Harmonic Measure, Infinite Kernels, and Symmetrization

John A. Velling ^{∗}

14 April 1995

Abstract

The vanishing of area for the infinite Nielsen kernel of an arbitrary open Riemann surface is shown to follow from iteration of a natural geometric operation on the unit disk. This operation compares the distribution of harmonic measure on the boundaries of two related simply connected domains, and is not yet sufficiently well understood.

§1. A Problem in Harmonic Measure

This paper is an attempt to interest the mathematical community in a problem which the author has found rather interesting. Consider the following —

Let D denote the unit disc in the complex plane, and S^{1} its boundary. We will want to think of D as furnished with the Poincaré metric, making it a complete metric space of constant curvature −1. Let I_{n}, n ∈ Z, be a collection of disjoint arcs in S^{1}, each with length < π. Let them be indexed by decreasing length. Each I_{n} is the boundary in S^{1} of a hyperbolic half-plane, H_{n}, whose other boundary component is a hyperbolic geodesic J_{n} inside of D. The indexing may then be thought of as the order of decreasing distance of J_{n} from 0 ∈ D. Since X = D \ ∪_{n}H_{n} is an intersection of hyperbolically convex regions, it is hyperbolically convex and hence simply connected. The condition on length implies that 0 is in the interior of X, whence we can map X conformally to D with 0 ↦ 0. Evidently X has rectifiable boundary, so that this Riemann mapping extends to ∂X and J_{n} → I′_{n}, a new arc in S^{1}.

∗Partially supported by NSF grant # 4401728 and PSC-CUNY grant #6-64131

The relationship between the lengths of the I_{n}, denoted by ℓ(I_{n}), and the lengths ℓ(I′_{n}) is far from clear. A lemma of Loewner implies that ℓ(∪_{n}I_{n}) ≤ ℓ(∪_{n}I′_{n}), but examples can easily be constructed where some of the I_{n} are shorter than the corresponding I′_{n}. One may also construct examples where I′_{1} is not the longest of the I′_{n}, so that there is no immediate geometric interpretation of the index on the I′_{n}. The issue of interest is then the following

Problem: Is ℓ(I_{1}) ≤ ℓ(I′_{1})?

It is believed that the answer to this is ‘yes’. We will first present a nice consequence of an affirmative answer, and then several pieces of evidence in support of this conclusion.

§2. Several Consequences of an Affirmative Answer

Let R_{0} be a finite type Riemann surface with boundary, and R_{−∞} denote its infinite Nielsen kernel (the meaning of which will be reviewed below). The following is a theorem of J. Cao [2]:

Theorem: [2] If R_{0} is a finite type Riemann surface with boundary, R_{−∞} has no interior.

Based on the positive resolution of the above problem, we may offer an extension of the above theorem.

‘Theorem’ 1: For any Riemann surface R_{0} of second kind there is no interior to R_{−∞}.

Our proof, by much the same reasoning as Cao’s but with emphasis on Schwarz’s lemma, shows that the topology of R_{0} is inessential.

We also obtain the following stronger version of theorem 1, which we distinguish because of the brevity of the respective proofs.

‘Theorem’ 2: For any Riemann surface R_{0} of second kind, R_{−∞} has no area.

Before demonstrating the above results, we explain several terms.

§3. Riemann surfaces of the second kind and Nielsen kernels

A Riemann surface of the second kind is conformally a quotient of D = {z ∈ C : |z| < 1} by a group orientation preserving Möbius transformations preserving D and acting discontinuously on S^{1} = ∂D. We will give such surfaces hyperbolic geometry inherited from D as a model of H^{2}. The groups acting here are said to be of the second kind, and the domain of discontinuity in S^{1} for such a group is a union of disjoint arcs. Examples of such surfaces include infinite volume hyperbolic surfaces of finite type.

Henceforth let R_{0} = D/Γ_{0} be a Riemann surface of the second kind and assume for the moment that the group is non-elementary. The Nielsen kernel of such a Riemann surface, denoted by R_{−1}, is a new Riemann surface of the same topological type as R_{0}. It is constructed as follows: Let {I_{k}} denote the collection of arcs in the domain of discontinuity for the action of Γ_{0} on S^{1}. Denote by J_{k} the hyperbolic geodesic spanning the corresponding I_{k}. On deleting from D the hyperbolic halfplanes bounded by the I_{k} and J_{k} we are left with a connected, simply connected, Γ_{0}-invariant domain in D. The quotient of this domain by Γ_{0} is conformally the desired surface R_{−1}. The Nielsen kernel of R_{0} is empty when Γ_{0} is elementary. For more background on this, see any of Halpern’s papers concerning this: [7], for example. We will henceforth assume Γ_{0} is non-elementary.

One may iterate this procedure to obtain a sequence of Riemann surfaces R_{0}, R_{−1}, R_{−2}, … with R_{n−1} the Nielsen kernel of R_{n} (n ≤ 0). There are thus natural injections ι_{nm} : R_{n} → R_{m} obtained from the above procedure, so that ∂(ι_{(n−1)n}(R_{n−1})) ⊂ R_{n} is a collection of simple geodesics (with respect to the hyperbolic geometry of R_{n}) homotopic to ideal boundary components of R_{n}. Evidently ι_{nm} ∘ ι_{ml} = ι_{nl}. Identifing R_{n} with all of its images ι_{nm}(R_{n}), we see that {R_{k}}_{−∞}^{n−1} is a nested sequence of open sets in R_{n}. The intersection of these will be called R_{−∞}, the infinite Nielsen kernel. As a topological space with an induced conformal structure (even though R_{−∞} may not be a Riemann surface) from the embeddings into the R_{n}, R_{−∞} is independent of the R_{n} in which it is embedded.

§4. Proofs of the ‘Theorems’

This development assumes the an affirmative answer to the problem. We will also make use of the following versions of Schwarz’s lemma:

Lemma 1: Let A, B be subdomains of D such that 0 ∈ B ⊂ A. If f : D⟶^{1−1}A ⊂ D and g : A → B with f(0) = g(0) = 0, then |f′(0)| ≥ |(g∘f)′(0)|.

Lemma 2: Let f : D⟶^{1−1}D with f(0) = 0 and D_{r} = {rz : |z| ∈ D} ⊂ f(D) for some r < 1. Then |f′(0)| > r.

Proof of ‘Theorem’ 1: Let R_{0} be any non-elementary Riemann surface of the second kind. Let p ∈ R_{n} ⊂ R_{m} ⊂ R_{0} for fixed n. Let ι̃_{nm} : D → D be the lifts of ι_{nm} : R_{n} → R_{m} such that 0 is always a lift of p. By the problem and lemma 1, there exists a constant c < 1 such that |ι̃′_{n(n−1)}(0)| < c. (One may take as A in lemma 1 the domain D\ (the largest halfplane, as viewed from p ∈ R_{0}, deleted on the creation of R_{−1}). Thus |ι̃′_{n1}(0)| < c^{n}. The theorem now follows by the contrapositive of lemma 2. □

To prove ‘theorem’ 2 we will need

Lemma 3: Let D(n) be a family of disks indexed by the nonnegative integers, and f_{n} : D(n) → D(n − 1) be univalent injections with f_{n}(0) = 0. Assume there is an r < 1 such that f_{n}(D(n)) always misses a piece of definite area in D_{r} ⊂ D(n − 1). With g_{n} = f_{1} ∘ f_{2} ∘ ⋯ f_{n}, 0 is not a point of Lebesgue density for ∩_{1}^{∞}g_{n}(D(n)) ⊂ D(0).

Proof of Lemma 3: Our hypothesis of missing area in D_{r} at each step implies, by the Koebe \frac{1}{4}-theorem that there is some c < 1 so that f′_{n}(0) < c for all n. We suppose that there is an ϵ > 0 such that, for every n > 0, X_{n−1} = D_{r} \ f_{n}(D(n)) ⊂ D(n − 1) has area > ϵ(πr^{2}). Then the Koebe distortion theorem tells us that

\frac{∫_{X_{i}} |g′_{i}|^{2} dx dy}{∫_{B_{r}(0)} |g′_{i}|^{2} dx dy} ≥ \frac{(1 − r)^{8}}{(1 + r)^{8}} · \frac{∫_{X_{i}} 1 dx dy}{∫_{B_{r}(0)} 1 dx dy} ≥ \frac{(1 − r)^{8}}{(1 + r)^{8}} ϵ,

i.e., inside of arbitrarily small balls about 0 (since |g′_{n}(0)| < c^{n} → 0) a definite proportion of the area is missing. □

Proof of ‘Theorem’ 2: We apply the lemma 4 at each point of R_{−∞} ⊂ ⋯ R_{−2} ⊂ R_{−1} ⊂ R_{0} by lifting to the universal covers of the R_{n} (n ∈ Z_{−}) to see that no point of R_{−∞} is a point of Lebesgue density. The problem furnishes the area deficiency needed in Lemma 3. □

It appears that one can continue in this vein to obtain bounds on the local Hausdorff dimension of R_{−∞}, since not only are we excising a substantial piece of area at every level but these pieces are very regular. One would thus be able to bound the Hausdorff dimension of the infinite Nielsen kernel for finite type surfaces strictly away from 2.

§5. A Comment on the infinite Nielsen extension

One further consequence of a positive resolution of our conjecture is the ability to state conclusively that the infinite Nielsen extension (reviewed below) cannot arise from certain naive constructions.

The operation of taking Nielsen kernels has an inverse. Thus if we have a Riemann surface, R_{0}, of the second kind we may find its Nielsen extension, R_{1}, into which R_{0} naturally injects as Nielsen kernel. For a full description of this process, see [4], [5], or [6], for example. There are again special cases when R_{0} = D/Γ_{0} for Γ_{0} elementary, so we still assume that Γ_{0} is non-elementary.

With this assumption, R_{1} is a surface of the second kind, of the same topological type as R_{0}, and we may iterate the extension operation. We obtain a sequence of surfaces — R_{1}, R_{2}, … — where we may consider R_{n} ⊂ R_{n+1}. Thus we also have the limiting surface R_{+∞} = ∪_{n∈Z}R_{n}. This is called the infinite Nielsen extension of R_{0}. Bers [1] showed that when R_{0} is of finite type, i.e. of finite genus with a finite number of boundary curves and punctures, R_{+∞} is a surface of the same topological type as R_{0} but it is quasiconformally distinct — it has punctures where R_{0} has boundary curves. In particular, it is still hyperbolic. The Nielsen extension map, with R_{0} ↦ R_{+∞}, is surjective to the space of punctured surfaces of the appropriate topological type [6]. We will further restrict ourselves to this setting where R_{0} is of finite type.

It is reasonable to ask if either the infinite Nielsen extension, with R_{0} ↦ R_{+∞}, or the inverse, which identifies the R_{0} giving rise to a specific punctured surface as R_{+∞}, occur in a more naive geometric way. Two possibilities for such in this latter case are:

1. \underline{horocyclic excision}: With R_{+∞} = D/Γ_{+∞}, every puncture on R_{+∞} corresponds to an equivalence class of parabolic fixed points for the action of Γ_{+∞}. We may obtain surfaces with boundary curves where R_{+∞} has punctures by cutting along horocycles about the punctures.

2. \underline{disk excision}: In [3], Haas showed that for a finite genus Riemann surface with boundary curves, R_{0}, there is a unique extension to a surface of the same genus such that the boundary curves of R_{0} are boundaries of geometric disks. (Geometry is spherical, planar, or hyperbolic depending on genus.) We obtain a punctured surface by removing the centers of these disks, and may reverse this process by excising geometric disks about the centers.

These operations are clearly conformally natural. We ask if the R_{0} giving rise to R_{+∞} through the infinite Nielsen extension may be obtained from R_{+∞} by either of these excisions. In both cases, as a corollary of theorem 1, the answer is seen to be ‘no’. This is because the complement of a locally finite collection of disjoint disks in any of D, C, or Ĉ necessarily has interior. Thus we have

Corollary: The family of R_{0} yielding a fixed R_{+∞} by the infinite Nielsen extension is obtained neither by the horocyclic extension nor by the hyperbolic disk extension.

§6. Some Supporting Evidence.

To obtain an affirmative answer to the problem, it would suffice, by subordination and a normal families argument, to establish the following conjecture —

Conjecture: Let I_{1}, I_{2}, …, I_{n} be a finite collection of disjoint arcs filling out S^{1}, each of length < π, and ordered by decreasing length. Let I′_{k} be as described in §1. Then ℓ(I_{1}) ≤ ℓ(I′_{1}), with strict inequality unless the I_{k} are all of the same length and fill out S_{1} (i.e. the full symmetric case).

We can see that this is true for Möbius images of the full symmetric case as follows.

First note that in the full symmetric case, we may assume that the endpoints of our arcs are at the n-th roots of 1, ρ_{n}^{k} = e^{\frac{2πki}{n}}. Let I_{k} be the arc in S^{1} from ρ_{n}^{k−1} to ρ_{n}^{k}, J_{k} the hyperbolic geodesic between these two points, and H_{k} the hyperbolic halfplane bounded between I_{k} and J_{k}. There is a Riemann map from D \ ∪_{1}^{n}H_{k} to D preserving 0 and the ρ_{n}^{k}, mapping the radius from 0 to ρ_{n}^{k} (call this R_{k}) bijectively to itself. Let f denote this map. Let p ∈ D be a point in the hyperbolic triangle bounded by R_{0}, R_{1}, and J_{1}. Now let h_{1} be the harmonic function on D which is 1 on I_{1} and 0 on the rest of S^{1}, so that h_{2} = h_{1} ∘ f is harmonic on D \ ∪_{1}^{n}H_{k}. In this setting, our problem is equivalent to showing that h_{1}(p) ≤ h_{2}(p), with equality holding only when p = 0.

This is now straightforward by subordination. Since h_{1} < 0 in D, we have h_{1} < h_{2} on J_{1}. Also, since h_{1} is increasing radially on R_{0} and R_{1}, and f is an expansion with respect to the hyperbolic metric (Schwarz-Pick lemma), h_{1} ≤ h_{1} ∘ f = h_{2} on R_{0} and R_{1}, with equality only at the origin. Subordination now gives the result in this special case.

One may also show that the above conjecture is true for any four disjoint arcs filling S^{1}, using elliptic functions. A general proof, however, eludes this author.

References

[1] Bers, L., Nielsen extensions of Riemann surfaces, Ann. Acad. Sci. Fenn. 2 (1976), 17–22.

[2] Cao, J., The Bers–Nielsen kernels and souls of open surfaces with negative curvature, preprint.

[3] Haas, A., Linearization and mappings into pseudocircle domains, Trans. AMS 282 (1984), 415–429.

[4] Halpern, N., On the infinite Nielsen extension of a torus, Comp. Var. 4 (1984), 111–117.

[5] —, Some remarks on Nielsen extensions of Riemann surfaces, Mich. Math. J. 30 (1983), 65–68.

[6] —, A theorem on infinite Nielsen extensions, Proc. AMS 97 (1986), 459–462.

[7] —, Two theorems on Nielsen cores, Comp. Var. 8 (1987), 123–127.