aa r X i v : . [ m a t h . M G ] S e p QUASISYMMETRIC KOEBE UNIFORMIZATION
S. MERENKOV AND K. WILDRICK
Abstract.
We study a quasisymmetric version of the classical Koebe uni-formization theorem in the context of Ahlfors regular metric surfaces. In par-ticular, we prove that an Ahlfors 2-regular metric surface X homeomorphic toa finitely connected domain in the standard 2-sphere S is quasisymmetricallyequivalent to a circle domain in S if and only if X is linearly locally connectedand its completion is compact. We also give a counterexample in the countablyconnected case. Introduction
Uniformization problems are amongst the oldest and most important problemsin mathematical analysis. A premier example is the measurable Riemann mappingtheorem, gives a robust existence theory for quasiconformal mappings in the plane.A quasiconformal mapping between domains in a Euclidean space is a homeomor-phism that sends infinitesimal balls to infinitesimal ellipsoids of uniformly boundedellipticity. The theory of quasiconformal mappings has been one of the most fruitfulin analysis, yielding applications to hyperbolic geometry, geometric group theory,complex dynamics, partial differential equations, and mathematical physics.In the past few decades, many aspects of the theory of quasiconformal mappingshave been extended to apply to abstract metric spaces. A key factor in these devel-opments has been the realization that in metric spaces with controlled geometry,the infinitesimal condition imposed by quasiconformal mappings actually implies astronger global condition called quasisymmetry [15]. The fact that quasisymmetricmappings are required to have good behavior at all scales makes them well suitedto metric spaces that a priori have no useful infinitesimal structure.A homeomorphism f : X → Y of metric spaces is quasisymmetric if there is ahomeomorphism η : [0 , ∞ ) → [0 , ∞ ) such that if x, y, z are distinct points of X ,then d Y ( f ( x ) , f ( y )) d Y ( f ( x ) , f ( z )) ≤ η (cid:18) d X ( x, y ) d X ( x, z ) (cid:19) . The homeomorphism η is called a distortion function of f . If we wish to emphasizethat a quasisymmetric mapping has a particular distortion function η , we will callit η -quasisymmetric.Despite the highly developed machinery for quasiconformal analysis on metricspaces, an existence theory for quasisymmetric mappings on metric spaces analo-gous to that of conformal mappings on Riemann surfaces has only recently beenexplored. The motivation for such results arises from geometric group theory [5], Mathematics Subject Classification. the dynamics of rational maps on the sphere [7], and the analysis of bi-Lipschitzmappings and rectifiable sets in Euclidean space [4].More than a decade after foundational results of Tukia and V¨ais¨al¨a in dimensionone [24], Bonk and Kleiner [5] gave simple sufficient conditions for a metric spaceto be quasisymmetrically equivalent to the standard 2-sphere S . Theorem 1.1 (Bonk–Kleiner) . Let X be an Ahlfors -regular metric space home-omorphic to the sphere S . Then X is quasisymmetrically equivalent to the sphere S if and only if X is linearly locally connected. The condition that X is linearly locally connected (LLC), which heuristicallymeans that X does not have cusps, is a quasisymmetric invariant. Ahlfors 2-regularity, which states that the two-dimensional Hausdorff measure of a ball isuniformly comparable to the square of its radius, is not. See Section 3 for precisedefinitions. A version of Theorem 1.1 for metric spaces homeomorphic to the planewas derived in [26], and a local version given in [27].In this paper, we seek a version of Theorem 1.1 for domains in S . The moti-vation for our inquiry comes from the Kapovich–Kleiner conjecture of geometricgroup theory, described in Section 2, and from analogous classical conformal uni-formization theorems onto circle domains.A circle domain is a domain Ω ⊆ S such that each component of S \ Ω is either around disk or a point. In 1909 [18], Koebe posed the following conjecture, known asthe Kreisnormierungsproblem: every domain in the plane is conformally equivalentto a circle domain. In the 1920’s [19], Koebe was able to confirm his conjecture inthe finitely connected case.
Theorem 1.2 (Koebe’s uniformization onto circle domains) . Let Ω ⊆ C be adomain with finitely many complementary components. Then Ω is conformallyequivalent to a circle domain. We first state a quasisymmetric version of Koebe’s theorem, which we attain asa consequence of our main result. Denoting the completion of a metric space X by X , we define the metric boundary of X by ∂X = X \ X . We say that a componentof ∂X is non-trivial if it contains more than one point. Theorem 1.3.
Let ( X, d ) be an Ahlfors -regular metric space that is homeomor-phic to a domain in S , and such that ∂X has finitely many non-trivial components.Then ( X, d ) is quasisymmetrically equivalent to a circle domain if and only if ( X, d ) is linearly locally connected and the completion X is compact. Theorem 1.3 is only quantitative in the sense that the distortion function of thequasisymmetric mapping may be chosen to depend only on the constants associatedto the various conditions on X and the ratio of the diameter of X to the minimumdistance between components of ∂X .In 1993, He and Schramm confirmed Koebe’s conjecture in the case of countablymany complementary components [13]. In full generality the conjecture remainsopen. A key tool in He and Schramm’s proof was transifinte induction on the rankof the boundary of a domain Ω in S , which measures the complexity with whichcomponents of the boundary converge to one another. The rank of a collection ofboundary components is defined via a canonical topology on the set of componentsof the boundary; see Section 4. It is shown there that if a metric space ( X, d )is homeomorphic to a domain Ω in S and is linearly locally connected, then the UASISYMMETRIC KOEBE UNIFORMIZATION 3 natural topology on the set of components of the metric boundary is homeomorphicto the natural topology on the set of boundary components of Ω. This allows usto define rank as in the classical setting. We denote the topologized collection ofcomponents of ∂X by C ( X ).In the following statement, which is our main result, we consider quasisymmetricuniformization onto circle domains Ω with the property that C (Ω) is uniformlyrelatively separated , meaning there is a uniform lower bound on the relative distance △ ( E, F ) = dist(
E, F )min { diam( E ) , diam( F ) } , between any pair of non-trivial boundary components. This condition is appearsnaturally in both classical quasiconformal analysis and geometric group theory.Moreover, we employ annular linear local connectedness (ALLC), which is morenatural than the LLC condition in this setting. Theorem 1.4.
Let ( X, d ) be a metric space, homeomorphic to a domain in S ,such that the closure of the collection of non-trivial components of ∂X is countableand has finite rank. Moreover, suppose that (1) ( X, d ) is Ahlfors -regular, (2) setting, for each integer k ≥ , n k = sup card { E ∈ C ( X ) : E ∩ B ( x, r ) = ∅ and − k r < diam E ≤ − k +1 r } , where the supremum is taken over all x ∈ X and < r < X , it holdsthat ∞ X k =0 n k − k < ∞ . Then ( X, d ) is quasisymmetrically equivalent to a circle domain such that C (Ω) isuniformly relatively separated if and only if X has the following properties: (3) the completion X is compact, (4) ( X, d ) is annularly linearly locally connected, (5) C ( X ) is uniformly relatively separated. Theorem 1.4 is quantitative in the sense that the distortion function of thequasisymmetric mapping may be chosen to depend only on the constants associatedto the various conditions on X , and vice-versa.A key tool in our proof is the following similar uniformization result of Bonk,which is valid for subsets of S [1]. Theorem 1.5 (Bonk) . Let { S i } i ∈ N be a collection of uniformly relatively separateduniform quasicircles in S that bound disjoint Jordan domains. Then there is aquasisymmetric homeomorphism f : S → S such that for each i ∈ N , the set f ( S i ) is a round circle in S . This result, which is also quantitative, allows us to conclude that if a metricspace (
X, d ) as in the statement of Theorem 1.4 satisfies conditions (1)-(5), thenit is quasisymmetrically equivalent to a circle domain with uniformly relativelyseparated boundary circles as soon as there is any quasisymmetric embedding of X into S . Thus, producing such an embedding is the main focus of this paper.It is of great interest to know if conditions (1) and (2) can be replaced withconditions that are quasisymmetrically invariant. By snowflaking, i.e., raising the S. MERENKOV AND K. WILDRICK metric to power 0 < α <
1, the sphere S = R ∪ {∞} in one direction only, say,in the direction of x -axis, one produces a metric space homeomorphic to S thatsatisfies all the assumptions of Theorem 1.4 (and Theorem 1.1) except for Ahlfors2-regularity, but fails to be quasisymmetrically equivalent to S . On the other hand,not every quasisymmetric image of S is Ahlfors 2-regular, as is seen by the usualsnowflaking of the standard metric on S .Our second main result is the existence of a metric space satisfying all assump-tions of Theorem 1.4, except for condition (2), that fails to quasisymmetricallyembed in S . Theorem 1.6.
There is a metric space ( X, d ) , homeomorphic to a domain in S ,with the following properties • ∂X has rank , • X is Ahlfors -regular, • the completion X is compact, • X is annularly linearly locally connected, • the components of ∂X are uniformly relatively separated, • there is no quasisymmetric embedding of X into S . We now outline the proof of Theorem 1.4 and the structure of the paper. InSection 4 we establish a topological characterization of the boundary componentsof a metric space as the space of ends of the underlying topological space, at least inthe presense of some control on the geometry of the space. This allows us to developa notion of rank, and in Section 6, a theory of cross-cuts analogous to the classicaltheory. A key tool in this development is the following purely topological statement:every domain in S is homeomorphic to a domain in S with totally disconnectedcomplement. This folklore theorem is proven in Section 5. In Section 7, we usecross-cuts and a classical topological recognition theorem for S to uniformize theboundary components of X . The resulting theorem, which generalizes [26, Theorem1.3], may be of independent interest: Theorem 1.7.
Suppose that X is a metric space that is homeomorphic to a domainin S , has compact completion, and satisfies the λ - LLC condition for some λ ≥ .Then each non-trivial component of ∂X is a topological circle satisfying the λ ′ - LLC condition for some λ ′ ≥ depending only on λ . In particular, if the space X is additionally assumed to be doubling, then each non-trivial component of ∂X isquasisymmetrically equivalent to S with distortion function depending only on λ and the doubling constant. Suppose that (
X, d ) is an Ahlfors 2-regular and ALLC metric space that is home-omorphic to a domain in S . Section 8 shows that the completion X is homeomor-phic to the closure of an appropriately chosen circle domain. In Section 9 we provegeneral theorems implying that each non-trivial component of ∂X has Assouaddimension strictly less than 2, and hence, up to a bi-Lipschitz mapping, is theboundary of a planar quasidisk. We describe a general gluing procedure in Section10, and use it to “fill in” the non-trivial components of ∂X . The resulting space b X is ALLC, but not always Ahlfors 2-regular. In order to guarantee this, we imposecondition (2) of Theorem 1.4. The assumption of finite rank allows us to reduceto the case that there are only finitely many components of ∂X . We show that b X is homeomorphic to S , and apply Theorem 1.1. The problem is now planar, and UASISYMMETRIC KOEBE UNIFORMIZATION 5 the above mentioned theorem of Bonk now yields the desired result. Section 11summarizes our work and provides a formal proof of Theorems 1.3 and 1.4.Section 12 is dedicated to proving Theorem 1.6. We conclude in Section 13 withdiscussion of related open problems.1.1.
Acknowledgements.
We wish to thank Mario Bonk, Peter Feller, PekkaKoskela, John Mackay, Daniel Meyer, Raanan Schul, and Jeremy Tyson for usefulconversations and critical comments. Some of the research leading to this worktook place while the second author was visiting the University of Illinois at Urbana-Champaign and while both authors were visiting the State University of New Yorkat Stony Brook. We are very thankful for the hospitality of those institutions. Also,the first author wishes to thank the Hausdorff Research Institute for Mathematicsin Bonn, Germany, for its hospitality during the Rigidity program in the Fall 2009.2.
Relationship to Gromov hyperbolic groups
In geometric group theory, there is a natural concept of a hyperbolic group due toGromov. These abstract objects share many features of Kleinian groups, including aboundary at infinity. The boundary ∂ ∞ G of a Gromov hyperbolic group is equippedwith a natural family of quasisymmetrically equivalent metrics. The structure ofthis boundary is categorically linked to the structure of the group: a large-scale bi-Lipschitz mapping between Gromov hyperbolic groups induces a quasisymmetricmapping between the corresponding boundaries at infinity, and vice versa.One of the premier problems in geometric group theory is Cannon’s conjecture,which states that for every Gromov hyperbolic group G with boundary at infinity ∂ ∞ G homeomorphic to S , there exists a discrete, co-compact, and isometric actionof G on hyperbolic 3-space. In other words, if a Gromov hyperbolic group has thecorrect boundary at infinity, then it arises naturally from the corresponding situ-ation in hyperbolic geometry. By a theorem of Sullivan [23], Cannon’s conjectureis equivalent to the following statement: if G is a Gromov hyperbolic group, then ∂ ∞ G is homeomorphic to S if and only if ∂ ∞ G is quasisymmetrically equivalent to S . If ∂ ∞ G is homeomorphic to S , then each natural metric on ∂ ∞ G is LLC andAhlfors Q -regular for some Q ≥
2. If Q = 2, then Theorem 1.1 confirms Cannon’sconjecture. Indeed, even stronger statements are known [6].Closely related to Cannon’s conjecture is the Kapovich–Kleiner conjecture, whichstates that for every Gromov hyperbolic group G with boundary at infinity ∂ ∞ G homeomorphic to the Sierpi´nski carpet, there exists a discrete, co-compact, andisometric action of G on a convex subset of hyperbolic 3-space with totally geodesicboundary. The Kapovich–Kleiner conjecture is implied by Cannon’s conjecture,and is equivalent to the following statement: if G is a Gromov hyperbolic group,then ∂ ∞ G is homeomorphic to the Sierpi´nski carpet if and only if ∂ ∞ G is qua-sisymmetrically equivalent to a round Sierpi´nski carpet, i.e., to a subset of S thatis homeomorphic to the Sierpi´nski carpent and has peripheral curves that are roundcircles.Theorem 1.4 can be seen as a uniformization result for domains that might ap-proximate a Sierpi´nski carpet arising as the boundary of a hyperbolic group. If ∂ ∞ G is homeomorphic to the Sierpi´nski carpet, then it is ALLC and the peripheral cir-cles are uniformly separated uniform quasicirlces. Recent work of Bonk establishedthe Kapovich–Kleiner conjecture in the case that ∂ ∞ G can be quasisymmetricallyembedded in S ; see [1]. As noted by Bonk–Kleiner in [2], this is true when the S. MERENKOV AND K. WILDRICK
Assouad dimension of ∂ ∞ G is strictly less than two, a hypothesis analgous to con-dition (2) in Theorem 1.4. This observation and its proof provided ideas that willbe used in Section 10. 3. Notation and basic results
Metric Spaces.
We are often concerned with conditions on a mapping orspace that involve constants or distortion functions. These constants or distortionfunctions are refered to as the data of the conditions. A theorem is said to be quantitative if the data of the conclusions of theorem depend only on the data ofthe hypotheses. In the proof of quantitative theorems, given non-negative quantities A and B , we will employ the notation A . B if there is a quantity C ≥
1, dependingonly on the data of the conditions in the hypotheses, such that A ≤ CB . We write A ≃ B if A . B and B . A .We will often denote a metric space ( X, d ) by X . Given a point x ∈ X and anumber r >
0, we define the open and closed balls centered at x of radius r by B ( X,d ) ( x, r ) = { y ∈ X : d ( x, y ) < r } and B ( X,d ) ( x, r ) = { y ∈ X : d ( x, y ) ≤ r } . For 0 ≤ r < R , we denote the open annulus centered at x of inner radius r andouter radius R by A ( X,d ) ( x, r, R ) = { y ∈ X : r < d ( x, y ) < R } . Note that when r = 0, this corresponds to B ( X,d ) ( x, R ) \{ x } . Where it will not cause confusion, we denote B ( X,d ) ( x, r ) by B X ( x, r ), B d ( x, r ),or B ( x, r ). A similar convention is used for all other notions which depend implicitlyon the underlying metric space.We denote the completion of a metric space X by X , and define the metricboundary of X by ∂X = X \ X . These notions are not to be confused with theirtopological counterparts.For ǫ >
0, the ǫ -neighborhood of a subset E ⊆ X is given by N ǫ ( E ) = [ x ∈ E B ( x, ǫ ) . The diameter of E is denoted by diam( E ) , and the distance between two subsets E, F ⊆ X is denoted by dist( E, F ) . If at least one of E and F has finite diameter,then the relative distance of E and F is defined by △ ( E, F ) = dist(
E, F )min { diam E, diam F } , with the convention that △ ( E, F ) = ∞ if at least one of E and F has diameter 0. Remark . If f : X → Y is a quasisymmetric homeomorphism of metric spaces,and E and F are subsets of X , then △ ( f ( E ) , f ( F )) ≃ △ ( E, F ) . This is easily seen using [14, Proposition 10.10].Let S = { ( x, y, z ) ∈ R : x + y + z = 1 } be the standard 2-sphere equippedwith the restriction of the Euclidean metric on R . We say that Ω is a domain in S if it is an open and connected subset of S . We always consider a domain in S as already metrized, i.e., equipped with the restriction of the standard sphericalmetric. UASISYMMETRIC KOEBE UNIFORMIZATION 7
Dimension and measures.
A metric space X is doubling if there is a con-stant N ∈ N such that for any x ∈ X and r >
0, the ball B ( x, r ) can be coveredby at most N balls of radius r/
2. This condition is quantitatively equivalent tothe existence of constants α ≥ C ≥ X is ( α, C ) -homogeneous ,meaning that for every x ∈ X , and 0 < r ≤ R , the ball B ( x, R ) can be cov-ered by at most C ( R/r ) α balls of radius r . The infinimum over all α such that X is ( α, C )-homogeneous for some C ≥ Assouad dimension of X .Hence, doubling metric spaces are precisely those metric spaces with finite Assouaddimension.In a doubling metric space, some balls may have lower Assouad dimension thanthe entire space. To rule out this kind of non-homogeneity, one often employs amuch stricter notion of finite-dimensionality. The metric space ( X, d ) is
Ahlfors Q -regular , Q ≥
0, if there is a constant K ≥ x ∈ X and0 < r < diam X ,(3.1) r Q K ≤ H QX ( B ( x, r )) ≤ Kr Q , where H QX denotes the Q -dimensional Hausdorff measure on X . It is quantitativelyequivalent to instead require that (3.1) hold for all open balls of radius less that2 diam X . The existence of any Borel regular outer measure on X that satisfies(3.1) quanitatively implies that X is Ahlfors Q -regular; see [14, Exercise 8.11]. Remark . Suppose that (
X, d ) is Ahlfors Q -regular, Q ≥
0. Given S ⊆ ∂X , thespace ( X ∪ S, d ) is again Ahlfors Q -regular, quantitatively. This is proven as in [26,Lemma 2.11].3.3. Connectivity conditions.
Here we describe various conditions that controlthe existence of “cusps” in a metric space by means of connectivity. The basicconcept of such conditions arose from the theory of quasiconformal mappings inthe plane, where they play an important role as invariants.Let λ ≥
1. A metric space (
X, d ) is λ -linearly locally connected ( λ -LLC) if forall a ∈ X and r >
0, the following two conditions are satisfied:(i) for each pair of distinct points x, y ∈ B ( a, r ), there is a continuum E ⊆ B ( a, λr ) such that x, y ∈ E ,(ii) for each pair of distinct points x, y ∈ X \ B ( a, r ), there is a continuum E ⊆ X \ B ( a, r/λ ) such that x, y ∈ E .Individually, conditions ( i ) and ( ii ) are referred to as the λ -LLC and λ -LLC conditions, respectively.The LLC condition extends in a particularly nice way to the completion of ametric space. We say that a metric space ( X, d ) is λ - g LLC if for all a ∈ X and r > x, y ∈ B X ( a, r ), there is an embedding γ : [0 , → X such that γ (0) = x , γ (1) = y , γ | (0 , ⊆ X , and im γ ⊆ B X ( a, λr ) , (ii) For each pair of distinct points x, y ∈ X \ B X ( a, r ), there is an embedding γ : [0 , → X such that γ (0) = x , γ (1) = y , γ | (0 , ⊆ X , and im γ ⊆ X \ B X ( a, r/λ ) . Individually, conditions ( i ) and ( ii ) are referred to as the λ - g LLC and λ - g LLC conditions, respectively. S. MERENKOV AND K. WILDRICK
If a metric space X is λ - g LLC, then it is also λ -LLC. The next proposition statesthat the two conditions are quantitatively equivalent for the spaces in considerationin this paper. Proposition 3.3.
Let i ∈ { , } , and let ( X, d ) be a locally compact and locallypath-connected metric space that satisfies the λ - LLC i condition. Then X is λ ′ - ] LLC i , where λ ′ depends only on λ .Proof. The key ingredient is the following statement: If U ⊆ X is an open subsetof X , and E ⊆ U is a continuum, then any pair of points x, y ∈ E are contained inan arc in U . The details are straightforward and left to the reader. (cid:3) Let λ ≥
1. A metric space (
X, d X ) is λ -annularly linearly locally connected ( λ -ALLC) if for all points a ∈ X and all 0 ≤ r < R , each pair of distinct points in theannulus A ( a, r, R ) is contained in a continuum in the annulus A ( a, r/λ, λR ) . The ALLC condition forbids local cut-points in addition to ruling out cusps. Forexample, the standard circle S is LLC but not ALLC. In our setting, the ALLCcondition is a more natural assumption, and is in some cases equivalent to the LLCcondition.We omit the proofs of the following three statements. The first is based ondecomposing an arbitrary annulus into dyadic annuli. The second uses the fact thatin a connected space, any distinct pair of points is contained in annulus around somethird point. The third states that the ALLC condition extends to the boundary asin Proposition 3.6. Lemma 3.4.
Let λ ≥ . Suppose that a connected metric space ( X, d ) satisfies thecondition that for all points a ∈ X and all r > , each pair of disctinct points inthe annulus A ( a, r, r ) is contained in a continuum in the annulus A ( a, r/λ, λr ) . Then X satisfies the λ - ALLC condition.
Lemma 3.5.
Suppose that ( X, d ) is a connected metric space that satisfies the λ - ALLC condition. Then ( X, d ) satisfies the λ - LLC condition.
Proposition 3.6.
Suppose that ( X, d ) is a locally compact and locally path-connectedmetric space that satisfies the λ - ALLC condition. Then there is a quantity λ ′ ≥ ,depending only on λ , such that for all a ∈ X and ≤ r < R , for each pair ofdistinct points x, y ∈ A X ( a, r, R ) , there is an embedding γ : [0 , → X such that γ (0) = x , γ (1) = y , γ | (0 , ⊆ X , and im γ ⊆ A X ( a, r/λ ′ , λ ′ R ) . There is a close connection between the ALLC condition and the uniform relativeseparation of the components of the boundary of a given metric space. The followingstatement addresses only circle domains, but a more general result is probably valid.We postpone the proof until Section 5.
Proposition 3.7.
Let Ω be a circle domain. Then Ω satisfies the ALLC condition ifand only if the components of ∂ Ω are uniformly relatively separated, quantitatively. In the case of metric spaces that are homeomorphic to domains in S with finitelymany boundary components, the LLC-condition may be upgraded to the ALLC-condition. Again, we postpone the proof until Section 5. UASISYMMETRIC KOEBE UNIFORMIZATION 9
Proposition 3.8.
Let ( X, d ) be a metric space homeomorphic to a domain in S ,and assume that the boundary ∂X has finitely many components. If ( X, d ) is λ - LLC , λ ≥ , then it is Λ - ALLC , where Λ depends on λ and the ratio of the diameterof X to the minimum distance between components of ∂X . The space of boundary components of a metric space
In this section we assume that (
X, d ) is a connected, locally compact metricspace with the additional property that the completion X is compact. Note thatas X is locally compact, it is an open subset of X . Hence ∂X is closed in X andhence compact.4.1. Boundary components and ends.
Of course, the topological type of ∂X depends on the specific metric d . However, the goal of this section is to showthat under a simple geometric condition, the collection C ( X ) of components of ∂X depends only on the topological type of X .We define an equivalence relation ∼ on ∂X by declaring that x ∼ y if and onlyif x and y are contained in the same component of ∂X . Then there is a bijectionbetween C ( X ) and the quotient ∂X/ ∼ , and hence we may endow C ( X ) with thequotient topology. Since ∂X is compact, the space C ( X ) is compact as well. Givena compact set K ⊆ X and a component U of X \ K , denote C ( K, U ) = { E ⊆ C ( X ) : E ⊆ ( ∂U ∩ ∂X ) } . Let U ( X ) denote the collection of sequences { x i } ⊆ X with the property thatfor every compact set K ⊆ X , there is a number N ∈ N and a connected subset U of X \ K such that { x i } i ≥ N ⊆ U. Define an equivalence relation e ∼ on U ( X ) by { x i } e ∼ { y i } if and only if the sequence { x , y , x , y , . . . } is in U ( X ). An equivalence class E defined by e ∼ is called an end of X , and we denote the collection of ends of X by E ( X ).Given a compact subset K ⊆ X and a component U of X \ K , define E ( K, U ) = { [ { x i } ] : { x i } ∈ U ( X ) and ∃ N ∈ N such that { x i } i ≥ N ⊆ U } , That this set is well-defined follows from the definition of the equivalence relationon U ( X ). Let B be the collection of all such sets. Proposition 4.1.
The collection B generates a unique topology on E ( X ) such thatevery open set is a union of sets in B .Proof. We employ the standard criteria for proving generation [22, Section 13]. As X is connected, taking K = ∅ shows that B contains E ( X ). Thus it suffices to showthat given compact subsets K and K of X and components U and U of X \ K and X \ K respectively, and given an end E ∈ E ( K , U ) ∩ E ( K , U ) , there is a compact set K and a component U of X \ K such that(4.1) E ∈ E ( K, U ) ⊆ ( E ( K , U ) ∩ E ( K , U )) . Let { x i } ∈ U ( X ) represent the end E . By definition of U ( X ), there is a com-ponent U of X \ K , where K = K ∪ K , such that { x i } i ≥ N ⊆ U for some N ∈ N .This implies that E ∈ E ( K, U ). By assumption, there is a number M ∈ N such that { x i } i ≥ M is contained in U ∩ U . Since U is a connected subset of X \ K and X \ K , it follows that U ⊆ U ∩ U . This implies (4.1). (cid:3) We set the topology on E ( X ) to be that given by Proposition 4.1. It followsquickly from the definitions that E ( X ) is a Hausdorff space. Remark . As ends are defined purely in topological terms, a homeomorphism h : X → Y to some other topological space Y induces a homeomorphism φ : E ( X ) →E ( Y ). This homeomorphism is natural in the sense that a sequence { x i } ∈ U ( X )represents the end E ∈ E ( X ) if and only { h ( x i ) } ∈ U ( Y ) represents the end φ ( E ) ∈E ( Y ).To relate the ends of a metric space to the components of its metric boundary,we need an elementary result regarding connectivity. Let ǫ > x and y bepoints in any metric space Z . An ǫ -chain connecting x to y in Z is a sequence x = z , z , . . . , z n = y of points in Z such that d ( z j , z j +1 ) ≤ ǫ for each j = 0 , . . . , n − Lemma 4.3.
Let x and y be points in a metric space Z . If x and y lie in the samecomponent of Z , then for every ǫ > there is an ǫ -chain in Z connecting x to y .The converse statement holds if Z is compact. An end always defines a unique boundary component.
Proposition 4.4. If { x i } and { y i } are Cauchy sequences representing the sameend of X , then they represent points in the same component of ∂X .Proof. Denote by x and y the points in ∂X defined by { x i } and { y i } , respectively.By Lemma 4.3, in order to show that x and y are contained in a single componentof ∂X , it suffices to find an ǫ -chain in ∂X connecting x to y for every ǫ >
0. Tothis end, fix ǫ >
0. Let K be the compact subset of X defined by(4.2) K = X \N X ( ∂X, ǫ/ . By assumption, we may find N ∈ N so large that x N and y N lie in a connected subsetof X \ K , and that d ( x, x N ) and d ( y, y N ) are both less than ǫ/
3. By Lemma 4.3,we may find an ǫ/ x N = z , . . . , z n = y N in X \ K . By (4.2), for each j = 1 , . . . , n − z ′ j ∈ ∂X such that d ( z j , z ′ j ) < ǫ/
3. The triangleinequality now implies that x, z ′ , . . . , z ′ n − , y is an ǫ -chain in ∂X , as required. (cid:3) Remark . Proposition 4.4 allows us to define a map Φ : E ( X ) → C ( X ) as follows.Let E ∈ E ( X ) and let { x i } i ∈ N ∈ U ( X ) be a sequence representing E . As X iscompact, we may find a Cauchy subsequence { x i j } of { x i } that represents a pointin some boundary component E ′ ∈ C ( X ). Set Φ( E ) = E ′ . This is well defined byProposition 4.4.We now consider when a boundary component E ∈ C ( X ) corresponds to an end.For subspaces of S , this is always the case. The key tool in the proof of this is thefollowing purely topological fact, which is mentioned in the proof of [5, Lemma 2.5]. Proposition 4.6.
Each domain Ω in S may be written as a union of open andconnected subsets Ω ⊆ Ω ⊆ . . . of Ω such that for each i ∈ N , the closure of Ω i is a compact subset of Ω and ∂ Ω i is a finite collection of pairwise disjoint Jordancurves. UASISYMMETRIC KOEBE UNIFORMIZATION 11
Proposition 4.7.
Let Ω be a domain in S . If x and y are points in the samecomponent of ∂ Ω , then any Cauchy sequences representing x and y are in U (Ω) and represent the same end of Ω .Proof. The metric boundary ∂ Ω coincides with the usual topological boundary ofΩ in S . Let { Ω i } denote the exhaustion of Ω provided by Proposition 4.6. Since x and y are in the same component of ∂ Ω, for each i ∈ N they belong to a singlesimply connected component U i of S \ Ω i . Let { x i } and { y i } be Cauchy sequencesrepresenting x and y respectively.Suppose that K is a compact subset of Ω. We may find i ∈ N so large thatΩ i ⊇ K . Then U i ∩ Ω does not intersect K . Moreover, there is a number N ∈ N such that ( { x i } i ≥ N ∪ { y i } i ≥ N ) ⊆ U i ∩ Ω . It now suffices to show that U i ∩ Ω is connected. Let a and b be points in U i ∩ Ω.We may find j ≥ i so large that a and b are contained in Ω j . The set U i ∩ Ω j is a simply connected domain with finitely many disjoint closed topological disksremoved from its interior, and is therefore path-connected. Thus a and b may beconnected by a path inside U i ∩ Ω j , and hence inside U i ∩ Ω. (cid:3) The proof given above does not even pass to metric spaces that are merelyhomeomorphic to a domain in S , as it need not be the case that the completion ofsuch a space embeds topologically in S . However, under an additional assumptioncontrolling the geometry of X , we can give a different proof. Proposition 4.8.
Suppose that X satisfies the LLC condition. If x and y arepoints in the same component of ∂X , then any Cauchy sequences representing x and y are in U ( X ) and represent the same end of X . Lemma 4.9.
Suppose that X satisfies the λ - LLC condition for some λ ≥ . Let E be a connected subset of ∂X and let ǫ > . Then N X ( E, ǫ ) ∩ X is contained ina connected subset of N X ( E, λǫ ) ∩ X .Proof. It suffices to show that if x and y are points in N X ( E, ǫ ) ∩ X , then there isa continuum containing x to y inside of N X ( E, λǫ ) ∩ X . Let x ′ and y ′ be pointsin E such that d ( x, x ′ ) < ǫ and d ( y, y ′ ) < ǫ . By Lemma 4.3, there is an ǫ -chain x ′ = z ′ , . . . , z ′ n = y ′ in E . For each j = 1 , . . . , n −
1, find a point z j ∈ X such that d ( z j , z ′ j ) < ǫ . The triangle inequality implies that for j = 0 , . . . , n − z j +1 ∈ B X ( z j , ǫ ) . Repeatedly applying the λ -LLC condition and concatenating now yields the de-sired result. (cid:3) Proof of Proposition 4.8.
Let x and y be points in a connected subset E of ∂X , andlet { x i } and { y i } be Cauchy sequences in X corresponding to x and y , respectively.Let K be a compact subset of X . As X is compact, we may find ǫ > E, K ) > λǫ . Let N ∈ N be so large that { x i } i ≥ N ∪ { y i } i ≥ N ⊆ N X ( E, ǫ ) ∩ X. Lemma 4.9 now implies the desired results. (cid:3)
Remark . Proposition 4.8 is not true without some control on the geometryof X . The following example was pointed out to us by Daniel Meyer. Let ( r, θ, z )denote cylindrical coordinates on R , and set X = { ( r, θ, z ) : ( r − + z = 1 }\{ (0 , , } . Equipped with the standard metric inherited from R , the space X is homeomorphicto a punctured disk, and hence has two ends. However, the metric boundary of X consists only of the point { (0 , , } .The following statement is the main result of this section. We note that in thecase that X is a domain in S , the statement is mentioned in [13]. Theorem 4.11.
Suppose that X is either a domain in S or satisfies the LLC condition. Then the map Φ : E ( X ) → C ( X ) defined in Remark 4.5 is a homeomor-phism that is natural in the sense that a Cauchy sequence { x i } ∈ U ( X ) representsthe end E ∈ E ( X ) if and only if it represents a point on the boundary component φ ( E ) ∈ C ( X ) . In the proof of the following lemma, we consider only the case that X satisfiesthe LLC condition. If X is a domain in S , a proof is easily constructed usingProposition 4.6. Lemma 4.12.
Suppose that X is a domain in S or satisfies the LLC condition.Let K be a compact subset of X and let U be a component of X \ K . Then thefollowing statements hold: (i) for each x ∈ U ∩ ∂X , there is a number δ > such that B X ( x, δ ) ∩ X ⊆ U , (ii) if E ∈ C ( X ) intersects ∂U ∩ ∂X , then E ∈ C ( K, U ) , (iii) the set C ( K, U ) is open in C ( X ) .Proof. We assume that X satisfies the λ -LLC condition for some λ ≥ δ > a, b ∈ B X ( x, δ ) ∩ X such that a ∈ U and b is in some other compo-nent of X \ K . Using the λ -LLC condition to connect a to b inside of B X ( a, λδ )produces a point of K in the ball B X ( x, λδ ). Letting δ tend to 0 produces acontradiction with the assumption that K is a compact subset of X .Statement (i) implies that collection { E ∩ ∂V ∩ ∂X : V is a component of X \ K } consists of pairwise disjoint open subsets of E . Hence the connectedness of E provesstatement (ii).Now, recall that C ( X ) is endowed with the quotient topology. Hence, by state-ment (ii), in order to show that C ( K, U ) is open in C ( X ), it suffices to show that S E ∈C ( K,U ) E is open in ∂X . This follows from statements (i) and (ii). (cid:3) Proof of Theorem 4.11.
Proposition 4.7 or 4.8 shows that Φ is injective. Given E ′ ∈ C ( X ) and a Cauchy sequence { x i } i ∈ N representing a point in E ′ , Proposition4.7 or 4.8 also state that { x i } i ∈ N is in U ( X ) and hence represents an end E ∈ E ( X ).By definition, this implies that Φ( E ) = E ′ , and so Φ is surjective.We now check that the bijection Φ is a homeomorphism. Since C ( X ) is compactand E ( X ) is Hausdorff, this is true if Φ − is continuous. Hence, by Lemma 4.12 (iii) UASISYMMETRIC KOEBE UNIFORMIZATION 13 and the defintion of the toplogy on E ( X ), it suffices to show that for any compactset K ⊆ X and any component U of X \ K ,Φ( E ( K, U )) = C ( K, U ) . Let E be an end in E ( K, U ). By definition, the limit of any Cauchy sequencerepresenting E lies in ∂U ∩ ∂X . Hence, Φ( E ) intersects ∂U ∩ ∂X , and so Lemma4.12 (ii) shows that Φ( E ) ∈ C ( K, U ). Now, let E ′ ∈ C ( K, U ) and choose a Cauchysequence { x i } i ∈ N representing a point in E ′ . Lemma 4.12 (i) implies that thereis N ∈ N such that { x i } i ≥ N is contained in U . Again, Proposition 4.7 or 4.8states that { x i } is in U ( X ) and hence represents an end E . Thus, by definition, E ∈ E ( K, U ) and Φ( E ) = E ′ . (cid:3) Rank.
We breifly recall the notion of rank as discussed in [13]. Let T be acountable, compact, and Hausdorff topological space. Set T = T , and for each n ≥
1, set T n , to be the set of non-isolated points of T n − , and endow T n withthe subspace topology. This process can be continued using transfinite inductionto define T α for each ordinal α , though we will not have need for this. For eachordinal α , the space T α is again countable, compact, and Hausdorff. By the Bairecategory theorem, there is a unique ordinal α such that T α is finite and non-empty;this ordinal is defined to be the rank of T .Let ( X, d ) be a metric space that is either a domain in S or satisfies the LLC -condition. By Theorem 4.11, the space of boundary components C ( X ) is homeomor-phic to the space of ends E ( X ). As mentioned above, the former is clearly compactand the latter clearly Hausdorff, hence both are compact Hausdorff spaces. Hence,if S is closed and countable subset of C ( X ), the rank of S is defined.5. Domains with totally disconnected complement
The simplest possible structure of a boundary component is that it consists of asingle point. The aim of this section is to show that if we are only concerned withthe topological properties, we may always assume this is the case.
Proposition 5.1.
Every domain in S is homeomorphic to a domain in S thathas totally disconnected complement. To prove Proposition 5.1, we employ the theory of decomposition spaces [9].A decomposition of a topological space S is simply a partition of S . The non-degenerate elements of a decomposition are those elements of the partition thatcontain at least two points. The decomposition space S/G associated to a decom-position G of a topological space S is the topological quotient of S obtained by,for each g ∈ G , identifying the points of g . A decomposition G is an upper semi-continuous decomposition if each element is compact, and given any g ∈ G andany open set U ⊆ S containing g , there is another open set V containing g withthe property that if g ′ ∈ G intersects V , then g ′ ⊆ U . If G is an upper semi-continuous decomposition of a separable metric space S , then S/G is a separableand metrizable space. [9, Proposition I.2.2].We will use one powerful theorem from classical decomposition space theory. Itidentifies decompositions of S that are homeomorphic to S itself [21]. Theorem 5.2 (Moore) . Suppose that G is an upper semi-continuous decompositionof S with the property that for each g ∈ G , both g and S \ g are connected. Then S/G is homeomorphic to S . We also employ a powerful theorem of classical complex analysis. It states thatany domain in S can be mapped conformally (and hence homeomorphically) to a slit domain , i.e., to a domain in S that is either complete, or whose complementarycomponents are points or compact horizontal line segments in R [11, V.2]. Theorem 5.3.
Any domain in S is conformally equivalent to a slit domain. Lemma 5.4. If Ω ⊆ S is a slit domain, then the components of R \ Ω containingat least two points form the non-degenerate elements of an upper semi-continuousdecomposition of S .Proof. Let G denote the decomposition of S whose non-degenerate elements arethose components of R \ Ω that contain at least two points. Let g ∈ G , and let U ⊆ S be an open set containing g . Without loss of generality we may assumethat g = [0 , × { } and that U is an open and bounded subset of R containing g .Set l = max { x : ( x, ∈ R \ U and x < } and r = min { x : ( x, ∈ R \ U and x > } . Since Ω ∩ U is open and g is a compact and connected subset of U , there are closed,non-degenerate intervals L, R ⊆ R such that( L × { } ) ⊆ (( l, × { } ) ∩ Ω ∩ U and ( R × { } ) ⊆ ((1 , r ) × { } ) ∩ Ω ∩ U. If G is not upper semi-continuous, then for every n ∈ N , we may find some horizontalline segment g n ∈ G and points ( x n , y n ) , ( x ′ n , y n ) ∈ R such that( x n , y n ) ∈ g n ∩ N ( g, /n ) and( x ′ n , y n ) ∈ g n ∩ ( S \ U ) . After passing to a subsequence, we may assume that ( x n , y n ) tends to a point of g .Moreover, passing to another subsequence if needed, we may assume that eitherlim sup n →∞ x ′ n ≤ l or lim inf n →∞ x ′ n ≥ r. We consider the latter case; a similar argument applies in the former. For suffi-ciently large n , the point x ′ n is greater than any point in R , while x n is less thanany point in R . By the connectedness of g n , we conclude that there is a point( z n , y n ) ∈ g n with z n ∈ R . After passing to yet another subsequence, we mayassume that ( z n , y n ) converges to a point in R × { } . This is a contradiction as Ωis open. See Figure 1. (cid:3) PSfrag replacements ( l, L g = [0 , × { } UR ( r, g n Figure 1.
If upper semi-continuity fails.
UASISYMMETRIC KOEBE UNIFORMIZATION 15
Proof of Proposition 5.1.
Let Ω be a domain in S . By Theorem 5.3, there is ahomeomorpism h : Ω → Ω , where Ω is a slit domain. By Lemma 5.4, the com-ponents of S \ Ω with at least two points form the non-degenerate elements of anupper semi-continuous decomposition G of S . As each element of G is either apoint or a compact line segment in R , the hypotheses of Theorem 5.2 are satisfied.Thus there is a homeomorphism h : S /G → S . Let π : S → S /G denote thestandard projection map. By definition π | Ω is a homeomorphism and π ( S \ Ω ) istotally disconnected. Thus h ◦ π ◦ h : Ω → S is a homeomorphism, and the imageof Ω under this map has totally disconnected complement. (cid:3) The ends of a domain in S with totally disconnected complement are particularlyeasy to understand: they are in bijection with the points of the complement, whichare precisely the boundary components. Proposition 5.5.
Suppose Ω is a domain in S with totally disconnected comple-ment. Then there is a homeomorphism φ : E (Ω) → ∂ Ω with the property that asequence { x n } ∈ U (Ω) represents the end E ∈ E (Ω) if and only if it converges to φ ( E ) .Proof. Since a totally disconnected subset of S cannot have interior, we see that S \ Ω = ∂ Ω . Moreover, it is clear that C (Ω) and ∂ Ω are naturally homeomorphic.Hence Theorem 4.11 provides the desired homeomorphism. (cid:3)
The following statement transfers the work of this section to the general setting.For the remainder of this section, we assume that (
X, d ) is a metric space that hascompact completion and is homeomorphic to a domain in S . Corollary 5.6.
Suppose that X satisfies the LLC condition or is a domain in S .Then there is a continuous surjection h : X → S such that h | X is a homeomorphismonto a domain Ω with totally disconnected complement. Moreover, the map h isconstant on each boundary component E ∈ C ( X ) , and for any ǫ > , there is ǫ ′ > such that (5.1) h − ( B S ( h ( E ) , ǫ ′ )) ⊆ N X ( E, ǫ ) . Finally, h induces a homeomorphism from C ( X ) to ∂ Ω .Proof. We address only the case that X satisfies the LLC condition. We haveassumed that X is homeomorphic to a domain in S . Hence by Proposition 5.1,there is a homeomorphism h : X → Ω, where Ω ⊆ S is a domain with totallydisconnected complement.By Theorem 4.11 and Remark 4.2, there is a homeomorphism φ : C ( X ) → E (Ω)with the property that a Cauchy sequence { x i } ∈ U ( X ) converges to a point of theboundary component E ∈ C ( X ) if and only if { h ( x i ) } represents the end φ ( E ).Moreover, Proposition 5.5 provides a homeomorphism φ : E (Ω) → ∂ Ω with theproperty that a sequence { y i } ∈ U ( X ) represents the end E ∈ E (Ω) if and onlyif it converges to the boundary point φ ( E ) ∈ ∂ Ω. We define the extension of h to ∂X by setting h ( x ) = φ ◦ φ ( E ), where E ∈ C ( X ) is the boundary componentcontaining x ∈ ∂X . The naturality properties of φ and φ ensure that h : X → S so defined is continuous. As h | X is a homeomorphism onto Ω and ∂ Ω = S \ Ω, thedefinitions show that the extended map is a surjection.Now, let E ∈ C ( X ) and ǫ >
0. By construction (or from the fact that h | X is ahomeomorphism and S \ Ω is totally disconnected), the set h ( E ) consists of a single point in ∂ Ω. Suppose that there is no ǫ ′ > x n ∈ X \N X ( E, ǫ )such that { h ( x n ) } converges to h ( E ) ∈ ∂ Ω. By compactness and the fact that h | X is a homeomorphism onto Ω, the sequence { x n } has a limit point x ∈ ∂X \N X ( E, ǫ ).This means that x lies in some boundary component F = E . However, the conti-nuity of h implies that h ( F ) = h ( x ) = h ( E ), which contradicts the fact that φ ◦ φ is injective.Corollary 4.11 implies that C ( X ) is naturally homeomorphic to E ( X ), and Re-mark 4.2 shows that E ( X ) is naturally homeomoprhic to E (Ω). Proposition 5.5 nowyields the final statement of the theorem. (cid:3) As an application of Proposition 5.1, we prove Proposition 3.7, which relates theALLC condition and the relative separation of boundary components, and Propo-sition 3.8, which improves the LLC condition to the ALLC condition.
Proof of Proposition 3.7.
Let Ω be a circle domain, and suppose that there is anumber c > { E i } i ∈ I of ∂ Ω satisfy(5.2) △ ( E i , E j ) ≥ c whenever i = j ∈ I . Fix Λ ≥ c − < −
1. We will show that Ωis Λ-ALLC. Let p ∈ Ω and r >
0. It suffices to show that A Ω ( p, r/ Λ , r )is connected.Let h : Ω → S be the continuous surjection provided by Corollary 5.6. Thecomplement of h (Ω) is a compact and totally disconnected set, and hence hastopological dimension 0 [17, Section II.4]. The definition of Λ and (5.2) guaranteethat there is at most one index i ∈ I such that E i intersects both B S ( p, r/ Λ)and S \ B ( p, r ) . This implies that h ( A Ω ( p, r/ Λ , r )) is the complement, in thedomain bounded by two Jordan curves that touch at no more than one point, of aset of topological dimension 0. It is therefore connected [17, Theorem IV.4]. SeeFigure 2.PSfrag replacements pr/ Λ 2Λ r h h ( p ) Figure 2.
The ALLC property of circle domains.We leave the converse statement as an exercise for the reader, as it is not neededin this paper. (cid:3)
UASISYMMETRIC KOEBE UNIFORMIZATION 17
In the proof of Proposition 3.8, we will use the following separation theorem ofpoint-set topology [17, Section V.9].
Theorem 5.7 (Janiszewski) . Suppose that A and B are closed subsets of S suchthat card( A ∩ B ) ≤ . If y and z are points of S that lie in the same component of S \ A and in the same component of S \ B , then y and z lie in the same componentof S \ ( A ∪ B ) . Lemma 5.8.
Let ( X, d ) be a metric space homeomorphic to a domain in S thatsatisfies the LLC -condition, and let A and B be disjoint closed subsets of X . Let C be the collection of components that intersect A ∩ B , and assume that card C ≤ . If u, v ∈ X are in the same component of X \ A and in the same component of X \ B ,then they are in the same component of X \ ( A ∪ B ) .Proof. Let h : X → S be the continuous surjection provided by Corollary 5.6. Inparticular h | X is a homeomorphism onto a domain Ω = S \ T , where T is a closedtotally disconnected set. Moreover, h maps each component of ∂X to a distinctpoint of T . Hence, our assumptions imply that h ( A ) and h ( B ) are closed subsetsof S such that h ( A ) ∩ h ( B ) is either empty or a single point of T . Set e A = h ( A ) and e B = h ( B ) ∪ T. Then e A and e B are again closed subsets of S whose intersection is either empty ora single point, and S \ e A ⊇ Ω \ h ( A ) and S \ e B = Ω \ h ( B ) . Thus, if u and v are points of X that are in the same component of X \ A and in thesame component of X \ B , then h ( u ) and h ( v ) are in the same component of S \ e A and in the same component of S \ e B . Janiszewski’s Theorem now implies that h ( u )and h ( v ) are in the same component of S \ ( e A ∪ e B ) = Ω \ ( h ( A ) ∪ h ( B )) . Since h | X is a homeomoprhism onto Ω, this yields the desired result. (cid:3) Proof of Proposition 3.8.
By Lemma 3.4, it suffices to consider a point x ∈ X andradius r >
0, and suppose that y and z are points of A ( x, r, r ). Denote by δ > ∂X , and setF s = diam X/δ . ByProposition 3.3 we may assume that X satisfies the λ - g LLC condition for some λ ≥ r < δ/ (4 λ ). Set A = S X ( x, λr ) and B = B X ( x, r/ (2 λ )) . The λ - g LLC-condition implies that y and z lie in the same component of X \ A (namely, the component containing x ) and in the same component of X \ B . Therestriction on r implies that there is at most one component E of ∂X that satisfies E ∩ ( A ∪ B ) = ∅ . Hence, by Lemma 5.8, the points y and z are in the same componentof X \ ( A ∪ B ). Since ( X, d ) is locally path connected, this implies that they arecontained in a continuum in A ( x, r/ (2 λ ) , λr ) . Since A ( x, r, r ) is empty if r > diam X , we may now assume that δ/ (4 λ ) ≤ r ≤ diam X. The ^ LLC condition provides an embedding γ : [0 , → X such that γ (0) = y , γ (1) = z , and im γ ⊆ B ( x, λr ). If there is no t ∈ [0 ,
1] such that γ ( t ) ∈ B ( x, r/ (16 λs )), then the proof is complete. Otherwise, let t = min { t ∈ [0 ,
1] : γ ( t ) ∈ S ( x, r/ (8 λs )) } ,t = max { t ∈ [0 ,
1] : γ ( t ) ∈ S ( x, r/ (8 λs )) } . Since γ is an embedding, X is connected, and y, z ∈ A ( x, r, r ), we see that t < t .As r/ (8 λs ) < δ/ (4 λ ), we may apply the first case considered above to the points γ ( t ) and γ ( t ), producing a continuum Γ ⊆ A ( x, r/ (16 λ s ) , r/ (4 s )) that contains γ ( t ) and γ ( t ). Now, the continuum γ ([0 , t ]) ∪ Γ ∪ γ ([ t , ⊆ A ( x, r/ (16 λ s ) , λr )contains y and z . (cid:3) Crosscuts
In this section, we assume that X is a metric space that has compact completion,is homeomorphic to a domain in S , and satisfies the λ -LLC condition for some λ ≥ crosscut is an embedding γ : [0 , → X such that γ ([0 , ∩ ∂X = γ (0) ∪ γ (1) . Note that if γ is a crosscut, then γ | (0 , is a proper embedding. If γ (0) and γ (1)are distinct points that belong to the same component E of ∂X , then we say that γ is an E -crosscut .The following proposition shows that under the assumption of the LLC condi-tion, E -crosscuts behave as they do in the case of a circle domain in S . Proposition 6.1.
Let E be a component of ∂X , and let γ be an E -crosscut. Then X \ im γ has precisely two components. Denoting these components by U and V , itholds that: (i) for every ǫ > , there is a closed set K ⊆ X with dist( E, K ) > suchthat U \ K is a connected subset of N X ( E, ǫ ) ∩ U , and the same statementis valid for V , (ii) the sets U \ im γ and V \ im γ are the components of X \ im γ , and U ∩ V =im γ . (iii) the sets U ∩ E and V ∩ E are connected.Proof. By Proposition 3.3, we may assume that X in fact satisfies the λ - g LLC condition.Let h : X → S be the continuous surjection provided by Corollary 5.6, and letΩ = h ( X ). Since h | X is a homeomorphism, the map h ◦ γ : (0 , ֒ → Ω is a properembedding. As h ( E ) is a single point and h is continuous,(6.1) lim t → h ◦ γ ( t ) = h ( E ) = lim t → h ◦ γ ( t ) . This implies that the continuous map h ◦ γ : [0 , → Ω ∪ { h ( E ) } defines a Jordancurve. The Jordan curve theorem states that S \ im( h ◦ γ ) consists of two disjointdomains e U and e V , each homeomorphic to R , that have common boundary im( h ◦ γ ).Then U := h − ( e U ∩ Ω) and V := h − ( e U ∩ Ω) are disjoint non-empty open setssatisfying U ∪ V = X \ im γ . As h | X is a homeomorphism, in order to show that U and V are the components of X \ im γ , we need only show e U ∩ Ω and e V ∩ Ω UASISYMMETRIC KOEBE UNIFORMIZATION 19 are connected. Since S \ Ω is compact and totally disconnected, it has topologicaldimension 0 [17, Section II.4]. Thus e U ∩ Ω homeomorphic to the complement in R of a set of topological dimension 0, and hence is connected [17, Theorem IV.4].The same proof applies to e V ∩ Ω.We procede to the proof of (i). Let ǫ >
0. By Corollary 5.6, we may find ǫ ′ > h − ( B S ( h ( E ) , ǫ ′ )) ⊆ N X ( E, ǫ ) . As im( h ◦ γ ) is a Jordan curve, by Schoenflies’ theorem there is a homeomorphism H : S → S such that H ( e U ) is a standard ball in S whose boundary contains thepoint H ◦ h ( E ). By the continuity of H − and (6.2) we may find ǫ ′′ > h − ◦ H − ( B S ( H ◦ h ( E ) , ǫ ′′ )) ⊆ N X ( E, ǫ ) . The set B S ( H ◦ h ( E ) , ǫ ′′ ) ∩ H ( e U )is the non-empty intersection of two standard balls in S and hence is itself home-omorphic to R . As before, [17, Section II.4 and Theorem IV.4] imply that B S ( H ◦ h ( E ) , ǫ ′′ ) ∩ H ( e U ) ∩ H (Ω)is connected. It now follows from (6.3), the fact that h | X is a homeomorphism ontoΩ, and the definition of U that h − ◦ H − ( B S ( H ◦ h ( E ) , ǫ ′′ )) ∩ U is a connected subset of N X ( E, ǫ ). We set K = X \ ( h − ◦ H − ( B S ( H ◦ h ( E ) , ǫ ′′ ))) . The continuity of H ◦ h now shows that dist( E, K ) >
0. An analogous proof appliesto V .We next address (ii). It follows from the definitions that U \ im γ and V \ im γ are non-empty, closed in X \ im γ , and satisfy( U \ im γ ) ∪ ( V \ im γ ) = X \ im γ. Moreover, as U \ im γ is the topological closure in X \ im γ of the connected set U ,it is connected. Similarly, V \ im γ is connected. Hence U \ im γ and V \ im γ are thecomponents of X \ im γ .As the common boundary of e U and e V is im h ◦ γ , we see that U ∩ V ⊇ im γ .Suppose there is a point z ∈ X \ im γ that is an accumulation point of both U and V .Let 0 < ǫ < dist( z, im γ ) /λ . By assumption we may find points u ∈ U ∩ B X ( z, ǫ )and v ∈ V ∩ B X ( z, ǫ ). The λ - g LLC condition now implies that u and v can beconnected in X \ im γ , a contradiction. Hence U ∩ V = im γ .To prove (iii), we show that U ∩ E is connected; the corresponding statementfor V is proven in the same way. If U ∩ E is not connected, then we may finddisjoint, non-empty, and compact sets A and B such that A ∪ B = U ∩ E . Fix0 < ǫ < dist( A, B ) /
2. We first claim that there is a number δ > N X ( E, δ ) ∩ U ⊆ N X ( A ∪ B, ǫ ) ∩ U. If this claim is false, then for every n ∈ N there are points u n ∈ U and x n ∈ E such that d ( x n , u n ) < /n and dist( u n , A ∪ B ) ≥ ǫ . Since E is compact, there is a subsequence of { x n } that converges to a point x ∈ E . It follows that thecorresponding subsequence of { u n } converges to x as well. Hence x ∈ U ∩ E butdist( x, A ∪ B ) ≥ ǫ , a contradiction. This proves the claim.By (i), there is a closed set K ⊆ X such that dist( E, K ) > U \ K is aconnected subset of N X ( E, δ ) ∩ U, and hence, by the claim, of N X ( A ∪ B, ǫ ) ∩ U. However, since dist(
E, K ) >
0, we may find points a ∈ N X ( A, ǫ ) ∩ ( U \ K ) and b ∈ N X ( B, ǫ ) ∩ ( U \ K ) . This is a contradiction since dist(
A, B ) > ǫ . (cid:3) Uniformization of the boundary components
In this section, we assume that X is a metric space that has compact completion,is homeomorphic to a domain in S , and satisfies the full λ - g LLC condition for some λ ≥ E ∈ C ( X ) with at least two points ishomeomorphic to S and satisfies the λ ′ -LLC condition, for some λ ′ ≥ λ . To do so, we employ a recognition theorem of point set topology: a metricspace is homeomorphic to S if and only if it is a locally connected continuum suchthat removal of any one point results in a connected space, while the removal of anytwo points results in a space that is not connected [25]. This section adapts [26,Section 4] to our setting; we omit certain proofs that need little or no translation. Proposition 7.1.
Let E be a component of ∂X , and let p ∈ E . Then E \{ p } isconnected.Proof. As a single point set and the empty set are connected, we may assume that E has at least three points. Let x and y be arbitrary distinct points of E \{ p } .It suffices to show that there is a connected subset of E \{ p } that contains x and y . By the g LLC -condition, there is an E -crosscut γ with γ (0) = x and γ (1) = y . By Proposition 6.1 (ii), there is a unique component U of X \ im γ such that U ∩ E does not contain the point p . Then x and y are contained U ∩ E , which byProposition 6.1 (iii) is a connected subset of E \{ p } . (cid:3) Proposition 7.2.
Let E be a component of ∂X with at least two points. If p, q ∈ E are distinct points, then E \{ p, q } is not connected.Proof. See [26, Proposition 4.11]. Here, Proposition 6.1 plays the role of [26, Lemma4.10]. (cid:3)
We now consider the local connectivity of boundary components of X . We willneed a few technical lemmas. Lemma 7.3.
Let E be a component of ∂X , and suppose that γ and γ ′ are E -crosscuts with the property that there is a compact interval I ⊆ (0 , such that γ ( t ) = γ ′ ( t ) for all t / ∈ I . Then there is a closed subset K ⊆ X with dist( E, K ) > such that if points p and q in X \ K are in a single component of X \ im γ , then theyare in a single component of X \ im γ ′ .Proof. See [26, Lemma 4.12]. (cid:3)
We will briefly need a notion of transversality. Let Y be a topological spacehomeomorphic to a domain in S , and let α, β : (0 , → Y be embeddings. Given UASISYMMETRIC KOEBE UNIFORMIZATION 21 x ∈ im α ∩ im β , we say that α and β intersect transversally at x if there is an openneighborhood U of x and a homeomorphism h : U → R such that h ( U ∩ im α ) isthe x -axis and h ( U ∩ im β ) is the y -axis. Lemma 7.4.
Let a , b , p , and q be distinct points on a Jordan curve α ⊆ S , andlet U ⊆ S be a simply connected domain with boundary α . If V is any open subsetof U containing α , then there are embeddings α ab : [0 , → V and α pq : [0 , → V connecting a to b and p to q respectively, such that either im α ab ∩ im α pq = ∅ , or α ab and α pq have a single intersection, and that intersection is in U and is transverse.Proof. By the Sch¨onflies theorem, there is a homeomorphism H : S → S suchthat H ( α ) is a round circle. Then H ( V ) is open in H ( U ) and it contains a roundannulus that has H ( α ) as a boundary component. Clearly H ( a ), H ( b ), H ( p ), and H ( q ) may be connected as desired inside this annulus. Taking inverse images under H now yields the desired result. (cid:3) Lemma 7.5.
Let a , b , p , and q be distict points of a component E of ∂X , and let γ ab and γ pq be E -crosscuts connecting a to b and p to q , respectively. If p and q arecontained in a single component of X \ γ ab , then a and b are contained in a singlecomponent of X \ γ pq .Proof. The fact that the points a , b , p , and q are all distinct implies that K =im γ ab ∩ im γ pq is a compact subset of X . If K is empty, then γ ab connects a to b without intersecting γ pq , as desired. Hence we may assume that K = ∅ .By Corollary 5.6, there is a continuous surjection h : X → S where h | X is ahomeomorphism onto a domain Ω in S with totally disconnected complement.Let Ω ⊆ Ω ⊆ . . . be the exhaustion of Ω guaranteed by Proposition 4.6. Since a , b , p , and q are all elements of the same boundary component E , for each i ∈ N thereis a single simply connected component U i of S \ Ω i containing h ( { a, b, p, q } ). Fix i ∈ N so large that h ( K ) ⊆ Ω i . Then we may find parameters t , t , s , s ∈ (0 , t = min { t ∈ [0 ,
1] : h ◦ γ ab ( t ) ∈ ∂U i } ,t = max { t ∈ [0 ,
1] : h ◦ γ ab ( t ) ∈ ∂U i } ,s = min { s ∈ [0 ,
1] : h ◦ γ pq ( t ) ∈ ∂U i } ,s = max { s ∈ [0 ,
1] : h ◦ γ pq ( t ) ∈ ∂U i } . Since Ω i has only finitely many boundary components and h ( K ) is a compactsubset of Ω i , we may find a relatively open neighborhood V of ∂U i in S \ U i suchthat V ⊆ Ω i \ h ( K ). By Lemma 7.4, there are embeddings α ab : [0 , → V and α pq : [0 , → V connecting h ◦ γ ab ( t ) to h ◦ γ ab ( t ) and h ◦ γ pq ( s ) to h ◦ γ pq ( s )respectively, such that either im α ab ∩ im α pq = ∅ , or α ab and α pq have a singletransversal intersection.Let e γ ab be the path defined by concatenating γ ab | [0 ,t ] , h − ◦ α ab , and γ ab | [ t , . Similarly define e γ pq . Then either im e γ ab and im e γ pq are disjoint, or they have a singleintersection, and that intersection is transversal and located in X . In the formercase, a and b are in a single component of X \ e γ pq , and so Lemma 7.3 provides thedesired result. In the latter case, the transversality and Proposition 6.1 (ii) implythat p and q are not contained in a connected subset of X \ e γ ab . Lemma 7.3 showsthat this is a contradiction. (cid:3) Proposition 7.6.
Let E be a component of ∂X . Then E satisfies the λ - LLC condition. In particular, E is locally connected.Proof. We may assume that E has at least two points. Let p ∈ E , and r >
0. Itsuffices to find a continuum F such that B X ( p, r ) ∩ E ⊆ F ⊆ B X ( p, λ r ) ∩ E. As E itself is connected, we may assume that there is some point q ∈ E \ B X ( p, λ r ) . The g LLC condition provides an E -crosscut γ pq connecting p to q . Let U and V bethe components of X \ im γ pq , and set A = U ∩ E and B = V ∩ E . By Proposition6.1 (iii), A and B are connected. As { p, q } = A ∩ B and d ( p, q ) > λ r , we may finddistinct points a ∈ A and b ∈ B such that d ( p, a ) = 2 λ r and d ( p, b ) = 2 λ r . The λ - g LLC condition provides a crosscut γ ab connecting a to b with im γ ab ⊆ B X ( p, λ r ) . By Proposition 6.1 (ii), there is a unique component W of X \ im γ ab with p ∈ W ∩ E .Set F := W ∩ E . Applying Proposition 6.1 (iii) again, we see that the set F isconnected.We first show that F ⊆ B X ( p, λ r ) ∩ E. Suppose that there is a point x ∈ F \ B X ( p, λ r ). By the λ - g LLC condition, there is a path connecting x to q withoutintersecting B X ( p, λ r ). This implies that p and q are in the same componentof X \ γ ab . However, by Proposition 6.1 (ii), the points a and b lie in differentcomponents of X \ im γ pq . This contradicts Lemma 7.5.We now show that B X ( p, r ) ∩ E ⊆ F. Since γ ab is continuous, we may findparameters 0 < t a < < t b < γ ab ([0 , t a ])) ≤ λ r and diam( γ ab ([ t b , ≤ λ r. Set a ′ = γ ab ( t a ) and b ′ = γ ab ( t b ). Then a ′ , b ′ ∈ X \ B X ( p, λ r ) , and so the λ - g LLC condition provides an embedding γ a ′ b ′ : [0 , → X such that γ a ′ b ′ (0) = a ′ , γ a ′ b ′ (1) = b ′ , and im γ a ′ b ′ ⊆ X \ B X ( p, λr ) . The set S = γ ab ([0 , t a ]) ∪ im γ a ′ b ′ ∪ γ ab ([ t b , B X ( p, λr ) , and is the image of a path in X . Since the imageof a path in X is arc-connected, we may find a crosscut γ ′ connecting a to b withim γ ′ ⊆ S . Furthermore, we may find a compact interval I ⊆ (0 ,
1) such that γ ′ ( t ) = γ ab ( t ) for all t ∈ [0 , \ I . Suppose that there is a point x ∈ B X ( p, r ) ∩ E that is not contained in F . Then x and p are in different components of X \ im γ ab . By Lemma 7.3, this implies that x and p are in different components X \ im γ ′ .However, the λ - g LLC condition shows that x and p may be connected by an arccontained in B X ( p, λr ). This is a contradiction. (cid:3) Proof of Theorem 1.7.
We suppose that X is a metric space homeomorphic to adomain in S , has compact completion, and satisfies the λ -LLC condition, λ ≥ X in fact satisfies the λ ′ - g LLC condition for some λ ′ dependingonly on λ . Let E be a component of ∂X with at least two points. As X is compact, E is a continuum. Hence, Propositions 7.1, 7.2, and 7.6 along with the recognitiontheorem of [25] show that E is a topological circle. Proposition 7.6 shows that E is4 λ ′ -LLC . The desired statement now follows from [26, Proposition 4.15] and thecharacterization of quasicircles given in [24]. (cid:3) UASISYMMETRIC KOEBE UNIFORMIZATION 23
Remark . Let E be a component of ∂X , let γ be an E -crosscut, and let U and V be the components of X \ im γ . By Proposition 6.1 (ii) and (iii), the sets U ∩ E and V ∩ E are connected and have intersection { γ (0) , γ (1) } . Since Theorem 1.7 impliesthat E is a topological circle, we may conclude that ( U ∩ E ) \ im γ and ( U ∩ E ) \ im γ are the components of E \ im γ .8. Topological uniformization of the completion
In this section, in which the notation and assumptions are as in the previoussection, we give the following topological uniformization for the completion of X ,at least in the case of finitely many boundary components. Theorem 8.1.
Let N ∈ N , and assume that ∂X has N components, all of whichare non-trivial. Then the completion X is homeomorphic to the closure of a circledomain that has N boundary components, all of which are non-trivial.Remark . The homeomorphism type of a metric space X and the homeomor-phism type of the boundary ∂X do not, in general, determine the homemorphismtype of the completion X . The following example demonstrates this. The real pro-jective plane R P , formed by identifying antipodal points of S , can be metrizedas a subset of R endowed with the standard metric. Let X ⊆ R P ⊆ R be theimage of the sphere minus the equator under the indentification of antipodal points.Then X is homeomorphic to the disk, and ∂X is homeomorphic to the image ofthe equator under the indentification of antipodal points, i.e., it is homeomorphicto the circle. However, the completion X is homeomorphic to all of R P , and notthe closed disk. The LLC condition prevents this phenomena from occurring in thesetting we are most interested in.The proof of Theorem 8.1 relies on the following topological characterization ofthe closed disk, due to Zippin [28]. Here, given a topological space Y , an embedding α : [0 , → Y is said to span a subset J of Y if im α ∩ J = { α (0) , α (1) } and α (0) = α (1). Note that an embedding spans a component E of the boundary ∂X if and only if it is an E -crosscut. Theorem 8.3 (Zippin) . A locally connected and metrizable continuum Y is home-omorphic to the closed disk D if and only if Y contains a topological circle J withthe following properties • there is an embedding α : [0 , → Y that spans J , • for every embedding α : [0 , → Y spanning J , the set Y \ im α is not con-nected, • for every embedding α : [0 , → Y spanning J and every closed subset I ( [0 , , the set Y \ α ( I ) is connected.Remark . The homeomorphism constructed in the proof of Theorem 8.3 mapsthe Jordan curve J onto the unit circle. Lemma 8.5.
Let E ∈ C ( X ) be an isolated and non-trivial component of ∂X , andlet h : X → S be the continuous surjection provided by Corollary 5.6. Then for allsufficiently small ǫ > , the set U = h − ( B S ( h ( E ) , ǫ )) is homeomorphic to a closed annulus. Proof.
Corollary 5.6 states that h induces a homeomorphism from C ( X ) to ∂ ( h ( X )).Since E is isolated, it follows that U ∩ ∂X = E when ǫ is sufficiently small.Fix such an ǫ , and set p = h ( E ) ∈ S and α = h − ( S S ( p, ǫ )) . Then α is aJordan curve in X . Equip U a ( S \ B S ( p, ǫ ))with the disjoint union topology, and consider the quotient topological space Y = (cid:16) U a ( S \ B S ( p, ǫ )) (cid:17) /x ∼ h ( x ) , x ∈ α . The result will follow once it is shown that Y is homeomorphic to the closed disk,with π ( E ) corresponding the unit circle. To do so, we employ Theorem 8.3.Let π denote the usual projection map onto Y , and note that π | U and π | S \ B S ( p , ǫ ) are embeddings, and π | α is a two-to-one map. After some effort, it can be seenthat Y is compact, second-countable, and regular. The g LLC-condition on X alsoimplies that Y is connected and locally path-connected. By Urysohn’s metrizationtheorem, it is also metrizable. Moreover, as h | X is a homeomorphism, the set U \ E is homeomorphic to B S ( p, ǫ ) \{ p } . Thus, elementary point-set topology shows that Y \ π ( E ) is homeomorphic to S \{ p } , i.e., to the plane.By Theorem 1.7, the set π ( E ) is a topological circle in Y . The existence ofan embedding α : [0 , → Y that spans π ( E ) follows from the g LLC condition on X . We next check that any embedding α : [0 , → Y that spans π ( E ) disconnects Y . Towards a contradiction, suppose that Y \ α is connected. Since Y is locallypath-connected and Hausdorff, it follows that Y \ α is arc-connected.We assume that the image of α intersects the circle π ( α ); the argument in thecase that this does not occur is a simpler version of what follows. Set t = min { t ∈ [0 ,
1] : α ( t ) ∈ π ( α ) } and t = max { t ∈ [0 ,
1] : α ( t ) ∈ π ( α ) } . Then 0 < t ≤ t <
1. Denote p = α (0) ∈ π ( E ) ,q = α (1) ∈ π ( E ) ,p ′ = α ( t ) ∈ π ( α ) ,q ′ = α ( t ) ∈ π ( α ) , and let A be an arc of the topological circle π ( α ) with endpoints p ′ and q ′ . It ispossible that A is a single point. Let α ′ : [0 , → Y be an embedding such that α ′ ( t ) = α ( t ) when t ∈ [0 , t ] ∪ [ t , α ′ : [ t , t ] → Y is an embeddingparameterizing A . Then im( α ′ ) ⊆ π ( U ), and hence α ′ defines an E -crosscut e α in X .By Remark 7.7, we may find points x and y in E that are in different componentsof X \ e α . Since im α ∩ π ( E ) = π (im e α ∩ E ) , the points π ( x ) and π ( x ) are in Y \ α .Our assumption now provides an embedding γ : [0 , → Y \ α such that γ (0) = π ( x ) and γ (1) = π ( y ). Since Y \ π ( E ) is homeomorphic to the plane R , Sch¨onfliestheorem implies there is a homeomorphism Φ : Y \ π ( E ) → R that sends α | (0 , tothe real line. Since π ( x ) and π ( y ) are in π ( E ), the embedding Φ ◦ γ | (0 , is proper.Moreover, its image does not intersect the real line. Let B be a ball in R that UASISYMMETRIC KOEBE UNIFORMIZATION 25 contains the compact set Φ( π ( α )). As Φ ◦ γ | (0 , is proper, we may find points s , s ∈ (0 ,
1) such that if t / ∈ ( s , s ), then Φ ◦ γ ( t ) / ∈ B . From the geometry of R ,we see that there is a path β : [ s , s ] → R \ ( B ∪ R × { } ) . Then im β does not intersect im(Φ ◦ α ′ ) . Let γ ′ denote the concatenation of γ | [0 ,s ] ,Φ − ◦ β , and γ | [ s , . Then im γ ′ contains π ( x ) and π ( y ) and is contained in π ( U ) \ α ′ .It follows that im( π − ( γ ′ )) connects x to y inside of X \ e α , contradicting the defini-tion of x and y . See Figure 3.PSfrag replacements pq p ′ q ′ π ( x ) π ( y ) α π ( α ) π ( E ) γ A Bβ Φ Φ( γ )Φ( α ) Φ( A )Φ( p ′ ) Φ( q ′ ) Figure 3.
The proof of Lemma 8.5.Finally, we check that if I ( [0 ,
1] is closed, then α ( I ) does not separate the Y . Since Y is locally path connected, it suffices to show that any pair of points x, y ∈ Y \ π ( E ) can be connected without intersecting α ( I ). This follows easily fromthe Sch¨onflies theorem. (cid:3) We will also need the following well-known topological statement, the proof ofwhich is left to the reader.
Lemma 8.6.
Let ǫ > and p ∈ S . Given any homeomorphism φ : S S ( p, ǫ ) → S S ( p, ǫ ) , there is a homeomorphism Φ : A S ( p, ǫ, ǫ ) → A S ( p, ǫ, ǫ ) such that Φ | S S ( p,ǫ ) = φ and Φ | S S ( p, ǫ ) is the identity mapping. Given a metric space Y homeomorphic to a domain in S , we denote by N ( Y ) ⊆C ( Y ) the collection of non-trivial components of ∂Y , and by I ( Y ) ⊆ N ( Y ) thecollection of non-trivial components of ∂Y that are isolated as points in C ( Y ).Theorem 8.1 is a special case of the following result. Theorem 8.7.
Denote e X = X ∪ [ E ∈I ( X ) E . Then there is a circle domain Ω ′ ⊆ S and a homeomorphism e h : e X → Ω ′ ∪ [ F ∈N (Ω ′ ) F . Moreover, Ω ′ and e h may be chosen so that e h induces a homeomorphism from C ( X ) to C (Ω ′ ) .Proof. Let h : X → S denote the continuous surjection provided by Corollary 5.6.Let E ∈ I ( X ). By Lemma 8.5, there is a number ǫ E > h E : h − ( B S ( h ( E ) , ǫ E / → A S ( h ( E ) , ǫ E / , ǫ E / . Denote the domain of h E by U E , and set V E = h − ( B S ( h ( E ) , ǫ E )) . We may assumethat the Jordan curve β E := h − ( S S ( h ( E ) , ǫ E )) is mapped onto S S ( h ( E ) , ǫ E ).Moreover, we may choose the number ǫ E so small that V E ∩ ∂X = E, and that theresulting collection { V E } E ∈I ( X ) is pairwise disjoint.According to Lemma 8.6, for each E ∈ I ( X ), there is a homeomorphism Φ E of A S ( h ( E ) , ǫ E / , ǫ E ) to itself that agrees with h E ◦ h − on S S ( h ( E ) , ǫ E /
2) and isthe identity on S S ( h ( E ) , ǫ E ).Define Ω ′ = h ( X ) \ [ E ∈I ( X ) B S ( h ( E ) , ǫ E / . Then the collection of non-trivial boundary components of Ω ′ is given by N (Ω ′ ) = { S S ( h ( E ) , ǫ E / } E ∈I ( X ) . The map e h : e X → Ω ′ ∪ (cid:16)S F ∈N (Ω ′ ) F (cid:17) defined by e h ( x ) = h ( x ) x / ∈ S E ∈I ( X ) V E , Φ E ◦ h ( x ) x ∈ V E \ U E ,h E ( x ) x ∈ U E now yields the desired homeomorphism. See Figure 4.PSfrag replacements U E V E h E h h ( E ) h ( E ) ǫǫ/ ǫ/ E Figure 4.
Topological uniformization of the completion.The final assertion follows from the construction and the fact that there is anatural homeomorphism from C ( h ( X )) to C (Ω ′ ). (cid:3) UASISYMMETRIC KOEBE UNIFORMIZATION 27
9. ALLC and porous quasicircles
A subset Z of a metric space ( X, d ) is C -porous , C ≥
1, if for every z ∈ Z and0 < r ≤ diam X , there is a point x ∈ X such that B (cid:16) x, rC (cid:17) ⊆ B ( z, r ) \ Z. Porous subsets are small compared to the ambient space in a quantitative sense.See [10, Section 5.8] and [3, Lemma 3.12] for a proof of the following well-knowntheorem.
Theorem 9.1.
Let ( X, d ) be an Ahlfors Q -regular metric space, Q > , and let Γ be a subset of X . Then Γ is C -porous for some C ≥ if and only Γ is ( α, C ′ ) -homogeneous for some α < Q and C ′ ≥ , quantitatively. Theorem 9.2.
Let ( X, d ) be an ALLC metric space, and let Γ ⊆ X be a quasicircle.Then Γ is porous in X , quanitatively.Proof. We suppose that (
X, d ) is Λ-ALLC, Λ ≥
1, and that Γ ⊆ X is a Jordancurve satisfing the following condition: there is λ ≥ x and y on Γ diam( I ) ≤ λd ( x, y ) , where I is a component of Γ \{ x, y } of minimal diameter. This condition is quanti-tatively equivalent to the assumption that Γ is a quasicircle; see [24].Let z ∈ Γ and 0 < r ≤ diam X . We consider three cases. Case 1: < r < (diam Γ) / (4Λ) . We may find a point w ∈ Γ such that d ( z, w ) ≥ r . We may also find points u, v ∈ Γ such that { z, v, w, u } is cyclically orderedon Γ, d ( z, u ) = r/ (4Λ) = d ( z, v ), and if J ( z ) is the component of Γ \{ u, v } thatcontains z , then J ( z ) ⊆ B ( z, r/ (4Λ)) . The Λ-ALLC condition on the space X implies the existence of a continuum α ⊆ A ( z, r/ (8Λ ) , r/
2) that contains u and v . Let J ( u ) be the component ofΓ \{ z, w } containing u , and let I ( u ) = J ( u ) ∪ { z, w } . Define J ( v ) and I ( v ) similarly.We claim that dist( v, I ( u )) ≥ r/ (8 λ Λ) . If not, then the λ -three point conditionimplies that either z or w is within a distance of r/ (8Λ) of v , which is not the case.Now, the connectedness of α implies that there is a point x ∈ α such thatdist( x, I ( u )) = rs , where s = 8(2 λ + 1)Λ . Suppose that there is a point y ∈ I ( v ) such that d ( x, y ) < r/s . Then dist( y, I ( u )) < r/s , and hence the λ -three point condition implies that either z or w is within adistance 2 λr/s of y , and hencedist( x, { w, z } ) < λrs + rs = r . Combined with the facts that x ∈ α ⊆ A ( z, r/ (8Λ ) , r/
2) and d ( z, w ) ≥ r , thisyields a contradiction. Hence dist( x, Γ) ≥ r/s , and so the fact that d ( x, z ) < r/ B (cid:16) x, rs (cid:17) ⊆ B ( z, r ) \ Γ . Case 2: ≤ r ≤ diam X . We may find a point x ∈ X such that d ( x, z ) = r/
4. Since diam Γ ≤ r/ z ∈ Γ, we see thatdist( x, Γ) ≥ d ( x, z ) − diam Γ ≥ r/ . Hence B ( x, r/ ⊆ B ( z, r ) \ Γ . Case 3: diam Γ / (4Λ) ≤ r < . In this case, r < diam Γ4Λ . Thus, Case 1 implies that there is a point x ∈ X such that B (cid:16) x, r s (cid:17) ⊆ B (cid:16) z, r (cid:17) \ Γ ⊆ B ( z, r ) \ Γ . (cid:3) We combine the results of this section with those of the previous sections in thefollowing statement.
Corollary 9.3.
Let ( X, d ) be a doubling metric space that is homeomorphic to adomain in S , has compact completion, and satisfies the ALLC condition. Theneach component of the boundary ∂X is a porous subset of X , quantitatively.Proof. By Lemma 3.5, the space X is also LLC, quantitatively. Theorem 1.7 nowimplies that each component of the boundary ∂X is a quasicircle, quantitatively.Thus Theorem 9.2 yields the desired result. (cid:3) The following theorem states that, up to bi-Lipschitz equivalence, having As-souad dimension strictly less than 2 characterizes quasicircles in S among the classof all quasicircles [16]. Theorem 9.4 (Herron–Meyer) . Let (Γ , d Γ ) be a metric circle. The following state-ments are equivalent, quantitatively. • Γ is a quasicircle that is ( α, C ) -homogeneous for some ≤ α < and C ≥ . • Γ is bi-Lipschitz equivalent to a quasicircle in S .Remark . Combined with the classical theory of planar quasiconformal map-pings, Theorem 9.4 has the following consequence, which we will use in the proofof our main result. Let Γ be a quasicircle that is ( α, C )-homogeneous for some1 ≤ α < C ≥
1. Then there is a domain D Γ ⊆ S such that(i) D Γ is Ahlfors 2-regular,(ii) D Γ is quasisymmetrically equivalent to D ,(iii) the boundary ∂D Γ is bi-Lipschitz equivalent to Γ.10. Gluing
We now describe a process for gluing together metric spaces along bi-Lipschitzequivalent subsets. Parts of the basic construction maybe found in [8], and relateddeeper results are included in [12].For the remainder of this section, we let I be a possibly uncountable index set,which we extend by one symbol 0 to create the index set I = I ∪{ } . We consider acollection { ( X i , d i ) } i ∈ I of compact metric spaces and a pairwise disjoint collection { E i } i ∈ I of continua in X such that there is a number L ≥ i ∈ I , there exists an L -bi-Lipschitz homeomorphism f i : E i → f i ( E i ) ⊆ X i . UASISYMMETRIC KOEBE UNIFORMIZATION 29
The basic gluing construction.
We first consider the disjoint union e Z = a i ∈ I X i . We then consider the set Z obtained by gluing each space X i to X via f i , i.e., Z is the quotient of e Z by the equivalence relation ∼ generated by the condition thatfor all i ∈ I , if x ∈ E i , then x ∼ f i ( x ).We wish to define a natural metric on Z . To do so, we define an auxilliarydistance function for points z, w ∈ e Z by e d ( z, w ) = ( d i ( z, w ) z, w ∈ X i , ∞ otherwise . We now define a distance function d on Z by setting, for all equivalence classes a, b ∈ Z , d ( a, b ) = inf n X k =1 e d ( z k , z ′ k ) , where the infimum is taken over all sequences z = z , z ′ , . . . , z n , z ′ n in e Z such that z ∈ a , z ′ n ∈ b , and if n >
1, then π ( z ′ k ) = π ( z k +1 ) for all i = 1 , . . . , n −
1. We saythat such a sequence z is an admissible sequence from a to b if the e d -length of z l e d ( z ) = n X k =1 e d ( z k , z ′ k )is finite, and if for any k = 1 , . . . , n − z ′ k = z k +1 . The triangle inequality implies that the infimum in the definition of d ( a, b ) may betaken over all admissible sequences. Proposition 10.1.
The distance function d is a metric on the set Z , and if points a and b of Z have representatives z a and z b satisfying e d ( z a , z b ) < ∞ , then (10.2) e d ( z a , z b ) L ≤ d ( a, b ) ≤ e d ( z a , z b ) . Proof.
Let a, b ∈ Z . The definitions quickly imply that d ( a, b ) = d ( b, a ), that d ( a, a ) = 0, and that d satisfies the triangle inequality.Before showing that d ( a, b ) = 0 implies that a = b , we prove (10.2). Let z a and z b be representatives of a and b respectively, such that e d ( z a , z b ) < ∞ . Thesecond inequality in (10.2) follows from the fact that z a , z b is an admissible sequenceconnecting a to b . Towards a proof of the first inequality, let z = z , z ′ , . . . , z n , z ′ n be an admissible sequence connecting a to b . We consider only the case that z a and z b are in X ; the other cases are handled similarly. It suffices to assume that z = z a and z ′ n = z b and to show that l e d ( z ) ≥ e d ( z a , z b ) /L . For each k = 1 , . . . , n , let i k ∈ I be the index such that z k , z ′ k ∈ X i k . Since z is admissible, we may assumethat if i k = 0, then z k , z ′ k ∈ f i k ( E i k ) . Since each f i k is an L -bi-Lipschitz mapping, the triangle inequality implies that l e d ( z ) = n X k =1 d i k ( z k , z ′ k ) ≥ X { k : i k =0 } d ( z k , z ′ k ) + X { k : i k =0 } d ( f − i k ( z k ) , f − i k ( z ′ k )) L ≥ d ( z a , z b ) L = e d ( z a , z b ) L , as desired.Now, suppose that d ( a, b ) = 0. If there are representatives z a and z b of a and b respectively that satisfy e d ( z a , z b ) < ∞ , then (10.2) shows that z a = z b , and hence a = b . Suppose no such representatives exist. Then we may assume without loss ofgenerality that a has a representative z a ∈ X i \ f i ( E i ) for some i ∈ I , and that b hasno representative in X i . Then any admissible sequence from a to b has e d -length atleast dist d i ( z a , f i ( E i )). Since f i ( E i ) is compact, we conclude that d ( a, b ) is positive,a contradiction. (cid:3) Remark . The metric d defines a topology on Z , which we will refer to as themetric topology on Z . There is another natural topology on Z , which we will referto as the quotient topology on Z . It is obtained as follows. First, we equip e Z withthe disjoint union topology, i.e., a set A ⊆ e Z is open if and only for each i ∈ I ,the set A ∩ X i is open in X i . Note that a e d -ball in e Z of finite radius is open. Thenwe consider the quotient topology on Z arising from the equivalence relation ∼ ,i.e., the maximal topology on Z in which the standard projection map π : e Z → Z continuous. It is not hard to check that every open set in the metric topologyon Z is open in the quotient topology on Z . Moreover, if card I < ∞ , then thetopologies coincide. Simple examples show that the topologies may differ if this isnot the case.The following proposition states that away from the gluing sets, the space Z islocally isometric to e Z . Proposition 10.3.
Let r > , and suppose that a ∈ Z satifies (10.3) dist d a, π [ i ∈ I E i !! ≥ r. Then there is unique representative z b of each b ∈ B d ( a, r ) , and π − : B d ( a, r ) → B e d ( z a , r ) is a well-defined bijective isometry.Proof. Let b and c be points of B d ( a, r ). Then Proposition 10.1 and (10.3) implythat if z b and z c are representatives of b and c respectively, then(10.4) dist e d { z b , z c } , [ i ∈ I ( f i ( E i ) ∪ E i ) ! ≥ r. This immediately implies that b and c have unique representatives z b and z c , re-spectively. Since d ( b, c ) < r , there is admissible sequence z from b to c that has e d -length less than 2 r . The conditions (10.1) and (10.4) now imply that z = z b , z c , UASISYMMETRIC KOEBE UNIFORMIZATION 31 and hence that d ( b, c ) = e d ( z b , z c ). A similar argument shows that z b and z c are in B e d ( z a , r ). These facts together imply the desired statement. (cid:3) Preservation of the
ALLC condition.
In this and the following subsection,we make the following assumption on the spaces in the collection { X i } i ∈ I .(A) there is a constant C ≥ d i ( X i ) ≤ C diam d i f i ( E i ) for all i ∈ I .Heuristically, this assumption means that the spaces ( X i , d i ) are “flat”.We now show that if each space in the collection { X i } i ∈ I satisfies the ALLCcondition with a uniform constant, then the glued space ( Z, d ) also satisfies theALLC condition, quantitatively.
Lemma 10.4.
Suppose that there is a constant λ ≥ such that for each i ∈ I ,the space ( X i , d i ) is λ - ALLC . Then there is a quantity Λ ≥ , depending only onthe data, with the following property. Let i ∈ I , a ∈ Z , and r > . Then at leastone of the following two statements holds: (i) the annulus A d ( a, r, r ) is contained in π ( X i ) , (ii) there is a point b ′ ∈ A d ( a, r/ Λ , r ) ∩ π ( E i ) such that for each point b ∈ A d ( a, r, r ) ∩ π ( X i ) , there is a continuum E containing b and b ′ satisfying E ⊆ A d ( a, r/ Λ , r ) ∩ π ( X i ) . Proof.
We first assume that a / ∈ π ( X i ), and will show that the second statementabove holds. We consider two sub-cases. Case 1: r > LCλ dist d ( a, π ( E i )) . Let a ′ be a point of π ( E i ) such that d ( a, a ′ ) < r LCλ .
The triangle inequality implies that(10.5) A d ( a, r, r ) ⊆ A d (cid:16) a ′ , r , r (cid:17) . If diam d i f i ( E i ) < r/ (2 C ), then Proposition 10.1 and condition (A) show that π ( X i ) ⊆ B d (cid:0) a ′ , r (cid:1) . By (10.5), this now implies that A d ( a, r, r ) ∩ π ( X i ) is emptyand hence the claim is vacuously true. Thus we may assume that(10.6) diam d i f i ( E i ) ≥ r C .
Let z a ′ be the representative of a ′ in f i ( E i ). The connectedness of f i ( E i ) and (10.6)imply that there is a point z b ′ ∈ f i ( E i ) such that d i ( z a ′ , z b ′ ) = r C .
Set b ′ = π ( z b ′ ) . Let b ∈ A d ( a, r, r ), and denote by z b the representative of b ∈ X i .Proposition 10.1, the above equality, and (10.5) show that z b , z b ′ ∈ A d i (cid:16) z a ′ , r C , Lr (cid:17) . The λ -ALLC condition in X i now provides a continuum E ′ containing z b and z ′ b such that E ′ ⊆ A d i (cid:16) z a ′ , r Cλ , Lλr (cid:17) . Proposition 10.1 shows that π ( E ′ ) ⊆ A d (cid:16) a ′ , r CLλ , Lλr (cid:17) . The triangle inequality now shows that π ( E ′ ) ⊆ A d (cid:16) a, r CLλ , (3 Lλ + 1) r (cid:17) . Clearly π ( E ′ ) is a continuum containing b and b ′ . Case 2: r ≤ LCλ dist d ( a, π ( E i )). We may assume that there is a point b ∈ A d ( a, r, r ) ∩ π ( X i ), for otherwise statement (ii) above is vacuously true. Thedefinition of an admissible chain shows that this implies the existence of a point b ′ ∈ π ( E i ) satisfying d ( a, b ′ ) < r and d ( b, b ′ ) < r . Given a point b ∈ A d ( a, r, r ) ∩ π ( X i ), Proposition 10.1 shows that we may find representatives z b and z b ′ in X i of b and b ′ respectively such that d i ( z b , z b ′ ) < Lr . By Lemma 3.5, the λ -ALLCcondition in X i provides a continuum E ′ ⊆ B d i ( z b , Lλr ) that connects b and b ′ .Proposition 10.1 implies that π ( E ′ ) ⊆ B d ( b, Lλr ) ⊆ B d ( a, (4 Lλ + 2) r ) . The restriction that r ≤ LCλ dist d ( a, π ( E i )) implies that B d ( a, r/ (32 LCλ ) doesnot intersect π ( E i ). The definition of an admissible sequence now shows that B d ( a, r/ (32 LCλ ) does not intersect π ( X i ). Thus π ( E ′ ) is a continuum contain-ing b and b ′ and satisfying π ( E ′ ) ⊆ A d (cid:16) a, r LCλ , (4 Lλ + 2) r (cid:17) . Now, we assume that a ∈ π ( X i ), that the first statement above does not hold,and that the second statement above is not trivially true. That is, we assume that A d ( a, r, r ) intersects both π ( X i ) and Z \ π ( X i ).We claim that A d ( a, r/ (4 LC ) , r ) intersects π ( E i ). Since π ( E i ) is connected, ifthis is not the case, then either π ( E i ) ⊆ B d ( a, r/ (4 LC )) or π ( E i ) ⊆ Z \ B d ( a, r ).If the first possibility occurs, then condition (A) yields a contradiction with theassumptions that A d ( a, r, r ) meets π ( X i ) and that a ∈ π ( X i ). If the secondpossibility occurs, then the assumption that A d ( a, r, r ) meets Z \ π ( X i ) and thedefinition of admissible chain yield a contradiction.Thus we may find a point b ′ ∈ A d ( a, r/ (4 LC ) , r ) ∩ π ( E i ). That this pointsatisfies the requirements of the second statement of lemma is left to the reader. (cid:3) Lemma 10.5.
Suppose that there is a constant λ ≥ such that for each i ∈ I ,the space ( X i , d i ) is λ - ALLC . There is a quantity Λ ≥ , depending only on thedata, with the following property. Let a ∈ Z and r > . If u and v are points in A d ( a, r, r ) ∩ π ( X ) , then there is a continuum E containing u and v satisfying E ⊆ A d (cid:16) a, r Λ , r (cid:17) . Proof.
We claim that Λ = 4 Lλ + 2 fulfills the requirements of the lemma. If a ∈ π ( X ), this follows from Proposition 10.1 and the λ -ALLC condition on X ;the details are left to the reader. Hence we assume that a / ∈ π ( X ), and set s = Lλ + 12 . First, we consider the case that r ≤ s dist d ( a, π ( X )). Then B d ( a, r/s ) does notintersect π ( X ), and so the existence of the desired continuum follows from Propo-sition 10.1 and Lemma 3.5; again, the details are left to the reader. UASISYMMETRIC KOEBE UNIFORMIZATION 33
Now suppose that r > s dist d ( a, π ( X )). Then there is a point a ′ ∈ π ( X ) suchthat d ( a, a ′ ) = dist d ( a, π ( X )) . The triangle inequality yields A d ( a, r, r ) ⊆ A d (cid:18) a ′ , ( s − rs , (2 s + 1) rs (cid:19) . Let z u , z v , and z a ′ be representatives of u , v , a ′ , respectively, that are containedin X . Proposition 10.1 and the λ -ALLC condition in X imply that there is acontinuum E ′ ⊆ X containing z u and z v satisfying E ′ ⊆ A d (cid:18) z a ′ , ( s − rλs , (2 s + 1) Lλrs (cid:19) . Proposition 10.1, the triangle inequality, and the definition of s now show that π ( E ′ ) ⊆ A d (cid:18) a ′ , ( s − rLλs , (2 s + 1) Lλrs (cid:19) ⊆ A d (cid:16) a, r Lλ , (2 Lλ + 2) r (cid:17) , proving the claim in this case as well. (cid:3) Theorem 10.6.
Suppose that there is a constant λ ≥ such that for each i ∈ I ,the space ( X i , d i ) is λ - ALLC . Then ( Z, d ) is Λ - ALLC , where Λ ≥ depends onlyon the data.Proof. Let a ∈ Z and r >
0. We show that each pair of points u, v ∈ A d ( a, r, r ) iscontained in continuum E ⊆ Z satifying E ⊆ A d (cid:16) a, r Λ , r (cid:17) , where Λ ≥ z u ∈ X i u and z v ∈ X i v of u and v , respectively. If i u = i v ∈ I , then Lemma 10.4 provides the desired continuum. If i u = i v = 0,then Lemma 10.5 provides the desired continuum. If i u = i v and neither are 0,we employ Lemma 10.4. If the first possibility in Lemma 10.4 holds, the desiredcontinuum is easily constructed using Proposition 10.1. If the second possibilityholds, we are provided with continua E u and E v that connect u and v to points u ′ ∈ π ( X ) and v ′ ∈ π ( X ) respectively, where E u ∪ E v ⊆ A d ( a, r/ Λ , r ) . If i u = 0, then we instead set u ′ = u , and similarly define v ′ = v if i v = 0.Applying Lemma 10.5 to u ′ and v ′ and concatenating now produces the desiredcontinuum. (cid:3) Preservation of Ahlfors regularity.
In this subsection only, we add tocondition (A) two assumptions on the geometry of the base space X . First, weassume the uniform relative separation of the gluing sets:(B) there is a constant c > △ ( E i , E j ) ≥ c for all i = j ∈ I .We also assume, without loss of generality, that c ≤ k ≥
0, set n k := sup card (cid:26) i ∈ I : E i ∩ B d ( z, r ) = ∅ and 2 − k < diam d ( E i ) r ≤ − k +1 (cid:27) , where the supremum is taken over all z ∈ X and 0 < r ≤ d X . We assumethat there are numbers Q > ≤ M ≤ ∞ such that(C) P k ∈ N n k (2 − k ) Q ≤ M. Theorem 10.7.
Suppose that there is a constant K ≥ such that for each i ∈ I ,the space ( X i , d i ) is Ahlfors Q -regular with constant K . Then ( Z, d ) is Ahlfors Q -regular, quantitatively.Proof. As noted in the definition of Ahlfors regularity, it suffices to show that for a ∈ Z and 0 < r ≤ d Z, (10.7) H Q ( Z,d ) ( B d ( a, r )) ≃ r Q . First suppose that a satisfiesdist d a, π [ i ∈ I E i !! ≥ r. Then Proposition 10.3 implies that H Q ( Z,d ) ( B d ( a, r )) = H Q ( e Z, e d ) ( B e d ( z a , r )) = H Q ( X i ,d i ) ( B d i ( z a , r )) , where z a ∈ X i is the unique representative of a . It also follows that r can be nolarger than twice the diameter of ( X i , d i ). Since ( X i , d i ) is Ahlfors Q -regular, thedesired estimate follows.Next we suppose that a ∈ π ( E i ) for some i ∈ I . Let z a ∈ E i and z a = f i ( z a ) ∈ f i ( E i ) ⊆ X i be the representatives of a . Set B = B d ( a, r ) ∩ π ( X ). Proposition10.1 implies that π | X is an L -bi-Lipschitz mapping, and that π (cid:0) B e d ( z a , r ) (cid:1) ⊆ B ⊆ π (cid:0) B e d ( z a , Lr ) (cid:1) . Note that by the triangle inequality, Proposition 10.1, and condition (A),diam d Z ≃ diam d X . Hence, we may apply the Ahlfors Q -regularity of ( X , d ) to see that(10.8) H Q ( Z,d ) ( B ) ≃ H Q ( X ,d ) ( B d ( z a , r )) ≃ r Q . Thus H Q ( Z,d ) ( B d ( a, r )) & r Q . We now work towards an upper bound for H Q ( Z,d ) ( B d ( a, r )). For j ∈ I set B j = B d ( a, r ) ∩ π ( X j ) , and let J ⊆ I be the set of indices such that B j = ∅ .Furthermore, for k ∈ Z , define J k = (cid:26) j ∈ J : 2 − k < diam d ( E j ) Lc − r ≤ − k +1 (cid:27) . Now, we may write B d ( a, r ) = B ∪ [ k ∈ Z [ j ∈ J k B j . By definition, B j ⊆ π ( X j ) for any j ∈ J . As before, by Proposition 10.1, π | X j isan L -bi-Lipschitz mapping. Thus the Ahlfors Q -regularity of each X j and condition(A) imply that(10.9) H Q ( Z,d ) ( B j ) ≤ H Q ( Z,d ) ( π ( X j )) ≃ H Q ( X j ,d j ) ( X j ) ≃ (diam d E j ) Q . UASISYMMETRIC KOEBE UNIFORMIZATION 35
Fix j ∈ J . We claim that E j ∩ B d ( z a , Lr ) = ∅ . By definition, we may find apoint b ∈ π ( X j ) and an admissible chain z from a to b of e d -length less than r . Itfollows from the definitions that there is an admissible sub-chain z ′ connecting a to a point b ′ ∈ π ( E j ), and the e d -length of z ′ is also less than r . Let z b ′ be therepresentative of b ′ in E j . By Proposition 10.1, it holds that d ( z a , z b ′ ) < Lr, proving the claim. Since we have assumed that c ≤
1, the claim implies thatcard J k ≤ n k for each integer k ≥ j ∈ J with theproperty that diam d E j > Lc − r , and hencecard [ k ≤− J k ≤ . Suppose j ∈ J is an index with the above property, and let a j be a point of B j .Then by the triangle inequality, B j ⊆ B d ( a j , r ) ∩ π ( X j ) . Proposition 10.1 implies that π | X j is an L -bi-Lipschitz mapping onto π ( X j ).Hence the Ahlfors Q -regularity of X j implies that(10.10) H Q ( Z,d ) ( B j ) ≤ H Q ( Z,d ) ( B d ( a j , r ) ∩ π ( X j )) . r Q . Thus, inequalities (10.8), (10.9), and (10.10), along with condition (C), implythat H Q ( Z,d ) ( B d ( a, r )) ≤ H Q ( Z,d ) ( B ) + X k ∈ Z X j ∈ J k H Q ( B j ) . r Q + ∞ X k =0 X j ∈ J k diam d ( E j ) Q . r Q + r Q ∞ X k =0 card( J k )(2 − k ) Q ! . r Q , as desired. Note that this upper bound is also valid when r > d Z .Finally, we consider the full case thatdist d a, π [ i ∈ I E i !! < r. We may find an index i ∈ I and a point b ∈ π ( E i ) such that d ( a, b ) < r. Thusthe triangle inequality and the previous case show that H Q ( Z,d ) ( B d ( a, r )) ≤ H Q ( Z,d ) ( B d ( b, r )) . r Q . To get the desired lower bound, we consider two subcases. If(10.11) dist d a, π [ i ∈ I E i !! < r/ , then, as above, we may find an index i ∈ I and a point b ∈ π ( E i ) such that d ( a, b ) < r/
2. The triangle inequality and the previous case now show that H Q ( Z,d ) ( B d ( a, r )) ≥ H Q ( Z,d ) ( B d ( b, r/ & r Q . If (10.11) does not hold, then setting r ′ = r/
6, we see thatdist d a, π [ i ∈ I E i !! ≥ r ′ , and we may apply the first case considered in the proof to conclude that H Q ( Z,d ) ( B d ( a, r )) ≥ H Q ( Z,d ) ( B d ( a, r ′ )) & ( r ′ ) Q ≃ r Q , as desired. (cid:3) putting it together In this section, we synthesize the results of the previous sections to produce aproof of our main result. We begin by setting up an induction.Let Y be a metric space. Given a subset S of C ( Y ), denote by cl S the topo-logical closure of S in C ( Y ). Let N ( Y ) ⊆ C ( Y ) denote the collection of non-trivialcomponents of ∂X , and let I ( Y ) denote the points of N ( Y ) that are isolated pointsof C ( Y ). Lemma 11.1.
Let ( X, d ) be a metric space, homeomorphic to a domain in S , suchthat conditions (1) - (5) of Theorem 1.4 hold. Then ( X, d ) bi-Lipschitzly embeds intoa metric space ( Z, d Z ) that is homeomorphic to a domain in S , satisfies condi-tions (1) - (5) of Theorem 1.4 quantitatively, and such that C ( Z ) is homeomorphicto C ( X ) \I ( X ) .Proof. We leave the verification of the quantitativeness of the statement to thereader, as it follows easily from the quantitativeness of the results proven thus far.Denote I ( X ) = { E i } i ∈ I . Since ( X, d ) has compact completion, the index set I has cardinality no larger than countably infinite.Fix i ∈ I . Since X is Ahlfors 2-regular, it is doubling. Lemma 3.5 implies that X is LLC. Hence, Theorem 1.7 and Corollary 9.3 imply that E i is a quasicircle that isporous in X . Remark 9.5 provides an Ahlfors 2-regular quasidisk D i ⊆ S with theproperty that there is a bi-Lipschitz map f i : E i → ∂D i . It is easily seen by using[14, Proposition 10.10] that the ALLC property is preserved by quasisymmetricmappings. Hence, D i is ALLC. Note that none of the data of the conditionsdiscussed in this paragraph depend on i .We now apply the results of Section 10. Let X be the completion X , and foreach i ∈ I set X i = D i . We employ the bi-Lipschitz maps f i defined above asthe gluing maps f i : E i → f i ( E i ) ⊆ X i . The conditions stated at the beginning ofSection 10 are met by construction, and hence we may consider the resulting gluedmetric space (
Z, d Z ).We first show that Z is homeomorphic to a domain in S . The proof is similarin spirit to that of Theorem 8.7. Denote e X = X ∪ [ i ∈ I E i ! . UASISYMMETRIC KOEBE UNIFORMIZATION 37
By Theorem 8.7, there is a circle domain Ω ′ ⊆ S and a homeomorphism e h : e X → Ω ′ ∪ [ F ∈N (Ω ′ ) F . Denote the image of e h by f Ω ′ . Since Ω ′ is a circle domain, for each i ∈ I , we may write e h ( E i ) = S S ( p i , r i ) , where p i ∈ S and r i >
0. Since e h induces a homeomorphismfrom C ( X ) to C ( f Ω ′ ), for each i ∈ I there is a number ǫ i > B S ( p i , r i + ǫ i ) ∩ ∂ Ω ′ = S S ( p i , r i )and such that the resulting collection { B S ( p i , r i + ǫ i ) } i ∈ I is pairwise disjoint. More-over, the set Ψ = Ω ′ ∪ [ i ∈ I B S ( p i , r i ) ! is a domain in S .Proposition 10.1 implies that for each i ∈ I , there is a bi-Lipschitz homeomor-phism ι i : X i → π ( X i ). Fix i ∈ I . There is a homeomorphism g i : X i → B S ( p i , r i ).By Lemma 8.6, there is a homeomorphism Φ i of A S ( p i , r i , r i + ǫ i ) to itself such thatΦ i coincides with the identity on S S ( p i , r i + ǫ i ) and coincides with g i ◦ ι − i ◦ ι ◦ e h − on S S ( p i , r i ). Now, the map H : Z → Ψ defined by H ( a ) = e h ◦ ι − ( a ) a ∈ π ( X ) \ (cid:16)S i ∈ I ι ◦ e h − ( B S ( p i , r i + ǫ i )) (cid:17) , Φ i ◦ e h ◦ ι − ( a ) a ∈ ι ◦ e h − ( A S ( p i , r i , r i + ǫ i )) ,g i ◦ ι − i ( a ) a ∈ π ( X i ) , is the desired homeomorphism. See Figure 5.PSfrag replacements p i p i ǫ i + r i r i r i Φ i X X i π ( X ) π ( X i ) ι ι i g i E i e h Figure 5.
Filling in holes.We now verify that Z satisfies conditions (1)-(5) of Theorem 1.4. Condition (A),which was imposed at the beginning of Subsection 10.2, is equivalent to the assertionthat diam D i . diam ∂D i , which follows from the fact that D i is a quasidisk or from the fact that D i is planar. Conditions (B) and (C), which were imposed atthe beginning of Subsection 10.3, follow from the assumptions (5) and (2) in thestatement of Theorem 1.4.Hence, Theorems 10.6 and 10.7 imply that ( Z, d Z ) is ALLC and Ahlfors 2-regular. That Z satisfies the remaining conditions (2), (3), (5) of Theorem 1.4and that C ( Z ) is homeomorphic to C ( X ) \I ( X ) follow from the construction andProposition 10.1; we leave the details to the reader. (cid:3) Proof of Theorem 1.4.
The necessity of conditions (3)-(5) follows easily from thebasic properties of quasisymmetric mappings and Proposition 3.7.Now, let (
X, d ) be a metric space, homeomorphic to a domain in S , such thatthe closure of the non-isolated components of C ( X ) is countable and has finite rank,and such that conditions (1)-(5) hold. We will show that ( X, d ) is quasisymmet-rically equivalent to a circle domain whose collection of boundary components areuniformly relatively separated. Again, we leave the issue of quantitativeness to thereader.We first reduce to the case that cl N ( X ) = C ( X ). Let T = C ( X ) \ cl N ( X ).Consider the subspace e X of X defined by e X = X ∪ [ E ∈T E ! . Let h : X → S be the continuous surjection provided by Corollary 5.6. Since each E ∈ T is trivial, Corollary 5.6 implies that the map h | e X is a homeomorphism.Moreover, h induces a homeomorphism of C ( X ) onto the totally disconnected set S \ h ( X ). Hence, the set { h ( E ) } E ∈T is open in S \ h ( X ). It follows that theimage h ( e X ) is a domain in S . Remark 3.2 and Propostion 3.6 imply that e X is Ahlfors 2-regular and ALLC, quantitatively. Moreover, the space e X clearlysatisfies the remaining assumptions of Theorem 1.4, since C ( e X ) = C ( X ) \T andhence cl N ( e X ) = cl N ( X ) = C ( e X ) . Thus, if Theorem 1.4 is valid for spaces such that the non-trivial boundarycomponents are dense in the space of all boundary components, then applying thetheorem to e X and restricting the resulting quasisymmetric mapping to X provesthe theorem for X .We now assume without loss of generality that cl N ( X ) = C ( X ). Our assump-tions now imply that C ( X ) is countable and has finite rank, and we proceed byinduction on the rank. If the rank of C ( X ) is 0, then every boundary compo-nent is isolated. By Lemma 11.1, there is a bi-Lipschitz embedding ι : X ֒ → Z where ( Z, d Z ) is complete, homeomorphic to a domain in S , and satisfies condi-tions (1)-(5). Condition (3) implies that Z is compact, and hence homeomorphicto S . Lemma 3.5 implies that Z is LLC. Bonk and Kleiner’s uniformization re-sult, Theorem 1.1, now provides a quasisymmetric homeomoprhism f : Z → S . ByProposition 10.1, there is a bi-Lipschitz embedding ι : X ֒ → Z . The composition f ◦ ι is a quasisymmetric homeomorphism onto its image Ω := f ◦ ι ( X ). Thismapping extends to a quasisymmetric homeomorphism of X onto Ω [14, Propo-sition 10.10]. As discussed in the proof of Lemma 11.1, each of the componentsΓ , . . . , Γ N of ∂X is a quasicircle. Hence ∂ Ω consists of finitely many quasicircles
UASISYMMETRIC KOEBE UNIFORMIZATION 39 { f ◦ ι (Γ ) , . . . , f ◦ ι (Γ N ) } , and by Remark 3.1min i = j ∈{ ,...,N } △ ( f ◦ ι (Γ i ) , f ◦ ι (Γ j )) ≃ min i = j ∈{ ,...,N } △ (Γ i , Γ j ) . Thus Bonk’s uniformization result in S , Theorem 1.5, provides a quasisymmetrichomeomorphism g : S → S with the property that g ◦ f ◦ ι ( X ) is a circle domain.Again, the relative separation of the boundary components of this domain is con-trolled by Remark 3.1. This completes the proof in the case that the rank of C ( X )is 0.We now assume that the desired statement is true in the case that the rank of C ( X ) is an integer k ≥
1, and suppose that the rank of C ( X ) is k + 1. Theorem11.1 now states that X bi-Lipschitzly embedds into a metric space ( Z, d Z ) thatis homeomorphic to a domain in the sphere, satisfies conditions (1)-(5), and suchthat C ( Z ) has rank k . By induction, Z is quasisymmetrically equivalent to a circledomain. The remainder of the proof proceeds as in the base case. (cid:3) Proof of Theorem 1.3.
This follows from Proposition 3.8 and Theorem 1.4, afternoting that the minimal relative separation of components of the boundary is con-trolled by the ratio of the minimal distance between components of the boundaryto the diameter of the space. (cid:3)
Remark . We have defined a circle domain to be a subset of S ; one mayalso consider circle domains in R , which need not have compact completion. Ananalogous version of Theorem 1.4 for such domains can easily be derived fromTheorem 1.4 and the techniques of [26, Section 6].12. The counter-example
In this section, we prove Theorem 1.6.The desired space (
X, d ) is obtained as follows. First we define a sequence ofmultiply connected domains ( Q n ). Let Q denote the open unit square (0 , × (0 ,
1) in the plane. Let Q be the domain obtained by removing the vertical linesegment { / }× [1 / , /
4] from Q . We define Q n +1 by subdividing Q n into 2 n × n dyadic subsquares of equal size in the obvious way, and replacing each square inthe subdivision by a copy of Q that has been scaled by 1 / n . See Figure 6 for Q , Q , and Q . We denote by Q n the completion of Q n in the path metric d Q n induced from the plane. The induced metric on the completion is denoted by d Q n .The boundary components of Q n give rise to the metric boundary components of( Q n , d Q n ). The components of the metric boundary of ( Q n , d Q n ) that correspondto the boundary components of Q n other than the outer boundary will be calledthe slits of Q n .Next we define a planar domain Q inductively as follows. We start with Q andreplace the left-lower subsquare (0 , / × (0 , /
2) by a copy of Q that has beenscaled by a factor of 1 /
2. The resulting domain is denoted by R . Then we replace R ∩ ((0 , / × (0 , / Q that has been scaled by a factor of 1 / R . We continue in this fashion and at n thstep we replace R n − ∩ ((0 , / n ) × (0 , / n )) by a copy of Q n +1 that has beenscaled by a factor of 1 / n . The resulting domain is denoted by R n . The countablyconnected domain that results after infinitely many such replacements is denotedby Q , see Figure 7. Figure 6.
Domains Q , Q and Q . Figure 7.
Domain Q .The desired metric space ( X, d ) is the domain Q endowed with the path metric d Q induced from the plane. We denote by π X the projection of X onto Q , i.e., themap that identifies the points in X that correspond to the same point of Q . Themap π X is clearly 1-Lipschitz since the path metric on Q dominates the Euclideanmetric.Alternatively, the space X can be defined as an inverse limit of the sequence ofmetric spaces ( R n , p mn ) , m ≤ n . Here R n is the completion of R n in the pathmetric induced from the plane and p mn is the projection of R n onto R m thatidentifies the points of R n that correspond to the same point of R m . The map π X is then the natural projection of X onto Q . The extended metric d on X nowsatisfies the equation d ( p, q ) = lim d R n ( p n , q n ) , where { p n ∈ R n } and { q n ∈ R n } are sequences corresponding to p and q in theinverse limit system.To establish the desired properties of the space X , we consider the slit carpet S that has been studied in [20]. The space S is the inverse limit of the system( Q n , π mn ) , m ≤ n , where π mn is the projection of Q n onto Q m that identifies thepoints on the slits of Q n that correspond to the same point of Q m . See Figure 8.We make S into a metric space by endowing it with the metric d S ( p, q ) = lim d Q n ( p n , q n ) , UASISYMMETRIC KOEBE UNIFORMIZATION 41
Figure 8.
Slit carpet S .where { p n ∈ Q n } and { q n ∈ Q n } are sequences corresponding to p and q in theinverse limit system. The space S is a geodesic metric space which is a metricSierpi´nski carpet , i.e., a metric space homeomorphic to the well known standardSierpi´nski carpet, see [20, Lemma 2.1]. The inverse limits of slits of Q n form thefamily of peripheral circles of S , i.e., embedded simple closed curves whose removaldoes not separate S . The natural projection π S of S onto Q factors throughthe projection π S → X of S onto X , i.e., π S = π X ◦ π S → X . The projection π S → X is clearly 1-Lipschitz. Lemma 12.1.
There exists a constant c > such that for every p ∈ X and every ≤ r < X ) , there exists q ∈ Q with B Q ( q, c · r ) ⊆ π X ( B X ( p, r )) ⊆ B Q ( π X ( p ) , r ) . Proof.
Fix p ∈ X and 0 < r < X ). The second inclusion follows since π X is1-Lipschitz. To show the first inclusion, we use the corresponding property of S proved in [20]. Since π S → X is 1-Lipschitz, for any p ′ ∈ π − S → X ( p ) we have π S → X ( B S ( p ′ , r )) ⊆ B X ( p, r ) . By [20, Lemma 2.2], there exists c > p ′ as above, there is q ∈ Q with B Q ( q, c · r ) ⊆ π S ( B S ( p ′ , r )) . Combining these inclusions with the factorization of π S yields the desired inclusion. (cid:3) The following lemma implies that the Lipschitz map π X is David–Semmes reg-ular, see [10, Definition 12.1]. Lemma 12.2.
There exists C ≥ such that for every q ∈ Q and r > , thepreimage π − X ( B ( q, r )) can be covered by at most C balls in X of radii at most C · r .Proof. By Lemma 2.3 in [20], there exists C ≥ π − S ( B ( q, r )) can becovered by at most C balls with radii at most C · r . If ( B ( p ′ i , r i )) is a family of suchballs in S , then ( B ( π S → X ( p ′ i ) , r i )) is the desired family in X . (cid:3) Lemma 12.3.
The metric space X is homeomorphic to the planar domain Q andsatisfies conditions (1) and (3) - (5) of Theorem 1.4. Moreover, the rank of ∂X is .Proof. The first assertion of the lemma follows from the fact that Q is locallygeodesic. We leave it to the reader to verify that each non-trivial component of ∂X is a scaled copy of S , that the collection of components of ∂X is uniformlyrelatively separated, and that C ( X ) has a single limit point, which implies that therank of ∂X is 1. The completion X is compact since S is compact and π S → X iscontinuous.We now establish the Alhfors 2-regularity of X . Since the boundary of X haszero 2-measure, it is enough to show Ahlfors 2-regularity of the completion X . Let B X ( p, r ) be any ball with 0 < r < X ). Then, since π X is 1-Lipschitz, H ( B X ( p, r )) ≥ H ( π X ( B X ( p, r ))) , and by the first inclusion in Lemma 12.1, the right-hand side is at least r /C forsome C ≥ π X ( B X ( p, r )) by open balls ˜ B i of radii ˜ r i at most some δ > B X ( p, r ) by balls B j of radii r j at most C · δ with X j r j ≤ C X i ˜ r i . Since π X ( B X ( p, r )) is contained in the Euclidean ball of radius r , the Ahlfors reg-ularity now follows.We will check the ALLC condition in several steps. We first check the LLC condition. Let B X ( p, r ) be an arbitrary ball and let x, y ∈ B X ( p, r ). Since X isendowed with the path metric induced from the plane and d ( x, y ) < r , there is acurve γ in X that connects x and y and such that its length is at most 2 r . Thus E = γ is the desired continuum contained in B ( p, r ), i.e., X satisfies the 3-LLC condition.Now we show that X satisfies the LLC condition. Let B ( p, r ) be any ball in X and let x, y ∈ X \ B ( p, r ). We may assume that r ≤
1. Let v x denote a continuum in X that contains x and projects by π X one-to-one onto a closed vertical interval I x that satisfies the following properties. The end points of I x are π X ( x ) and π X ( x ′ )for some x ′ ∈ v x with π X ( x ′ ) contained in the boundary of Q , so that the lengthof I x is not larger than the Euclidean distance from π X ( p ) to the horizontal sideof the boundary of Q that contains π X ( x ′ ). We define v y and y ′ similarly. Itfollows from the choice of v x and v y that the distance from p to v x , respectively v y , is at least r/
2. Indeed, suppose the distance from p to, say, v x is less than r/
2. Then there exists x ′′ ∈ v x so that d X ( p, x ′′ ) < r/
2. Since d X is the inducedmetric on the completion of ( Q, d Q ) and d Q is the path metric induced from theplane, d X ( x, x ′′ ) equals the Euclidean distance between π X ( x ) and π X ( x ′′ ), and d X ( p, x ′′ ) is at least the Euclidean distance between π X ( p ) and π X ( x ′′ ). From thechoice of v x we conclude that d X ( x, x ′′ ) ≤ d X ( p, x ′′ ). The triangle inequality yieldsa contradiction.The points x ′ and y ′ are contained in some closed horizontal intervals h x and h y respectively, on the outer boundary of X (i.e., the metric boundary componentof ( Q, d Q ) that corresponds to the boundary of Q ) whose lengths are r/
4. The
UASISYMMETRIC KOEBE UNIFORMIZATION 43 distances from h x and h y to p are then at least r/
4. If v x ∪ v y ∪ h x ∪ h y is connected,we are done. Otherwise, let l and l ′ denote the two complementary componentsof h x ∪ h y in the outer boundary of X . We claim that at least one of l or l ′ isat a distance at least r/ p . Indeed, the sum of the distances from p to l and l ′ must be at least r/ Q separating π X ( v x ∪ h x ) from π X ( v y ∪ h y ) must be at least r/
4. Thus either v x ∪ v y ∪ h x ∪ h y ∪ l or v x ∪ v y ∪ h x ∪ h y ∪ l ′ is the desired continuum E in X in the LLC conditionwith λ = 8.The next step is to establish the ALLC property for X . Let A X ( p, r ) denote B X ( p, r ) \ B X ( p, r ) and let x, y ∈ A ( p, r ). We may assume that r ≤
1. If 1 / ≤ r ≤
1, then the continuum E found in the proof of the LLC condition works toconclude ALLC in this case, because the diameter of E is at most 2 and thus itis at most 4 r . If 0 < r < /
2, the proof of the existence of a desired continuumfollows the lines similar to those in the proof of LLC , but we first need to localizethat argument. Indeed, first we can find a unique n ∈ N such that 1 / ≤ n r < n is at least three. We consider thedyadic subdivision D of Q into squares of side length 4 / n . Let s be the interiorof a square in this subdivision. We denote by s X the preimage of s under π X . Letalso ∂s X denote the metric boundary of s X , i.e., the closure of s X in X less s X .Note that from the definitions of X and π X it follows that ∂s X is the union of fourclosed arcs, each isometric via π X to a side of the boundary of s .From the choice of n and the fact that π X is 1-Lipschitz, it follows that theprojection π X ( B X ( p, r )) can be covered by four squares from D . Moreover, theycan be chosen to be the first generation dyadic subsquares of a single square of sidelength 8 / n , not necessarily dyadic. Let F be the family of the interiors of thesefour squares and let K denote the closure in X of ∪ s ∈F ( s X ) . The set K is compact and it contains B X ( p, r ). The contour of K , denoted c ( K ),is the closure in X of the set of all points q such that q belongs to ∂s X for a unique s ∈ F . The contour c ( K ) is thus a union of closed arcs each of which is isometricto a horizontal or vertical side of the boundary of s for some s ∈ F . Since elementsof F are interiors of the first generation dyadic subsquares of a single square, it iseasy to see that c ( K ) is connected, and thus it is a continuum. Considering variouscombinatorial possibilities for c ( K ) one can easily conclude that c ( K ) does nothave global cut points. The rest of the proof of the ALLC property for X followsessentially the same lines as the proof of the LLC condition where the boundaryof Q should be replaced by c ( K ). The diameter of the resulting continuum E isat most 16 / n ≤ r .Finally, given a point p ∈ X , a radius r >
0, and points x, y ∈ A X ( p, r, r ),we modify the above continuum E to obtain one in X as follows. If E does notpass through the point p in X that projects to (0 , X , and thus E canbe modified slightly so that these boundary components are avoided. If E doespass though p , then p = p , and we can first modify E in an arbitrarily smallneighborhood of p to avoid this point and then apply the above. (cid:3) If (
X, d ) is a metric space and λ >
0, we denote by λX the metric space ( X, λ · d X ).The following lemma and its corollary show that the space X has a weak tangent space that contains S , see [8, Chapters 7, 8], [10, Chapter 9] for the backgroundon Gromov–Hausdorff convergence and weak tangent spaces. Lemma 12.4.
The slit carpet S is the Gromov–Hausdorff limit of the sequence ( X n = 2 n π − X ([0 , / n ] × [0 , / n ])) . Proof.
We use Theorem 7.4.12 from [8]. Let ǫ > N ∈ N be chosen sothat 1 / N +1 < ǫ . The 1-skeleton graph of the dyadic subdivision ˜ D N of Q into2 N +1 × N +1 subsquares pulls back to a graph D N in S via π S and graphs D N,n in each X n via π X ◦ − n . This means that D N is a graph embedded in S and D N,n is a graph embedded in X n such that the sets of vertices are the sets of preimagesof the vertices of ˜ D N by π S and π X ◦ − n respectively. Two vertices are connectedby an edge in D N or D N,n if and only if they are connected by an edge in ˜ D N . Notethat there are pairs of distinct vertices of D N , respectively D N,n , that get mappedto the same vertex of ˜ D N . We do not connected such pairs of vertices by an edge.If n ≥ N , the graphs D N and D N,n are identical. Since the vertices of these graphsform ǫ -nets, the lemma follows. (cid:3) The following corollary is immediate.
Corollary 12.5.
The completion of X in the path metric has a weak tangent spacethat contains S . A metric Sierpi´nski carpet S is called porous if there exists C ≥ p ∈ S and 0 < r ≤ diam( S ), there exists a peripheral circle J in S with J ∩ B ( p, r ) = ∅ and rC ≤ diam( J ) ≤ C · r. Lemma 12.6.
The slit carpet S cannot be quasisymmetrically embedded into thestandard plane R .Proof. The metric Sierpi´nski carpet S is porous, see [20, Proposition 2.4], andits peripheral circles are uniform quasicircles, in fact they are isometric to circles.Assume that there is a quasisymmetric embedding f : S ֒ → R . An easy appli-cation of Proposition 10.8 in [14] implies that the image f ( S ) is a porous metricSierpi´nski carpet in the restriction of the Euclidean metric. The peripheral circlesof S , and hence the boundaries of the complementary components of f ( S ), areuniform quasicircles. Applying Theorem 9.2 to each complementary component of f ( S ) now shows that f ( S ) is porous as a subset of R . Theorem 9.1 now statesthat the Assouad dimension, and hence the Hausdorff dimension, of f ( S ) is strictlyless than two. On the other hand, according [20, Proposition 2.4], S is Ahlfors2-regular, and it contains a curve family of positive 2-modulus [20, Lemma 4.2].This contradicts [14, Theorem 15.10]. (cid:3) Proof of Theorem 1.6.
By Lemma 12.3, it suffices to show that X cannot be qua-sisymmetrically embedded into the standard 2-sphere S . Suppose such a qua-sisymmetric embedding f : X → S can be found. It extends to a quasisymmetricembedding f of the completion X into S [14, Proposition 10.10]. The map f theninduces a quasisymmetric embedding of every weak tangent space of X into thestandard plane. Corollary 12.5 provides a weak tangent space of X that contains S . This contradicts Lemma 12.6. (cid:3) UASISYMMETRIC KOEBE UNIFORMIZATION 45
Open questions
Question . In Theorem 1.4, can the assumption that ∂X have finite rank be re-moved? It seems likely that this is the case. An affirmative answer is implied by anaffirmative answer to the following question. Let ( X, d ) be a metric space, homeo-morphic to a domain in S , that satisfies conditions (1)-(5) of Theorem 1.4, and hasno isolated trivial boundary components. Consider the glued space ( Z, d Z ) formedfrom X and a collection of quasidisks { D i } corresponding to (all) the componentsof ∂X , as in the proof of Theorem 1.4. Is it true that ( Z, d Z ) is homeomorphic to S ? Question . Suppose that (
X, d ) is a metric space, homeomorphic to a domainin S , that satisfies the ALLC condition. Are the components of ∂X uniformlyrelatively separated, quanitatively? Using the techniques of Section 5, it can beshown that the answer is “yes” in the case that ( X, d ) is a domain in S . Wesuspect that the answer is “no” in general. Question . Is there a quantitative statement, analogous to Theorem 1.4, thatuniformizes onto the class of all circle domains? Given a particular circle domainΩ that does not have uniformly relatively separated boundary components, canone give sufficient intrinsic conditions for a metric space to be quasisymmetricallyequivalent to Ω?
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Department of Mathematics, University of Illinois at Urbana-Champaign, 1409 WGreen Street, Urbana, IL 61801, USA
E-mail address : [email protected] Mathematisches Institut, Universit¨at Bern, Sidlerstrasse 5, 3012 Bern, Switzerland
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