Documenta Math. 413 Realizing Countable Groups As Automorphism Groups of Riemann Surfaces

Every countable group can be realized as the full automorphism group of a Riemann surface as well as the full group of isometries of a Riemannian manifold. 2000 Mathematics Subject Classification: Primary 30F99, Secondary 20F29,20B27,32M05

Theorem.Let G be a (finite or infinite) countable group.Then there exists a (connected) Riemann surface M such that G is isomorphic to the group Aut O (M ) of all holomorphic automorphisms of M .Moreover, there exists a Riemannian metric h on M such that Aut O (M ) equals the group of all isometries of (M, h).
Our strategy is as follows: Using Galois theory of coverings, we first construct a Riemann surface M 1 on which G acts.Then we remove a discrete subset S ⊂ M 1 to kill excess automorphisms.However, we have to show that passing from M 1 to M 1 \ S we do not risk enlarging the automorphism group, i.e., we will show that every automorphism from M 1 \ S extends to M 1 .For this purpose we employ the Freudenthal's theory of topological ends.Finally, hyperbolicity of the Riemann surface is exploited to ensure that there is a hermitian metric of constant negative curvature such that the group of all holomorphic automorphisms coincides with the group of all isometries.Let us remark that by uniformization theory it is well-known that the following is the list of all Riemann surface with positive-dimensional automorphism group and that their automorphism groups are well-known: (See [1] for this and other basic facts on Riemann surfaces.)Therefore our result yields a complete characterization which groups may occur as automorphism group of a Riemann surface.The above list furthermore has the following consequence which we will use later on: Fact.Let M be a Riemann surface with non-commutative fundamental group π 1 (M ).Then Aut O (M ) is discrete and acting properly discontinuously on M .In particular, every orbit is closed.

Hyperbolic Riemannian Surfaces
A Riemannian surface M is "hyperbolic" (in the sense of Kobayashi) if and only if its universal covering is isomorphic to the unit disk.In this case the Poincaré metric on the unit disk induces a unique hermitian metric of constant Gaussian curvature −1 on M .Every holomorphic automorphism of M is an isometry and conversely every isometry is either holomorphic or antiholomorphic.Thus the group of holomorphic automorphisms of M is a subgroup of index 1 or 2 in the group of isometries and the group of isometries coincides with the group of all holomorphic or antiholomorphic diffeomorphisms of M .
2 Galois theory of coverings Proposition 1.Let G be a countable group.Let M 0 be a Riemann surface whose fundamental group is not finitely generated.Then there exists an unramified covering M 1 → M 0 such that there is an effective G-action on M 1 , Aut O (M 1 ) is discrete and acting properly discontinuously on M 1 .
Proof.Let F ∞ be a free group with countably infinitely many generators α 1 , α 2 , . ... By standard results on Riemann surfaces (see e.g.[1]) we have Since G is countable, there is a surjective group homomorphism ζ : F ∞ → G. Furthermore, we may require that α 1 , α 2 ∈ ker ζ.Then we obtain a short exact sequence of groups where N is non-commutative, because it contains a free group with two generators (viz.α 1 and α 2 ).By Galois theory of coverings, this implies that there exists an unramified covering M 1 → M 0 with π 1 (M 1 ) N and an effective G-action on M 1 .Finally, discreteness of Aut O (M 1 ) as well as the action of Aut O (M 1 ) being properly discontinuous is implied by the "Fact" established above.

Topological ends
Let us recall the basic facts from the theory of ends as developed by Freudenthal [3].Let X be a locally compact topological space.Then the set of "ends" e(X) is defined by e(X) = lim Thus, if K n is an exhaustion of X by an increasing sequence of compact subsets, then every end ∈ e(X) can be represented by a sequence U n of connected components of X \ K n with U n ⊃ U n+1 .For every connected component W n,i of X \ K n we now define E n,i as the set of ends which can be represented with U n = W n,i .Now X = X ∪ e(X) becomes a compact topological space as follows: As a basis of the topology we take the family of all open subsets of X together with V n,i = W n,i ∪ E n,i for all n, i.
Then every proper continuous map between locally compact topological spaces X and Y extends to a continuous maps between X and Ȳ .In particular, every homeomorphism of X extends to a homeomorphism of X.
Definition.An end of a Riemann surface X is called a "puncture" if there is an open neighbourhood W of in X such that • there is a homeomorphism ξ : We now prove that the ends of a certain special class of Riemann surfaces cannot be punctures.Let X 0 = D \ A and let X 1 → X 0 be an unramified covering.
Then no end of X 1 is a puncture.
Proof.First we show that A is indeed closed.Let vol(B n ) = πr 2 n denote the euclidean volume.Since the balls B n are disjoint, we have vol(B n ) ≤ π and therefore lim r n = 0.It follows that for every s < 1 there is a natural number N such that |a n | − r n > s for all n ≥ N .Therefore A = ∪ n B n is actually a locally finite union, and closedness of the B n implies that A is closed.Now let us assume that there exists an end which is a puncture.The natural embeddings X 0 → D → C composed with the projection π : X 1 → X 0 yield a bounded holomorphic function f on X 1 .Let ∈ e(X 1 ) be a puncture with a connected open neighbourhood W as in the above definition of the notion "puncture".Then the Riemann extension theorem implies that f extends through .In other words, lim x→ f (x) = a exists.Evidently a is contained in the closure of X 0 in C.However, the boundary ∂X 0 is given by ∂X 0 = ∂D ∪ (∪ n ∂B n ), and the openness of holomorphic maps implies that a cannot lie on either ∂D or on one of the sets ∂B n .Therefore a ∈ X 0 .Now choose contractible open neighbourhoods U of a in X 0 and V of in W such that V \ { } ⊂ π −1 (U ).Since π : X 1 → X 0 is an unramified covering and U is simply-connected, we obtain π −1 (U ) G × U where G is equipped with the discrete topology.Being connected, V * = V \ { } must be contained in one connected component of π −1 (U ).This implies the following: If a n is a sequence in V * such that lim π(a n ) = a, then there is a point p ∈ X 1 with lim a n = p.But this contradicts the fact that by construction there is sequence a n in V * with lim a n = ∈ X 1 and lim π(a n ) = a.Thus this case can be ruled out as well, i.e., there cannot exist an end which is a puncture.

Proof of the theorem
Proof.Let a n be a diverging sequence in D = {z ∈ C : |z| < 1} and r n ∈ R >0 such that all the closed balls B n = {z ∈ C : |z −a n | ≤ r n } are disjoint subsets of D. Let A = ∪ n B n .Then A is a closed subset of D and the fundamental group of X 0 = D\A is not finitely generated.Hence, by prop. 1 there is an unramified covering π : X 1 → X 0 with an effective G-action on X 1 .Let Aut O (X 1 ) denote the group of all holomorphic automorphisms of X 1 and A the group of all diffeomorphisms of X 1 which are either holomorphic or antiholomorphic.Again by prop. 1 we may assume that Aut O (X 1 ) is discrete and acting properly discontinuously.Since Aut O (X 1 ) is of finite index in A, the group A is likewise discrete and acting properly discontinuously on X 1 .Now, for every g ∈ A \ {e} the fixed point set Let h be the unique hermitian metric of constant Gaussian curvature −1 on X and I its isomorphism group.We claim that I = Aut O (X) G.To show this, it suffices to show that every holomorphic or antiholomorphic automorphism of X extends to a holomorphic or antiholomorphic automorphism of X 1 .If φ is a holomorphic or antiholomorphic automorphism of X, it is in particular a selfhomeomorphism and therefore extends to a homeomorphism φ of the compact topological space X = X ∪ e(X) (where e(X) is the set of ends as explained in §3.above).Now e(X) = e(X 1 ) ∪ S. Evidently every end of X given by a point of S is a puncture as defined in §3.On the other hand, due to prop. 2 none of the ends of X 1 is a puncture.Now φ| e(X) is a permutation of the elements of e(X) which stabilizes the set of those ends which are punctures.Hence φ(S) = S. Thus φ extends to a continuous self-map of X 1 = X ∪ S.However, a continuous map which is holomorphic or antiholomorphic everywhere except for some isolated points, is necessarily holomorphic resp.antiholomorphic everywhere (by Riemann extension theorem).Hence every φ ∈ I of X extends to a holomorphic or antiholomorphic automorphism of X 1 .Consequently Aut O (X) = I = {g ∈ A : g(S) = S} = G.

Proposition 2 .
Let a n be a diverging sequence in D = {z ∈ C : |z| < 1} and r n ∈ R >0 such that all the closed balls B n = {z ∈ C : |z − a n | ≤ r n } are disjoint subsets of D. Then A = ∪ n B n is a closed subset of D.