Annales Academiæ Scientiarum Fennicæ Series A. I. Mathematica Volumen 19, 1994, 205–246

ABELIAN NONDISCRETE CONVERGENCE GROUPS IN THE PLANE

A. Hinkkanen and G. J. Martin

University of Illinois at Urbana–Champaign, Department of Mathematics Urbana, IL 61801, U.S.A.; aimo@symcom.math.uiuc.edu The University of Auckland, Department of Mathematics Auckland, New Zealand

Abstract. Let g generate a nondiscrete convergence group in the 2-sphere S^{2} . We show that then there exists a homeomorphism f of S^{2} onto itself such that (f^{−1} ∘g∘f)(z) equals e^{2πiα}z or e^{2πiα}/z̄ where α is an irrational number. In combination with known results this implies that every sense-preserving cyclic convergence group in S^{2} is topologically conjugate to a group of Möbius transformations. Further, we prove that any abelian nondiscrete convergence group on S^{2} is topologically conjugate to a Möbius group.

1. Introduction and results

1.1. The essential topological properties of uniformly quasiconformal groups are captured by the notion of a convergence group, defined by Gehring and Martin in [4, p. 335]. We say that a group G of homeomorphisms of the n-dimensional sphere S^{n} onto itself is a convergence group if it has the following convergence property: every sequence of elements of G contains a subsequence, say g_{j}, such that

(i) g_{j} → g and g_{j}^{−1} → g^{−1} uniformly on S^{n}, where g is a homeomorphism; or (ii) there are x_{0}, y_{0} ∈ S^{n} (possibly x_{0} = y_{0} ) such that g_{j} → x_{0} and g_{j}^{−1} → y_{0}

locally uniformly on S^{n} \ {y_{0}} and S^{n} \ {x_{0}}, respectively.

Any group of K -quasiconformal, or, for n = 1, of K -quasisymmetric functions, for a fixed K ≥ 1, is a convergence group. A group G is discrete if it does not contain a sequence of distinct elements converging to the identity mapping uniformly on S^{n}; otherwise, G is nondiscrete.

We say that the group G (or the function g) is topologically conjugate to a Möbius group (or to a Möbius transformation) if there is a homeomorphism f of S^{n} onto itself such that f^{−1} ∘ G ∘ f (or f^{−1} ∘ g ∘ f) is a Möbius group (or transformation). Of course, a Möbius group is any group of Möbius transformations, and we allow the elements of G as well as Möbius transformations to be sensereversing. We denote the identity mapping by Id and write g^{0} = Id, g^{j} = g ∘g^{j−1} and g^{−j} = (g^{−1})^{j} for j ≥ 1.

1991 Mathematics Subject Classification: Primary 30C62.

Research of both authors partially supported by the U.S. National Science Foundation.

1.2. We consider under what circumstances a cyclic convergence group G on S^{n} generated by g ≠ Id is topologically conjugate to a Möbius group, or, equivalently, the question of when such a function g is conjugate to a Möbius transformation. Suppose first that n = 1. If G is discrete then g is conjugate to an elliptic, parabolic, hyperbolic or orientation reversing Möbius transformation of S^{1} . If G is nondiscrete then g is also conjugate to a Möbius transformation h which we may take to be an irrational rotation z ↦ e^{2πiα}z where α is an irrational number. For these results, see [5] and [21].

Suppose then that n ≥ 2, and let G be discrete. Gehring and Martin ([4, p. 340]) classified the functions g that can occur as generators into three classes:

(i) g is called elliptic if g has finite order; (ii) g is parabolic if g has a unique fixed point x_{0} and then g^{j}(x) → x_{0} as j → ∞

or j → −∞ locally uniformly on S^{n} \ {x_{0}}; (iii) g is loxodromic if g has exactly two fixed points x_{1} and x_{2} and then, say,

g^{j}(x) → x_{1} and g^{−j}(x) → x_{2} as j → ∞, locally uniformly on S^{n} \ {x_{2}} and S^{n} \ {x_{1}}, respectively.

When n ≥ 3, we refer to [4, pp. 354–356] for discussion and references. Suppose that n = 2. Kerékjártó [14] proved that a sense-preserving loxodromic function is topologically conjugate to the Möbius transformation h(z) = 2z. The same proof applies to sense-reversing loxodromic functions and shows that they are conjugate to h(z) = 2z̄. Furthermore, a result of Sperner [20] and Kerékjártó [13] shows that a sense-preserving parabolic mapping is topologically conjugate to h(z) = z + 1. In [6], it was proved that a sense-reversing parabolic function is conjugate to h(z) = z̄ + 1.

A theorem due in part to Brouwer, Kerékjártó, and Eilenberg [3] shows that an elliptic element g in S^{2} is topologically conjugate to an orthogonal transformation, and thus, to h(z) = cz or h(z) = c/z̄ where c is a root of unity.

1.3. The question remains of what can be said in the situation that g generates a nondiscrete convergence group. This is settled by the following theorem, which is one of our main results.

Theorem 1. Let g generate a nondiscrete convergence group in S^{2} . Then there is a homeomorphism f of S^{2} onto itself such that

(1.1)  (f^{−1} ∘ g ∘ f)(z) = e^{2πiα}z

if g is sense-preserving, while

(1.2)  (f^{−1} ∘ g ∘ f)(z) = e^{2πiα}/z̄

if g is sense-reversing, where α is an irrational number.

Thus such a sense-preserving g is conjugate to an irrational rotation.

All of the above together yields the following consequence.

Corollary 1. If g generates a convergence group in S^{2} , then g is topologically conjugate to a Möbius transformation.

It follows that among the homeomorphisms of S^{2} , the functions topologically conjugate to Möbius transformations are exactly the ones that generate cyclic groups with the convergence property expressed in the definition of a convergence group, which is analogous to the characteristic property of normal families of analytic functions.

We note that in spite of this result for cyclic groups, not every discrete convergence group in S^{2} containing only sense-preserving elements is topologically conjugate to a Möbius group, as has been pointed out in [4, Theorem 7.31, p. 356]. For various results on noncyclic discrete convergence groups on S^{2} , see [16].

Our second main result is the following extension of Theorem 1.

Theorem 2. Let G be an abelian nondiscrete convergence group on S^{2} . Then there is a homeomorphism f of S^{2} onto itself such that f^{−1} ∘ G ∘ f is a Möbius group.

Let G be a (not necessarily cyclic) nondiscrete convergence group on S^{2} . We say that x ∈ S^{2} is in the limit set L^{∗}(G) of G if and only if there is y ∈ S^{2} (possibly y = x) and a sequence g_{j} ∈ G such that g_{j} → x locally uniformly in S^{2} \ {y} as j → ∞. (Note that the more usual definition of L^{∗}(G) by means of properly discontinuous action would lead to L^{∗}(G) being always equal to S^{2} .) The proof of [4, Theorems 4.5 and 4.9] shows that either L^{∗}(G) = S^{2} , or L^{∗}(G) is a perfect nowhere dense subset of S^{2} , or L^{∗}(G) consists of at most two points. In the last case we say that G is elementary. By Theorem 1, a nondiscrete cyclic convergence group has an empty limit set and is thus elementary.

In general one might ask what can be said about the structure of an elementary convergence group G. If L^{∗}(G) consists of exactly one point, then the example in [4, p. 356] shows that G need not be topologically conjugate to a Möbius group if G is discrete. A simple modification of that same example, obtained by replacing the Fuchsian group acting on the unit disk by the group of all Möbius transformations of the unit disk onto itself, gives a counterexample for nondiscrete groups.

However, it seems to us that when L^{∗}(G) is empty or consists of two points, positive conjugacy results might be obtained by means of introducing a certain metric on S^{2} that is invariant under G. As the development of the theory of this metric involves a lot of detailed work, we will postpone it to another paper.

Throughout the paper, we denote the complex plane by C and identify S^{2} with \overline{C} = C ∪ \{∞\}, whenever convenient. We denote by H(S^{2}) the group of all homeomorphisms of S^{2} onto itself. We write B(z_{0}, r) = {z ∈ C : |z − z_{0}| < r}, B(r) = B(0, r), S(z_{0}, r) = {z ∈ C : |z − z_{0}| = r} and S(r) = S(0, r) whenever z_{0} ∈ C and r > 0. We also write S^{1} for S(0, 1). We denote the real axis by R and write \overline{R} for the extended real axis R ∪ \{∞\}. The set of integers is denoted by Z.

We next note the following result.

Theorem 3. Let G be a closed abelian subgroup of H(S^{2}) with only sensepreserving elements and with the convergence property. Then G is isomorphic as a topological group to one of the following:

(1) R, R^{2} , R ⊕ Z, R ⊕ Z_{m}, C \ {0}, S^{1} , S^{1} ⊕ Z, or

(2) Z, Z^{2} , Z_{m}, Z ⊕ Z_{m}, Z_{2} ⊕ Z_{2} ,

the group operation being multiplication on C \ {0}, S^{1} , and the S^{1} -part of S^{1} ⊕ Z, and addition otherwise. The possibilities in (1) occur if and only if G is nondiscrete.

Here Z_{m} = Z/(mZ), where m ≥ 2. The group Z_{2} ⊕Z_{2} , also denoted by D_{2} , is the Vierergruppe. For example, {z ↦ λz : λ > 0} is isomorphic to the additive group R. Not all closed nondiscrete abelian subgroups of H(S^{2}) are convergence groups. One example of this is

G = {(x, y) ↦ (sx, ty) : s, t ∈ R \ {0}}.

To prove Theorem 3, we note that if G is as in Theorem 3, then G is topologically conjugate to a Möbius group. If G is nondiscrete, this follows from Theorem 2 above, and if G is discrete, the same follows by combining [4, Theorem 5.15, p. 343], and [16, Theorem 4.3, p. 397]. Considering what kind of closed abelian sense-preserving Möbius groups there can be, we arrive at the list given in (1) and (2). For discrete groups, all possibilities (up to conjugacy by a Möbius transformation, which does not affect the conclusion of Theorem 3) are listed in [2, pp. 84–90]. Note that all abelian Möbius groups are elementary ([2, pp. 69, 70, 83]). For nondiscrete groups, the possibilities can be gathered from the proof of Theorem 2. Alternatively, they may be obtained from the description of elementary Möbius groups with only sense-preserving elements as given in [2, p. 84], considering the possibilities that can occur when the group is, in addition, abelian and closed. This proves Theorem 3.

In the proof of Theorem 2, we shall make use of a result of Kerékjártó ([12, p. 115]) that states the following. If h_{t} is a one-parameter family of sensepreserving homeomorphisms of S^{2} onto itself, all of them, other than the identity, fixing the same point and no other point, then there is a homeomorphism f of \overline{C} such that (f^{−1}∘h_{t}∘f)(z) = z+t for all real t. Later on in the proof of Theorem 2, we shall need a similar result for a one-parameter family of loxodromic mappings. As it seems to us that this result has not been obtained in previous literature, we shall prove it here. We state it separately for possible later reference.

Theorem 4. Let h_{t} be a one-parameter family of loxodromic mappings of S^{2} . Thus each h_{t} (for t ∈ R) is a sense-preserving homeomorphism of S^{2} onto itself, each h_{t} for t ≠ 0 can be topologically conjugated to a loxodromic or hyperbolic Möbius transformation, and h_{t+u} = h_{t}∘h_{u} for all real t and u. Further, the mapping of R into H(S^{2}) taking t onto h_{t} is a continuous function of t. Then there is a homeomorphism f of S^{2} onto itself such that (f ∘ h_{t} ∘ f^{−1})(z) = e^{t}z for all z ∈ \overline{C} = S^{2} and all real t.

If each h_{t} commutes with every rotation z ↦ cz, where |c| = 1, then we may choose f so that f also commutes with every such rotation.

The proof of Theorem 4 will be given in Subsection 6.13.

Finally, we want to observe the following two interesting corollaries of our work here, suggested by the referee.

Corollary 2. There is a neighbourhood 𝒰 of the identity mapping in H(S^{2}) such that no nontrivial convergence group of S^{2} is contained in 𝒰 .

Thus every convergence group has the “no small subgroups” property. It is clear that in order to establish the result it is enough to consider only cyclic groups. A nondiscrete cyclic group is necessarily a conjugate of an irrational rotation by Corollary 1. The sphere is then foliated by invariant topological circles, one of which must have large diameter. It is not too difficult to see that the result follows by consideration of the nearly transitive action of the cyclic group on this circle. If the cyclic group is discrete and either parabolic or loxodromic the result is immediate. There remains only the case that the cyclic group has finite order. Again however the existence of an invariant foliation by topological circles, since the map is topologically conjugate to a rotation, leads directly to a proof. Indeed we see that the neighbourhood 𝒰 can be taken to be of the form

𝒰 = {f ∈ H(S^{2}) : sup{q(x, f(x)) : x ∈ S^{2}} < ε}

for some suitable ε > 0 where q(x, y) denotes the chordal distance between x and y.

The following result is a consequence of a group having no small subgroups (see [10, Theorem 6, p. 95]).

Corollary 3. Any closed nondiscrete convergence group of S^{2} contains a nontrivial one-parameter subgroup.

We would like to thank the referee for his detailed comments. In particular, he suggested the argument given in Subsection 6.7, which is perhaps simpler than our original argument for this case, he pointed out a difficulty in our original argument for Case II in Section 6, which prompted us to provide the present proof, and he suggested the previous two corollaries.

We would also like to thank Pekka Tukia for useful suggestions.

2. Outline of the proof of Theorem 1

2.1. As the proof of Theorem 1 involves several steps that sometimes require lengthy proofs, we formulate two lemmas that essentially imply Theorem 1. Then we prove these lemmas in the sections that follow. We assume throughout that g generates a nondiscrete convergence group G on S^{2} .

Lemma 1. Suppose that g is topologically conjugate to the function z ↦ e^{2πiα}z for some irrational α whenever g is sense-preserving. Then g is topologically conjugate to z ↦ e^{2πiα}/z̄ for some irrational α whenever g is sense-reversing.

In view of Lemma 1, we may assume from now on that g is sense-preserving. One of the most difficult auxiliary results we need to prove is the following.

Lemma 2. If g is as in Theorem 1 and is sense-preserving, then there is no sequence m_{j} → ∞ such that g^{m_{j}} → x_{0} and g^{−m_{j}} → y_{0} uniformly on compact subsets of S^{2} \ {y_{0}} and S^{2} \ {x_{0}}, respectively.

Thus any sequence of iterates of g contains a subsequence, say g^{k_{j}}, such that g^{k_{j}} → h and g^{−k_{j}} → h^{−1} uniformly on S^{2} where h is a homeomorphism.

2.2. Recall that H(S^{2}) is the group of all homeomorphisms of S^{2} onto itself. For x, y ∈ S^{2} we write q(x, y) for the chordal distance of x and y induced by the Euclidean metric in R^{3} . For f_{1}, f_{2} ∈ H(S^{2}) we set

d(f_{1}, f_{2}) = sup{q(f_{1}(x), f_{2}(x)) : x ∈ S^{2}}.

The metric d induces the topology of uniform convergence on S^{2} , the usual topology of H(S^{2}).

Following Kerékjártó, we say that h ∈ H(S^{2}) is regular at x ∈ S^{2} if for each ε > 0 there is δ > 0 such that if q(x, y) < δ then q(h^{n}(x), h^{n}(y)) < ε for all integers n. Kerékjártó proved in [14, p. 250], that if h is a nonperiodic sense-preserving homeomorphism of S^{2} that is regular at each point of S^{2} , then h is topologically conjugate to a rotation z ↦ e^{2πiα}z, where α is irrational. To complete the proof of Theorem 1, it thus suffices to show that g is regular on S^{2} .

If g is as in Theorem 1 and is sense-preserving, pick x ∈ S^{2} and ε > 0. If g is not regular at x, then there is a sequence of points y_{k} with q(x, y_{k}) → 0 and of integers m_{k} with |m_{k}| → ∞ such that

q(g^{m_{k}}(x), g^{m_{k}}(y_{k})) ≥ ε

for all k. By Lemma 2, we may pass to a subsequence without changing notation and assume that g^{m_{k}} → h uniformly on S^{2} where h is a homeomorphism. But then g^{m_{k}}(x) → h(x) and g^{m_{k}}(y_{k}) → h(x), which gives a contradiction. Thus g is regular at x. This proves Theorem 1, subject to Lemmas 1 and 2.

Remark. It appears that well-known results of Montgomery and Zippin on compact connected transformation groups in R^{3} (and S^{2} ), such as [17, Theorem 6.7.1, p. 260], do not directly apply here, for even if Lemma 2 implies that the closure of G in H(S^{2}) is compact, there seems to be no obvious way to prove that it is also connected. In retrospect, of course, it is seen that this closure is indeed connected.

3. Proof of Lemma 1

Let the assumptions of Lemma 1 be satisfied, and let g be sense-reversing. We may perform a preliminary conjugation and assume that g^{2}(z) = e^{2πiα}z where α is irrational. Then g commutes with e^{2πiα}z so that

(3.1)  g(cz) = cg(z) for all z ∈ C

whenever c = e^{2πinα} for some integer n, and thus, by continuity, whenever |c| = 1. Thus g maps each circle S(r) onto a circle S(u(r)) where u is a homeomorphism of (0, ∞) onto itself. Also u ∘ u = Id since g^{2}(z) = e^{2πiα}z.

If u is increasing, then (3.1) implies that g is sense-preserving, which is a contradiction. Thus u is strictly decreasing and has a unique fixed point t on (0, ∞). We let v be any increasing homeomorphism of [t, ∞) onto [1, ∞) so that v(t) = 1. For 0 < r < t, we set v(r) = 1/v(u(r)). Thus v is an increasing homeomorphism of (0, ∞) onto itself, and (v ∘ u ∘ v^{−1})(r) = 1/r for all r > 0.

By (3.1), for all r > 0 there is ψ(r) with |ψ(r)| = 1 such that

g(re^{iθ}) = u(r)ψ(r)e^{iθ}

for all r and θ. Clearly ψ is a continuous function of r on (0, ∞). We have ψ(r)ψ(u(r)) = e^{2πiα} for all r > 0 since g^{2}(z) = e^{2πiα}z. In particular, ψ(t)^{2} = e^{2πiα}. We define f(0) = 0, f(∞) = ∞, f(re^{iθ}) = v(r)e^{iθ} for 0 < r ≤ t, and

f(re^{iθ}) = v(r)ψ(r)\overline{ψ(t)} e^{iθ} for t < r < ∞,

so that f is a homeomorphism of S^{2} onto itself. One can verify that

(f ∘ g ∘ f^{−1})(z) = ψ(t)/z̄.

This proves Lemma 1.

4. Results from two-dimensional topology

4.1. We shall often need to refer to convergence of continua. Recall that a subset C of S^{2} is called a continuum if C is closed and connected and contains at least two points. If C and C_{i} for i ≥ 1 are continua in S^{2} , we say that C_{i} converges to C as i → ∞, and write C_{i} → C, if any neighbourhood of any point of C intersects all but finitely many C_{i}, and if any point outside C has a neighbourhood that intersects only finitely many C_{i}. A basic result that can be found in [22, Theorem 7.1, p. 8 and (9.12), p. 12], states the following.

Lemma 3. Any sequence C_{i} of continua, contained in a fixed compact subset of the plane, contains a subsequence converging to a nonempty closed connected set C. If each C_{i} has diameter at least ε, for some fixed positive ε, then C is a continuum of diameter at least ε.

4.2. The next lemma is a purely topological result which is easy to believe and which should be known, but for which we have not been able to find a direct reference. Recall that a set Γ is locally connected at z_{0} ∈ Γ if any neighbourhood U_{1} of z_{0} contains a neighbourhood U_{2} of z_{0} such that Γ ∩ U_{2} is contained in a single component of Γ ∩ U_{1} ([22, p. 15]).

When we talk about straight lines on S^{2} , we are identifying S^{2} with \overline{C}.

Lemma 4. Let D be a simply connected domain in S^{2} , set Γ = ∂D and suppose that Γ is not locally connected at z_{0} ∈ Γ. Then Γ contains a continuum of points ζ, contained in a preassigned neighbourhood of z_{0} , with the following properties. There are parallel lines L_{1} and L_{2} , which we can choose to intersect a preassigned neighbourhood of ζ, containing ζ in the open strip domain S determined by them, and there are distinct components C_{i} of S ∩ Γ such that the disjoint continua \overline{C}_{i} tend to a continuum 𝒞 containing ζ and such that if i < j < k, then C_{j} separates C_{i} from C_{k} and from 𝒞 in S. The continuum 𝒞 is contained in the component of \overline{S} ∩ Γ containing ζ. Each set \overline{C}_{i} as well as 𝒞 intersects both L_{1} and L_{2} . If the coordinate axes are chosen so that L_{1} and L_{2} are parallel to the y -axis, then for i ≥ 1 and j = 1, 2, there are points in \overline{C}_{i} ∩ L_{j} with y -coordinate y_{ij} and points in 𝒞 ∩ L_{j} with y -coordinate y_{j} such that y_{ij} strictly increases to y_{j} as i → ∞, for j = 1, 2, or such that y_{ij} strictly decreases to y_{j} as i → ∞, for j = 1, 2.

Furthermore, any neighbourhood W_{0} of ζ contains a neighbourhood W of ζ such that among the components of W ∩ D, there are distinct components V_{i} for i ≥ 1 containing points w_{i} that tend to ζ as i → ∞.

If D_{0} is a domain with D_{0} ⊂ D and ∂D ⊂ ∂D_{0} , then there are distinct components V_{i} of W ∩ D with w_{i} ∈ V_{i} ∩ D_{0} such that w_{i} → ζ as i → ∞.

We remark that 𝒞 need not be the closure of any component C of Γ ∩ S but may be larger than any such closure. For instance, Γ ∩ S might be a horizontal segment while 𝒞 could contain an additional vertical segment of L_{1} and/or L_{2} .

To get an example, suppose that

Γ = {iy : −1 ≤ y ≤ 1} ∪ {x + iy : y = sin(1/x), 0 < x ≤ 1}

in Lemma 4. Then Γ is not locally connected at the origin. For the lines L_{1} and L_{2} , we may take the horizontal lines y = ±\frac{1}{2} , so that S is the strip \{x + iy : |y| < \frac{1}{2}\}. Clearly \overline{S} ∩ Γ has infinitely many components, and they cluster to 𝒞 = \{iy : |y| ≤ \frac{1}{2}\}, which is the component of \overline{S} ∩ Γ containing the origin. In this case 𝒞 happens to be the closure of a component of S ∩ Γ. The continuum of points ζ referred to in Lemma 4 can be taken to be the closed line segment from −iy_{0} to iy_{0} , where 0 < y_{0} < 1.

4.3. Let Γ be a continuum. Recall that a point z_{0} ∈ Γ is called a cut point if Γ \ {z_{0}} is not connected. We shall need to show that various continua are Jordan curves. For this purpose, we quote the converse of Jordan’s curve theorem ([22, Statement (2.4), p. 34]).

Lemma 5. Let Γ be a continuum in S^{2} . Let D be one of the components of S^{2} \ Γ, and suppose that Γ = ∂D. If, in addition, Γ is locally connected at each of its points and if Γ has no cut points, then Γ is a Jordan curve.

We will use Lemma 4 to prove that a given continuum Γ is locally connected. It is also necessary to have a criterion that guarantees that Γ has no cut points.

Lemma 6. Let Γ be a continuum in S^{2} , and let D and E be distinct components of S^{2} \ Γ. If Γ = ∂D = ∂E then Γ has no cut points.

Note that S^{2} \ Γ might have more than two components, and yet Γ might be the boundary of each of them (cf. Lakes of Wada [8, p. 143]).

Proof of Lemma 6. Let z_{0} be a cut point of Γ, and let x and y be points in distinct components of Γ\{z_{0}}. We apply [23, Theorem IV.5.6, p. 112], taking the closed set F in that theorem to be {z_{0}}. We conclude that there is a continuum γ ⊂ D ∪ {z_{0}} that separates x and y. Since x, y ∈ ∂E, there are points of E in two distinct components of S^{2} \ γ. Thus E, being connected, intersects γ, which is impossible since γ ⊂ D ∪ {z_{0}}. Hence Γ has no cut points, and Lemma 6 is proved.

4.4. Proof of Lemma 4. Let the assumptions of Lemma 4 be satisfied. Then Γ is not locally connected, and in particular, Γ is a continuum (rather than a point or the empty set). We shall assume that z_{0} ≠ ∞, leaving the simple modifications required in the case z_{0} = ∞ to the reader.

By [22, Theorem 10.2, p. 13], there is a disk neighbourhood U of z_{0} contained in a preassigned neighbourhood of z_{0} , and an infinite sequence C′_{i} of distinct components of Γ ∩ \overline{U} converging to a nondegenerate limit continuum C′ which contains z_{0} and is disjoint from each C′_{i}. We have C′ ⊂ Γ, and indeed C′ is contained in the component of Γ ∩ \overline{U} containing z_{0} . Note that there are points z′_{i} ∈ C′_{i} with z′_{i} → z_{0} as i → ∞.

In the rest of the proof, we shall make various assumptions about the existence of limits that can be justified by passing to subsequences. For the sake of convenience, we shall assume that this is done automatically but without changing notation. Also we may say that certain statements that are true for all large i, are true for all i.

Since Γ is connected, we have C′_{i} ∩ ∂U ≠ ∅ for all i. We find ζ′_{i} ∈ C′_{i} ∩ ∂U with ζ′_{i} → ζ′ ∈ C′ ∩ ∂U as i → ∞. Let L_{1} and L_{2} be parallel lines orthogonal to the line segment from z_{0} to ζ′ such that z_{0} and ζ′ are outside \overline{S} and indeed lie in distinct components of C \ \overline{S}, where S is the open strip domain bounded by L_{1} and L_{2} . Each C′_{i} intersects both L_{1} and L_{2} . Let C_{i} be a component of C′_{i} ∩ S whose closure intersects both L_{1} and L_{2} . Note that \overline{C}_{i} → 𝒞 where 𝒞 is contained in a component of C′ ∩ \overline{S}, and that the continuum 𝒞 intersects both L_{1} and L_{2} . Without loss of generality, we assume that L_{1} and L_{2} are vertical. For j = 1, 2, let y_{ij} be the y -coordinate of a point in \overline{C}_{i} ∩ L_{j}. Then y_{ij} → y_{j} as i → ∞, where y_{j} is the y -coordinate of a point in 𝒞 ∩ L_{j}. For any distinct i and k, we have

(y_{i1} − y_{1})(y_{i2} − y_{2}) > 0  and  (y_{i1} − y_{k1})(y_{i2} − y_{k2}) > 0

since C′_{i}, C′_{k} and C′ are disjoint. Thus we may assume, for example, that y_{i1} and y_{i2} are strictly decreasing sequences.

For a point ζ that is to have the properties specified in Lemma 4, we could take any point in 𝒞 ∩ S ⊂ S ∩ Γ, and so it is clear that there is a continuum of such points. Given ζ and a neighbourhood W_{0} of ζ, which we may assume to be a disk, we can replace L_{1} and L_{2} by some other vertical lines to ensure that they intersect W_{0} , and we still obtain continua \overline{C}_{i} with all of the above properties. We assume that this has been done without changing notation. It is now obvious that if i < j < k, then C_{j} separates C_{i} from C_{k} and from 𝒞 in S.

There are points ζ_{i} ∈ C_{i} with ζ_{i} → ζ as i → ∞. Let U_{i} = B(ζ_{i}, ϱ_{i}) be a neighbourhood of ζ_{i} disjoint from Γ \ C_{i}, where ϱ_{i} → 0. Since ζ_{i} ∈ ∂D_{0} , there is a point α_{i} ∈ U_{i} ∩ D_{0} .

Now W = W_{0} ∩ S is a neighbourhood of ζ contained in W_{0} , and α_{i}, ζ_{i} ∈ W for all i.

We claim that if k > n, then α_{2n} and α_{2k} lie in distinct components of W ∩ D. If not, we can join α_{2n} and α_{2k} by a polygonal arc in S ∩ D. But this arc must intersect C_{2n+1} , which is a contradiction. Thus W ∩ D has distinct components V_{i} for i ≥ 1 with w_{i} = α_{2i} ∈ V_{i} ∩ D_{0} and α_{2i} → ζ. This proves Lemma 4.

5. Proof of Lemma 2

5.1. To prove Lemma 2, suppose that g^{m_{j}} → x_{0} and g^{−m_{j}} → y_{0} locally uniformly on S^{2} \ {y_{0}} and S^{2} \ {x_{0}}, respectively. Since g is sense-preserving, it follows from [15, Statement 12.1, p. 157], that g has at least one fixed point z_{0} in S^{2} . One can verify that z_{0} = x_{0} or z_{0} = y_{0} , and by replacing g by g^{−1} , if necessary, we may assume that g(x_{0}) = x_{0} . Recall that there is a sequence n_{j} → ∞ such that g^{n_{j}} → Id and g^{−n_{j}} → Id uniformly on S^{2} .

5.2. As the rest of the proof of Lemma 2 is rather complicated, it is convenient to first outline the proof, stating intermediate results as sublemmas. After that we proceed to prove the sublemmas.

Sublemma 1. If U is a nonempty proper open subset of S^{2} and if p is a positive integer, then we cannot have \overline{g^{p}(U)} ⊂ U.

Sublemma 2. We have x_{0} = y_{0} , and x_{0} is the unique fixed point of g. Similarly, x_{0} is the only fixed point of g^{q} for any nonzero integer q. If g^{k_{j}} is another sequence with k_{j} → ∞ such that g^{k_{j}} → x_{1} and g^{−k_{j}} → y_{1} locally uniformly on S^{2} \ {y_{1}} and S^{2} \ {x_{1}}, respectively, then x_{1} = y_{1} = x_{0} .

5.3. Our eventual aim is to find a Jordan curve Γ containing x_{0} such that g^{k}(Γ) = Γ for some positive integer k. Suppose that such a curve Γ has been found, and let h be a homeomorphism of S^{2} = \overline{C} onto itself taking Γ onto \overline{R} with h(x_{0}) = ∞. Then g̃ = h ∘ g^{k} ∘ h^{−1} generates a nondiscrete convergence group, maps R onto itself, and fixes ∞. Since x_{0} is the unique fixed point of g^{k}, it follows that g̃ has no fixed points on R. Thus either g̃(x) > x for all x ∈ R, or g̃(x) < x for all x ∈ R. In either case,

(5.1)  g̃^{n}(x) → ∞

as n → ∞, for each real x.

Since g^{n_{j}} → Id, we have g^{kn_{j}} = (g^{n_{j}})^{k} → Id, and so g̃^{n_{j}} → Id as j → ∞. This contradicts (5.1), and it follows that the assumption that g^{m_{j}} → x_{0} must have been false. This then proves Lemma 2.

The curve Γ will arise as the common boundary of certain domains. To define the relevant open sets, we first need some more terminology and preliminary results.

5.4. The orbit Gx of x ∈ S^{2} is defined as the set Gx = {g^{n}(x) : n ∈ Z}. The sets G^{+}x = {g^{n}(x) : n ≥ 0} and G^{−}x = {g^{n}(x) : n ≤ 0} are called half orbits. We may assume that x_{0} = ∞ and also consider g as a homeomorphism of C onto itself. Then by a result of Homma and Kinoshita ([9, Theorem 5, p. 370]), there is a point x_{1} ∈ S^{2} \ {x_{0}} (and, indeed, an everywhere dense set of such points x_{1} ) for which G^{+}x_{1} is not dense in S^{2} .

Sublemma 3. No point x ∈ S^{2} has a dense orbit under G.

The following technical result is necessary.

Sublemma 4. If U ⊂ S^{2} \{x_{0}} is a nonempty open set with g(U) = U, then x_{0} ∈ ∂U. If U has only finitely many components U_{1}, …, U_{n}, then x_{0} ∈ ∂U_{i} for all i.

5.5. Choose x_{1} ∈ S^{2} \ \{x_{0}\} and suppose that x_{2} ≠ x_{0} and x_{2} ∉ \overline{Gx_{1}} . By conjugating g by a Möbius transformation, if necessary, we may assume that x_{2} and x_{0} are antipodal points so that q(x_{0}, x_{2}) = 2. For any y ∈ S^{2} we write

ϱ(y) = sup{q(x_{0}, g^{n}(y)) : n ∈ Z} ≥ q(x_{0}, y).

The choice of x_{1} and x_{2} implies that ϱ(x_{1}) < 2. Choose r so that ϱ(x_{1}) < r < 2. We define V = {y ∈ S^{2} : ϱ(y) > r} and E = {y ∈ S^{2} : q(x_{0}, y) > r}.

Sublemma 5. The set V is a nonempty open set, and in fact

(5.2)  V = g(V) = ⋃_{n=−∞}^{∞} g^{n}(E).

Furthermore, V has only finitely many components, and they are of the form

V_{i} = g^{i−1}(V_{1}) for 1 ≤ i ≤ N

for some N ≥ 1, where V_{1} is the component of V containing E and g^{N}(V_{1}) = V_{1} .

We have x_{0} ∈ ∂V_{i} for all i. Furthermore, we have ϱ(x) = r for all x ∈ ∂V \ {x_{0}}.

Sublemma 6. The set W = S^{2} \ \overline{V} is a nonempty open set. If W_{1} is a component of W then g^{M}(W_{1}) = W_{1} for some M ≥ 1, and x_{0} ∈ ∂W_{1} . Furthermore, any such component W_{1} is simply connected. We have ∂W_{1} ⊂ ∂W ⊂ ∂V .

Choose such a component W_{1} of W, and write Γ_{1} = ∂W_{1} . Then Γ_{1} is a continuum containing x_{0} , and g^{M}(Γ_{1}) = Γ_{1} . Next let 𝒱 be the component of S^{2} \ \overline{W}_{1} containing V_{1} , and let 𝒲 be the component of S^{2} \ \overline{𝒱} containing W_{1} .

Sublemma 7. The domains 𝒱 and 𝒲 are simply connected and have the same boundary Γ. We have x_{0} ∈ Γ and g^{MN}(Γ) = Γ. The continuum Γ has no cut points. We have Γ ⊂ Γ_{1} ⊂ ∂V so that ϱ(x) = r for all x ∈ Γ \ {x_{0}}.

The sole purpose of taking the complement so many times to obtain the sets W, 𝒱 and 𝒲 from V has been to ensure that we get two domains with the same boundary.

With the aid of Lemma 5, we now show that Γ is a Jordan curve. Since g^{MN}(Γ) = Γ, the proof of Lemma 2 can then be completed as explained in Subsection 5.3 above.

Sublemma 8. The continuum Γ is locally connected at each of its points. Thus Γ is a Jordan curve.

This completes the proof of Lemma 2, subject to Sublemmas 1–8.

5.6. Proof of Sublemma 1. Suppose that U is a nonempty proper open subset of S^{2} and that p is a positive integer such that F = \overline{g^{p}(U)} ⊂ U. Then F is a fixed compact proper subset of U, and for m ≥ 2, we have g^{pm}(U) ⊂ F ⊂ U. But g^{pn_{j}} → Id uniformly on S^{2} while g^{pn_{j}}(U) ⊂ F, which is a contradiction. This proves Sublemma 1.

5.7. Proof of Sublemma 2. Recall that g(x_{0}) = x_{0} . If x_{0} ≠ y_{0} , let U be a spherical disk centred at x_{0} such that y_{0} ∉ \overline{U}. Since g^{m_{j}} → x_{0} , there is j_{0} such that if j ≥ j_{0} and p = m_{j} then Γ = g^{p}(∂U) ⊂ U. Let the components of S^{2} \ Γ be D_{1} and D_{2} where \overline{D}_{1} ⊂ U. By Sublemma 1, we cannot have g^{p}(U) = D_{1} . Thus g^{p}(U) = D_{2} and so x_{0} = g^{p}(x_{0}) ∈ D_{2} .

Since y_{0} ∈ S^{2} \ \overline{U}, we have, with p = m_{j},

(5.3)  g^{−p}(S^{2} \ U) ⊂ g^{−p}(S^{2} \ D_{1}) = g^{−p}( \overline{D}_{2}) = \overline{U}.

Since S^{2} \ U is a compact subset of S^{2} \ {x_{0}}, we have g^{−m_{j}} → y_{0} uniformly on S^{2} \ U. This contradicts (5.3), and it follows that y_{0} = x_{0} .

Thus if z_{0} ≠ x_{0} , then g^{m_{j}}(z_{0}) → x_{0} . So we cannot have g(z_{0}) = z_{0} , and it follows that x_{0} is the unique fixed point of g.

Since g^{k} is a homeomorphism for any k ≥ 1 and g^{k}(x_{0}) = x_{0} , we have g^{k}(z_{0}) ≠ x_{0} if z_{0} ≠ x_{0} . Fix q ≥ 2 and write m_{j} = qk_{j} + r_{j} where 0 ≤ r_{j} ≤ q − 1. If z_{0} ≠ x_{0} and g^{q}(z_{0}) = z_{0} , then g^{m_{j}}(z_{0}) = g^{r_{j}}(z_{0}) so that g^{m_{j}}(z_{0}) belongs to a finite set not containing x_{0} . Thus g^{m_{j}}(z_{0}) does not tend to x_{0} . This contradiction shows that x_{0} is the unique fixed point of g^{q} for any q ≥ 1 and hence for any nonzero q.

If g^{k_{j}} → x_{1} and g^{−k_{j}} → y_{1} then at least one of x_{1} and y_{1} must be equal to x_{0} . If, for example, x_{1} = x_{0} , we argue as above to deduce that y_{1} = x_{0} also. This proves Sublemma 2.

5.8. Proof of Sublemma 3. Recall that there is a point x_{1} ∈ S^{2} \ {x_{0}} such that the half orbit G^{+}x_{1} is not dense in S^{2} . Suppose that x_{2} ∈ S^{2} and that Gx_{2} is dense in S^{2} . Since Gx_{0} = {x_{0}}, we have x_{2} ≠ x_{0} . We may or may not have x_{2} = x_{1} . Since Gx_{2} is dense, we have a_{j} = g^{k_{j}}(x_{2}) → x_{1} while a_{j} ≠ x_{1} for each j, for some sequence k_{j} with k_{j} → ∞ or k_{j} → −∞. Similarly, given any x_{3} ∈ S^{2} , we have b_{j} = g^{ℓ_{j}}(x_{2}) → x_{3} and b_{j} ≠ x_{3} , for some sequence ℓ_{j}. Choose x_{3} so that x_{3} ≠ x_{0} and so that G^{+}x_{1} fails to intersect some neighbourhood U of x_{3} . For each j, choose q_{j} = n_{p(j)} − k_{j} + ℓ_{j} where p(j) is chosen so that p(j + 1) > p(j) and q_{j} → ∞. Since g^{n_{p(j)}} → Id, we have g^{q_{j}}(a_{j}) = g^{n_{p(j)}}(b_{j}) → x_{3} .

By passing to a subsequence and using the definition of a convergence group together with Sublemma 2, we may assume that g^{q_{j}} → h uniformly on S^{2} , where h is a homeomorphism, or that g^{q_{j}} → x_{0} locally uniformly in S^{2} \ {x_{0}}. The latter alternative cannot occur since a_{j} → x_{1} ≠ x_{0} and g^{q_{j}}(a_{j}) → x_{3} ≠ x_{0} . Thus g^{q_{j}} → h, which implies that g^{q_{j}}(x_{1}) → h(x_{1}) and that g^{q_{j}}(a_{j}) → h(x_{1}). Hence h(x_{1}) = x_{3} and so x_{3} ∈ \overline{G^{+}x_{1}} , which gives a contradiction. It follows that Gx is not dense for any x and Sublemma 3 is proved.

5.9. Proof of Sublemma 4. If U is as in Sublemma 4, pick x ∈ U and note that g^{m_{j}}(x) ∈ U and g^{m_{j}}(x) → x_{0} . Thus x_{0} ∈ \overline{U} and so x_{0} ∈ ∂U. If U has only the components U_{1}, …, U_{n}, note that g maps each U_{i} onto some U_{j}. Thus the U_{i} can be partitioned into cycles of components with the following property. If U_{1}, …, U_{k}, for example, is such a cycle, and if the components have been renumbered in a suitable way, then g^{i−1}(U_{1}) = U_{i} for 1 ≤ i ≤ k, and g^{k}(U_{1}) = U_{1} .

If x_{0} ∉ \overline{U}_{1} then

x_{0} ∉ \overline{U}_{1} ∪ ⋯ ∪ \overline{U}_{k} = ⋃_{j=−∞}^{∞} g^{j}( \overline{U}_{1}),

which is impossible since g^{m_{j}} → x_{0} in some disk contained in U_{1} . Thus x_{0} ∈ ∂U_{1} and so x_{0} ∈ ∂U_{i} for 1 ≤ i ≤ k. The same argument applies to all the other cycles, and we deduce that x_{0} ∈ ∂U_{i} for 1 ≤ i ≤ n. This proves Sublemma 4.

5.10. Proof of Sublemma 5. The definition of ϱ readily implies that V is equal to the right hand expression in (5.2), and this representation of V shows that g(V) = V . Thus (5.2) holds. Let the components of V be denoted by V_{i} for i ≥ 1, and let V_{1} be the component of V containing E.

Since g(V) = V , it follows that g maps any V_{i} onto some V_{k}. Clearly any V_{i} is contained in, and thus coincides with, g^{k}(V_{1}) for some k. Recall that g^{±n_{j}} → Id. As V_{1} is open and g^{±n_{j}} → Id, if n_{j} is large enough, then g^{±n_{j}}(V_{1}) intersects V_{1} and thus g^{±n_{j}}(V_{1}) = V_{1} . Therefore V has only finitely many components, say V_{1}, …, V_{N}, which we may number so that V_{j} = g^{j−1}(V_{1}) for 1 ≤ j ≤ N while g^{N}(V_{1}) = V_{1} . Note that if q is the smallest positive integer such that g^{−q}(V_{1}) = V_{1} , then the components V_{1}, g^{−1}(V_{1}), …, g^{−(q−1)}(V_{1}) are distinct, and g^{q}(V_{1}) = g^{q}(g^{−q}(V_{1})) = V_{1} . Thus q = N, and the components g^{−1}(V_{1}), g^{−2}(V_{1}), …, g^{−(q−1)}(V_{1}) coincide with g^{N−1}(V_{1}), g^{N−2}(V_{1}), …, g(V_{1}).

By Sublemma 4, we have x_{0} ∈ ∂V_{i} for 1 ≤ i ≤ N.

Suppose that x ∈ ∂V \ {x_{0}}. Then x ∉ V so that ϱ(x) ≤ r. Since x ∈ ∂V , there are points w_{n} ∈ V tending to x with ϱ(w_{n}) > r. For each n, there is g^{k_{n}} such that q(x_{0}, g^{k_{n}}(w_{n})) > r. We pass to a subsequence, apply Sublemma 2, and find that g^{k_{n}} → h uniformly on S^{2} , where h is a homeomorphism, since we cannot have g^{k_{n}} → x_{0} uniformly in a neighbourhood of x. Thus g^{k_{n}}(w_{n}) → h(x) and so q(x_{0}, h(x)) ≥ r. Therefore q(x_{0}, g^{k_{n}}(x)) → r, and consequently ϱ(x) ≥ r. Hence ϱ(x) = r, as required. This proves Sublemma 5.

5.11. Proof of Sublemma 6. Clearly W = S^{2} \ \overline{V} is open. To show that W is nonempty, it suffices to prove that x_{1} ∈ W. Since ϱ(x_{1}) < r, we have x_{1} ∉ V . By the last statement of Sublemma 5, we also have x_{1} ∉ ∂V . Thus x_{1} ∈ W.

Since g(V) = V by Sublemma 4, we have g( \overline{V} ) = \overline{V} and so g(W) = W. Thus any g^{p} maps a component of W onto another such component. Let W_{1} be a component of W. Since g^{n_{j}} → Id and W_{1} is open, we have g^{n_{j}}(W_{1}) ∩ W_{1} ≠ ∅ and thus g^{n_{j}}(W_{1}) = W_{1} for all large j. Hence there is a smallest positive integer M such that g^{M}(W_{1}) = W_{1} , and then the components W_{1}, g(W_{1}), …, g^{M−1}(W_{1}) are distinct. Applying Sublemma 4 to

U = ⋃_{i=1}^{M} g^{i−1}(W_{1}),

we deduce that x_{0} ∈ ∂g^{i−1}(W_{1}) for 1 ≤ i ≤ M. In particular, x_{0} ∈ ∂W_{1} for any component W_{1} of W.

Since V has N components, the domain W_{1} is of connectivity at most N. Let Γ_{1}, …, Γ_{k} be the components of ∂W_{1} , where k ≤ N and x_{0} ∈ Γ_{1} . If Γ_{1} = {x_{0}} then k ≥ 2 and, since any iterate of g^{M} maps each Γ_{i} onto some Γ_{j}, it follows that

g^{jM}(Γ_{2} ∪ ⋯ ∪ Γ_{k}) = Γ_{2} ∪ ⋯ ∪ Γ_{k}

for each j ≥ 1. But Γ_{2} ∪ ⋯ ∪ Γ_{k} is a compact subset of S^{2} \ {x_{0}} on which g^{Mk_{j}} → x_{0} for some sequence k_{j} → ∞. (For if m_{j} = Mk_{j} +ℓ_{j} where 0 ≤ ℓ_{j} < M then, since g^{m_{j}} → x_{0} , we also have g^{Mk_{j}} = g^{m_{j}} ∘ g^{−ℓ_{j}} → x_{0} .) This contradiction shows that Γ_{1} is a continuum. For the same reason, it follows that Γ_{2}∪⋯∪Γ_{k} = ∅, so that Γ_{1} = ∂W_{1} , and W_{1} is simply connected. Obviously ∂W_{1} ⊂ ∂W ⊂ ∂V . This proves Sublemma 6.

5.12. Proof of Sublemma 7. Since \overline{W}_{1} and \overline{𝒱} are connected, it follows that 𝒱 and 𝒲 are simply connected, being components of the complement of a connected closed set. Let us write Γ = ∂𝒱 . We have x_{0} ∈ Γ since x_{0} ∈ ∂V_{1} (by Sublemma 5) and since x_{0} ∈ \overline{W}_{1} (by Sublemma 6) so that x_{0} ∉ 𝒱 . Clearly Γ ⊂ ∂W_{1} = Γ_{1} .

Since g^{M}(W_{1}) = W_{1} by Sublemma 6, we have g^{M}(S^{2} \\overline{W}_{1}) = S^{2} \\overline{W}_{1} . Thus g^{M} maps 𝒱 onto a component of S^{2} \ \overline{W}_{1} , and so does g^{kM} for any k ≥ 1. But V_{1} ⊂ 𝒱 and g^{N}(V_{1}) = V_{1} by Sublemma 5 so that g^{MN}(V_{1}) = V_{1} also. It follows that g^{MN}(𝒱 ) = 𝒱 . Therefore g^{MN}(Γ) = Γ, as required.

Since g^{MN}(𝒱 ) = 𝒱 , g^{M}(W_{1}) = W_{1} , and W_{1} ⊂ 𝒲 , we see in the same way that g^{MN}(𝒲 ) = 𝒲 . Clearly ∂𝒲 ⊂ ∂𝒱 = Γ. We claim that ∂𝒲 = Γ. If not, pick z_{0} ∈ Γ \ ∂𝒲 . Then z_{0} has a connected neighbourhood U with U ∩ \overline{𝒲} = ∅ while U ∩ 𝒱 ≠ ∅.

Since Γ ⊂ ∂W_{1} , we have z_{0} ∈ ∂W_{1} and so U ∩ W_{1} ≠ ∅. But W_{1} ⊂ 𝒲 so that U ∩ 𝒲 ≠ ∅. Since also U ∩ \overline{𝒲} = ∅, we get a contradiction. It follows that ∂𝒲 = Γ, as asserted.

Since 𝒱 and 𝒲 are simply connected domains with ∂𝒱 = ∂𝒲 = Γ, they are distinct components of S^{2} \ Γ. Now Lemma 6 implies that Γ has no cut points. We clearly have Γ = ∂𝒱 ⊂ ∂W_{1} = Γ_{1} . By Sublemma 6, we further have ∂W_{1} ⊂ ∂V , and by Sublemma 5, we have ϱ(x) = r for all x ∈ ∂V \ {x_{0}} and hence for all x ∈ Γ \ {x_{0}}. This proves Sublemma 7.

5.13. Proof of Sublemma 8. We apply Lemma 14 with D = 𝒱 and note that by Sublemma 7, the continuum Γ has no cut points. Thus Lemma 5 implies that Γ is a Jordan curve provided that Γ is locally connected at each of its points. To prove this, it suffices, by [22, Statement 10.4, p. 14], to show that Γ is locally connected at each point of Γ \ {x_{0}}. To get a contradiction, we now assume that z_{1} ∈ Γ \ {x_{0}} and that Γ is not locally connected at z_{1} .

We may assume that x_{0} = ∞ ∈ \overline{C} = S^{2} . Choose ε ∈ (0, \frac{1}{2}q(x_{0}, z_{1})) and set

W_{0} = {y ∈ S^{2} : q(z_{1}, y) < ε}.

By Lemma 4, there are parallel lines L_{1} and L_{2} , which we may assume to be vertical, intersecting W_{0} , and distinct components C_{i} of S ∩ Γ, where S is the open strip domain between L_{1} and L_{2} , with the properties described in Lemma 4. All the \overline{C}_{i} as well as their limit continuum 𝒞 are contained in a fixed compact subset E_{1} of S^{2} \ {x_{0}} = C. We may and will take E_{1} to be the cap

E_{1} = {y ∈ S^{2} : q(x_{0}, y) ≥ ε_{1}}

where ε_{1} is sufficiently small and in particular, 0 < ε_{1} < ϱ(x_{1})/2 < r/2. We further take ε_{1} so small that the \overline{C}_{i} and 𝒞 are contained in the smaller set {y ∈ S^{2} : q(x_{0}, y) ≥ 2ε_{1}}.

We define the set H_{1} , not necessarily a group, by

H_{1} = \{g^{n}, g^{−n} : n ∈ Z and q(x_{0}, g^{n}(z)) ≥ \frac{1}{2}ε_{1} for some z ∈ E_{1}\} ⊂ G.

We claim that there is r_{2} > 0 such that if h ∈ H_{1} and z_{1} ∈ E_{1} then q(x_{0}, h(z_{1})) ≥ r_{2} . If not, then there are g^{k_{j}} ∈ H_{1} and z_{j} ∈ E_{1} such that z_{j} → w ∈ E_{1} and q(x_{0}, g^{k_{j}}(z_{j})) → 0. Since for each j there is w_{j} ∈ E_{1} such that q(x_{0}, g^{k_{j}}(w_{j})) ≥ \frac{1}{2}ε_{1} or q(x_{0}, g^{−k_{j}}(w_{j})) ≥ \frac{1}{2}ε_{1} , we cannot have g^{k_{j}} → x_{0} and g^{−k_{j}} → x_{0} uniformly on E_{1} even if we pass to a subsequence. Thus, by Sublemma 2, we may assume that g^{k_{j}} → h uniformly on S^{2} where h is a homeomorphism. But h(x_{0}) = x_{0} = h(w), which gives a contradiction. Thus there is a number r_{2} with the required property. We may assume that r_{2} ≤ ε_{1}/2.

We set E_{2} = {y ∈ S^{2} : q(x_{0}, y) ≥ r_{2}} and claim that H_{1} is equicontinuous on E_{2} with respect to the chordal metric q. If not, there are sequences a_{i}, b_{i} ∈ E_{2} and g^{k_{i}} ∈ H_{1} , and a positive number ε_{2} , such that a_{i} → b ∈ E_{2} and b_{i} → b as i → ∞ while q(g^{k_{i}}(a_{i}), g^{k_{i}}(b_{i})) ≥ ε_{2} for all i. By passing to a subsequence, we may assume that g^{k_{i}} → h where h is a homeomorphism, since we cannot have g^{k_{i}} → x_{0} and g^{−k_{i}} → x_{0} uniformly even on E_{1} . But then g^{k_{i}}(a_{i}) → h(b) and g^{k_{i}}(b_{i}) → h(b), which gives a contradiction. Hence H_{1} is equicontinuous on E_{2} .

So if δ is a given positive number, there is δ′ > 0 such that

(5.4)  q(φ(z), φ(w)) < 10^{−2} min{ε_{1}, δ}

whenever z, w ∈ E_{2} , φ ∈ H_{1} and q(z, w) ≤ δ′.

Let L_{3} be the vertical line halfway between L_{1} and L_{2} . Choose δ_{0} > 0 so small that any two of the sets L_{i} ∩ E_{2} for 1 ≤ i ≤ 3 have chordal distance not less than 2δ_{0} from each other.

We now choose δ so small that

(5.5)  q(φ(z), φ(w)) < 10^{−2} min{ε_{1}, δ_{0}}

whenever z, w ∈ E_{2} , φ ∈ H_{1} and q(z, w) ≤ δ, and then we pick δ′ accordingly so that δ′ < ε_{1} and so that also (5.4) holds. We further assume that δ + r < 2, the diameter of S^{2} in the chordal metric.

5.14. Suppose that ζ_{1}, ζ_{3} ∈ L_{i} ∩ E_{1} and ζ_{2} ∈ L_{j} ∩ E_{1} , where i ≠ j and i, j ∈ {1, 2, 3}. Suppose that g^{k_{j}} ∈ H_{1} and that g^{k_{j}} → h uniformly on S^{2} as j → ∞, where h is a homeomorphism. We claim that

(5.6)  q(h(ζ_{1}), h(ζ_{2})) ≥ δ.

If, in addition, q(ζ_{1}, ζ_{3}) ≤ δ′, then

(5.7)  q(h(ζ_{1}), h(ζ_{3})) ≤ 10^{−2} min{ε_{1}, δ} = δ_{1},

say.

To prove (5.6), suppose that it does not hold. Then q(g^{k_{j}}(ζ_{1}), g^{k_{j}}(ζ_{2})) < δ for all large j. Since ζ_{1}, ζ_{2} ∈ E_{1} , we have g^{k_{j}}(ζ_{i}) ∈ E_{2} for i = 1, 2. Since g^{k_{j}} ∈ H_{1} , we also have g^{−k_{j}} ∈ H_{1} , by the definition of H_{1} . Now (5.5), applied to z = g^{k_{j}}(ζ_{1}), w = g^{k_{j}}(ζ_{2}) and φ = g^{−k_{j}}, gives q(ζ_{1}, ζ_{2}) < 10^{−2}δ_{0} , which contradicts the definition of δ_{0} . This proves (5.6).

If q(ζ_{1}, ζ_{3}) ≤ δ′, then by (5.4), we have

q(g^{k_{j}}(ζ_{1}), g^{k_{j}}(ζ_{3})) < δ_{1}

for all j. This gives (5.7) as j → ∞.

5.15. Fix i. There is a component Δ_{i} of S \ (C_{i−1} ∪ C_{i+1}) containing C_{i}. The domain Δ_{i} is simply connected, being a component of the complement of the connected set L_{1} ∪ L_{2} ∪ \overline{C_{i−1}} ∪ \overline{C_{i+1}} . Also Δ_{i} is bounded since C_{i} separates C_{i−1} from C_{i+1} in S.

Since C_{i} ⊂ Δ_{i} and \overline{C}_{i} ∩ L_{j} ≠ ∅, we have L_{j} ∩ ∂Δ_{i} ≠ ∅ for j = 1, 2. We claim that for all large i, we have

(5.8)  q(L_{j} ∩ ∂Δ_{i}) < δ′  for j = 1, 2,

where q(E) denotes the diameter of the set E in the chordal metric. Suppose that j = 1. By Lemma 4 and symmetry, we may assume, without loss of generality, that there are points ζ_{i} ∈ \overline{C}_{i} ∩ L_{1} with y -coordinate y_{i1} strictly decreasing to y_{1} as i → ∞, where y_{1} is the y -coordinate of a point in 𝒞 ∩ L_{1} . Then y_{1} is the maximal y -coordinate of any point in 𝒞 ∩ L_{1} (since 𝒞 and the \overline{C}_{i} are disjoint and contained in \overline{S} and since \overline{C}_{i} → 𝒞 ). Furthermore, if t_{i} is the maximal y - coordinate of any point in \overline{C}_{i} ∩ L_{1} then t_{i} → y_{1} also as i → ∞, since \overline{C}_{i} → 𝒞 . Now if u_{i} is the minimal y -coordinate of any point in \overline{C}_{i} ∩ L_{1} , then by the above, we have t_{i−1} − u_{i+1} → 0 as i → ∞. Clearly L_{1} ∩ ∂Δ_{i} is contained in the closed interval of L_{1} lying between the points of L_{1} with y -coordinates t_{i−1} and u_{i+1} . This proves (5.8) for j = 1, and the proof for j = 2 is similar.

If j ≥ i + 2, then Δ_{i} ∩ Δ_{j} = ∅. For if not, then Δ_{i} ∪ Δ_{j} is connected so that C_{i} and C_{j} can be joined in Δ_{i} ∪ Δ_{j} and hence in S \ C_{i+1} . This is impossible since C_{i+1} separates C_{i} from C_{j} in S, by Lemma 4. Thus Δ_{i} ∩ Δ_{j} is empty.

5.16. Pick a point z_{i} ∈ C_{i} ∩ L_{3} , so that z_{i} ∈ Δ_{i}. Since z_{i} ∈ Γ, we have ϱ(z_{i}) = r by Sublemma 7. Choose a sequence g^{k_{j}} such that q(x_{0}, g^{k_{j}}(z_{i})) → r. By Sublemma 2, we may assume that g^{k_{j}} → h_{i} and g^{−k_{j}} → h_{i}^{−1} uniformly on S^{2} , where h_{i} is a homeomorphism. Then q(x_{0}, h_{i}(z_{i})) = r while q(x_{0}, h_{i}(z)) ≤ r for all z ∈ Γ. Since z_{i} ∈ E_{1} and r > ε_{1}/2, we have g^{k_{j}} ∈ H_{1} for all large j.

Fix i so large that (5.8) holds. For j = 1, 2, there is ζ_{j} ∈ L_{j} ∩ ∂Δ_{i} with ζ_{j} ∈ \overline{C}_{i}. Hence ζ_{j} ∈ Γ∩E_{1} also. Suppose that ζ′_{j} ∈ L_{j} ∩∂Δ_{i}. Then q(ζ_{j}, ζ′_{j}) < δ′ by (5.8). Also ζ′_{j} ∈ E_{1} since δ′ < ε_{1} . By (5.7), we obtain

(5.9)  q(h_{i}(ζ_{j}), h_{i}(ζ′_{j})) ≤ δ_{1} ≤ 10^{−2}δ.

Next, by (5.6) we have

(5.10)  q(h_{i}(z_{i}), h_{i}(ζ′_{j})) ≥ δ

for any such point ζ′_{j}, and in particular if ζ′_{j} = ζ_{j}. We define

E_{3} = {y ∈ S^{2} : q(x_{0}, y) ≤ r or q(h_{i}(ζ_{1}), y) ≤ δ_{1} or q(h_{i}(ζ_{2}), y) ≤ δ_{1}}.

Then h_{i}(z_{i}) ∈ ∂E_{3} by (5.9) and (5.10). Since ∂Δ_{i} \ (L_{1} ∪ L_{2}) ⊂ \overline{C_{i−1}} ∪ \overline{C_{i+1}} ⊂ Γ, we further have h_{i}(∂Δ_{i}) ⊂ E_{3} .

Recall that δ + r < 2 = q(x_{0}, x_{2}) and define E_{4} = {y ∈ S^{2} : q(x_{2}, y) < 2 − δ − r}. Then E_{4} ⊂ S^{2} \ E_{3} . Since h_{i}(∂Δ_{i}) ⊂ E_{3} , we thus have either S^{2} \ E_{3} ⊂ h_{i}(Δ_{i}) or (S^{2} \ E_{3}) ∩ h_{i}(Δ_{i}) = ∅. Since h_{i}(z_{i}) ∈ h_{i}(Δ_{i}) ∩ ∂E_{3} , the open set h_{i}(Δ_{i}) must intersect S^{2} \ E_{3} , and so S^{2} \ E_{3} ⊂ h_{i}(Δ_{i}). We conclude that the set E_{4} , which is independent of i, is contained in h_{i}(Δ_{i}), and therefore h_{i}^{−1}(E_{4}) ⊂ Δ_{i} ⊂ E_{1} for all large i.

For each i, there is some iterate of g, say g^{p_{i}}, such that q(g^{p_{i}}(y), h_{i}^{−1}(y)) < 2^{−i} for all y ∈ S^{2} . In particular, g^{p_{i}}(E_{4}) ⊂ E_{2} for all large i. By passing to a subsequence and using Sublemma 2, we may assume that g^{p_{i}} → ψ uniformly on S^{2} , where ψ is a homeomorphism. Then h_{i_{ℓ}}^{−1} → ψ uniformly on S^{2} , for some sequence i_{ℓ} that tends to infinity as ℓ → ∞. By passing to a further subsequence, we may assume that i_{ℓ+1} − i_{ℓ} ≥ 2 for all ℓ so that Δ_{i_{ℓ}} and Δ_{i_{m}} are disjoint for all distinct ℓ and m. In particular, if U_{ℓ} = h_{i_{ℓ}}^{−1}(E_{4}) then U_{ℓ} ∩ U_{m} = ∅ whenever ℓ ≠ m.

We claim that ψ(x_{2}) ∈ U_{ℓ} for all large ℓ. This gives a contradiction and completes the proof of Sublemma 8. If the claim is false, we may pass to another subsequence and assume that ψ(x_{2}) ∉ U_{ℓ} for any ℓ.

Recall that x_{0} = ∞, and let γ be a circle in E_{4} ⊂ C centred at x_{2} . Then

δ_{2} = min{q(ψ(x_{2}), ψ(y)) : y ∈ γ} > 0,

and so by uniform convergence,

min{q(ψ(x_{2}), h_{i_{ℓ}}^{−1}(y)) : y ∈ γ} > δ_{2}/2

for all large ℓ. Thus the Jordan curve γ_{ℓ} = h_{i_{ℓ}}^{−1}(γ) then lies outside the disk

E_{5} = {y ∈ S^{2} : q(ψ(x_{2}), y) < δ_{3}} ⊂ C

if 0 < δ_{3} < δ_{2}/2 and δ_{3} < q(ψ(x_{2}), x_{0}). We note that ψ(x_{2}) ≠ ψ(x_{0}) = x_{0} = h_{i_{ℓ}}^{−1}(x_{0}) ∉ γ_{ℓ}.

Let E_{6} be the open disk containing x_{2} with ∂E_{6} = γ. Then ψ(x_{2}) ∉ h_{i_{ℓ}}^{−1}(E_{6}) ≡ U′_{ℓ} ⊂ U_{ℓ}. Since U′_{ℓ} coincides with one of the components of S^{2} \γ_{ℓ}, we deduce that U′_{ℓ} ∩ E_{5} = ∅. Thus h_{i_{ℓ}}^{−1}(x_{2}) ∉ E_{5} for all large ℓ. Since h_{i_{ℓ}}^{−1}(x_{2}) → ψ(x_{2}) as ℓ → ∞, we obtain a contradiction. Hence ψ(x_{2}) ∈ U_{ℓ} for all large ℓ, as asserted. This proves Sublemma 8, and also completes the proof of Lemma 2. Thus the proof of Theorem 1 is also complete.

6. Proof of Theorem 2

6.1. Let G be an abelian nondiscrete convergence group on S^{2} . We denote by \overline{G} the closure of G in H(S^{2}) so that h ∈ \overline{G} if and only if there is a sequence g_{n} ∈ G such that g_{n} tends to the homeomorphism h uniformly on S^{2} . Clearly \overline{G} is a group, and it is nondiscrete if G is. As in [5, Lemma 5, p. 93], we see that \overline{G} is a convergence group if G is. It clearly suffices to find a function f such that f^{−1} ∘ \overline{G} ∘ f is a Möbius group. On the other hand, if f^{−1} ∘ G ∘ f is a Möbius group, then so is f^{−1} ∘ \overline{G} ∘ f, so that we are not putting any extra restrictions on f by considering \overline{G}. Therefore we may and will assume that G is closed, that is, G = \overline{G}. In view of Theorem 1 and the above, we may assume that G is not contained in the closure of a cyclic group.

We write fix(g) for the set of fixed points of g ∈ G. Since g ∘ h = h ∘ g and h^{−1} exists for g, h ∈ G, it is easily seen that for all g, h ∈ G,

(6.1.1)  fix(g) = h(fix(g)).

We divide the argument into two cases as follows: I. There is a parabolic element g in G; and II. There is no parabolic element in G.

6.2. Case I. We may assume that fix(g) = \{∞\}, identifying S^{2} and \overline{C}. If h ∈ G then h fixes ∞ by (6.1.1). Let H denote the index one or two subgroup of sense-preserving elements of G. If h ∈ G and if h has at most two fixed points, then since fix(h) = g(fix(h)) by (6.1.1), it follows that fix(h) = {∞} also. By Corollary 1, this is true whenever h ≠ Id and h is not conjugate to c/z̄ where |c| = 1, and in particular whenever h ∈ H \{Id}. Thus each function in H \{Id} is parabolic. So if h_{1}, h_{2} ∈ H and h_{1}(z) = h_{2}(z) for some z ∈ C then h_{1}^{−1} ∘ h_{2} ∈ H and (h_{1}^{−1} ∘ h_{2})(z) = z so that h_{1}^{−1} ∘ h_{2} = Id, and hence h_{1} = h_{2} . We shall often make use of this fact.

If h_{1}, h_{2} ∈ G \ H then h_{1} ∘ h_{2}^{−1} ∈ H. Thus either G = H or

(6.2.1)  G = H ∪ {h ∘ h_{1} : h ∈ H}

for a fixed but arbitrary h_{1} ∈ G \ H.

First we prove that H is topologically conjugate to a Möbius group. Clearly H is nondiscrete and locally compact. Also H contains no small subgroups, that is, Id has a neighbourhood U in H such that {Id} is the only subgroup of H contained in U. For example, we may take

U = {h ∈ H : d(h, Id) < 1},

where the metric d is as defined in Subsection 2.2, noting that the diameter of S^{2} in the chordal metric q is equal to 2. For if J ≠ {Id} is a subgroup of H and h ∈ J \ {Id}, then h is parabolic with fixed point at infinity so that for all large n,

d(h^{n}, Id) ≥ q(h^{n}(0), 0) > 1.

Thus h^{n} ∉ U, so that J is not contained in U.

By [10, Theorem 6, p. 95], the group H contains a nontrivial one-parameter family of elements. That is, corresponding to each real number t there is a function h_{t} ∈ H such that h_{t} ∘ h_{u} = h_{t+u} for all real t and u, and the mapping that takes t onto h_{t} is a continuous function of the real axis R into H(S^{2}). Also, h_{t} ≠ Id for some t. Obviously, h_{0} = Id and h_{−t} = (h_{t})^{−1} .

We now claim that h_{t} ≠ Id whenever t ≠ 0, which then immediately implies that h_{t} ≠ h_{u} whenever t ≠ u. For suppose that h_{t} = Id for some t ≠ 0. Then (h_{t/n})^{n} = h_{t} = Id. Thus h_{t/n} cannot be parabolic, and so h_{t/n} = Id. It follows that h_{rt} = Id for each rational number r. The continuity of the map u ↦ h_{u} now implies that h_{u} = Id for all u, which is a contradiction. Thus h_{t} ≠ Id for t ≠ 0 and h_{t} ≠ h_{u} if t ≠ u.

We next note that if g_{n} ∈ H and g_{n} → x_{0} and g_{n}^{−1} → y_{0} locally uniformly in \overline{C} \ \{y_{0}\} and \overline{C} \ \{x_{0}\}, respectively, then x_{0} = y_{0} = ∞. Since each g_{n} fixes ∞, it follows that at least one of x_{0} and y_{0} is ∞. Replacing g_{n} by g_{n}^{−1} , if necessary, we may assume that x_{0} = ∞. Suppose that y_{0} ≠ ∞. Then if t ≠ 0, ε > 0 and ε is small enough we have g_{n} → ∞ uniformly in h_{t}(B(y_{0}, ε)) since h_{t}(y_{0}) ≠ y_{0} . But since for z ∈ B(y_{0}, ε), we have

h_{t}(g_{n}(z)) = g_{n}(h_{t}(z)) → ∞ as n → ∞ ,

it follows that g_{n} → ∞ uniformly in B(y_{0}, ε). But g_{n}^{−1} → y_{0} uniformly in B(y_{0}, ε). Thus

Id = g_{n} ∘ g_{n}^{−1} → ∞ as n → ∞

uniformly in B(y_{0}, ε), which is a contradiction. Thus y_{0} = ∞, as required, and so x_{0} = y_{0} = ∞.

6.3. Next we verify that the h_{t} satisfy a continuity condition considered by Kerékjártó [12]. For z ∈ C, define Γ(z) = {h_{t}(z) : t ∈ R}. If Γ(z) ∩ Γ(w) ≠ ∅ then h_{t}(z) = h_{u}(w) for some t, u ∈ R, so that h_{s}(z) = h_{s+u−t}(w) for all s ∈ R. Thus Γ(z) ⊂ Γ(w) and similarly Γ(w) ⊂ Γ(z) so that Γ(z) = Γ(w). So two sets Γ(z) are disjoint unless they coincide.

Suppose that z_{n} → w ∈ C as n → ∞. We want to prove that then Γ(z_{n}) → Γ(w) uniformly in the chordal metric. Then it will follow from a result of Kerékjártó ([12, p. 115]) that there is a homeomorphism f_{1} of \overline{C} such that f_{1}^{−1} ∘ h_{t} ∘ f_{1} = T_{t} for all real t, where T_{t}(z) = z + t for all complex t and z. Obviously, f_{1}(∞) = ∞.

Set

δ_{n} = sup ({q(x, Γ(w)) : x ∈ Γ(z_{n})} ∪ {q(x, Γ(z_{n})) : x ∈ Γ(w)})

where q(x, Γ(w)) = inf{q(x, y) : y ∈ Γ(w)}. We claim that δ_{n} → 0 as n → ∞. If not, we may pass to a subsequence and assume that there are real numbers t_{n} such that

(6.3.1)  q(h_{t_{n}}(z_{n}), y) ≥ ε > 0

for all n and for all y ∈ Γ(w), or that

(6.3.2)  q(h_{t_{n}}(w), y_{n}) ≥ ε > 0

for all n and for all y_{n} ∈ Γ(z_{n}), and further that the sequence t_{n} converges, possibly to ∞ or −∞. If t_{n} → u ∈ R, we have h_{t_{n}} → h_{u} uniformly on \overline{C} so that h_{t_{n}}(z_{n}) → h_{u}(w) ∈ Γ(w). This contradicts (6.3.1). If |t_{n}| → ∞, write t_{n} = k_{n}+θ_{n} where k_{n} ∈ Z and 0 ≤ θ_{n} < 1. We may assume that θ_{n} → θ ∈ [0, 1] so that h_{θ_{n}} → h_{θ}. Since h_{k_{n}} = h_{1}^{k_{n}} and h_{1} fixes ∞ and is conjugate to the translation z ↦ z + 1, we have h_{k_{n}}(z) → ∞ and so h_{t_{n}}(z) → ∞ locally uniformly in C. But then q(h_{t_{n}}(z_{n}), h_{t_{n}}(w)) → 0, which contradicts (6.3.1) since h_{t_{n}}(w) ∈ Γ(w). We deal with (6.3.2) in the same way. This proves that δ_{n} → 0 as n → ∞.

6.4. Define H_{1} = f_{1}^{−1} ∘ H ∘ f_{1} and 𝒯_{1} = {T_{t} : t ∈ R} ⊂ H(S^{2}). Then 𝒯_{1} ⊂ H_{1} . If H_{1} = 𝒯_{1} , it follows that H is topologically conjugate to a group of translations. So we may and will now assume that H_{1} ≠ 𝒯_{1} . We say that g_{1}, g_{2} ∈ H_{1} are equivalent if g_{1} = T_{t} ∘ g_{2} for some real t. Clearly this defines an equivalence relation, and the equivalence classes form an abelian group Γ. The group Γ acts on \overline{R} as follows. If h ∈ Γ, h′ ∈ H_{1} and h′ is a representative of h, we set h(t) = Im h′(it) for t ∈ R, and h(∞) = ∞. Clearly h is then a well-defined function on \overline{R}, continuous on R. If Im h′(it) = Im h′(iu), and if s = Re{h′(it) − h′(iu)}, then, since H_{1} is abelian, we have

h′(iu + s) = h′(iu) + s = Re h′(it) + i Im h′(it) = h′(it)

so that iu + s = it since h′ is a homeomorphism. Thus s = 0 and hence h′(it) = h′(iu) and so t = u. Therefore h is one-to-one. We have |h(t)| → ∞ as t → ±∞ since

C = h′(C) = {h′(it + u) : t, u ∈ R} = {h′(it) + u : t, u ∈ R}.

Thus h defines a homeomorphism of \overline{R} onto itself, and we denote the group of such homeomorphisms by J. Since H_{1} ≠ 𝒯_{1} , the group J is not trivial.

Alternatively, we may note that each g_{1} ∈ H_{1} commutes with T_{t} for all real t and therefore maps each horizontal line onto another such line. Identifying every horizontal line with its y -coordinate, we obtain the group action of Γ on R. Two elements of H_{1} give rise to the same element of the resulting group J if and only if they differ by T_{t} for some real t.

We claim that J is a convergence group on \overline{R}, which we identify with S^{1} . For if g_{n} ∈ J, we may choose the corresponding sequence of representatives g_{1n} ∈ H_{1} of the classes in Γ that give rise to the g_{n}, so that Re g_{1n}(0) = 0, and pass to a subsequence such that g_{1n} → ĝ ∈ H_{1} uniformly on \overline{C}, or g_{1n} → ∞ and g_{1n}^{−1} → ∞ locally uniformly in C as n → ∞, since H has the corresponding property and f_{1}(∞) = ∞. In the former case, g_{n} → g̃ given by g̃(t) = Im ĝ(it), and in the latter case, g_{n} → ∞, locally uniformly on R. The same argument shows that J is closed.

If h ∈ J and h(t) = t for some t ∈ R then there is h′ ∈ H_{1} with h′(it) = a+it for some real a. Then

it = (T_{−a} ∘ h′)(it)

so that T_{−a} ∘ h′ = Id, the only element of H_{1} that has a finite fixed point. Thus h′ = T_{a} so that Im h′(it) = t and h(t) = t for all real t. Hence h = Id, and so the elements of J \ {Id} have no finite fixed points.

6.5. Now J is an abelian convergence group on \overline{R} with parabolic elements only (note that J may or may not be discrete), so that by [5, Theorem 1, p. 88], there is a homeomorphism F_{2} of \overline{R} fixing infinity such that F_{2}^{−1} ∘ J ∘ F_{2} is a group of translations. Now define a homeomorphism f_{2} of \overline{C} by

f_{2}(t + iu) = t + iF_{2}(u) when t, u ∈ R

and by f_{2}(∞) = ∞. If g_{1} ∈ H_{1} then

(f_{2}^{−1} ∘ g_{1} ∘ f_{2})(t + iu) = t + Re g_{1}(iu) + i(F_{2}^{−1} ∘ h ∘ F_{2})(u) = t + Re g_{1}(iu) + i(u + a),

say, where h ∈ J corresponds to the equivalence class of g_{1} and F_{2}^{−1} ∘h∘F_{2} = T_{a}. Thus if H_{2} = f_{2}^{−1} ∘ H_{1} ∘ f_{2} , then the action of each element of H_{2} in the direction of the imaginary axis amounts to a translation. Note that

f_{2}^{−1} ∘ T_{t} ∘ f_{2} = T_{t} for all real t,

so that 𝒯_{1} ⊂ H_{2} .

Since J_{2} = F_{2}^{−1} ∘ J ∘ F_{2} is a closed group of translations, it is cyclic unless it is equal to 𝒯 , the group of all translations of R. In either case, if g_{1}, g_{2} ∈ H_{2} correspond to the same element of J_{2} then g_{2} = T_{t} ∘g_{1} for some real t. Thus each element of H_{2} can be uniquely represented in the form T_{t} ∘ φ_{u} for some suitable t ∈ R and φ_{u} ∈ H_{2} , where the map x ↦ x + t lies in J_{2} , Re φ_{u}(0) = 0, and Im φ_{u}(0) = u, hence Im φ_{u}(z) = u + Im z for all z. We have φ_{u} ∘ φ_{v} = T_{t} ∘ φ_{u+v} where t = t(u, v) ∈ R depends on u and v only.

6.6. Since φ_{u} commutes with each T_{t} and increases the imaginary part of the variable by u, we may write

(6.6.1)  φ_{u}(x + iy) = x + ψ_{u}(y) + i(y + u)

where the real-valued function ψ_{u} is continuous in y with ψ_{u}(0) = 0. We look for a homeomorphism f_{3} of \overline{C} fixing ∞ of the form

(6.6.2)  f_{3}(x + iy) = x + Q(y) + iy,

where Q is a real-valued continuous function of y, that would conjugate H_{2} to a group of translations. Clearly f_{3}^{−1} ∘ T_{t} ∘ f_{3} = T_{t} for all t ∈ R, no matter how Q is chosen.

We have

{f_{3}^{−1} ∘ φ_{u} ∘ f_{3}}(x + iy) = x + Q(y) + ψ_{u}(y) − Q(y + u) + i(y + u),

which is a translation if and only if

(6.6.3)  Q(y + u) − Q(y) = ψ_{u}(y) + K(u)

where K(u) depends on u only and not on y.

If J_{2} ≠ 𝒯 , we may assume that J_{2} is generated by the map x ↦ x + 1. Then φ_{n} = T_{t(n)} ∘ φ_{1}^{n} for all n ∈ Z where t(n) ∈ R, so that H_{2} is generated 𝒯_{1} and φ_{1} . We choose K(1) = 0 and define Q(y) = ψ_{1}(0)y for 0 ≤ y ≤ 1. Then the condition

Q(y + 1) − Q(y) = ψ_{1}(y) for all y ∈ R

extends Q to a continuous function on R. We conclude that H_{2} is conjugate to a group of translations in this case.

6.7. Suppose that J_{2} = 𝒯 . The following argument, which is perhaps simpler than our original one for this case, was suggested by the referee. We first claim that for any g ∈ H_{2} there is a unique h ∈ H_{2} with h^{2} = g. For if g ∈ H_{2} then g is given by g(t + iu) = t + Re g_{1}(iu) + i(u + a) for some real a and some g_{1} ∈ H_{1} that are related as in Subsection 6.5. Since J_{2} = 𝒯 , there is g_{2} ∈ H_{1} such that F_{2}^{−1} ∘ η ∘ F_{2} = T_{a/2} where η ∈ J corresponds to the equivalence class of g_{2} . We define h ∈ H_{2} by h(t+iu) = t+b+Re g_{2}(iu)+i(u+ \frac{1}{2}a) where the real parameter b is still at our disposal. We find that

h^{2}(t + iu) = t + 2b + Re g_{2}(iu) + Re g_{2}(i(u + \frac{1}{2}a)) + i(u + a).

We choose b so that

Re g_{1}(0) = 2b + Re g_{2}(0) + Re g_{2}(\frac{1}{2}ai).

Then h^{2}(0) = g(0) so that g^{−1} ∘ h^{2} ∈ H_{2} fixes the origin. But then g^{−1} ∘ h^{2} = Id since H_{2} is a topological conjugate of H and H has this property, and so g = h^{2} , as required.

Next, if h_{1}^{2} = g also for some h_{1} ∈ H_{2} then the translation in J_{2} corresponding to h_{1} must be T_{a/2} , the same as for h. Thus there is a real number c such that the maps h and T_{c} ∘h_{1} agree at the origin and hence everywhere in the plane. Thus g = h^{2} = T_{2c} ∘ h_{1}^{2} = T_{2c} ∘ g, and so c = 0 and h = h_{1} . Hence the solution h ∈ H_{2} to g = h^{2} is unique.

As we noted after (6.2.1), H is locally compact and contains no small subgroups, so that the topological conjugate H_{2} of H has the same properties. We shall show that any neighbourhood of {Id} in H_{2} contains an element not in 𝒯_{1} . Then by [10, Lemma 40, p. 129] (see [10, Theorem 7, p. 95], for the definition of the set K in Lemma 40), there is a (compact) neighbourhood U of {Id} in H_{2} with the following property: the set of points of the form X(1), where X(t) defines a one-parameter subgroup of H_{2} such that {X(t) : |t| ≤ 1} ⊂ U, is a (closed) neighbourhood of {Id} in H_{2} . Choosing such a point X(1) ∈ U \ 𝒯_{1} , we find a nontrivial one-parameter family X(t) in H_{2} that {X(t) : t ∈ R} ∩ 𝒯_{1} = {Id}. For if t ≠ 0 and X(t) = T_{u} then by the existence and uniqueness of square roots just proved we see that X(rt) is a translation in 𝒯_{1} for all dyadic rational numbers r. By continuity, this is then true for all real r, which gives a contradiction when r = 1/t.

Now to show that any neighbourhood of {Id} in H_{2} contains a function not in 𝒯_{1} , suppose, to get a contradiction, that some neighbourhood U of {Id} in H_{2} is contained in 𝒯_{1} . Since 𝒯_{1} ≠ H_{1} and hence 𝒯_{1} ≠ H_{2} , there is h ∈ H_{2} \ 𝒯_{1} . By the above, for each k ≥ 1, we can define h_{k} ∈ H_{2} with h_{k}^{2^{k}} = h. Since h ∉ 𝒯_{1} , we have h_{k} ∉ 𝒯_{1} . It now suffices to show that some subsequence of h_{k} tends to Id as k → ∞. Choose a subsequence h_{k_{j}} satisfying the definition of a convergence group.

Recall that

h_{k}(x + iy) = (x + b_{k}(y)) + i(y + 2^{−k}a)

where a ∈ R\{0} is fixed (and depends on h only) and b_{k} is a continuous function of y. Now in any case, b_{k_{j}}(y) tends to a limit (either a finite-valued function or the constant ±∞) locally uniformly in y as j → ∞. In case of a finite limit function, say β(y), we have h_{k}(x+iy) → (x+iy)+β(y) ≡ κ(x+iy), and since κ ∈ H_{2} and κ(0) = T_{β(0)}(0), we have κ = T_{β(0)} so that β(y) = β(0) for all y. Suppose that b_{k_{j}}(0) → b where b ∈ [−∞, ∞]. With a_{j} = b_{k_{j}}(0), we have (h_{k_{j}} ∘ T_{−a_{j}})(0) → 0 and hence g_{k_{j}} ≡ h_{k_{j}} ∘ T_{−a_{j}} tends to a homeomorphism in H_{2} fixing the origin. Thus g_{k_{j}} → Id. Note that g_{k}(x + iy) = (x + b_{k}(y) − b_{k}(0)) + i(y + 2^{−k}a). Now h_{k_{j}} = T_{a_{j}} ∘ g_{k_{j}}. Suppose that b ≠ 0. (We may have b = ±∞.) Then

h(0) = h_{k_{j}}^{2^{k_{j}}} (0) = (T_{2^{k_{j}} a_{j}} ∘ g_{k_{j}}^{2^{k_{j}}})(0) = 2^{k_{j}}a_{j} + g_{k_{j}}^{2^{k_{j}}} (0).

For 0 ≤ p < 2^{k_{j}}, the points g_{k_{j}}^{p}(0) have their y -coordinates between 0 and a. When j is large enough, |b_{k_{j}}(y) − b_{k_{j}}(0)| < min{1, |b|/3} for |y| ≤ |a| since g_{k_{j}} → Id, while furthermore |a_{j}| > min{2, 2|b|/3} and a_{j} will have the same sign as b. It follows that

| Re g_{k_{j}}^{2^{k_{j}}} (0)| < 2^{k_{j}} min{1, |b|/3}

so that | Re h(0)| = | Re h_{k_{j}}^{2^{k_{j}}} (0)| > 2^{k_{j}} min{1, |b|/3} → ∞ as j → ∞, which is a contradiction. Thus b = 0, so that β(y) is finite and β(y) = β(0) = b = 0 for all y. This shows that h_{k_{j}} → Id as j → ∞. This also completes the proof of the existence of the subgroup X(t).

Write h_{t} instead of X(t). We have h_{1} ∉ 𝒯_{1} so that a = Im h_{1}(0) ≠ 0. By the argument at the beginning of Subsection 6.7 and by continuity, Im h_{t}(0) = at for all real t. By reparametrization, we may assume that a = 1. Thus if g ∈ H_{2} , there are real t and u such that g(0) = T_{t} ∘ h_{u}(0) and so g = T_{t} ∘ h_{u}. Hence (cf. the end of Subsection 6.5) each element of H_{2} can be uniquely represented in the form T_{t} ∘ h_{u} for some real t and u. We may write

h_{r}(x + iy) = x + χ(r, y) + i(y + r).

The condition h_{r+s} = h_{r} ∘ h_{s} implies that

(6.7.1)  χ(r + s, y) = χ(r, y) + χ(s, r + y)

for all real y, r, s.

We now define a function f of R^{2} into itself by

f(x + iy) = (h_{y} ∘ T_{x})(0) = h_{y}(x) = x + χ(y, 0) + iy.

Clearly f is a homeomorphism of R^{2} onto itself (and hence extends to a homeomorphism of S^{2} ). The map f commutes with each T_{t}. We further have

(h_{u} ∘ f)(x + iy) = h_{u}(x + χ(y, 0) + iy) = x + χ(y, 0) + χ(u, y) + i(y + u).

We define T_{iu} by T_{iu}(x + iy) = x + i(y + u). We have

\begin{aligned}(f ∘ T_{iu})(x + iy) & = f(x + i(y + u)) = x + χ(u + y, 0) + i(u + y) \\ & = (h_{u} ∘ f)(x + iy) \end{aligned}

since χ(u + y, 0) = χ(y, 0) + χ(u, y) as we see by replacing r, s, y by y, u, 0 in (6.7.1). Hence f^{−1} ∘ T_{t} ∘ f = T_{t} and f^{−1} ∘ h_{u} ∘ f = T_{iu} so that f^{−1} ∘ H_{2} ∘ f is a group of translations, in fact the group of all translations of C. We have now proved in all cases (arising when considering Case I) that H is topologically conjugate to a group of translations containing 𝒯_{1} .

6.8. We remark that an alternative proof in the case J_{2} = 𝒯 can be given along the following lines. This, in fact, was our original proof, and will appear in [7]. In this case consider ψ_{u}(v) defined by (6.6.1). Since φ_{u} ∘ φ_{v} = φ_{v} ∘ φ_{u}, we have

(6.8.1)  ψ_{u}(s) + ψ_{v}(s + u) = ψ_{v}(s) + ψ_{u}(s + v)

for all u, v, s ∈ R, as a calculation shows. Taking s = 0 in (6.8.1) and recalling that ψ_{u}(0) = 0, we get

ψ_{u}(v) = ψ_{v}(u).

Furthermore, for any C > 0 there is M > 0 such that |ψ_{u}(v)| ≤ M whenever |u| ≤ C and |v| ≤ C; and ψ_{u_{n}}(v_{n}) → 0 whenever u_{n}, v_{n} are real sequences with u_{n} → 0 and v_{n} → 0 as n → ∞.

Under these assumptions, it is shown in [7] that ψ_{u}(v) is of the form

ψ_{u}(v) = Q(u + v) − Q(u) − Q(v)

for some continuous function Q with Q(0) = 0. We may then use Q in the definition of f_{3} , and (6.6.3) is obviously satisfied.

6.9. Let us now set f_{4} = f_{1} when H_{1} = 𝒯_{1} and f_{4} = f_{1} ∘f_{2} ∘f_{3} otherwise so that H_{4} = f_{4}^{−1} ∘ H ∘ f_{4} is a Möbius group containing 𝒯_{1} . Set G_{4} = f_{4}^{−1} ∘ G ∘ f_{4} so that G_{4} contains H_{4} . If G = H, there is nothing else to prove. Otherwise, G is given by (6.2.1), so that

(6.9.1)  G_{4} = H_{4} ∪ {h ∘ g_{0} : h ∈ H_{4}}

for some g_{0} ∈ G_{4} \ H_{4} .

Recall that 𝒯_{1} = {T_{s} : s ∈ R} and write 𝒯_{2} = {T_{s} : s ∈ C}. Since H_{4} is closed, it is clearly equal to 𝒯_{1} or 𝒯_{2} , or is generated by 𝒯_{1} and T_{iα} for some positive α. If g_{0} fixes β ∈ C then, since H_{4} is abelian, g_{0} fixes T_{s}(β) whenever T_{s} ∈ H_{4} .

If H_{4} = 𝒯_{2} then g_{0}(s) = g_{0}(s + 0) = s + g(0) for all s ∈ C so that g_{0} is sense-preserving, which is a contradiction. Thus H_{4} ≠ 𝒯_{2} .

By Corollary 1, the function g_{0} is conjugate to z̄ + 1 or 2z̄ or c/z̄ where |c| = 1. Thus, if fix(g_{0}) contains at least three points, then fix(g_{0}) is a Jordan curve. So if H_{4} = 𝒯_{1} then either fix(g_{0}) = {∞} and g_{0} is conjugate to z̄ + 1, or fix(g_{0}) is a horizontal line and g_{0} is conjugate to 1/z̄, hence to z̄. If H_{4} ≠ 𝒯_{1} and fix(g_{0}) ≠ {∞}, then fix(g_{0}) contains infinitely many horizontal lines, which is impossible. Thus if H_{4} ≠ 𝒯_{1} then fix(g_{0}) = {∞} and g_{0} is conjugate to z̄ + 1.

6.10. If H_{4} = 𝒯_{1} and fix(g_{0}) is a horizontal line, we may assume that fix(g_{0}) = \overline{R}. Set U = \{z : Im z > 0\} and L = \{z : Im z < 0\}. Since g_{0} is sense-reversing, g_{0} interchanges U and L. Also g_{0}(z +t) = g_{0}(z)+t for all z ∈ C and t ∈ R. So if we define

f_{5}(z) = z  when z ∈ L ∪ \overline{R}

and

f_{5}(z) = g_{0}(z̄)  when z ∈ U ,

then f_{5} is a homeomorphism of \overline{C} that fixes ∞. Since g_{0} is conjugate to z̄, we have g_{0}^{2} = Id. Thus f_{5}^{−1}(z) = z for z ∈ L ∪ \overline{R} and f_{5}^{−1}(z) = \overline{g_{0}(z)} for z ∈ U. Now one can verify that

(f_{5}^{−1} ∘ g_{0} ∘ f_{5})(z) = z̄

and

(f_{5}^{−1} ∘ T_{s} ∘ f_{5})(z) = z + s

for all z ∈ \overline{C} and all s ∈ R. Hence f_{5}^{−1} ∘ G_{4} ∘ f_{5} is a Möbius group, and so is f_{6}^{−1} ∘ G ∘ f_{6} where f_{6} = f_{4} ∘ f_{5} .

6.11. If g_{0} is conjugate to z̄ + 1, set

Γ_{t} = {g_{0}(z) : Re z = t} for t ∈ R

so that the Γ_{t} are disjoint open Jordan arcs whose union is C. Since g_{0} commutes with T_{t}, it follows that

Γ_{t} = {z + t : z ∈ Γ_{0}} for all real t.

Thus Γ_{0} has a unique point of intersection with each horizontal line, and the map t ↦ Im g_{0}(it) is a homeomorphism of R onto itself. Since g_{0} is sense-reversing and the curves Γ_{t} move to the right as t increases, the map Im g_{0}(it) is a decreasing function of t and therefore has a fixed point. Hence there are t, u ∈ R such that g_{0}(it) = u + it. But then T_{−u} ∘ g_{0} ∈ G_{4} \ H_{4} fixes it and hence fixes each point on a line and is conjugate to z̄. We may use (6.9.1) with g_{0} replaced by any element of G_{4} \ H_{4} , in particular by T_{−u} ∘ g_{0} . Now the argument above shows that G_{4} is conjugate to a Möbius group when H_{4} = 𝒯_{1} .

Finally, if H_{4} ≠ 𝒯_{1} but G_{4} ≠ H_{4} , we have found an element of G_{4} \ H_{4} that fixes each point on a line, which contradicts what was said about the case H_{4} ≠ 𝒯_{1} at the end of Subsection 6.9.

This completes the proof of Theorem 2 in Case I.

6.12. Case II. We reduce Case II to Case I by means of a suitable transformation. We essentially conjugate G by the exponential function but we need to be careful about the technical details. Let H be the index one or two subgroup of G consisting of the sense-preserving elements of G. Note that H is a closed nondiscrete abelian convergence group. We first consider the problem of conjugating H to a Möbius group. Pick g_{0} ∈ H \ {Id}. Then g_{0} is conjugate to 2z or cz where |c| = 1, and has therefore exactly two fixed points, which we may assume to be 0 and ∞. We claim that each g ∈ H fixes 0 and ∞. If not, then by (6.1.1), g interchanges 0 and ∞. Thus fix(g) = {a, b} where a, b ∉ {0, ∞}. Now any h ∈ H \ {Id} fixes or interchanges each of the pairs a, b and 0, ∞. Since a hyperbolic or loxodromic map fixes two points and interchanges no pairs of points, h must be elliptic. Since h^{2} fixes each of 0, ∞, a and b, and h^{2} is elliptic, we have h^{2} = Id. If g_{n} is a sequence of distinct functions in H with g_{n} → Id uniformly on S^{2} then for all large n, we cannot have g_{n}(0) = ∞. Thus g_{n} fixes 0 and ∞. Similarly, g_{n} fixes a and b. But since g_{n} is elliptic, we have g_{n} = Id. This contradiction shows that each function in H \ {Id} fixes both 0 and ∞ and hence fixes no other point.

We say that an element of H \ {Id} is elliptic if it can be topologically conjugated to an elliptic Möbius transformation, which is therefore of finite order unless it is an irrational rotation. An element is hyperbolic (loxodromic, respectively) if it can be topologically conjugated to a hyperbolic (loxodromic, respectively) Möbius transformation. Thus each element of H \ {Id} is elliptic or hyperbolic or loxodromic.

We note that if g_{1}, g_{2} ∈ H \ {Id} are elliptic then also g_{1}^{−1} and g_{1} ∘ g_{2} are elliptic (or g_{1} ∘g_{2} = Id). This is clear for g_{1}^{−1} . Concerning g_{1} ∘g_{2} , we may assume that a preliminary conjugation has been performed which takes g_{1} to an elliptic Möbius transformation, and so we assume that g_{1} actually is an elliptic Möbius transformation. If g_{1}^{n} = g_{2}^{m} = Id then (g_{1} ∘ g_{2})^{mn} = Id since H is abelian so that g_{1} ∘ g_{2} is elliptic. So we may assume that one of g_{1} and g_{2} , say g_{1} , is an irrational rotation, say g_{1}(z) = cz where c = e^{2πiβ} for some irrational number β. Since g_{1} and g_{2} commute, it follows as in (3.1) that g_{2} , and hence g_{1} ∘ g_{2} , maps every circle centred at the origin onto itself. Thus g_{1} ∘ g_{2} cannot be hyperbolic, loxodromic (or parabolic, which is ruled out by membership in H anyway), and is therefore elliptic. Thus the elliptic elements together with Id form a subgroup of H, which we denote by E.

Suppose that H = E. We shall show that then H contains a function g that is topologically conjugate to an irrational rotation. Suppose that this is not the case. Then every element of H \{Id} is of finite order. Since H is nondiscrete and hence infinite, it follows that the orders of the elements of H must be unbounded. To see this, it suffices to show that for any positive integer N, the subgroup H_{N} of H generated by all the elements of H of order N, is finite. We note that by the commutativity on H, if g = g_{1} ∘ … ∘ g_{k} ∈ H_{N} where g_{j}^{N} = Id, then g^{N} = g_{1}^{N} ∘ … ∘ g_{k}^{N} = Id so that the order of g is at most N. Now to get a contradiction, suppose that H_{N} is infinite. Then there is a sequence h_{n} of distinct elements of H_{N} satisfying the definition of a convergence group. If h_{n} → h where h is a homeomorphism, then the functions g_{n} = h_{n+1} ∘ h_{n}^{−1} ∈ H_{N} tend to Id uniformly on S^{2} while each g_{n} has order at most N. But by a theorem of Newman [18], there is ε = ε(N) > 0 such that for each n, we have q(x_{n}, g_{n}(x_{n})) > ε for some x_{n} ∈ S^{2} . This gives a contradiction. Thus there must be points x_{0}, y_{0} ∈ S^{2} such that h_{n} → x_{0} and h_{n}^{−1} → y_{0} locally uniformly in S^{2} \ {y_{0}} and S^{2} \ {x_{0}}, respectively. Since each h_{n} fixes 0 and ∞, we have {x_{0}, y_{0}} = {0, ∞}, say x_{0} = ∞ and y_{0} = 0. However, h_{n}(1) is bounded. To see this, note that by the uniform convergence on {z : |z| = 1}, if h_{n}(1) → ∞, then h_{n}({z : |z| = 1}) → ∞. Thus, as h_{n}(0) = 0, h_{n}(∞) = ∞, for all sufficiently large n, the image h_{n}(B(1)) contains the unit disk B(1) as a relatively compact subset. So if μ_{n} is a homeomorphism fixing 0 and ∞ such that μ_{n}^{−1} ∘ h_{n} ∘ μ_{n} = v_{n} is a rotation about the origin, say v_{n}(z) = cz where c is a root of unity, we see that there is a compact set K_{n} = μ_{n}(\overline{B(1)}), which is the closure of a Jordan domain containing the origin, such that K_{n} is contained in the interior of v_{n}(K_{n}). This is seen to be impossible, considering any point of K_{n} which is at a maximal distance from the origin among all points of K_{n}. So h_{n}(1) is bounded, which contradicts the fact that also h_{n}(1) → x_{0} = ∞. We conclude that H_{N} is finite, as asserted, and so the orders of the elements of H are unbounded.

Choosing elements h_{n} of H of order at least n and replacing h_{n} by a suitable iterate of h_{n} that has the same order as h_{n}, if necessary, without changing notation, we find a sequence h_{n} such that if h_{n} is topologically conjugate to a rotation around the origin by the angle 2πα_{n} then α_{n} → α as n → ∞ where α is an irrational number. Then h_{n}(1) is bounded away from {0, ∞}. This follows by considering rotations μ_{n}^{−1}∘h_{n}∘μ_{n} = v_{n} as in the previous paragraph. Therefore by passing to a subsequence we may assume that h_{n} converges to a limit, uniformly on the entire sphere, which will necessarily be a homeomorphism g and g ∈ H = E as H is closed. Now g must be elliptic and not of finite order. To see this last claim we let us suppose that g is elliptic of finite order, say N. Because of the uniform convergence it is clear that g_{n} = h_{n}^{N} → Id as n → ∞. Now each g_{n} is the topological conjugate of a finite order rotation through angle 2Nπα_{n} → 2Nπα ≢ 0 (mod 2π). Choose η ∈ (0, 1) so that η ≡ Nα (mod 1). Choose the smallest integer m so that mη > 1. Then for all n the points {h_{n}^{j}(1) : 0 ≤ j ≤ m} lie on an invariant Jordan curve for h_{n} winding once around the origin (for instance the image of a round circle under the topological conjugacy) and as mη > 1 the sum of the angular displacements

∠(1, 0, g_{n}(1)) + ∠(g_{n}(1), 0, g_{n}^{2}(1)) + … + ∠(g_{n}^{m−1}(1), 0, g_{n}^{m}(1)) > 2π.

Thus at least one of the angular displacements is greater than 2π/m.

We have shown that there is a number m such that for every n there is an integer j with 0 ≤ j ≤ m such that the angular displacement ∠(g_{n}^{j}(1), 0, g_{n}^{j+1}(1)) is at least 2π/m. But the uniform convergence to the identity mapping should imply that this displacement tends uniformly to 0, which is not the case. Hence g, as claimed above, has infinite order.

We may then assume, in view of Theorem 1, that a preliminary conjugation has been performed so that g(z) = c_{0}z where c_{0} = e^{2πiα} for some irrational number α. Since H is closed, it follows that the map ℛ_{c} given by ℛ_{c}(z) = cz belongs to H for every complex number c with |c| = 1. Now if h ∈ H then, since g ∘ h = h ∘ g, we have, as in (3.1), h(cz) = ch(z) whenever |c| = 1. Thus h maps each circle S(r) centred at the origin onto some such circle S(u(r)). Here u(r) is a homeomorphism of (0, ∞) onto itself so that u is strictly increasing since h fixes 0 and ∞. If u(r) > r or u(r) < r for some r > 0 then clearly we cannot have h^{n_{j}} → Id for any sequence of integers n_{j} → ∞, so that h cannot be elliptic. Thus u(r) = r for all r > 0. It follows that |h(z)| = |z| and h(z) = zh(|z|)/|z| for all z ∈ S^{2} \ {0, ∞}. In particular, if h(1) = a then |a| = 1 so that the map ℛ\overline{a} belongs to H. But (ℛ\overline{a} ∘ h)(1) = 1 so that ℛ\overline{a} ∘ h = Id, and hence h = ℛ_{a}, so that h is a Möbius transformation. Thus, after the preliminary conjugation, we have made H into a Möbius group, as required. This completes our treatment of the case H = E. For future reference, we note that the above argument in this paragraph also shows that if E contains all rotations ℛ_{c} then E contains no other functions.

Suppose then that H ≠ E. Suppose first that E is not a finite cyclic group. We claim that E contains a function topologically conjugate to an irrational rotation. If not, then E contains functions of arbitrarily large finite order, by the argument that we used in the case H = E. Choosing elements h_{n} of E of order at least n and replacing h_{n} by a suitable iterate of h_{n} that has the same order as h_{n}, if necessary, without changing notation, we find a sequence h_{n} such that if h_{n} is topologically conjugate to a rotation around the origin by the angle 2πα_{n} then α_{n} → α as n → ∞ where α is an irrational number. As before, we see that h_{n}(1) is bounded away from 0 and ∞. By passing to a subsequence we may therefore assume that h_{n} → h ∈ H where h is a homeomorphism. Now it is seen by the same argument as in the case H = E above, that if h ∈ E then h is topologically conjugate to an irrational rotation. If h ∉ E then h is loxodromic and h^{k}(1) → ∞, say, as k → ∞. For each k there is n = n(k) such that q(h_{n}^{k}(1), h^{k}(1)) < 1/k so that h_{n(k)}^{k}(1) → ∞. But h_{n(k)}^{k} ∈ E is conjugate to a rotation around the origin (by the angle 2πkα_{n(k)} ). Hence, by the same argument as before, we see that h_{n(k)}^{k}(1) remains bounded away from 0 and ∞, which gives a contradiction. This shows that indeed h ∈ E.

Thus we may perform a preliminary conjugation of H and assume that H contains an irrational rotation so that E contains and therefore, as noted before, coincides with the group of all rotations.

Suppose that H also contains a sequence of distinct loxodromic elements tending to the identity. Now H contains no small subgroups. For if g ∈ H \ {Id} is not elliptic then g fixes 0 and ∞ and g is topologically conjugate to the Möbius transformation z ↦ 2z, by the theorem of Kerékjártó [14] referred to in the introduction. Hence g^{n}(1) → ∞ as n → ∞ or as n → −∞, and so the subgroup of H generated by g cannot be contained in the neighbourhood U = {h ∈ H : d(h, Id) < 1/2} of the identity. If g is elliptic then g is an elliptic Möbius transformation fixing 0 and ∞ so that again it is clear that there is some fixed neighbourhood U of Id independent of g that does not contain the group generated by g. Now by [10, Theorem 6, p. 95], H contains a nontrivial oneparameter family h_{t}, and by [10, Theorem 5, p. 93], we may in fact assume that some nontrivial h_{t} is loxodromic. Then it clearly follows that h_{t} is loxodromic whenever t ≠ 0.

Since h_{n} = h_{1}^{n}, the points h_{n}(1) cluster to both 0 and ∞, so that {|h_{t}(1)| : t ∈ R} = {t : t > 0}. We claim that H is generated by E and the maps h_{t} for t ∈ R. For if h ∈ H then there is c with |c| = 1 and t ∈ R such that h(1) = ch_{t}(1). Since h^{−1} ∘ ℛ_{c} ∘ h_{t} ∈ H fixes the point 1 ∉ {0, ∞}, it follows that h = ℛ_{c} ∘ h_{t}.

By the last statement of Theorem 4, there is a homeomorphism f of S^{2} onto itself that conjugates each rotation ℛ_{c} onto itself, and conjugates each h_{t} to the dilation z ↦ e^{t}z. Hence f^{−1} ∘ H ∘ f is a Möbius group.

Suppose that H does not contain a sequence of distinct loxodromic elements tending to the identity. (We are still assuming that H ≠ E and that E is the group of all rotations.) Now if h ∈ H \ E then, since g ∘ h = h ∘ g, we have, as in (3.1), h(cz) = ch(z) whenever |c| = 1. Thus h maps each circle S(r) centred at the origin onto some such circle S(u(r)). Here u(r) = u_{h}(r) is a homeomorphism of (0, ∞) onto itself so that u is strictly increasing since h fixes 0 and ∞. For different choices of h we get different values for u_{h}(1) and at least one of h and h^{−1} gives a value of u_{h}(1) > 1. Some h = h_{0} gives the smallest value of u_{h}(1) > 1 as otherwise there is a sequence of distinct loxodromic elements tending to the identity. For the same reason the set of values of u_{h}(1) > 1 has no finite limit point. We further note that if h, k ∈ H and if there is r > 0 such that u_{h}(r) = u_{k}(r) then h = ℛ_{c} ∘ k for some c with |c| = 1 so that u_{h} = u_{k}. Otherwise, we have either u_{h}(r) < u_{k}(r) for all r > 0, or u_{h}(r) > u_{k}(r) for all r > 0. Similarly, if h ∈ H \ E, we have u_{h}(r) > r for all r > 0, or u_{h}(r) < r for all r > 0.

Denote the map u_{h} for h = h_{0} by u_{0} . We shall now prove that every possible value of u_{h}(1) is of the form u_{0}^{n}(1) for some integer n where u_{0}^{n} denotes the n^{th} iterate of u_{0} . Concerning the values of u_{h}(1) > 1, suppose that the first N (where N ≥ 1) have been proved to be of the form u_{0}^{n}(1) where 1 ≤ n ≤ N. Let the next value be u_{g}(1) where g ∈ H. By definition, u_{0}^{N}(1) < u_{g}(1) ≤ u_{0}^{N+1}(1). If u_{g}(1) < u_{0}^{N+1}(1), note that u_{0}^{N}(r) < u_{g}(r) < u_{0}^{N+1}(r) for all r > 0, and consider κ = g ∘ h_{0}^{−N} ∈ H. We have u_{κ} = u_{g} ∘ u_{0}^{−N}, and u_{g} and u_{0} commute. Hence 1 < u_{κ}(1) < u_{0}(1), which is a contradiction. We deduce that u_{g}(1) = u_{0}^{N+1}(1), as required.

We claim that every element h of H is of the form ℛ_{c}∘h_{0}^{n} for some integer n. Choose n so that u_{h}(1) = u_{0}^{n}(1) and hence u_{h} = u_{0}^{n}. As noted above, there is c with |c| = 1 such that h(1) = ch_{0}^{n}(1). Consequently, h = ℛ_{c} ∘ h_{0}^{n}, as required.

There is an increasing homeomorphism k of (0, ∞) onto itself such that (k^{−1} ∘ u_{0} ∘ k)(x) = ax where a > 0 and a ≠ 1 (we may take a = 2, in fact). Define f(re^{iθ}) = k(r)e^{iθ}, and let f fix 0 and ∞. Then f is a homeomorphism of S^{2} onto itself that conjugates each ℛ_{c} to itself and satisfies (f^{−1} ∘h_{0} ∘f)(z) = acz for some c with |c| = 1. Thus f^{−1} ∘ H ∘ f is a Möbius group.

Suppose then that H ≠ E and that E is a finite cyclic group. Now clearly H contains no small subgroups (and is locally compact), and so, by [10, Theorem 6, p. 95], H contains a nontrivial one-parameter family h_{t}. Since H is nondiscrete, it must contain a sequence of distinct loxodromic elements tending to the identity. Now by [10, Theorem 5, p. 93], we may in fact assume that some nontrivial h_{t} is loxodromic. Then it clearly follows that h_{t} is loxodromic whenever t ≠ 0. By our Theorem 4, still to be proved, we may perform a preliminary conjugation of H and assume that h_{t}(z) = e^{t}z for all real t. However, we prefer to change notation and assume that for all real positive t, the mapping h_{t}(z) = tz belongs to H.

Suppose now that g ∈ H \ {h_{t} : t > 0}. Since g(tz) ≡ tg(z) for all t > 0, it follows that g maps rays (from 0 to ∞) onto rays. Hence g determines a homeomorphism k of the circle S^{1} onto itself via k(e^{iθ}) = e^{iψ} if and only if g({te^{iθ} : t > 0}) = {te^{iψ} : t > 0}. We claim that the mappings of S^{1} so obtained when g goes through all elements of H, form a convergence group on S^{1} . (Note that if g = h_{t} then k = Id.) Clearly the set K of these maps k is a group. Note that if k fixes a point so that g maps some ray onto itself then there is z ≠ 0, ∞ such that g(z) = tz for some t > 0. Then t^{−1}g ∈ H fixes z so that t^{−1}g = Id and g = h_{t}. Thus k = Id. So every element of K \ {Id} is elliptic.

Suppose that k_{n} ∈ K and that g_{n} ∈ H gives rise to k_{n}. We may replace g_{n} by tg_{n} for any t > 0 without changing k_{n}. Thus we may choose t depending on n so that |g_{n}(1)| = 1. Since H is a convergence group, we may pass to a subsequence without changing notation and assume that g_{n} converges to g. Since g_{n} fixes 0 and ∞ and |g_{n}(1)| = 1, it follows that the limit function g must be a homeomorphism to which g_{n} converges uniformly on the sphere, and then g ∈ H since H is closed. Thus g gives rise to k ∈ K and clearly k_{n} → k uniformly on S^{1} . Every element of K is topologically conjugate to a Möbius transformation by [5], for example, and as we have seen above, every such Möbius transformation must be elliptic or the identity. Since H is abelian, so is K. Thus K is nondiscrete or finite, and in both cases there is a homeomorphism p of S^{1} onto itself such that K′ = p^{−1} ∘ K ∘ p is again a group of elliptic Möbius transformations, by [5]. Since H and K are abelian, so is K′. Thus all the elements of K′, when viewed as Möbius transformations of the unit disk, have the same fixed point in the disk, and by choosing p in an appropriate way (replacing p by p∘M for a suitable Möbius transformation M of the unit disk, if necessary), we may assume that this fixed point is the origin. Hence each element k′ of K′ is of the form k′(e^{iθ}) = ce^{iθ} where |c| = 1. The group K′, being a closed abelian group of rotations, is either a finite cyclic group, or contains every rotation ℛ_{exp(2πiα)} given by ℛ_{exp(2πiα)}(z) = e^{2πiα}z.

We claim that K′ is a finite cyclic group on the basis that E is a finite cyclic group. To get a contradiction, suppose that K′ contains all rotations around the origin. Suppose that every element of E has order at most N, and consider an element k′ of K′ of order 2N, say. Then the corresponding element k of K has order 2N, and if g ∈ H corresponds to k, then g^{2N} fixes every ray so that g^{2N} = h_{t} for some t > 0. Set h(z) = g(z)/t^{1/(2N)} so that h ∈ H. Then, since H is abelian, we have h^{2N} = Id while clearly h^{j} ≠ Id if 1 ≤ j < 2N. Thus h is elliptic of order 2N, which is a contradiction. It follows that K′, and hence K, is a finite cyclic group.

Now define a homeomorphism f_{1} of the sphere onto itself fixing 0 and ∞ by f_{1}(re^{iθ}) = rp(e^{iθ}). Then f_{1}^{−1} ∘ h_{t} ∘ f_{1} = h_{t} for all t > 0. Further, if g ∈ H gives rise to k ∈ K and if (p^{−1} ∘ k ∘ p)(e^{iθ}) = ce^{iθ} where |c| = 1, then

(6.12.1)  (f_{1}^{−1} ∘ g ∘ f_{1})(re^{iθ}) = cre^{iθ}|g(p(e^{iθ}))|.

Now H′ = f_{1}^{−1} ∘ H ∘ f_{1} contains every h_{t}, and each element of H′ fixes 0 and ∞, and no element of H′ \ {Id} has any other fixed points.

Suppose then that K′ is a finite cyclic group generated by ℛ_{exp(2πi/n)} corresponding to a function g′ = f_{1}^{−1} ∘ g ∘ f_{1} ∈ H′ where g ∈ H. Then c = e^{2πi/n} in (6.12.1). We note that now H is generated by g and the dilations h_{t} for t > 0. For if h ∈ H then the action of h on rays is the same as for some g^{j}. Thus g^{−j} ∘h maps each ray onto itself so that g^{−j} ∘ h = h_{t} for some t > 0. Next, g^{n} fixes each ray so that g^{n} = h_{t} for some t > 0. Replacing g by g ∘ h_{u} and hence replacing g′ by g′ ∘ h_{u} where u^{n} = 1/t, we may assume that g^{n} = (g′)^{n} = Id.

We define f(re^{iθ}) = F (θ)re^{iθ} and choose F (θ) > 0 for 0 ≤ θ ≤ 2π/n so that f will define a homeomorphism of the sector {re^{iθ} : r > 0, 0 ≤ θ ≤ 2π/n} onto itself satisfying

|g′(rF (0))| = |g′(f(r))| = rF (0)|g(p(1))| = rF (2π/n)

for all r > 0. That is, we require that F (2π/n) = F (0)|g(p(1))|. We use the equation

F (θ + 2π/n) = F (θ)|g(p(e^{iθ}))|

to extend the definition of F to all θ. This leads to the requirement that

F (0) = F (2π) = F (0) ∏_{j=0}^{n−1}|g(p(e^{2πij/n}))|,

which reads

(6.12.2)  1 = ∏_{j=0}^{n−1}|g + (p(e^{2πij/n}))|.

Applying (6.12.1) n times starting with re^{iθ} = 1, we obtain (6.12.2) since (g′)^{n} = Id. Now f conjugates each h_{t} onto itself while f^{−1} ∘ g′ ∘ f = ℛ_{exp(2πi/n)} . Thus we find an f such that f^{−1} ∘ H′ ∘ f is a Möbius group. This completes the proof that we can conjugate H to a Möbius group.

6.13. Proof of Theorem 4. Let the assumptions of Theorem 4 be satisfied. We may assume that every h_{t} fixes 0 and ∞, and then for any t ≠ 0, the function h_{t} has no fixed points in C \ {0}. For any z ∈ C \ {0}, all the points h_{t}(z) are distinct. For if not, then there are real distinct t and u with h_{t}(z) = h_{u}(z). But then h_{u−t} fixes z, a contradiction. Hence {h_{t}(z) : t ∈ R} is an open Jordan arc. We claim that this arc has the endpoints 0 and ∞. As n → ∞, the points h_{n}(z) = h_{1}^{n}(z) tend to 0 or ∞, say to ∞. We want to show that lim_{t→∞} h_{t}(z) = ∞. Suppose that t(j) = n(j) + ε(j) → ∞ where n(j) ∈ Z and 0 ≤ ε(j) < 1. To get a contradiction, suppose that h_{t(j)}(z) does not tend to ∞. By passing to a subsequence, we may assume that h_{t(j)}(z) → α ∈ C and that ε(j) → ε ∈ [0, 1]. By the assumptions of Theorem 4, we have h_{ε(j)} → h_{ε} uniformly on S^{2} . Hence

h_{t(j)}(z) = h_{ε(j)}(h_{n(j)}(z)) → h_{ε}( lim_{j→∞} h_{n(j)}(z)) = h_{ε}(∞) = ∞

as j → ∞, which gives a contradiction. It follows that lim_{t→∞} h_{t}(z) = ∞. Similarly, lim_{n→−∞} h_{n}(z) = 0 and so lim_{t→−∞} h_{t}(z) = 0.

Now define a family of mappings on C as follows. Suppose that z ∈ C. Then the points h_{t}(e^{z}) for t ∈ R define a Jordan arc through e^{z} = h_{0}(e^{z}). There is a unique arc of points k_{t}(z) in the plane such that k_{0}(z) = z and e^{k_{t}(z)} = h_{t}(e^{z}) for all real t. This defines k_{t}(z). Let m be any integer. Then the curves h_{t}(e^{z}) and h_{t}(e^{z+2πim}) are the same, and it follows from the definition above that k_{t}(z + 2πim) = k_{t}(z) + 2πim for all t. It easily follows from the definition that for any fixed real t, the map k_{t} is continuous in z. The map k_{t} is one-to-one, for if k_{t}(z) = k_{t}(w) then h_{t}(e^{z}) = h_{t}(e^{w}) so that e^{z} = e^{w} and so z = w + 2πim for some integer m. But then k_{t}(z) = k_{t}(w) + 2πim so that m = 0 and so z = w, as required. Also k_{t}(C) = C. For if w ∈ C then there is ζ ∈ C \ {0} such that h_{t}(ζ) = e^{w}. Choose any z ∈ C with e^{z} = ζ. Then

{k_{t}(z + 2πim) : m ∈ Z} = {w + 2πin : n ∈ Z}

so that w ∈ k_{t}(C), as required. Thus each k_{t} is a homeomorphism of C onto itself, and clearly k_{0} = Id.

If t ≠ 0, z ∈ C, m ∈ Z, and k_{t}(z) = z + 2πim, then by the definition above, we have

h_{t}(e^{z}) = e^{k_{t}(z)} = e^{z+2πim} = e^{z},

which is a contradiction. In particular, taking m = 0, we see that k_{t} has no finite fixed points when t ≠ 0. Further if t, u are distinct real numbers and z ∈ C then k_{t}(z) ≠ k_{u}(z). For if k_{t}(z) = k_{u}(z) then h_{t}(e^{z}) = h_{u}(e^{z}), a contradiction. Hence the points k_{t}(z) for t ∈ R form a Jordan arc γ(z).

We claim that k_{t+u} = k_{t} ∘ k_{u} for all real t and u. For this purpose, pick z ∈ C, denote γ(z) by γ, and write

Γ(e^{z}) = {h_{t}(e^{z}) : t ∈ R} = {e^{k_{t}(z)} : t ∈ R}.

Now k_{u}(z) ∈ γ, and h_{u}(e^{z}) = e^{k_{u}(z)} ∈ Γ(e^{z}). But now Γ(h_{u}(e^{z})) = Γ(e^{z}). Hence k_{t}(k_{u}(z)) is obtained from k_{u}(z) by lifting an inverse image (under e^{z}) of Γ(h_{u}(e^{z})) going through k_{u}(z). But this must be the arc γ itself. Thus k_{t}(k_{u}(z)) is the unique point w on γ with e^{w} = h_{t}(e^{k_{u}(z)}). But k_{t+u}(z) ∈ γ and

e^{k_{t+u}(z)} = h_{t+u}(e^{z}) = h_{t}(h_{u}(e^{z})) = h_{t}(e^{k_{u}(z)}).

It follows that k_{t+u}(z) = k_{t}(k_{u}(z)), and so k_{t+u} = k_{t} ∘k_{u}, as claimed. This shows that the set G = {k_{t} : t ∈ R} is a group, and in fact, an abelian group.

If t_{n} ∈ R and t_{n} → ∞ or t_{n} → −∞, it is now seen that k_{t_{n}} → ∞ locally uniformly in C. If t_{n} → t ∈ R then clearly k_{t_{n}} → k_{t} locally uniformly in C. Thus G is an abelian closed convergence group, and every element of G \ {Id} is parabolic. Furthermore, since k_{t}(z + 2πim) ≡ k_{t}(z) + 2πim, it follows that the group G′ generated by G and the mappings z ↦ z + 2πim for m ∈ Z, is also an abelian closed convergence group, and by the property k_{t}(z) ≠ z + 2πim (for m ∈ Z and t ≠ 0) established above, every element of G′ \ {Id} is parabolic.

It cannot be the case that Re k_{t}(z) = Re z for some t ≠ 0 and all z with Re z = r for some fixed real r. For suppose that this identity holds. Then |h_{t}(e^{z})| = |e^{k_{t}(z)}| = e^{Re k_{t}(z)} = e^{Re z} = |e^{z}| for this t and all such z. Then |h_{t}(w)| = |w| whenever |w| = e^{r}. This contradicts the assumption that h_{t} is loxodromic so that the orbit of any point e^{z} under the iterates of h_{t} clusters to 0 and ∞. This proves our assertion. In particular, if t ≠ 0, we cannot have k_{t}(z) ≡ z + 2πim for any integer m.

For any z ∈ C, we have {Re k_{t}(z) : t ∈ R} = R. By continuity, this follows as soon as {Re k_{t}(z) : t ∈ R} is unbounded above and below. This is the case since e^{Re k_{n}(z)} = |h_{n}(e^{z})| and since h_{n}(e^{z}) for n ∈ Z clusters to 0 and ∞.

By the part of Theorem 2 for groups with only parabolic elements, which we have proved already without any reference to Theorem 4, there is a homeomorphism f_{1} of \overline{C} fixing infinity such that each element of G^{0} = f_{1}^{−1} ∘ G′ ∘ f_{1} , other than the identity, is a parabolic Möbius transformation fixing infinity, that is, a translation. Since k_{t+u} = k_{t} ∘ k_{u} for all real t and u, we may clearly choose f_{1} so that f_{1}^{−1} ∘ k_{t} ∘ f_{1} = T_{t} for all real t where T_{t}(z) = z + t as before. Write g = f_{1}^{−1} ∘ T_{2πi} ∘ f_{1} ∈ G^{0} . Then g commutes with T_{t} for all real t while g^{m} is not equal to T_{t} for any real t, for any nonzero integer m. Thus g = T_{a} where a is a nonreal complex number.

We now define f_{2}(z) = az/(2πi), f = f_{1} ∘ f_{2} , and G′′ = f_{2}^{−1} ∘ G^{0} ∘ f_{2} = f^{−1} ∘ G′ ∘ f. Then f^{−1} ∘ k_{t} ∘ f = T_{2πit/a} while f^{−1} ∘ T_{2πi} ∘ f = T_{2πi}.

Thus G′′ contains the maps z ↦ z + 2πim for all m ∈ Z, and f satisfies f(z + 2πim) = f(z) + 2πim for all z ∈ C and all m ∈ Z. We define a_{t} ∈ C by (f^{−1} ∘ k_{t} ∘ f)(z) ≡ z + a_{t}, so that a_{t} = 2πit/a. Then a_{t+u} = a_{t} + a_{u} for all real t and u. We have a_{t} ≡ ct where the constant c is not purely imaginary.

Now define a homeomorphism F of C \ {0} onto itself by F (z) = e^{f(log z)} . This is well-defined, no matter which value of log z is used, and is continuous by local considerations. It clearly maps C \ {0} onto itself. If F (z) = F (w) then f(log z) = f(log w) + 2πim = f(log w + 2πim) for some m ∈ Z. But then log z = log w + 2πim so that z = w. Thus F is one-to-one, and so F defines a homeomorphism of C \ {0} onto itself. Clearly F extends to a homeomorphism of S^{2} onto itself.

Suppose that z ∈ C \ {0}. We have

(F^{−1} ∘ h_{t} ∘ F)(z) = exp {f^{−1}(log[h_{t}(e^{f(log z)})])}
= e^{f^{−1}(k_{t}(f(log z)))} = e^{log z+a_{t}} = e^{a_{t}}z = e^{ct}z.

Write c = c_{1} + ic_{2} where c_{1} and c_{2} are real. We replace F (z) by the function F_{1} defined by F (z) = F_{1}(z) when z ∈ {0, ∞} and F_{1}(re^{iθ}) = F (r^{c_{1}}e^{i(θ+β(r))}) for z = re^{iθ} ∈ C \ {0}. Here β(r) can be any function satisfying β(e^{t}r) − β(r) ≡ c_{2}t (mod 2π) whenever r > 0 and t is real. One possible choice is β(r) = c_{2} log r. It is now easy to verify that then (h_{t}∘F_{1})(z) = F_{1}(e^{t}z) so that (F_{1}^{−1}∘h_{t}∘F_{1})(z) = e^{t}z.

Suppose finally that each h_{t} commutes with every rotation around the origin. We claim that then each k_{t} commutes with the translation T_{iu} for every real u. To see this, note that if γ = e^{iu} for a fixed real u and if h_{t}(γz) = γh_{t}(z) for all z, then for any fixed z ∈ C, we have

e^{k_{t}(z)+iu} = γe^{k_{t}(z)} = γh_{t}(e^{z}) = h_{t}(γe^{z}) = h_{t}(e^{z+iu}) = e^{k_{t}(z+iu)}

so that (k_{t}(z) + iu) − k_{t}(z + iu) = 2πim for some integer m. For t = 0 this holds with m = 0 so that by continuity with respect to t, for fixed z and u, the same identity holds with m = 0 for all real t. This shows that k_{t} commutes with T_{iu}. If now Re k_{t}(z) = Re z for some t ≠ 0 and some z, then Re k_{t}(z + iu) = Re(k_{t}(z) + iu) = Re(z + iu) for all real u. As we have seen, this leads to a contradiction. Hence T_{iu} ∘ k_{t} has no finite fixed points when (t, u) ≠ (0, 0).

Next we may, in the above proof, replace G′ by the group generated by the k_{t} and the T_{iu} for all real t and u. We still find a homeomorphism f_{1} such that G^{0} = f_{1}^{−1} ∘ G′ ∘ f_{1} is a group of translations, while f_{1}^{−1} ∘ k_{t} ∘ f_{1} = T_{t} for all t. We pass to G′′ as before using a map f_{2} , and set f = f_{1} ∘ f_{2} . The mappings f^{−1} ∘ T_{iu} ∘ f = q_{u} form a one-parameter family of translations so that they are given by q_{u} = T_{bu} for some nonzero complex number b. Since q_{2π} = T_{2πi}, we have b = i. Hence q_{u} = T_{iu} and so f^{−1} ∘ T_{iu} ∘ f = T_{iu}. Defining F as before, we find that for any γ = e^{iu} with |γ| = 1 we have

(F^{−1} ∘ ℛ_{γ} ∘ F)(z) = e^{log z+iu} = γz.

If a further transformation is needed and F_{1}(re^{iθ}) = F (r^{c_{1}}e^{i(θ+log r)}), then

(ℛ_{γ} ∘ F_{1})(z) = F (r^{c_{1}}e^{i(θ+u+log r)}) = F_{1}(re^{i(θ+u)}) = F_{1}(γz),

as required for F_{1} to conjugate each ℛ_{γ} onto itself. This completes the proof of Theorem 4.

6.14. We return to the proof of Theorem 2 in Case II. We have shown that the subgroup H of G consisting of the sense-preserving elements of G can be topologically conjugated to a Möbius group. Let us therefore assume, without changing notation, that H, in fact, is a Möbius group consisting of mappings of the form z ↦ cz where c ∈ C \ {0}. Then there is a closed nondiscrete group J′ of translations T_{a} of C, containing T_{2πin} for all integers n, such that H = {z ↦ e^{a}z : T_{a} ∈ J′}.

Since J′ is closed and nondiscrete, it contains a one-parameter family of translations. In fact, the set E = {g̃(0) : g̃ ∈ J′} is a closed set in C and E is an additive group which contains points arbitrarily close to 0 as well as all the points 2πin for n ∈ Z. We claim that if E ≠ C then

(6.14.1)  E = {tc_{1} + nc_{2} : t ∈ R, n ∈ Z}

where c_{1}, c_{2} ∈ C with c_{1} ≠ 0 and Re(c̄_{1}c_{2}) = 0. For suppose that a_{n} ∈ E \ {0} and a_{n} → 0. If a_{n} belongs to line L_{n} through the origin, we may assume that the L_{n} tend to a limit line L. If w ∈ L and ε > 0, there is a line L_{n} such that the distance of w from L_{n} is less than ε/2 and such that |a_{n}| < ε/2. Since ma_{n} ∈ E for all m ∈ Z, there is β ∈ E with |β − w| < ε. Since ε was arbitrary and E is closed, we have w ∈ E, and so L ⊂ E (cf. an argument of Baker in [1, p. 285]). We may write L = {tc_{1} : t ∈ R} where c_{1} ≠ 0. If L ≠ E ≠ C and c_{2} ∈ E \ L then nc_{2} + tc_{1} ∈ E for all n ∈ Z and t ∈ R. We may thus only consider those c_{2} ∈ E \ L that are perpendicular to L as vectors in R^{2} , that is, those with Re(c̄_{1}c_{2}) = 0. These elements c_{2} of E on the line M through 0 orthogonal to L either form a discrete set of the form {mc′_{2} : m ∈ Z} for some c′_{2} ≠ 0, or M ⊂ E, in which case E = C. Thus (6.14.1) holds. Note that if E = L then (6.14.1) holds with c_{2} = 0.

Thus H consists of mappings M_{c} where M_{c}(z) = cz, such that if V is defined by H = {M_{c} : c ∈ V } then either V = C \ {0} or

(6.14.2)  V = {c_{3}^{n}e^{c_{1}t} : t ∈ R, n ∈ Z}

where c_{3} = e^{c_{2}} ≠ 0.

6.15. If G ≠ H then G is generated by H and some h ∈ G \ H. Also G is abelian. If V = C \ {0} then

(6.15.1)  h(cz) = ch(z) for all z ∈ C and c ∈ V,

and taking z = 1 we see that h is sense-preserving, which is a contradiction. Let V be as in (6.14.2). Suppose first that |c_{3}| = |e^{c_{1}}| = 1. Then

V = {c ∈ C : |c| = 1}

since c_{1} ≠ 0. Now (6.15.1) is the same as (3.1), so that the proof of Lemma 1 shows that G is topologically conjugate to a Möbius group.

Suppose then that |c_{3}| ≠ 1 or |e^{c_{1}}| ≠ 1. Now (6.14.2) and (6.15.1) with z = 1 show that h fixes 0 and ∞. If c_{2} = 0 then, since 2πi ∈ E, we have c_{1} = ib where b is real. But then |c_{3}| = |e^{c_{1}}| = 1, which is a contradiction. Thus c_{2} ≠ 0, and in view of the structure of the set E, we may assume first that c_{1} and c_{2} are orthogonal vectors in R^{2} and then, since c_{1} can be multiplied by a nonzero real number without changing E, that c_{1} = ic_{2} . Since 2πi ∈ E, there are t_{0} ∈ R and n_{0} ∈ Z such that (n_{0} + it_{0})c_{2} = 2πi, that is,

c_{2} = 2π(t_{0} + in_{0})(t_{0}^{2} + n_{0}^{2})^{−1}.

Replacing c_{2} by −c_{2} , if necessary, we may assume that n_{0} ≥ 0. A calculation shows that either n_{0} ≠ 0 and

(6.15.2)  V = {ω^{k} exp[t(n_{0} − it_{0})] : t ∈ R, k ∈ Z and 0 ≤ k < n_{0}}

where ω = exp{2πi/n_{0}}, or n_{0} = 0 ≠ t_{0} and

(6.15.3)  V = {exp[u(n + it)] : n ∈ Z, t ∈ R}

where u = 2π/t_{0} . In the latter case, V contains the unit circle (take n = 0), so that (6.15.1) and the proof of Lemma 1 show that G is topologically conjugate to a Möbius group. (In fact, it may be that (6.15.3) implies some contradiction but we need not be concerned about that.) More precisely, to get (6.15.2), note that every α ∈ E is of the form α = (n + it)c_{2} = 2πi(n + it)/(n_{0} + it_{0}). Choose k ∈ Z with 0 ≤ k < n_{0} so that k ≡ n (mod n_{0} ). Then choose u = 2π(nt_{0}−tn_{0})/(n_{0}(n_{0}^{2}+t_{0}^{2})), write e^{α} in terms of k and u, and denote u again by t. This gives (6.15.2).

6.16. So we may assume that V is as in (6.15.2). Since h is sense-reversing and fixes 0 and ∞, it follows that the image h(S(1)) of the unit circle covered in the positive direction, when projected radially by the map z ↦ z/|z| onto a path Γ on the unit circle, is homotopic to the circle covered once in the negative direction. On the other hand, if Γ starts at α = h(1)/|h(1)|, then it goes through ωα, ω^{2}α, …, ω^{n_{0}−1}α in this order since h(ω^{k}) = ω^{k}h(1) by (6.15.1) and (6.15.2). The part of Γ from α to ωα is homotopic to S(1) covered m times plus an arc of angular measure 2π/n_{0} covered in the positive direction. Since h(ω^{k}z) = ω^{k}h(z), it follows by calculating the winding number of Γ around the origin that −1 = 1 + n_{0}m, that is, n_{0}m = −2, so that n_{0} = 1 or n_{0} = 2.

We have h^{2} ∈ H, so h^{2} = M_{c} for some c ∈ V . There is d ∈ V with d^{2} = ±c. Replacing h by h ∘ M_{1/d} if necessary, we may assume that h^{2} = Id or h^{2} = M_{−1} , the latter case occurring at most when n_{0} = 2.

6.17. Suppose that h^{2} = Id, and for r > 0, set D_{0}(r) = B(r) ∪ h(B(r)), let U(r) be the unbounded component of \overline{C} \ \overline{D_{0}(r)}, and set D(r) = C \ \overline{U(r)}. Then D(r) is a Jordan domain and Γ(r) = ∂D(r) is a Jordan curve. This is clear if S(r) and h(S(r)) have at most one point of intersection, and otherwise follows from a theorem of Kerékjártó ([11, Hilfssatz I, p. 87]; see also [19, Example, p. 168]) since Γ(r) is the boundary of one of the components of

\overline{C} \ \{S(1) ∪ h(S(1))\}.

Also h maps U(r) and hence Γ(r) and D(r) onto itself. The map h | Γ(r) is sensereversing and has exactly two fixed points, say a(r) and b(r), that divide Γ(r) into two arcs that are interchanged by h (this follows since h is sense-reversing and fixes 0 and ∞).

We may write the elements of V as e^{t(1−it_{0})} or ±e^{t(1−it_{0}/2)} depending on if n_{0} = 1 or n_{0} = 2. In each case, we may take

(6.17.1)  a(r) = rc(r)a(1)  and  b(r) = rc(r)b(1)

where c(r) = exp{−(log r)it_{0}} or c(r) = exp{−(log r)it_{0}/2} so that |c(r)| = 1. If n_{0} = 2, we further have b(r) = −a(r) for all r, since h(−z) = −h(z) and so D(r) = {−z : z ∈ D(r)}. More precisely, if c ∈ V then by (6.15.1) we have h(B(|c|r)) = {cz : z ∈ h(B(r))} and so D(|c|r) = {cz : z ∈ D(r)}, hence Γ(|c|r) = {cz : z ∈ Γ(r)}. This together with (6.15.1) gives (6.17.1).

We now define a homeomorphism f_{3} of \overline{C} fixing 0 and ∞ as follows. Let F be a sense-preserving homeomorphism of Γ(1) onto S(1) with F (a(1)) = 1 and F (b(1)) = −1, for example, the boundary value map of a suitable conformal mapping of D(1) onto B(1). Set Γ_{1} = F^{−1}(S(1) ∩ {w : Im w ≥ 0}) and Γ_{2} = Γ(1) \ Γ_{1} . Suppose that z ∈ Γ(r). If n_{0} = 1 there is a unique c ∈ V with |c| = r. If n_{0} = 2, there are two points, say ±c, that belong to V and have modulus r. Then we choose c = exp{(log r)(1 − it_{0}/2)}. Now z/c ∈ Γ(1). If z/c ∈ Γ_{1} , we set f_{3}(z) = |c|F (z/c). If z/c ∈ Γ_{2} , we set f_{3}(z) = \overline{f_{3}(h(z))}. Thus f_{3} | Γ(r) is a homeomorphism of Γ(r) onto S(r). We have f_{3}(z) = \overline{f_{3}(h(z))} also when z/c ∈ Γ_{1} since then z = h(w) where w/c ∈ Γ_{2} and since h^{2} = Id. Thus (f_{3} ∘ h ∘ f_{3}^{−1})(z) = z̄ for all z.

Clearly the curves Γ(r) are disjoint and their union is C \ {0}. Thus f_{3} is one-to-one and onto, and continuous at 0 and at ∞. For c_{0}, c ∈ V and z/c ∈ Γ_{1} , we have f_{3}(z) = |c|F (z/c) and

f_{3}(c_{0}z) = |cc_{0}|F {(c_{0}z)/(c_{0}c)} = |cc_{0}|F (z/c) = |c_{0}|f_{3}(z).

If z/c ∈ Γ_{2} then f_{3}(z) = \overline{f_{3}(h(z))} while

f_{3}(c_{0}z) = \overline{f_{3}(h(c_{0}z))} = \overline{f_{3}(c_{0}h(z))} = \overline{|c_{0}|f_{3}(h(z))} = |c_{0}| \overline{f_{3}(h(z))}.

Thus f_{3}(c_{0}z) = |c_{0}|f_{3}(z) for all z and so

f_{3} ∘ M_{c_{0}} ∘ f_{3}^{−1} = M_{|c_{0}|}.

It follows that f_{3} conjugates G to a Möbius group.

It follows easily from the definitions that f_{3} is continuous on C \ {0} also and hence on \overline{C}. Since \overline{C} is compact, f_{3} is a homeomorphism. Thus G is topologically conjugate to a Möbius group.

6.18. If h^{2} = M_{−1} then z = h^{2}(z) = −z for any fixed point z of h so that fix(h) = {0, ∞}. But we may proceed as above and define

D_{0}(r) = ⋃_{n=0}^{3} h^{n}(B(r)) = D′_{0}(r) ∪ h^{2}(D′_{0}(r))

where D′_{0}(r) = B(r) ∪ h(B(r)). Let the unbounded component of \overline{C} \ \overline{D′_{0}(r)} be U_{1}(r) so that U_{1}(r) is a Jordan domain as above, and set D_{1}(r) = C \ \overline{U_{1}(r)} and D_{2}(r) = D_{1}(r) ∪ h^{2}(D_{1}(r)). Then the unbounded components of \overline{C} \ \overline{D_{0}(r)} and \overline{C} \ \overline{D_{2}(r)} coincide. Thus, if that component is denoted by U(r), then U(r) is a Jordan domain and Γ(r) = ∂U(r) is a Jordan curve contained in C \ {0}. Since h^{4} = Id, the map h takes each of D_{0}(r), U(r) and Γ(r) onto itself. Now h | Γ(r) is sense-reversing and has two fixed points so that h has two fixed points in C \ {0}, which is a contradiction. Hence the case h^{2} = M_{−1} cannot occur.

This proves Theorem 2 in Case II. Thus the proof of Theorem 2 is complete.

246 A. Hinkkanen and G. J. Martin

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Received 12 March 1991