Abstract
We study the problem of lifting an Abelian group H of automorphisms of a closed Riemann surface S (containing anticonformals ones) to a suitable Schottky uniformization of S (that is, when H is of Schottky type). If H+ is the index two subgroup of orientation preserving automorphisms of H and R = S/H+, then H induces an anticonformal automorphism ? : R ? R. If ? has fixed points, then we observe that H is of Schottky type. If ? has no fixed points, then we provide a sufficient condition for H to be of Schottky type. We also give partial answers for the excluded cases.
Highlights
Let us consider a closed Riemann surface S
Retrosection theorem [2], [12] asserts that S can be uniformized by a suitable Schottky group G, that is, a geometrically finite Kleinian group without parabolic transformations and isomorphic to a free group of finite rank
In [10] we have provided a necessary and sufficient condition for a group H of automorphisms to be of Schottky type
Summary
Let us consider a closed Riemann surface S. Since Ω/Ge is closed, we have hE, T i = hE, z 7→ −1/zi and, in particular, we may assume T (z) = −1/z In this way, for each elliptic transformation E ∈ Ge+ there is an imaginary reflection T ∈ Ge −Ge+ such that T E = ET and that hE, T i uniformizes a real projective plane with exactly one branch value (either of order 3 or 5). Arguments similar to the example constructed above can be suitable modified in order to have a closed Riemann surface S with an anticonformal automorphism F : S → S so that S/F is the connected sum of two real projective planes and exactly 3 different branch values, say of orders 3, 5 and 7, and not of Schottky type. For #Set2 ≥ 6 theorem 7 in general fails
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