Quantitative quasisymmetric uniformization of compact surfaces
Abstract.
Bonk and Kleiner showed that any metric sphere which is Ahlfors 2regular and linearly locally contractible is quasisymmetrically equivalent to the standard sphere, in a quantitative way. We extend this result to arbitrary metric compact orientable surfaces.
2010 Mathematics Subject Classification:
Primary 30C65; Secondary 30C62, 51F991. Introduction and statement of results
Through isothermal coordinates, every Riemannian metric on a compact orientable surface determines a Riemann surface structure on . By the classical uniformization theorem, carries a conformally equivalent Riemannian metric of constant curvature (in the case of a sphere), (for a torus), or (for higher genus surfaces). Hence, the original Riemannian metric on can be conformally deformed to a metric of constant curvature.
The purpose of this note is to extend the above discussion to certain classes of possibly nonsmooth distances on compact orientable surfaces. In this setting we have a metric, but no smooth structure, and the appropriate category replacing the class of conformal mappings is the class of quasisymmetric mappings.
In a metric space we will denote the distance between and by , or if the metric space is clear from the context. Let and be metric spaces. An embedding , i.e., a homeomorphism from onto its image , is a quasisymmetric embedding if there is a homeomorphism such that for all triples of distinct points , , and in ,
The homeomorphism is called a distortion function for the mapping . If a quasisymmetric embedding has a distortion function , it will be called an quasisymmetric embedding. The inverse of an quasisymmetric map is quasisymmetric with . A proof of this fact can be found in the book of Heinonen [Hei01, Proposition 10.6], which also serves as an excellent introduction to the theory of quasisymmetric mappings on metric spaces. A quasisymmetric mapping distorts relative distances by a controlled amount. While it may distort distances, it must do so moreorless isotropically.
A motivating example arises from Cannon’s Conjecture in geometric group theory. The boundary at infinity of a Gromov hyperbolic group carries a natural family of visual metrics, any two of which are quasisymmetrically equivalent. For this reason, one might say that the metric on the boundary of a Gromov hyperbolic group is defined only up to quasisymmetry. Cannon’s Conjecture can be phrased as follows: If the boundary of a Gromov hyperbolic group is homeomorphic to the sphere , then each visual metric on is quasisymmetrically equivalent to the standard metric on . If Cannon’s conjecture is true, then a Gromov hyperbolic group whose boundary is homeomorphic to acts discretely and cocompactly by isometries on 3dimensional hyperbolic space. See [Bon06] and the references therein for a more detailed discussion of this problem.
The following quasisymmetric uniformization theorem of Bonk and Kleiner [BK02] led to significant progress on Cannon’s Conjecture:
Theorem 1 (BonkKleiner).
Let be an Ahlfors regular metric space that is homeomorphic to . Then is quasisymmetrically equivalent to equipped with the smooth metric of constant curvature if and only if is linearly locally contractible.
A metric space is linearly locally contractible, if there is a constant such that for each point and radius , the ball is contractible inside the dilated ball . The metric space is Ahlfors regular if there is a constant such that for each and radius , the twodimensional Hausdorff measure satisfies
Theorem 1 is quantitative in the sense that the quasisymmetric map can be chosen to have a distortion depending only on the data of , i.e., the constants and appearing in the linear local contractibility and Ahlfors regularity conditions.
The linear local contractibility condition is a quantitative quasisymmetric invariant, i.e., if is quasisymmetric and is linearly locally contractible with constant , then is linearly locally contractible with a constant that depends only on and . In contrast, Ahlfors regularity is not a quasisymmetric invariant.
Given a Gromovhyperbolic group with homeomorphic to , each visual metric on is linearly locally contractible. However, it is not known whether there always is an Ahlfors regular visual metric on . Simple examples show that Theorem 1 fails without the assumption of Ahlfors regularity.
Problems outside of geometric group theory, such as the search for an intrinsic characterization of up to biLipschitz equivalence (note that Ahlfors regularity is a biLipschitz invariant), led to the development of versions of Theorem 1 for noncompact simply connected surfaces [Wil08] and for a large class of planar domains [MW13], as well as a local version [Wil10]. In this work, we provide a version of Theorem 1 that applies to all orientable compact surfaces, including those of higher genus. This is part of an ongoing program to complete the classification of Ahlfors regular and linearly locally contractible metric surfaces up to quasisymmetry.
Theorem 2.
Let be a metric compact orientable surface. Assume that is Ahlfors regular and linearly locally contractible. Then there exists a Riemannian metric of constant curvature , , or on such that the identity map is quasisymmetric with a distortion function depending only on the data of .
Here the data of consists of the topological genus of , and the constants in the Ahlfors regularity and linear local contractibility conditions.
Except for the statement of dependence of the distortion function only on the data of , Theorem 2 follows immediately from the local uniformization theorem of [Wil10] and an elementary localtoglobal result for quasisymmetric mappings due to Tukia and Väisälä [TV80, Theorem 2.23]. While a quantitative proof is significantly more involved, it provides much more information about the “space” of quasisymmetric structures on surfaces. For example, let us suppose that satisfies the assumptions of Theorem 2 and is homeomorphic to a torus. All smooth Riemannian metrics of constant curvature on the torus are quasisymmetrically equivalent, but not with a uniform distortion function. In essence, the quantitative Theorem 2 allows us to assign to a point in the Riemann moduli space of the torus that is “roughly optimal”. Finding a smooth metric on the torus that is quasisymmetrically equivalent to with “minimal” quasisymmetric distortion seems to be both a difficult and illdefined question, but one of obvious interest.
Our proof of Theorem 2 proceeds as follows. Let be a metric surface as given in the theorem. By the local uniformization result of [Wil10], has an atlas of uniformly quasisymmetric mappings. Our first step is to create a compatible conformal atlas, giving the structure of a Riemann surface. This step can be thought of as creating “quasiisothermal” coordinates; while the chart transitions are conformal, the chart mappings themselves are only quasisymmetries. The classical uniformization theorem then provides a globally defined conformal homeomorphism to a Riemann surface that is the quotient of the sphere, plane, or disk by an appropriate group of Möbius transformations. The Riemann surface inherits a Riemannian metric of constant curvature , , or , respectively. Although conformal maps are locally quasisymmetric, this does not immediately give us global information about the metric properties of the uniformizing map . We show that the uniformizing mapping is in fact globally quasisymmetric with a distortion function that depends only on the data of . The key idea in this step is a type of Harnack inequality. The claim in the theorem then follows by taking to be the pullback of the metric under .
2. Background and preliminary results
2.1. Notation
In a metric space , we will denote the distance between points and of by or when the space is understood. We denote by
the open ball of radius centered at and by the corresponding closed ball. For an open or closed ball of radius and a number , the notation denotes the same type of ball with the same center and radius .
We denote the complex plane with the standard Euclidean metric by , and the disk model of hyperbolic space equipped with the standard hyperbolic metric of curvature by . The open ball of radius centered at is denoted by in and by in , i.e.,
Note that is the Euclidean radius for , whereas it is the hyperbolic radius for .
2.2. Quasisymmetric mappings
Let be an embedding of metric spaces. For and , define
If is an quasisymmetric embedding, then for and ,
We will need a statement about the equicontinuity of quasisymmetric maps, which is a slightly more general version of [Hei01, Corollary 10.27].
Lemma 3.
Let be an quasisymmetric embedding of metric spaces and let and be positive real numbers. If and , then
is a modulus of continuity for .
Proof.
Given with , there exists such that . Then
If , the trivial estimate yields the desired estimate. ∎
Note that the modulus of continuity provided by Lemma 3 is not scaleinvariant. If the metrics on and are scaled by the same quantity, the resulting modulus of continuity may still change; see subsection 2.4.
The following result, which is a slight variation of [TV80, Theorem 2.23], gives a localtoglobal result for quasisymmetric homeomorphisms between compact spaces in terms of Lebesgue numbers. We include a proof for the reader’s convenience. Recall that is a Lebesgue number for a covering if for every set with there exists such that .
Theorem 4 (TukiaVäisälä).
Let be a homeomorphism between compact connected metric spaces and . Suppose that

is a finite open covering of with Lebesgue number .

satisfies the implication

there is a quasisymmetric distortion function such that for each , the restricted mapping is quasisymmetric.
Then is quasisymmetric with distortion function depending only on and the ratios and .
Proof.
Let , , and be distinct points of . Set
We consider four cases.

Suppose that Then is contained in some , and so

Suppose that but . Since is connected with , there exists with . Then there exists such that and
This implies that

Suppose that but By the same argument as in the previous case, there exists with , as well as an index such that . This gives
which implies that

Suppose that Then and , so
Combining these estimates on in terms of yields the desired result. ∎
2.3. Conformality and quasisymmetry
A conformal mapping between domains in is quasisymmetric when restricted to a relatively compact subdomain , with quasisymmetric distortion depending only on and . This fact, which should be compared to Koebe’s distortion theorem, will play a key role in the proof of Theorem 12. A more general statement is even true: one may consider quasiconformal mappings in higher dimensional Euclidean spaces, although the quasisymmetric distortion then also depends on the maximal quasiconformal dilatation. See [Väi81, Theorem 2.4] for a proof of this fact, which we will use in the following form:
Proposition 5.
Let be a conformal embedding. For each , the restriction is a quasisymmetric embedding with distortion function that depends only on .
An easy consequence of Proposition 5 is the following.
Corollary 6.
There is a universal constant such that for any conformal embedding ,
Proof.
By Proposition 5, there is a universal quasisymmetric distortion function for . Hence, if , then
Thus, choosing so small that fulfills the requirements of the statement. ∎
We will also need a hyperbolic version of this result. Recall that the hyperbolic disk is equipped not just with a conformal structure, but also with the standard hyperbolic metric. Our proof employs Koebe’s distortion theorem for specificity, but could also be carried out using Proposition 5 alone.
Proposition 7.
Let be a conformal embedding of the Euclidean unit disk into the hyperbolic plane. Then the restriction is a quasisymmetric embedding with a universal distortion function.
Proof.
The map is conformal in , so by Koebe’s distortion theorem [Dur83], the image contains the Euclidean disk , and the image of the Euclidean disk is contained in the Euclidean disk . This implies
and
It follows from the definition of the hyperbolic metric that the identity mapping
is, up to scaling, a biLipschitz embedding with universal constants, so it is quasisymmetric with universal distortion. Since the composition of quasisymmetric maps is quasisymmetric, quantitatively, the result follows from Proposition 5. ∎
The following statement is an easy corollary of Proposition 7.
Corollary 8.
There is a universal constant such that for any conformal embedding ,
Proof.
By Proposition 7, there is a universal quasisymmetric distortion function for . Hence, if , then
Choosing so small that , we get the claim of the corollary. ∎
For the remainder of the paper, for convenience we define .
2.4. The data of , scalings, and normalization
Let be a metric space that is Ahlfors regular with constant , linearly locally contractible with constant , and homeomorphic to a compact orientable surface. We will refer to and as the data of .
For , we may form a new metric space by multiplying the original metric on by . Then is again Ahlfors regular and linearly locally contractible, and has the same data as . Moreover, the identity mapping from to is quasisymmetric with . Hence, in the proof of Theorem 2, we may scale the domain as we see fit.
For the remainder of this article, let be a metric space that is Ahlfors regular with constant , linearly locally contractible with constant , homeomorphic to a compact orientable surface, and has unit diameter. When we state that a quantity depends only the data of , we are implicitly assuming this normalization.
The main reason for this convention (aside from notational convenience) is that Lemma 3 is not scaleinvariant. Without this normalization, we would not be able to say that certain moduli of continuity depend only on the data of .
3. Finding a conformal structure
By [Wil10, Theorem 4.1], possesses a quasisymmetric atlas:
Theorem 9.
There is a quantity and a quasisymmetric distortion function , each depending only on the data of , such that for each there is a neighborhood of such that

,

there exists an quasisymmetric map with ,
The original theorem in [Wil10] does not include the normalization . However, since is quasisymmetric, the basic distortion estimates [Hei01, Proposition 10.8] imply that where depends only on . As the Möbius transformation has quasisymmetric distortion depending only , we may assume that . Accordingly, given a pair where is a homeomorphism, we will call the center of .
We use the atlas provided by Theorem 9 to produce a conformal atlas on that is adapted to its metric. We have separated the construction into two lemmas. The first is purely metric; the second modifies the output of the first.
Lemma 10.
Let be given. Then there exists a quasisymmetric distortion function , radii , , and a positive integer , such that the following statements hold:

There exists an atlas of , where each mapping is an quasisymmetric homeomorphism with center denoted by .

The collection is pairwise disjoint.

The collection covers .

For each , it holds that , and
Moreover, depends only on the data of , while , , and depend only on the data of and .
Note that the collection forms an open cover of for which is a Lebesgue number; cf. Theorem 4. Moreover, Lemma 3 implies that for each , the restriction and its inverse have moduli of continuity that depend only on and the data of .
Proof.
For each , let be the quasisymmetric mapping provided by Theorem 9 with , so that
Applying Lemma 3 to and its inverse, we see that there is a common modulus of continuity for all of the mappings , depending only on the data of .
Hence, there is a radius and a number , each depending only on and the data of , such that for each ,
(1) 
Let be a maximal separated set in . Then the open balls are pairwise disjoint, while the open balls cover . Since is Ahlfors regular and we have assumed that , this implies that is comparable to and therefore depends only on and the data. Moreover, as , for each , it holds that . In particular is also a cover of . This shows that is the desired atlas. ∎
We now adapt the atlas given in Lemma 10 so that the transition mappings are conformal. This step is similar to the proof that a quasiconformal structure on a surface has a compatible conformal structure; see [Kuu67] and [Can69].
Lemma 11.
There exists a quasisymmetric distortion function , radii , , and a positive integer , all depending only on the data of , such that the following statements hold:

There exists an atlas of , where each mapping is an quasisymmetric homeomorphism with center denoted by .

The collection is pairwise disjoint.

The collection covers .

For each , it holds that , and

The transition maps are conformal wherever defined.
Proof.
We begin by letting be a number which will be determined below and will depend only on the data of . Consider the atlas provided by Lemma 10. Let us say that this atlas is given by quasisymmetric charts where depends only on the data of . The associated radii , and the number of charts depend only on the data of and .
Since is connected, the charts can be relabeled to satisfy
for .
Define by setting . We will iteratively construct for as follows. For , we assume that the already constructed map is quasisymmetric on , has center , and that if , then the transition functions are conformal where defined.
In the following, we will write for indices in .
For , define . Then is quasisymmetric and hence quasiconformal on . Therefore, the complex dilatation of is welldefined (up to a.e. equivalence) on . Given another index , it holds that on . It therefore follows from the conformality of that a.e. on the intersection . This shows that there exists a measurable Beltrami coefficient with a. e. on for each index , and on . We extend to the whole plane by for . By the Measurable Riemann Mapping Theorem there exists a unique quasiconformal map , normalized by , , with complex dilatation a.e. By the symmetry of and the chosen normalization, , and so . We define . The transformation formula for Beltrami coefficients (see, e.g., [LV73, IV.5.1]) shows that if , then is conformal. Moreover, .
We claim that for each , the mapping has a quasisymmetric distortion function that depends only on the data of ; as this is true of , it suffices to prove the same of , and we may also assume that . As a normalized quasiconformal selfmap of the unit disk, is quasisymmetric with distortion controlled by the maximal dilatation , see e. g. [Väi81, Theorem 2.4]. By an inductive argument, it is easy to show that this dilatation is bounded by a constant depending only on (which depends only on the data of ) and the number of charts (which depends on the data and ). However, the bound is actually independent of , by the following argument.
Fix . By the uniform quasisymmetry of , there is a quantity , depending only on the data of , such that for all indices with , the complex dilatation of satisfies For , define
If , then . If , then , and hence
The transformation formula for Beltrami coefficients now shows that for almostevery point ,
Since we see that As discussed above, we may now conclude that each of the mappings has a quasisymmetric distortion function that depends only on the data of .
We have now seen the atlas of satisfies conditions (1) and (5) of the statement. Moreover, setting , the conditions (2) and (3) follow directly from the corresponding statements for the atlas . Note that only condition (4) involves the constant . We now show how to choose so that condition (4) is satisfied.
Let us make the convention that is the identity. As discussed above, the mappings are uniformly quasisymmetric with a distortion function depending only on the data of . Hence, by Lemma 3, there is a common modulus of continuity for all of the mappings that depends only on the data of . Since is a universal constant, we may choose depending only on the data of such that for each , it holds that Having so chosen , the radius depends only on the data of , and so we may also choose depending only on the data of such that for for each , This establishes condition (4). ∎
3.1. Uniformizing to a standard metric
The atlas given by Lemma 11 determines a conformal structure on the compact orientable surface , i.e., the pair determines a Riemann surface. By the classical uniformization theorem, is conformally equivalent to a standard Riemann surface , where denotes the standard Riemann surface structure on the sphere, the plane, or the unit disk, and is a discrete group of Möbius transformations acting freely and properly discontinuously on . The standard spherical, plane, or hyperbolic Riemannian metric on then descends to a Riemannian metric of constant curvature , , or on , compatible with the conformal structure. We fix a uniformizing conformal homeomorphism , and equip with the distance function arising from the Riemannian metric of constant curvature. Denote by the quotient mapping from to . Recall that may be expressed as where the infimum is taken over all smooth paths in such that the projected path connects and . A priori, it is not clear how the properties of the distance or the map depend on the original metric space . The following statement is the main result of this paper, and completes the proof of Theorem 2.
Theorem 12.
The uniformizing map is quasisymmetric, with distortion depending only on the data of .
Proof of Theorem 12..
In the case that , then itself must be homeomorphic to , and so Theorem 1 implies Theorem 12. We consider the remaining planar and hyperbolic cases together.
Define by . Then is a conformal homeomorphism that lifts to a conformal embedding .
We use a sequence of claims to complete the proof.
Claim 1.
Let . The projection maps isometrically onto In other words, for each , .
Proof of Claim 1.
For , define
If , then is independent of and depends only on the group . If , then is bounded below by the minimal translation distance of a nonidentity element of , and is a Lipschitz function of , see e. .g. [Bea83, Section 7.35]. These facts imply that for any and ,

if , then is an isometry,

if is injective, then .
Since is injective, it follows that . Corollaries 6 and 8 now complete the proof of this claim. ∎
Claim 2.
For each , the mapping is quasisymmetric with distortion that depends only on the data of .
Proof of Claim 2.
By Propositions 5 and 7, the mapping restricted to is quasisymmetric with a universal distortion function. By Claim 1, this is also true of . Since we may write , and is quasisymmetric where depends only on the data, it follows that for each , the mapping is quasisymmetric with distortion that depends only on the data of . According to Lemma 11, , implying the claim. ∎
Claim 3.
There is a constant depending only on the data of such that for each ,
(2) 
Proof of Claim 3.
As covers and , it holds that also covers . Since is a homeomorphism, this implies that As is connected, it follows that
(3) 
Recall that the number of charts depends only on the data of .
Consider indices and in , and suppose that . Then , and so On the other hand, , and so , implying . Moreover, since is a quasisymmetric embedding with universal distortion function, there is a quantity depending only on the ratio of to , and hence only on the data of , such that
Since in addition,
we have now shown that the quantities
are all comparable with constants that depend only on the data of . The same is true with the roles of and reversed.
Since the open sets cover the connected space , for any pair of indices , there is a sequence of distinct indices of length at most so that for each Since depends only on the data of , (3) proves the claim. ∎
We complete the proof of Theorem 12 by employing Theorem 4. We consider the covering of given by . By Claim 2, the mapping is quasisymmetric on each element of this cover with a distortion function that depends only on the data of . Moreover, Lemma 11 implies that this cover has Lebesgue number . Suppose that satisfy . We may find indices and in such that and . Then , so and