Topological properties of quasiconformal automorphism groups

Let G⊊C\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$G \subsetneq \mathbb {C}$$\end{document} be a bounded, simply connected domain in C\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathbb {C}$$\end{document}, and denote by Q(G):=f:G⟶G|fis a quasiconformal mapping ofGontoG\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} Q(G):= \left\{ f: G \longrightarrow G \ |\ f \text { is a quasiconformal mapping of } G \text { onto } G \ \right\} \end{aligned}$$\end{document}the quasiconformal automorphism group of G. In a canonical manner, the set Q(G) carries the structure of a (non–abelian) group with respect to composition of mappings. Moreover, we endow the set Q(G) with the topology of uniform convergence by the supremum metric dsup(f,g):=supz∈Gf(z)-g(z),f,g∈Q(G).\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} d_{\sup }(f, g):= \sup \limits _{z \in G} \left| f(z) - g(z) \right| , f, g \in Q(G). \end{aligned}$$\end{document}In this paper, we present results concerning topological properties of Q(G) such as completeness, separability, path–connectedness, discreteness and compactness.


Introduction
Our idea of studying Q(G) was initiated by a paper of Gaier (1984, see [4]) about the conformal automorphism group which was motivated by the question, whether the identity id G of the domain G can uniformly be approximated by conformal automorphisms of G. Naturally, the same question arises for quasiconformal automorphisms of G using thereby the results about RðGÞ as a subspace of Q(G).
Group-theoretic and topological information can be transferred between quasiconformal automorphism groups by conjugation: Theorem 1 Let F be a conformal mapping from the unit disk D onto G. Then the mapping Furthermore, the mapping U is continuous if and only if the domain G has prime ends only of the first kind.
Proof We only have to show the statement about continuity of U.For this we use the known fact that G has prime ends only of the first kind if and only if the Riemann mapping F is uniformly continuous.First, assume G has prime ends only of the first kind.Then, analogously to the proof of Satz 2b) in Gaier (1984, [4]) we get also for ðf n Þ n in QðDÞ converging uniformly on D to f 2 QðDÞ, h n : where x F denotes the modulus of continuity of F. As F is uniformly continuous on D this yields the continuity of U. Secondly, assume U is continuous.Then, the assertion directly follows by Satz 2b) in Gaier (1984, [4]) since also the restriction of U to RðDÞ with U RðDÞ ð Þ¼RðGÞ is continuous.h Information on the theory of prime ends, originally due to Carathéodory, can be found e.g. in Pommerenke (1992, [9]).
As in the case of RðGÞ with Gaier (1984, cf. [4, Hilfssatz 4, p. 234]) also the space Q(G) becomes somehow topologically homogenous by the right multiplication, i.e. the operation f À! f g for fixed g 2 QðGÞ: Proposition 2 The right multiplication in Q(G) is an isometry.Gaier (1984) concluded that RðGÞ is a topological group if G has prime ends only of the first kind (cf.[4, p. 235, Satz 5a)]).In the quasiconformal setting, it turns out that a related statement remains valid as well: Proof By the same arguments as Gaier (1984) used for RðGÞ (cf.[4,Satz 4]) we get the characterization that Q(G) is a topological group if and only if Now let [ 0; f 2 QðGÞ and ðu n Þ n2N QðGÞ converge to id G .The domain G has prime ends only of the first kind such that due to Theorem 1 there exists such a sequence.
Since G is a Jordan domain, f extends homeomorphically to G, and, by the compactness of G, f is uniformly continuous on G, i.e. there is a d [ 0 such that For sufficiently large n we get ju n ðzÞ À zj\d for all z 2 G and therefore Gaier (1984) showed that RðGÞ is always complete, regardless of the boundary structure of G (cf. [4, Satz 1b), p. 229]).However, in this section we will show that Q (G) is always incomplete even for domains with prime ends only of the first kind.For the proof we use a special family of quasiconformal mappings: Definition 1 Let q 2 Cð½0; 1Þ be a strictly increasing function of the interval [0, 1] onto itself.Then the mapping Obviously, the mapping q provides the deviation of f q from being conformal as shown by the following result (to be found in Astala et. al. 2008, cf.[1, Section 2.6, pp.28-29]): Lemma 4 For each piecewise C 1 -mapping q 2 Cð½0; 1Þ as in Definition 1 the corresponding radial stretching f q belongs to QðDÞ with complex dilatation l f q ðzÞ ¼ z z Á jzjq 0 ðjzjÞ À qðjzjÞ jzjq 0 ðjzjÞ þ qðjzjÞ for almost every z 2 D, where q 0 denotes the (a.e.existing) derivative of q.
The radial stretching visualized in Fig. 1 is build with 123 Topological properties of quasiconformal...
Proof First, the case G ¼ D will be treated.Consider the sequence of strictly increasing and continuous piecewise linear self mappings of [0, 1] defined by which converges uniformly on [0, 1] to the non-injective limit function The corresponding sequence of radial stretchings f q n À Á n2N belongs to QðDÞ because of Lemma 4 and due to the uniform convergence of ðq n Þ n2N we get q n ðxÞ À e qðxÞ j j À! n!1 0 where analogously f e q ðzÞ :¼ e qðjzjÞ z jzj with f e q ð0Þ :¼ 0. However, the limit function f e q is not injective and therefore does not belong to QðDÞ.This shows the incompleteness of the metric space QðDÞ.For a general domain G, let z 0 2 G be a fixed inner point and let B$G be an open ball centered at z 0 in G. Clearly the sequence ðf q n Þ n of the proof's first part together with its limit function f e q can be transferred to B via conformal equivalence, denoted by g n and g q , respectively.Then, define quasiconformal automorphisms h n of G by and likewise for h q and g q .Now, the sequence h n ð Þ n2N converges uniformly to the non-injective limit function h q on G. h In [11] Volynec (1992) shows that in case of the space of conformal automorphisms the topological property of separability geometrically means that the boundary cannot be complicated in the sense of prime end theory, i.e.G has prime ends only of the first kind.This equivalence can be extended to the space of quasiconformal automorphisms: Theorem 6 The space Q(G) is separable if and only if G has prime ends only of the first kind.
Proof If the metric space Q(G) is separable, then, in particular, the subspace RðGÞ is separable, which is equivalent to G having prime ends only of the first kind by Volynec (1992) [11,Theorem 3,p. 201].For the other direction, we first consider the case G ¼ D. Each mapping f 2 QðDÞ extends homeomorphically to D, see e.g.Lehto (1987, cf. [6, p. 13]), such that QðDÞ is a subspace of the metric space C b ðDÞ of bounded continuous complex-valued functions on D. It is known from topology (see e.g.Conway (1990), cf.[3, Theorem 6.6, p. 140]) that C b ðDÞ is separable.This implies the separability of QðDÞ.Finally, in the general case for a domain G with prime ends only of the first kind, the separability of Q(G) follows from Theorem 1, since Q(G) is the continuous image of the separable space QðDÞ by U. h Naturally, this result raises the question for concrete countable and dense subsets of Q(G) in the case that G has prime ends only of the first kind, and how such automorphisms may be constructed, especially in the case of QðDÞ.

Path-connectedness of Q(G)
Dealing with the topological question about the decomposition of Q(G) into pathcomponents we first encounter the result of Schmieder (1986, cf.[10]) that the subspace RðGÞ is path-connected if and only if G has prime ends only of the first kind.For the whole space Q(G) we cannot show the same equivalence, but at least that also the whole space Q(G) consists of only one path-component in case of G having a simple boundary.In order to prove the corresponding Theorem 10 below we need some preparation.
At first, for the construction of paths in Q(G) it is natural to use liftings of paths from RðGÞ: Lemma 7 Let G be a domain having prime ends only of the first kind.Then for every f 2 QðGÞ and every r 2 RðGÞ, the mappings f and r f can be joined by a path in Q (G).
Definition 2 Let G(C be a simply connected domain and Obviously, by the Measurable Riemann Mapping Theorem for each Beltrami coefficient l 2 B L 1 ðGÞ, i.e. open unit ball of L 1 ðGÞ, there exists a unique normalized quasiconformal mapping f : G À! D with l f ¼ l almost everywhere in G.The main idea for proving the Theorem 10 below is to build intermediate paths in B L 1 ðGÞ and then to conclude from pointwise on uniform convergence for the corresponding normalized quasiconformal mappings.Therefore, we finally need Proposition 8 (Branner and Fagella (2014) [2, Theorem 1.30(b), p. 43]) Let K be an open subset of R, G(C be a Jordan domain, z 1 ; z 2 2 G distinct points and ðl t Þ t2K be a family in B L 1 ðGÞ.Suppose t À! l t ðzÞ is continuous for every fixed z 2 G (whenever defined).Moreover, assume there exists k\1 such that kl t k L 1 ðGÞ k for all t 2 K, and denote by f t : G À! D the quasiconformal mapping normalized w.r.t z 1 ; z 2 and satisfying l f t ¼ l t almost everywhere in G. Then t À! f t ðzÞ is continuous for every fixed z 2 G. and Proposition 9 (Näkki and Palka (1973) [8,Corollary 4.4,p. 432]) Let G 0 C be a domain with finitely many boundary components which is finitely connected on the boundary.Furthermore, let ðf n Þ n2N be a sequence of K-quasiconformal mappings of a domain G onto G 0 converging pointwise in G to a homeomorphism f.Then the sequence ðf n Þ n converges uniformly on G to f.By means of the results above we now can prove our theorem about pathconnectedness in Q(G).
Theorem 10 The space Q(G) is path-connected if the domain G has prime ends only of the first kind.
Proof First, the case G ¼ D will be considered.
Let f ; g 2 QðDÞ with complex dilatations l f ; l g 2 B L 1 ðDÞ.W.l.o.g, assume that l f 6 ¼ l g ; otherwise, g ¼ r f for some r 2 RðDÞ by the Measurable Riemann Mapping Theorem and f and g can be joined by a path in QðDÞ by Lemma 7.For an arbitrary, but fixed z Ã 2 D, we find r f ; r g 2 RðDÞ such that ðr f f Þð0Þ ¼ 0 ¼ ðr g gÞð0Þ and ðr f f Þðz Ã Þ [ 0 and ðr g gÞðz Ã Þ [ 0 with l r f f ¼ l f resp.l r g g ¼ l g almost everywhere in D. Now Lemma 7 provides paths c f ; c g : ½0; 1 À! QðDÞ joining f with r f f and g with r g g.
Next, let K :¼ ðÀa; 1 þ aÞ for some fixed a [ 0 and using the straight line we construct a new path C in B L 1 ðDÞ by which satifies sup t2K kl t k L 1 ðDÞ \1.For each t 2 K there exists a unique normalized (w.r.t.0; z H ) / t 2 QðDÞ with l / t ¼ l t a. e. in D and / t ¼ r f f for t 2 ðÀa; 0Þ and / t ¼ r g g for t 2 ð1; 1 þ aÞ.Obviously, by the definition of C the mapping t À! l t ðzÞ is continuous for each z 2 D such that the same is true for t À! / t ðzÞ by Proposition 8.
Due to the uniform boundedness of kl t k L 1 ðDÞ for all t 2 K the whole family ð/ t Þ t2K QðDÞ is K-quasiconformal for a suitable K.Then, by Proposition 9, pointwise convergent sequences / t n À Á n are even uniformly convergent on D such that the induced mapping H : K À! QðDÞ; t À! / t is continuous with respect to the topology of uniform convergence on QðDÞ.
Eventually, appropriately combining the paths c f , c g and H yields a path form f to g. Hence QðDÞ is path-connected.
Finally, for an arbitrary domain G having prime ends only of the first kind, the bijective mapping U : QðDÞ À! QðGÞ is continuous by Theorem 1. Thus Q(G) is path-connected as well.h Naturally, the question for the corresponding counterpart rises: Does G necessarily have prime ends only of the first kind if the space Q(G) is path-connected?If this question should happen to be answered negatively, which domain with at least one prime end not of the first kind could have a path-connected space of quasiconformal automorphisms?

Discreteness of Q(G)
For the space of conformal automorphisms RðGÞ in [4] Gaier (1984) gives examples for both discrete and non-discrete spaces which intimately depend on the structure of the boundary of the underlying domain G.
For the much larger space Q(G), however, we get that the subgroup

This implies now
Theorem 12 The space Q(G) does not have any isolated element.
Using Proposition 11 we can find a sequence ðf n Þ n2N Q 1 ðGÞ with pairwise distinct elements 6 ¼ f converging to f uniformly on G.Then, for each g 2 QðGÞ by the isometry of right multiplication in Q(G) (cf.Proposition 2) the sequence ðf n f À1 gÞ n QðGÞ converges to g without meeting g. h Naturally, the question raises how to construct explicit quasiconformal automorphisms uniformly approximating id G , especially in the case of G having at least one prime end not of the first kind, since then RðGÞ might be discrete following Gaier (1984, cf.can homeomorphically be mapped onto ½1; þ1Þ in the evident way.Therefore, if K is restricted to a compact set, also the corresponding subset of M will be compact in QðDÞ. This example suggests the conjecture that the maximal dilatation of a compact subset of Q(G) is necessarily uniformly bounded from above.However, Lehto and Virtanen (1973, cf. [7]) construct an example for non-good approximation of

Fig. 1
Fig. 1 Radial stretching on a cartesian grid in D from the following result of Kiikka (1983, cf.[5, Theorem 1, p. 252]).Proposition 11 Let G; G 0 C be domains, let f : G À! G 0 be a K-quasiconformal mapping and [ 0. Then there exists e K ! 1 and a e K -quasiconformal C 1diffeomorphism e f : G À! G 0 such that f ðzÞ À e f ðzÞ \ for every z 2 G.