z ) = −∞ iszero, for an arbitrary Euclidean rectifiable path L in U , the integral (cid:82) L (cid:112) λ ( z ( s )) ds is well de-fined (could be ∞ ). One defines the distance d U on U between two points to be the infimum ofthe lengths of paths between them in λ ( z ) | dz | . There may be some points whose d U -distanceto any other point is infinite. These are called points at infinity and they form an isolated setin U . Let ˜ U be the complement of the set of points at infinity. It is proved by Reshetnyak that ( ˜ U , d U ) is a surface of bounded curvature whose curvature measure is w . The main theorem ofReshetnyak’s conformal geometry of surfaces of bounded curvature is Theorem 7.1.2 in [29]. Theorem 10.3 (Reshetynyak) . Let ( S, d ) be a surface of bounded curvature. Then for anypoint p ∈ S , there exists a neighborhood U (cid:48) of p and an open disk U in the plane together witha Borel measure w such that ( U (cid:48) , d | U (cid:48) ) is isometric to ( ˜ U , d U ) . Let φ : ( U (cid:48) , d | U (cid:48) ) → ( ˜ U , d U ) be an orientation preserving isometry produced in the abovetheorem. Then { ( U (cid:48) , φ ) } forms the analytic charts on the surface ( S, d ) .In conclusion, the metrics in surfaces with bounded curvature can be treated as Riemanniandistance derived from Riemannian metrics by relaxing the smoothness condition.Therefore, in a surface of bounded curvature ( S, d ) whose area measure is m and the under-lying conformal structure is C , we can use the path metric d and area measure m to computethe extremal length of a curve family Γ . In particular, we have the estimate EL (Γ , S, C ) ≥ l d (Γ) m ( S ) . IRCLE DOMAIN 37
This estimate has been used extensively in the previous sections on ( ∂C H ( Y ) , d P∂C H ( Y ) ) .Finally, for a compact set Y ⊂ S and Σ = ∂C H ( Y ) , we claim that the two induced pathmetrics d P Σ and d E Σ on Σ produce the same complex structure. In particular, this shows for anycurve family Γ in Σ , EL (Γ , Σ , d P Σ ) = EL (Γ , Σ , d E Σ ) . To see the claim, following Alexandrov[1], one constructs a sequence of convex hyperbolic polyhedral surfaces converging uniformlyon compact sets to ( ∂C H ( Y ) , d P∂C H ( Y ) ) . The convexity implies that (27) holds. Now on poly-hedral surfaces, the conformal structures induced from d P and d E are the same since these twometrics are conformal in H . The work of Reshetnyak ([29], Theorems 7.3.1, p112) shows thatif a sequence of bounded curvature surfaces converge uniformly to a bounded curvature surfacesuch that (27) holds, then the isothermal coordinates (with appropriate normalization) convergeto the isothermal coordinate of the limit surface. Therefore these two conformal structures on ∂C H ( Y ) are the same. R EFERENCES [1] Alexandrov, A. D.,
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