Annales Academiæ Scientiarum Fennicæ Mathematica Volumen 30, 2005, 135–142

A SIMPLY CONNECTED, HOMOGENEOUS DOMAIN THAT IS NOT A QUASIDISK

Geir Arne Hjelle

Norwegian University of Science and Technology, Department of Mathematical Sciences NO-7491 Trondheim, Norway; hjelle@math.ntnu.no

Abstract. We construct a simply connected, homogeneous domain that is not a quasidisk. This shows that a theorem by Sarvas (1985) can not be generalized to simply connected domains instead of Jordan domains.

1. Introduction

We let 𝒬(K) denote the family of K -quasiconformal self-mappings of the Riemann sphere \overline{C}. In particular 𝒬(1) denotes the family generated by the Möbius group

\{φ(z) = \frac{az + b}{cz + d} : ad − bc ≠ 0\}

and the reflection f(z) = z̄. We refer to [5] for further information on quasiconformal mappings. Contrary to usual convention we will allow quasiconformal mappings to be orientation-reversing as well as orientation-preserving. This peculiarity will be discussed at the end of the article.

A set S ⊂ \overline{C} is said to be homogeneous with respect to a family ℱ of mappings, if for each pair of points z_{1}, z_{2} ∈ S there exists a mapping f ∈ ℱ such that

f(S) = S and f(z_{1}) = z_{2}.

It is easy to show that there is a family ℬ of conformal mappings such that the open unit disk, D, is homogeneous with respect to it. For example take

(1)  ℬ = \{φ(z) = λ\frac{z − a}{1 − āz} : a ∈ D, |λ| = 1\}.

By using translations, dilations and reflections, it can be shown that any disk is homogeneous with respect to 𝒬(1). The converse can also be shown to be true [2].

2000 Mathematics Subject Classification: Primary 30C62.

Research supported by grants from the Research Council of Norway, project #155060 and the Norwegian University of Science and Technology.

Theorem 1. A domain D is a disk if and only if it is homogeneous with respect to the family 𝒬(1).

The mappings φ in (1) distorts the boundary of D. If we require the boundary to be left untouched, a conformal map will no longer do the job. Instead we rely on a theorem of Teichmüller. He pointed out that given a compact set E in the unit disk, any point in E can be mapped to any other point in E by a K -quasiconformal mapping f. Moreover, f can be chosen to fix the boundary of D [8].

We will now look at the connection between homogeneity and quasidisks. In 1977 Timo Erkama proved the following result for the boundary [1].

Theorem 2. A simply connected domain D is a quasidisk if and only if ∂D is homogeneous with respect to the family 𝒬(K) for some fixed K.

A few years later Jukka Sarvas proved a similar theorem concerning homogeneity of the domain [7].

Theorem 3. A Jordan domain D is a quasidisk if and only if it is homogeneous with respect to the family 𝒬(K) for some fixed K.

2. A simply connected, homogeneous domain

In Theorem 2 a simply connected domain is sufficient, while a Jordan domain is crucial in Sarvas’ proof of Theorem 3. We will now show that the hypothesis of the latter theorem can not be weakened to also include simply connected domains. This result has been known for some time and a preprint by Näkki and Palka on the matter exists [6]. Their construction is based on the same technique used in Example 4.6 of [3] and some results from the field of Kleinian groups. We will present an example which does not rely on this theory.

Theorem 4. There exists a simply connected domain D, homogeneous with respect to 𝒬(K) for some fixed K, that is not a quasidisk.

We start by constructing the domain. Let r_{1} and r_{2} be reflections in the circles |z + 2| = 1 and |z − 2| = 1, respectively, and let t denote the translation t(z) = z + i. Define G_{0} = {r_{1}, r_{2}, t, t^{−1}} and let G be the group generated by the elements of G_{0} . Finally denote by E the closed set

E = \{z = x + iy : |x| ≤ 2 − \sqrt{1 − y^{2}}, |y| ≤ \frac{1}{2}\}.

See Figure 1. The domain D will then be given by the union

D = ⋃_{g∈G} g(E).

Figure 1. The set E, the two circles we reflect in and some of the other components of D.

The domain D will be an infinite “strip” and is indicated in Figure 2.

Let us introduce some more notation. A set P = g(E) with g ∈ G is called a component of D. If g = g_{n} ∘ ⋯ ∘ g_{1} and n is minimal, P is called an nth generation component and is denoted P^{(n)} . We also define

(2)  D_{n} = int( ⋃_{i≤n} P^{(i)}).

Observe that we can introduce a metric d_{c} on the set of components in the following way: Let P_{1}^{(i)} = φ(E) and P_{2}^{(j)} = ψ(E) with φ, ψ ∈ G, and let n be the generation of P = ψ ∘ φ^{−1}(E). Define d_{c}(P_{1}, P_{2}) = n. Also note that for P^{(n)} = g(E) = g_{n} ∘ ⋯ ∘ g_{1}(E) the union of the n components

⋃_{i=1}^{n} g_{n} ∘ ⋯ ∘ g_{i}(E)

is path connected. Thus, d_{c}(P_{1}, P_{2}) = n and the minimality of n implies that the shortest path from P_{1} to P_{2} crosses n − 1 other components.

Before we go on to show that D is a simply connected, homogeneous domain, we prove two key lemmas.

Lemma 1. D_{n} as defined in (2) is a simply connected domain for all n = 0, 1, 2, ….

Proof. We use induction. Clearly D_{0} = int(E) is simply connected. Assume D_{n−1} is simply connected. We will show that \overline{D}_{n−1} ∪P^{(n)} is simply connected for all P^{(n)} . Then also D_{n} will be simply connected as two different nth generation components do not have any point in common that is not already in \overline{D}_{n−1} .

Figure 2. The domain D and its components. To the left is a blown-up image showing more detail.

Obviously \overline{D}_{n−1} ∪P^{(n)} is connected. We must make sure that the addition of P^{(n)} does not split the boundary into two or more parts, or equivalently we must see that the set \overline{D}_{n−1} ∩ P^{(n)} is connected. Seeking a contradiction we assume that \overline{D}_{n−1} ∩ P^{(n)} is disconnected. P^{(n)} is surrounded by four components of generation n − 1 or n + 1. We can illustrate the situation as follows:

The black lines mark \overline{D}_{n−1} ∩ P^{(n)} . Let g = g_{n} ∘ ⋯ ∘ g_{1} be such that P^{(n)} = g(E) with g_{i} ∈ G_{0} . Then

because there are no components of generation −1. Thus there are φ_{j} ∈ G_{0} such that P_{j}^{(1)} = φ_{j}(E) for j = 1, 2. Moreover, φ_{1} and φ_{2} can be chosen such that either φ_{1} = t and φ_{2} = t^{−1} , or φ_{1} = r_{1} and φ_{2} = r_{2} . But then

(3)  P_{j} = P_{j}^{(n−1)} = g(φ_{j}(E)) = g_{n} ∘ ⋯ ∘ g_{1} ∘ φ_{j}(E).

The components P_{j} are of generation n − 1, hence some of the terms in (3) must cancel for both j = 1, 2. To make the remaining discussion less abstract, assume g_{1} = t and φ_{1} = t, φ_{2} = t^{−1} . The other cases may be treated equally.

Clearly P_{2} = g_{n} ∘ ⋯ ∘ g_{2} ∘ t ∘ t^{−1}(E) = g_{n} ∘ ⋯ ∘ g_{2}(E) is of generation n − 1. Let h^{−1} = g_{n} ∘ ⋯ ∘ g_{2} and consider h(P_{1}) = h ∘ g_{n} ∘ ⋯ ∘ g_{2} ∘ t ∘ t(E) = t ∘ t(E). We want to find the shortest path from h(E) to t ∘ t(E) and hence calculate d_{c}(h(E), t ∘ t(E)). As g = g_{n} ∘ ⋯ ∘ g_{1} is minimal, we know that the shortest path from h(E) to t(E) goes through E and that d_{c}(h(E), t(E)) = n. This implies that h(E) does not lie above E, or in other words that h(E) lies below the line y = \frac{1}{2} . It is then easy to realize that d_{c}(h(E), t ∘ t(E)) = n + 1 by considering Figure 2. The reason is that a shortest path would use the components t^{n}(E), n ∈ Z as much as possible.

This contradicts the assumption that P_{1} is of generation n − 1 and therefore also that \overline{D}_{n−1} ∪ P^{(n)} is disconnected.

Lemma 2. D is contained in the infinite strip {z : | Re z| < 2}.

Proof. Let L and R denote the two halfplanes L = {z ∈ C : Re z ≤ −2} and R = {z ∈ C : Re z ≥ 2}. We will show that g(L ∪ R) does not meet E for any g ∈ G. Let Λ be the set of all disks g(R), g ∈ G, that touch L (excluding R itself) and define L_{0} = r_{1}(R) ∈ Λ. We claim that L_{0} is the biggest disk in Λ. Any disk L_{λ} ∈ Λ can be written

L_{λ} = t^{n_{m+1}} ∘ r_{1} ∘ t^{n_{m}} ∘ ⋯ ∘ r_{1} ∘ t^{n_{1}}, n_{i} ∈ Z.

Represent a disk L_{λ} by the pair (a, ϱ) where a = Im z_{λ} and z_{λ} is the center of L_{λ}. The value ϱ denotes the radius of L_{λ}. To calculate ϱ observe that the reflection r_{1} takes the disk (a, ϱ) to (1/a, ϱ/a^{2}). Define the sequence ((a_{i}, ϱ_{i}))_{i=0}^{m} by

L_{0} ∼ (a_{0}, ϱ_{0}) ⟶^{r_{1}∘t^{n_{1}}}(a_{1}, ϱ_{1}) ⟶ ⋯ ⟶^{r_{1}∘t^{nm}}(a_{m}, ϱ_{m}).

As the translation t^{n_{m+1}} does not affect the radius of L_{λ} we have ϱ = ϱ_{m}. Obviously a_{i} ∈ Q so write a_{i} = b_{i}/c_{i} with b_{i}, c_{i} ∈ Z and let a_{0} = b_{0} = 0 and c_{0} = 1. Then

a_{i} = \frac{1}{a_{i−1} + n_{i}} = \frac{1}{\frac{b_{i−1}}{c_{i−1}} + n_{i}} = \frac{c_{i−1}}{b_{i−1} + c_{i−1}n_{i}}

or b_{i} = c_{i−1} and c_{i} = b_{i−1} + c_{i−1}n_{i}. We now claim that ϱ_{i} = ϱ_{0}/c_{i}^{2} and prove this by induction. For i = 0 the claim is obvious. Assume the relation holds for i = k − 1. Then for i = k,

ϱ_{k} = \frac{ϱ_{k−1}}{(a_{k−1} + n_{k})^{2}} = ϱ_{k−1} · a_{k}^{2} = \frac{ϱ_{0}}{c_{k−1}^{2}} · \frac{b_{k}^{2}}{c_{k}^{2}} = \frac{ϱ_{0}}{c_{k−1}^{2}} · \frac{c_{k−1}^{2}}{c_{k}^{2}} = \frac{ϱ_{0}}{c_{k}^{2}} .

As c_{m} ∈ Z we get that ϱ = ϱ_{m} is either infinite, or smaller than or equal to ϱ_{0} which is the radius of L_{0} . The case when ϱ is infinite is when L_{0} is mapped on the halfplane R. Since R is not in Λ, L_{0} is the biggest disk in Λ.

Using that L_{0} is both the biggest disk in Λ and the disk in Λ closest to the point z = 2, shows that the bulb L̃_{0} containing L_{0} is the bulb in ⋃_{g∈G} g(L ∪ R) coming closest to E. Now define the sequence of points (z_{i}) by

z_{0} = −2, z_{1} = r_{1}(2), z_{2} = r_{1} ∘ r_{2}(−2), z_{3} = r_{1} ∘ r_{2} ∘ r_{1}(2), ….

Then lim_{i→∞} z_{i} will be the point in L̃_{0} furthest away from z = −2 ∈ L. Let d_{i} be the distance d_{i} = |2 − z_{i}|. Then

d_{0} = 4, d_{i} = 4 − \frac{1}{d_{i−1}} , i ≥ 1.

Thus lim_{i→∞} d_{i} is the attracting fixed point of f = 4 − 1/f, which is 2 + \sqrt{3} . So the bulb L̃_{0} reaches a distance of 4 − (2 + \sqrt{3} ) = 2 − \sqrt{3} < 1 away from z = −2 and hence does not meet E.

Having proved these lemmas, we are ready to prove Theorem 4. We first show that D is open. Let F be the union of images of E under the identity and the sixteen mappings

r_{j}, t, t^{−1}, r_{j} ∘ t, r_{j} ∘ t^{−1}, t ∘ r_{j}, t^{−1} ∘ r_{j}, r_{j} ∘ t ∘ r_{j}, r_{j} ∘ t^{−1} ∘ r_{j}

for j = 1, 2. See Figure 1. Take F_{0} to be the interior of F. Then F_{0} is an open set which properly contains E. Hence we can write D = ⋃_{g∈G} g(F_{0}) and it follows that D is open, because it is the union of open sets.

To see that D is simply connected observe that D = ∪D_{n} and D_{n} ⊆ D_{n+1} , where D_{n} was defined in (2). As each D_{n} is a simply connected domain, the complements D_{n}^{∗} = \overline{C} \ D_{n} will be closed, compact and connected sets. Since (D_{n}^{∗})_{n} is a decreasing sequence, D_{n}^{∗} ⊆ D_{n−1}^{∗}, with D^{∗} = ∩D_{n}^{∗}, D^{∗} is also closed, compact and connected. Hence D is simply connected.

Next we prove the homogeneity of D. By Riemann’s mapping theorem F_{0} ⊆ D can be mapped onto the unit disk by a conformal mapping f. E is a compact subset of F_{0} , so f(E) is a compact subset of f(F_{0}) = D. So by Teichmüller’s theorem [8], for any pair of points z_{1}, z_{2} ∈ f(E) there exists a K -quasiconformal self-mapping h of \overline{C} which is the identity outside f(F_{0}) and maps z_{1} to z_{2} . From the construction of D, for any point z ∈ D there exists a mapping g ∈ G and a point w ∈ E such that z = g(w). Altogether, it follows that D is homogeneous with respect to the family

ℱ = G ∘ f^{−1} ∘ {h} ∘ f ∘ G ⊆ 𝒬(K).

To complete the proof, we show that D is not locally connected at ∞ ∈ ∂D. It will follow that D is not a Jordan domain and hence not a quasidisk. We construct a neighborhood of ∞ in the following way: For n ∈ N, let

C_{n} = {z : | Re z| ≤ 2n, | Im z| ≤ n}.

Then C_{n}^{∗} is a neighborhood of ∞. Fix n ≥ 1. For any neighborhood U ⊆ C_{n}^{∗} there is a number m ∈ N such that the points z_{1} = (n + m)i and z_{2} = −(n + m)i are in U ∩ D. However, the two points can not be connected in C_{n}^{∗} ∩ D because D is contained in the strip {z : | Re z| < 2}. Thus D is not locally connected at ∞.

3. Remarks

To end things off we provide a few comments. First of all, for the construction to work as it is, we must admit 𝒬(K) to contain both orientation-preserving and orientation-reversing mappings. Usually a K -quasiconformal mapping is orientation-preserving by definition. In particular this is the case in the theorems of Erkama and Sarvas.

The example is however easily modified to use a family of orientation-preserv-ing mappings only. The orientation is reversed each time we apply a reflection, r_{j}. So for g ∈ G to be orientation-preserving, it must consist of an even number of reflections.

Let Ẽ = E ∪ r_{1}(E). In the example we found for each z ∈ D a mapping g ∈ G and a point w ∈ E with z = g(w). If g consists of an even number of reflections, define g̃ = g and w̃ = w. If not, let g̃ = g ∘ r_{1} and w̃ = r_{1}(w). Then g̃ ∈ G is an orientation-preserving mapping such that z = g̃(w̃) for some point w̃ ∈ Ẽ. Ẽ will be a compact subset of some F̃_{0} ⊂ D. Thus the same construction of a homogeneous family as before can be carried out, but now with Ẽ, g̃ and w̃ instead of E, g and w.

As a second comment, we remark that the same construction can be done in more generality. Define r_{1} and r_{2} to be the reflections in the circles |z+b| = a and |z −b| = a, and t to be the translation t(z) = z +ia. Then for any 0 < a < b < ∞ and with

E = \{z = x + iy : |x| ≤ b − \sqrt{a^{2} − y^{2}} , |y| ≤ \frac{1}{2}a\}

a similar example can be carried out.

Acknowledgement. This article is based on work done in my diploma thesis [4], supervised by Kari Hag. The idea of the construction in the example was suggested by Frederick W. Gehring. I thank them both for their help and support.

References

[1] Erkama, T.: Quasiconformally homogeneous curves. - Michigan Math. J. 24, 1977, 157– 159.

[2] Erkama, T.: Möbius automorphisms of plane domains. - Ann. Acad. Sci. Fenn. Ser. A I Math. 10, 1985, 155–162.

[3] Gehring, F. W., and B. P. Palka: Quasiconformally homogeneous domains. - J. Anal. Math. 30, 1976, 172–199.

142 Geir Arne Hjelle

[4] Hjelle, G. A.: Quasidisks – examples and counterexamples. - Diploma Thesis, Norwegian University of Science and Technology (NTNU), 2002.

[5] Lehto, O., and K. I. Virtanen: Quasiconformal Mappings in the Plane. - Springer-Verlag, Berlin, 1973.

[6] Näkki, R., and B. P. Palka: Preprint.

[7] Sarvas, J.: Boundary of a homogeneous Jordan domain. - Ann. Acad. Sci. Fenn. Ser. A I Math. 10, 1985, 511–514.

[8] Teichmüller, O.: Ein Verschiebungssatz der quasikonformen Abbildung. - Deutsche Mathematik 7, 1944, 336–343.

Received 8 January 2004