Symmetries of compact Riemann surfaces

Lecture Notes in Mathematics Editors: J.-M. Morel, Cachan F. Takens, Groningen B. Teissier, Paris 2007 Emilio Bujala...

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Lecture Notes in Mathematics Editors: J.-M. Morel, Cachan F. Takens, Groningen B. Teissier, Paris

2007

Emilio Bujalance · Francisco Javier Cirre José Manuel Gamboa · Grzegorz Gromadzki

Symmetries of Compact Riemann Surfaces

123

Emilio Bujalance

José Manuel Gamboa

Facultad de Ciencias, UNED Matemáticas Fundamentales C/ Senda del Rey 9 28040 Madrid Spain [email protected]

Universidad Complutense Madrid Facultad de Matemáticas, UCM ´ Departamento de Algebra Plaza de las Ciencias 3 28040 Madrid Spain [email protected]

Francisco Javier Cirre Facultad de Ciencias, UNED Matemáticas Fundamentales C/ Senda del Rey 9 28040 Madrid Spain [email protected]

Grzegorz Gromadzki University of Gdansk Department of Mathematics Wita Stwosza 57 80-952 Gdansk Poland [email protected]

ISBN: 978-3-642-14827-9 e-ISBN: 978-3-642-14828-6 DOI: 10.1007/978-3-642-14828-6 Springer Heidelberg Dordrecht London New York Lecture Notes in Mathematics ISSN print edition: 0075-8434 ISSN electronic edition: 1617-9692 Library of Congress Control Number: 2010935187 Mathematics Subject Classification (2010): 30F, 14H, 20H, 20D, 57M c Springer-Verlag Berlin Heidelberg 2010 This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilm or in any other way, and storage in data banks. Duplication of this publication or parts thereof is permitted only under the provisions of the German Copyright Law of September 9, 1965, in its current version, and permission for use must always be obtained from Springer. Violations are liable to prosecution under the German Copyright Law. The use of general descriptive names, registered names, trademarks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. Cover design: SPi Publisher Services Printed on acid-free paper springer.com

´ To Alvaro and my grandchildren To my family ´ To Belén, Alvaro and Irene To the memory of my parents

Preface

The content of this monograph is situated in the intersection of important branches of mathematics like the theory of one complex variable, algebraic geometry, low dimensional topology and, from the point of view of the techniques used, combinatorial group theory. The main tool comes from the Uniformization Theorem for Riemann surfaces, which relates the topology of Riemann surfaces and holomorphic or antiholomorphic actions on them to the algebra of classical cocompact Fuchsian groups or, more generally, non-euclidean crystallographic groups. Foundations of this relationship were established by A. M. Macbeath in the early sixties and developed later by, among others, D. Singerman. Another important result in Riemann surface theory is the connection between Riemann surfaces and their symmetries with complex algebraic curves and their real forms. Namely, there is a well known functorial bijective correspondence between compact Riemann surfaces and smooth, irreducible complex projective curves. The fact that a Riemann surface has a symmetry means, under this equivalence, that the corresponding complex algebraic curve has a real form, that is, it is the complexification of a real algebraic curve. Moreover, symmetries which are non-conjugate in the full group of automorphisms of the Riemann surface, correspond to real forms which are birationally non-isomorphic over the reals. Furthermore, the set of points fixed by a symmetry is homeomorphic to a projective smooth model of the real form. The monograph consists of an extensive Introduction, a compilation of basic results in the Preliminaries, four principal Chapters and a short Appendix on asymmetric Riemann surfaces. After the Preliminaries, in Chap. 2, we focus our attention on the quantitative results concerning upper bounds for the number of conjugacy classes of symmetries. We divide our study into three cases, according to the nature of the set of points fixed by the symmetries. Namely we distinguish whether this set is empty or not and, accordingly, consider just symmetries with fixed points, just symmetries without fixed points and finally hybrid configurations allowing both types of symmetries simultaneously. Chapter 3 can be seen as a variation on the classical Harnack theorem, that states that the set of points fixed by a symmetry of a Riemann surface of genus g has at most g + 1 connected components, all of them being closed Jordan curves, called ovals in Hilbert’s terminology introduced in the nineteenth century. We first deal with the problem of finding the total number of ovals of a specified vii

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number of non-conjugate symmetries. We next consider the same problem for all the symmetries (conjugate or not) of a Riemann surface. We finally deal with the total number of ovals of a pair of symmetries in terms of the order of its product and the genus of the surface. The monograph is actually devoted to the symmetries of Riemann surfaces of genus at least two since they are the ones uniformized by the hyperbolic plane. The theory of symmetries of the remaining surfaces, that is, the Riemann sphere and the tori, is well-known for a long time but, for the sake of completeness and the reader’s convenience, we devote the main part of Chap. 4 to this subject. We also outline the classification of the symmetry types of hyperelliptic Riemann surfaces as being the double covers of the Riemann sphere. Finally, Chap. 5 is dedicated to the symmetries of Riemann surfaces with large groups of automorphisms. Such surfaces are important since on the one hand they are determined by a 2-generator presentation of their groups of automorphisms, and on the other hand they can be defined over the algebraic numbers due to the celebrated theorem of Belyi from the late seventies. Furthermore, by a recent result of B. Köck and D. Singerman, these algebraic numbers can be chosen to be reals if the surface is symmetric. The foundations for the study of symmetries of such surfaces were established by Singerman, who found necessary and sufficient algebraic conditions in terms of the mentioned above generating pair for such a surface to be symmetric. In the first section, apart from Singerman’s proof, we give formulae to compute the number of ovals of these symmetries, to which we refer as Singerman symmetries. Using these formulae we deal, in the next two sections, with the significant families of Macbeath-Singerman and Accola-Maclachlan and Kulkarni surfaces. Finally we describe the symmetries of the last two families by means of algebraic formulae. Acknowledgments. The authors are very grateful to the three referees for their very helpful, deep and accurate comments concerning the first version of the monograph. We are also very grateful to Dr Ewa Kozłowska-Walania for her careful reading of the final version and the number of conspicuous and helpful comments. Emilio Bujalance is partially supported by Spanish MTM2008-00250. Francisco Javier Cirre is partially supported by Spanish MTM2008-00250. José Manuel Gamboa is partially supported by Spanish MTM2008-00272, Proyecto Santander Complutense PR34/07-15813 and GAAR Grupos UCM 910444. Grzegorz Gromadzki is supported by the Sabbatical Grant SAB2005-0049 of the Spanish Ministry of Education and by the Research Grant NN201 366436 of the Polish Ministry of Sciences and Higher Education. Madrid, Gdańsk, April 2010

Emilio Bujalance Francisco Javier Cirre José Manuel Gamboa Grzegorz Gromadzki

Contents

1

Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . 1.1 NEC Groups and Their Signatures . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . 1.2 Normal Subgroups of NEC Groups . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . 1.3 Centralizers of Reflections. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . 1.4 Uniformization and Automorphism Groups of Riemann and Klein Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . 1.4.1 Maximal NEC Groups .. . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . 1.4.2 Teichmüller Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . 1.5 Symmetric Riemann Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . 1.5.1 Algebraic Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . .

1 1 6 8 11 13 14 15 18

2 On the Number of Conjugacy Classes of Symmetries of Riemann Surfaces .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . 2.1 Conjugacy Classes of Involutions in 2-Groups . . . . . . . . .. . . . . . . . . . . . . . . . . 2.2 Symmetries with Non-Empty Set of Fixed Points . . . . . .. . . . . . . . . . . . . . . . . 2.3 Symmetries with Empty Set of Fixed Points . . . . . . . . . . .. . . . . . . . . . . . . . . . . 2.4 Symmetries of Surfaces Admitting a Fixed Point Free Symmetry . . . . .

21 21 23 29 31

3 Counting Ovals of Symmetries of Riemann Surfaces . . . . .. . . . . . . . . . . . . . . . . 3.1 Enumeration of Ovals of Symmetries at Large . . . . . . . . .. . . . . . . . . . . . . . . . . 3.2 Total Number of Ovals of Non-Conjugate Symmetries . . . . . . . . . . . . . . . . . 3.3 Total Number of Ovals of all Symmetries of a Riemann Surface.. . . . . . 3.4 Total Number of Ovals of a Couple of Symmetries . . . .. . . . . . . . . . . . . . . . .

33 33 34 41 55

4 Symmetry Types of Some Families of Riemann Surfaces . . . . . . . . . . . . . . . . . 4.1 Symmetry Type of the Riemann Sphere . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . 4.2 Symmetry Types of Tori . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . 4.2.1 Symmetric Tori . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . 4.3 Symmetry Types of Hyperelliptic Riemann Surfaces . .. . . . . . . . . . . . . . . . . 4.3.1 A Geometric Method . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . 4.3.2 An Example .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . .

65 65 68 73 82 85 87

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5 Symmetry Types of Riemann Surfaces with a Large Group of Automorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . 91 5.1 Some General Results. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . 91 5.2 Symmetry Types of Macbeath–Singerman Surfaces . . .. . . . . . . . . . . . . . . . . 95 5.3 Symmetry Types of Accola–Maclachlan and Kulkarni Surfaces . . . . . . .110 5.3.1 Number of Ovals of the Symmetries .. . . . . . . . . . .. . . . . . . . . . . . . . . . .110 5.3.2 Separating Character of the Symmetries.. . . . . . .. . . . . . . . . . . . . . . . .120 5.4 Algebraic Formulae for the Symmetries . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . .125 6 Appendix . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . .145 6.1 Compact Riemann Surfaces Without Symmetries .. . . . .. . . . . . . . . . . . . . . . .145 References .. . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . .151 Index . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . .157

Introduction

By a symmetry σ of a compact Riemann surface S we mean an antianalytic involution σ : S → S. A Riemann surface which admits a symmetry is called symmetric. Under the well known functorial bijective correspondence between compact Riemann surfaces and smooth, irreducible, complex projective curves, symmetric surfaces correspond to curves definable over the field R of real numbers. If σ : S → S is a symmetry then the pair (S, σ) is usually called a real algebraic curve, see the foundational monograph [4] by Alling and Greenleaf to justify this definition. Some topological features of the real curve (S, σ) can be obtained from its associated symmetry σ. For instance, the set of real points of the curve is homeomorphic to the fixed point set Fix(σ) of the symmetry. In addition, symmetries which are non-conjugate within the full group Aut(S) of automorphisms of S correspond to real curves which are non-isomorphic over the real numbers but are isomorphic over the complex numbers. With a language closer to the one we will use here, let us show an example of two non-birationally R-isomorphic real algebraic curves whose complexifications are birationally C-isomorphic. Let us consider for t = 0, 1, the degree 3 homogeneous polynomial Ft (x, y, z) = y 2 z − x(x2 + (−1)t z 2 ). An easy computation shows that at any point in the complex projective plane P2 (C), the partial derivatives of Ft are not simultaneously zero. So, for t = 0, 1, each set St = {[x : y : z] ∈ P2 (C) : Ft (x, y, z) = 0} admits a structure of compact Riemann surface. In fact, S0 and S1 are birationally C-isomorphic as complex algebraic curves via the isomorphism ϕ : S0 → S1 ; [x : y : z] → [ξx : ξ 4 y : ξ 3 z], where ξ = eiπ/4 . However, their sets of R-rational points, that is, the real curves S0 (R) and S1 (R), are not birationally R-isomorphic. Indeed, both are smooth but S0 (R) is connected while S1 (R) has two connected components. The paper [32] by Cirre and Gamboa presents many other examples of non-isomorphic real algebraic curves with isomorphic complexifications. xi

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These phenomena lead naturally to the following problems we deal with in this monograph: (1) Is a complex smooth algebraic curve C definable over the reals? (2) Assume that this question has an affirmative answer. How many nonbirationally R- isomorphic real algebraic curves admit C as its complexification? The projective smooth models of such real curves are usually called the real forms of C. (3) What can be said about the topology of the real forms of C? The expository work by Gromadzki [51] can be understood as the first attempt to survey the known answers to these questions. Because of the methods to be used, it seems convenient to translate these questions into a more suitable language. To that end we use the terminology introduced at the beginning. In particular, the first of the above problems reads off: is a compact Riemann surface symmetric? Let σ and τ be symmetries of the compact Riemann surface S. The pairs (S, σ) and (S, τ ) are real forms of S; they are said to be isomorphic if there exists an automorphism ϕ of S such that σ = ϕ ◦ τ ◦ ϕ−1 . In this way the second problem to be treated is the counting of the number of conjugacy classes of symmetries with respect to the group Aut(S) of analytic and antianalytic automorphisms of the Riemann surface S. Finally, the topological type of a symmetry σ of S is determined, together with the genus of S, by the number of connected components, or ovals (in the nineteenth century Hilbert’s terminology) of the fixed point set Fix(σ) = {p ∈ S : σ(p) = p} and the connectedness character of its complement S \ Fix(σ) in S. More precisely, the triple (g, k, ε) is said to be the topological type of a symmetry σ of a genus g surface S if the set Fix(σ) has k connected components, and ε = 1 or ε = 0 according to whether S \Fix(σ) is connected or not. We say that σ is non-separating if ε = 1 and separating otherwise. A classical result due to Harnack [59] and Weichold [127] states that the necessary and sufficient conditions for a triple to be admissible, that is, to be the topological type of some symmetry σ, are the following: 1≤k ≤g+1

if ε = 0 with g + 1 ≡ k (mod 2);

0≤k≤g

if ε = 1.

The pair (k, ε) is usually codified by the symbol +k if ε = 0 and −k if ε = 1. It is called the species of the symmetry σ and denoted by sp(σ). It has to be mentioned that the orbit space Xσ = S/σ of the compact Riemann surface S under the symmetry σ is usually called a compact Klein surface. The fixed point set Fix(σ) is homeomorphic to the topological boundary of Xσ . Both are in fact homeomorphic to the set of real points of the projective smooth irreducible real algebraic curve associated to the symmetry σ. If this set is empty then we say that the corresponding real curve (S, σ) is purely imaginary. These pairs correspond to complex algebraic curves which can be defined over the reals but have no R-rational points. It is well known that the set S \ Fix(σ) is either connected or

Introduction

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it has two connected components. In the first case, i.e., if σ is non-separating, then the orbit space Xσ = S/σ is non-orientable, while in the separating case Xσ is orientable. An expository account of the functorial correspondence between real algebraic curves and Klein surfaces can be found in [45], see also the condensed versions [103, 104] by Natanzon. As we shall see throughout this monograph, a fundamental component to approach the problems mentioned above is the knowledge of the full automorphism group Aut(S) of the analytic and antianalytic automorphisms of S and its subgroup Aut+ (S) consisting of the analytic ones. Moreover, to determine the topology of a given symmetry σ of S, the centralizer C(Aut(S), σ) of σ in Aut(S) plays a fundamental role. Although automorphism groups do not constitute the core of this work, we will need them very frequently. It is worth mentioning that the factor group C(Aut(S), σ)/σ is isomorphic to the group of automorphisms of the Klein surface S/σ . There is a vast literature concerning groups of automorphisms of such surfaces. Among them we should mention [22], [58], [79]–[90], [93], the pioneering papers [114] and [40] and the exceptionally complete work [105]. We now describe briefly the content of this monograph. We also quote the contributions of different authors to the development of the employed techniques and related topics. Although in Chap. 4 we study the symmetries of the sphere and the tori, we will mainly be concerned with compact Riemann surfaces of genus bigger than one. By the Uniformization Theorem, such a surface S can be presented as the orbit space of the hyperbolic plane H under the action of a surface Fuchsian group Γ. Moreover, using covering theory, it can be proved that each automorphism group of S = H/Γ is a factor group Λ/Γ, where Λ is a non-euclidean crystallographic (NEC in short) group containing Γ as a normal subgroup. The key point now is that the algebraic structure of both Fuchsian and NEC groups is well known and this is why we devote Sect. 1.1 to the presentation of some basic facts about these groups. The above shows that in order to move ahead with the combinatorial approach of the study of symmetries of Riemann surfaces it is essential to understand the relation between the presentations of two NEC groups Γ and Λ, where the first is a normal subgroup of the second one. This task is mainly due to E. Bujalance, who developed in a series of papers [10–12] at the beginning of the eighties, an efficient method to solve this problem based on surgery of fundamental regions. It is also worth mentioning the article by J. A. Bujalance [27] concerning this problem. These results appear, without proofs, in Sect. 1.2. One of the main elements in the combinatorial approach to the study of symmetries of compact Riemann surfaces is the analysis of the centralizers of hyperbolic reflections in NEC groups. Singerman found in his Ph. D. Thesis [115], see also [119], the isomorphism type of centralizers of reflections in NEC groups. Going a bit more into the details of Singerman’s proof, explicit generators of these groups can be obtained, see the papers [48, 51] by G. Gromadzki. We present them in Sect. 1.3.

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Section 1.4 concerns uniformization of compact Riemann and Klein surfaces by means of Fuchsian and NEC groups, respectively, and its consequences. We pay special attention to the explanation of the notions of maximal Fuchsian or NEC groups and maximal signatures, and the relation between them. Although we have not included proper proofs of the results we will be using throughout the monograph, for which the reader is referred to [22, Chap. 5], we present carefully the main concepts. To finish this preliminary chapter, we explain the basics about symmetries in Sect. 1.5. We recall the notions of topological type and species of a symmetry and the classical Harnack-Weichold necessary and sufficient conditions for a given triple to be the topological type of some symmetry. We also approach the problem of deciding whether a Riemann surface is symmetric. This depends, in general, on its analytic type. However, there is an exception, pointed out by Singerman, who showed in [118] that if the group Aut+ (S) of analytic automorphisms of S is large enough then the symmetrical character of S depends only on the group Aut+ (S). Moreover, Singerman obtained a necessary and sufficient condition for the surface S to be symmetric and here we provide a slightly different proof of his criterion. Surfaces S with large analytic automorphism group Aut+ (S) are rather special and, perhaps, the most interesting ones. In particular they are Belyi surfaces since Aut+ (S) can be uniformized by a triangle Fuchsian group. This implies, by Belyi’s Theorem, see [7], that S can be defined by polynomial equations whose coefficients are algebraic numbers. Furthermore, by the recent results of Köck and Singerman [66] and Köck and Lau [67] on symmetric Riemann surfaces with large group of automorphisms, these algebraic numbers can be chosen to be real. Chapter 2 is devoted to quantitative aspects of the theory; we deal with the problem of finding the number of conjugacy classes of symmetries of Riemann surfaces. The study of symmetries that fix points comes back to the seminal work of Natanzon [95] who proved, using deep topological methods, that a Riemann sur√ face of genus g has at most 2( g + 1) non-conjugate symmetries that fix points. Moreover, he showed that this upper bound is attained for each value g of the form g = (2n−1 − 1)2 . Later on, Bujalance, Gromadzki and Singerman proved in [24] that these are the only values of g for which Natanzon’s bound is sharp. Moreover, if the bound is attained then all the symmetries are non-separating. In the same article the authors found an upper bound for the number of conjugacy classes of separating symmetries of a surface of genus g. At a first sight this bound seems to be a strictly increasing function of the genus, but later on it was discovered that this is so only up to some extent. Indeed, Gromadzki and Izquierdo proved in [53] that a Riemann surface of even genus has at most four non-conjugate symmetries that fix points. This result was extended to surfaces of odd genus by Bujalance, Gromadzki and Izquierdo in [23]. In that paper, and for each odd genus, the authors found sharp upper bounds for the number of such symmetries. We reprove these results in Sect. 2.2 of this chapter. The search of an upper bound for the number of conjugacy classes of fixed point free symmetries is much more involved. In Sect. 2.3 we provide an upper bound

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valid for those surfaces which have no symmetry with fixed points. The bound, which depends only on the 2-adic part of g − 1, was obtained originally in [18] and it was shown to be attained for infinitely many values of g. Finally, in Sect. 2.4 we obtain an upper bound for the number of conjugacy classes of symmetries of a genus g surface allowing both fixed point free symmetries and symmetries with ovals. Once more it turns out that this bound depends only on the 2-adic part of g − 1. Chapter 3 deals with several enumerations of ovals of the symmetries of a Riemann surface. Section 3.1 is crucial for the rest of the monograph; its main result allows us to find the number of ovals of a symmetry of a Riemann surface S from the algebraic structure of the full automorphism group Aut(S) and from the topological type of the action of Aut(S) on S. It was originally established in [49]. As we mentioned, a Riemann surface of even genus has at most four non-conjugate symmetries and, as an application of the result just quoted, Gromadzki and Izquierdo found in [54] the maximal total number of ovals of such extremal configuration of symmetries. The problem of finding the maximal number of ovals of a fixed number k of nonconjugate symmetries of a Riemann surface of genus g has been investigated by many authors throughout the years. However, it has been solved in its full generality just recently [56]. The first results, concerning low values of k, were obtained by Natanzon in [96, 100, 105], where he showed that an upper bound for such number is 2g + 2k−1 for k = 2, 3, 4 and characterized the pairs (g, k) for which this bound is attained. Later on, Singerman in [121] showed that for each non-negative integer k there exist infinitely many values of g for which there exists a Riemann surface of genus g admitting k non-conjugate symmetries having 2g − 2 + 2k−3(9 − k) ovals in total. In his work, Singerman also conjectured that this is in fact the best possible upper bound. This was shown by Gromadzki in [50] to be false for k > 9 by showing that, for k ≥ 9, the maximal possible number of ovals is 2g − 2 + 2r−3 (9 − k), where r is the smallest positive integer for which k ≤ 2r−1 . Moreover, this bound is attained, for arbitrary k ≥ 9, for infinitely many values of g. Later on Natanzon proved in [107] that Singerman’s conjecture is true under the additional assumption that the symmetries are separating. The presentation of these results is the main goal of Section 3.2. It is worth mentioning that Singerman’s conjecture was found to be true for k = 9 in [50] and it was conjectured to be also true for k in range 5 ≤ k ≤ 8. This has recently been answered in the affirmative by Gromadzki and Kozłowska-Walania in [56]. Section 3.3 concerns the total number of ovals of all symmetries of a Riemann surface. Recall that a simple closed curve on a Riemann surface S is said to be an oval of S if it is an oval of some symmetry of S. Let S be the number of ovals of S and let ν(g) be the maximum of S where S runs over all Riemann surfaces of genus g. Using topological methods, Natanzon proved in [105] that ν(g) ≤ 42(g−1), and Gromadzki improved this bound in [49] by using combinatorial methods. We present the complete proofs of these results in this section.

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Finally, Sect. 3.4 is devoted to the study of pairs of symmetries of Riemann surfaces. A lot of work has been done in this topic, and we include in this section some of the most relevant results. Once more the first and fundamental steps in this kind of questions are due to Natanzon, who classified topologically in [102] pairs of commuting symmetries. Natanzon in [105] and later on Bujalance, Costa and Singerman in [21], found an upper bound for the total number of ovals of two symmetries in terms of the genus of the surface and the order of their product. A finer bound, which involves the number of points fixed by the product of these symmetries, has been obtained by Gromadzki and Kozłowska-Walania in [55]. On the other hand, it was proved in [21] that two symmetries σ1 and σ2 of a genus g Riemann surface S having k1 and k2 ovals, where k1 + k2 ≥ g + 3, always commute. In a recently published paper by Kozłowska-Walania [69], this bound has been proved to be optimal to guarantee the commutativity of each pair of symmetries of S, with one exception in each genus g > 2. Another interesting result concerning pairs of symmetries was obtained by Bujalance and Costa, who calculated in [19] upper bounds for the degree of hyperellipticity of the product of two commuting symmetries. These upper bounds vary according to the separating character of the symmetries and they depend just on the numbers of their ovals. A nice improvement has been published by KozłowskaWalania in [68], where the upper bounds for the degree of hyperellipticity are substituted by its precise values. Izquierdo and Singerman showed in [63] that the existence of a symmetry whose number of ovals is extremal, that is, either 0 or g + 1 where g is the genus of the surface, imposes restrictions on the number of ovals of any other symmetry of the same surface. They also found extra restrictions if the separating character of the symmetries is considered. Later on, Costa and Izquierdo [34] showed that for every admissible triple (g, k, ε) there exists a genus g surface admitting symmetries σ and τ with topological types (g, k, ε) and (g, 1, 1), respectively. This result has a deep consequence: the locus of symmetric Riemann surfaces of fixed genus g ≥ 2 is a connected subspace of the moduli space Mg of Riemann surfaces of genus g. Of course this result is not new, but what is new is its proof. Klein conjectured it and Seppälä provided a modern and complete proof in [111] by using strong deformation of curves. The study of the number of connected components of distinguished subspaces of Mg is a recurrent theme in algebraic geometry. In fact the connectedness of the most important subspaces is rather exceptional, as it was shown, for example, by Buser, Seppälä and Silhol in [28]. In this article the authors study the subset of the moduli space of stable curves of genus bigger than one consisting of curves admitting a given finite group as a group of analytic automorphisms. They prove that this subset is always compact, is not connected in general, and it is connected for the group of order 2. In the same vein, it is worth mentioning that in the already quoted paper [34], Costa and Izquierdo proved the disconnectedness of the subspace of p-gonal Riemann surfaces of genus g for fixed values of p and g. This extends an earlier theorem by Gross and Harris [59] only valid for p = 3.

Introduction

xvii

The class of p-gonal surfaces has attracted the interest of many authors. In what concerns symmetries, we quote here the result of Costa and Izquierdo in [35] where they study the symmetries of cyclic p-gonal Riemann surfaces by means of Fuchsian and NEC groups. To finish, it is worth mentioning the paper by Bujalance, Costa and Gromadzki [20], where the behaviour of symmetries with maximal number of ovals under non-ramified coverings is studied. Chapter 4 is devoted to the presentation of classical selected examples. To begin with, we study the Riemann sphere Σ in Sect. 4.1. It is elementary to show that the maps σ1 : z → z¯ and σ2 : z → −1/¯ z are symmetries of Σ and that they are the only ones, up to analytic conjugation. Section 4.2 is devoted to classify the symmetries of the tori, for which we follow closely the approach by Alling [3]. Each torus is presented as the orbit space C/L for a suitably arranged lattice L. The symmetrical character of the torus and the topological type of its symmetries are expressed in terms of the lattice L. As it is classical, the analysis requires the cases of square or hexagonal lattices to be treated separately. In Section 4.3 we explain how the complete classification of the symmetries of hyperelliptic Riemann surfaces was obtained by the authors of this monograph in their previous work [14]. This work is too extensive even to be completely summarized here, but we explain an example in detail, showing how both the combinatorial approach and the use of algebraic equations, combined with a topological method, are fruitful in this case. In Chap. 5 we deal with symmetries of surfaces S whose group of analytic automorphisms Aut+ (S) is large enough. Following [52], we call these symmetries Singerman symmetries. As mentioned above, the symmetrical character of such surfaces depends only on Aut+ (S). In Sect. 5.1 we give formulae for the number of ovals of the symmetries of such surfaces in terms of the orders of the isotropy groups of some automorphisms acting on Aut+ (S), and the orders of some distinguished elements in Aut+ (S). These results constitute a fundamental component in the development of the next sections of this chapter. The understanding of the symmetries of the so called Macbeath-Singerman surfaces is the goal of Sect. 5.2. These are genus g surfaces admitting the projective special linear group PSL(2, q), where q is a prime power, as its group of analytic automorphisms of the maximal order 84(g − 1). Klein [65] was the first to discover the existence of such surfaces, as he showed that the group of analytic automorphisms of the genus 3 surface S = {[x : y : z] ∈ P2 (C) : x3 y + y 3 z + z 3 x = 0}, known as the Klein quartic, is the projective special group PSL(2, 7) of order 168. Macbeath [72] proved much later the existence of a unique Riemann surface of genus 7 on which the group PSL(2, 8) of order 504 acts as its full group of analytic automorphisms. Following ideas of Singerman from [118], we show that all Macbeath-Singerman surfaces are symmetric. We also determine the number of symmetries they admit, which we call Macbeath-Singerman symmetries, and the topological type of each

xviii

Introduction

of them. Remarkably, all of them are non-separating. These results were proved for the first time by Broughton, Bujalance, Costa, Gamboa and Gromadzki in [8]. The proof we present here is quite different and relies heavily on the results of the previous section of this chapter. In the 1960’s, Accola [1] and Maclachlan [77] proved, independently, that for every integer g ≥ 2 there is a compact Riemann surface Xg of genus g whose automorphism group has order 8g + 8. It is called the Accola-Maclachlan surface and it is defined by the polynomial equation y 2 = x2g+2 −1. The result is interesting as 8g + 8 is the largest order of an automorphism group that can be attained for every genus g. Much later, Kulkarni [70] considered the question of uniqueness of the surfaces attaining this bound. It turns out that the Accola-Maclachlan surface Xg is the unique one if g ≡ 0, 1, 2 (mod 4) and g sufficiently large. However, for large enough g ≡ 3 (mod 4), Kulkarni also proved that, in addition to Xg , there exists exactly one other surface, called Kulkarni surface, of genus g whose automorphism group also has order 8g + 8. In Sect. 5.3 we show that these surfaces are symmetric and, moreover, we determine the number of conjugacy classes of symmetries they admit and the topological type of each of them. As in the example of Sect. 5.2, the proof proposed here relies on the results in Sect. 5.1 and it is quite different from the original one which appeared in [9]. It must be pointed out that the examples selected to this chapter are in some sense exceptional because it has been possible to decide successfully the separating character of each symmetry. But, of course, they are not the only ones. In their paper [2], Akbas and Singerman not only calculated the number of ovals of the symmetries of the modular surfaces X0 (N ) = H/Γ0 (N ), but also showed that they are separating for N = 2, 3, 5, 7, 13 and non-separating for all other primes N . The situation is slightly worse for the symmetries of the modular surfaces X(N ) = H/Γ(N ). All of them are non-separating in case N ≡ 3 (mod 4) is prime but, as far as we know, there is no general answer for primes N ≡ 1 (mod 4). Another interesting example, that we do not explain in the monograph, is due to Tyszkowska [126], who obtained sharp upper bounds for the number of ovals of the symmetries of the Belyi surfaces admitting PSL(2, p) as its group of automorphisms. Section 5.4 is devoted to finding polynomial equations of the sets of points fixed by the symmetries of families of Riemann surfaces studied in the precedent ones. The key point is the Galois theory of finite coverings, as explained to the authors by P. Turbek. In fact Turbek is responsible for the original finding of equations of the symmetries of the Accola-Maclachlan surfaces occurring in [9], but in this monograph we have chosen a more geometrical approach. However, the presentation of the part of this section concerning defining equations of the sets of points fixed by the symmetries of the Kulkarni surfaces follows closely Turbek’s article [124]. It is convenient to explain a little bit the method employed. We begin with a plane model of our Riemann surface S, possibly with singularities, defined as the zero set in C2 of a polynomial P ∈ C[X, Y ]. A symmetry σ of S can be seen as an involution of the quotient field EP of the coordinate ring C[X, Y ]/(P ) of S. We look for a different polynomial Q ∈ R[X, Y ] which also defines S. Then the quotient fields

Introduction

xix

EP and EQ are isomorphic via a birational isomorphism, say ϕ : EP → EQ . With respect to these new coordinates, the symmetry σ = ϕ−1 ◦ σ ◦ ϕ acts as complex conjugation: σ (i) = −i ; σ (X) = X ; σ (Y ) = Y, √ where i = −1. Observe that the fixed points of σ are the real solutions of the equation Q(X, Y ) = 0. We finish this work about symmetries with a few words about asymmetric surfaces, that is, surfaces admitting no symmetry. Such surfaces have recently played an important role in the study of deformations and moduli of complex surfaces, as in the paper [29] by Catanese, where the author finds a counterexample to a conjecture of Friedman and Morgan relating diffeomorphisms and deformations of such complex surfaces. Let Mg be the moduli space of complex isomorphism classes of complex algebraic curves of genus g ≥ 2. Since Mg is a quasiprojective variety defined in some projective space Pn (C) by means of polynomials with real (in fact rational) coefficients, complex conjugation induces an anticonformal involution σg∗ : Mg → Mg . Let MR g be the complex moduli space of real algebraic curves of genus g, which consists of complex isomorphism classes of complex algebraic curves that are defined by real polynomials. It is clear that the set Fix(σg∗ ) of points fixed by σg∗ contains R ∗ MR g but, as observed by Clifford Earle in [40], the inclusion Mg ⊂ Fix(σg ) is proper. The asymmetric curves are precisely those whose isomorphism classes occur in the difference Fix(σg∗ ) \ MR alä showed in [110] that every asymmetric g . Sepp¨ curve is in fact a covering of a real algebraic curve. It is classical that for any integer g > 2 there exists a compact Riemann surface of genus g whose group of analytic automorphisms is trivial. Indeed, Greenberg proved in [47] that outside a proper analytic subset of the Teichmüller space, all compact Riemann surfaces of genus g ≥ 3 have the identity as its only analytic automorphism. However, it is not easy to construct examples of such surfaces. It is worth mentioning the paper by Mednykh [91] who constructed, for each pair of integers (p, r), where p > 3 is prime and r ≥ 2p, a fundamental region of a Fuchsian group which uniformizes a compact Riemann surface of genus g = (p− 1)(r − 1)/2 with trivial automorphism group. Later on, Everitt in [42] found new examples for all g > 2, using Schreier coset graphs for subgroups of triangle groups. Combining covering theory with Galois theory of algebraic function fields in one variable, Turbek [123, 125] provided defining equations of compact Riemann surfaces with trivial group of analytic automorphisms. In the same vein, Earle in [40] was the first to find examples of pseudo-real Riemann surfaces, that is, surfaces without symmetries but with orientation reversing automorphisms. Later on, Bujalance and Turbek constructed in [26] algebraic equations of the elements of an infinite family of pseudo-real Riemann surfaces. The construction we present in Chap. 6 is a particular case of the one in [26]. More recently, Bujalance, Conder and Costa in [17] have shown that there exist pseudo-real Riemann surfaces of genus g for each g ≥ 2 and, furthermore, that the

xx

Introduction

maximum number of automorphisms of such a surface is 12(g − 1). This bound turns out to be sharp for infinitely many values of g. Another instance of pseudo-real surfaces occurs in [13], where Riemann surfaces of even genus g with an orientation reversing automorphism of order 2g are studied. These surfaces constitute a family of real dimension three and “most” (but not all) of them are asymmetric. In fact, a defining algebraic equation depending on three real parameters can be given for each such surface and it turns out that those which are symmetric depend just on two parameters.

Chapter 1

Preliminaries

In this chapter we present some preliminary results concerning, mostly, the representation of a compact Riemann S as the orbit space of the hyperbolic plane H under the action of a surface Fuchsian group Γ. This is a consequence of the Uniformization Theorem for Riemann surfaces. The main point here is that if S = H/Γ is symmetric then its full automorphism group Aut(S) is a factor group Λ/Γ where Λ is a proper NEC group containing Γ as a normal subgroup. We will recall the algebraic structure of NEC groups, which is codified by their signatures, and emphasize the relation between the signatures of a pair of NEC groups Γ and Λ, where the first one is a normal subgroup of the second one. Special attention will be paid to maximal signatures and maximal NEC groups, since they are closely related to full automorphism groups of surfaces. Of course, this chapter also contains the basics about the main protagonists of this monograph, namely, symmetric surfaces and their symmetries.

1.1 NEC Groups and Their Signatures The Uniformization Theorem for Riemann surfaces says that every compact Riemann surface S of genus bigger than one is the orbit space of the hyperbolic plane H under the action of a certain subgroup of the group Aut+ (H) of analytic selfhomeomorphisms of H. Since H is simply connected, every analytic automorphism of S can be lifted to an analytic self-homeomorphism of H. Analogously, every orientation reversing automorphism of S (a symmetry, for instance) can be lifted to an antianalytic self-homeomorphism of H. We start this section with a description of the group Aut(H) of analytic and antianalytic self-homeomorphisms of H. As a trivial consequence of the maximal modulus principle, it follows that Aut(H) is the following disjoint union: az + b f :z→ with {a, b, c, d} ⊂ R and ad − bc > 0 ∪ cz + d a¯ z+b ∪ f :z→ with {a, b, c, d} ⊂ R and ad − bc < 0 . c¯ z+d

Aut(H) =

E. Bujalance et al., Symmetries of Compact Riemann Surfaces, Lecture Notes in Mathematics 2007, DOI 10.1007/978-3-642-14828-6 1, c Springer-Verlag Berlin Heidelberg 2010

1

2

1 Preliminaries

The elements of Aut(H) are called automorphisms of H. The first set of this union is the subgroup Aut+ (H) of analytic automorphisms of H. It consists of the orientation preserving hyperbolic isometries of H, while the orientation reversing ones are those in the second set. Let GL(2, R) be the group of 2 × 2 non-singular matrices with real entries. It is clear that the mapping GL(2, R) → Aut(H) ; A = where fA : H → C ; z →

ab cd

→ fA

⎧ az + b ⎪ ⎪ ⎪ cz + d if det A > 0, ⎨ ⎪ ⎪ z+b ⎪ ⎩ a¯ if det A < 0, c¯ z+d

is a group epimorphism. Its kernel {λI2 : λ ∈ R∗ }, where I2 is the identity matrix, is the center C(GL(2, R)) of GL(2, R). Therefore we identify Aut(H) with the factor group PGL(2, R) = GL(2, R)/C(GL(2, R)). Consequently, Aut(H) is a topological group and it makes sense to talk about its discrete subgroups. Definitions 1.1.1. Let Γ be a subgroup of Aut(H). (1) We say that Γ is a non-euclidean crystallographic group (shortly, NEC group) if it is a discrete subgroup and the orbit space H/Γ is compact. (2) An NEC group Γ is said to be a Fuchsian group if it is contained in Aut+ (H). Otherwise Γ is said to be a proper NEC group. (3) Given a proper NEC group Γ, its subgroup Γ+ = Γ ∩ Aut+ (H), consisting of its orientation preserving elements, is called its canonical Fuchsian subgroup. Obviously, [Γ : Γ+ ] = 2, and Γ+ is the unique subgroup of index 2 in Γ contained in Aut+ (H). If Γ is an NEC group then the orbit space H/Γ can be endowed with a dianalytic structure, see [4, Theorem 1.8.4]. A fundamental region for Γ can be constructed as a convex bounded hyperbolic polygon with a finite number of sides. A modification of the region and a suitable labelling of the edges provides the following canonical surface symbol: (+)

α1 β1 α1 β1 . . . αg βg αg βg ξ1 ξ1 . . . ξr ξr ε1 γ10 . . . γ1s1 ε1 . . . εk γk0 . . . γksk εk

if H/Γ is orientable, or (−)

α1 α∗1 . . . αg α∗g ξ1 ξ1 . . . ξr ξr ε1 γ10 . . . γ1s1 ε1 . . . εk γk0 . . . γksk εk

otherwise. A primed edge is paired to the corresponding unprimed edge by means of an orientation preserving automorphism while a starred edge is paired to the corresponding unstarred edge through an orientation reversing automorphism.

1.1 NEC Groups and Their Signatures

3

Taking into account the surface symbol, it is possible to obtain the following presentation of the group Γ: generators: – x1 , . . . , xr (elliptic elements), – c10 , . . . , c1s1 , . . . , ck0 , . . . , cksk (reflections), – e1 , . . . , ek (orientation preserving elements, usually hyperbolic and in some cases elliptic), – a1 , b1 , . . . , ag , bg (hyperbolic translations) in case (+), – d1 , . . . , dg (glide reflections) in case (−), and relations: – – – – –

i = 1 for i = 1, . . . , r, xm i cisi = e−1 i ci0 ei for i = 1, . . . , k, c2ij−1 = c2ij = (cij−1 cij )nij = 1 for i = 1, . . . , k and j = 1, . . . , si , −1 −1 −1 x1 · · · xr e1 · · · ek a1 b1 a−1 1 b1 · · · ag bg ag bg = 1 in case (+), 2 2 x1 · · · xr e1 · · · ek d1 · · · dg = 1 in case (−).

Throughout the monograph, a set of generators as the above one will be called a set of canonical generators of Γ. The first presentations for NEC groups appeared in [128] and their structure was clarified by the introduction of signatures in [73]. Definitions 1.1.2. Let g, k be non-negative integers and let mi for i = 1, . . . , r, and nij for i = 1, . . . , k and j = 1, . . . , si , be integers ≥ 2. (1) An abstract signature is a collection of symbols and non-negative integers of the form s = (g; ±; [m1 , . . . , mr ]; {(n11 , . . . , n1s1 ), . . . , (nk1 , . . . , nksk )}).

(1.1)

The non-negative integer g is called the orbit genus of s. If the sign “+” appears then we write sign(s) = “+”; otherwise sign(s) = “−”. The integers m1 , . . . , mr are called the proper periods of s and the nij are called the link periods of the period cycle (ni1 , . . . , nisi ). An empty set of proper periods, (i.e., r = 0), will be denoted by [−], an empty period cycle (i.e., si = 0) by (−), and the fact that s has no period cycles (i.e., k = 0) by {−}. (2) Given an NEC group Γ with the above presentation, the signature (1.1) is defined as its signature s(Γ), the sign of s(Γ) being “+” in the case (+) and “−” otherwise. The orbit genus of s(Γ) is usually called the orbit genus of Γ. (3) Since a Fuchsian group contains no orientation reversing elements, its signature has no period cycles and its sign is always “+”. Hence we may drop such data and in the sequel the signature of a Fuchsian group will be represented simply by (g; m1 , . . . , mr ). Signatures of the form (0; k, , m), that we abbreviate as [k, , m], are called triangle Fuchsian signatures. Fuchsian groups with such signatures are called triangle Fuchsian groups.

4

1 Preliminaries

(4) A signature of the form s = (0; +; [−]; {(n1 , . . . , ns )}) will be abbreviated as (n1 , . . . , ns ), when no confusion can arise. If s = 3 then s is called triangle NEC signature. NEC groups with these signatures are called triangle NEC groups. In the obvious manner, a presentation of an NEC group Γ can be read from its signature. In fact, signatures give a procedure to classify NEC groups up to isomorphism, as it was proved by Macbeath [73] and Wilkie [128]. Proposition 1.1.3. Let Γ be an NEC group with signature s = s(Γ) as in (1.1). Let Γ be another NEC group with signature s = s(Γ ) = (g ; ±; [m1 , . . . , mr ]; {(n11 , . . . , n1s1 ), . . . , (nk 1 , . . . , nk s )}). k

Let us write Ci = (ni1 , . . . , nisi ) and Ci = (ni1 , . . . , nis ). Then Γ and Γ are i isomorphic as abstract groups if and only if (1) (2) (3) (4)

sign(s) = sign(s ), g = g , r = r , k = k and si = si for i = 1, . . . , k, mi = mϕ(i) for a permutation ϕ of {1, . . . , r}, if sign(s) = “+” then there exists a permutation φ of {1, . . . , k} such that one of the following conditions holds true: (4a) Ci is a cyclic permutation of Cφ(i) for each i ∈ {1, . . . , k} or (4b) Ci is a cyclic permutation of the inverse of Cφ(i) for each i ∈ {1, . . . , k},

(5) if sign(s) = “−” then there exists a permutation φ of {1, . . . , k} such that Ci is a cyclic permutation of either Cφ(i) or of the inverse of Cφ(i) , for each i ∈ {1, . . . , k}. If the sign is “+” then the corresponding period cycles are all paired in the same way (either all directly or all inversely). If the sign is “−” then some period cycles may be paired directly and some inversely. In addition to this algebraic information, the signature of an NEC group Γ also provides topological information about the canonical projection H → H/Γ, see Proposition 1.1.4 below. Let Γ be an NEC group and consider the canonical projection f : H → H/Γ. At each point p ∈ H, the map f behaves locally as z → z m (see [4, Sect. 5] for a more rigorous statement). The integer m is the ramification index of f at p; we say that f is ramified at p if m > 1. The ramified points of f are precisely those fixed by orientation preserving elements of Γ. The elliptic elements of Γ fixing p constitute a cyclic group, whose order is the ramification index of f at such a point. Moreover, it turns out that all the points in the same fiber as p have the same ramification index. We say that f (p) is a branch point of f with branching order m. The next result collects the topological interpretation of s(Γ), as proved by Wilkie in [128]. Proposition 1.1.4. Let Γ be an NEC group with signature (1.1) and let S = H/Γ. Then: (1) g is the topological genus of S.

1.1 NEC Groups and Their Signatures

5

(2) sign(s(Γ)) = “+” if and only if S is orientable. (3) The integers m1 , . . . , mr are the branching orders with respect to the canonical projection H → H/Γ of the r conic points lying in the interior of S. (4) The integer k is the number of connected components of the boundary of S. (5) The integers ni1 , . . . , nisi are the branching orders with respect to the canonical projection H → H/Γ of the si corner points lying on the i-th connected component of the boundary of S. For simplicity, g, k and “ ± ” are called the topological data of the projection H → H/Γ, whilst the integers mi and nij are its branching data. This proposition shows that the knowledge of the topological and branching data of the projection H → H/Γ is equivalent to that of the signature of Γ. The next lemma gives the relation between the elements of finite order in an NEC group (reflections and elliptic isometries) and a set of canonical generators. It is well known but, for the sake of completeness, we present a proof here. Lemma 1.1.5. Any reflection of an NEC group is conjugate to one of its canonical reflections. Any elliptic element is conjugate either to a power of some of its canonical elliptic generators or to a power of the product of two consecutive canonical reflections. Proof. Let be the axis of a reflection c in an NEC group Γ. Then there exists γ ∈ Γ for which γ () meets an edge of a given fundamental region F of Γ. Since γ () is the set of fixed points of γ cγ −1 , the last is a canonical reflection, which proves the first part of the statement. Similarly, if p is the fixed point of an elliptic element x then γ (p) ∈ F for some γ ∈ Γ and, on the other hand, it is the fixed point of xγ = γ xγ −1 . Therefore γ (p) is a vertex of F and xγ is a power of some canonical elliptic generator of Γ or a power of the product of two consecutive canonical reflections.

Remark 1.1.6. It follows from the presentation of an NEC group that the number of conjugacy classes of reflections associated to a period cycle with v > 0 even link periods is v. If the period cycle is empty or all its link periods are odd then all reflections associated to such period cycle are pairwise conjugate. Definition 1.1.7. Let s be an abstract signature as given in (1.1) and define η = 2 if sign(s) = “+” and η = 1 otherwise. The area of s is defined to be ⎛

⎞ si r k 1 1 1 ⎠ Area(s) = 2π ⎝ηg + k − 2 + 1− + 1− . m 2 n i ij i=1 i=1 j=1 Part (1) in the following result justifies this definition, see [22]. Theorem 1.1.8. Let Γ be an NEC group with signature s(Γ). (1) The hyperbolic area of any fundamental region for Γ is Area(s(Γ)). It makes sense to call area of the NEC group Γ to such common value. We denote it by Area(Γ).

6

1 Preliminaries

(2) The signature s is the signature of some NEC group Γ if and only if Area(s) > 0 and sign(s) = “+” if g = 0. (3) If Γ is a subgroup of finite index of an NEC group Γ then Γ is also an NEC group and the so called Hurwitz–Riemann formula holds: Area(Γ ) = Area(Γ)[Γ : Γ ]. In what follows we shall say that an abstract signature s is an NEC (respectively Fuchsian) signature if it is the signature of an NEC (respectively Fuchsian) group.

1.2 Normal Subgroups of NEC Groups A fundamental tool in the combinatorial study of compact Riemann surfaces and their automorphisms is the relation between the signature of an NEC group and the signatures of its normal subgroups of finite index. In a series of papers [10–12] written at the beginning of the eighties, E. Bujalance developed a method to deal with this problem based on surgery of fundamental regions. It is also worth mentioning the article by J. A. Bujalance [27], where normal subgroups of even index are considered, and the papers by Hoare [60] and Singerman [116], who employed a rather different approach. We refer the interested reader to [22], where unified proofs together with the references to the original papers can be found. Throughout this section we shall assume that Γ and Λ are NEC groups with Γ a normal subgroup of Λ of index N. The signature of Λ will have the general form given in (1.1). Our goal is to determine the signature of Γ, which clearly depends on how Γ is embedded in Λ. We will describe the relations between the signs, the sets of proper periods and the sets of period cycles of s(Λ) and s(Γ). Once we know them, the relation between the orbit genera of both groups is a straightforward consequence of the Hurwitz–Riemann formula. To deal with the problem of finding the links between the signs we need to introduce some notions. A canonical generator of Λ not belonging to Γ is said to be a proper generator of Λ with respect to Γ. An element of Λ which is expressable as a composition of proper generators of Λ is said to be a word of Λ with respect to Γ. Finally, such a word is said to be orientable if it represents an orientation preserving element and non-orientable otherwise. With these definitions at hand we can state the following result. Theorem 1.2.1. With the above notations, we have: (1) If N is odd then the signs of s(Λ) and s(Γ) coincide. (2) If N is even then s(Γ) has sign “ − ” if and only if Γ contains either a glide reflection of the canonical generators of Λ or a non-orientable word in Λ. Concerning the sets of proper periods the following theorem holds.

1.2 Normal Subgroups of NEC Groups

7

Theorem 1.2.2. Let {(cij−1 , cij ) : i ∈ I, j ∈ Ji } be the set of all pairs of consecutive canonical reflections in Λ \ Γ. Let let qij be the order of the image in Λ/Γ of the product cij−1 cij , for i ∈ I and j ∈ Ji , and let p denote the order of the image in Λ/Γ of the canonical elliptic generator x of Λ corresponding to the proper period m . Then the proper periods of s(Γ) are the following: . . . , m /p , nij /qij , .N/2q . . . .ij., nij /qij : = 1, . . . , r, i ∈ I, j ∈ Ji ]. [m /p , N/p Finally we shall present results concerning the sets of period cycles. If N is odd then all reflections of Λ belong to Γ, which makes things much easier. Theorem 1.2.3. Let N be odd. Then each period cycle C = (n1 , . . . , ns ) of s(Λ) induces N/ period cycles of s(Γ), all of them having the form (n1 , . . . , ns , . . ., n1 , . . . , ns ), where is the order in Λ/Γ of the image of the canonical hyperbolic generator of Λ corresponding to the period cycle C. The case of even N is more involved and we divide it into two subcases. Theorem 1.2.4. Let N be even, let C = (n1 , . . . , ns ) be one of the period cycles of s(Λ) and let be defined as in the previous theorem. Assume that all canonical reflections corresponding to C belong to Γ. Then C produces N/ period cycles in s(Γ) of the form (n1 , . . . , ns , . . ., n1 , . . . , ns ) if the sign of s(Λ) is “ − ”. If the sign is “+” then there exists a non-negative integer N1 ≤ N/ such that C produces N1 period cycles in s(Γ) of the above form and N/ − N1 period cycles in s(Γ) of the form (ns , . . . , n1 , . . ., ns , . . . , n1 ). The last result deals with the remaining case, that is, not all reflections of a given period cycle of s(Λ) belong to Γ. Theorem 1.2.5. Let N be even, let C = (n1 , . . . , ns ) be a non-empty period cycle of s(Λ) and let {e, c0 , . . . , cs } be the set of canonical generators of Λ corresponding to C. Assume that the set J = {(i, j) ∈ {1, . . . , s} × {0, . . . , s − 1} : i ≤ j, ci−1 , cj+1 ∈ Γ, ci , . . . , cj ∈ Γ} is not empty. Denote by n(i, j) the order of the image of ci−1 cj+1 in Λ/Γ. Then, for each pair (i, j) ∈ J, the numbers ni and nj are even and s(Γ) has N/2n(i, j) period cycles, each of them consisting of n(i, j) copies of the periods nj+1 /2, nj , nj−1 , . . . , ni+1 , ni /2, ni+1 , . . . , nj , where the quotients nj+1 /2 and ni /2 are omitted if they are equal to 1.

8

1 Preliminaries

Remark 1.2.6. The proper periods and period cycles of s(Γ) are exactly those induced by the ones of s(Λ) and described in the above theorems. There are no more proper periods or period cycles in s(Γ). An immediate consequence of the above results is the following corollary, originally proved by Singerman [119]. We provide a different proof which, however, is not a consequence of the above results. The corollary is essential in the study of symmetric Riemann surfaces since, up to certain extent, it describes how the symmetry type of a Riemann surface S (see Definition 1.5.5) depends just on the topological data of the action on S of the group Aut+ (S) of all analytic automorphisms of S, see Sect. 1.5.1. Corollary 1.2.7. If s(Γ) = (g; ±; [m1 , . . . , mr ]; {(n11 , . . . , n1s1 ), . . . , (nk1 , . . . , nksk )}) is the signature of the NEC group Γ then the signature of its canonical Fuchsian subgroup Γ+ is s(Γ+ ) = (ηg + k − 1; m1 , m1 , . . . , mr , mr , n11 , . . . , n1s1 , . . . , nk1 , . . . , nksk ), where η = 2 if sign(s(Γ)) = “+” and η = 1 otherwise. Proof. We have to look for conjugacy classes in Γ+ of elliptic elements of Γ. Let γ ∈ Γ \ Γ+ . Each canonical elliptic element xi of Γ and its conjugate xγi belong to Γ+ and they are not conjugate there. Indeed, let p be the fixed point of xi γ + −1 and assume that xα γ and therei = xi for some α ∈ Γ . Then p is fixed by α fore α−1 γ = xm for some m ∈ Z, a contradiction. Furthermore, for each δ ∈ Γ, i the element xδi is conjugate in Γ+ either to xi or to xγi , according to whether δ belongs to Γ+ or not. Moreover, cij−1 cij ∈ Γ+ and for any γ ∈ Γ \ Γ+ we have (cij−1 cij )γ = (cij cij−1 )γcij−1 . Finally, the orbit genus of Γ+ can be calculated using the Hurwitz–Riemann formula. So the result follows.

Remark 1.2.8. Corollary 1.2.7 also follows directly from a more general result of Singerman [116] about the relation between the signatures of two Fuchsian groups Λ1 and Λ2 , where Λ1 is a subgroup of Λ2 but not necessarily a normal one.

1.3 Centralizers of Reflections In order to classify topologically a symmetry σ of a compact Riemann surface we will have to count the number of connected components of the set of points fixed by σ. To that end, the description of centralizers of reflections in an NEC group plays a crucial role, see Theorem 3.1.1. Singerman found in his Ph. D. Thesis [115] the isomorphism type of centralizers of reflections in NEC groups. In this section we show how to find explicit generators for these centralizers.

1.3 Centralizers of Reflections

9

Lemma 1.3.1. Let c0 , . . . , cs , e be a set of canonical generators corresponding to a period cycle (n1 , . . . , ns ) of an NEC group Λ. If all ni are even then the centralizer C(Λ, ci ) of ci in Λ equals (1) ci ⊕ (ci−1 ci )ni /2 ∗ (ci ci+1 )ni+1 /2 = Z2 ⊕ (Z2 ∗ Z2 ) if s = 0, i = 0, (2) c0 ⊕ (ecs−1 e−1 c0 )ns /2 ∗ (c0 c1 )n1 /2 = Z2 ⊕ (Z2 ∗ Z2 ) if s = 0, i = 0, if s = 0. (3) c0 ⊕ e = Z2 ⊕ Z Proof. Observe that in each case the centralizer C(Λ, ci ) contains the group in the statement. We shall prove the converse inclusions. For i = 0, let γ i−1 , γ i , γ i+1 be the edges of a fundamental region F corresponding to ci−1 , ci , ci+1 . Let be the hyperbolic line containing γ i . Then for λ ∈ Λ, the product λci λ−1 is a reflection with axis λ(). So Λ centralizes ci if and only if λ() = , or equivalently, if and only if λ(F) is adjacent to . The element (ci−1 ci )ni /2 (ci ci+1 )ni+1 /2 is the composition of two half-turns with respect to the ends of γ i and so it is a hyperbolic isometry with axis and whose translation length equals twice the hyperbolic length of γ i . Now, composing some power of (ci−1 ci )ni /2 (ci ci+1 )ni+1 /2 with ci (ci ci+1 )ni+1 /2 we produce an element λ ∈ Λ such that λ(F) is an arbitrary face adjacent to and lying on the same side as F with respect to . On the other hand, composing this last with ci we produce an element μ ∈ Λ such that μ(F) is an arbitrary face adjacent to and lying on the opposite side as F with respect to . The case i = 0 is similar; actually one must repeat the above arguments for the triple of reflections ecs−1 e−1 , c0 , c1 . For s = 0, the orientation preserving element e fixes . So F and e(F) have a common edge and both of them are adjacent to . Hence, for a suitable choice of k, ek (F) is an arbitrary face adjacent to and lying on the same side as F with respect to , while considering c0 ek (F) we obtain an arbitrary face lying on the other side as F with respect to . Therefore, when k runs over all integers and ε = 0 or 1, the product cε0 ek runs over all elements of the centralizer of c0 .

Using similar ideas we prove the following. Lemma 1.3.2. Let e, c0 , . . . , cs be a set of canonical generators corresponding to a period cycle (n1 , . . . , ns ) of an NEC group Λ and let C(Λ, ci ) be the centralizer in Λ of ci . (1) If s = 0 and all ni are odd then s−1 (ni+1 −1)/2 −1 . (ci+1 ci ) e C(Λ, c0 ) = c0 ⊕ i=0

(2) If ni , nj are even and ni+1 , . . . , nj−1 are odd, with i < j ≤ s, then C(Λ, ci ) = ci ⊕ (ci−1 ci )ni /2 ∗ x−1 (cj−1 cj )nj /2 x , where x = (cj−2 cj−1 )(nj−1 −1)/2 · · · (ci ci+1 )(ni+1 −1)/2 .

10

1 Preliminaries

(3) If ni , nj are even and ni+1 , . . . , ns , n1 , . . . , nj−1 are odd, with 1 ≤ j ≤ i, then C(Λ, ci ) = ci ⊕ ( (ci−1 ci )ni /2 ∗ x−1 (cj−1 cj )nj /2 x), where x=

j−1

(nj−t −1)/2 −1

(cj−1−t cj−t )

t=1

e

s−i−1

(cs−1−t cs−t )(ns−t −1)/2 .

t=0

Proof. The element β = e(cs−1 cs )(ns −1)/2 · · · (c1 c2 )(n2 −1)/2 (c0 c1 )(n1 −1)/2 belongs to C(Λ, c0 ), which gives one of the inclusions in (1). For the converse, let F be a fundamental region for Λ and let γ 0 , γ 1 , . . . , γ s be the part of the surface symbol for F corresponding to the period cycle (n1 , . . . , ns ). The axis of c0 splits into intervals being edges of the images of F abuting , each segment having a label from the surface symbol to which it belongs. Now for λ ∈ Λ, the conjugate λc0 λ−1 is a reflection with axis λ(). So Λ centralizes c0 if and only if λ() = , while the last is true if and only if λ(F) abuts on an edge labelled by γ 0 . Thus there is a bijective correspondence between segments of labelled by γ 0 and elements of C(Λ, c0 ). Finally, (ci ci+1 )(ni+1 −1)/2 ci (ci ci+1 )−(ni+1 −1)/2 = ci+1 and ecs e−1 = c0 . So the segment labelled by γ i is followed by γ i+1 for i = 0, . . . , s − 1 while γ s is followed by γ 0 . Therefore γ 0 , γ 1 , . . . , γ s , γ 0 are the labels of consecutive segments on and this labelling repeats on periodically. Hence c0 and β generate C(Λ, c0 ) indeed. Proofs of claims (2) and (3) are similar and we omit them.

Of particular interest in this monograph is the description of the centralizers of the canonical reflections of a triangle NEC group. For further reference, we display the corresponding result in Lemma 1.3.3. The fact that two reflections ci and cj are conjugate will be denoted by ci ∼ cj . Lemma 1.3.3. Let Λ be an NEC group with signature (k , , m ) and let c0 , c1 , c2 be a set of its canonical generators. Then, according to the parity of k , , m , the centralizers in Λ of the reflections c0 , c1 and c2 are the following: (1) For k = 2k, = 2 and m = 2m we have C(Λ, c0 ) = c0 ⊕ (c0 c1 )k ∗ (c0 c2 )m , C(Λ, c1 ) = c1 ⊕ (c0 c1 )k ∗ (c1 c2 ) , C(Λ, c2 ) = c2 ⊕ (c0 c2 )m ∗ (c1 c2 ) . (2) For k = 2k, = 2 and m = 2m + 1 we have c0 ∼ c2 and C(Λ, c0 ) = c0 ⊕ (c0 c1 )k ∗ (c2 c0 )m (c2 c1 ) (c0 c2 )m , C(Λ, c1 ) = c1 ⊕ (c0 c1 )k ∗ (c1 c2 ) .

1.4 Uniformization and Automorphism Groups of Riemann and Klein Surfaces

11

(3) For k = 2k, = 2 + 1 and m = 2m + 1 we have c0 ∼ c1 ∼ c2 and C(Λ, c0 ) = c0 ⊕ (c0 c1 )k ∗ (c2 c0 )m (c1 c2 ) (c1 c0 )k (c2 c1 ) (c0 c2 )m . (4) For k = 2k + 1, = 2 + 1 and m = 2m + 1 we have c0 ∼ c1 ∼ c2 and C(Λ, c0 ) = c0 ⊕ (c2 c0 )m (c1 c2 ) (c0 c1 )k .

1.4 Uniformization and Automorphism Groups of Riemann and Klein Surfaces A classical Riemann surface is a topological surface without boundary together with an analytic structure. This analytic structure makes the surface orientable. However, non-orientable surfaces or orientable surfaces with boundary may admit a dianalytic structure, which behaves in many aspects as the analytic structure of a classical Riemann surface. Roughly speaking, a dianalytic structure is given by an atlas whose transition functions are either analytic or antianalytic (a function is antianalytic if it is the composite of an analytic function with complex conjugation). The only way in which such a non-classical surface X arises is as the orbit space of a classical Riemann surface S under the action of an antianalytic involution. Such an orbit space is usually known as Klein surface. For the time being, this is the definition of Klein surface we shall use. The book [4] of Alling and Greenleaf and the extensive article [106] of Natanzon are excellent references for the basics on Klein surfaces. We define the genus of X as that of S. Although we shall study in detail the symmetries of the sphere and the tori, throughout this monograph, unless otherwise stated, the surfaces considered will be compact of genus ≥ 2. As said above, if Γ is an NEC group then the orbit space H/Γ admits a structure of compact Klein surface. The importance of NEC groups comes from the fact that, under certain restrictions, the converse is also true. To explain this we need the notion of surface NEC group. Definition 1.4.1. An NEC group Γ having signature s(Γ) = (g; ±; [−]; {(−), . k. ., (−)}), k ≥ 0 is said to be a surface group. If k > 0 then Γ is a bordered surface group. Equivalently, an NEC group Γ is a surface group if and only if it has no non-trivial orientation preserving element of finite order. It follows from the Uniformization Theorem that surface Fuchsian groups uniformize compact Riemann surfaces. Likewise, surface NEC groups uniformize compact Klein surfaces. In fact, let X = S/ σ be a compact Klein surface, where S is a compact Riemann surface and σ : S → S is an antianalytic involution. If S = H/Λ for some surface Fuchsian group Λ and c : H → H is a lifting of σ, then the group Γ = Λ, c generated by Λ and c is a surface NEC group such that X = H/Γ. This shows the following.

12

1 Preliminaries

Theorem 1.4.2. Let X be a compact Klein surface of genus ≥ 2. Then there exists a surface NEC group Γ such that X = H/Γ as Klein surfaces. Remark 1.4.3. (1) If X is the orbit space of the compact Riemann surface S under the action of the antianalytic involution σ and Γ is as above, then S = H/Γ+ as compact Riemann surfaces and σ = Γ/Γ+ , where Γ+ is the canonical Fuchsian subgroup of Γ. (2) If we write the Klein surfaces X and X as H/Γ and H/Γ respectively, then X and X are isomorphic if and only if Γ and Γ are conjugate in the group Aut(H). Theorem 1.4.2, together with the classification of NEC groups by means of signatures, opens the door to the combinatorial approach to the theory of Klein surfaces and their automorphism groups. We now summarize some general results concerning automorphisms of Klein surfaces due to May [82]. The full group of dianalytic automorphisms of the Klein surface X will be denoted by Aut(X). Theorem 1.4.4. Let Aut(X) be the full group of automorphisms of the Klein surface X = H/Γ and let N(Γ) be the normalizer of the surface NEC group Γ in Aut(H). Then, (1) N(Γ) is an NEC group. (2) Aut(X) N(Γ)/Γ. (3) A group G is a subgroup of Aut(X) if and only if it is isomorphic to a factor group Λ/Γ for some NEC group Λ containing Γ as a normal subgroup. Remark 1.4.5. If X is not a Klein surface but a Riemann surface S (in which case Γ is not a surface NEC group but a surface Fuchsian group) then the above holds true provided that Aut(S) stands for the full group of analytic and antianalytic automorphisms of S. Given an NEC group Λ, a factor group Λ/Γ, where Γ is a surface Fuchsian group, will be called a smooth factor. An epimorphism θ : Λ → G onto a finite group G, whose kernel is a surface Fuchsian group, will be called a smooth epimorphism. Observe that a necessary and sufficient condition for an epimorphism θ : Λ → G to be smooth is that θ preserves the orders of the elements of Λ of finite order. So an action of a finite group G on the Riemann surface H/Γ is defined by a smooth epimorphism θ : Λ → G where Λ is an NEC group and ker θ = Γ. Moreover, two such actions given by θ : Λ → G and θ : Λ → G are said to be topologically equivalent if there exists a commutative diagram Λ

ϕ

θ

θ

G

/ Λ

ψ

/ G

for some group isomorphisms ϕ : Λ → Λ and ψ : G → G .

1.4 Uniformization and Automorphism Groups of Riemann and Klein Surfaces

13

1.4.1 Maximal NEC Groups It is a rather difficult task to decide whether a given group of automorphisms of a Riemann surface is the full group of all its automorphisms. Signatures of NEC groups prove to be a useful tool in dealing with this problem. The key point in the solution is that almost all NEC signatures are maximal in some sense, and given such a maximal signature s there exists a maximal NEC group Λ with signature s. An NEC group is said to be maximal if it is not properly contained in another NEC group. In such a case, for every surface Fuchsian group Γ contained in Λ as a normal subgroup, the full automorphism group of the Riemann surface H/Γ is precisely Λ/Γ. Let us denote by dim(Λ) the dimension of the Teichmüller space of the NEC group Λ (see Sect. 1.4.2 below). Definition 1.4.6. An NEC signature s is said to be maximal if for every NEC group Λ containing an NEC group Λ with signature s the equality dim(Λ) = dim(Λ ) implies Λ = Λ . The definition of maximal Fuchsian signature is analogous. Remark 1.4.7. (1) Let s be the signature of a proper NEC group. If the signature s+ of its canonical Fuchsian subgroup is maximal then so is s. (2) Almost all Fuchsian signatures turn out to be maximal. A list of those which fail to be so was obtained by Greenberg in [47] and completed by Singerman in [117]. The list consists of nineteen pairs of signatures (s, s ) where s is nonmaximal and s is the signature of a Fuchsian group Δ properly containing a group Δ with signature s and such that dim(Δ) = dim(Δ ). If Δ is normal in Δ then (s, s ) is said to be a normal pair ; otherwise it is non-normal. The corresponding list of normal pairs of NEC signatures was obtained in [12], while the non-normal pairs were obtained by Estévez and Izquierdo in [41]. It must be pointed out that the maximality of the NEC signature s(Λ) does not imply the maximality of the NEC group Λ. However, the following two results, proved in [22, Chap. 5], should be mentioned. Theorem 1.4.8. If s is a maximal NEC signature then there exists a maximal NEC group Λ with s(Λ) = s. Corollary 1.4.9. Let G = Λ/Γ be a group of automorphisms of the Klein surface X = H/Γ, where Γ is a surface NEC group and Λ is an NEC group containing Γ as a normal subgroup. Assume that the signature s(Λ) of Λ is maximal. Then there exists a maximal NEC group Λ isomorphic to Λ, say via ϕ : Λ → Λ , and the full automorphism group of the Klein surface H/Γ , where Γ = ϕ(Γ), is G = Λ /Γ and the actions of G and G are topologically equivalent. Remark 1.4.10. This corollary is also true if H/Γ is a Riemann surface, in which case Γ is not a surface NEC group but a surface Fuchsian group.

14

1 Preliminaries

¨ 1.4.2 Teichmuller Spaces The last results in this section rely on rather deep mathematics: Teichmüller theory. Although we do not enter here into their proofs, we consider it convenient to present the basics about Teichmüller spaces in the context of NEC groups. The reader is referred to Sect. 4.7 in the book [113] by Seppälä and Sorvali and the paper [76] by Macbeath and Singerman. Let PGL(2, R) be the group of automorphisms of the hyperbolic plane H. Given an NEC group Λ, let R(Λ) be the set of group monomorphisms t : Λ → PGL(2, R) such that t(Λ) is also an NEC group. The automorphism group Aut(PGL(2, R)) of PGL(2, R) acts on R(Λ) by left multiplication, and the orbit space T(Λ) = R(Λ)/ Aut(PGL(2, R)) is the Teichmüller space of Λ. Since every automorphism α of PGL(2, R) is an inner automorphism, we see that the orbit [t] of t under this action is [t] = {t = αt : Λ → PGL(2, R) such that there exists m ∈ PGL(2, R) with t (λ) = mt(λ)m−1 for all λ ∈ Λ. }. In this definition m must be in Aut+ (H) in case Λ is a Fuchsian group. If Γ is a surface Fuchsian group of orbit genus g ≥ 2 then T(Γ) is the Teichmüller space of Riemann surfaces of genus g. If Δ is a Fuchsian group with signature s(Δ) = (g; m1 , . . . , mr ) then T(Δ) is a real cell of dimension dim(Δ) = 6g − 6 + 2r, as it was already known by Fricke and Klein [44]. If Λ is a proper NEC group then dim(Λ) = dim(Λ+ )/2, as proved by Keen in [64] and Singerman in [118]. There is also a well defined action of the automorphism group Aut(Λ) of Λ on the Teichmüller space T(Λ), given by right multiplication. This action is not effective since the inner automorphisms of Λ act trivially on T(Λ). The modular group of Λ is the factor group M(Λ) = Aut(Λ)/Inn(Λ), where Inn(Λ) is the normal subgroup of inner automorphisms. So the modular group M(Λ) acts on T(Λ), the quotient being called the moduli space of Λ. If Γ is a surface Fuchsian group of orbit genus g ≥ 2 then the moduli space of Γ is the moduli space Mg of Riemann surfaces of genus g. The interested reader is also referred to the articles by Earle [40], Buser, Seppälä and Silhol [28], Natanzon [95, 97] and Seppälä [110–112] to learn about this theory and how to use it to understand moduli spaces of real objects. It is worth mentioning also the papers [62], by Huisman and Lattarulo, where they study the moduli space of all real isotropic and Gaussian hyperelliptic curves and [31], where Cirre obtained a description of the five connected components of the moduli space of real algebraic curves of genus 2 by means of polynomial equalities and inequalities.

1.5 Symmetric Riemann Surfaces

15

1.5 Symmetric Riemann Surfaces We begin by introducing the notion of symmetry of a Riemann surface and defining the data that classify symmetries topologically. This leads to the notions of species of a symmetry and the symmetry type of a Riemann surface. Definitions 1.5.1. Let S be a compact Riemann surface. A symmetry of S is an antianalytic involution σ : S → S. A Riemann surface admitting a symmetry is called symmetric. In order to emphasize the real algebraic geometry counterpart of symmetries, these are sometimes called real structures of a Riemann surface. Let Fix(σ) be the fixed point set of a symmetry σ. It is well known that this set satisfies the following two properties. The first one is known as Harnack’s Theorem; we provide a proof in Theorem 1.5.3 below. A proof of the second property can be found, for instance, in [25]. (1) Fix(σ) consists of k disjoint Jordan curves, called ovals in Hilbert’s terminology, where 0 ≤ k ≤ g + 1. (2) S \ Fix(σ) consists of one connected component if the Klein surface S/ σ is non-orientable, and two otherwise. The symmetry σ is called non-separating, in the first case, and separating in the second one. These two properties play a key role in the topological nature of a symmetry since they classify symmetries up to homeomorphism. This justifies the following definition. Definition 1.5.2. The topological type of a symmetry σ of a Riemann surface S is the triple (g, k, ε), where g is the genus of S, k is the number of connected components of Fix(σ) and ε = 1 or ε = 0 according to whether S \ Fix(σ) is connected or not. We start this section with a short proof of the classical Harnack–Weichold Theorem, [59] and [127]. Theorem 1.5.3. A triple (g, k, ε) is the topological type of some symmetry of a compact Riemann surface of genus g if and only if 1≤k ≤g+1

if ε = 0 with g + 1 ≡ k (mod 2);

0≤k≤g

if ε = 1.

Proof. Let σ be a symmetry of a Riemann surface S of genus g ≥ 2. Then S = H/Γ for some surface Fuchsian group Γ and σ = Λ/Γ for some NEC group Λ containing Γ as a subgroup of index two. Clearly, Λ+ = Γ and so the signature of Λ is, by Corollary 1.2.7, s(Λ) = (g ; ±, [−]; {(−), . k. ., (−)}),

16

1 Preliminaries

where k = g + 1 − ηg with η = 2 if sign (s(Λ)) = “ + ” and η = 1 otherwise. In addition ε = 0 means that η = 2, and ε = 1 implies g = 0. All of this proves the “only if ” part. Conversely, assume that the triple (g, k, ε) satisfies the conditions in the statement of the theorem. Assume first that ε = 0 and let us consider an NEC group Λ with signature s(Λ) = (g ; +; [−]; {(−), . k. ., (−)}), where g = (g + 1 − k)/2. Then the epimorphism θ : Λ → Z2 = a induced by the assignment θ(ci ) = a; θ(aj ) = θ(bj ) = θ(ei ) = 1,

for i = 1, . . . , k and j = 1, . . . , g

satisfies that ker θ is a surface Fuchsian group having orbit genus g. Thus S = H/ ker θ is a Riemann surface of genus g having a symmetry of topological type (g, k, 0). For ε = 1, let g = g + 1 − k and let Λ be an NEC group with signature (g ; −; [−]; {(−), . k. ., (−)}). Let θ : Λ → Z2 = a be the epimorphism induced by the assignment θ(ci ) = a, θ(dj ) = a, θ(ei ) = 1 for i = 1, . . . , k and j = 1, . . . , g . Again its kernel ker θ is a surface Fuchsian group of orbit genus g and therefore S = H/ ker θ is a Riemann surface of genus g having a symmetry of type (g, k, 1). This completes the proof.

For example, the possible topological types of the symmetries of a genus two Riemann surface are (2, 1, 0), (2, 3, 0), (2, 0, 1), (2, 1, 1) and (2, 2, 1). The topological classification of a symmetry σ depends just on the number k of ovals of Fix(σ) and on the separating character of Fix(σ). These two data constitute the so called species of σ. Definition 1.5.4. The species of a symmetry σ, denoted by sp(σ), is +k if its topological type is (g, k, 0) and −k if its topological type is (g, k, 1). Let Aut(S) denote the group of analytic and antianalytic self-homeomorphisms of the Riemann surface S. It is called the automorphism group of S, and its elements automorphisms of S. Let Aut+ (S) be its subgroup consisting of the analytic automorphisms. Observe that with this notation, a symmetry of S is an involution σ ∈ Aut(S) \ Aut+ (S). Two symmetries σ and τ are said to be conjugate if σ = f ◦ τ ◦ f −1 for some automorphism f (either analytic or antianalytic, see Remark 1.5.6) of S. Clearly, two conjugate symmetries have the same species. This motivates the following definition. Definition 1.5.5. The symmetry type of S is the unordered list of species of all conjugacy classes of symmetries of S.


17

Remark 1.5.6. Notice that two symmetries σ and τ of S are conjugate via a dianalytic automorphism if and only if they are conjugate via an analytic automorphism. Indeed, if τ = ϕσϕ−1 where ϕ is antianalytic then τ = ψσψ −1 where ψ = ϕσ is analytic. This allows us, if necessary, to focus just on analytic conjugation of symmetries. The following is a characterization of the symmetric nature of a Riemann surface in terms of NEC groups. It is an immediate consequence of Theorem 1.4.4 applied to the case of Riemann surfaces. Corollary 1.5.7. The Riemann surface S = H/Γ is symmetric if and only if there exists a proper NEC group Λ containing Γ as a subgroup of index 2. As a consequence of this corollary, for each symmetry σ of the Riemann surface S = H/Γ, there exists a proper NEC group Λ containing Γ as a subgroup of index 2 such that σ = Λ/Γ. Notice that under these conditions, Γ coincides with the canonical Fuchsian subgroup Λ+ of Λ. Therefore Λ contains no orientation preserving elements of finite order, because Λ+ is a surface Fuchsian group. In other words, Λ is a surface NEC group. It turns out that the species of σ can be read from the signature of Λ. For example, the number of period cycles of s(Λ) coincides with the number of boundary components of the Klein surface H/Λ = (H/Γ)/(Λ/Γ) = S/ σ (see Proposition 1.1.4), which is the number of disjoint Jordan curves of Fix(σ). Moreover, the sign of the signature of Λ coincides with the sign of the species of σ, since, as said above, S \ Fix(σ) has two components if and only if H/Λ is an orientable Klein surface. Finally, the formulae for the areas of s(Γ) and s(Λ), together with the Hurwitz– Riemann formula, imply that the genus of H/Λ equals g + 1 − k if H/Λ is nonorientable and (g + 1 − k)/2 otherwise, where g is the genus of S. Summarizing, we have the following theorem. Theorem 1.5.8. Let Γ be a surface Fuchsian group and let S = H/Γ be a compact Riemann surface of genus g ≥ 2 which admits a symmetry σ. Write σ = Λ/Γ, where Λ is a proper NEC group containing Γ as a subgroup of index 2. Then the species of σ can be read from the signature of Λ in the following way: ⎧ 0 ⎪ ⎪ ⎨ sp(σ) = k ⎪ ⎪ ⎩ −k

if s(Λ) = (g + 1; −; [−]; {−}); if s(Λ) = ((g + 1 − k)/2; +; [−]; {(−), . k. ., (−)}); if s(Λ) = (g + 1 − k; −; [−]; {(−), . k. ., (−)}).

The following Lemma 1.5.9 will be useful in the sequel in order to determine whether a given symmetry fixes points or not. Lemma 1.5.9. Let σ be a symmetry of a compact Riemann surface S = H/Γ with Aut(S) = Λ/Γ. Let θ : Λ → Aut(S) be the corresponding smooth epimorphism with ker θ = Γ. Let us write σ = θ(λ) for some λ ∈ Λ. Then

18

1 Preliminaries

(1) If σ is fixed point free then λ is a glide reflection. (2) If σ fixes points in S then λ can be chosen to be a reflection (and hence conjugate to a canonical reflection). Proof. Observe that λ is an orientation reversing element of Λ and so it is either a glide reflection or a reflection. Claim (1) is obvious since the points fixed by a reflection λ projects onto points in S fixed by θ(λ). As to claim (2), let q ∈ S be a point fixed by σ and let z ∈ H be a point which projects onto q. The equality σ(q) = q means that λ preserves the Γ-orbit of z, and so λ(z) = γ(z) for some γ ∈ Γ. Then γ −1 ◦ λ is an orientation reversing element in Λ (because Γ is Fuchsian) which fixes points in H. Therefore γ −1 ◦ λ is a reflection c in Λ and θ(c) = θ(λ) = σ.

1.5.1 Algebraic Conditions What makes a compact Riemann surface symmetric? Let us look for necessary algebraic conditions on the group Aut+ (S) of analytic automorphisms of a compact Riemann surface S for it to be symmetric. Let S and Aut+ (S) be represented respectively as H/Γ and Δ/Γ for some Fuchsian groups Γ and Δ, where this last contains the first as a normal subgroup. Let us view Δ as an abstract group K and let H be the corresponding normal subgroup of K isomorphic to Γ. By Corollary 1.2.7, there is only a finite number of groups L1 , . . . , Ln which can be realized as NEC groups and which contain K as a subgroup of index 2. From the proof of this result we also know the way in which such embeddings K ⊆ Li look like. Now it is an entirely algebraic matter to decide which Li contains H as a normal subgroup. Let us call such Li algebraically admissible for S, and let us say that Li is conformally admissible if in addition it can be realized as an NEC group Λi containing Δ. It is clear that the existence of algebraically admissible groups is a necessary condition for S to be symmetric. The converse is not true: the existence of such a group L may not be sufficient for S to be symmetric, for reasons of conformal nature. This is so because, as we already pointed out, dim(Λ) = dim(Λ+ )/2 for the Teichmüller dimensions of a proper NEC group Λ and its canonical Fuchsian subgroup Λ+ . Simply saying this means that there are “more” Fuchsian groups isomorphic to K than NEC groups isomorphic to L. However it is worth mentioning that in such a situation a conformal structure on the underlying topological surface can be defined so that the new Riemann surface S is symmetric, its group Aut+ (S ) is isomorphic to Aut+ (S) and the actions of both groups are topologically equivalent. In fact, any NEC group Λ realizing L contains normal subgroups Δ and Γ isomorphic to Δ and Γ respectively and we have a commutative diagram: Δ Δ/Γ

∼

/ Δ

∼

/ Δ /Γ


19

where the horizontal arrows are isomorphisms and the vertical ones are the corresponding canonical projections. In this monograph we shall see that the conformal structure of S usually plays a modest role in both the qualitative and the quantitative study of symmetries, just up to determine Aut+ (S) and to decide which of the algebraically admissible groups are actually conformally admissible. So in certain qualitative and quantitative studies of symmetries of Riemann surfaces, the matter depends on Aut(S) and on the topological characteristics of this action. The signature of Λ and the epimorphism θ : Λ → Aut(S) take care of both. In one word, and this is a general philosophy of our approach in this monograph, “most of it is an algebraic question”. Fortunately, the above quoted necessary condition for the surface H/Γ to be symmetric is also sufficient when Δ is a triangle group, since dim(Δ) = 0. In such a case any algebraically admissible group is also conformally admissible. Assume that Δ has signature [k, , m] and let {x1 , x2 , x3 } be a canonical set of generators. Now a Riemann surface S admitting a group G isomorphic to K/H as the group of analytic automorphisms corresponds to a pair (a, b) of generators of G of orders k and respectively and whose product has order m; the surface S can be written as S = H/Γ, where Γ = ker θ for a group homomorphism θ : Δ → G induced by the assignment θ(x1 ) = a, θ(x2 ) = b. To continue, we shall need the following result due to Singerman [118]. As the original proof contains a harmless gap, we provide a somewhat different one here. Theorem 1.5.10. Let S be a Riemann surface corresponding to a generating pair (a, b) where a, b and ab have orders k, and m respectively. Then, S is symmetric if and only if the mapping ϕ : a → a−1 , b → b−1 or ϕ : a → b−1 , b → a−1 induces an automorphism of G = a, b = Aut+ (S). Proof. Let us write S = H/Γ with Γ a surface Fuchsian group, and G = Δ/Γ = Aut+ (S) where Δ is a Fuchsian group with signature [k, , m] containing Γ as a normal subgroup. If S is symmetric then there exists a proper NEC group Λ containing Δ as a subgroup of index 2 and Γ as a normal subgroup. Then Λ/Γ = Aut(S). According to Corollary 1.2.7, there are two possibilities for the signature of Λ. If the periods of Δ are pairwise different then s(Λ) = (k, , m), but if two periods of Δ coincide, say k = , then either s(Λ) = (k, , m) or s(Λ) = (0; +; [k]; {(m)}). Assume first that Λ has signature (k, , m) and let {c0 , c1 , c2 } be a set of canonical generators of Λ. Then x1 = c0 c1 , x2 = c1 c2 , x3 = c2 c0 (1.2) constitute a set of canonical generators for Δ. Note that and c1 x2 c1 = x−1 c1 x1 c1 = x−1 1 2 .

(1.3)

Then, for a = Γx1 , b = Γx2 and v = Γc1 ∈ Λ/Γ we have av = a−1 and bv = b−1 . Therefore, the assignment a → a−1 , b → b−1 induces in fact an automorphism of G. Observe that Aut(S) = Λ/Γ is a semidirect product G Z2 .

20

1 Preliminaries

Suppose now that s(Λ) = (0; +; [k]; {(m)}) and let {x, e, c0 , c1 } be a set of canonical generators of Λ. Then x1 = x,

x2 = c0 x−1 c0 ,

x3 = c0 c1

(1.4)

can be chosen to constitute a set of canonical generators for Δ. Again it is easy to check that c0 x1 c0 = x−1 and c0 x2 c0 = x−1 (1.5) 2 1 . Now for a = Γx1 , b = Γx2 and v = Γc0 ∈ Λ/Γ we have av = b−1 and bv = a−1 , and therefore the assignment a → b−1 , b → a−1 induces an automorphism of G. Suppose conversely that the assignment a → a−1 , b → b−1 induces an automorphism of G. As dim(Δ) = 0, there exists an NEC group Λ with signature (k, , m) containing Δ. Now a word w = w(a, b) is the identity in Λ/Γ if and only if w(x1 , x2 ) belongs to Γ, and analogously, w(a−1 , b−1 ) = 1 if and only if w = −1 −1 w(x−1 , b → b−1 induces an automorphism of G, we 1 , x2 ) ∈ Γ. So, as a → a −1 see from (1.3) that for w = w(x1 , x2 ) ∈ Γ, also c1 w(x1 , x2 )c1 = w(x−1 1 , x2 ) ∈ Γ and therefore Γ is a normal subgroup of Λ. In this case, the image of c1 in the factor group Λ/Γ can be chosen as a symmetry of S. Finally, assume that the assignment a → b−1 , b → a−1 induces an automorphism of G. Then k = and there exists an NEC group Λ with s(Λ) = (0; +; [k]; {(m)}) containing Δ as a subgroup of index 2. Also now w = w(x1 , x2 ) −1 belongs to Γ if and only if the same happens to w(x−1 2 , x1 ). Thus, using the above −1 −1 equalities (1.5) we see that c0 w(x1 , x2 )c0 = w(x2 , x1 ) ∈ Γ for every w ∈ Γ and therefore Γ is a normal subgroup of Λ. So the surface H/Γ is symmetric because the image of c0 in Λ/Γ is a symmetry of S.

Remark 1.5.11. From equalities (1.2) and (1.4) above it follows easily that for a symmetric Riemann surface S with Aut+ (S) = a, b, its automorphism group Aut(S) is the semidirect product Aut(S) = Aut+ (S) Z2 = a, b t, where t acts as the automorphism given in the proof of Theorem 1.5.10, and the epimorphism from Λ onto Aut+ (S) Z2 is defined as follows: if s(Λ) = (k, , m) then θ : Λ → Aut+ (S) Z2 ; c0 → at ; c1 → t ; c2 → tb, whilst for s(Λ) = (0; +; [k]; {(m)}) we have θ : Λ → Aut+ (S) Z2 ; x → a ; e → a−1 ; c0 → t ; c1 → t(ab)−1 .

Chapter 2

On the Number of Conjugacy Classes of Symmetries of Riemann Surfaces

As said in the Introduction, under the correspondence between compact Riemann surfaces and smooth irreducible complex projective algebraic curves, the fact that a Riemann surface S is symmetric means that the corresponding complex curve C can be defined over the field R of real numbers. This is why such a symmetry is often called a real form of C. Symmetries which are non-conjugate in the automorphism group Aut(S) of S correspond to non-isomorphic real forms of C. In this chapter we shall pay attention to quantitative results concerning the number of conjugacy classes of symmetries. We will distinguish cases according to whether the sets of fixed points of the symmetries are empty or not. We start with a study of conjugacy classes of involutions in 2-groups at large.

2.1 Conjugacy Classes of Involutions in 2-Groups Given an abstract group G, it makes sense to say that an involution x ∈ G is a “symmetry” provided that a concept of orientation in such a group is defined. This is done in the following definitions. Definitions 2.1.1. Let G be an abstract group. (1) G is said to be abstractly orientable if there exists an epimorphism α : G → Z2 = {±1}. In such a case, α is an orientation of the group G. If an orientation α is chosen then we say that G is abstractly oriented. (2) Let α be an orientation of G. An element x ∈ G is orientation preserving (respectively orientation reversing) with respect to the orientation α if α(x) = +1 (respectively α(x) = −1). Examples of orientable groups are provided by proper NEC groups and groups of automorphisms of symmetric Riemann surfaces. Lemma 2.1.2. Let G be a 2-group containing a cyclic group ZN = x as a subgroup of index 2r . Then G has at most 2r+1 − 1 conjugacy classes of involutions. Furthermore, if G is abstractly oriented and x preserves the orientation then G has at most 2r conjugacy classes of orientation reversing involutions. E. Bujalance et al., Symmetries of Compact Riemann Surfaces, Lecture Notes in Mathematics 2007, DOI 10.1007/978-3-642-14828-6 2, c Springer-Verlag Berlin Heidelberg 2010

21

22

2 Number of Conjugacy Classes of Symmetries

Proof. Let 2

2

2

2

2

ZN = H0 ≤ H1 ≤ H2 ≤ · · · ≤ Hr−1 ≤ Hr = G be a subnormal series for G and let xi ∈ Hi \ Hi−1 for i = 1, . . . , r. Then each element g ∈ G can be uniquely represented as g = xε xε11 · · · xεrr for some integers ε ∈ {0, . . . , N − 1} and εi ∈ {0, 1}. Let us denote w = xε11 · · · xεrr and observe that there are 2r − 1 non-trivial elements of this form. We shall show that for any such element w = 1 there are at most 2 conjugacy classes of involutions among the elements of the set {w, xw, x2 w, . . . , xN −1 w}. This will complete the proof of the first part of the lemma since for w = 1 this set has one involution. Observe that w may not be an element of order 2 and, furthermore, among these elements there may not even exist elements of order 2. Assume then that there are at least two elements xk w and x w of order 2 and assume also that k and are chosen so that k > and m = k − is minimal. We shall show that each involution xn w is conjugate either to xk w or to x w. We have 1 = (xk w)2 = xm (x w)2 w−1 xm w = xm w−1 xm w. So wx−m = xm w and therefore x+sm w has order 2 for each integer s. Moreover, xsm (x w)x−sm = x+2sm w,

(2.1)

xsm (xk w)x−sm = xk+2sm w = x+(2s+1)m w.

(2.2)

Now let xn w be an arbitrary element of order 2. Then n = + tm + j for some integers t, j, where 0 ≤ j < m, and since both xn w and x+tm w have order 2, it follows by the minimality of m that j = 0. Thus xn w = x+tm w which, by (2.1) and (2.2), is conjugate to x w if t is even, and it is conjugate to xk w if t is odd. This shows the first part of the lemma. Assume now that G is abstractly oriented and that x preserves the orientation. Then half of the 2r elements w = xε11 · · · xεrr reverses orientation and the other half preserves it. For each of the 2r−1 orientation reversing ones, the (at most two) conjugacy classes of involutions in the set {w, xw, x2 w, . . . , xN −1 w} are the only ones that reverse the orientation. This shows the second part of the lemma. Let DN be a dihedral group and let x, y ∈ DN be two generating involutions. If DN is abstractly oriented and x and y reverse the orientation then ZN = xy is a subgroup of DN of index 2 generated by an orientation preserving element. So, as a consequence of the above Lemma 2.1.2, we get the following result. Corollary 2.1.3. Let G be a 2-group containing a dihedral group DN as a subgroup of index 2r . Then G has at most 2r+2 − 1 conjugacy classes of involutions. Furthermore if G is abstractly oriented and DN is generated by two involutions which reverse the orientation then G has at most 2r+1 conjugacy classes of orientation reversing involutions. The next technical lemma deals with 2-groups of automorphisms of a Riemann surface. Together with Lemma 2.1.2 and Corollary 2.1.3, it will play a key role in the sequel.

2.2 Symmetries with Non-Empty Set of Fixed Points

23

Lemma 2.1.4. Let S be a Riemann surface of genus g ≥ 2, and let 2r−1 be the largest power of 2 dividing g −1. Let G be a 2-group of automorphisms of S of order 2t and assume that t ≥ r + 1. Then G contains a cyclic or a dihedral subgroup of index 2r . Proof. Let us write S = H/Γ for some surface Fuchsian group Γ and G = Λ/Γ for some NEC group Λ containing Γ as a normal subgroup. Assume that Λ has signature s(Λ) = (h; ±; [m1 , . . . , mν ]; {(n11 , . . . , n1s1 ), . . . , (nk1 , . . . , nksk )}).

(2.3)

We claim that s(Λ) has either a proper period or a link period. In fact, by the Hurwitz–Riemann formula we have ⎛ ⎞ si ν k g−1 mi − 1 nij − 1 ⎠ = 2t−r ⎝ηh − 2 + k + + 2r−1 m 2nij i i=1 i=1 j=1 where either η = 1 or η = 2 depending on the sign of s(Λ). Since (g − 1)/2r−1 is odd and t − r ≥ 1, the expression in brackets cannot be an integer. So there must be a non-trivial mi or nij , as claimed. Moreover, since G is a 2-group, all periods of Λ are powers of 2, since otherwise Γ would have elements of finite order. It follows that mi ≥ 2t−r for some i or nij ≥ 2t−r−1 for some i, j. Assume first that Λ has a proper period m ≥ 2t−r ; in this case the image x in G of an elliptic generator of Λ of order m is still an element of order m and so for m = m/2t−r , the element xm generates a cyclic subgroup of G of index 2r . Assume now that Λ has a link period n ≥ 2t−r−1 ; in this case the images c and c in G of two consecutive reflections of Λ whose product has order n are involutions, since otherwise Γ would be a proper NEC group. Moreover, for n = n/2t−r−1 , the element (cc )n has order 2t−r−1 and so c and (cc )n generate a dihedral subgroup of G of index 2r . This completes the proof. Remark 2.1.5. The proof shows that in fact G contains a cyclic subgroup generated by an orientation preserving element or a dihedral subgroup generated by two orientation reversing elements, of index 2r in both cases. Remark 2.1.6. Let G+ denote the subgroup of G consisting of its orientation preserving elements. With the notations in the above proof of Lemma 2.1.4, the existence of a proper period or a link period in the signature of Λ shows that G+ acts on S with fixed points.

2.2 Symmetries with Non-Empty Set of Fixed Points The quantitative study of conjugacy classes of symmetries started with a seminal result of Natanzon [95] who proved, using topological methods, that a complex √ algebraic curve of genus g ≥ 2 has at most 2( g + 1) non-isomorphic real forms

24


with real points. He also showed that this bound is attained for infinitely many values of g, those being of the form (2n − 1)2 . Here we go further, namely, we determine the maximal number of conjugacy classes of symmetries with fixed points that a compact Riemann surface S of genus g ≥ 2 can admit. Assume that σ1 , . . . , σk are representatives of the conjugacy classes of symmetries of S. Since each σi belongs to a Sylow 2-subgroup of Aut(S) and all Sylow 2-subgroups are conjugate, we may assume that all these symmetries generate a 2-group G. We now establish a fundamental result on this topic, whose first proof appeared in [23]. Theorem 2.2.1. Let S be a Riemann surface of genus g ≥ 2 and let us write g = 2r−1 u+1 with u odd. Then the maximum number of non-conjugate symmetries with fixed points that S admits is 2r+1 . Furthermore, this bound is attained if and only if u ≥ 2r+1 − 3. Proof. Let k be the number of conjugacy classes of symmetries with fixed points of S. As we observed above, we can choose representatives of these classes such that they generate a 2-group, say of order 2t . If t ≤ r then k < 2t ≤ 2r and so the first part of the statement is proved in this case. If t ≥ r + 1 then the first part is a direct consequence of Lemma 2.1.4, Corollary 2.1.3 and Lemma 2.1.2. Let now S = H/Γ be a Riemann surface with the maximum number 2r+1 of conjugacy classes of symmetries with fixed points and let G be a 2-group generated by 2r+1 representatives of these classes. Let us write G = Λ/Γ for some NEC group Λ with signature (2.3). Let C1 , . . . , Cn be the different period cycles of Λ involving these symmetries, and assume that C1 , . . . , Cm are non-empty and Cm+1 , . . . , Cn are empty. Observe that n > 0 because they are symmetries with fixed points. As each empty period cycle involves at most one symmetry we see that C1 , . . . , Cm involve at least 2r+1 − (n − m) symmetries. As each non-empty period cycle Ci of length si involves at most si non-conjugate symmetries, see Remark 1.1.6, we get s1 + · · · + sm ≥ 2r+1 − n + m. We shall show that Area(Λ) ≥ 2π

2r 2r+1 − 3 − 4 |G|

.

Observe that each term (1/2)(1 − 1/nij ) occurring in the formula of Area(Λ) is not smaller than 1/4 because nij ≥ 2. Since G is generated by 2r+1 orientation reversing involutions, we see that |G| ≥ r+2 2 . In particular, we may repeat the proof of Lemma 2.1.4 to show that Λ has a proper period ≥ |G|/2r or a link period ≥ |G|/2r+1 . In the first case s 1 + · · · + sm 2r + Area(Λ) ≥ 2π n − 2 + 1 − |G| 4 r+1 r+1 2r 2r + 3n + m − 4 −3 2 2 − > 2π − . ≥ 2π 4 |G| 4 |G|


25

If Λ has a link period ≥ |G|/2r+1 then m > 0 and 2r s 1 + · · · + sm − 1 1 + − Area(Λ) ≥ 2π n − 2 + 4 2 |G| r+1 r+1 2r 2r + 3n + m − 7 −3 2 2 − ≥ 2π − . ≥ 2π 4 |G| 4 |G| So, in both cases, 4π(g − 1) = Area(Γ) = |G|Area(Λ) ≥ 2π

2r 2r+1 − 3 − 4 |G|

|G|.

Since |G| ≥ 2r+2 we get g − 1 ≥ (2r+1 − 3)

|G| − 2r−1 ≥ (2r+1 − 3)2r−1 − 2r−1 = 2r−1 (2r+1 − 4). 8

Therefore u = (g − 1)/2r−1 ≥ 2r+1 − 4. However, u is odd by assumption and consequently u ≥ 2r+1 − 3. Conversely, let g = 2r−1 u + 1 with u ≥ 2r+1 − 3 and let s = u + 3. Consider a maximal NEC group Λ with signature (0; +; [−]; {(2, s+1 . . . , 2)}), and let {c0 , . . . , cs+1 } be a canonical set of generators of Λ. Let us consider the group G = Zr+2 = x1 ⊕ · · · ⊕ xr+2 and let a1 , . . . , a2r+1 be the involutions in G 2 whose length in x1 , . . . , xr+2 is odd. We define a homomorphism θ : Λ → G by choosing θ(ci ) ∈ {a1 , . . . , a2r+1 } for 0 ≤ i ≤ s + 1 so that θ(ci ) = θ(ci+1 ) for 0 ≤ i ≤ s, and such that θ is in fact an epimorphism. Observe that this is indeed possible because s ≥ r + 1. Clearly, ker θ is a surface Fuchsian group. The orbit space S = H/ ker θ is a Riemann surface of genus 2r−1 u + 1 (by the Hurwitz–Riemann formula) having G as its full group Aut(S) of automorphisms (by the maximality of Λ). Since the image under θ of each canonical reflection ci is a symmetry with fixed points we see that S has 2r+1 non-conjugate symmetries with fixed points. Every even value of g can be written as 2r−1 u + 1 with r = 1 and u odd. In this way we obtain the main result in [53] as a corollary of Theorem 2.2.1. Corollary 2.2.2. A Riemann surface of even genus g has at most 4 non-conjugate symmetries with fixed points. Furthermore this bound is attained for every even genus g ≥ 2.

26


Remark 2.2.3. Given an arbitrary integer g ≥ 2, there is an integer r ≥ 1 and an odd integer u ≥ 1 such that g = 2r−1 u + 1. Fix r ≥ 1 and consider all values of g of this form. Observe that the numbers g − 1 are just the solutions of the congruence x ≡ 2r−1 (mod 2r ). Suppose that u ≥ 2r+1 − 3. Then g ≥ 2r−1 (2r+1 − 3) + 1 = 22r − 3 · 2r−1 + 1 > 22r − 4 · 2r−1 + 1 = (2r − 1)2 . √ Henceforth 2( g + 1) > 2r+1 and thus, for the values of g corresponding to √ u ≥ 2r+1 − 3, the bound 2( g + 1) obtained by Natanzon in [95] for the number of non-conjugate symmetries with fixed points is not sharp. On the other hand, if u ≤ 2r+1 − 5 then g ≤ 2r−1 (2r+1 − 5) + 1 = 22r − 5 · 2r−1 + 1 < 22r − 4 · 2r−1 + 1 = (2r − 1)2 . √ Hence 2( g + 1) < 2r+1 in this case and so, for the values of g corresponding to r+1 u≤2 − 5, Natanzon’s bound is better than the one in Theorem 2.2.1. We now calculate sharp bounds for the remaining values of g. Notation 2.2.4. For each integer g ≥ 2 we denote by μf (g) the maximal number of conjugacy classes of symmetries with fixed points that a genus g Riemann surface can admit. With this notation, Theorem 2.2.1 can be stated as μf (g) = 2r+1

for g = 2r−1 u + 1 with u odd and u ≥ 2r+1 − 3.

Next we shall calculate the remaining values of this function, as stated in [23]. To that end we fix r ≥ 2 (since the case r = 1 is solved in Corollary 2.2.2) and consider the function 2s−4 − 1 f (s) = , 2r−s which is strictly increasing for s > 3. Since f (4) = 0 and f (r + 3) = 2r+2 − 8 we see that for each odd positive integer u < 2r+2 − 7 there exists a unique integer s ∈ {4, . . . , r + 2} such that f (s) < u ≤ f (s + 1), that is, 2s−4 − 1 2s−3 − 1 < u ≤ . 2r−s 2r−s−1 The next theorem shows that μf (g) depends on this value of s; in fact, it shows that μf (g) = min{2r−s+2 u + 4, 2s−1 }. Theorem 2.2.5. Let g = 2r−1 u + 1, where r ≥ 2 and u < 2r+2 − 7 is odd. Let s be defined as above. Then


μf (g) =

⎧ ⎪ r−s+2 ⎪ u+4 ⎨2

if

2s−4 − 1 2s−3 − 1 < u ≤ ; 2r−s 2r−s

⎪ ⎪ ⎩ 2s−1

if

2s−3 − 1 2s−2 − 2 < u ≤ . 2r−s 2r−s

27

Proof. Let S be a Riemann surface of genus g = 2r−1 u + 1 where u < 2r+2 − 7 such that S has k non-conjugate symmetries with fixed points. First we shall show that k is not greater than the proposed value for μf (g). As before, by Sylow theory, we may assume that the k symmetries generate a 2-group G, say of order 2t . Observe that the product of two of these symmetries is an orientation preserving element of G which generates a cyclic subgroup of index ≤ 2t−1 . So Lemma 2.1.2 yields k ≤ 2t−1 . (2.4) Let us write S = H/Γ and G = Λ/Γ for some surface Fuchsian group Γ and a proper NEC group Λ containing Γ as a normal subgroup. It is easy to see, using arguments similar to those in the proof of Theorem 2.2.1, that Area(Λ) ≥ π(k − 4)/2. So, by the Hurwitz–Riemann formula, 4π(g − 1) = |G|Area(Λ) ≥ 2t−1 π(k − 4), which gives k ≤ 2r−t+2 u + 4. (2.5) Let us suppose first that (2s−4 − 1)/2r−s < u ≤ (2s−3 − 1)/2r−s .

(2.6)

If t ≥ s then k ≤ 2r−s+2 u + 4 by (2.5). If t < s then k ≤ 2t−1 ≤ 2s−2 by (2.4), and so k < 2r−s+2 u + 4, because 2s−2 < 2r−s+2 u + 4 by (2.6). We now suppose that (2s−3 − 1)/2r−s < u ≤ (2s−2 − 2)/2r−s .

(2.7)

If t ≥ s + 1 then k ≤ 2r−s+1 u + 4 ≤ 2s−1 , where we have used (2.5) for the first inequality and (2.7) for the second. If t ≤ s then k ≤ 2t−1 ≤ 2s−1 by (2.4). To finish the proof we consider an arbitrary integer s ∈ {4, . . . , r + 2} and an arbitrary odd integer u in the range (2s−4 − 1)/2r−s < u ≤ (2s−2 − 2)/2r−s . Let G = Zs2 = x1 ⊕ · · · ⊕ xs . Let A be the set consisting of the 2s−1 involutions of G which can be written as words of odd length in x1 , . . . , xs . Let us write k = 2r−s+2 u + 4 (k ≥ 5) and let Λ be a maximal NEC group with signature (0; +; [−]; {(2, . k. ., 2)}). Observe that k > s because u > (2s−4 − 1)/2r−s . Hence there exists an epimorphism θ : Λ → G such that the image θ(ci ) of each canonical reflection belongs to A and θ(ci ) = θ(ci+1 ). In addition, if k ≤ 2s−1 then θ can be defined so that the k canonical reflections c1 , . . . , ck are mapped onto distinct elements of A.

28


Then ker θ is a surface Fuchsian group and the orbit space S = H/ ker θ is a Riemann surface of genus 2r−1 u + 1 having G as its full group of automorphisms. Now, if k ≤ 2s−1 , which happens if and only if u ≤ (2s−3 − 1)/2r−s, then {θ(ci ) : i = 1, . . . , k} are representatives of the conjugacy classes of symmetries with fixed points in S. On the other hand, if k > 2s−1 , which happens if and only if u > (2s−3 − 1)/2r−s, then the 2s−1 elements in A are representatives of the conjugacy classes of symmetries with fixed points in S. Remark 2.2.6. The values of u in the range 2r+1 − 3 ≤ u < 2r+2 − 7 are covered by both Theorems 2.2.1 and 2.2.5. Let us check that the formulae of μf (g) given by these theorems coincide for these values of u. First, Theorem 2.2.1 gives directly μf (g) = 2r+1 . To apply the formula of Theorem 2.2.5 we observe that the value of the parameter s corresponding to those u in the range 2r+1 − 3 ≤ u < 2r+2 − 7 is s = r + 2. So, μf (g) = min{2r−s+2 u + 4, 2s−1 } = min{u + 4, 2r+1 } = 2r+1 , because u + 4 ≥ 2r+1 + 1. Example 2.2.7. The function g → μf (g) is not increasing because μf (g) = 4 for all even values of g (see Corollary 2.2.2). However, if we write g = 2r−1 u + 1 and fix a value of r then the function u → μf (2r−1 u + 1) is increasing (but not strictly) as a function of u. It attains the maximal value 2r+1 for u = 2r+1 − 3 and remains constant from that moment onwards. We illustrate this in Table 2.1, where the pairs (g, μf (g)) are computed for small values of r. Table 2.1 Values of the pair (g, µf (g)) where g = 2r−1 u + 1 with u odd for small values of r r=1

r=2

r=3

r=4

(2, 4) (4, 4) (6, 4) (8, 4) (10, 4) (12, 4) (14, 4) (18, 4) (20, 4) (22, 4) (24, 4) (26, 4) (28, 4) (30, 4) (32, 4) (34, 4)

(3, 5) (7, 7) (11, 8) (15, 8) (19, 8) (23, 8) (27, 8) (31, 8) (35, 8) (39, 8) (43, 8) (47, 8) (51, 8) (55, 8) (59, 8) (63, 8)

(5, 6) (13, 8) (21, 9) (29, 11) (37, 13) (45, 15) (53, 16) (61, 16) (69, 16) (77, 16) (85, 16) (93, 16) (101, 16) (109, 16) (117, 16) (125, 16)

(9, 8) (25, 10) (41, 14) (57, 16) (73, 16) (89, 16) (105, 17) (121, 19) (137, 21) (153, 23) (169, 25) (185, 27) (201, 29) (217, 31) (233, 32) (249, 32)

2.3 Symmetries with Empty Set of Fixed Points

29

2.3 Symmetries with Empty Set of Fixed Points In this section we shall deal with symmetries without fixed points. These symmetries correspond to the so called purely imaginary curves, that is, complex algebraic curves which can be defined over the reals but have no R-rational points. For an arbitrary value of g ≥ 2, let μi (g) denote the maximal number of conjugacy classes of fixed point free symmetries that can be admitted by a Riemann surface S of genus g which has no symmetry with fixed points. Theorem 2.3.1. Let us write g = 2r−1 u + 1 with u odd. Then μi (g) ≤ 2r . Furthermore, this bound is attained whenever u ≥ 2r + 1. Proof. Let S be a compact Riemann surface of genus g having no symmetry with fixed points and let G be a 2-group of automorphisms of S generated by representatives of all the conjugacy classes of fixed point free symmetries. Let us write S = H/Γ and G = Λ/Γ, where Γ has signature (g; −) and Λ is a proper NEC group. Since S has no symmetry with fixed points, Λ contains no reflection, and so its signature is (h; −; [m1 , . . . , mv ]; {−}) for some h ≥ 1. Let 2s be the largest proper period in s(Λ), if any (observe that each proper period mi is a power of 2). Then, by the Hurwitz–Riemann formula, 4π(g − 1) = |G|2π(h − 2 + m/2s ) for some non-negative integer m. Since g − 1 = 2r−1 u we get u=

m |G| |G| h − 2 + = r+s (2s (h − 2) + m) . r s 2 2 2

This yields that the order of G divides 2r+s because u is odd. The image in G of the elliptic element of order 2s is an orientation preserving element which generates a cyclic subgroup of index 2r . Hence μi (g) ≤ 2r by Lemma 2.1.2. To prove the second part, let u ≥ 2r + 1 and let Λ be a maximal NEC group with signature (h; −; [2, 2, 2]; {−}), where h = (u + 1)/2 ≥ r + 1. Take G = Zr+1 with generating basis {z1 , . . . , zr+1 }, and let θ : Λ → G be the epimorphism 2 given by θ(di ) = zi for 1 ≤ i ≤ r + 1, θ(di ) = z1 for r + 2 ≤ i ≤ h and θ(x1 ) = z1 z2 , θ(x2 ) = z2 z3 , θ(x3 ) = z1 z3 . Then Γ = ker θ is a surface Fuchsian group and X = H/Γ is a Riemann surface of genus g = 2r−1 u + 1, without symmetries with fixed points and admitting 2r conjugacy classes of fixed point free symmetries. The results are more precise if we restrict our considerations to Riemann surfaces whose full group Aut(S) acts without fixed points, that is, no automorphism of S (either analytic or antianalytic) fixes points in S. These are the surfaces S for which the normal covering S → S/ Aut(S) is unramified. For each g ≥ 3, let μw i (g) denote the maximal number of conjugacy classes of symmetries that a genus g Riemann surface whose full group Aut(S) acts without fixed points may admit. Observe that μw i (g) does not make sense for g = 2 since all surfaces of genus 2 are hyperelliptic and the hyperelliptic involution fixes points.

30


r−1 Theorem 2.3.2. Let us write g = 2r−1 u + 1 with u odd. Then μw . i (g) ≤ 2 Assume that g ∈ / {3, 5}. Then the bound is attained if and only if u ≥ r − 2. For w g ∈ {3, 5} we have μw i (3) = 1 and μi (5) = 2.

Proof. Let S be a genus g Riemann surface such that Aut(S) acts fixed point freely and let G be a 2-group of automorphisms of S generated by representatives of the conjugacy classes of its symmetries. Let us write S = H/Γ and G = Λ/Γ, where Γ has signature (g; −) and Λ is a proper NEC group. Since the automorphisms in G act fixed point freely, the group Λ contains no reflection and no elliptic element; hence s(Λ) = (h; −; [−]; {−}) for some integer h > 2. By the Hurwitz–Riemann formula, 2π|G|(h − 2) = 4π(g − 1) = 2r+1 πu, which implies that |G| divides 2r because G is a 2-group and u is odd. The product of two symmetries generates a dihedral group of index ≤ 2r−2 and so Corollary 2.1.3 r−1 yields μw . i (g) ≤ 2 Assume that g ∈ / {3, 5}. If this bound is attained then |G| = 2r , h − 2 = u and, by Lemma 2.1.2, no element of G has order greater than two. So G = Zr2 . Moreover, since G is generated by the cosets Γdi , where d1 , . . . , dh form a set of canonical generators of Λ, it follows that h ≥ r and so u ≥ r − 2. Conversely, if u ≥ r − 2 then u + 2 ≥ 4 since otherwise g = 3 or 5. So we may take a maximal NEC group Λ with signature (u + 2; −; [−]; {−}). Let {d1 , . . . , du+2 } be a set of canonical generators of Λ. Take G = Zr2 with generating basis {z1 , . . . , zr }, and let θ : Λ → G be the epimorphism induced by the assignment θ(di ) = zi for 1 ≤ i ≤ r and θ(dj ) = z1 for r + 1 ≤ j ≤ u + 2. Then ker θ is a surface Fuchsian group and S = H/ ker θ is a Riemann surface of genus g = 2r−1 u + 1 with exactly 2r−1 conjugacy classes of symmetries whose full group Aut(S) acts fixed point freely. If the bound were attained for g = 3 or g = 5 then, with the above notations, s(Λ) = (3; −; [−]; {−}), which is not a maximal signature. Indeed, according to the list of normal pairs of NEC signatures given in [12], for each Λ with the above signature, there exists an NEC group Λ with signature s(Λ ) = (0; +; [2, 2, 2]; {(−)}) containing Λ as a normal subgroup of index 2. Up to automorphisms in Λ and Λ , there is a unique embedding of Λ in Λ , given by d1 = x1 c, d2 = cx2 and d3 = x2 cx3 x2 , see [12, Proposition 4.8], where {x1 , x2 , x3 , c} is a set of canonical generators of Λ . Using this embedding it is easy to see that any smooth epimorphism θ : Λ → G, where G = Z22 if g = 3 and G = Z32 if g = 5, can be extended to a smooth epimorphism θ : Λ → G where G = Z32 if g = 3 and G = Z42 if g = 5. Hence ker θ = ker θ and so the Riemann surface H/ ker θ = H/ ker θ admits automorphisms with fixed points, namely, the images under θ of the elliptic w elements of Λ . This is a contradiction and so μw i (3) < 2 and μi (5) < 4. Let us consider now a maximal NEC group Λ with signature (4; −; [−]; {−}) and define the epimorphisms θ1 : Λ → Z2 = σ by θ1 (di ) = σ for 1 ≤ i ≤ 4 and θ2 : Λ → Z22 = σ1 , σ2 by θ2 (d1 ) = θ2 (d2 ) = σ1 and θ2 (d3 ) = θ2 (d4 ) = σ2 . The group Aut(Sj ) of the Riemann surface Sj = H/ ker θj acts fixed point freely and has one conjugacy class of symmetries if j = 1 and two if j = 2. This yields w μw i (3) = 1 because S1 has genus 3, and μi (5) ≥ 2 because S2 has genus 5.

2.4 Symmetries of Surfaces Admitting a Fixed Point Free Symmetry

31

We finally show that μw i (5) < 3. Suppose, to get a contradiction, that there exists a Riemann surface S = H/Γ of genus 5 with three conjugacy classes of symmetries and let G be a 2-group generated by representatives of them. Observe that |G| ≥ 23 . Writing G = Λ/Γ and using that G acts fixed point freely, we get, by the Hurwitz–Riemann formula, that s(Λ) = (3; −; [−]; {−}) and |G| = 8. Then G = Z32 and we may repeat the above arguments to show that the action of G extends to a group G which does not act fixed point freely.

2.4 Symmetries of Surfaces Admitting a Fixed Point Free Symmetry In the previous sections we have studied either collections of symmetries having fixed points or collections of fixed point free symmetries of surfaces that do not admit symmetries with fixed points. Now we shall study hybrid configurations. In this section we calculate the maximal number of conjugacy classes of symmetries that can be admitted by a Riemann surface S of genus g which has a fixed point free symmetry. The computation of this bound was also carried out in [18]. It turns out that this bound is the same as in Theorem 2.2.1 (for symmetries with fixed points) but now it is attained for a wider range of genera. Theorem 2.4.1. Let S be a compact Riemann surface of genus g admitting a fixed point free symmetry. Let us write g = 2r−1 u + 1 with u odd. Then the number of conjugacy classes of symmetries of S is at most 2r+1 . Furthermore this bound is attained whenever u ≥ r − 2. Proof. Let G be a 2-group of automorphisms of S generated by representatives of the conjugacy classes of symmetries of S. Let us write |G| = 2t . Clearly, G has at most 2t−1 conjugacy classes of symmetries. If t ≥ r + 1 (otherwise there is nothing to prove) then Lemma 2.1.4 yields that G contains either a cyclic or a dihedral subgroup of index 2r . In the first case the number of conjugacy classes of symmetries in G is ≤ 2r by Lemma 2.1.2 (see also Remark 2.1.5) and ≤ 2r+1 in the second one by Corollary 2.1.3. Therefore S has at most 2r+1 conjugacy classes of symmetries. In fact, the above shows that this bound is attained only if G contains a dihedral subgroup of index 2r and the subgroup G+ of orientation preserving elements acts with fixed points on S, see Remark 2.1.6. Suppose now that u ≥ r − 2. Let s = u + 3 ≥ 4 and take a maximal NEC group Λ with signature (0; +; [−]; {(2, s+1 . . . , 2)}). Let {c0 , . . . , cs+1 } be a canonical set of generators for Λ and let G = Zr+2 = x1 ⊕ · · · ⊕ xr+2 . Since s ≥ r + 1, 2 the assignment ci → xi for 1 ≤ i ≤ s + 1, where the indices of xi are modulo r + 1, induces an epimorphism θ : Λ → G. In fact, ker θ is a surface Fuchsian group and so S = H/Γ is a Riemann surface; its genus equals 2r−1 u + 1 and its full automorphism group Aut(S) = Zr+2 has 2r+1 symmetries which are pairwise 2 non-conjugate. In addition, S admits fixed point free symmetries since, for instance,

32


the image under θ of the glide reflection c1 c2 c3 is one of them. Hence the bound 2r+1 is achievable when u ≥ r − 2. Recall from Theorem 2.2.1 that 2r+1 is also the upper bound for the number of conjugacy classes of symmetries with fixed points. As a consequence of Theorems 2.2.1 and 2.4.1 we get the following. Corollary 2.4.2. The maximum number of non-conjugate symmetries (of any type) that a Riemann surface of genus g may admit is 2r+1 , where 2r−1 is the largest power of 2 dividing g − 1. Corollary 2.4.3. A Riemann surface with the maximum number of non-conjugate symmetries with fixed points admits no symmetry with empty set of fixed points.

Chapter 3

Counting Ovals of Symmetries of Riemann Surfaces

Throughout this chapter we will be involved in several enumerations of ovals of the symmetries of a Riemann surface S. Recall that by an oval of a symmetry σ we mean a connected component of the fixed point set Fix(σ) of σ. The first section is short but crucial for the rest of the monograph. Theorem 3.1.1 allows us to find the number σ of ovals of σ from the algebraic structure of the full automorphism group Aut(S) and from the topological features of the action of Aut(S) on S. It was established for the first time in [49]. In Sect. 3.2 we present the current state of a problem investigated by many authors throughout the years, namely, to calculate the maximal number of ovals of a fixed number k of non-conjugate symmetries of a Riemann surface of genus g. Section 3.3 concerns the total number of ovals of all symmetries of a Riemann surface. We present Gromadzki’s upper bound [49] for the invariant ν(g), defined as the maximum of the number of ovals of S, where S runs over all compact Riemann surfaces of genus g. The fourth and last section of this chapter is devoted to the study of pairs of symmetries of Riemann surfaces, a topic on which much work has been done. The reader may find in [55] an upper bound for the total number of ovals of two symmetries of a surface S in terms of the genus of the surface, the order of their product and the number of points fixed by this product.

3.1 Enumeration of Ovals of Symmetries at Large We begin by showing how to calculate the number of ovals of a symmetry σ of a Riemann surface S in terms of the group G = Aut(S) and the topological features of the action of G on S. This number of ovals will be denoted by σ. Let us write S = H/Γ and G = Λ/Γ, where Λ is a proper NEC group containing the surface Fuchsian group Γ as a normal subgroup. Let θ : Λ → G with ker θ = Γ be the canonical smooth epimorphism. Recall that if the symmetry σ fixes points then it is conjugate to θ(ci ) for some canonical reflection ci of Λ, see Lemma 1.5.9. The next theorem, originally proved in [49], gives a formula to calculate σ. In this formula the centralizer of an element h in an abstract group H is denoted by C(H, h). E. Bujalance et al., Symmetries of Compact Riemann Surfaces, Lecture Notes in Mathematics 2007, DOI 10.1007/978-3-642-14828-6 3, c Springer-Verlag Berlin Heidelberg 2010

33

34

3 Counting Ovals of Symmetries of Riemann Surfaces

Theorem 3.1.1. With the above notations, the number of ovals of a symmetry σ with fixed points is σ = [ C(G, θ(c)) : θ(C(Λ, c)) ], where c runs over all non-conjugate canonical reflections of Λ whose images under θ are conjugate in G to σ. Proof. Let us write σ = Γσ /Γ for some NEC subgroup Γσ of Λ. On the one hand, Fix(σ) is homeomorphic to the boundary of S/σ, whilst on the other hand the number of ovals of Fix(σ) coincides with the number of empty period cycles of s(Γσ ), because the Riemann surfaces S/σ and H/Γσ are isomorphic. Hence, we have to count the reflections of Λ belonging to Γσ but non-conjugate there. Observe that Γσ = θ−1 (σ). So, as σ has fixed points, it is conjugate to θ(ci ) for some canonical reflection ci of Λ; without loss of generality we can assume that θ(ci ) = σ because conjugate symmetries have the same number of ovals. Now, given w ∈ Λ, its conjugate cw i belongs to Γσ if and only if w belongs to θ−1 (C(G, θ(ci ))), the inverse image of the centralizer of θ(ci ) in G, which we shall denote by Ci . In particular we see that Ci normalizes Γσ and so, for v, w ∈ Ci , the −1 reflections cvi and cw v ∈ C(Λ, ci )Γσ . i of Γσ are conjugate in Γσ if and only if w As a consequence, the conjugates of ci give rise to [Ci : C(Λ, ci )Γσ ] = [C(G, σ) : θ(C(Λ, ci ))] empty period cycles in Γσ . −1 Let now cw = Ci . Furj ∈ Γσ for some j = i and w ∈ Λ. Then wCj w v −1 ∈ Ci . Finally, given u, u ∈ Ci and v = thermore, cj ∈ Γσ if and only if vw uw, v = u w, the reflections cvj and cvj are conjugate in Γσ if and only if v −1 v ∈ C(Λ, cj )w−1 Γσ w = C(Λ, cj )Γ, which means that u−1 u ∈ wC(Λ, cj )Γw−1 . So the conjugates of cj give rise to [Ci : wC(Λ, cj )Γw−1 ] = [Cj : C(Λ, cj )Γ] = [C(θ(Λ), θ(cj )) : θ(C(Λ, cj ))] empty period cycles in Γσ , and therefore the result follows.

In what follows the index wi = [C(G, θ(ci )) : θ(C(Λ, ci ))] will be called the contribution of ci to σ. Usually we shall say that ci contributes to σ with wi ovals.

3.2 Total Number of Ovals of Non-Conjugate Symmetries In this section we shall find a bound for the maximum number of ovals that k nonconjugate symmetries of a Riemann surface can admit. For non-conjugate separating symmetries such a bound was found by Natanzon [96] using methods different from ours. Our result is more general as we find here the bound without any assumption

3.2 Total Number of Ovals of Non-Conjugate Symmetries

35

on separability. However, it works only for k ≥ 9. As to the cases k ≤ 9 we should mention the article by Gromadzki and Izquierdo [54], where they showed that the number of ovals that three non-conjugate symmetries of a Riemann surface of even genus g can admit does not exceed 2g + 3, and the article by Gromadzki and Kozłowska-Walania [56], with an analogous study for k = 5, 6, 7 and 8. The next lemma is clearly important in virtue of Theorem 3.1.1. Lemma 3.2.1. Let y1 and y2 be two involutions of a finite group G and let n denote the order of their product. Then, the order of the centralizer C(G, yi ) of yi in G does not exceed 2|G|/n for i = 1, 2. Proof. Let H be the dihedral group generated by y1 and y2 and observe first that C(H, yi ) equals either Z2 or Z2 ⊕ Z2 , according to n being odd or even. Fix a system R of representatives for the cosets of G/H. Then each element g of G can be represented as g = yx for some y ∈ H and x ∈ R, both uniquely determined. Now assume that two elements g = yx and g = y x belong to C(G, yi ). Then y y −1 ∈ H and y y −1 = g g −1 ∈ C(G, yi ). Thus g g −1 ∈ C(H, yi ) and so the lemma follows because the order of C(H, yi ) does not exceed 4 for i = 1, 2. We now prove the following elementary lemma of a combinatorial nature. Lemma 3.2.2. Assume that k ≥ 3 labels are effectively used to label s points situated on a circle in such a way that there is no pair of consecutive points with the same label. Then at least k − 1 points have neighbours with different labels. Proof. We shall prove the lemma by induction on s. As the cases s = 3 and s = 4 are trivial, we assume that s ≥ 5. There is nothing to prove if no point has neighbours with the same label; in this case the s points have neighbours with different labels. So assume that there are three consecutive points i − 1, i and i + 1, say with labels 1, k and 1 respectively, and consider the configuration T = {1, . . . , i − 1, i + 2, . . . , s} obtained by deleting the points i and i + 1. Assume first that at least one of these points has label k. Then, by the inductive hypothesis, t ≥ k − 1 points have neighbours with different labels. If, in the new configuration, the point i − 1 has neighbours with the same label then in the former configuration these t points have neighbours with different labels whilst, if i − 1 has neighbours with different labels then in the former configuration t − 1 of these points and one among the points i − 1 and i + 1 has neighbours with different labels. If none of the points in the set T has label k then we have a configuration of s − 2 points on a circle labeled by k − 1 labels. For k = 3, necessarily s is even and the points i − 1 and i + 1 have neighbours with different labels in the initial configuration. So assume that k > 3. Then, by the inductive hypothesis, k − 2 points of T have different labels. This way the assertion follows because in this case these points and the point i + 1 have neighbours whose labels in the former configuration are different. We are in a position to explain the main result in [50], which gives a bound for the maximal number of ovals that k non-conjugate symmetries of a Riemann surface can admit.

36


Theorem 3.2.3. Let σ1 , . . . , σk , with k ≥ 9, be non-conjugate symmetries with fixed points of a Riemann surface S of genus g ≥ 2 such that G = Aut(S) is a 2-group generated by σ1 , . . . , σk . Then σ1 + · · · + σk ≤ 2(g − 1) +

9−k |G|. 8

Proof. Let us write S = H/Γ and G = Λ/Γ where Γ is a surface Fuchsian group and Λ is a proper NEC group containing Γ as a normal subgroup. Let θ : Λ → G be the canonical epimorphism. We write the signature of Λ as s(Λ) = (g ; ±; [m1 , . . . , mr ]; {C1 , . . . , Cm , (−), . . ., (−)}),

(3.1)

where Ci = (ni1 , . . . , nisi ) and denote s = s1 + · · · + sm . Observe that every proper period mi and every link period nij are powers of 2 because G is a 2-group. Observe also that the images under θ of all canonical reflections of Λ generate G. For short, the sum σ1 + · · · + σk will be denoted by Σ. Assume first that none of σ1 , . . . , σk is central in G. Then |C(G, σi )| ≤ |G|/2 for i ≤ k. So any canonical reflection of Λ corresponding to an empty period cycle of s(Λ) contributes with at most |G|/4 ovals to Σ, by Theorem 3.1.1 and part (3) of Lemma 1.3.1. On the other hand, a reflection corresponding to a non-empty period cycle of s(Λ) contributes with at most |G|/8 ovals to Σ, by Theorem 3.1.1 and part either (1) or (2) of Lemma 1.3.1. Hence Σ ≤ (2 + s)|G|/8. Now, as Area(Λ) ≥ 2π(m + − 2 + s/4), we get, by the Hurwitz–Riemann formula, that g − 1 ≥ (4m + 4 − 8 + s)|G|/8. Moreover, we claim that 8m + 6 + s > k + 7. Indeed, since k ≤ + s (see Remark 1.1.6), we have 8m + 6 + s − k − 7 ≥ 8m + 5 − 7, which is positive since for m = 0 also s = 0 and so ≥ k ≥ 9. This way we get Σ≤

(2 + s)|G| (8m + 8 − 16 + 2s)|G| (16 − 6 − 8m − s)|G| = + 8 8 8 (9 − k)|G| . ≤ 2(g − 1) + 8

From now we will assume that some of the symmetries σ1 , . . . , σk , say z, is a central element of G. Let us prove now that the theorem holds true if m = 0. In this case, by Theorem 3.1.1 and part (3) of Lemma 1.3.1, we have Σ ≤ |G|/2, and, by the Hurwitz–Riemann formula, 2g − 2 ≥ |G|( − 2). Consequently, |G| (4 − )|G| = |G|( − 2) + 2 2 (9 − k)|G| (16 − 4)|G| < 2(g − 1) + ≤ 2(g − 1) + 8 8

Σ≤

because 4 − k > 7 as ≥ k ≥ 9. Thus, in what follows we assume m > 0.


37

Let us show now that we can assume that = 0. Suppose that = 0 and consider an NEC group Λ with signature . . ., (−)}) s = (g ; ±; [m1 , . . . , mr ]; {(2, 2, 2, 2, n11, . . . , n1s1 ), C2 , . . . , Cm , (−), −1 which has one empty period cycle less than s(Λ). By a little abuse of the notations, for the sake of technical simplicity, we denote in the same way as in the group Λ some of the canonical generators of Λ ; namely those generators which correspond to “pieces” of the signature of Λ in the signature of Λ and for the sake of terminological convenience we shall refer to these generators of Λ as old generators. To be more precise, this means that the hyperbolic generators of Λ are a1 , b1 , . . . , ag , bg or d1 , . . . , dg , according to whether sign(s ) = “+” or sign(s ) = “−”, the elliptic generators are x1 , . . . , xr , the generators corresponding to the first non-empty period cycle are e1 , c0 , c1 , c2 , c3 , c10 , c11 , . . . , c1s1 , the generators corresponding to the remaining non-empty period cycles are ei , ci0 , ci1 , . . . , cisi , whilst the generators corresponding to empty period cycles are em+1 , cm+1 , . . . , em+−1 , cm+−1 . Furthermore, according to this convention, c0 , c1 , c2 and c3 , are the new generators, whilst the remaining are the old ones. We may assume, without loss of generality, that θ(c10 ) = z and we shall consider the following two cases separately: (1) θ(cm+ ) = z,

(2) θ(cm+ ) = z.

Case 1: θ(cm+ ) = z. We define an epimorphism θ : Λ → G via the following assignment: on the old canonical generators but e1 , the map θ acts as θ, and we define θ (e1 ) = θ(e1 · · · em+ )θ(e2 · · · em+−1 )−1 , θ (c0 ) = θ (e1 c1s1 e−1 1 ), θ (c1 ) = θ (c3 ) = z, and θ (c2 ) = θ(cm+ ).

Then Γ = ker θ is a surface Fuchsian group. Indeed, by Theorem 1.2.2 its signature has no proper periods, by Theorem 1.2.5 it has no link periods, and finally, by Theorem 1.2.1, its sign is “ + ”. Let S = H/Γ . As Area(Λ) = Area(Λ ) both S and S have the same genus. Let Σ be the sum of the number of ovals of representatives of all conjugacy classes of symmetries with fixed points of the surface S . We shall prove now that Σ ≤ Σ . As the images under θ of all, except c10 , old canonical reflections corresponding to non-empty period cycles and their neighbours coincide with their images under θ it follows, from Theorem 3.1.1 and either (1) or (2) of Lemma 1.3.1, that each of these reflections contributes with at least as many ovals to Σ as to Σ. Similarly, by Theorem 3.1.1 and part (3) of Lemma 1.3.1, old reflections corresponding to empty

38


period cycles contribute to Σ with at least as many ovals as to Σ. Hence, to prove the inequality Σ ≤ Σ , it suffices to show that c10 , c0 , c1 , c2 and c3 contribute all together to Σ with at least as many ovals as cm+ and c10 contribute to Σ. Let w10 denote the contribution of c10 to Σ. Then c10 contributes to Σ either with w10 or w10 /2 ovals according to whether θ(c10 c11 )n11 /2 = z or not. Similarly, c0 contributes to Σ either with w10 or w10 /2 ovals according to whether θ(c1s1 −1 c1s1 )n1s1 /2 = z or not. Consequently, the reflections c10 and c0 contribute to θ (c10 ) with at least as many ovals as c10 contributes to θ(c10 ). Assume now that cm+ contributes with wm+ ovals to θ(cm+ ). Then c2 contributes to Σ with either wm+ ovals if θ(em+ ) = 1 or wm+ /2 ovals if θ(em+ ) = 1. In the first case we are done. In the second one, let n and n be the orders of the products θ (c0 )θ (c2 ) and θ (c2 )θ (c10 ), respectively, and let n = max{n , n }. Then, by Lemma 3.2.1, the order of the centralizer of θ(cm+ ) is not bigger than 2|G|/n and so cm+ contributes to Σ with at most |G|/n ovals, i.e., wm+ ≤ |G|/n, whilst now c1 and c3 contribute with |G|/4n + |G|/4n ≥ |G|/2n ≥ wm+ /2 ovals to z. This proves the inequality Σ ≤ Σ also in this case. Case 2: θ(cm+ ) = z. We define an epimorphism θ : Λ → G as follows; on c0 and on all old canonical generators, θ acts as in the previous case. Moreover, we define θ (c1 ) = θ (c3 ) = θ(cm+ ) and θ (c2 ) = θ(c10 ). Using again Theorems 1.2.1, 1.2.2 and 1.2.5 it follows straightforwardly that Γ = ker θ is a surface Fuchsian group and hence S = H/Γ is a Riemann surface. Its genus coincides with that of S by the Hurwitz–Riemann formula. Let Σ denote again the sum of the numbers of ovals of representatives of all the conjugacy classes of symmetries with fixed points of the surface S . As in the precedent case we shall prove that Σ ≤ Σ . Also here each old canonical reflection but c10 contributes to Σ with as many ovals as it contributes to Σ. The new reflection c2 contributes to Σ with no less ovals than c10 contributes to Σ. Moreover, cm+ contributes to θ(cm+ ) with either |G|/4 or |G|/2 ovals according to θ(em+ ) = 1 or θ(em+ ) = 1. In the first case we see that Σ ≤ Σ as also c3 contributes with |G|/4 ovals to Σ . If θ(em+ ) = 1, then θ (e1 ) = θ(e1 ). Hence, in this case θ (c0 ) = θ(c10 ) and therefore both c1 and c3 contribute with |G|/4 ovals to Σ . Thus, again Σ ≤ Σ . This way we have substituted Λ by another NEC group Λ whose signature has one period cycle less that s(Λ) and Σ ≤ Σ . After repeating the argument we can assume from the very beginning that Λ has no empty period cycles, i.e., its signature has the form s(Λ) = (g ; ±; [m1 , . . . , mr ]; {(n11 , . . . , n1s1 ), . . . , (nm1 , . . . , nmsm )}). The next step is to show that, actually, we can assume that m = 1, that is, Λ has just one period cycle. Suppose that m > 1. Observe first that we can assume that θ(c1s1 ) = z and θ(c20 ) = z. Let Λ be an NEC group with signature (g ; ±; [m1 , . . . , mr ]; {(n11 , . . . , n1s1 , 2, 2, n21 , . . . , n2s2 , 2, 2), C3 , . . . , Cs }).


39

Observe that s(Λ ) has one period cycle less than s(Λ). Moreover, the reflections of Λ corresponding to the first period cycle are c10 , . . . , c1s1 , c0 , c20 , . . . , c2s2 , c1 , c2 and it is easily seen that Area(Λ) = Area(Λ ). Let us define an epimorphism θ : Λ → G via the following assignment: except on e1 , the epimorphisms θ and θ coincide when acting on all old canonical generators and, furthermore, we define θ (e1 ) = θ(e1 )θ(e2 ), θ (c0 ) = θ (c1 ) = z and θ (c2 ) = θ (e−1 1 )θ(c10 )θ (e1 ).

Using once more Theorems 1.2.1, 1.2.2 and 1.2.5 we realize that Γ = ker θ is a surface Fuchsian group. Then S = H/Γ is a Riemann surface of genus g. Let Σ be the sum of the numbers of ovals of representatives of all the conjugacy classes of symmetries with fixed points of the surface S . Let us check that Σ ≤ Σ . Indeed, all old canonical reflections, but c10 and c20 , contribute to Σ with at least as many ovals as they contribute to Σ. Let us denote by wiΣ the contribution of ci0 to Σ. Then wiΣ = qi /4ki , where qi is the order of the centralizer of θ(ci0 ) in G and ki is the order of nisi /2 the product θ(ci0 ci1 )ni1 /2 θ(e−1 ei ) for i = 1, 2. In particular i (cisi −1 cisi ) Σ wi ≤ qi /4. On the other hand, since the elements θ (c10 c11 )n11 /2 θ (e1 c1 c2 e−1 1 ) and θ (c1s1 −1 c1s1 )n1s1 /2 θ (c1s1 c0 ) have order 2, we see that c10 and c1s1 contribute with no less ovals to Σ than c10 contributes to Σ. Similarly, it is easy to check that c20 and c2s2 contribute to Σ with no less ovals than c20 contributes to Σ. This proves our claim Σ ≤ Σ . Therefore, after repeating the process, we conclude that it is sufficient to deal with the case of an NEC group Λ with signature s(Λ) = (g ; ±; [m1 , . . . , mr ]; {(n1 , . . . , ns )}). Observe that the Hurwitz–Riemann formula yields s ≤ 8(g − 1)/|G| + 4.

(3.2)

Let c0 , . . . , cs denote the corresponding canonical reflections. We may assume that θ(c0 ) is a central symmetry of S and so in particular θ(c0 ) = θ(cs ). Consider c0 , . . . , cs−1 as s points on a circle labelled by the symbols θ(c0 ), . . . , θ(cs−1 ) respectively. By Lemma 3.2.2, at least for k − 1 indices in range 0 ≤ i1 < · · · < ik−1 ≤ s − 1, we have θ(cij −1 ) = θ(cij +1 ), where the indices are taken modulo s. Now, if nij > 2 or nij +1 > 2 then θ(cij ) is not central in G, and so |C(G, θ(cij ))| ≤ |G|/2. Therefore cij contributes to Σ with at most |G|/8 ovals. If both nij = nij +1 = 2 then |θ(C(Λ, cij ))| ≥ 8 and thus also now cij contributes with at most |G|/8 ovals to Σ. The remaining canonical reflections contribute to Σ with no more than |G|/4 ovals. Consequently, using inequality (3.2) we get (k − 1)|G| (s − k + 1)|G| s|G| (1 − k)|G| + = + 8 4 4 8 (9 − k)|G| (1 − k)|G| = 2(g − 1) + . ≤ 2(g − 1) + |G| + 8 8

Σ≤

This completes the proof.

40


Remark 3.2.4. We will see in Theorem 3.2.6 that the bound obtained in Theorem 3.2.3 is sharp for k ≥ 9. However, the proof of Theorem 3.2.3 works for k ≥ 3, as is easy to see. Corollary 3.2.5. Let σ1 , . . . , σk , with k ≥ 9, be non-conjugate symmetries with fixed points of a Riemann surface S of genus g ≥ 2. Then σ1 + · · · + σk ≤ 2(g − 1) + 2r−3 (9 − k), where r is the smallest positive integer for which k ≤ 2r−1 . Proof. As we are looking for the number of ovals of {σ1 , . . . , σk } and conjugate symmetries have the same number of ovals, we can assume, using Sylow theorem, that {σ1 , . . . , σk } generate a 2-subgroup G of Aut(S). Clearly, k ≤ |G|/2 and, moreover, |G| ≥ 2r by the definition of r. Write S = H/Γ and G = Λ/Γ for a surface Fuchsian group Γ and a proper NEC group Λ containing it as a normal subgroup. Write the signature of Λ as (3.1) in the proof of Theorem 3.2.3. As s + ≥ k ≥ 9, it follows from [22, Theorem 2.4.7] and [41] that s(Λ) is a maximal NEC signature. Hence, by [22, Theorem 5.1.2], there exists a maximal NEC group Λ isomorphic to Λ, say via ϕ : Λ → Λ . Let S = H/Γ , where Γ = ϕ(Γ). Then Aut(S ) = Λ /Γ and ϕ induces a group isomorphism ϕ : Λ/Γ → Λ /Γ . Therefore τ1 = ϕ(σ 1 ), . . . , τk = ϕ(σ k ) are nonconjugate symmetries of S . The group generated by each σi can be written as σi = Λi /Γ for some proper NEC subgroup Λi of Λ containing Γ as a subgroup of index 2. Here σi is the number of period cycles of s(Λi ). Thus σi = τi because τi = ϕ(Λi )/Γ . Therefore G ∼ = Aut(S ) is a 2-group and, by Theorem 3.2.3, σ1 + · · · + σk = τ1 + · · · + τk ≤ 2(g − 1) + (9 − k)|G|/8. Since |G| ≥ 2r and 9 − k ≤ 0, the corollary follows.

The next result shows that the bound obtained in Corollary 3.2.5 is attained for infinitely many values of g. We present here the proof given in [50]. Theorem 3.2.6. Let k ≥ 9 be an integer and let r be the smallest positive integer satisfying k ≤ 2r−1 . Then, for each t ≥ k − 3 there exists a Riemann surface S of genus g = 2r−2 t + 1 having k non-conjugate symmetries σ1 , . . . , σk with fixed points such that σ1 + · · · + σk = 2(g − 1) + 2r−3 (9 − k). Proof. Let G = Zr2 = z1 ⊕ · · · ⊕ zr , and let Λ be a maximal NEC group with s(Λ) = (0; +; [−]; {(2, .2s . ., 2)}), where s = t + 2 ≥ k − 1. Let {a1 , . . . , a2r−1 } be all elements of order 2 in G which have odd length with respect to the set

3.3 Total Number of Ovals of all Symmetries of a Riemann Surface

41

of generators {z1 , . . . , zr }, and assume that a1 , . . . , ar generate G. Since r is the smallest positive integer such that k ≤ 2r−1 we deduce that r ≤ k and so the assignment ⎧ ⎪ ⎪ ⎨a1 θ(ci ) = aj+2 ⎪ ⎪ ⎩a k

for i = 2j,

0 ≤ j ≤ s,

for i = 2j + 1,

0 ≤ j ≤ k − 2,

for i = 2j + 1,

k − 1 ≤ j ≤ s − 1,

induces an epimorphism θ : Λ → G whose kernel Γ is a surface Fuchsian group. Hence S = H/Γ is a Riemann surface having k non-conjugate symmetries with fixed points. Its genus, by the Hurwitz–Riemann formula, equals 2r−2 t + 1. We see that for each 0 ≤ j ≤ k − 2 the reflection c2j contributes with 2r−3 ovals to a1 , whilst each one of the remaining 2s − k + 1 non-conjugate canonical reflections of Λ contributes with 2r−2 ovals to the corresponding symmetry. As a result σ1 + · · · + σk = 2r−3 (k − 1) + 2r−2 (2s − k + 1) = 2r−1 s + 2r−3 (1 − k) = 2(g − 1) + 2r + 2r−3 (1 − k) = 2(g − 1) + 2r−3 (9 − k), which is the equality we were looking for.

3.3 Total Number of Ovals of all Symmetries of a Riemann Surface A simple closed curve on a Riemann surface S is said to be an oval of S if it is an oval of some symmetry of S. Let S be the number of ovals on S and let ν(g) be the maximum of S where S runs over all Riemann surfaces of genus g. Using topological methods, Natanzon proved in [100] and [101], that ν(g) ≤ 42(g − 1). Here, using combinatorial methods as before, we sharpen this result. To begin with, we state and prove the following Lemma 3.3.1, which is a consequence of part (2) in Lemma 1.3.3. Lemma 3.3.1. Let Λ be an NEC group with triangle signature (2k, 2 + 1, 2m) and let {c0 , c1 , c2 } be a system of canonical generators of Λ. Then C(Λ, c0 ) = c0 ⊕ (c0 c1 )k ∗ (c0 c2 )m ; C(Λ, c1 ) = c1 ⊕ (c0 c1 )k ∗ (c2 c1 ) (c0 c2 )m (c1 c2 ) . Proof. By choosing c0 = c2 , c1 = c0 and c2 = c1 we get a system of canonical generators for the presentation of Λ associated to the signature (2m, 2k, 2 + 1). So the lemma follows from part (2) in Lemma 1.3.3.

42


In the same vein as Theorem 3.1.1 we have the following result, see [49]. Theorem 3.3.2. Let S be a Riemann surface represented as H/Γ for some surface Fuchsian group Γ. Let Λ be an NEC group and let θ : Λ → Aut(S) be a group epimorphism with ker θ = Γ. Then

(1) S = [Aut(S) : θ(C(Λ, c))], where c runs over a set of representatives of conjugacy classes of canonical reflections of Λ. (2) 2S ≤ k |Aut(S)|, where k is the number of non-conjugate canonical reflections of Λ. (3) Suppose that every period cycle of Λ has an even link period. Then 4S ≤ t |Aut(S)|, where t is the number of even link periods of s(Λ). Proof. (1) Each oval of S corresponds to a reflection of Λ and so to the conjugate cw j of some canonical reflection cj ∈ Λ by some w ∈ Λ. Observe that if the reflections cj and c are conjugate in Λ then for each w ∈ Λ there exists v ∈ Λ v such that cw = cj . Thus, to calculate S we have to count down the ovals contributed by elements of the form cw , where c runs over a set of representatives of conjugacy classes of canonical reflections of Λ. Moreover, given v, w ∈ Λ one can show, as in the proof of Theorem 3.1.1, v −1 that cw w ∈ C(Λ, cj )Γ. j and cj give rise to the same oval if and only if v Consequently, the reflections conjugate to cj give rise to exactly [Λ : C(Λ, cj )Γ] = [Λ/Γ : C(Λ, cj )Γ/Γ] = [θ(Λ) : θ(C(Λ, cj )] distinct ovals. Finally, as the contributions of non-conjugate canonical reflections cj and cj are disjoint, the proof of the first part is finished. (2) This is an evident consequence of part (1), since each group θ(C(Λ, c)) is a proper subgroup of Aut(S). (3) As two canonical reflections of Λ corresponding to odd link periods are conjugate, in the formula of part (1) we have to sum up over all canonical reflections corresponding to even link periods. For each such reflection c let c be a consecutive canonical reflection such that cc has even order n. Then both c and (cc )n/2 are in C(Λ, c). Moreover, θ(c) = 1 = θ(cc )n/2 since Γ has no elements of finite order. Furthermore, θ(c) = θ(cc )n/2 since otherwise the product (cc )n/2 c would be an orientation reversing element of Γ. Therefore |θ(C(Λ, c))| ≥ 4 and so the result follows immediately from part (1). We shall use Theorem 3.3.2 to estimate an upper bound for the function ν. Throughout the rest of this section we write the genus g surface as S = H/Γ for some surface Fuchsian group Γ and denote G = Aut(S). Moreover, we denote by θ : Λ → G a group epimorphism with ker θ = Γ, where Λ is an NEC group. (3.3.3) A first bound for S in terms of Λ and g. Let us write the signature of Λ as s(Λ) = (g ; ±; [m1 , . . . , mr ]; {C1 , . . . , Cm , (−), . . ., (−)}),


43

where Ci = (ni1 , . . . , nisi ), and denote s = s1 + · · · + sm . Thus, Area(Λ) ≥ 2π(ηg + m + − 2 + r/2 + s/4)

(3.3)

and therefore |G| ≤

ηg

2(g − 1) . + 3/4 + m − 2 + r/2 + ( + s)/4

So by part (2) in Theorem 3.3.2 we get S ≤

(g − 1)( + s) |G|( + s) ≤ . 2 ηg + 3/4 + m − 2 + r/2 + ( + s)/4

(3.4)

We are ready to present Gromadzki’s proof, see [49], of the inequality ν(g) ≤ 12(g −1) for most values of the genus g ≥ 2. More precisely, we have the following. Theorem 3.3.4. Let g ≥ 4 be an integer with g ∈ / {5, 7, 9}. Then ν(g) ≤ 12(g − 1). Moreover, ν(g) = 12(g − 1) if g = 8m2 + 1 with m ≥ 2. Finally, ν(2) = 24, ν(3) = 36, ν(5) = 72, ν(7) = 126 and ν(9) = 100. Proof. Using the previous notations, it is enough to study the case ηg + 3/4 + m+ r/2 < 2 since otherwise, by (3.4), we have S ≤ 4(g − 1). In particular ηg < 2. If ηg = 1 then m = 0 and, since ≥ 1, we get r = 0 and = 1, i.e., Area(Λ) = 0, an absurdity. Thus g = 0 and so 3 + 4m + 2r < 8. In particular < 3 and if = 2 then m = r = 0, i.e., Area(Λ) = 0, an absurdity. Suppose that = 1, which implies 2m + r ≤ 2, and in particular either m = 0 or m = 1. In the first case r = 2 and either m1 > 2 or m2 > 2, because Area(Λ) > 0. But in such a situation Area(Λ) ≥ π/3 and so S ≤

|G| 4π(g − 1) ≤ ≤ 12(g − 1). 2 Area(Λ)

Therefore, if = 1 then also m = 1 and this implies, by (3.3), Area(Λ) ≥ sπ/2. Consequently, 4π(g − 1) S ≤ ≤ 8(g − 1). Area(Λ) So we may assume that = 0 and 2m + r < 4. But m + = k ≥ 1, that is m = 1 and r ≤ 1. We conclude that it is enough to study the case in which s(Λ) = (0; +; [m1 , . . . , mr ]; {(n1 , . . . , ns )}), where r ≤ 1.

44


First, let s = 1. Then r = 1 because Area(Λ) > 0 and so s(Λ) = (0; +; [m]; {(n)}). Note that Area(Λ) = 2π(1/2 − 1/m − 1/2n), which implies m ≥ 3. If n = 2 then m ≥ 5 and thus Area(Λ) ≥ π/10. Then, by part (3) in Theorem 3.3.2, S ≤

|G| 4π(g − 1) ≤ = 10(g − 1). 4 4π/10

If n = 3 then m ≥ 4 and so Area(Λ) ≥ π/6. Thus |G| ≤ 24(g − 1) and therefore S ≤ 12(g − 1) by part (2) in Theorem 3.3.2. If n = 4 then m ≥ 3. So Area(Λ) ≥ π/12 and consequently S ≤ |G|/4 ≤ 12(g − 1). Now let n = 5. Then, for m ≥ 4 we get S ≤

|G| 20 ≤ (g − 1). 2 3

For n = 5 and m = 3, |G| = 30(g − 1). Let c0 and c1 be the canonical reflections of Λ and let e be the connecting generator. Notice that c0 , e(c0 c1 )2 ∈ C(Λ, c0 ) and θ(c0 ) = θ(e(c0 c1 )2 ). Furthermore, θ(e(c0 c1 )2 ) = 1, because |θ(e)| = 3 while |θ((c0 c1 )2 )| = 5. Hence, |θ(C(Λ, c0 ))| ≥ 4 and therefore S ≤

|G| < 8(g − 1). 4

For n = 6 we have S ≤ |G|/4 ≤ 6(g − 1). Finally, S ≤ |G|/2 < 11(g − 1) for n ≥ 7. This finishes the analysis of the case s = 1. Now let s = 2. Again r = 1, because Area(Λ) > 0, and so s(Λ) = (0; +; [m]; {(n1 , n2 )}). Note that in this case

1 1 1 1 − . + Area(Λ) = 2π 1 − m 2 n1 n2

If m ≥ 3 then Area(Λ) ≥ π/3 and so, by part (2) in Theorem 3.3.2, S ≤ |G| =

4π(g − 1) ≤ 12(g − 1). Area(Λ)

Assume then that m = 2. If both n1 , n2 are odd then Area(Λ) ≥ π/3 and again by part (2) in Theorem 3.3.2, S ≤ |G| =

4π(g − 1) ≤ 12(g − 1). Area(Λ)


45

If n1 is even and n2 is odd then Area(Λ) ≥ π/6. Therefore, by (3) in Theorem 3.3.2, S ≤ |G|/2 ≤ 12(g − 1) because Area(Λ) > 0. In such a case Area(Λ) ≥ π/4 and this implies S ≤ |G|/2 ≤ 8(g − 1). Thus, also the case s = 2 is finished. The case s = 3 is more complicated and will be considered later on. Assume now that s ≥ 4. Then for r = 1, Area(Λ) ≥ π(s − 2)/2. Thus, |G| ≤ 8(g − 1)/ (s − 2) and therefore S ≤

s|G| 4s ≤ (g − 1) ≤ 8(g − 1). 2 s−2

Hence we can assume r = 0, i.e., s(Λ) = (0; +; [−]; {(n1 , . . . , ns )}), which we shall abbreviate as s(Λ) = (n1 , . . . , ns ), and whose area is

s 1 Area(Λ) = π s − 2 − n i=1 i

.

If all ni are odd then Area(Λ) ≥ 2π/3. Consequently, |G| ≤ 6(g − 1) and S ≤ 3(g − 1). Thus, assume that some ni is even and observe that, as in this case Area(Λ) ≥ π(s − 4)/2, one has |G| ≤ 8(g − 1)/(s − 4) and therefore, for s > 5, and by (3) in Theorem 3.3.2, S ≤

2s (g − 1) ≤ 10(g − 1). s−4

So let s = 4 and let c0 , c1 , c2 , c3 be the set of canonical reflections of Λ. If each link period ni is odd then Area(Λ) ≥ 2π/3. Hence, |G| ≤ 6(g − 1) and so S ≤ 12(g − 1) by (2) in Theorem 3.3.2. If the number of odd link periods is either 2 or 3 then Area(Λ) ≥ π/3. Hence, |G| ≤ 12(g − 1) and thus S ≤ 6(g − 1), using (3) in Theorem 3.3.2 once more. So we can assume that at least three of the link periods, say n1 , n2 , n3 , are even. Now if n4 is odd and ni ≥ 4 for some i ≤ 3 then Area(Λ) ≥ 5π/12. Therefore |G| ≤ (48/5)(g − 1) and so S ≤ 3|G|/4 < 8(g − 1). If n4 is even and at least two of the link periods are ≥ 4 then Area(Λ) ≥ π/2, and consequently, by (3) in Theorem 3.3.2, S ≤ |G|/ ≤ 8(g − 1). Henceforth, for s = 4 it just remains to analyze the signature s(Λ) = (2, 2, 2, n) where, necessarily, n ≥ 3 because Area(Λ) > 0. Note that Area(Λ) = π(n − 2)/2n, which implies |G| =

8n (g − 1). n−2

Now the reflections c0 , c1 , c2 ∈ C(Λ, c1 ), and θ(c0 ) = θ(c1 ) = θ(c2 ) and θ(c0 c2 ) = θ(c1 ).

46


Moreover, θ(c0 ) = θ(c2 ) since otherwise (c0 c3 )2 ∈ Γ would be an element of finite order. Thus, |θ(C(Λ, c1 ))| ≥ 8 and in the same way one can prove that |θ(C(Λ, c2 ))| ≥ 8. Therefore, by Theorem 3.3.2, S ≤

3|G| 6n 2|G| 2|G| + = ≤ (g − 1) ≤ 12(g − 1) 8 4 4 n−2

for n ≥ 4. Also for n = 3, S ≤

|G| 4n 2|G| |G| + = ≤ (g − 1) = 12(g − 1). 8 4 2 n−2

Henceforth it only remains to study the case s = 3. If r = 1 then Area(Λ) ≥ π/2 and so, by (2) in Theorem 3.3.2, S ≤ 3|G|/2 ≤ 12(g − 1). So we have proved that S ≤ 12(g − 1) except for s(Λ) = (k, , m). Note that in this case 1 1 1 Area(Λ) = π 1 − − − . k m By [73] we can assume, without loss of generality, that k ≤ ≤ m. Let c0 , c1 , c2 be a system of canonical reflections for Λ. If k, and m are odd then Area(Λ) ≤ 4π/15 unless s(Λ) = (3, 3, 5). In the first cases, by (2) in Theorem 3.3.2, S ≤

|G| 15 ≤ (g − 1). 2 2

In the last case |G| = 30(g − 1). However, the centralizer C(Λ, c0 ) contains both c0 and (c2 c0 )2 (c1 c2 )(c0 c1 ). Furthermore, θ(c0 ) = θ((c2 c0 )2 (c1 c2 )(c0 c1 )), because otherwise Γ would contain an orientation reversing element. Moreover, since θ(c0 c2 ) = 1, it follows that θ((c2 c0 )2 (c1 c2 )(c0 c1 )) = 1. Thus |θ(C(Λ, c0 ))| ≥ 4 and therefore S ≤ (15/2)(g − 1), as all reflections of Λ are conjugate. Assume now that some of the integers k, , m is even. As before we can show, using part (3) in Theorem 3.3.2, that S ≤ 12(g − 1) for k ≥ 3. Thus, suppose that k = 2 and recall that we have assumed ≤ m. Observe first that for ≥ 8, Area(Λ) ≥ π/4 and so S ≤

3|G| ≤ 12(g − 1). 4

If = 7 then Area(Λ) ≥ 3π/14. Therefore |G| ≤ (56/3)(g − 1) and S ≤ 2|G|/4 < 10(g − 1). Let = 6. Then Area(Λ) = (m − 3)π/3m and this implies |G| =

12m (g − 1). m−3


47

Therefore, for odd m, part (2) in Theorem 3.3.2 implies that S ≤

6m (g − 1) < 11(g − 1). m−3

Let now m be even. Then, by (2) in Theorem 3.3.2, S ≤ 3|G|/4 ≤ 12(g − 1) whenever m ≥ 12. Thus, we just have to analyze the cases m ∈ {6, 8, 10}. Here we have c0 c1 , (c1 c2 )3 ∈ C(Λ, c1 ) and their images under θ are distinct, since otherwise θ(c0 c2 ) = θ(c1 c2 )4 , which is impossible, as the first of these two elements has order m while the second one has order 3. Furthermore, θ(c1 ) ∈ θ(c0 c1 ), θ(c1 c2 )3 . So |θ(C(Λ, c1 ))| ≥ 8. Similarly, one can show that |θ(C(Λ, c0 ))| ≥ 8. Thus S ≤

6m 2|G| |G| + = (g − 1) ≤ 12(g − 1). 8 4 m−3

For = 5, Area(Λ) = (3m − 10)π/10m, that is, |G| =

40m (g − 1). 3m − 10

If m is odd then m ≥ 5, hence S ≤ |G|/4 ≤ 10(g − 1). If m is even then S ≤ 2|G|/4 ≤ 12(g − 1) whenever m ≥ 8. Therefore we have to consider the case m = 6. Here |G| = 30(g − 1). Now c0 c1 , (c0 c2 )3 ∈ C(Λ, c0 ) and again their images under θ are distinct since otherwise θ(c2 c1 ) = θ(c0 c2 )2 which is impossible, as the first of these two elements has order 5 while the second one has order 3. As before we deduce that |θ(C(Λ, c0 ))| ≥ 8 and therefore S ≤

|G| |G| + < 12(g − 1). 8 4

So it remains to deal only with the cases = 3 and = 4. We shall see below that the group G of all automorphisms of a Riemann surface with a which has more than 12(g − 1) ovals is a factor group of some group G certain specific presentation which makes it finite. In virtue of the Hurwitz–Riemann imposes restrictions on the genera of the corresponding formula, the order of G Riemann surfaces. In most cases the calculation of the order is easily done. In one particular case, however, the use of a Computer Algebra System has been necessary. We acknowledge that all the results obtained have been checked by using the GAP Program (Groups, Algorithms and Programming) developed by J. Neubüsers group at Aachen [46].

48


For = 4, Area(Λ) = (m − 4)π/2m, that is, |G| =

16m (g − 1). m−4

If |θ(C(Λ, c1 ))| < 8 then, by either (1) or (2) in Lemma 1.3.3, θ(c0 c1 ) = θ(c1 c2 )2 , and therefore θ(c0 c2 ) = θ(c1 c2 )3 . Thus θ(c0 c2 )4 = 1, which implies that m divides 4. Hence Area(Λ) ≤ 0, which is false. Therefore |θ(C(Λ, c1 ))| ≥ 8. Now if m is even, similar arguments show that |θ(C(Λ, c0 ))| ≥ 8. So S ≤

8m 2|G| |G| + = (g − 1) ≤ 12(g − 1) 8 4 m−4

for m ≥ 12. For m = 10, |θ(C(Λ, c2 ))| ≥ 8, since otherwise θ(c0 c2 )5 = θ(c1 c2 )2 and so θ(c0 c2 )4 = θ(c1 )θ(c2 c0 )θ(c1 ), which implies θ(c0 c2 )5 = 1, an absurdity. Similarly, |θ(C(Λ, c2 ))| ≥ 8 for m = 6, 8. Therefore, for m = 8, 10, S ≤

3 |G| ≤ 12(g − 1). 8

Therefore, we just have to deal with the case m = 6. Observe that |G| = 48(g − 1) and S ≤ 18(g − 1), as we already know that |θ(C(Λ, ci ))| ≥ 8 for i = 0, 1, 2. If |θ(C(Λ, c1 ))| > 8 for each index i = 0, 1, 2, then |θ(C(Λ, ci ))| ≥ 12 and S ≤ 3|G|/12 = 12(g − 1). Assume then that |θ(C(Λ, ci ))| = 8 for some i = 0, 1, 2. If |θ(C(Λ, c0 ))| = 8 then, by part (1) of Lemma 1.3.3, θ(c0 c1 )θ(c0 c2 )3 is an element of order 2, and so G is a factor group of the group with presentation c0 , c1 , c2 | c20 , c21 , c22 , (c0 c1 )2 , (c1 c2 )4 , (c0 c2 )6 , ((c0 c1 )(c0 c2 )3 )2 , which can be easily shown to have order 96. So either g = 2 or g = 3. However, later on we shall show that there exist Riemann surfaces of genera 2 and 3 with 24 and 36 ovals, respectively. Now if |θ(C(Λ, c1 ))| = 8 then θ(c0 c1 )θ(c1 c2 )2 is an element of order 2 and thus G is a factor group of the group with presentation c0 , c1 , c2 | c20 , c21 , c22 , (c0 c1 )2 , (c1 c2 )4 , (c0 c2 )6 , ((c0 c1 )(c1 c2 )2 )2 , which again can be shown to have order 48 and so g = 2 in this case. Finally, assume that |θ(C(Λ, ci ))| > 8, for i = 0, 1 and |θ(C(Λ, c2 ))| = 8. Then S ≤ 2|G|/12 + |G|/8 = 14(g − 1). As before we claim that G is a factor group of the group with presentation c0 , c1 , c2 | c20 , c21 , c22 , (c0 c1 )2 , (c1 c2 )4 , (c0 c2 )6 , ((c1 c2 )2 (c0 c2 )3 )2 , which can be shown to have order 192. So the only values that g can attain in this case are g = 2, 3 or 5. But later on we shall see that there exists a Riemann surface of genus 5 with 72 ovals.


49

So let m be odd. Then, for m ≥ 9, S ≤

|G| |G| + < 11(g − 1) 8 4

and therefore the only cases left are m = 5 and m = 7. Assume first that m = 7. Then |θ(C(Λ, c0 ))| ≥ 8. Indeed, if this were not the case then by (2) of Lemma 1.3.3, θ(c0 c1 ) = θ(c2 c0 )3 θ(c1 c2 )2 θ(c0 c2 )3 = θ(c0 c2 )3 θ(c1 )θ(c0 c2 )θ(c1 )θ(c0 c2 )3 . Therefore θ(c1 c2 ) and θ(c0 c2 ) are conjugate, which is impossible as they have distinct orders. Consequently, S ≤

4m 2|G| ≤ (g − 1) < 10(g − 1). 8 m−4

Finally, let us assume that m = 5. Then, similarly as above, we can show that |θ(C(Λ, c0 ))| ≥ 8. We already know that |θ(C(Λ, c1 ))| ≥ 8 and we shall show that |θ(C(Λ, c1 ))| ≥ 16. Indeed, if this is not the case then θ(c0 c1 )θ(c1 c2 )2 has either order 2 or order 3. However, one can easily show that in the first case θ(c1 c2 )2 = 1 while in the second one θ(c1 c2 ) = 1, and both equalities are false. Moreover, if |θ(C(Λ, c1 ))| = 16 then S ≤ |G|/16 + |G|/8 = 15(g − 1) and G is a factor group of the group with presentation c0 , c1 , c2 | c20 , c21 , c22 , (c0 c1 )2 , (c1 c2 )4 , (c0 c2 )5 , ((c0 c1 )(c1 c2 )2 )4 , which can be shown to have order 320. Therefore g equals either 2, 3 or 5 in this case. Thus, assume that |θ(C(Λ, c1 ))| > 16. Then |θ(C(Λ, c1 ))| ≥ 20 and we shall look now for the order of θ(C(Λ, c0 )). If |θ(C(Λ, c0 ))| = 8 then S ≤ |G|/20 + with presentation |G|/8 = 14(g − 1) and G is a factor group of the group G c0 , c1 , c2 | c20 , c21 , c22 , (c0 c1 )2 , (c1 c2 )4 , (c0 c2 )5 , ((c0 c1 )(c2 c0 )2 (c2 c1 )2 (c0 c2 )2 )2 , which can be checked to have order 640. So here g = 2, 3, 5 or 9. For g = 2, 3, 5 there exist Riemann surfaces with more than 14(g − 1) ovals. So let g = 9. Then However, (c0 c1 )(c1 c2 )2 represents in G an element of order 8. Hence G = G. |θ(C(Λ, c1 ))| = 32 by (2) of Lemma 1.3.3 and thus for g = 9 there exists a Riemann surface having |G|/8 + |G|/32 = (25/2)(g − 1) = 100 ovals. Finally, if |θ(C(Λ, c0 ))| > 8 then |θ(C(Λ, c0 ))| ≥ 12, and this implies S ≤

|G| |G| + < 11(g − 1). 12 20

Thus we are led to analyze just what happens if s(Λ) = (2, 3, m), whose area is Area(Λ) = (m − 6)π/6m, and so |G| =

24m (g − 1). m−6

50


Let first m = 2m for some m ≥ 4. We shall show that |θ(C(Λ, ci ))| ≥ 8 for i = 0, 1. Indeed, if |θ(C(Λ, c0 ))| < 8 then, by Lemma 3.3.1, θ(c0 c1 ) = θ(c0 c2 )4 , which implies θ(c2 c1 ) = θ(c0 c2 )3 and so θ(c0 c2 )9 = 1, a contradiction. Now if |θ(C(Λ, c1 ))| < 8 then, again by Lemma 3.3.1, θ(c0 c1 ) = θ(c2 c1 )θ(c2 c0 )m θ(c1 c2 ). Using the defining relations for Λ we can easily show that

θ(c0 c2 )2 = θ(c1 )θ(c0 c2 )m +1 θ(c1 ). However, the last implies that θ(c0 c2 )4 = θ(c1 )θ(c0 c2 )2 θ(c1 ) and so θ(c0 c2 )6 = 1, which is false. Thus, for m ≥ 12, S ≤

2|G| ≤ 12(g − 1). 8

Therefore, in case of even m, it remains to deal with the cases m = 8 and m = 10. Assume first that m = 8. We know that |θ(C(Λ, ci ))| ≥ 8 for i = 0, 1, which implies S ≤ 2|G|/8 = 24(g − 1). If |θ(C(Λ, c0 ))| = 8 then G is a factor group with presentation of the group G c0 , c1 , c2 | c20 , c21 , c22 , (c0 c1 )2 , (c1 c2 )3 , (c0 c2 )8 , ((c0 c1 )(c0 c2 )4 )2 . as the last has order 96. Thus, here S has genus 2. FurtherConsequently, G = G, more, (c0 c1 )(c2 c1 )(c0 c2 )4 (c1 c2 ) represents in G an element of order 2 and so, in virtue of Lemma 3.3.1, |θ(C(Λ, c1 ))| = 8. Hence, there exists a Riemann surface of genus g = 2 with 2|G|/8 = 24 ovals. Assume therefore that |θ(C(Λ, c0 ))| > 8, which implies that |θ(C(Λ, c0 ))| > 12, since otherwise (θ(c0 c1 )θ(c0 c2 )4 )3 = 1 and from the defining relations for Λ we deduce that θ(c1 c2 ) = 1, a contradiction. Henceforth, |θ(C(Λ, c0 ))| ≥ 16. with Now, if |θ(C(Λ, c1 ))| = 8 then G is a factor group of the group G presentation c0 , c1 , c2 | c20 , c21 , c22 , (c0 c1 )2 , (c1 c2 )3 , (c0 c2 )8 , ((c0 c1 )(c2 c1 )(c0 c2 )4 (c1 c2 ))2 , which can be shown to have order 384. Thus, S has genus g = 2, 3 or 5 as |G| = 96(g − 1). We have already shown that there exists a Riemann surface of genus while (c0 c1 )(c0 c2 )4 represents in G g = 2 with 24 ovals. If g = 5 then G = G, an element of order 4 and thus, in virtue of Lemma 3.3.1, |θ(C(Λ, c0 ))| = 16. So there exists a Riemann surface of genus g = 5 with |G|/8 + |G|/16 = 18(g − 1) = 72 ovals. Finally, we shall show that there exists a Riemann surface of genus g = 3 with 18(g − 1) = 36 ovals. Indeed, one can show that the group G with presentation c0 , c1 , c2 | c20 , c21 , c22 , (c0 c1 )2 , (c1 c2 )3 , (c0 c2 )8 , [((c0 c1 )(c1 c2 ))2 , ((c1 c2 )(c0 c1 ))2 ]


51

has order 192. Moreover, (c0 c1 )(c0 c2 )4 and (c0 c1 )(c2 c1 )(c0 c2 )4 (c1 c2 ) represent in G elements of order 4 and 2, respectively, and thus, in virtue of Lemma 3.3.1, |θ(C(Λ, c0 ))| = 16 and |θ(C(Λ, c1 ))| = 8. Consequently, the corresponding surface has |G|/16 + |G|/8 = 18(g − 1) = 36 ovals. Hence we can assume that |θ(C(Λ, c1 ))| ≥ 12. Indeed, if |θ(C(Λ, c1 ))| ≥ 16 then 2|G| = 24 = 12(g − 1). S ≤ 16 Thus, let |θ(C(Λ, c1 ))| = 12. Then S ≤

|G| |G| + = 28 = 14(g − 1) 16 12

with presentation and G is a factor group of the group G c0 , c1 , c2 | c20 , c21 , c22 , (c0 c1 )2 , (c1 c2 )3 , (c0 c2 )8 , ((c0 c1 )(c2 c1 )(c0 c2 )4 (c1 c2 ))3 , which was checked, using the mentioned GAP program, to have order 4896. But using GAP once more we see that (c0 c1 )(c0 c2 )4 represents in G an element of order 9. Therefore |θ(C(Λ, c0 ))| ≥ 36 and so S =

|G| |G| + < 21 < 11(g − 1). 12 36

which may be smooth factors of Λ The orders of the possible factor groups of G of order 288 the element are 1632, 288 and 96. However, in a factor group G of G 4 (c0 c1 )(c0 c2 ) would still represent an element of order 9 and therefore also in this with orders 96 case we would have S < 11(g − 1). In those factor groups G of G and 1632, the product (c0 c1 )(c0 c2 )4 would represent an element of order 3, which we already showed to be impossible. This finishes the discussion of the case m = 8 and so we shall assume now that m = 10. Here |G| = 60(g − 1) and, as |θ(C(Λ, ci ))| ≥ 8 for i = 0, 1, we have S ≤

2|G| = 15(g − 1). 8

So we can assume that g = 2, 5. If |θ(C(Λ, ci ))| > 8 then |θ(C(Λ, ci ))| ≥ 12 for i = 0, 1, which implies S ≤ 2|G|/12 = 10(g − 1). So let |θ(C(Λ, c0 ))| = 8. with presentation Then G is a factor group of the group G c0 , c1 , c2 | c20 , c21 , c22 , (c0 c1 )2 , (c1 c2 )3 , (c0 c2 )10 , ((c0 c1 )(c0 c2 )5 )2 . has order 240. We have assumed that g = 2, 5 and we shall see In this case G that also g = 3. Otherwise, if g = 3 then |G| = 120. Now c0 c1 and c1 c2 rep elements generating a subgroup H which either equals G or it has resent in G

52


order 60. However, in the former case Γ would contain an orientation reversing element, which is impossible. So H has order 60. Clearly c0 c1 and c1 c2 represent in H elements of order 2 and 3, respectively, whose product is an element of order 10. But then it is easy to show that H contains a normal subgroup of order 2, which is obviously impossible. If |θ(C(Λ, c1 ))| = 8 then G is a factor group of the group with presentation G c0 , c1 , c2 | c20 , c21 , c22 , (c0 c1 )2 , (c1 c2 )3 , (c0 c2 )10 , ((c0 c1 )(c2 c1 )(c0 c2 )5 (c1 c2 ))2 of order 720. So g = 2, 3, 4, 5, 7 or 13. We already know that there exist Riemann surfaces of genera g = 2, 3, 5 with more than 15(g − 1) ovals. Later on we shall show that there exists a Riemann surface of genus g = 7 with 126 ovals. So we just However, have to deal with the cases g = 4 and g = 13. In the latter case G = G. an element of order 6. Hence |θ(C(Λ, c0 ))| = 24 by (c0 c1 )(c0 c2 )5 represents in G Lemma 3.3.1, and thus S = 10(g − 1). If g = 4 then |G| = 120 and, as in the case |θ(C(Λ, c0 ))| = 8, one can show that this is impossible. Finally, let m = 2m + 1 be odd. Then |θ(C(Λ, c0 ))| ≥ 8 since otherwise, by (3) of Lemma 1.3.3,

θ(c0 c1 ) = θ(c0 c2 )m +1 θ(c1 c2 )θ(c1 c0 )θ(c2 c1 )θ(c0 c2 )m . But then, using the defining relations for Λ, one can show as before that θ(c1 c2 ) and θ(c0 c2 )2 are conjugate, which implies that θ(c0 c2 )6 = 1, a contradiction. Hence, |θ(C(Λ, c0 ))| ≥ 8 and therefore S ≤

3m |G| = (g − 1) ≤ 9(g − 1) 8 m−6

whenever m ≥ 9. So it remains to consider the case m = 7. Here S ≤ 21(g − 1) and the bound is attained if and only if |θ(C(Λ, c0 ))| ≥ 8. This condition is equivalent to θ((c0 c1 )(c2 c0 )3 (c1 c2 )(c1 c0 )(c2 c1 )(c0 c2 )3 )2 = 1. Therefore, the equality S = 21(g − 1) is equivalent to the fact that G is a factor with presentation group of the group G c0 , c1 , c2 | c20 , c21 , c22 , (c0 c1 )2 , (c1 c2 )3 , (c0 c2 )7 , ((c0 c1 )(c2 c0 )3 (c1 c2 )(c1 c0 )(c2 c1 )(c0 c2 )3 )2 , which can be shown to have order 1008. Now c0 c1 and c1 c2 represent elements which either equals G or has order 504. However, in the generating a subgroup of G former case, Γ would contain an orientation reversing element, which is an absurd. and this proves that there exists a Riemann surface of genus g = 7 with So G = G 21(g − 1) = 126 ovals.


53

Now if |θ(C(Λ, c0 ))| > 8 then |θ(C(Λ, c0 ))| ≥ 12 and so S ≤

2m |G| = (g − 1) ≤ 14(g − 1). 12 m−6

However, the last bound is attained if and only if θ((c0 c1 )(c2 c0 )3 (c1 c2 )(c1 c0 )(c2 c1 )(c0 c2 )3 ) with presentation has order 3, or equivalently, if G is a factor group of the group G c0 , c1 , c2 | c20 , c21 , c22 , (c0 c1 )2 , (c1 c2 )3 , (c0 c2 )7 , ((c0 c1 )(c2 c0 )3 (c1 c2 )(c1 c0 )(c2 c1 )(c0 c2 )3 )3 , and therefore which can be shown to have order 336. As before we claim that G = G S has genus 3. Finally, if |θ(C(Λ, c0 ))| > 12 then |θ(C(Λ, c0 ))| ≥ 16 and this implies S ≤

|G| < 11(g − 1). 16

This finishes the proof of the first part of the theorem. Now we will show that the bound 12(g − 1) is attained for infinitely many values of g. Let us consider the group Ω = Z2 ∗ Z3 = x, y | x2 , y 3 and let M be the subgroup of Ω generated by A = [x, y] and B = [x, y −1 ]. A straightforward computation gives Ax = A−1 , B x = B −1 , Ay = A−1 B, B y = A−1 . Therefore M is a normal subgroup of Ω and by the Kurosh subgroup theorem [71], M is free. Now we shall prove that M is non-cyclic. Indeed, otherwise we would have A = C α and B = C β for some integers α and β and some C ∈ M . Then A−1 B = C β−α . Now, as M = C is a cyclic infinite normal subgroup of Ω, either C y = C or C y = C −1 . In the first case A = (A−1 B)y

−1

= (C y

−1

)β−α = C β−α = A−1 B.

(3.5)

Therefore, B = A2 and so (A−1 B)2 = (Ay )2 = (A2 )y = B y = A−1 . Hence, A = B 2 and consequently B 4 = A2 = B, that is B 3 = 1. So B = 1 which implies A = 1, a contradiction.

54


In case C y = C −1 we have C y A = (A−1 B)y

−1

= (C y

−1

−1

= C −1 , and in this way

)β−α = (C −1 )β−α = C α−β = AB −1 .

Whence [x, y −1 ] = B = 1 and this implies A = [x, y] = 1, which is an absurd. Now let K be the normal closure in M of the set L = {Ak , B k , [A, B]2 , [A, [A, B]], [B, [A, B]]}. Clearly, the factor group M/K is a metabelian group of order 2k 2 , provided that k is even. We claim that if, in addition, k is a multiple of 4 then K is a normal subgroup of Ω. Indeed, it suffices to show that W x , W y ∈ K for any element W ∈ L. For −1 W = Ak it is clear that W x = (Ak ) ∈ K and a straightforward computation shows that W y = (A−1 B)k ≡ A−k B k [A, B]k(k−1)/2 mod K. Therefore, if 4 divides k then W y ∈ K. One can deal similarly with the remaining elements of L. It is straightforward to check that in the factor group G = Ω/K, the commutator [A, B] is conjugate to (xy)6 , whilst [A, [A, B]] and [B, [A, B]] are conjugate to (xy)6 (yx)6 . Therefore, G has the following presentation x, y | x2 , y 3 , (xy)12 , [x, y]k , (xy)6 (yx)6

(3.6)

and xy represents in G an element of order 12. Thus, G is a smooth factor group Δ/Γ of a Fuchsian group Δ with signature [2, 3, 12] and so G acts as a group of automorphisms on a Riemann surface S = H/Γ which, by the Hurwitz–Riemann formula, has genus g = (k 2 /2) + 1 = 8m2 + 1, because k = 4m is a multiple of 4. Clearly the assignment x → x−1 , y → y −1 induces an automorphism of G. So, by Theorem 1.5.10, the surface S is symmetric and therefore Γ is a normal subgroup of an NEC group Λ with signature (2, 3, 12), as the last is the signature of all Fuchsian groups containing Δ. Let us write Λ = c0 , c1 , c2 | c20 , c21 , c22 , (c0 c1 )2 , (c1 c2 )3 , (c0 c2 )12 and denote by a, b and c the images of c0 , c1 and c2 , respectively, under the canon = Λ/Γ. Observe that x1 = c0 c1 and x2 = c1 c2 ical projection from Λ onto G are canonical elliptic generators of Δ of orders 2 and 3, respectively, and so we can assume that x = ab and y = bc. Finally, (ab)(ac)6 and (ab)(bc)(ac)6 (cb) are elements of order 2 as both are conjugate to (xy)6 x. Therefore, by (2) of Lemma 1.3.3, |θ(C(Λ, ci ))| = 8 for i = 0, 1 and so, by Theorem 3.3.2, S has 12(g − 1) ovals. This completes the proof.

3.4 Total Number of Ovals of a Couple of Symmetries

55

has the following presentation: It is worth noting that G a, b, c | a2 , b2 , c2 , (ab)2 , (bc)3 , (ac)12 , (acb)2k , ((ac)6 b)2 .

(3.7)

Indeed, the elements x = ab and y = bc generate, in the group defined in (3.7), a normal subgroup of index 2 having the presentation given in (3.6).

3.4 Total Number of Ovals of a Couple of Symmetries In this section we shall find, using Theorem 3.1.1, a bound for the total number of ovals of a pair of symmetries of a Riemann surface of genus g, in terms of g and the order and number of points fixed by their product. The first precedent of this kind of result goes back to Natanzon [102], who classified pairs of commuting symmetries. Afterwards, Bujalance and Costa studied in [19] pairs of symmetries of p-hyperelliptic Riemann surfaces. Some of the results we present here have been recently showed by Kozłowska-Walania [69]. Let S = H/Γ be a Riemann surface and let θ : Λ → Aut(S) be a smooth epimorphism with ker θ = Γ. Recall that if a symmetry σ of S and the image θ(ci ) of a canonical reflection ci of Λ are conjugate then we denote the index [C(G, θ(ci )) : θ(C(Λ, ci ))] by wi and we call it the contribution of ci to σ. Lemma 3.4.1. Let Dn = Λ/Γ be a dihedral group of automorphisms of a Riemann surface S = H/Γ generated by two non-central symmetries σ and τ , and let C = (n1 , . . . , ns ) be a period cycle of the signature of the NEC group Λ. (1) If n is odd then the reflections corresponding to C contribute to σ and τ with at most 2 ovals in total. (2) If n is even then the reflections corresponding to C contribute to σ and τ with at most t ovals in total, where t is the number of even link periods if s ≥ 1 and some ni is even, and with at most 2 ovals in total in the remaining cases. Proof. Let θ : Λ → Dn be the canonical epimorphism. The case of odd n is an immediate consequence of Theorem 3.1.1, since all canonical reflections c corresponding to the period cycle C are conjugate, the centralizer C(Dn , θ(c)) has order two and c ∈ C(Λ, c). Now, for even n, the centralizer of any non-central element of Dn has order 4. Since ci ∈ C(Λ, ci ), necessarily the contribution wi of θ(ci ) satisfies wi ≤ 2, and since σ and τ are non-conjugate, we may assume that either s ≥ 2 or s = 1 and n1 is even. If c belongs to two odd link periods then we can suppose that c contributes neither to σ nor to τ . If c belongs to an even link period n and the product cc has order n then (cc )n /2 ∈ C(Λ, c). Now θ((cc )n /2 c) = 1, because ker θ is a Fuchsian group, and therefore θ(C(Λ, c)) has order 4. We give now a new proof of an upper bound, first estimated by Bujalance, Costa and Singerman in [21], for the total number of ovals of two symmetries of a Riemann

56


surface. The original proof used a deep method due to Hoare [60], while the proof here is based on Theorem 3.1.1 and follows ideas of Kozłowska-Walania in [69]. Theorem 3.4.2. Let σ and τ be two symmetries of a Riemann surface S of genus g whose product has order n. Then ⎧ 2(g − 1) ⎪ ⎨ + 4 if n is odd; n σ + τ ≤ ⎪ ⎩ 4g + 2 if n is even. n Proof. Let t = σ + τ denote the total number of ovals of σ and τ and let Dn = σ, τ . Hence Dn = Λ/Γ for some surface Fuchsian group Γ and an NEC group Λ with signature s(Λ) = (h; ±; [m1 , . . . , mr ]; {C1 , . . . , Ck , (n1 ), . . . , (n ), (−), .m. ., (−)}), where n1 , . . . , n are odd and each period cycle Ci = (ni1 , . . . , nisi ) with either si ≥ 2 or si = 1 and ni1 is even. Let s = s1 + · · · + sk . Observe first that either r = 0 or s + = 1. Otherwise s(Λ) would have a unique link period, say n0 , and so, by Corollary 1.2.7, the signature of the canonical Fuchsian group of Λ would be s(Λ+ ) = (h ; n0 ). As Λ+ /Γ = Zn , the relation x1 [a1 , b1 ] · · · [ah , bh ] = 1 in Λ+ would give θ(x1 ) = 1 for the canonical smooth epimorphism θ : Λ → Dn , which is impossible because Γ = ker θ is a surface Fuchsian group. Let us first analyze the case of odd n. We know that t ≤ 2k + 2 + 2m, by Lemma 3.4.1, and so 2π(g − 1) = Area(Λ) ≥ 2π(k + + m − 2) ≥ π(t − 4), n which implies t ≤ 4 + 2(g − 1)/n. Let now n be even. Observe that, as both σ and τ have fixed points, either k > 0 or + m ≥ 2. So the only possibilities are: (1) (2) (3) (4) (5) (6)

k ≥ 2, k = 0 and ≥ 2, k = = 0 and m ≥ 2, k = 0, = m = 1, k = 1 and at least one of h, r, , m is positive, k = 1 and h = r = = m = 0.

By Lemma 3.4.1, we have 2 + 2m + s ≥ t. Thus r s 2π(g − 1) = Area(Λ) ≥ 2π ηh + k + + m − 2 + + + n 3 2 4 +m t r + + , ≥ 2π ηh − 2 + k + + 2 2 3 4


57

where η = 2 if the sign of s(Λ) is “+” and η = 1 otherwise. Using that either r = 0 or s + = 1, it is easy to see that in all cases (1)–(5), we have ηh + k +

3 r +m + + ≥ . 2 2 3 2

Thus 2π(g − 1) = Area(Λ) ≥ π n

t −1 , 2

which gives t≤

4g 4(g − 1) +2< + 2. n n

Therefore we can assume that Λ is an NEC group with signature s(Λ) = (0; +; [−]; {(n1 , . . . , ns )}). Observe now that for n = 2 the desired bound obviously holds in virtue of Harnack– Weichold theorem. So let n = 2. The images of the elements xi = ci−1 ci generate Zn and they satisfy the relation x1 · · · xs = 1. So, if one of the link periods is odd, then either there is another odd link period or a link period ni ≥ 6. In the first case we have 2π(g − 1) s−2 2 π 4 = Area(Λ) ≥ 2π −1 + + = s−2− . n 4 3 2 3 Thus, by Lemma 3.4.1, t≤s−2≤

4 4 4g 4g − + < + 2. n n 3 n

In the second case one gets, similarly, t≤s−1≤

4 4g 4g − +2< + 2. n n n

Therefore, we can assume that n ≥ 4 and all link periods of s(Λ) are even. If there are more than three link periods ni ≥ 4 then s−4 3 π(s − 2) 2π(g − 1) = Area(Λ) ≥ 2π −1 + + ≥ n 4 2 2 and, consequently, t=s≤

4g 4 4g − +2< + 2. n n n

58


If all link periods are equal to 2 except n1 and n2 then n1 = n2 = n and so the bound holds. Therefore we can assume that all link periods are equal 2 except ni , ni , ni , and that 4 ≤ ni ≤ ni ≤ ni . Now, if ni ≥ 6 then 2π(g − 1) s−3 5 π(s − 2) = Area(Λ) ≥ 2π −1 + + ≥ n 4 4 2 and by Lemma 3.4.1 we have t≤s≤

4 4g 4g − +2< + 2. n n n

On the other hand, if ni = 4 then also n = 4 and arguing as above we get t≤

4 4g 4g − +3= + 2. n n n

So finally let ni = 4 and ni = ρ ≥ 6. Observe that ρ ≥ n/4 and s−3 3 1 s−3 3 1 2π(g − 1) = Area(Λ) ≥ 2π −1+ + +1− = 2π + − . n 4 8 ρ 4 8 ρ Consequently, t≤

4 3 4 4g 3 3 4g 4g − + + ≤ + + ≤ + 2. n n 2 ρ n ρ 2 n

Our next result shows that the precedent upper bounds are sharp in all cases. Theorem 3.4.3. The upper bounds occurring in Theorem 3.4.2 are attained for all arithmetically admissible values of g and n. Proof. Let first n be an odd divisor of g − 1. Consider an NEC group Λ with signature s(Λ) = (0; +; [−]; {(−), (g−1)/n . . . . . . .+. .2, (−)}) and the smooth epimorphism θ : Λ → Dn = σ, τ | σ 2 , τ 2 , (στ )n induced by the assignment θ(ei ) = 1 for i = 1, . . . , (g − 1)/n + 2 and θ(ci ) =

σ if i is odd; τ if i is even.

As every period cycle contributes with one oval to each of the symmetries σ and τ , we deduce that 2(g − 1) + 4. σ + τ = n


59

Now let n be an even divisor of 4g. Consider an NEC group Λ with signature s(Λ) = (0; +; [−]; {(2, . s. ., 2, n, n)}) whose canonical reflections are denoted c0 , . . . , cs , cs+1 , where s = 4g/n. Let θ : Λ → Dn = σ, τ be the smooth epimorphism induced by the following assignment: •

If s is even then θ(c2i ) = σ, θ(c2i+1 ) = τ (στ )n/2−1 θ(cs ) = σ, θ(cs+1 ) = τ.

•

for 0 ≤ i ≤ s/2 − 1,

If s is odd then θ(c2i ) = σ, θ(c2i+1 ) = τ (στ )n/2−1 θ(cs+1 ) = τ.

for 0 ≤ i ≤ (s − 1)/2,

Therefore, the reflections of Λ corresponding to the unique period cycle of s(Λ) contribute with 4g/n + 2 ovals to σ + τ . Theorem 3.4.2 provides bounds of the form [2(g − 1)/n] + 4 and [4g/n] + 2, where [ · ] denotes the integer part function, for the total number of ovals of two symmetries. We now study such bounds in detail. For instance, we show that the first bound is attained only for n dividing g − 1. As announced above the following Theorems 3.4.4, 3.4.5 and 3.4.7 are due to Kozłowska-Walania [69]. Theorem 3.4.4. Let σ and τ be two symmetries of a genus g Riemann surface S, whose product στ has odd order n. If n does not divide g − 1 then

2(g − 1) σ + τ ≤ + 3. n Even more, for each odd integer n ≥ 3 this bound is attained for infinitely many values of g. Proof. Let t = σ + τ denote the total number of ovals of σ and τ and write σ, τ = Dn = Λ/Γ for some surface Fuchsian group Γ and an NEC group Λ with signature s(Λ) = (h; ±; [m1 , . . . , mr ]; {C1 , . . . , Ck , (n1 ), . . . , (n ), (−), .m. ., (−)}), where each Ci = (ni1 , . . . , nisi ) and si ≥ 2. Now, as Area(Λ) = 2π(g − 1)/n and n does not divide g − 1, it follows that s(Λ) has either link periods or proper periods. Observe that if r = 0 then s(Λ) has at least two link periods. Otherwise, by Corollary 1.2.7, the signature of the canonical Fuchsian group of Λ would be

60


s(Λ+ ) = (h ; n0 ); but there is no smooth epimorphism from a Fuchsian group with this signature onto the cyclic group Zn , see the proof of Theorem 3.4.2. So Λ has either a proper period or at least two link periods. Then 2π(g − 1) 1 = Area(Λ) > 2π k++m−2+ ≥ π(2(k++m) − 3) ≥ π(t−3) n 2 which in turn gives t ≤ [2(g − 1)/n] + 3 because t is an integer. To finish we show that for any non-negative integer m there exists a Riemann surface S of genus g = n(m + 1) admitting two symmetries σ and τ whose product has order n and having [2(g − 1)/n] + 3 ovals in common. Let us consider an NEC group with signature (0; +; [−]; {(−), m+1 . . . , (−), (n, n)}), and let θ : Λ → σ, τ = Dn be the smooth epimorphism induced by the assignment θ(ei ) = 1 for 1 ≤ i ≤ m + 2; θ(ci0 ) = σ for 1 ≤ i ≤ m + 1, θ(cm+2,0 ) = θ(cm+2,2 ) = σ;

θ(cm+2,1 ) = τ.

Then ker θ is a surface Fuchsian group and the Riemann surface S = H/ ker θ has genus n(m + 1) by the Hurwitz–Riemann formula. Theorem 3.1.1 yields that both σ and τ have m + 2 ovals and so 2(g − 1) σ + τ = 2m + 4 = + 3. n In contrast to the previous theorem, the bound [4g/n] + 2 for even n in Theorem 3.4.2 without any divisibility conditions on the pair (n, g) cannot be improved. Theorem 3.4.5. For each even integer n > 4 there are infinitely many values of g for which n does not divide 4g and such that there exists a Riemann surface of genus g admitting two symmetries, whose product has order n, and with [4g/n] + 2 ovals in total. Proof. Let Λ be an NEC group with signature s(Λ) = (0; +; [−]; {(−), (2, .2m . ., 2)}) and consider the epimorphism θ : Λ → Dn = σ, τ | σ 2 , τ 2 , (στ )n induced by the assignment θ(e1 ) = θ(e2 ) = 1, θ(c10 ) = σ and which maps the reflections corresponding to the unique non-empty period cycle of s(Λ) alternatively to τ and (στ )n/2−1 σ. The orbit space S = H/ ker θ is a


61

Riemann surface of genus g = mn/2 + 1 which admits two symmetries having, by Theorem 3.1.1, 2m + 2 = [4g/n] + 2 ovals in total. Corollary 3.4.6. Let S be a surface of genus g. Let σ and τ be two symmetries of S such that σ = g + 1 − q and τ = g + 1 − q for some integers q and q satisfying g ≥ q + q + 1. Then σ and τ commute. Proof. The total number of ovals of these symmetries satisfies σ + τ = 2g + 2 − q − q ≥ g + 3. Suppose that σ and τ do not commute. Then the order n of στ is greater than 2. By Theorem 3.4.2 we get, for even n, g + 3 ≤ σ + τ ≤

4g + 2 ≤ g + 2, n

a contradiction. For odd n, g + 3 ≤ σ + τ ≤ and so g ≤ 1, which is not our case.

2(g − 1) 2(g − 1) +4≤ +4 n 3

By Harnack’s theorem, σ ≤ g + 1 for each symmetry σ of a genus g surface. The symmetries attaining such upper bound are called M -symmetries. This is why a symmetry σ with g + 1 − q ovals is called an (M − q)-symmetry. Now, we shall show that for each genus g ≥ 3, and with only one exception for the values of q and q , the lower bound q + q + 1 for the genus g in Corollary 3.4.6 is in fact the minimal lower bound for g which guarantees the commutativity of an (M − q)-symmetry and an (M − q )-symmetry of a Riemann surface of genus g. Theorem 3.4.7. Let g, q and q be integers such that 2 ≤ g ≤ q + q and q, q ≤ g. There exists a Riemann surface of genus g admitting an (M − q)-symmetry σ and an (M − q )-symmetry τ such that στ = τ σ if and only if either g = 2 or g > 2 and {q, q } = {1, g}. Proof. The exceptional case was considered by Natanzon [108], who proved that two symmetries with g > 2 and 1 ovals, respectively, always commute. On the other hand, for g = 2 and {q, q } = {1, 2}, we can choose n = 8 in Theorem 3.4.2 to obtain a Riemann surface of such genus 2 with two noncommuting symmetries having one and two ovals. So in what follows let {q, q } = {1, g} and we assume that q ≤ q . We define s = g − q, s = g − q and we distinguish several cases according to the value of q + q − g (mod 4).

62


For q + q − g ≡ 0 (mod 4) consider an NEC group Λ with signature

s(Λ) = (h; −; [−]; {(2, . s. ., 2, 4, 2, .s. ., 2, 4)}) where h = (q+q −g)/4, and the epimorphism θ : Λ → D4 = σ, τ | σ 2 , τ 2 , (στ )4 induced by the assignment θ(e) = 1, θ(di ) = σ for 1 ≤ i ≤ h and such that the consecutive canonical reflections corresponding to the unique period cycle of s(Λ) are mapped onto

σ, τ στ, σ, τ στ, . . . , σ(στ )2s , τ, στ σ, τ, στ σ, . . . , τ (στ )2s , σ. s+1

s +1

By the Hurwitz–Riemann formula the surface S = H/ ker θ has genus g and by Theorem 3.1.1 it admits two non-commuting symmetries σ and τ having g + 1 − q and g + 1 − q ovals, respectively. For q + q − g ≡ 2 (mod 4) consider an NEC group Λ with signature

s(Λ) = (h; −; [2]; {(2, . s. ., 2, 4, 2, .s. ., 2, 4)}) where h = (q + q − 2 − g)/4, and the epimorphism θ : Λ → D4 defined as in the previous case, with the extra data θ(x) = θ(e) = (στ )2 . As before the genus g Riemann surface S = H/ ker θ admits two non-commuting symmetries σ and τ having g + 1 − q and g + 1 − q ovals, respectively. Now, let q + q − g ≡ 3 (mod 4). Consider an NEC group Λ with signature

s(Λ) = (h; −; [4]; {(2, . s. ., 2, 4, 2, .s. ., 2, 4)}) where h = (q + q − 3 − g)/4, and the epimorphism θ : Λ → D4 which sends the consecutive canonical reflections corresponding to the unique period cycle of the signature s(Λ) to:

σ, τ στ, σ, τ στ, . . . , σ(στ )2s , τ, στ σ, τ, στ σ, . . . , τ (στ )2s , τ στ s+1

s +1

and θ(x) = στ , θ(e) = τ σ. Again, the genus g surface H/ ker θ admits two noncommuting symmetries σ and τ having g + 1 − q and g + 1 − q ovals, respectively. If q + q − g ≡ 1 (mod 4) and g < q + q − 1 consider an NEC group Λ with signature

s(Λ) = (h; −; [2, 4]; {(2, . s. ., 2, 4, 2, .s. ., 2, 4)}) where h = (q + q − 5 − g)/4, and the epimorphism θ : Λ → D4 induced by the assignment that acts on the consecutive canonical reflections corresponding to the unique period cycle of the signature s(Λ) as in the previous case, and


63

θ(x1 ) = (στ )2 , θ(x2 ) = θ(e) = στ . As before, the genus g surface S = H/ ker θ admits two non-commuting symmetries σ and τ having g + 1 − q and g + 1 − q ovals, respectively. Finally for g = q + q − 1 let us take q ≥ 2 and let Λ be an NEC group with signature s(Λ) = (0; +; [−]; {(2, q−2 . . . , 2, 4, 2, q. −2 . . , 2, 4, 4, 4)}) and the epimorphism θ : Λ → D4 induced by the assignment which maps the reflections corresponding to the only period cycle of s(Λ) as follows:

σ, τ στ, σ, τ στ, . . . , σ(στ )2q , τ, στ σ, τ, στ σ, . . . , τ (στ )2q , σ, τ, σ. q−1

q −1

Once more the genus g surface S = H/ ker θ admits two non-commuting symmetries σ and τ having g + 1 − q and g + 1 − q ovals, respectively.

Chapter 4

Symmetry Types of Some Families of Riemann Surfaces

This chapter is devoted to three selected examples of families of surfaces whose symmetries can be completely classified. The last means to calculate the number of conjugacy classes of symmetries, to count the number of ovals of each of them and to determine the separating character of each symmetry. The first two sections are devoted to the sphere and the tori, which require specific methods since they are not uniformized by the hyperbolic plane. In the third section we study the symmetries of the hyperelliptic Riemann surfaces. We classified them completely in our joint memoir [14]. However this work is too extensive to be summarized here. Hence we just explain the necessary tools to attack the problem and we develop an example in detail. To begin with, we classify the symmetries in the genus 0 case, that is, the symmetries of the Riemann sphere.

4.1 Symmetry Type of the Riemann Sphere The group of analytic automorphisms of the Riemann sphere Σ = C ∪ {∞} is the group of Möbius transformations az + b Aut+ (Σ) = : a, b, c, d ∈ C, ad − bc = 1 . cz + d The value at ∞ of the Möbius transformation m(z) = (az + b)/(cz + d) is a/c if c = 0, m(∞) = ∞ if c = 0. Let σ1 : Σ → Σ denote the extension to Σ of complex conjugation z → z¯, defined by σ1 (∞) = ∞. Recall that a map σ : Σ → Σ is an antianalytic automorphism of Σ if the composition σ ◦ σ1 is an analytic automorphism of Σ. Thus there exist a, b, c, d ∈ C such that ad − bc = 1 and σ(z) =

a¯ z+b c¯ z+d

for each z ∈ Σ.

E. Bujalance et al., Symmetries of Compact Riemann Surfaces, Lecture Notes in Mathematics 2007, DOI 10.1007/978-3-642-14828-6 4, c Springer-Verlag Berlin Heidelberg 2010

65

66

4 Symmetry Types of Some Families of Riemann Surfaces

A symmetry of Σ is an antianalytic involution of Σ. Two significant examples of symmetries of Σ are complex conjugation σ1 and the antipodal map σ2 defined by σ1 : Σ → Σ ; z → z¯ ;

σ2 : Σ → Σ ; z →

−1 . z¯

Our goal in this section is to prove that each symmetry of Σ is conjugate either to σ1 or to σ2 . Before doing this, we recall without proof some well known facts about semilinear endomorphisms of complex vector spaces. Definition and Proposition 4.1.1. Let E be a complex vector space. (1) A map f : E → E is a semilinear endomorphism if f (au + bv) = a ¯f (u) + ¯bf (v) for all a, b ∈ C and all u, v ∈ E. (2) The matrix Mf (B) = (aij ) of the semilinear endomorphism f with respect to the basis B = {u1 , . . . , un } of E is defined by the equalities f (uj ) =

n

aij ui , for 1 ≤ j ≤ n.

i=1

(3) Let B1 = {u1 , . . . , un } and B2 = {v1 , . . . , vn } be two bases of E and let C(B1 , B2 ) = (cij ) ∈ Mn (C) be the basis change matrix defined by uj =

n

cij vi , for 1 ≤ j ≤ n.

i=1

Then the following equality holds true: C(B1 , B2 )Mf (B1 ) = Mf (B2 )C(B1 , B2 ). (4) Let g : E → E be another semilinear endomorphism. Then f ◦ g is a linear endomorphism of E whose matrix with respect to the basis B is Mf ◦g (B) = Mf (B)Mg (B). Theorem 4.1.2. Every symmetry of the Riemann sphere Σ is conjugate either to σ1 or to σ2 . Moreover, Σ/ σ1 is the closed disk and Σ/ σ2 is the real projective plane. In particular, σ1 and σ2 are not conjugate. Proof. Let σ : Σ → Σ ; z → (a¯ z + b)/(c¯ z + d) be a symmetry. Denote A=

ab cd

and I =

1 0 0 1

4.1 Symmetry Type of the Riemann Sphere

67

and consider the semilinear endomorphism 2

2

fA : C → C ;

z1 z2

→ A

z¯1 , z¯2

that satisfies MfA (E) = A, where E is the standard basis of C2 . Since σ 2 is the identity map and det(A) = 1, either fA2 or −fA2 is the identity endomorphism. This implies, by part (4) in Proposition 4.1.1, that AA¯ = ±I. We now distinguish two cases: Case 1: AA¯ = I, that is, fA2 is the identity. Since fA is not a dilatation, there exists a vector u ∈ C2 such that {u, fA(u)} is a basis of C2 . Thus also B = {v1 = u + fA (u), v2 = i(u − fA (u))} is a basis of C2 , where i =

√ −1. Then,

fA (v1 ) = fA (u) + fA2 (u) = fA (u) + u = v1 , fA (v2 ) = −i(fA (u) − fA2 (u)) = i(−fA (u) + u) = v2 . Therefore MfA (B) = I. The basis change matrix C = C(B, E) =

c11 c12 c21 c22

¯ Let ϕ : Σ → Σ satisfies, by part (3) in Proposition 4.1.1, the equality I = C −1 AC. be the analytic automorphism ϕ(z) =

c11 z + c12 . c21 z + c22

The equality C = AC¯ means that the antianalytic automorphism ϕ◦σ1 , with matrix ¯ Therefore C, coincides with the antianalytic automorphism σ ◦ ϕ, with matrix AC. ϕ−1 ◦ σ ◦ ϕ = σ1 , that is, σ is conjugate to complex conjugation. Case 2: AA¯ = −I, that is, −fA2 is the identity. Any non-zero vector u ∈ C2 is not an eigenvector of fA . Otherwise fA (u) = λu for some λ ∈ C and this would imply ¯ A (u) = λλu ¯ = |λ|2 u, −u = fA2 (u) = fA (λu) = λf that is, |λ|2 = −1, an absurdity. So B1 = {w1 = u, w2 = fA (u)} is a basis of C2 . Moreover, fA (w1 ) = w2 and fA (w2 ) = fA2 (w1 ) = −w1 , i.e., MfA (B1 ) =

0 −1 . 1 0

68


Therefore the basis change matrix C1 = C(B1 , E) =

c11 c12 c21 c22

satisfies MfA (B1 ) = C −1 AC¯ by part (3) in Proposition 4.1.1. Hence, the analytic automorphism c11 z + c12 ϕ(z) = c21 z + c22 of Σ satisfies σ(ϕ(z)) = ϕ(σ2 (z)). So in this case σ is conjugate to the antipodal map. Finally, let Δ be the closed disk of radius 1 centered at the origin of C. It is easy to see that the map Δ → Σ/ σ1 ; z →

i, [(z + i)/(iz + 1)]σ1 if z = if z = i, ∞

is a homeomorphism. On the other hand, let S2 be the unit sphere in R3 and let πN : S2 → Σ be the extension of the stereographic projection from the north pole (0, 0, 1) ∈ S2 that −1 maps this point to ∞. Let τ = πN ◦ σ2 ◦ πN . Obviously the orbit spaces S2 / τ and Σ/ σ2 are homeomorphic. Moreover, for each point p ∈ S2 let us denote by [p]∼ ∈ P2 (R) the associated projective point, and let [p]τ be its class modulo τ . A straightforward computation shows that the map P2 (R) → S2 / τ ; [p]∼ → [p]τ is a homeomorphism, and so Σ/ σ2 is the real projective plane.

The closed disk is an orientable surface with one boundary component, that is, the species of the symmetry σ1 is +1. The real projective plane is a non-orientable surface with empty boundary; that is, σ2 has species 0. Consequently, Theorem 4.1.2 yields the following. Corollary 4.1.3. The symmetry type of the Riemann sphere is {+1, 0}.

4.2 Symmetry Types of Tori Let S be a Riemann surface of genus one. It is well known (see [92, Chap. II]) that, up to analytic isomorphism, S is the orbit space S = C/L where L = Zω1 + Zω2 = {rω1 + sω2 : r, s ∈ Z}

4.2 Symmetry Types of Tori

69

is the lattice generated over Z by two R-linearly independent complex numbers ω1 and ω2 . The set RL = {λω1 + μω2 : 0 ≤ λ, μ < 1} is called a fundamental parallelogram of L. We start this section with a description of the analytic automorphisms of a complex torus. This is a necessary step in order to classify symmetries up to conjugacy. We follow closely the exposition in [3, Chap. 12], [39] and [92, Chap. 3]. We first consider analytic maps in general. Proposition 4.2.1. Let S1 and S2 be two tori defined by the lattices L1 and L2 , respectively, and let f : S1 → S2 be a non-constant analytic map. (1) There exist a, b ∈ C with a = 0 such that aL1 ⊂ L2 and f : S1 → S2 ; z + L1 → (az + b) + L2 . (2) The index [L2 : aL1 ] of aL1 as a subgroup of L2 is the degree of f . In particular, f is an isomorphism if and only if L2 = aL1 . Proof. (1) The map f is unramified because the Euler characteristic of both S1 and S2 equals 0. In particular, f is a local homeomorphism. Let πi : C → Si ; z → z + Li be the corresponding universal covering, for i = 1, 2. The domain of the covering f ◦ π1 : C → S2 is simply connected and so it is isomorphic, as a covering, to π2 . Therefore there exists a holomorphic function F : C → C such that π2 ◦ F = f ◦ π1 . We shall prove that F (z) = az + b where a ∈ C \ {0} and b ∈ C satisfies f (0 + L1 ) = b + L2 . In such a case, we will have f (z + L1 ) = f (π1 (z)) = π2 (F (z)) = π2 (az + b) = (az + b) + L2 for each z + L1 ∈ S1 . Moreover, for each 1 ∈ L1 (a1 +b)+L2 = π2 (a1 +b) = π2 (F (1 )) = f (π1 (1 )) = f (0+L1 ) = b+L2 . Therefore a1 ∈ L2 and so aL1 ⊂ L2 . To show that F (z) = az + b it suffices to see that the derivative F is a bounded (and so a constant) function. Fix a point ∈ L1 and consider the holomorphic function F : C → C ; z → F (z + ) − F (z). Its image F (C) is contained in L2 . In fact, π1 (z + ) = π1 (z) for each z ∈ C, and so π2 (F (z)) = π2 (F (z + )) − π2 (F (z)) = f ◦ π1 (z + ) − f ◦ π1 (z) = 0.

70


Since C is connected and L2 is discrete, it follows that F is constant. Thus its derivative vanishes identically, that is, F (z + ) = F (z) for all (z, ) ∈ C × L1 . Therefore F (C) = F (P) where P is the closure of a fundamental parallelogram of L1 . Since P is compact, F is a bounded function, as desired. So F (z) = az + b for some constant numbers a, b. Observe that a = 0 because f is not constant. Moreover, b satisfies b + L2 = π2 (b) = π2 (F (0)) = f (π1 (0)) = f (0 + L1 ). This concludes the proof of part (1). (2) Let k = [L2 : aL1 ] and let z1 , . . . , zk ∈ L2 be such that L2 =

k

(zi + aL1 ).

i=1

Let us define ξi = a−1 (zi − b) + L1 ∈ S1 for 1 ≤ i ≤ k. It is easy to check that f −1 (0) = {ξ1 , . . . , ξk }. Then k = card f −1 (0) = deg(f ) because f is unramified.

As a consequence of Proposition 4.2.1 we get the following description of the analytic automorphisms of a complex torus. Corollary 4.2.2. Every analytic automorphism f : S → S of the complex torus S = C/L is of the form f : S → S ; z + L → (az + b) + L with |a| = 1 such that aL = L and b ∈ C. Proof. It is clear that f is an automorphism, with inverse f −1 : S → S ; z + L → a−1 (z − b) + L. So it only remains to show that a is unimodular. Let ∈ L be an element of minimal length in L \ {0}. Since a ∈ aL = L we have || ≤ |a| = |a||| and so |a| ≥ 1. Conversely, |a| ≤ 1 using that a−1 L = L.

Observe that, for any b ∈ C, the translation tb : z +L → (z +b)+L is an analytic automorphism of the torus C/L. Since L is discrete, b can be chosen so that rb ∈ /L for any r ∈ Z. For such values of b, the order of tb is infinite. In particular, this shows that, unlike surfaces of higher genus, each torus has infinitely many analytic automorphisms. Moreover, the complex number b can be chosen so that the order of the automorphism tb is n for any positive integer n. However, if we restrict to those automorphisms which fix 0 + L (which correspond therefore to homomorphisms of the group structure of C/L) then the situation changes significantly, as Proposition 4.2.5 shows. We first introduce two special kinds of lattices that will play an important role in the sequel.


71

Definitions 4.2.3. Let L ⊂ C be a lattice generated over Z by two R-linearly independent complex numbers. (1) The lattice L is said to be a square lattice if it admits orthogonal generators over Z of the same length. (2) The lattice L is said to be a hexagonal lattice if it admits generators over Z of the same length forming an angle of π/3. As an application of Proposition 4.2.1 we have the following. Proposition 4.2.4. (1) Every torus C/L1 where L1 is a square lattice is isomorphic to C/L(i) where L(i) = Z + Zi. (2) Every torus C/L2 where L2 is a hexagonal lattice is isomorphic to C/L(eπi/3 ) where L(eπi/3 ) = Z + Zeπi/3 . Proof. (1) Let us write L1 = Zω1 + Zω2 . We may assume, after relabeling the generators if necessary, that ω2 = iω1 . Then f : C/L1 → C/L(i) ; z + L1 → ω1−1 z + L(i) is a well defined isomorphism. The proof of part (2) is analogous and we omit it.

Another consequence of Proposition 4.2.1 is the following (see [3, Chap. 9] and [92, Chap. III]): each torus is analytically isomorphic to S = C/L(ω), where L(ω) = Z + Zω and the complex number ω can be chosen to satisfy the following conditions |ω| ≥ 1 if Re(ω) ≥ 0, Im(ω) > 0, −1 < 2Re(ω) ≤ 1 with (4.1) |ω| > 1 if Re(ω) < 0, where Re(ω) and Im(ω) stand, respectively, for the real and imaginary parts of ω. In the sequel we will assume that any lattice L is of the form L = L(ω) where ω satisfies these conditions. The next proposition shows that, as said above, there are very few automorphisms of the complex torus C/L which fix 0 + L. Proposition 4.2.5. Let f : S → S be a non-trivial analytic automorphism of the complex torus S = C/L such that f (0 + L) = 0 + L. Then f : z + L → az + L, where a is a primitive nth -root of unity for n = 2, 3, 4 or 6. Moreover, if L is a square lattice then a4 = 1, and if L1 is a hexagonal lattice then a6 = 1. Proof. We may assume that L = L(ω) for some ω satisfying conditions (4.1). Let us determine the set M = M (ω) = { ∈ L \ {0} : || is minimal}.

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Let ∈ M and write = r + sω with r, s ∈ Z. Clearly, || ≤ 1 because 1 ∈ L. Therefore 1 ≥ ||2 = (r + sω)(r + s¯ ω ) = r2 + 2Re(ω)rs + |ω|2 s2 . If s = 0 then 1 ≥ r2 and so r = ±1. Assume now that s = 0. We claim that |ω| = 1. Otherwise, using that −1 < 2Re(ω) ≤ 1, we get 1 ≥ r2 + 2Re(ω)rs + |ω|2 s2 > r2 + 2Re(ω)rs + s2 ≥ ≥ r2 − |rs| + s2 = (|r| − |s|)2 + |rs|. Thus |r| = |s| and rs = 0, a contradiction. So ω = eiθ with π/3 ≤ θ < 2π/3 and so −1 < 2 cos θ ≤ 1. Observe that rs ≤ 0 since otherwise 1 ≥ r2 +2rs cos θ+s2 > r2 − rs + s2 = (r − s)2 + rs ≥ rs ≥ 1, a contradiction. So rs ≤ 0 and hence 1 ≥ r2 + 2rs cos θ + s2 ≥ r2 + rs + s2 = (r + s)2 + |rs|. The unique integer solutions (r, s) with s = 0 to this inequality are (0, ±1) and ±(1, −1). They correspond to ±ω and ±(1 − ω). However, |1 − ω|2 = (1 − cos θ)2 + sin2 θ = 2 − 2 cos θ, and so the unique value of θ ∈ [π/3, 2π/3) for which |1 − ω| ≤ 1 is θ = π/3. This proves that ⎧ if ω = eiθ with θ = π/3; ⎨ {±1, ±ω} M = {±1, ±ω, ±(1 − ω)} if ω = eπi/3 ; ⎩ {±1} otherwise. Observe that in all cases || = 1 for each ∈ M. Let now f : S → S be an automorphism such that f (0 + L) = 0 + L. We know, by Corollary 4.2.2, that f : z + L → az + L where aL = L and |a| = 1. In particular this implies that multiplying by a permutes (cyclically) the points in M, that is, aM = M and so a · 1 = a ∈ M. If M = {±1} then a = ±1. If M = {±1, ±ω} then either a = ±1 or a = ±ω; in the last case, ±ω 2 = aω ∈ M and so ω 2 = −1, that is, ω = i = ±a. • If ω = eπi/3 then ω − 1 = e2πi/3 and so a ∈ M = {±1, ±eπi/3, ±e2πi/3 }. • •

Therefore, if a = ±1 then a is either a primitive 4th-root of unity (in which case ω = i and L is a square lattice) or a primitive 3rd-root or 6th-root of unity (in which case L is a hexagonal lattice). Conversely, if L is a square lattice then ω = i and a4 = 1, whilst if L is a hexagonal lattice then ω = eπi/3 and a6 = 1.


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Remark 4.2.6. Every torus admits a non-trivial automorphism. In fact, since L = −L for every lattice L, the transformation h : z + L → −z + L is an analytic involution of the torus C/L. Moreover, the automorphisms of C/L fixing 0 + L constitute a cyclic group of order 4 if L is square, of order 6 if L is hexagonal and of order 2 otherwise.

4.2.1 Symmetric Tori Let L be a lattice of the form L = L(ω) = Z+Zω where ω satisfies conditions (4.1). Our first goal is to characterize the symmetric tori C/L(ω) in terms of the complex number ω. For each symmetric torus we will also determine its symmetries and the species of each of them. The next proposition can be seen as the counterpart, for antianalytic maps, of Proposition 4.2.1. To prove it we adapt some of the already used arguments. Proposition 4.2.7. Let σ : S → S be a symmetry of the complex torus S = C/L where L = Z + Zω and ω ∈ C satisfies conditions (4.1). (1) There exist a, b ∈ C with |a| = 1, aL = L and a¯b + b ∈ L such that σ : S → S ; z + L → (a¯ z + b) + L. (2) (3) (4) (5) (6)

If a ∈ {±1} then Re(ω) ∈ {0, 1/2}. If |ω| > 1 then a ∈ {±1} and so Re(ω) ∈ {0, 1/2}. If L is a square lattice then a ∈ {±1 ± i}. If L is a hexagonal lattice then a ∈ {±1, ±eπi/3, ±e2πi/3 }. If |ω| = 1 and L is neither square nor hexagonal then a ∈ {±ω}.

Proof. (1) Let π : C → S ; z → z + L be the universal covering of S. As σ is locally a homeomorphism, the composite σ ◦ π : C → S is also a covering. Since its domain is simply connected, it must be isomorphic to the universal covering of S. Hence there exists a continuous function f : C → C such that π ◦ f = σ ◦ π. Since π is a holomorphic map and σ is antianalytic, it follows that f is antianalytic too, see [5, Theorem 1]. Let κ : C → C ; z → z¯ be complex conjugation. Then g = f ◦ κ : C → C is an analytic function that satisfies π ◦ g = (π ◦ f ) ◦ κ = σ ◦ (π ◦ κ) = σ ◦ p, where p = π ◦ κ : C → C/L ; z → z¯ + L. We first show that the derivative g is a bounded function. Once this is proved, g would be constant and so g(z) = az + b, where a ∈ C \ {0} and b ∈ C satisfies σ(0 + L) = b + L. In such a case we will have f (z) = a¯ z + b and so σ(z + L) = σ(π(z)) = π(f (z)) = π(a¯ z + b) = (a¯ z + b) + L.

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Consider the lattice K = κ(L) and let T be the torus T = C/K. The covering p : C → S factorizes throughout T as p = κ ¯ ◦ π where ¯ : T → S ; z + K → z¯ + L. π : C → T ; z → z + K and κ Then, the map τ = σ ◦ κ ¯ : T → S satisfies the equality κ ◦ π ) = σ ◦ p = π ◦ g. τ ◦ π = σ ◦ (¯ Fix a point k ∈ K and consider the holomorphic function gk : C → C ; z → g(z + k) − g(z), with image gk (C) ⊂ L. In fact, π (z + k) = π (z) for each z ∈ C and so π(gk (z)) = π(g(z + k)) − π(g(z)) = τ ◦ π (z + k) − τ ◦ π (z) = 0. Consequently, since C is connected and L is discrete, gk is constant. Thus its derivative vanishes identically, that is g (z + k) = g (z) for all (z, k) ∈ C × K. Therefore, g (C) = g (P) where P is the closure of a fundamental parallelogram of K. Since P is compact, the derivative g is a bounded function and so it is constant. This yields the formula for σ in the statement. Observe that a = 0 since otherwise g would be constant and so would p. We now prove the other conditions for a and b in part (1). Since σ has order two we get z + L = σ 2 (z + L) = |a|2 z + a¯b + b + L and so (|a|2 − 1)z + a¯b + b ∈ L for all z ∈ L. The discreteness of L yields |a| = 1 and so a¯b + b ∈ L. Now, for each ∈ L we have (a¯ + b) + L = π(a¯ + b) = π(f ()) = σ(π()) = σ(0 + L) = b + L. So a¯ ∈ L and hence aL ⊂ L. Conjugating this expression and using that a ¯ = 1/a we get L ⊂ aL. This concludes the proof of part (1). (2) Since a¯ ω ∈ aL = L and by assumption a = ±1, we get ω ¯ ∈ L. So also ω+ω ¯ = 2Re(ω) belongs to L and, by conditions (4.1), 2Re(ω) ∈ L ∩ (−1, 1] = {0, 1}. That is, Re(ω) ∈ {0, 1/2}. In what follows we write a = r + sω, with r, s ∈ Z, and so 1 = |a|2 = (r + sω)(r + s¯ ω ) = r2 + 2Re(ω)rs + |ω|2 s2 .


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(3) Since |a| = 1 it is enough to check that s = 0. Assume to a contrary, that s = 0. Recall that −1 < 2Re(ω) ≤ 1 by conditions (4.1). So, as |ω| > 1, we get 1 = r2 + 2Re(ω)rs + |ω|2 s2 > r2 + 2Re(ω)rs + s2 ≥ ≥ r2 − |rs| + s2 = (|r| − |s|)2 + |rs|. Thus |r| = |s| and rs = 0, that is, r = s = 0 and therefore a = 0, a contradiction. (4) By Proposition 4.2.4 we may assume that ω = i. Then Re(ω) = 0 and so 1 = s2 + r2 . Consequently, either s = 0 and r = ±1 or r = 0 and s = ±1. This means that a ∈ {±1, ±i}. (5) Again by Proposition 4.2.4, we may assume that ω = eπi/3 and so 2Re(ω) = 1. Thus, 1 = r2 + 2Re(ω)rs + |ω|2 s2 = r2 + rs + s2 = (r + s/2)2 + 3s2 /4, which implies s2 = 0 or 1. Hence a = ±1 if s = 0, and r(r + s) = 0 if s2 = 1. Therefore a = ±ω if r = 0 and a = ±(ω − 1) = ±ω 2 if r + s = 0. Consequently, a ∈ {±1, ±eπi/3 , ±e2πi/3 }. (6) Now Re(ω) ≥ 0 by conditions (4.1) and, in fact, Re(ω) ∈ (0, 1/2) because L is neither a square nor a hexagonal lattice. Then rs ≤ 0, since otherwise 1 = r2 + 2Re(ω)rs + |ω|2 s2 = r2 + 2Re(ω)rs + s2 > s2 + r2 ≥ 2, a contradiction. In fact rs = 0 because rs < 0 implies 1 = r2 + 2Re(ω)rs + |ω|2 s2 = r2 + 2Re(ω)rs + s2 > r2 + rs + s2 = (r + s)2 − rs = (r + s)2 + |rs|. This means r = s = 0, again a contradiction. Therefore either r = 0 and s = ±1 or r = ±1 and s = 0. In the last case a = ±1, which implies, by part (3), that Re(ω) ∈ {0, 1/2} against the hypothesis. Therefore a = ±ω.

As a consequence of this Proposition 4.2.7, we get the following characterization of the symmetric tori C/L(ω) in terms of ω. Corollary 4.2.8. Let ω ∈ C satisfy conditions (4.1) and let L = Z + Zω. The torus S = C/L is symmetric if and only if either |ω| = 1 or |ω| > 1 with Re(ω) ∈ {0, 1/2}. Proof. The necessity of the conditions in the statement follows from part (3) in Proposition 4.2.7. For the converse, suppose first that |ω| > 1 and Re(ω) = 0 or Re(ω) = 1/2. In the first case ω ¯ = −ω ∈ L, and in the second one ω ¯ = 1 − ω ∈ L. Thus L is invariant under complex conjugation and it follows that σ : S → S ; z + L → z¯ + L

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is a symmetry of S. On the other hand, if |ω| = 1 then S admits the symmetry σ : S → S ; z + L → ω z¯ + L because ω ω ¯ = 1 ∈ L.

Next we classify, up to conjugation, all symmetries of a given torus. In order to do that it is convenient to know the explicit formula for f ◦ σ ◦ f −1 where σ is a symmetry and f is an automorphism which, by Remark 1.5.6, may be assumed to be analytic. Proposition 4.2.9. Let L ⊂ C be a lattice and let f : S → S ; z + L → (cz + d) + L and σ : S → S ; z + L → (a¯ z + b) + L be, respectively, an analytic automorphism and a symmetry of S = C/L. (1) Then ¯ + L. f ◦ σ ◦ f −1 : S → S ; z + L → (ac2 z¯ + bc + d − ac2 d) (2) Suppose that L is neither square nor hexagonal. Then ¯ + L. z + bc + d − ad) f ◦ σ ◦ f −1 (z + L) = (a¯ Proof. Composing f with the translation C/L → C/L ; z +L → (z −d)+L yields the automorphism z + L → cz + L of C/L which fixes 0 + L. So c is a primitive nth -root of unity for n = 2, 3, 4 or 6, by Proposition 4.2.5. Using this fact, the proof of part (1) is a straightforward computation. For the second part, notice that Proposition 4.2.5 also implies that c2 = 1 whenever L is neither a square nor a hexagonal lattice.

To begin with, we classify the symmetries of those tori C/L(ω) with |ω| > 1. Proposition 4.2.10. Let ω ∈ C satisfy conditions (4.1) with |ω| > 1 such that Re(ω) = 0 or Re(ω) = 1/2. Let L = Z + Zω and S = C/L. (1) If Re(ω) = 0 then each symmetry of S is conjugate to one of the following: σ1 : S → S ; z + L → z¯ + L;

σ2 : S → S ; z + L → (¯ z + 1/2) + L;

σ3 : S → S ; z + L → −¯ z + L;

σ4 : S → S ; z + L → (−¯ z + ω/2) + L.

The surfaces S/ σ1 and S/ σ3 are closed annuli, while S/ σ2 and S/ σ4 are Klein bottles.


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(2) If Re(ω) = 1/2 then each symmetry of S is conjugate to one of the following: σ1 : S → S ; z + L → z¯ + L;

σ3 : S → S ; z + L → −¯ z + L.

Both surfaces S/ σ1 and S/ σ3 are Möbius strips. (3) The symmetries above are pairwise non-conjugate. Proof. Given a symmetry σ of S there exist, by part (3) in Proposition 4.2.7, complex numbers a, b with a ∈ {1, −1} such that σ : S → S ; z + L → (a¯ z + b) + L. Conjugating σ by the automorphism f : S → S ; z + L → (z − b/2) + L we get f ◦ σ ◦ f −1 : z + L → a¯ z+

1 ¯ ab + b + L. 2

We may assume, after translating by an element of L, that (a¯b + b)/2 lies in the fundamental parallelogram RL = {r + sω : 0 ≤ r, s < 1} of L. Since a¯b + b belongs to L (by part (1) in Proposition 4.2.7), we see that a¯b + b is one of the four vertices of RL , that is, a¯b + b ∈ {0, 1, ω, 1 + ω}. We distinguish cases according to whether Re(ω) = 0 or Re(ω) = 1/2. (1) Re(ω) = 0. If a = 1 then a¯b + b = 2Re(b), which is a real number. So a¯b + b = 0 or 1. In the first case, σ is conjugate to σ1 ; in the second one, σ is conjugate to σ2 . If a = −1 then a¯b + b = 2Im(b)i, which is purely imaginary. So a¯b + b = 0 or ω. In the first case σ is conjugate to σ3 ; in the second one, σ is conjugate to σ4 . Next we must determine the topology of the orbit spaces S/ σi for 1 ≤ i ≤ 4. Since S has genus one, each surface S/ σi is either an annulus or a Möbius strip or else a Klein bottle. Therefore the topology of each S/ σi is determined by the number (two, one or zero, respectively) of its boundary components, that is, by the number of ovals of the fixed point set Fix(σi ) of each symmetry σi . Let z = r+sω with 0 ≤ r, s < 1 be a point of the fundamental parallelogram RL . It is straightforward to check that a point z + L is fixed by σ1 : z + L → z¯ + L if and only if 2sω ∈ L, that is, 2s ∈ Z. But s ∈ [0, 1) and so either s = 0 or s = 1/2. This means that Fix(σ1 ) consists of the projection under the covering map π : C → S of the two horizontal segments [0, 1) × {0} and [0, 1) × {ω/2} of R. It is clear that the projections are two disjoint ovals in S; so Fix(σ1 ) has two ovals and S/ σ1 is indeed an annulus. An analogous argument shows that Fix(σ3 ) consists of the projection under π of the two vertical segments {0} × [0, ω) and {1/2} × [0, ω). The projections are also disjoint and therefore S/ σ3 is an annulus.

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On the other hand, both σ2 and σ4 are fixed point free and so both S/ σ2 and S/ σ4 are Klein bottles. (2) Re(ω) = 1/2. If a = 1 then the same argument as above shows that σ is conjugate to σ1 or σ2 . But now σ1 and σ2 are conjugate. Indeed, the automorphism g : S → S ; z + L → (z + Im(ω)i/2) + L satisfies g ◦ σ2 ◦ g −1 = σ1 , as is easy to check. If a = −1 then a¯b + b = 2Im(b)i ∈ iR. As Re(ω) = 1/2, the unique vertex of the fundamental parallelogram RL that lies in the imaginary axis is 0. So a¯b + b = 0 and therefore σ is conjugate to σ3 : z + L → −¯ z + L. Next we show that S/ σ1 and S/ σ3 are Möbius strips, or equivalently, their boundaries are connected and non-empty. We study just the first case. Let us write z = r + sω with 0 ≤ r, s < 1. It is straightforward to check that a point z + L is fixed by σ1 : z + L → z¯ + L if and only if s(ω − ω ¯ ) ∈ L, that is, 2sIm(ω)i ∈ L. Since 2Im(ω)i = 2ω − 1 belongs to L and there is no point of L in the open segment joining 0 and 2Im(ω)i, we see that s = 0. Therefore Fix(σ1 ) consists of the projection under π : C → S of the horizontal segment [0, 1) × {0} of the fundamental parallelogram R. This shows that S/ σ1 is indeed a Möbius strip. (3) We have already proved that σ1 (respectively σ3 ) in (1) is not conjugate to σ2 (respectively σ4 ) because S/ σ1 and S/ σ3 are closed annuli and S/ σ2 and S/ σ4 are Klein bottles. On the other hand, it follows from part (2) in Proposition 4.2.9 that σ1 and σ3 in (1) and (2) are not conjugate. By the same reason σ2 and σ4 in (1) are not conjugate. The classification of the symmetries of the tori C/L(ω) where L(ω) is neither square nor hexagonal and with |ω| = 1 is the following. Proposition 4.2.11. Let ω ∈ C satisfy conditions (4.1) with |ω| = 1 and such that 0 < Re(ω) < 1/2. Let L = Z + Zω and S = C/L. Then each symmetry of S is conjugate to one of the following: σ1 : S → S ; z + L → ω z¯ + L;

σ2 : S → S ; z + L → −ω z¯ + L,

and both orbit spaces S/ σ1 and S/ σ2 are Möbius strips. Moreover, σ1 and σ2 are not conjugate. Proof. The last part is an immediate consequence of part (2) in Proposition 4.2.9. For the first one we repeat the same arguments as in the beginning of the proof of Proposition 4.2.10 to get that any symmetry of S is conjugate to σ : z + L → a¯ z+

1 ¯ ab + b + L where a¯b + b ∈ {0, 1, ω, 1 + ω}. 2

Here a = ±ω by part (6) in Proposition 4.2.7. Let us write ω = eiθ with π/3 < θ < π/2. We distinguish cases according to whether a = ω or a = −ω.


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If a = ω then a¯b + b = eiθ ¯b + b = eiθ/2 eiθ/2¯b + e−iθ/2 b ∈ Reiθ/2 . So a¯b + b = 0 or 1 + ω since the argument of the other points in {0, 1, ω, 1 + ω} is different to θ/2. Thus σ : z + L → ω z¯ + L or σ : z + L → (ω z¯ + (1 + ω)/2) + L, respectively. But these two symmetries are conjugate via the automorphism z+L → z + (ω − 1)/4 + L, as is easy to check. If a = −ω then, writing β = ¯beiθ/2 , we get a¯b + b = −eiθ¯b + b = −eiθ/2 ¯beiθ/2 − be−iθ/2

= ei(π+θ/2) β − β¯ = ei(π+θ/2) 2Im(β)i ∈ Rei(θ−π)/2 . So a¯b + b = 0 since the argument of the other points in {0, 1, ω, 1 + ω} is different to (θ − π)/2. Thus σ(z + L) = −ω z¯ + L = σ2 (z + L). To finish we have to prove that S/ σ1 and S/ σ2 are Möbius strips or, equivalently, their boundaries are connected and non-empty. Both cases are similar and so we just deal with S/ σ1 . Let us write z = r + sω with 0 ≤ r, s < 1. It is straightforward to check that z + L is fixed by σ1 : z + L → z¯ + L if and only if (r − s)(1 − ω) ∈ L. This implies that r − s is an integer and necessarily r = s because |r − s| < 1. Therefore Fix(σ1 ) consists of the projection under π : C → S of the diagonal of the fundamental parallelogram R joining 0 and 1 + ω. This shows that S/ σ1 is indeed a Möbius strip.

To finish this section we classify the symmetries of the special tori, that is, those tori S = C/L where the lattice L is either square or hexagonal. By Proposition 4.2.4 the lattice L may be assumed to be L = Z + Zi if L is a square lattice and L = Z + Zeπi/3 if L is a hexagonal lattice. Proposition 4.2.12. Let L = Z + Zi and S = C/L. Then each symmetry of S is conjugate to one of the following: σ1 : z + L → z¯ + L;

σ2 : z + L → (¯ z + 1/2) + L,

σ3 : z + L → i¯ z + L.

Moreover, S/ σ1 is a closed annulus, S/ σ2 is a Klein bottle and S/ σ3 is a Möbius strip. In particular, the symmetries σ1 , σ2 and σ3 are pairwise nonconjugate.

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Proof. Repeating the same arguments as in the beginning of the proof of Proposition 4.2.10 we get that any symmetry of S is conjugate to σ : z + L → a¯ z+

1 ¯ ab + b + L 2

where a¯b + b ∈ {0, 1, i, 1 + i}.

Here a ∈ {±1, ±i} by part (4) in Proposition 4.2.7. If a = 1 then a¯b + b = ¯b + b = 2Re(b) ∈ R. So a¯b + b = 0 or 1. In the first case σ = σ1 and in the second, σ = σ2 . If a = −1 then a¯b + b = −¯b + b = 2Im(b)i ∈ iR. So a¯b + b = 0 or i. In the first case σ : z + L → −¯ z + L; in the second, σ : z + L → (−¯ z + 1/2) + L. However, conjugating by the automorphism z + L → iz + L we see that the first symmetry is conjugate to σ1 and the second one to σ2 . If a = i then a¯b + b = i¯b + b = eπi/4 (¯beπi/4 + be−πi/4 ) ∈ Reπi/4 . So a¯b + b = 0 or 1 + i, that is, σ(z + L) = i¯ z + L = σ3 (z + L)

or σ(z + L) = (i¯ z + (1 + i)/2) + L.

However, conjugating by the automorphism z + L → (z + (i − 1)/4) + L we see that both symmetries are conjugate. If a = −i then a¯b + b = −i¯b + b = e3πi/2¯b + b = e3πi/4 (¯be3πi/4 + be−3πi/4 ) ∈ 3πi/4 Re . So a¯b + b = 0, that is, σ(z + L) = −i¯ z + L. However, this symmetry is conjugate to σ3 via the automorphism z + L → −iz + L, as is easy to check. As to the topological types of the orbit spaces S/ σi , the calculations used to compute the species of symmetries σ1 and σ2 in Proposition 4.2.10 show that the fixed point set of σ1 has 2 connected components, while the fixed point set of σ2 is empty. Thus S/ σ1 is a closed annulus and S/ σ2 is a Klein bottle. Finally, by repeating the arguments employed in the proof of Proposition 4.2.11, we deduce that S/ σ3 is a Möbius strip.

Proposition 4.2.13. Let S = C/L where L = Z + Zω and ω = eπi/3 . Then each symmetry of S is conjugate to one of the following: σ1 : S → S ; z + L → z¯ + L;

σ2 : S → S ; z + L → ω z¯ + L.

Each orbit space S/ σi is a Möbius strip and the symmetries σ1 and σ2 are nonconjugate. Proof. The same arguments as in the beginning of the proof of Proposition 4.2.10 yields that any symmetry of S is conjugate to z + L → a¯ z+

1 ¯ ab + b + L 2

where a¯b + b ∈ {0, 1, ω, 1 + ω}.


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Here a = ekπi/3 = ω k with k ∈ {0, . . . , 5} by part (5) in Proposition 4.2.7. Observe that a¯b + b = ekπi/6 ¯bekπi/6 + be−kπi/6 ∈ Rekπi/6 . So if k = 3, 4 or 5 then a¯b + b = 0 since the other points in {0, 1, ω, 1 + ω} have argument different to kπ/6. If k = 0, 1 or 2 then, in addition to 0, the term a¯b + b can also attain the value 1, 1 + ω or ω, respectively. This yields nine symmetries in total and our goal is to show that each of them is conjugate either to σ1 or to σ2 . For those symmetries with a¯b + b = 0 we do as follows: there exists c = enπi/3 , for a suitable n, such that ac2 = 1 if a = ekπi/3 with k even and ac2 = ω if k is odd. For such a value of c, the transformation fc : z + L → cz + L is an automorphism of S (by Proposition 4.2.5) which conjugates the symmetry z + L → a¯ z + L to σ1 if k is even and to σ2 if k is odd. As to the symmetry z + L → (ω 2 z¯ + ω/2) + L, it is easy to check that it is conjugate to z + L → (¯ z + 1/2) + L via the automorphism fc with c as above. Finally, the symmetries z+L → (¯ z +1/2)+L and z+L → (ω z¯+(ω+1)/2)+L are conjugate, via the automorphism f : S → S given by z + L → (z + ω/2) + L, to σ1 and σ2 respectively. This proves the first part of the statement. Moreover, Proposition 4.2.9 yields that σ1 and σ2 are non-conjugate because c2 = ω for all c such that c6 = 1. To finish we will prove that S/ σ1 is a Möbius strip; we leave the details of the remaining case S/ σ2 to the reader. Let us write z = r+sω with 0 ≤ r, s < 1. It is straightforward to check √ that z+L is fixed by σ1 : z + L → z¯ + L if and only if s(ω − ω ¯ ) ∈ L, that is, s 3i ∈ L. This implies s = 0. Consequently, Fix(σ1 ) consists of the projection under π : C → S of the horizontal segment [0, 1) × {0} of the fundamental parallelogram R. This shows that S/ σ1 is indeed a Möbius strip.

As a consequence of the results in this section we obtain Corollary 4.2.14, which describes the symmetry type of each torus. However, it must be pointed out that the above results provide us with more information than just the symmetry type since they describe explicitly a symmetry which realizes each species. Corollary 4.2.14. Let L = Z + Zω be a lattice with ω satisfying conditions (4.1). Then the symmetry type of the torus S = C/L is: (1) (2) (3) (4) (5)

{0, 0, +2, +2} if |ω| > 1 and Re(ω) = 0; {−1, −1} if |ω| > 1 and Re(ω) = 1/2; {−1, −1} if |ω| = 1 and 0 < Re(ω) < 1/2; {−1, 0, +2} if ω = i; {−1, −1} if ω = eπi/3 .

For any other value of ω the torus is not symmetric.

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4.3 Symmetry Types of Hyperelliptic Riemann Surfaces Hyperelliptic Riemann surfaces have attracted the attention of geometers since long time ago. The study of their symmetries began with the works of Natanzon [98, 99] at the end of the seventies of the past century, and Maskit [78] in 1995. They both restricted themselves to the case of symmetries whose number of ovals is maximal. Later on, the authors of this monograph classified all symmetries of each hyperelliptic surface without any assumption on the number of their ovals, [14]. This work is too extensive even to survey it here. Hence we present in this section the main ingredients to understand why the computation of the symmetry types of hyperelliptic surfaces is achievable. We will also explain an illustrative example in detail. It is worth mentioning that the results are also described in an algebraic way. That is, the surfaces are given by means of defining polynomial equations and the symmetries are described as the composite of complex conjugation with birational transformations. A Riemann surface is called cyclic p-gonal if it is a cyclic p-covering of the Riemann sphere. The 2-gonal surfaces are, therefore, the hyperelliptic ones. Symmetries on trigonal and cyclic p-gonal surfaces, for p an odd prime, have been studied by Costa and Izquierdo in [35, 36]. The proofs in these two articles are based on the theory of NEC groups, in contrast to the more geometric approach we follow here with hyperelliptic surfaces. We first recall the notion of hyperellipticity. Definition 4.3.1. A Riemann surface S of genus g ≥ 2 is hyperelliptic if any of the following equivalent conditions holds: (1) There exists a meromorphic function πS : S → Σ = C ∪ {∞} of degree 2. (2) There exists an analytic automorphism ρS : S → S of order 2 with 2g + 2 fixed points. (3) There exists an analytic automorphism ρS : S → S of order 2 such that the orbit space S/ ρS has genus 0. The automorphism ρS is unique and it is called the hyperelliptic involution of S. It is central in the full group Aut(S) of analytic and antianalytic automorphisms of S. Hyperelliptic Riemann surfaces have the nice property of admitting an easy representation by polynomials. Indeed, it is well known that any such surface can be obtained, after desingularization, from the set of solutions in C2 of an equation of the form y 2 = P (x), where P ∈ C[x] is a monic polynomial whose roots in C are simple (see, for example, Sects. 1 and 4 of Chapter III in [92]). Let us fix some notations to be used later. Notations 4.3.2. A hyperelliptic Riemann surface S of genus g will be represented by the affine plane model S = {(x, y) ∈ C2 : y 2 = PS (x)},

4.3 Symmetry Types of Hyperelliptic Riemann Surfaces

83

where PS (x) = (x − e1 ) · · · (x − e2g+1+δ ) with ei = ej if i = j and δ = 0 or 1. In this model we identify the characteristic elements of a hyperelliptic surface. First, the projection πS onto the first coordinate πS : S → Σ ; (x, y) → x is a meromorphic function of degree 2. Its 2g + 2 branch points are thus the roots of PS together with ∞ if δ = 0. They constitute what we call (by abuse of language) the branch point set of S, which we denote by BS . With the above notations, BS =

if δ = 1, {e1 , . . . , e2g+2 } {e1 , . . . , e2g+1 , ∞} if δ = 0.

The automorphism of S interchanging the two sheets of πS , i.e., the hyperelliptic involution, has the following formula ρS : S → S ; (x, y) → (x, −y). In the sequel we will denote it by ρ if no confusion arises. Analytic isomorphisms between hyperelliptic Riemann surfaces are closely related to analytic automorphisms of the Riemann sphere, i.e. to Möbius transformations. We now describe this relation in terms of polynomial equations of such surfaces. Let S and T be hyperelliptic surfaces of genus g whose branch point sets are denoted by BS and BT , respectively. Every analytic isomorphism f : S → T induces a Möbius transformation fˆ : Σ → Σ which maps BS onto BT . In fact, fˆ is defined by the formula fˆ(πS (p)) = πT (f (p)) for any p ∈ S. f S

-

πS ? Σ

fˆ-

T πT ? Σ

Conversely, every Möbius transformation m : Σ → Σ which maps BS onto BT induces exactly two analytic isomorphisms f1 , f2 : S → T such that fî = m, for i = 1, 2. In fact, f2 = f1 ◦ ρS = ρT ◦ f1 . We call these isomorphisms liftings of m. Their formulae can be calculated explicitly.

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Liftings of Möbius Transformations Let us write S = {y 2 = PS (x)} and T = {y 2 = PT (x)}. Let m(x) =

ax + b with {a, b, c, d} ⊂ C and det m := ad − bc = 0 cx + d

be a Möbius transformation such that m(BS ) = BT . Then the formulae of its liftings, say f and f ◦ ρS , depend on whether ∞ ∈ BS or not and whether m fixes ∞ or not. These formulae, as they appear in [30], are the following. (1) If ∞ ∈ BS and m(∞) = ∞ then f (x, y) =

a g b a x+ , y· d d d

√ det m . d

(2) If ∞ ∈ BS and m(∞) = ∞ then f (x, y) =

y · cg ax + b , cx + d (cx + d)g+1

− det m · PT (a/c) .

(3) If ∞ ∈ / BS and m(∞) = ∞ then f (x, y) =

a g+1 b a x+ , y· . d d d

(4) If ∞ ∈ / BS and m(∞) = ∞ then f (x, y) =

y · cg+1 ax + b , P (a/c) . T cx + d (cx + d)g+1

If a Möbius transformation which maps BS onto BT is antianalytic, i.e., of the form (a¯ x + b)/(c¯ x + d), then the formulae of its liftings are obtained from the above ones just by replacing x and y for their complex conjugates x ¯ and y¯ respectively. In this section we are mainly interested in automorphisms of S, whose formulae are obtained by making T = S in the above. In this case, the relation between isomorphisms of hyperelliptic surfaces and Möbius transformations states that the group Aut(S) of automorphisms of S consists exactly of the liftings of those Möbius transformations (analytic or antianalytic) which preserve the branch point set BS . For notational convenience, we denote such a group by AutΣ (S): AutΣ (S) := {m ∈ Aut(Σ) : m(BS ) = BS }. Note that, algebraically, AutΣ (S) is nothing else but the factor group Aut(S)/ ρ.


85

According to the classification of the finite automorphism groups of the sphere, there are ten different classes of such groups which contain an antianalytic involution. They appear in a natural way as AutΣ (S) in the combinatorial study of Aut(S).

4.3.1 A Geometric Method Apart from the analytic methods used in Sects. 4.1 and 4.2 to compute the symmetry types of the Riemann sphere and tori, respectively, for surfaces of higher genus the most common method is the combinatorial study of NEC groups. However, the nice properties of hyperelliptic surfaces allow to develop a geometric method which turns out to be easier than the combinatorial one. The goal of this subsection is to explain this geometric approach. Assume that the hyperelliptic surface S admits a symmetry σ. Our purpose here is to compute the species of σ in terms of both its formula and the equation of S. This is summarized in Theorems 4.3.4 and 4.3.5 below, which are based on results in [57, Sect. 6], adapted to our point of view. In the same way as analytic isomorphisms induce Möbius transformations, the symmetry σ : S → S induces an antianalytic Möbius transformation σ : Σ → Σ, which also has order 2. We proved in Theorem 4.1.2 that if σ fixes points then it is conjugate to complex conjugation x → x ¯; otherwise it is conjugate to the antipodal map x → −1/¯ x. Let us assume first that σ is complex conjugation. In this case the polynomial PS defining the surface S has real coefficients since its roots are permuted by σ . Then one of the liftings of σ is σ : (x, y) → (¯ x, y¯) and so its fixed point set consists exactly of the points with real coordinates of the surface S = {y 2 = PS (x)}. It is then easy to see that if PS has 2k > 0 real roots then Fix(σ) consists of exactly k ovals. Moreover, the separating character of Fix(σ) depends on the number of real roots of PS or, more precisely, since ∞ may be a branch point, on the number of branch points of S fixed by σ . Indeed, if 2k is the number of branch points fixed by σ then ⎧ g + 1 if k = g + 1; ⎪ ⎪ ⎨ −k if 0 < k < g + 1; sp(σ) = ⎪ 1 if k = 0 and g is even; ⎪ ⎩ 2 if k = 0 and g is odd. The other lifting of σ is σ ◦ ρ : (x, y) → (¯ x, −¯ y). If σ fixes no branch point then PS is always positive on R; it follows that in this case Fix(σ ◦ ρ) is empty, that is, sp(σ ◦ ρ) = 0. If σ fixes some branch point we claim that the species of σ ◦ ρ coincides with the species of σ. To show this we use the obvious fact that if α and β are symmetries of two different Riemann surfaces S and T respectively, and f ◦ α = β ◦ f for some isomorphism f : S → T , then the species of α and β coincide.

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Lemma 4.3.3. If σ fixes some branch point then sp(σ ◦ ρ) = sp(σ). Proof. Let r be the greatest real root of PS and let m be the real Möbius transformation m : Σ → Σ ; x → −1/(x − r). Consider the hyperelliptic Riemann surface T whose branch point set is BT = m(BS ). Clearly, γ : T → T ; (x, y) → (¯ x, y¯) is a symmetry of T . Denoting by f a lifting of m it is easy to check that f ◦ σ ◦ ρ = γ ◦ f and so sp(σ ◦ ρ) = sp(γ ). It remains to see that sp(γ) = sp(σ). Since both symmetries γ and σ have the same formula (x, y) → (¯ x, y¯), it follows from the above that sp(γ) = sp(σ) if and only if the number of branch points of T fixed by γ coincides with the number of branch points of S fixed by σ . But this is clear because m ◦ σ ◦ m−1 = γ and so the branch points of T fixed by γ are the images by m of the branch points of S fixed by σ .

Theorem 4.3.4. Suppose that the branch point set of the hyperelliptic surface S of genus g is preserved by complex conjugation σ . Then S admits the symmetries defined by σ : S → S ; (x, y) → (¯ x, y¯) and σ ◦ ρ : S → S ; (x, y) → (¯ x, −¯ y). Let 2k be the number of branch points of S fixed by σ . (1) If k > 0 then sp(σ) = sp(σ ◦ ρ) =

g + 1 if k = g + 1, −k if k < g + 1.

(2) If k = 0 then sp(σ) =

1 if g is even, 2 if g is odd;

sp(σ ◦ ρ) = 0.

If the branch point set of S is preserved by an antianalytic involution σ different to complex conjugation but analytically conjugate to it, then we claim that the species of its liftings σ and σ ◦ ρ also depend only on the number of branch points fixed by σ . This can be proved similarly to the above lemma. If m ◦ σ ◦ m−1 = σ , then σ : T → T ; (x, y) → (¯ x, y¯) and σ ◦ ρ are symmetries of the hyperelliptic surface T whose branch point set is m(BS ). In fact, if f is a lifting of m then either f ◦ σ ◦ f −1 = σ (and so f ◦ σ ◦ ρ ◦ f −1 = σ ◦ ρ) or f ◦ σ ◦ f −1 = σ ◦ ρ (and so f ◦ σ ◦ ρ ◦ f −1 = σ). In both cases {sp(σ ), sp(σ ◦ ρ)} = {sp(σ), sp(σ ◦ ρ)}, which means that the species of σ and σ ◦ ρ depend only on the number of branch points of T fixed by σ . But the branch points of T fixed by σ are the images under m of the branch points of S fixed by σ . This proves our claim. Suppose now that σ is the antipodal map σ : Σ → Σ ; x → −1/¯ x. We claim that in this case the genus g must be odd since otherwise its liftings would not be involutions. Indeed, using the formulae of its liftings we see immediately that their squares are


87

σ 2 (x, y) = (σ ◦ ρ)2 (x, y) = x, y · (−1)g+1 .

So σ 2 = ρ for even g, i.e., σ is a pseudosymmetry in Singerman’s terminology from [120]. As to their species, since σ fixes no point, the same happens to its liftings. Therefore sp(σ) = sp(σ ◦ ρ) = 0. Summing up, we have shown the following theorem, which gives the species of the liftings of an antianalytic involution σ . If σ is conjugate to complex conjugation then we call it a reflection. Theorem 4.3.5. Let σ : Σ → Σ be an antianalytic involution preserving the branch point set of the hyperelliptic surface S of genus g. Let σ and σ ◦ ρ be its liftings and assume that both have order 2. (1) If σ is conjugate to the antipodal map then g is odd and sp(σ) = sp(σ ◦ ρ) = 0. (2) If σ is a reflection which fixes 2k > 0 branch points of S then sp(σ) = sp(σ ◦ ρ) =

g+1 −k

if k = g + 1, if k < g + 1.

(3) If σ is a reflection which fixes no branch point of S then {sp(σ), sp(σ ◦ ρ)} =

{1, 0} {2, 0}

if g is even, if g is odd.

Remark 4.3.6. Observe that in case (3) we do not know which one is the lifting with non-zero species, except if σ is complex conjugation (see Theorem 4.3.4). When dealing with polynomial equations we can exhibit explicit formulae of both liftings and so we can decide which one fixes points.

4.3.2 An Example Let S be a symmetric hyperelliptic Riemann surface of genus g and let AutΣ (S) be the group of Möbius transformations preserving its branch point set BS . This is a finite group of isometries of the Riemann sphere and so it has a nice geometric interpretation. For instance, AutΣ (S) can be the group of (orientation preserving or reversing) isometries of a regular cube, and it is easy to describe how its elements permute the 2g + 2 branch points of S. This allows us to describe the distribution of the branch points on the sphere and, in particular, to find out how many of them are fixed by a given symmetry, say σ . Once this is achieved, we can use Theorem 4.3.5 to compute the species of its liftings σ and σ ◦ ρ (provided they are involutions). The distribution of the branch points of S also provides an algebraic equation of S. With this equation at hand, the formulae of the automorphisms of S can be explicitly computed as liftings of the Möbius transformations in AutΣ (S). This way we get a presentation by means of generators and defining relations of the full group

88


Aut(S). It is now an exercise in finite group theory to determine the conjugacy classes of symmetries in Aut(S). Finally, this yields the symmetry type of S. Consequently, the computation of the symmetry types of hyperelliptic surfaces by means of this geometric method splits naturally into different cases according to the different types of groups AutΣ (S). We develop here in detail the case in which AutΣ (S) is a dihedral group of order 2n ≥ 4 generated by two reflections σ 1 and σ 2 (surfaces of this type are called surfaces of class I in [14]). For simplicity, we just consider the case in which the two fixed points of the rotation σ 2 ◦ σ 1 are branch points of S. Theorem 4.3.7. Let S be a hyperelliptic Riemann surface of genus g such that the induced group AutΣ (S) of Möbius transformations is dihedral of order 2n ≥ 4 generated by two reflections σ 1 and σ 2 . Assume that the two fixed points of the rotation σ 2 ◦ σ 1 are branch points of S. Then there exist non-negative integers r, p1 and p2 with 2r + p1 + p2 = 2g/n ≥ 3 such that S admits an algebraic equation of the form y2 = x ·

r

(xn − wjn )(xn − w ¯jn ) ·

j=1

p1

(xn − λnj ) ·

j=1

p2

(xn + μnj ),

j=1

where the roots of the polynomial on the right hand side are simple and λnj and μnj are positive real numbers for all j. Moreover, the full group Aut(S) of automorphisms of S is dihedral of order 4n generated by the symmetries σ1 : S → S ; (x, y) → (¯ x, y¯) and σ2 : S → S ; (x, y) → x ¯e2πi/n , y¯eπi/n . These are representatives of the unique two conjugacy classes of symmetries of S and their species are ⎧ ⎨ sp(σi ) =

g+1 −(pi + 1) ⎩ −(p1 + p2 + 2)/2

if pi = g (and hence n = 2), if n is even and pi < g, if n is odd,

for i = 1, 2. This yields the symmetry type of S, namely (1) {−1, g + 1} if pi = g for i = 1 or 2 (and hence n = 2), (2) {−(p1 + 1), −(p2 + 1)} if n is even and pi < g for i = 1 and 2, (3) {−(p1 + p2 + 2)/2, −(p1 + p2 + 2)/2} if n is odd. Proof. After conjugation by a Möbius transformation, we may choose the following reflections as generators for the dihedral group AutΣ (S): σ 1 : Σ → Σ ; x → x ¯ and σ 2 : Σ → Σ ; x → x ¯e2πi/n . 2 is the line {reπi/n : The fixed point set of σ 1 is the real axis R∪{∞} while that of σ r ∈ R}∪{∞}. The two fixed points of the rotation σ 2 ◦ σ1 : Σ → Σ ; x → x·e2πi/n


are 0 and Σ is

89

∞. It follows that a fundamental region F for the action of σ1 , σ2 in F = {reiθ : 0 ≤ r, 0 ≤ θ ≤ π/n} ∪ {∞}.

In particular, the branch point set BS of S consists of the orbits of points lying in this fundamental region. 2 consists (1) If w lies in the interior of F then its orbit under the action of σ1 , σ of the 2n roots of the polynomial (xn − wn )(xn − w ¯n ); observe that in this case wn is a complex number with positive imaginary part. (2) If λ = 0, ∞ is fixed by σ 1 (and hence it lies in the boundary of F) then its orbit consists of the n roots of the polynomial xn − λn ; observe that λn is a positive real number. (3) If ν = 0, ∞ is fixed by σ 2 (and hence it lies in the boundary of F) then its orbit consists of the n roots of the polynomial xn − ν n ; observe that ν n is a negative real number. (4) Finally, if α = 0 or ∞ then its orbit is {α} itself. Let r be the number of branch points lying in the interior of F and, for i = 1, 2, let σi ) and different from 0 and ∞. pi be the number of branch points lying in F ∩ Fix( Since we are assuming that both 0 and ∞ are also branch points, the total number of branch points of S is 2rn + p1 n + p2 n + 2. This yields the equality 2r + p1 + p2 = 2g/n in the statement of the theorem. In addition, the above also shows that a defining algebraic equation y 2 = PS (x) of S has the form given in the statement of the theorem, because the roots of PS (x) are precisely the finite branch points of S. The condition 2r + p1 + p2 ≥ 3 is necessary to assure that AutΣ (S) has no more automorphisms than those in σ1 , σ 2 and so AutΣ (S) is indeed dihedral of order 2n. In fact, if 2r + p1 + p2 = 1 or 2 then there exist Möbius transformations in AutΣ (S) \ σ1 , σ 2 . Explicitly, if p1 = 1 and r = p2 = 0 then the reflection defined as Σ → Σ ; x → λ21 /¯ x belongs to AutΣ (S) \ σ1 , σ 2 ; the case p2 = 1 and r = p1 = 0 is analogous. In case r = 1 and p1 = p2 = 0 the same happens to the reflection Σ → Σ ; x → |w1 |2 /¯ x. In case r = 0, p1 = p2 = 1 the transformation to be considered is Σ → Σ ; x → λ1 μ1 /x and finally, in case p1 = 2 and r = p2 = 0 (the case p2 = 2 and r = p1 = 0 is analogous) the transformation Σ → Σ ; x → λ1 λ2 /x works. With the algebraic equation of S at hand it is immediate to check that both maps σ1 : S → S ; (x, y) → (¯ x, y¯) and σ2 : S → S ; (x, y) → (¯ x · e2πi/n , y¯ · eπi/n ) are liftings of σ 1 and σ 2 respectively. Since σ1 and σ2 have order two and their composite σ2 ◦ σ1 : S → S ; (x, y) → (x · e2πi/n , y · eπi/n ) has order 2n, we see that they generate a dihedral group of order 4n which therefore coincides with the full group Aut(S). In particular, they are representatives of the unique two conjugacy classes of symmetries of Aut(S). Let us compute their species. To do this we use Theorem 4.3.5 and for that we need to calculate the number of branch points fixed by each σ i for i = 1, 2. We can use either the defining algebraic equation of S

90


or the geometric description of how the branch points are distributed in the above fundamental region F. Let us choose this last option. Assume first that n is even. Then the branch points fixed by σ 1 are 0, ∞, the p1 branch points lying in F ∩ Fix( σ1 ), say λ1 , . . . , λp1 , and the images of these last under the rotation ( σ2 ◦ σ 1 )n/2 : Σ → Σ ; x → −x, that is, −λ1 , . . . , −λp1 . This makes a total of 2p1 + 2 branch points lying on Fix( σ1 ). Hence, according to Theorem 4.3.5, we have sp(σ1 ) =

−(p1 + 1) if p1 + 1 < g + 1; g+1 if p1 + 1 = g + 1.

Observe that this last equality holds just for n = 2 and p2 = 0; in fact, if p1 = g then the equality 2r+p1 +p2 = 2g/n becomes 2r+p2 = g(2/n−1) which forces n = 2 and r = p2 = 0. As to the branch points fixed by σ 2 , they are 0, ∞, the p2 branch points lying in F ∩ Fix( σ2 ), say ν1 , . . . , νp2 , and the images of these last under the rotation ( σ2 ◦ σ 1 )n/2 : Σ → Σ ; x → −x, namely, −ν1 , . . . , −νp2 . This makes a total of 2p2 + 2 branch points lying on Fix( σ2 ). Hence, according to Theorem 4.3.5, sp(σ2 ) =

−(p2 + 1) if p2 + 1 < g + 1; g+1 if p2 + 1 = g + 1.

Again this last equality holds just for n = 2 and r = p1 = 0. Consequently, the symmetry type of S for n even is the one given in the statement of the theorem. Assume now that n is odd and let us keep the above meaning of λ1 , . . . , λp1 and ν1 , . . . , νp2 . In this case the branch points fixed by σ 1 are 0, ∞, λ1 , . . . , λp1 and the images of ν1 , . . . , νp2 under the rotation ( σ2 ◦ σ 1 )(n−1)/2 : Σ → Σ ; x → −x · e−πi/n . This makes a total of p1 + p2 + 2 branch points lying on Fix( σ1 ). Hence, according to Theorem 4.3.5, sp(σ1 ) = −(p1 + p2 + 2)/2. As to the branch points fixed by σ 2 , they are 0, ∞, ν1 , . . . , νp2 and the images of λ1 , . . . , λp1 under the rotation ( σ2 ◦ σ 1 )(n+1)/2 : Σ → Σ ; x → −x · eπi/n . This makes a total of p1 + p2 + 2 branch points lying on Fix( σ2 ). Hence, according to Theorem 4.3.5, sp(σ2 ) = −(p1 + p2 + 2)/2. Therefore, the symmetry type of S for n odd is {−(p1 + p2 + 2)/2, −(p1 + p2 + 2)/2}, as claimed in the statement.

Chapter 5

Symmetry Types of Riemann Surfaces with a Large Group of Automorphisms

In this chapter we study the symmetries of those surfaces S whose group Aut+ (S) of analytic automorphisms is large. As said in the introduction, the symmetries of these surfaces are topologically determined by Aut+ (S). We will explain in detail results that originally are due to the authors of this monograph in collaboration with Broughton [8, 9] and to Turbek alone [124]. More precisely, in this chapter we classify the symmetries of the Macbeath–Singerman, the Accola–Maclachlan and the Kulkarni surfaces. Our choice comes from the fact that the results are complete for these surfaces. However, the interested reader is invited to see also [15, 16, 94]. Many arguments are closely related to Gromadzki’s work in [48].

5.1 Some General Results Let us write the compact Riemann surface S as H/Γ for some surface Fuchsian group Γ and Aut+ (S) = Δ/Γ, where Δ is a Fuchsian group containing Γ as a normal subgroup. For the surfaces we shall deal with in this chapter, the group Δ has triangle signature. We shall find general formulae for the number of ovals of each symmetry of such surfaces. These formulae appeared in [48], see also the recent paper [52] for a refined version. Following Theorem 3.1.1 and Lemma 1.3.3 we divide this study into several cases according to the parity of the three proper periods of Δ. For the reader’s convenience, let us introduce some technical notions that we will use in the sequel. Given an automorphism ϕ of a group G, two elements x, y ∈ G are said to be ϕ-conjugate, and denoted by x ∼ϕ y

if x = wyϕ(w)−1

for some w ∈ G. This notion was introduced in the context of low dimension topology by Reidemeister in [109]. Observe that if ϕ is the identity then this notion coincides with the usual conjugacy. Recall also that the isotropy group Isotr(ϕ) of ϕ is the subgroup consisting of all elements of G fixed by ϕ: Isotr(ϕ) = {x ∈ G : ϕ(x) = x}. E. Bujalance et al., Symmetries of Compact Riemann Surfaces, Lecture Notes in Mathematics 2007, DOI 10.1007/978-3-642-14828-6 5, c Springer-Verlag Berlin Heidelberg 2010

91

92

5 Symmetry Types of Riemann Surfaces with a Large Group of Automorphisms

Finally, given an element y ∈ G we denote by ϕy and ϕy the automorphisms of G defined as ϕy (x) = ϕ(x)y = yϕ(x)y −1

and ϕy (x) = ϕ(xy ) = ϕ(yxy −1 ).

Let S be a symmetric Riemann surface corresponding to a generating pair (a, b). We know by Theorem 1.5.10 that Aut+ (S) admits an automorphism ϕ induced by a → a−1 , b → b−1 or by a → b−1 , b → a−1 . We shall call S a surface of the first type if ϕ is induced by the first assignment, and of the second type otherwise. We start with surfaces of the first type. Theorem 5.1.1. Let S be a symmetric Riemann surface of the first type whose group Aut+ (S) of analytic automorphisms is generated by a pair (a, b) of elements of order k and = 2 + 1, respectively, whose product has order m = 2m + 1. (1) The surface S has exactly one conjugacy class of symmetries with ovals. The number of its ovals is N/M , where N is the order of the isotropy group of ϕ in Aut+ (S) and k 2 a (ab)−m b ak b− (ab)m M= (ab)−m b ak

if k = 2k, if k = 2k + 1.

(2) The surface S has a symmetry without ovals if and only if there is an element x ∈ Aut+ (S) such that ϕ(x) = x−1 and x is not ϕ-conjugate to 1. Proof. Let us write S = H/Γ for some surface Fuchsian group Γ and G = Aut+ (S) = Δ/Γ, where Δ is a Fuchsian group with signature s(Δ) = [k , , m ]. By the proof of Theorem 1.5.10 there exists an NEC group Λ with signature s(Λ) = = Aut(S) = Λ/Γ. (k , , m ) containing Δ and Γ as normal subgroups so that G = G Z2 = By Remark 1.5.11 we know that the smooth epimorphism θ : Λ → G a, b t is given by θ(c0 ) = at, θ(c1 ) = t, θ(c2 ) = tb, where {c0 , c1 , c2 } is a set of canonical generating reflections of Λ. Let σ be a symmetry of S and write σ = Γσ /Γ where Γσ is a proper NEC group. If Γσ contains no conjugate to a canonical reflection then σ fixes no oval by Lemma 1.5.9. If Γσ contains a conjugate of a canonical reflection then σ fixes κ > 0 ovals, that is, the signature of Γσ has κ > 0 period cycles. Observe that c0 , c1 and c2 are pairwise conjugate since and m are odd. Therefore, the surface S has exactly one conjugacy class of symmetries with ovals. In addition, Theorem 3.1.1 yields at) : θ(C(Λ, c0 ))]. κ = [C(G, t) and C(G, at) have the same Since t and at are conjugate, the centralizers C(G, order. Now, the restriction to G of conjugation by t is precisely the automorphism ϕ. t) is twice the order of the isotropy group of It follows that the order of C(G, ϕ in G. Let us calculate now the order of θ(C(Λ, c0 )). Assume that k = 2k.

5.1 Some General Results

93

of the generators of Then, using Lemma 1.3.3 and noting that the images in G C(Λ, c0 ) have order 2, we obtain that θ(C(Λ, c0 )) has order 4M , where M is the order of the element θ (c0 c1 )k (c2 c0 )m (c1 c2 ) (c1 c0 )k (c2 c1 ) (c0 c2 )m = ak (ab)−m b ak b− (ab)m . This completes the first part of the proof for k even. The first part of the proof for k odd is similar and we omit it. Now, any symmetry σ has the form xt for some x ∈ G and, in fact, ϕ(x) = x−1 because (xt)2 = 1. Observe that σ fixes no oval if and only if σ is not conjugate to t. But xt is conjugate to t if and only x ∼ϕ 1, as is easy to see. This completes the proof of the theorem. As another application of Theorem 3.1.1, we state without proof Theorems 5.1.2 and 5.1.3. Theorem 5.1.2. Let S be a symmetric Riemann surface of the first type whose group Aut+ (S) of analytic automorphisms is generated by a pair (a, b) of elements of order k = 2k and = 2, respectively, whose product has order m = 2m. (1) The surface S admits three symmetries with ovals, σ1 , σ2 and σ3 , such that each symmetry with ovals of S is conjugate either to σ1 or to σ2 or else to σ3 . Furthermore, σ1 ∼ σ2 , σ2 ∼ σ3 and σ1 ∼ σ3 , respectively, if and only if a ∼ϕ 1, b ∼ϕ 1 and a ∼ϕ b±1 , respectively. Let M1 , M2 and M3 be the orders of ak (ab)m , ak b and (ab)m b , respectively, and let N1 , N2 and N3 be the orders of the isotropy groups of ϕa , ϕ and ϕb in Aut+ (S), respectively. Then, • • •

If σ1 , σ2 and σ3 are pairwise non-conjugate then σi has Ni /(2Mi ) ovals for i = 1, 2, 3. If σ1 , σ2 and σ3 are pairwise conjugate then each of them has N1 /(2M1 )+ N2 /(2M2 ) + N3 /(2M3 ) ovals. If (u, v, w) is a permutation of (1, 2, 3) such that σu ∼ σv and σw is nonconjugate to σu and σv then σu has Nu /2(Mu ) + Nv /(2Mv ) ovals and σw has Nw /(2Mw ) ovals.

(2) The surface S has a symmetry without ovals if and only if there is an element x ∈ Aut+ (S) such that ϕ(x) = x−1 and x is ϕ-conjugate neither to 1, nor to a±1 nor to b±1 . Theorem 5.1.3. Let S be a symmetric Riemann surface of the first type whose group Aut+ (S) of analytic automorphisms is generated by a pair (a, b) of elements of order k = 2k and = 2, respectively, whose product has order m = 2m + 1. (1) The surface S admits two symmetries with ovals, σ1 and σ2 , such that each symmetry with ovals of S is conjugate either to σ1 or to σ2 . Furthermore σ1 ∼ σ2 if and only if a ∼ϕ 1.

94


Let M1 and M2 be the orders of ak (ab)−m b (ab)m and ak b , respectively, and let N1 and N2 be the orders of the isotropy groups of ϕ and ϕa in Aut+ (S), respectively. Then, • •

If σ1 ∼ σ2 then σ1 has N1 /(2M1 ) + N2 /(2M2 ) ovals. If σ1 and σ2 are non-conjugate then σi has Ni /(2Mi ) ovals for i = 1,2.

(2) The surface S has a symmetry without ovals if and only if there is an element x ∈ Aut+ (S) such that ϕ(x) = x−1 and x is ϕ-conjugate neither to 1 nor to a±1 . If S is a Riemann surface of the second type and (a, b) is a generating pair then the assignment ϕ : a → b−1 , b → a−1 induces an automorphism. In particular a and b have the same order. For this type of surfaces we have the following. Theorem 5.1.4. Let S be a symmetric Riemann surface of the second type whose group Aut+ (S) of analytic automorphisms is generated by two elements a and b of order k whose product has order m. (1) The surface S has exactly one conjugacy class of symmetries with ovals. The number of its ovals is N/M , where N is the order of the isotropy group of ϕ in Aut+ (S) and 2 (ab)m/2 (ba)m/2 M= (ab)(m+1)/2 a

if m is even, if m is odd.

(2) The surface S has a symmetry without ovals if and only if there is an element x ∈ Aut+ (S) such that ϕ(x) = x−1 and x is not ϕ-conjugate to 1. Proof. Write S = H/Γ for some surface Fuchsian group Γ. Then G = Aut+ (S) = Δ/Γ, where Δ is a Fuchsian group with signature [k, k, m] containing Γ as a normal subgroup. By the proof of Theorem 1.5.10, there exists an NEC group Λ with signature s(Λ) = (0; +; [k]; {(m)}) containing Γ and Δ as normal subgroups and, by Remark 1.5.11, the group Γ is the kernel of the epimorphism = G Z2 = a, b t induced by the assignment θ:Λ→G θ(x) = a, θ(e) = a−1 , θ(c0 ) = t, θ(c1 ) = t(ab)−1 . Now c0 and c1 are conjugate in Λ and therefore the surface S has exactly one conjugacy class of symmetries with ovals. In order to count them, we have to find the number of empty period cycles in the signature of the subgroup Γ0 = θ−1 (t) = Γ, c0 . By Theorem 3.1.1, the number of ovals of this symmetry equals t)) is 2N , where N is the order of t)) : θ(C(Λ, c0 ))]. Now the order of C(G, [C(G, the isotropy group of ϕ in G and, by Lemmata 1.3.1 and 1.3.2, C(Λ, c0 ) =

c0 ⊕ (c0 c1 )m/2 ∗ e(c0 c1 )m/2 e−1 if m is even, if m is odd. c0 ⊕ e(c0 c1 )(m−1)/2

5.2 Symmetry Types of Macbeath–Singerman Surfaces

95

So for even m, the order of θ(C(Λ, c0 )) is four times the order of (ab)m/2 (ba)m/2 , whilst for odd m, it is twice the order of (ab)(m+1)/2 a. This completes the proof of the first part. The proof of the second part is analogous to that of Theorem 5.1.1.

5.2 Symmetry Types of Macbeath–Singerman Surfaces A finite group G is said to be a Hurwitz group if there exists a Riemann surface S of genus g ≥ 2 whose group of analytic automorphisms has order 84(g − 1) and it is isomorphic to G. In such a case the surface S is said to have a Hurwitz automorphism group and G is said to act as a Hurwitz group on the surface S. Hurwitz groups are generated by two elements a and b of orders 2 and 3, respectively, whose product has order 7, see [118, Sect. 4]. It is known that no surface of genus g = 2, 4, 5 or 6 has a Hurwitz automorphism group, whilst on the other hand Klein proved in [65] that the genus 3 Riemann surface S = {[x : y : z] ∈ P2 (C) : x3 y + y 3 z + z 3 x = 0} has the projective special linear group PSL(2, 7) of order 168 = 84(3 − 1) as a Hurwitz automorphism group. This surface is the so called Klein quartic. Macbeath proved in [72] the existence of a unique Riemann surface of genus 7 on which the group PSL(2, 8) of order 504 = 84(7 − 1) acts as a Hurwitz group. Later on, Macbeath characterized in [74] the values of the positive integers q for which the group PSL(2, q) acts as a Hurwitz group on some Riemann surface. These integers are usually called H-numbers. Macbeath’s characterization is the following, see [74, Theorem 8]. Theorem 5.2.1. A positive integer q is an H-number if and only if either q = 7 or q ≡ ±1 (mod 7) is prime, or q = p3 for some prime number p such that p ≡ ±2 (mod 7) or p ≡ ±3 (mod 7). A first result about the symmetric character of surfaces with large group of analytic automorphisms is the following. Theorem 5.2.2. Let p be a prime H-number and let S be a Riemann surface admitting PSL(2, p) as a Hurwitz automorphism group. Then S is symmetric. Proof. There exist elements a, b ∈ PSL(2, p) of orders 2 and 3 respectively, whose product c = ab has order 7 and such that a and b generate PSL(2, p). Let A, B, C ∈ SL(2, p) be representatives of a, b and c respectively, with C = AB. Since det(A) = det(B) = det(C) = 1 we have tr(A) = tr(A−1 );

tr(B) = tr(B −1 );

tr(C) = tr(C −1 ).

Let k be an algebraic closure of the finite field Fp of p elements. By [74, Theorem 3], there exists a matrix U ∈ SL(2, k) such that

96


U AU −1 = A−1

and U BU −1 = B −1 .

Let u ∈ PSL(2, k) be the class (mod ± I) of U . Hence, conjugation by u is an automorphism of PSL(2, p) which maps a to a−1 and b to b−1 . It follows from Theorem 1.5.10 that S is a symmetric surface of the first type. The surfaces occurring in Theorem 5.2.2 are called Macbeath–Singerman surfaces. Our goal in this section is to compute their symmetry types. We shall follow the presentation in [8], but we shall apply Theorem 5.1.1, which leads us to calculate the orders of some isotropy groups. To that end, we need some preparatory work. In what follows, p ≥ 7 will denote a prime number. Proposition 5.2.3. With the notations in the proof of Theorem 5.2.2, the element u ∈ PSL(2, k) has order 2. In particular, if u ∈ PSL(2, k) \ PSL(2, p) then the semidirect product PSL(2, p)u defined by uau−1 = a−1

and ubu−1 = b−1

is isomorphic to PGL(2, p). Proof. The argument is easier if u ∈ PSL(2, p). Indeed, it suffices to check that u2 g = gu2 for each g ∈ PSL(2, p), because PSL(2, p) is a simple group. Moreover, since a and b generate PSL(2, p), it is enough to show this equality for g = a and g = b. Recall that ugu−1 = g −1 for these values of g and so u2 gu−2 = u(ugu−1 )u−1 = ug −1 u−1 = (ugu−1 )−1 = g. Suppose now that u ∈ PSL(2, k) \ PSL(2, p). Let U ∈ SL(2, k) be a representative of u. Notice that v = u2 commutes with both a and b, and so vg = gv for each g ∈ PSL(2, p). Thus V = U 2 satisfies V C = ±CV for every C ∈ SL(2, p). By [122, Chap. 3, Sect. 6], there exist ω, τ ∈ k such that the matrix V is conjugate in SL(2, k) either to D=

ω 0 0 ω −1

or to

T =±

1 0 τ 1

.

Hence V = QXQ−1 where X denotes D or T , indistinctly, and Q=

q1 q2 q3 q4

is a matrix in SL(2, k). From the equality V C = ±CV it follows that X(Q−1 CQ) = ±(Q−1 CQ)X


97

for every C ∈ SL(2, p). Write C=

α1 α2 α3 α4

Then Q where

−1

CQ =

.

β 1 β2 β3 β4

,

β1 = q1 q4 α1 + q3 q4 α2 − q1 q2 α3 − q2 q3 α4 , β2 = q2 q4 α1 + q42 α2 − q22 α3 − q2 q4 α4 .

The matrix C can be chosen so that β1 = 0 and β2 = 0. Now, if X = D we get X(Q−1 CQ) =

β2 ω β1 ω β3 ω −1 β4 ω −1

while (Q−1 CQ)X =

β1 ω β2 ω −1 β3 ω β4 ω −1

.

Therefore ω = ω −1 and so u2 = v = 1, that is, u is an involution. On the other hand, if X = T then X(Q−1 CQ) =

β2 β1 β3 + τ β1 β4 + τ β2

while (Q−1 CQ)X =

β1 + τ β2 β2 β3 + τ β4 β4

.

Then τ = 0 and again u2 = v = 1. The last part follows easily from [37, Theorem 7.5].

Proposition 5.2.4. Let S be a Riemann surface admitting PSL(2, p) as a Hurwitz automorphism group. (1) The full group of dianalytic automorphisms Aut(S) of S is a semidirect product Aut(S) = PSL(2, p) Z2 . Moreover, with the notations in the proof of Theorem 5.2.2, Aut(S) =

PSL(2, p) ⊕ Z2 PGL(2, p)

if u ∈ PSL(2, p), if u ∈ PSL(2, k) \ PSL(2, p).

(2) Let us denote ϕu : PSL(2, p) → PSL(2, p) ; g → ugu−1

98


and let Isotr(ϕu ) = {g ∈ PSL(2, p) : ϕu (g) = g} be its isotropy group. (2.1) If u ∈ PSL(2, p) then the order of Isotr(ϕu ) coincides with the order of the centralizer in PSL(2, p) of the class (mod ± I) of the matrix M=

0 1 −1 0

∈ SL(2, p).

(2.2) If u ∈ PSL(2, k)\PSL(2, p) then the order of Isotr(ϕu ) is half the order of the centralizer in PGL(2, p) of the class (mod ±I) of the matrix N=

0 1 −δ 0

∈ GL(2, p),

where δ ∈ Fp is a non-square in Fp . Proof. (1) This part is an immediate consequence of Remark 1.5.11 and Proposition 5.2.3. (2.1) Let U ∈ SL(2, p) be a representative of u and consider the Fp -linear endomorphism x x 2 2 f : Fp → Fp ; → U , y y that satisfies f 2 = ε id for some ε = ±1. There exists a vector ω ∈ F2p such that B = {ω, εf (ω)} is a basis of F2p . The matrix of f with respect to B is Mf (B) =

0 1 ε 0

.

Moreover, ε = − det f = −1 and so Mf (B) is precisely the matrix M in the statement of the proposition. Hence there exists Q ∈ GL(2, p) such that U = QM Q−1 . Let m be the class (mod ± I) of M and let q be the class of Q (mod±I). Let Cm be the centralizer of m in PSL(2, p). An easy computation shows that the map Cm → Isotr(ϕu ) ; g → qgq −1 is a bijection, which proves our assertion. (2.2) By the first part we may assume that u ∈ PGL(2, p) \ PSL(2, p). Arguing as above, we deduce the existence of Q ∈ GL(2, p) such that U = QN Q−1 .


99

Let n be the class of N (mod ± I) and let q be the class (mod ± I) of Q. Let Cn be the centralizer of n in PGL(2, p). The result follows because the map Cn → Isotr(ϕu ) ; g → qgq −1

is two-to-one.

We are ready to compute, by using an elementary geometric argument, the order of the isotropy group of the automorphism ϕu . Proposition 5.2.5. Let u ∈ PGL(2, p). (1) If u ∈ PSL(2, p) then the order of the isotropy group of the automorphism ϕu of the group PSL(2, p) is | Isotr(ϕu )| =

p−1 p+1

if p ≡ 1 (mod 4), if p ≡ 3 (mod 4).

(2) If u ∈ PGL(2, p) \ PSL(2, p) then | Isotr(ϕu )| =

p+1 p−1

if p ≡ 1 (mod 4), if p ≡ 3 (mod 4).

Proof. (1) By part (2.1) in Proposition 5.2.4 the order of Isotr(ϕu ) is half the number of matrices C ∈ SL(2, p) such that CM = ±M C, where M= Writing C as

0 1 −1 0

C=

x y z t

.

,

the condition CM = ±M C, together with det C = 1, leads to

either

⎧ 2 ⎨ x + y2 = 1 x −t =0 ⎩ z +y =0

or

⎧ 2 ⎨ x + y 2 = −1 x + t = 0. ⎩ z −y = 0

Consider the conics Γ1 = {(x, y) ∈ F2p : x2 +y 2 = 1} and Γ2 = {(x, y) ∈ F2p : x2 +y 2 = −1}. Clearly, | Isotr(ϕu )| =

card(Γ1 ) + card(Γ2 ) . 2

100


Suppose now that p ≡ 1 (mod 4). Then there exists ξ ∈ Fp such that ξ 2 = −1 and so the linear isomorphism F2p → F2p ; (x, y) → (ξx, ξy) maps Γ1 onto Γ2 . Therefore | Isotr(ϕu )| = card(Γ1 ). The linear isomorphism F2p → F2p ; (x, y) → (x + ξy, x − ξy) maps Γ1 onto the hyperbola Γ3 = {(x, y) ∈ F2p : xy = 1}. Thus | Isotr(ϕu )| = card(Γ3 ) = p − 1, because Γ3 and F∗p = Fp \ {0} are bijective, via the map F∗p → Γ3 ; t → (t, t−1 ). Assume now that p ≡ 3 (mod 4). We choose a point (x0 , y0 ) ∈ Γ2 and consider the linear map F2p → F2p ; (x, y) → (y0 x + x0 y, x0 x − y0 y). This is a linear isomorphism which maps Γ1 onto Γ2 . Hence | Isotr(ϕu )| = card(Γ1 ). Moreover, since 1 + t2 = 0 for each t ∈ Fp , the inverse of the stereographic projection Fp → Γ1 \ {(1, 0)} ; t →

t2 − 1 1+t

, 2

2t 1 + t2

is a bijection. So | Isotr(ϕu )| = p + 1. (2) By part (2.2) in Proposition 5.2.4, the order of Isotr(ϕu ) is 1/4 the number of matrices D ∈ GL(2, p) such that DN = ±N D, where D=

x y z t

and N =

0 1 −δ 0

,

and either xt − yz = 1 or xy − zt = δ. The equalities DN = ±N D are equivalent to z = ∓δy and t = ±x. This, together with the condition on the determinant of D, leads us to consider the conics H1 = {(x, y) ∈ F2p : x2 +δy 2 = 1},

H2 = {(x, y) ∈ F2p : x2 +δy 2 = −1},

H3 = {(x, y) ∈ F2p : x2 +δy 2 = δ},

H4 = {(x, y) ∈ F2p : x2 +δy 2 = −δ},

that satisfy | Isotr(ϕu )| =

card(H1 ) + card(H2 ) + card(H3 ) + card(H4 ) . 4


101

The linear isomorphism F2p → F2p ; (x, y) → (y, δ −1 x) maps Hi+1 onto Hi for i = 1, 3. Therefore | Isotr(ϕu )| =

card(H1 ) + card(H2 ) . 2

If p ≡ 1 (mod 4) then there exists ξ ∈ Fp such that ξ 2 = −1. The linear isomorphism F2p → F2p ; (x, y) → (ξx, ξy) maps H1 onto H2 and so | Isotr(ϕu )| = card(H1 ) = p + 1, because, as 1 + δt2 is non-zero for each t ∈ Fp , the inverse of the stereographic projection from the point (1, 0) Fp → H1 \ {(1, 0)} ; t →

δt2 − 1 1 + δt

, 2

2t 1 + δt2

is a bijection. Suppose now that p ≡ 3 (mod 4). Hence −1 is not a square in Fp and so we may assume that δ = −1. This way we have bijections H2 → H1 ; (x, y) → (y, x) and F∗p → H1 ; t → (t + t−1 , t − t−1 ) and so | Isotr(ϕu )| = card(H1 ) = p − 1.

Remark 5.2.6. The above proof of Proposition 5.2.5 shows in particular that for an involution u ∈ PSL(2, p), we have p − 1 ≤ |C(PSL(2, p), u)| ≤ p + 1, where C(PSL(2, p), u) is the centralizer in PSL(2, p) of u. Corollary 5.2.7. Let p be an odd prime H-number and let S be a Macbeath– Singerman surface whose group of analytic automorphisms is presented as Aut+ (S) = PSL(2, p) = a, b | a2 , b3 , (ab)7 , . . . for some generating pair (a, b). Let M (a, b) be the order of the element a(ab)4 bab2 (ab)3 ∈ Aut+ (S).

102


(1) Suppose that Aut(S) = PSL(2, p) ⊕ Z2 . Then S admits exactly two conjugacy classes of symmetries. A representative of the first one fixes no oval while the number of ovals fixed by a representative of the second one is

(p − 1)/(2M (a, b))

if p ≡ 1 (mod 4),

(p + 1)/(2M (a, b))

if p ≡ 3 (mod 4).

(2) Suppose now that Aut(S) = PGL(2, p). Then S admits exactly one conjugacy class of symmetries. The number of ovals fixed by any of them is

(p + 1)/(2M (a, b))

if p ≡ 1 (mod 4),

(p − 1)/(2M (a, b))

if p ≡ 3 (mod 4).

Proof. The formulae are consequence of Theorem 5.1.1 and Proposition 5.2.5, taking into account that Aut(S)\Aut+ (S) contains either one or two conjugacy classes of involutions according to Aut(S) being PGL(2, p) or PSL(2, p)⊕Z2 respectively, see [38]. Remark 5.2.8. It is well known that Computer Algebra Systems like CAYLEY, MAGMA , GAP, MATLAB , MAPLE and MATHEMATICA find a presentation of the projective special linear group of the form PSL(2, p) = a, b | a2 , b3 , (ab)7 , . . . very quickly, and they compute the order M (a, b) of the element a(ab)4 bab2 (ab)3 associated to a given presentation of the required form. This enables us, by applying Corollary 5.2.7, to effectively calculate the number of ovals of each symmetry of a Macbeath–Singerman surface. For example, if p = 13 there exists, up to conformal equivalence, a unique Macbeath–Singerman surface S with Aut(S) = PSL(2, 13) ⊕ Z2 . By Corollary 5.2.7 it admits a fixed point free symmetry and, moreover, M (a, b) = 6; hence it admits also a symmetry with exactly one oval. On the other hand, there exist two non-isomorphic Macbeath–Singerman surfaces S1 , S2 such that Aut(Si ) = PGL(2, 13) for i = 1, 2. In this case all symmetries of each surface Si are conjugate and each of them has exactly one oval, because M (a, b) = 7. Recall that a symmetry σ of a Riemann surface S is said to be separating if S \ Fix(σ) is disconnected. Otherwise it is said that σ is non-separating. Although we have just remarked that M (a, b) can be calculated with the aid of a Computer Algebra System, in order to demonstrate the non-separating character of the involved symmetries, we will also need to show that M (a, b) ≥ 3. This is the goal of the next proposition. Proposition 5.2.9. Let p ≥ 7 be a prime number and let a presentation of the projective linear group be given by PSL(2, p) = a, b | a2 , b3 , (ab)7 , . . .. Then the order of the element a(ab)4 bab2 (ab)3 is at least 3.


103

Proof. Multiplying by −I if necessary, there exist representatives A, B ∈ SL(2, p) of a and b such that A2 = −I, B 3 = I and C 7 = I, where C = AB. We have to show that the order in SL(2, p) of the matrix M = A(AB)4 BAB 2 (AB)3 is not equal to 1, 2 or 4. This can be expressed in terms of the trace m = tr(M ). Clearly, if M has order 1 then m = 2. On the other hand, M 2 = mM − I, by Cayley–Hamilton theorem, and so mM = 2I if M has order 2, that is, m2 = 4. Moreover, if M has order 4 then I = M 4 = (mM − I)2 = m2 M 2 − 2mM + I = m2 (mM − I) − 2mM + I = m(m2 − 2)M + (1 − m2 )I, that is, m = 0. Therefore, it is enough to prove that m ∈ {−2, 0, 2}. To that end we first express m as a polynomial with respect to the trace γ of C. More precisely, we will show that the following equality holds: m = γ(γ 6 − 5γ 4 + 8γ 2 − 5).

(5.1)

As p ≥ 5, all elements of order 4 in SL(2, p) are conjugate, see [38]. Hence, we may assume that 0 1 A= −1 0 because A has order 4. On the other hand, by Cayley–Hamilton theorem, B 2 = βB − I with β = tr(B). Thus I = B 3 = B(βB − I) = β(βB − I) − B = (β 2 − 1)B − βI and so β = −1. Therefore, B 2 = −(B + I). Write M = A(AB)4 BAB 2 (AB)3 = AC 4 DB 2 C 3 , where D = BA. Notice that γ = tr(AB) = tr(BA) = tr(D) and so C 2 = γC − I,

C 3 = (γ 2 − 1)C − γI,

C 4 = (γ 3 − 2γ)C + (1 − γ 2 )I,

D2 = γD − I,

D3 = (γ 2 − 1)D − γI,

D4 = (γ 3 − 2γ)D + (1 − γ 2 )I.

104


Taking into account that AC = −B and BC = DB, the matrix M can be rewritten as

M = −A (γ 3 − 2γ)C + (1 − γ 2 )I D(B + I) (γ 2 − 1)C − γI

= (γ 3 − 2γ)B + (γ 2 − 1)A D (γ 2 − 1)DB − γB + (γ 2 − 1)C − γI . Moreover, since DC = BAAB = −B 2 = B + I and BDB = B 2 AB = −(B + I)AB = −(DB + C), we have

M = (γ 3 − 2γ)B + (γ 2 − 1)A (γ 3 − 2γ)DB + (γ 2 − 1)I − γD = −(γ 3 − 2γ)2 (DB + C) + (γ 3 − 2γ)(γ 2 − 1)B + γ(γ 3 − 2γ)(D + A) +(γ 3 − 2γ)(γ 2 − 1)(γC − I) + (γ 2 − 1)2 A − γ(γ 2 − 1)AD.

(5.2)

Notice that tr(AD) = tr(ABA) = tr(BA2 ) = tr(−B) = 1 and tr(DB) = tr(BAB) = tr(AB 2 ) = tr(−A(B + I)) = − tr(AB) − tr(A) = −γ. Thus tr(DB + C) = 0. We now deduce equality (5.1) after comparing the traces of both sides in equality (5.2): m = tr(M ) = (γ 3 − 2γ)(1 − γ 2 ) + γ 2 (γ 3 − 2γ) + +(γ 3 − 2γ)(γ 2 − 1)(γ 2 − 2) + γ(γ 2 − 1) = γ(γ 6 − 5γ 4 + 8γ 2 − 5). Next, we are going to prove that γ is a root of the polynomial f (T ) = T 3 + T 2 − 2T − 1. Since C 2 = γC − I we get C 6 = (γC − I)3 = (γ 5 − 4γ 3 + 3γ)C − (γ 4 − 3γ 2 + 1)I. Consequently, I = C 7 = C 6 C = (γ 6 − 5γ 4 + 6γ 2 − 1)C − (γ 5 − 4γ 3 + 3γ)I.

(5.3)


105

Thus, γ is a common root of the polynomials g(T ) = T 6 − 5T 4 + 6T 2 − 1 = (T 3 + T 2 − 2T − 1)(T 3 − T 2 − 2T + 1), h(T ) = T 5 − 4T 3 + 3T + 1 = (T 3 + T 2 − 2T − 1)(T 2 − T − 1) and so γ is a root of f = gcdFp [T ] (g, h). Therefore, using also equality (5.1), we deduce that γ is a common root in Fp of the polynomials f (T ) and m (T ) = T 7 − 5T 5 + 8T 3 − 5T − m. After dividing one gets m (T ) = (T 4 − T 3 − 2T 2 + T + 2)f (T ) − rm (T ) and so γ is a root of rm (T ) = 2T 2 + m − 2. Hence m ∈ {−2, 0, 2} and the proof is finished. Before proving that the symmetries of a Macbeath–Singerman surface whose group of analytic automorphisms is PSL(2, p), where p ≥ 7 is prime, are nonseparating, see Theorem 5.2.12, we need the next result concerning separating symmetries. Proposition 5.2.10. Let σ be a separating symmetry of a Riemann surface S. Let τ = σ be another symmetry of S commuting with σ and such that σ and τ have a common fixed point, and suppose that the fixed point sets of τ and σ share no connected component. Then h = τ ◦ σ has at most 2σ fixed points. Proof. Let S+ and S− be the connected components of S \ Fix(σ), and let v denote a common fixed point of σ and τ . Since τ commutes with σ, either it interchanges S+ and S− or it maps both of them to themselves. By analyzing the analytic map h locally around v we see that τ and σ behave like symmetries with respect to circles meeting at right angles, and this implies that S+ and S− are both mapped by τ to themselves. Consequently h(S+ ) = S− and h(S− ) = S+ . Therefore Fix(h) ⊂ Fix(σ) and so Fix(h) ⊂ Fix(τ ). Thus Fix(h) ⊂ Fix(τ ) ∩ Fix(σ) and, since the converse is evident, we get Fix(h) = Fix(τ ) ∩ Fix(σ). Finally, as the sets Fix(σ) and Fix(τ ) have no common connected component, h fixes at most two points on each oval of σ, as desired. The third result we will use in the proof of Theorem 5.2.12 is a classical formula due to Macbeath [75, Theorem 1] for counting the number of points fixed by an automorphism. We include a proof here for the sake of completeness. To state it we introduce some notations. Let S be a Riemann surface of genus ≥ 2 and let G be a group of automorphisms of S. Let Λ be a Fuchsian group and let θ : Λ → G be a

106


group epimorphism whose kernel is the surface Fuchsian group Γ that uniformizes S, that is, S = H/Γ. Let m1 , . . . , mr be the proper periods in the signature of Λ corresponding to the elliptic canonical generators x1 , . . . , xr of Λ. Given h ∈ G, let δi (h) = 1 if h is conjugate to a power of gi = θ(xi ) and δi (h) = 0 otherwise. Theorem 5.2.11. If h ∈ G is not the identity then the number of points of S fixed by h is given by the formula | Fix(h)| = |NG (h)|

r δi (h) i=1

mi

,

where NG (h) denotes the normalizer in G of the cyclic group generated by h. Proof. Since each elliptic generator xi fixes a unique point qi ∈ H, the set of points of H with non-trivial stabilizer under the action of Γ is the disjoint union of the orbits Γq1 ∪ · · · ∪ Γqr . Let h : H → H be the unique lifting of h. The stabilizer of h(qi ) is the subgroup h(xi ) h−1 . Denote by π : H → S the covering projection and let Qi = π (qi ) ∈ S for 1 ≤ i ≤ r. Then the points of S with non-trivial stabilizer in G are those of the form g(Qi ) with g ∈ G, the stabilizer of such a point being ggi g −1 . Note that h has a fixed point in the orbit GQi if and only if h is conjugate to a subgroup of gi , that is, δi (h) = 1. Thus all reduces to check that if this is the case then the number of fixed points of h in the orbit GQi is |NG (h)/mi . Observe that h is conjugate to a unique subgroup of gi . Moreover, h and any of its conjugates have the same number of fixed points in the orbit GQi . Thus, we m /d may assume that h = gi i where d is the order of h. This way, given g ∈ G, the automorphism h fixes the point g(Qi ) if and only if g ◦ h ◦ g −1 fixes Qi or, equivalently, g ∈ NG (h). But for each point Q ∈ GQi there are precisely mi elements g ∈ G such that Q = g(Qi ). Therefore, when we enumerate the elements g ∈ NG (h) we count each fixed point mi times, and hence, the number of fixed points of h in the orbit GQi is |NG (h)|/mi , as desired. Theorem 5.2.12. Let S be a Macbeath–Singerman surface whose group of analytic automorphisms is Aut+ (S) = PSL(2, p), where p is an odd prime H-number. Then each symmetry of S is non-separating. Proof. Since the result is obvious for symmetries without fixed points, we restrict ourselves to symmetries with ovals. Let Aut+ (S) = PSL(2, p) = a, b | a2 , b3 , (ab)7 , . . .. We have already proved the existence of a symmetry σ of S whose set Fix(σ) of fixed points is non-empty and such that Aut(S) = Aut+ (S) σ, with σaσ −1 = a−1 and σbσ −1 = b−1 .


107

We have seen in Corollary 5.2.7 that all symmetries of S with fixed points are conjugate. So it suffices to prove that σ is non-separating. To obtain a contradiction, suppose that σ is separating. Clearly τ = aσ ∈ Aut(S) \ Aut+ (S) and it is in fact a symmetry as τ 2 = aσaσ = aσaσ −1 = aa−1 = 1. Moreover, στ = τ σ, because στ = σaσ = σaσ −1 = a−1 = a = aσ 2 = τ σ. Therefore, by Corollary 5.2.7 and Proposition 5.2.10, | Fix(a)| ≤

p+1 . M (a, b)

On the other hand, from Theorem 5.2.11, | Fix(a)| =

|C(PSL(2, p), a)| , 2

where C(PSL(2, p), a) is the centralizer in PSL(2, p) of the involution a. Therefore, by Remark 5.2.6 and Proposition 5.2.9, we have p − 1 ≤ |C(PSL(2, p), a)| = 2| Fix(a)| ≤ a contradiction because p ≥ 7.

2(p + 1) 2(p + 1) ≤ , M (a, b) 3

Remark 5.2.13. Let p be an odd prime H-number and let S be a Macbeath– Singerman surface with Aut+ (S) = PSL(2, p). From Corollary 5.2.7 and Theorem 5.2.12 we would completely know the symmetry type of S once we could decide whether Aut(S) = PSL(2, p) ⊕ Z2 or Aut(S) = PGL(2, p). For a fixed surface S this is a rather difficult task. However, using results by Macbeath [74], Hall [58] and Conder [33] we shall compute, for a fixed p, the symmetry types of the family of all Macbeath–Singerman surfaces S with Aut+ (S) = PSL(2, p). Macbeath [74] proved that for each odd prime H-number p there exist exactly three conformally non-isomorphic Riemann surfaces on which PSL(2, p) acts as a Hurwitz group. Thus, the numbers d(p) and s(p) of non-isomorphic Riemann surfaces S having PSL(2, p) ⊕ Z2 and PGL(2, p), respectively, as the full automorphism group Aut(S), satisfy d(p) + s(p) = 3. Hence to finish the study of the symmetry types of Macbeath–Singerman surfaces one just needs to decide the value of, say, d(p).

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This was done, independently, by Hall [58] and Conder [33] and, to explain the employed method, we need the following: (5.2.14) Factorization of f1 (T ) in Fp [T ]. Let us see first that the polynomial f1 (T ) = T 3 + T 2 − 2T − 1 ∈ Fp [T ] occurring in (5.3) and having γ = tr(C) as a root, completely factorizes in the polynomial ring Fp [T ]. Since p is a prime H-number, p2 − 1 ∈ 7Z and so the cyclic multiplicative group ∗ Fp2 contains an element η of order 7. Observe that

T 3 f1 (T + T −1 ) = T 3 (T + T −1 )3 + (T + T −1 )2 − 2(T + T −1 ) − 1 = 1 + T + T2 + T3 + T4 + T5 + T6 =

T7 − 1 . T −1

Therefore μ1 = η + η −1 ∈ Fp2 is a root of f1 , because η 3 f1 (η + η −1 ) =

6

ηj =

j=1

η7 − 1 = 0. η−1

Since η 2 and η 3 have also order 7, both μ2 = η 2 + η −2 and μ3 = η 3 + η −3 are roots of f1 . Moreover, they are all distinct. In fact, suppose that η j + η −j = η k + η −k for some exponents j and k such that 1 ≤ j < k ≤ 3. After multiplying both sides by η k we get η 2k − η k+j − η k−j + 1 = 0, which is false. Indeed, the polynomials P (T ) = 1 + T + T 2 + T 3 + T 4 + T 5 + T 6 ∈ Z[T ] and Q(T ) = T 2k − T k+j − T k−j + 1 ∈ Z[T ] are coprime in Z[T ] because P (T ) is irreducible in Z[T ] and deg(Q) ≤ deg(P ). Thus, by Bezout’s identity, there exist polynomials R1 , R2 ∈ Z[T ] such that 1 = R1 (T )P (T ) + R2 (T )Q(T ). Since this is an identity with coefficients in Z, we can evaluate it at η to get a contradiction: 1 = R1 (η)P (η) + R2 (η)Q(η) = 0. Observe that each μj ∈ Fp , because it is invariant under every automorphism of Fp2 and Fp2 |Fp is a Galois field extension. Hence, f1 splits in Fp [T ] as f1 (T ) = (T − μ1 )(T − μ2 )(T − μ3 )

with each μj ∈ Fp .


Consider now the polynomials f2 (T ) = T p−1 − 1 =

109

(T ξ∈F∗ p

− ξ) and

f3 (T ) = f1 (T 2 ) = T 6 + T 4 − 2T 2 − 1 = (T 2 − μ1 )(T 2 − μ2 )(T 2 − μ3 ). W. Hall and M. Conder defined the invariant δ(p) = deg(gcdFp [T ] (f2 , f3 )), and it follows that δ(p) = 2κ(p), where κ(p) = card{1 ≤ j ≤ 3 : μj is a square in Fp }. 3 Observe that j=1 μj = 1 is a square in Fp and that the set of non-zero squares is an index 2 subgroup of F∗p . Hence, either κ(p) = 1 or κ(p) = 3; so either δ(p) = 2 or δ(p) = 6. With this terminology, the quoted result by Hall and Conder can be stated as follows. Proposition 5.2.15. With the above notations: (1) (2) (3) (4)

If If If If

δ(p) = 2 δ(p) = 2 δ(p) = 6 δ(p) = 6

and and and and

p ≡ 1 (mod p ≡ 3 (mod p ≡ 1 (mod p ≡ 3 (mod

4) 4) 4) 4)

then then then then

d(p) = 1. d(p) = 2. d(p) = 3. d(p) = 0.

Example 5.2.16. Let p = 29. To calculate δ(29) observe first that μ1 = 3, μ2 = 7 and μ3 = −11 are the roots in F29 of the polynomial f1 (T ) = T 3 + T 2 − 2T − 1. We use the quadratic reciprocity law to decide how many roots μj are squares in F29 . Indeed, 29 ≡ 2 (mod 3) is not a square in F3 and so μ1 is not a square in F29 . On the other hand, 29 ≡ 1 (mod 7) is a square in F7 and so μ2 is a square in F29 . Finally 29 ≡ 7 ≡ −4 (mod 11) is not a square in F11 . Thus −μ3 is not a square in F29 . Since −1 is a square in F29 , we conclude that μ3 is not a square in F29 . Therefore κ(29) = 1, that is, δ(29) = 2 and 29 ≡ 1 (mod 4). Hence d(29) = 1 and s(29) = 2. Let S1 , S2 and S3 be non-isomorphic Riemann surfaces on which the group PSL(2, 29) acts as a Hurwitz group of automorphisms. Then, up to reordering, Aut(S1 ) = PSL(2, 29) ⊕ Z2

and

Aut(S2 ) = Aut(S3 ) = PGL(2, 29).

Let (a1 , b1 ), (a2 , b2 ) and (a3 , b3 ) be generating pairs of PSL(2, 29) with a2i = b3i = (ai bi )7 = 1, providing the structures on S1 , S2 and S3 respectively. With the aid of MATHEMATICA we calculate the orders M (a1 , b1 ) = 14,

M (a2 , b2 ) = M (a3 , b3 ) = 15

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of the elements ai (ai bi )4 bi ai bi 2 (ai bi )3 . From Corollary 5.2.7 and Theorem 5.2.12 we deduce that S1 admits two conjugacy classes of symmetries, both nonseparating, the first one is fixed point free and the second one has exactly one oval. On the other hand, S2 and S3 admit a unique conjugacy class of symmetries, whose representatives are non-separating and have just one oval.

5.3 Symmetry Types of Accola–Maclachlan and Kulkarni Surfaces In the 1960’s, Accola [1] and Maclachlan [77] proved independently that for every integer g ≥ 2 there is a Riemann surface Xg of genus g whose automorphism group has order 8g + 8. It is called the Accola–Maclachlan surface. This result is interesting because 8g + 8 is the largest order of an automorphism group that can be uniformly constructed for every g. Much later, Kulkarni considered the question of uniqueness of these surfaces. In other words, is Xg the only surface of genus g whose automorphism group has order 8g + 8? It turns out, see [70], that this is so for g ≡ 0, 1, 2 (mod 4) and g sufficiently large. However, for large enough g ≡ 3 (mod 4), Kulkarni also proved that, in addition to the Accola–Maclachlan surface, there exists exactly one other surface of genus g whose automorphism group has order 8g + 8. We shall call it the Kulkarni surface and denote it by Yg . Moreover, he found the following presentations for the groups of analytic automorphisms of Xg and Yg : Aut+ (Xg ) = a, b | a2(g+1) , b4 , (ab)2 , ab2 a−1 b2 ; Aut+ (Yg ) = a, b | a2(g+1) , b4 , (ab)2 , b2 ab2 ag . Proposition 5.3.1. Both surfaces Xg and Yg are symmetric. Proof. It suffices to see, by Theorem 1.5.10, that the assignment a → a−1 , b → b−1 induces, in both cases, an automorphism of the group G of analytic automorphisms of either Xg or Yg . For Xg this is equivalent to saying that a−1 has order 2(g + 1), b−1 has order 4, a−1 b−1 has order 2 and b−2 is central in G. Since b−2 = b2 , just the third claim requires some care. Here, |a−1 b−1 | = |ba| = |ab| = 2. As to the surface Yg , it is enough to check that b−2 a−1 b−2 a−g = 1, which follows immediately from the last relator in the presentation of Aut+ (Yg ).

5.3.1 Number of Ovals of the Symmetries Our next goal is to calculate the number of ovals of the symmetries of the surfaces Xg and Yg . To that end we will apply Theorem 5.1.2. We start with the Accola– Maclachlan surface Xg .

5.3 Symmetry Types of Accola–Maclachlan and Kulkarni Surfaces

111

Number of Ovals of the Symmetries of the Accola–Maclachlan Surfaces First, we collect some information about the group Aut+ (Xg ). Proposition 5.3.2. Let Aut+ (Xg ) = a, b | a2(g+1) , b4 , (ab)2 , ab2 a−1 b2 . (1) Each element w ∈ Aut+ (Xg ) can be uniquely written as w = ai bj with 0 ≤ i ≤ 2g + 1 and 0 ≤ j ≤ 3. (2) For each integer i ∈ Z, the following equalities hold: ba2i+1 b = a−(2i+1)

and ba2i b = b2 a−2i .

(3) The element ag+2 b has order 2 if g is odd and order 4 if g is even. (4) The elements ag+1 b2 and ab3 have order 2. (5) Let ϕ be the automorphism of the group Aut+ (Xg ) induced by the assignment ϕ : a → a−1 , b → b−1 . Then a ϕ 1, b ϕ 1 and a ϕ b±1 . (6) Consider the automorphisms of the group Aut+ (Xg ) defined by ϕa : w → aϕ(w)a−1

and ϕb : w → b−1 ϕ(w)b.

Then | Isotr(ϕa )| = 8, | Isotr(ϕ)| = 4 and | Isotr(ϕb )| = 4(g + 1). (7) Consider the sets X = {x ∈ Aut+ (Xg ) : ϕ(x) = x−1 }; Y = {ai bj : 0 ≤ i ≤ 2g+1, j = 0, 2} and Z=

for even g, {b, b3 } 3 g+1 g+1 3 {b, b , a b, a b } for odd g.

Then X = Y ∪ Z. (8) The ϕ-conjugates to 1 in Aut+ (Xg ) are the elements of the form a2i b2j . The ϕ-conjugates to a±1 in Aut+ (Xg ) are the odd powers of a. The ϕ-conjugates to b±1 in Aut+ (Xg ) are the odd powers of b. In particular, ab2 ∈ X and it is ϕ-conjugate neither to 1, nor to a±1 , nor to b±1 . Proof. (1) It is easily seen that for different pairs (i, j) = (k, ) with 0 ≤ i, k ≤ 2g + 1 and 0 ≤ j, ≤ 3, the products ai bj and ak b are distinct elements in Aut+ (Xg ). Since the order of this last group is 8(g + 1), we are done.

112


(2) Note first that a2 ba2 = b because a2 ba2 = a(aba)a = ab−1 a = ab2 ba = b2 aba = b2 b−1 = b. This implies b−1 a2 b = a−2 , which we use to prove, by induction on i, the equality ba2i+1 b = a−(2i+1) , which is evident for i = 0. Hence, ba2i+1 b = ba2i−1 a2 b = (ba2i−1 b)b−1 a2 b = a−(2i−1) a−2 = a−(2i+1) . This proves the first equality for non-negative i. Consequently a2i+1 = b3 a−(2i+1) b3 = b2 (ba−(2i+1) b)b2 = (ba−(2i+1) b)b4 = ba−(2i+1) b and so the equality also holds for negative exponents. On the other hand, ba2i b = b2 a−2i . This is obvious for i = 0 and it was already proved for i = 1. Hence, by induction, ba2i+2 b = ba2i a2 b = (ba2i b)(b−1 a2 b) = (b2 a−2i )a−2 = b2 a−(2i+2) . It is easy to check that this equality also holds for negative values of i. (3) For odd g we get (ag+2 b)2 = ag+2 (bag+2 b) = ag+2 a−(g+2) = 1 and so ag+2 b has order 2. For even g, (ag+2 b)2 = ag+2 (bag+2 b) = ag+2 b2 a−(g+2) = b2 . Hence ag+2 b has order 4 in this case. (4) As b2 is a central element in Aut+ (Xg ), we have (ag+1 b2 )2 = a2(g+1) b4 = 1

and (ab3 )2 = abb2 abb2 = (ab)2 = 1.

(5) Suppose that a ∼ϕ 1. Then a = wϕ(w)−1 for some w = ai bj ∈ Aut+ (Xg ), that is, a = ai bj (a−i b−j )−1 = ai b2j ai . So b2j = a1−2i , a contradiction. Analogously, if b ∼ϕ 1 then b = wϕ(w)−1 for some w = ai bj ∈ Aut+ (Xg ), that is, b = ai bj (a−i b−j )−1 = ai b2j ai = a2i b2j . This implies a2i = b1−2j , which is false. Finally, suppose that a ∼ϕ b±1 , i.e., there exists w = ai bj ∈ Aut+ (Xg ) such that a = wb±1 ϕ(w)−1 = ai bj±1 bj ai = ai b2j±1 ai . Hence a1−2i = b2j±1 , again a contradiction.


113

(6) An element x = ai bj belongs to Isotr(ϕa ) if and only if ai bj = ϕa (x) = aϕ(x)a−1 = aa−i b−j a−1 = a1−i b−j a−1 or, equivalently, bj a2i−1 bj = a−1 . For even j this means a2i−1 = a−1 , that is, a2i = 1. Hence either i = 0 or i = g + 1. This way x ∈ {1, b2 , ag+1 , ag+1 b2 }. If j is odd then a−1 = bj a2i−1 bj = ba2i−1 b = a−(2i−1) or, equivalently, a2(i−1) = 1. Thus either i = 1 or i = g + 2 and so x ∈ {ab, ab3 , ag+2 b, ag+2 b3 }. Consequently, Isotr(ϕa ) = {1, b2, ag+1 , ag+1 b2 , ab, ab3 , ag+2 b, ag+2 b3 } and, in particular, | Isotr(ϕa )| = 8. As to the isotropy group of ϕ notice that x = ai bj belongs to Isotr(ϕ) if and only if ai bj = ϕ(ai bj ) = a−i b−j , that is, a2i = b−2j . Thus Isotr(ϕ) = {1, b2 , ag+1 , ag+1 b2 } and, in particular, | Isotr(ϕ)| = 4. To finish, x = ai bj belongs to Isotr(ϕb ) if and only if ai bj = ϕb (x) = b−1 ϕ(x)b = b−1 a−i b1−j or, equivalently, bai b = a−i b2(1−j) . If i is even then a−i b2 = a−i b2(1−j) , that is, j = 0 or 2. If i is odd then a−i = a−i b2(1−j) and so j = 1 or 3. Therefore | Isotr(ϕb )| = 4(g + 1). (7) Let x = ai bj with even j. Then ϕ(x) = a−i b−j = b−j a−i = x−1 and so x ∈ X. Hence Y ⊂ X. For odd j, the element x−1 = b−j a−i equals ϕ(x) = a−i b−j if and only if ai bj = bj ai , that is, ai b2 = bai b. If i is odd then bai b = a−i , by (2), and the equality ai b2 = a−i cannot hold. If i is even then ai b2 = bai b = a−i b2 , that is, either i = 0 or i = g + 1. Therefore X = Y ∪ Z. (8) An element x is ϕ-conjugate to 1 if there exists w = ai bj ∈ Aut+ (Xg ) such that x = wϕ(w)−1 = ai bj (a−i b−j )−1 = ai bj bj ai = a2i b2j . Analogously, x is ϕ-conjugate to a±1 if there exists w = ai bj ∈ Aut+ (Xg ) such that x = wa±1 ϕ(w)−1 = ai bj a±1 bj ai = a2i±1 . Finally, x is ϕ-conjugate to b±1 if there exists w = ai bj ∈ Aut+ (Xg ) such that x = wb±1 ϕ(w)−1 = ai bj b±1 bj ai = ai bai b2k = ai (bai b)b2k−1 = b2k+ε(i) with k = 0 or k = 1 and ε(i) = −1 if i is odd and ε(i) = 1 if i is even. In any case, x is an odd power of b.

114


Corollary 5.3.3. (1) The Accola-Maclachlan surface Xg admits exactly three pairwise non-conjugate symmetries σ1 , σ2 and σ3 with fixed points. The number of ovals of each symmetry is σ1 =

2 for odd g, 1 for even g,

σ2 = 1 and σ3 = g + 1.

(2) The surface Xg admits a fixed point free symmetry. There is one conjugacy class of fixed point free symmetries if g is even and two classes if g is odd. Proof. (1) The existence of exactly three pairwise non-conjugate symmetries with fixed points follows immediately from Theorem 5.1.2 and part (5) in Proposition 5.3.2. Moreover, by part (1) in Theorem 5.1.2 and parts (3) and (6) in Proposition 5.3.2 we get σ1 =

| Isotr(ϕa )| = 2|ag+2 b|

2 for odd g, 1 for even g.

Using (4) instead of (3) in Proposition 5.3.2 we obtain σ2 =

| Isotr(ϕ)| | Isotr(ϕb )| = 1 and σ = g + 1. = 3 2|ag+1 b2 | 2|ab3 |

(2) The first statement follows straightforwardly from Theorem 5.1.2 and part (8) in Proposition 5.3.2. The second one is an easy exercise in group theory, see [9].

Number of Ovals of the Symmetries of the Kulkarni Surfaces As to the symmetries of the Kulkarni surface Yg for g ≡ 3 (mod 4), we follow a similar strategy. We begin by collecting some properties of the group Aut+ (Yg ). Proposition 5.3.4. Let Aut+ (Yg ) = a, b | a2(g+1) , b4 , (ab)2 , b2 ab2 ag . (1) Each element w ∈ Aut+ (Yg ) can be uniquely written as w = ai bj with 0 ≤ i ≤ 2g + 1 and 0 ≤ j ≤ 3. (2) For each integer i ∈ Z the following equalities hold: ba2i+1 b = a(i+1)(g−1)−g

and ba2i = ai(g−1) b.

(3) The orders of ag+2 b, ag+1 b2 and ab3 are 2, 2 and 4, respectively. (4) Let ϕ be the automorphism of the group Aut+ (Yg ) induced by the assignment ϕ : a → a−1 , b → b−1 . Then a ϕ 1, b ϕ 1 and a ϕ b±1 .


115

(5) Consider the automorphisms of the group Aut+ (Yg ) defined by ϕa : w → aϕ(w)a−1

and ϕb : w → b−1 ϕ(w)b.

Then | Isotr(ϕa )| = 8, | Isotr(ϕ)| = 4 and | Isotr(ϕb )| = 2(g + 1). (6) Suppose that g ≡ 3 (mod 8). Then x = a(g+1)/2 b is ϕ-conjugate neither to 1 nor to a±1 nor to b±1 , but ϕ(x) = x−1 . (7) Suppose that g ≡ 7 (mod 8). Then each element x ∈ Aut+ (Yg ) satisfying ϕ(x) = x−1 is ϕ-conjugate either to 1 or to a±1 or to b±1 . (8) The centralizer of b2 in Aut+ (Yg ) is the subgroup C(Aut+ (Yg ), b2 ) = {ai bj : i ∈ 2Z, 0 ≤ i ≤ 2g, 0 ≤ j ≤ 3}. In particular, |C(Aut+ (Yg ), b2 )| = 4(g + 1). (9) The elements ag+1 and b commute and they generate an abelian group isomorphic to Z2 ⊕ Z4 whose only elements of order 2 are ag+1 , b2 and ag+1 b2 . Proof. (1) It is easily seen that for different pairs (i, j) = (k, ) with 0 ≤ i, k ≤ 2g + 1 and 0 ≤ j, ≤ 3, the products ai bj and ak b are distinct elements in Aut+ (Yg ). Since the order of this last group is 8(g + 1), we are done. (2) It is enough to prove both equalities, by induction, for i ≥ 0. They are obvious for i = 0. Moreover, b2 a−1 b2 = (b2 ab2 )−1 = ag and, consequently, ba2 b = babb3 ab = a−1 b3 ab = a−1 b2 (bab) = a−1 b2 a−1 = a−1 ag b2 = ag−1 b2 . Therefore, ba2i+1 b = (ba2 b)b3 a2i−1 b = (ag−1 b2 )b3 a2i−1 b = ag−1 (ba2i−1 b) = ag−1 ai(g−1)−g = a(i+1)(g−1)−g . For even exponents we have ba2i = ba2(i−1) a2 = a(i−1)(g−1) ba2 = a(i−1)(g−1) ag−1 b = ai(g−1) b. (3) For the first part, using (2) we get (ag+2 b)2 = ag+2 (bag+2 b) = ag+2 a(g+3)(g−1)/2−g = a(g+1)

2

/2

= 1.

For the second it suffices to check that (ag+1 b2 )2 = 1. Using (2) once more, 2 2 (ag+1 b2 )2 = ag+1 b(bag+1 )b2 = ag+1 ba(g −1)/2 b3 = ag+1 a(g −1)(g−1)/4 = 1, because the exponent (g 2 − 1)(g − 1)/4 is even.

116


As to ab3 note that (ab3 )2 = ab(b2 ab2 )b = a(bag+2 b) = a(g

2

−1)/2

.

2

Since (g − 1)/2 is odd, (ab3 )2 = 1. However, (ab3 )4 = ag −1 = 1 and so ab3 has order 4. (4) Suppose that a ∼ϕ 1. Then a = wϕ(w)−1 for some w = ai bj ∈ Aut+ (Yg ) or, equivalently, a = ai bj (a−i b−j )−1 = ai b2j ai , that is, b2j = a1−2i , which is false. Analogously, if b ∼ϕ 1 then b = wϕ(w)−1 for some w = ai bj ∈ Aut+ (Yg ), i.e., b = ai bj (a−i b−j )−1 = ai b2j ai . This implies, by (2), that an odd power of b belongs to the subgroup generated by a, which is an absurd. Finally, if a ∼ϕ b±1 then there exists w = ai bj ∈ Aut+ (Yg ) with a = wb±1 ϕ(w)−1 = ai bj±1 bj ai = ai b2j±1 ai and so b2j±1 = a1−2i , a contradiction. (5) An element x = ai bj belongs to Isotr(ϕa ) if and only if ai bj = ϕa (x) = aϕ(x)a−1 = aa−i b−j a−1 = a1−i b−j a−1 or, equivalently, bj a2i−1 bj = a−1 . For j = 0 this means a2i−1 = a−1 , that is, a2i = 1. Hence either i = 0 or i = g + 1. For j = 1 the condition is 1 = ba2i−1 ba, that is, ba2i−1 = ba or, equivalently, either i = 1 or i = g + 2. For j = 2 we must solve the equation b2 a2i−1 b2 = a−1 with 0 ≤ i ≤ 2g + 1, i.e., a2i−1 = b2 a−1 b2 = (b2 ab2 )−1 = ag , whose solutions are i = (g + 1)/2 and i = 3(g + 1)/2. Finally, for j = 3 we have b3 a2i−1 b3 = a−1 or, equivalently, a2i−1 = ba−1 b = a−g = ag+2 , that is, i = (g + 3)/2 and i = (3g + 5)/2. Therefore Isotr(ϕa ) = {1, ag+1 , ab, ag+2 b, a(g+1)/2 b2 , a3(g+1)/2 b2 , a(g+3)/2 b3 , a(3g+5)/2 b3 }. In particular, | Isotr(ϕa )| = 8. As to the isotropy group of ϕ, note that x = ai bj belongs to Isotr(ϕ) if and only if ai bj = a−i b−j , that is, a2i = b−2j = b2j . That means i = 0 or g + 1 and j = 0 or 2. Hence | Isotr(ϕ)| = 4 and in fact


117

Isotr(ϕ) = {1, ag+1 , b2 , ag+1 b2 }. Let us calculate now the elements x = ai bj ∈ Isotr(ϕb ). This is equivalent to saying that x = b−1 ϕ(x)b or, in other words, ai bj = b−1 a−i b−j b = b−1 a−i b1−j ⇐⇒ bai b = a−i b2(1−j) . This leads us to distinguish two cases, according to the parity of the exponent j. First, for odd j, we have a−i = bai b. By (2) this forces i to be odd, and hence a−i = bai b = a(i+1)(g−1)/2−g ⇐⇒ a(i−1)(g+1)/2 = 1 ⇐⇒ i ≡ 1 (mod 4). For even j we have a−i b = bai . Thus i must be even and a−i b = ai(g−1)/2 b, that is, ai(g+1)/2 = 1 or, equivalently, i ∈ 4Z. Thus Isotr(ϕb ) = {ai bj : i ≡ 1 (mod 4), 1 ≤ i ≤ 2g − 1, j = 1, 3} ∪ ∪ {ai bj : i ≡ 0 (mod 4), 1 ≤ i ≤ 2g − 1, j = 0, 2} and in particular | Isotr(ϕb )| = 2(g + 1). (6) We first check the equality ϕ(x) = x−1 , that is, a−(g+1)/2 b−1 = b−1 a−(g+1)/2 or, equivalently, ba(g+1)/2 = a(g+1)/2 b. Since the exponent (g + 1)/2 is even we get ba(g+1)/2 = a(g+1)(g−1)/4 b, by part (2), and so it suffices to prove the equality a(g+1)(g−1)/4 = a(g+1)/2 , that is, a(g+1)(g−3)/4 = 1. This last is evident because g ≡ 3 (mod 8). We must prove now that x is ϕ-conjugate neither to 1 nor to a±1 nor to b±1 . Suppose, to obtain a contradiction, that x ∼ϕ 1. Then there exists w = ai bj ∈ Aut+ (Yg ) such that x = wϕ(w)−1 , that is, a(g+1)/2 b = ai b2j ai which forces j to be odd. Then a(g+1)/2 b = ai b2 ai , and we distinguish two cases according to the parity of the exponent i. If i is odd then k = (i + 1)(g − 1)/2 − g is also odd and, by (2), a(g+1)/2 b = ai b2 ai = ai b(bai ) = ai bak b3 = ai−g+(k+1)(g−1)/2 b2 , a contradiction. If i is even then = i(g − 1)/2 is also even, and this implies a(g+1)/2 b = ai b(bai ) = ai ba b = ai+(g−1)/2 b2 , a contradiction again. Suppose that x ∼ϕ a±1 and let w = ai bj ∈ Aut(Yg ) such that x = wa±1 ϕ(w)−1 , that is, a(g+1)/2 b = ai bj a±1 bj ai . This is false because the right hand side of the last equality is an odd power of a. To check this it suffices to prove, by induction on j, that the product bj a±1 bj is an odd power of a. Indeed, given j ∈ Z there exists k ∈ Z such that bj a±1 bj = b(bj−1 a±1 bj−1 )b = ba2k−1 b = ak(g−1)−g and k(g − 1) − g is odd.

118


Finally, suppose that x ∼ϕ b±1 , and let w = ai bj ∈ Aut+ (Yg ) such that x = wb±1 ϕ(w)−1 , that is, there exists k ∈ Z with a(g+1)/2 b = ai bj b±1 bj ai = ai b2k−1 ai . For odd k this means that a(g+1)/2 b = ai bai and, by (2), the exponent i = 2 is even. Thus, a(g+1)/2 b = a2 ba2 = a2 a(g−1) b = a(g+1) b, and so a(g+1)/2 = a(g+1) . Therefore ag+1 = a2(g+1) = 1, a contradiction. (7) Let us describe explicitly the set X = {x ∈ Aut+ (Yg ) : ϕ(x) = x−1 }. Write x = ai bj . For j = 0 we have ϕ(x) = a−i = x−1 , that is, x ∈ X. In this case, either x ∼ϕ 1 or x ∼ϕ a. Indeed, if i = 2k then x = a2k = ak ϕ(ak )−1 , that is, x ∼ϕ 1, whilst if i = 2k + 1 then x = a2k+1 = ak aϕ(ak )−1 and so x ∼ϕ a. For j = 1 the equality ϕ(x) = x−1 reads a−i b−1 = b−1 a−i or, equivalently, i a b = bai . The last forces i to be even, say i = 2k. Therefore a2k b = ba2k = ak(g−1) b ⇐⇒ ak(g−3) = 1. Thus 2(g + 1) divides k(g − 3). Since 2(g + 1) ∈ 8Z but (g − 3) ∈ 8Z, necessarily k = 2 is even. Hence 1 = a2(g−3) = a2(g+1)−8 = a−8 and so either = 0 or = (g +1)/4, that is, either i = 0 or i = g +1. In the first case x = b is ϕ-conjugate to b; in the second one, x = ag+1 b is ϕ-conjugate to b as well because w = a2 yields wbϕ(w)−1 = a2 ba2 = a2 ag−1 b = x. For j = 2 the equality ϕ(x) = x−1 means a−i b−2 = b−2 a−i , that is, ai b2 = b2 ai and this is equivalent to ai = b2 ai b−2 = (b2 ab−2 )i = (b2 ab2 )i = a−gi ⇐⇒ ai(g+1) = 1 ⇐⇒ i ∈ 2Z. We already observed that even powers of a are ϕ-conjugate to 1. Finally let j = 3. Then ϕ(x) = x−1 if and only if a−i b−3 = b−3 a−i or, equivalently, ai b3 = b3 ai ; so ai b = b3 ai b2 = b(b2 ab2 )i = ba−gi . Hence i = 2k must be even and, in fact,


119

a2k b = ba−2kg = a−kg(g−1) b ⇐⇒ k(2 + g(g − 1)) ∈ (2g + 2)Z. In particular, k((g + 1)(g − 2) + 4) ∈ (g + 1)Z, that is, 4k ∈ (g + 1)Z. Thus k is even because g + 1 ∈ 8Z. On the other hand, let ξ ∈ Z be such that 4k = (g + 1)ξ. Then 2g + 2 divides η = k(2 + g(g − 1)) = k((g + 1)(g − 2) + 4) = (g + 1)(ξ + k(g − 2)) and so ξ + k(g − 2) is also even. Hence ξ is even and i = 2k = ξ(g + 1)/2, that is, either i = 0 or i = g + 1. We must show that both x = b3 and x = ag+1 b3 are ϕ-conjugate to b. The first is evident, and for the second it is enough to observe that a3 bϕ(a3 )−1 = a3 ba3 = a3 a2(g−1)−g b3 = ag+1 b3 . (8) Clearly, an element x = ai bj ∈ Aut+ (Yg ) commutes with b2 if and only if ai commutes with b2 . But b2 ai = ai b2 ⇐⇒ ai = b2 ai b2 = (b2 ab2 )i = a−gi , that is ai(g+1) = 1 or, equivalently, i is even, as desired. (9) We check only the equality ag+1 b = bag+1 . Notice that, by (2), we have bag+1 b−1 = a(g+1)(g−1)/2 = ag+1 because (g − 1)/2 is an odd integer. Corollary 5.3.5. (1) The Kulkarni surface Yg admits exactly three pairwise nonconjugate symmetries σ1 , σ2 and σ3 with fixed points. The number of ovals of each symmetry is

σ1 = 2, σ2 = 1 and σ3 =

g+1 . 4

(2) The surface Yg admits a fixed point free symmetry if and only if g ≡ 3 (mod 8). In such a case, all fixed point free symmetries of Yg are pairwise conjugate. Proof. (1) The existence of exactly three pairwise non-conjugate symmetries with fixed points follows immediately from Theorem 5.1.2 and part (4) in Proposition 5.3.4. Moreover, by part (1) in Theorem 5.1.2 and parts (3) and (5) in Proposition 5.3.4 we get σ1 =

8 | Isotr(ϕa )| = = 2. 2|ag+2 b| 4

120


By using again (3) and (5) in Proposition 5.3.4 we obtain σ2 =

2(g + 1) g+1 | Isotr(ϕ)| | Isotr(ϕb )| = 1 and σ = = . = 3 g+1 2 3 2|a b | 2|ab | 8 4

(2) The first statement is an immediate consequence of Theorem 5.1.2 and parts (6) and (7) in Proposition 5.3.4. The second one is an easy exercise in group theory, see [9].

5.3.2 Separating Character of the Symmetries Our goal now is to find out the separating or non-separating character of the symmetries of the Accola–Maclachlan and the Kulkarni surfaces. Obviously, the fixed point free symmetries are non-separating, so we just focus on symmetries with fixed points. As we observed in Theorem 1.5.3, Hurwitz proved in his classical article [59] that given a symmetry σ of a genus g Riemann surface the following conditions hold: 1 ≤ k + ε ≤ g + 1; k ≡ g + 1 (mod (2 − ε)), where k is the number of ovals of σ and ε = 0 or 1 according to σ being separating or not. In particular, (1) If k = g + 1 then ε = 0, and if k = g then ε = 1. (2) If k = 1 and g is odd then ε = 1. (3) If g ≡ 3 (mod 8) and k = (g + 1)/4 then ε = 1. Therefore, with the notations throughout this section, we get the following. Corollary 5.3.6. Let σ2 and σ3 be representatives of the conjugacy classes of symmetries of Xg with σ2 = 1 and σ3 = g + 1. Then σ3 is separating. Moreover, σ2 is non-separating if g is odd. Corollary 5.3.7. Let σ2 and σ3 be representatives of the conjugacy classes of symmetries of Yg with σ2 = 1 and σ3 = (g + 1)/4. Then σ2 is non-separating. Moreover σ3 is non-separating if g ≡ 3 (mod 8). (5.3.8) More on the separating character of the symmetries. To find out the separating character of the symmetry σ1 for both Xg and Yg , the symmetry σ2 for Xg and even g, and the symmetry σ3 for Yg and g ≡ 7 (mod 8), we must appeal to more subtle arguments. First we identify the symmetries σ1 , σ2 and σ3 of Xg as elements of the group Aut(Xg ) = Aut+ (Xg )ϕ Z2 , where ϕ is the automorphism of Aut+ (Xg ) induced by the assignment a → a−1 , b → b−1 and Aut+ (Xg ) = a, b | a2(g+1) , b4 , (ab)2 , ab2 a−1 b2 .


121

Let Δ be a Fuchsian group such that there exists a group epimorphism Δ → Aut+ (Xg ) whose kernel is the surface Fuchsian group Γ that uniformizes Xg , that is, Xg = H/Γ. Then the signature of Δ is [2(g + 1), 4, 2] and there exists a unique NEC group Λ containing Δ as a subgroup of index 2. By Theorems 1.2.1, 1.2.2 and 1.2.5, a presentation of Λ is the following Λ = c0 , c1 , c2 | c20 , c21 , c22 , (c0 c1 )2(g+1) , (c1 c2 )4 , (c0 c2 )2 and so, if we denote Z2 = t, the group Aut(Xg ) is the image of Λ under the epimorphism θ : Λ → Aut(Xg ) induced by the assignment c0 → at = σ1 ; c1 → t = σ2 ; c2 → tb = σ3 .

(5.4)

In particular, it is worth noting that σ1 σ3 = σ3 σ1 because σ1 σ3 = at2 b = ab has order 2. Then, as σ1 has two ovals and σ3 is an M -symmetry, that is, a symmetry with g + 1 ovals, the next result follows. Proposition 5.3.9. The symmetry σ1 of Xg for odd g is separating. This is an immediate consequence of the following lemma. Lemma 5.3.10. Let σ and τ be two commuting symmetries of a genus g Riemann surface X such that σ is an M -symmetry and τ has two ovals. Then τ is a separating symmetry. Proof. There exist a Fuchsian group Λ and an epimorphism θ : Λ → σ, τ whose kernel is a surface Fuchsian group uniformizing the surface X. By Theorems 1.2.1, 1.2.2, 1.2.5 and [27] the signature of Λ has the following form: s(Λ) = (h; ±; [2, . v. ., 2]; {(−), . t. ., (−), (2, .r.1., 2), . . . , (2, .r.s., 2)}) for some non-negative integers h, v, t and some even positive integers r1 , . . . , rs because, as ker θ is a surface Fuchsian group, the canonical reflections corresponding to a non-empty period cycle of s(Λ) must be mapped, alternatively, to σ and τ . This implies that r = r1 + · · · + rs ≤ 4, because τ has two ovals. On the other hand, the hyperbolic area of a fundamental region of Λ is Area(Λ) = 2π(ηh − 2 + v/2 + t + s + r/4), where η = 2 if sign(s(Λ)) = “+” and η = 1 otherwise. Now the Hurwitz–Riemann formula yields g + 3 = 2ηh + v + 2t + 2s + r/2.

122


But, as σ and τ have g + 3 ovals in total, we deduce from Theorem 3.1.1 that g + 3 ≤ 2t + r. After substituting, we get 2ηh + v + 2s ≤ r/2 ≤ 2. In particular, either s = 0 or s = 1. In the first case also r = 0, and so h = v = 0 and η = 2. In the second case, s = 1, that is, r = 4, h = v = 0 and η = 2. Thus, either s(Λ) = s1 or s2 , where s1 = (0; +; [−]; {(−), . t. ., (−)}) and s2 = (0; +; [−]; {(−), . t. ., (−), (2, 2, 2, 2)}), with t = (g + 3)/2 in the first case and t = (g − 1)/2 in the second one. Suppose first that s(Λ) = s1 , and let c0 , . . . , ct−1 be a set of canonical reflections of Λ. Since θ is surjective, it maps some canonical reflection to τ . In fact there is only one ci with this property; otherwise σ ≤ 2(t − 2) = g − 1, a contradiction. Thus, θ(cj ) = σ for j = i and so θ maps all connecting canonical generators of Λ onto (0, 0) ∈ Z2 ⊕ Z2 . Hence, applying Theorem 1.2.1 to the kernel of p ◦ θ, where p : Z2 ⊕ Z2 → Z2 ⊕ Z2 /τ is the canonical projection, we conclude that τ is a separating symmetry. Let us assume now that s(Λ) = s2 . Since ker θ is a surface group, the canonical reflections c0 , c1 , c2 , c3 and c4 of the non-empty period cycle in s2 are mapped by θ alternatively to σ and τ . Thus they contribute with two ovals to σ and also with two ovals to τ . Consequently, the reflections in the empty period cycles must contribute with g − 1 ovals to σ, and this forces θ(ei ) = (0, 0) for every connecting canonical generator ei of Λ. Using again Theorem 1.2.1, we deduce that τ is a separating symmetry. By combining Proposition 5.2.10 and Theorem 5.2.11 we obtain now a formula that yields the non-separating character of some symmetries of Xg and Yg . Proposition 5.3.11. Let G = Aut+ (S) with S = Xg or S = Yg , and consider the presentation of G given in the introduction of this section. Let σ, τ be two distinct commuting symmetries of S having a common fixed point such that σ is separating. For each x ∈ G let δx be either 1 if σ ◦ τ is conjugate to some power of x, or 0 otherwise. Then

|C(G, σ ◦ τ )|

δa δb δab + + ≤ 2σ. 2g + 2 4 2

Proof. Let h = σ ◦ τ . By Proposition 5.2.10 we have | Fix(h)| ≤ 2σ, and so it is enough to prove the equality

| Fix(h)| = |C(G, h)|

δa δb δab + + . 2g + 2 4 2

This is the formula in Theorem 5.2.11. Indeed, since σ and τ are commuting symmetries, h has order 2, and so NG (h) = C(G, h). Moreover, as we observed in


123

(5.3.8), the proper periods in the signature of a Fuchsian group Δ admitting a group epimorphism Δ → G whose kernel is the Fuchsian group Γ that uniformizes S are 2g + 2, 4 and 2, and we are done. Proposition 5.3.12. (1) The symmetry σ2 of Xg is non-separating for even g. (2) The symmetry σ3 of Yg for g ≡ 7 (mod 8) is non-separating. Proof. (1) We apply Proposition 5.3.11 with σ = σ2 and τ = σ3 ◦ σ2 ◦ σ3 . We must show first that they commute. Using the epimorphism defined by (5.4) we get σ ◦ τ = σ2 ◦ (σ3 ◦ σ2 ◦ σ3 ) = b2 , τ ◦ σ = (σ3 ◦ σ2 ◦ σ3 ) ◦ σ2 = tb2 t = b−2 = b2 , and so σ ◦ τ = τ ◦ σ. Moreover, σ and τ have a common fixed point, because σ ◦ τ is an analytic automorphism of Xg of finite order 2. On the other hand C(Aut+ (Xg ), σ ◦ τ ) = C(Aut+ (Xg ), b2 ) = Aut+ (Xg ), because b2 a = ab2 . Hence |C(Aut+ (Xg ), σ ◦ τ )| = 8(g + 1). Let us prove that δb = 1 and δa = δab = 0, where we are using the notations in Proposition 5.3.11. The first equality is obvious. As to δa , we see that if b2 were conjugate to some power ai of a then there would exist f ∈ Aut+ (Xg ) such that f b2 f −1 = ai , that is, b2 ∈ a, a contradiction. Thus δa = 0. Finally, if δab = 1 then there would exist f ∈ Aut+ (Xg ) such that f b2 f −1 = ab, and so a = b, which is an absurd. Suppose that σ2 is separating. Then, by Proposition 5.3.11,

2 = 2σ2 ≥ |C(Aut+ (Xg ), σ◦τ )|

δb δab 8(g + 1) δa + + = = 2(g+1), 2g + 2 4 2 4

a contradiction. (2) We now apply Proposition 5.3.11 with σ = σ3 and τ = σ2 ◦ σ3 ◦ σ2 . As before it is easily seen that σ and τ are commuting symmetries with a common fixed point. Moreover, σ ◦ τ = b2 and δa = δab = 0, δb = 1. By Proposition 5.3.4, |C(Aut+ (Yg )(σ ◦ τ ))| = 4(g + 1) and so, using Proposition 5.3.11 once more,

δ δb δab 4(g + 1) g+1 a = 2σ3 ≥ |C(Aut+ (Xg ), σ◦τ )| + + = = g+1, 2 2g + 2 4 2 4 a contradiction.

To finish the computation of the topological types of the symmetries of the Accola–Maclachlan and the Kulkarni surfaces we must calculate the separating character of the symmetry σ1 of Xg and Yg , the first just for even g. In this case we apply a different approach due to Hoare and Singerman, [61].

124


Proposition 5.3.13. (1) For even g the symmetry σ1 of Xg is separating. (2) The symmetry σ1 of Yg is non-separating. Proof. We prove just part (1) since the proof of (2) is essentially the same. Let K = σ1 , σ2 be the subgroup of Aut(Xg ) generated by σ1 and σ2 . From (5.4), we see that σ1 σ2 = (at)t = a is an element of order 2(g + 1). Hence K is the dihedral group of 4g + 4 elements. We shall use the obvious equalities σ1 (σ1 σ2 )i = (σ1 σ2 )2g+2−i σ1

and (σ2 σ1 )−1 = σ1 σ2 .

Let Γ1 be an NEC group with signature s(Γ1 ) = (0; +; [−]; {(2g + 2, 2g + 2, g + 1)}) containing the surface group Γ that uniformizes Xg . Let us denote by c0 , c1 and c2 the generating reflections of Γ1 satisfying (c0 c1 )2g+2 = (c1 c2 )2g+2 = (c2 c0 )g+1 = 1. Then Γ is the kernel of the group epimorphism Φ : Γ1 → K induced by the assignment c0 → σ1 ; c1 → σ2 ; c2 → σ1 (σ2 σ1 )2g . To prove this it suffices to use the Hurwitz–Riemann formula and to check that the orders of the products σ2 ◦ σ1 (σ2 σ1 )2g and σ1 (σ2 σ1 )2g ◦ σ1 are 2g + 2 and g + 1, respectively. The quoted result in [61] says that σ1 is separating if and only if the Schreier graph S of the set of cosets K/σ1 associated to the system of generators G = {Φ(cj ) : j = 0, 1, 2} of K is bipartite. The graph S has 2g + 2 vertices, denoted as vi = {(σ1 σ2 )i , (σ1 σ2 )i σ1 }, 0 ≤ i ≤ 2g + 1. Let us see how each Φ(cj ) links two vertices of S. First, Φ(c0 )(vi ) = σ1 (vi ) = {σ1 (σ1 σ2 )i , σ1 (σ1 σ2 )i σ1 } = {(σ2 σ1 )i−1 σ2 , (σ2 σ1 )i } = {(σ1 σ2 )2g+3−i σ2 , (σ1 σ2 )2g+2−i } = {(σ1 σ2 )2g+2−i , (σ1 σ2 )2g+2−i σ1 } = v2g+2−i . Analogously, Φ(c1 )(vi ) = σ2 (vi ) = {σ2 (σ1 σ2 )i , σ2 (σ1 σ2 )i σ1 } = {(σ2 σ1 )i σ2 , (σ2 σ1 )i+1 } = {(σ1 σ2 )2g+1−i , (σ1 σ2 )2g+1−i σ1 } = v2g+1−i .

5.4 Algebraic Formulae for the Symmetries

125

Finally, Φ(c2 )(vi ) = σ1 (σ2 σ1 )2g (vi ) = {σ1 (σ2 σ1 )2g−i , σ1 (σ2 σ1 )2g−i σ1 } = {(σ1 σ2 )2g−i , (σ1 σ2 )2g−i σ1 } = v2g−i . Therefore, Φ(cj )(vi ) = v2g+2−i−j

for 0 ≤ i ≤ 2g + 1 and j = 0, 1, 2.

Consequently, the set of vertices of the Schreier graph S, with the loops deleted, admits a bipartition V1 = {vi : 0 ≤ i ≤ g};

V2 = {vi : g + 1 ≤ i ≤ 2g + 1},

and so the symmetry σ1 of Xg is separating for even g.

We summarize the results of this section in terms of species (see Definition 1.5.4) as follows. Theorem 5.3.14. (1) The species of representatives of the three conjugacy classes of the symmetries with fixed points of the Accola–Maclachlan surface Xg and the Kulkarni surface Yg are displayed in the following table: Surface

g

Xg

odd

+2 −1 +(g + 1)

Xg

even

+1 −1 +(g + 1)

Yg

σ1 σ2

σ3

g ≡ 3 (mod 4) −2 −1 −(g + 1)/4

(2) The surface Xg admits one conjugacy class of fixed point free symmetries if g is even and two classes if g is odd. (3) The surface Yg admits a fixed point free symmetry if and only if g ≡ 3 (mod 8). In such a case, all fixed point free symmetries of Yg are pairwise conjugate.

5.4 Algebraic Formulae for the Symmetries It is well known that the Accola–Maclachlan and the Kulkarni surfaces admit a nice description by means of defining algebraic equations. Our goal in this section is to provide algebraic formulae for the symmetries and automorphisms of these surfaces when they are described by polynomial equations. We start with the Accola–Maclachlan surfaces.

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Algebraic Formulae for the Symmetries of the Accola–Maclachlan Surfaces Since the Accola–Maclachlan surface is hyperelliptic, we can use the geometric method explained in Sect. 4.3 to obtain algebraic formulae of its symmetries and automorphisms. Theorem 5.4.1. Let Xg : y 2 = x2g+2 − 1 be the Accola–Maclachlan surface of genus g. Its full automorphism group Aut(Xg ) has order 16(g+1) and presentation Aut(Xg ) = a, b, σ | a2g+2 , b4 , σ 2 , (ab)2 , ab2 a−1 b2 , (aσ)2 , (bσ)2 . Formulae for the generators a, b and σ in this plane model are

a : (x, y) → x · e

πi/(g+1)

, y , b : (x, y) →

1 iy , x xg+1

, σ : (x, y) → (¯ x, y¯).

A set of representatives of the conjugacy classes of symmetries of Xg and their species are the following: •

{σ, aσ, bσ, ab2 σ} if g is even, with sp(σ) = −1, sp(aσ) = 0, sp(bσ) = g + 1 and

•

sp(ab2 σ) = 1.

{σ, aσ, bσ, ab2 σ, ag+1 bσ} if g is odd, with sp(σ) = −1, sp(aσ) = 0, sp(bσ) = g+1, sp(ab2 σ) = 2 and sp(ag+1 bσ) = 0.

This yields the symmetry type of Xg . Proof. Let AutΣ (Xg ) be the group of Möbius transformations which preserve the branch point set of Xg . This group has order 8(g + 1) because Aut(Xg ) has order 16(g + 1). Observe that the 2g + 2 branch points of Xg are the (2g + 2)-th roots of unity, that is, ekπi/(g+1) for k = 0, . . . , 2g + 1. The set of these points is preserved under the action of the reflections σ i : Σ → Σ defined, for i = 1, 2, 3, by ¯, σ 1 : x → x

σ 2 : x → 1/¯ x and

σ 3 : x → x ¯ · eπi/(g+1) .

1 and σ 3 , and that these last two reflections It is easy to see that σ 2 commutes with σ generate a dihedral group of order 4g + 4. Therefore, σ 1 , σ 2 and σ 3 generate a group D2g+2 ⊕ Z2 of order 8g + 8, which must coincide with AutΣ (Xg ). This group has four conjugacy classes of symmetries, whose representatives are σ 1 , σ 2 , σ 3 and σ3 ◦ σ 1 )g+1 ◦ σ 2 : Σ → Σ ; x → −1/¯ x. σ 4 := (


127

Formulae for their liftings σi : Xg → Xg are x, y¯), σ1 : (x, y) → (¯ y 1 i¯ , g+1 , σ2 : (x, y) → x ¯ x ¯

σ3 : (x, y) → (¯ x · eπi/(g+1) , y¯), y −1 i¯ , g+1 . σ4 : (x, y) → x ¯ x ¯

Observe that σ1 , σ2 and σ3 are involutions, while σ4 is an involution if and only if g is odd. With these formulae at hand it is straightforward to check that the automorphisms a := σ3 ◦ σ1 , b := σ2 ◦ σ1 and σ := σ1 , which have the same formulae as those in the statement of the theorem, satisfy the defining relations of Aut(Xg ) given above. Using them, it is an easy exercise in group theory to determine that Aut(Xg ) has four conjugacy classes of symmetries if g is even, with representatives {σ, aσ, bσ, ab2 σ}, and five conjugacy classes if g is odd, with representatives {σ, aσ, bσ, ab2 σ, ag+1 bσ}. Let us compute their species by using Theorem 4.3.5. First, sp(ag+1 bσ) = 0 (if g is odd) since this symmetry is a lifting of the antipodal map x → −1/¯ x. To compute the species of the other symmetries we need to calculate the number of branch points fixed by each σ i for i = 1, 2, 3. Clearly, σ 1 fixes two branch points, σ 2 fixes all of them and σ 3 fixes none of them. Hence {1, 0} if g is even, sp(σ) = −1, sp(bσ) = g + 1 and {sp(aσ), sp(ab2 σ)} = {2, 0} if g is odd. Finally, it is easy to see that the points (x, y) of the form x = r · eπi/(2g+2) , y = is with r, s ∈ R such that s2 = r2g+2 + 1, are points in Xg which are fixed by ab2 σ. Therefore aσ is the fixed point free symmetry: 1 if g is even, 2 sp(aσ) = 0 and sp(ab σ) = 2 if g is odd, which finishes the proof of the theorem.

Algebraic Formulae for the Symmetries of the Kulkarni Surface Now we shall find defining equations for the symmetries with fixed points of the Kulkarni surface Yg , where g ≡ 3 (mod 4) is large enough. To that end we recall that Kulkarni proved in [70] that Yg is a (2g + 2)-cyclic covering of the Riemann sphere. In fact, it admits the following defining equation y 2g+2 = (x − 1)xg−1 (x + 1)g+2 . Recall also the presentation of the group G = Aut+ (Yg ) = a, b | a2(g+1) , b4 , (ab)2 , b2 ab2 ag of analytic automorphisms of Yg given at the beginning of Sect. 5.3.

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First we shall find explicit formulae for the elements of the group G. Second, observe that if C(x, y) denotes the field of rational functions of Yg , the symmetries of Yg are nothing else but the√field automorphisms of C(x, y) of order 2 that fix the real numbers but map i = −1 to −i. In particular, since the polynomial defining Yg has real coefficients, the automorphism σ of C(x, y) over R induced by the assignment σ(x) = x;

σ(y) = y;

σ(i) = −i

is a symmetry of Yg . Finally, we shall prove that {σ, a ◦ σ, σ ◦ b} are representatives of the three conjugacy classes of symmetries of Yg having fixed points. Throughout this section we will follow closely the approach and notations employed by Turbek in [124]. Our first goal is to obtain explicit formulae for the automorphisms a and b of Yg or, equivalently, of its function field C(x, y). For that we begin by studying the ramification data of the coverings associated to some distinguished subgroups of G. (5.4.2) Coverings associated to some subgroups of Aut+ (Yg ). Consider the subgroups H = a, b2 and K = a of G, which satisfy [G : H] = [H : K] = 2. We are going to study the ramified coverings Yg

(8g+8):1

/ Yg /G

(2g+2):1

Yg /K

2:1

2:1

/ Yg /H.

Let Δ be a triangle Fuchsian group with signature s(Δ) = [2g + 2, 4, 2]. Denote by x1 , x2 and x3 the elliptic canonical generators of Δ of orders 2g + 2, 4 and 2, respectively. The kernel of the epimorphism Δ → G induced by the assignment x1 → a ; x2 → b ; x3 → ab, is the surface Fuchsian group Γ which uniformizes the surface Yg . In this way we get a diagram / H/Δ H Yg = H/Γ

/ Yg /G = (H/Γ)/Δ/Γ.

The upper horizontal arrow ramifies over three branching points with ramification indices 2g+2, 4 and 2, and so the same holds true for the lower horizontal arrow. We denote by R0 , R∞ , R ∈ Yg /G the ramification points of the projection Yg → Yg /G, with ramification indices 2g + 2, 4 and 2, respectively. As to the covering Yg → Yg /K, it is easy to see that we may assume, without loss of generality, that the automorphism a generating K is defined by (x, y) →


129

(x, ξy), where ξ = eπi/g+1 . Of course, this is equivalent to saying that a is the automorphism of the function field C(x, y) of Yg defined by a(x) = x and a(y) = ξy. Therefore the covering Yg → Yg /K is nothing else but the projection Yg → Yg /K = Σ ; (x, y) → x, that ramifies over the points p−1 , p1 and p0 corresponding to x = −1, x = 0 and x = 1 in Σ. Using elementary methods, as in [43] or [92, pp. 73, 74], it is easily checked that the respective ramification indices are 2g + 2 2g + 2 2g + 2 = 2g + 2, = g + 1, = 2g + 2. gcd(2g +2, g +2) gcd(2g +2, g −1) gcd(2g +2, 1) Of course, since both Yg /G and Yg /H are covered by the sphere Yg /K, they are also spheres. To understand the covering Yg /H → Yg /G let us denote by r1 , . . . , rk its ramification indices. Then, by the Hurwitz–Riemann formula, k

(1 − 1/ri ) 2g(Yg /H) − 2 = [G : H] 2g(Yg /G) − 2 + i=1

and, since g(Yg /H) = g(Yg /G) = 0 and each ri divides the degree of the map Yg /H → Yg /G, which equals 2, we get −2 = 2(−2 + k/2), i.e., k = 2. Clearly, the point R0 , that corresponds to a fixed point of the automorphism a, is unbranched with respect to Yg /H → Yg /G, and so both R and R∞ are branching points whose ramification index equals 2. Therefore there exist exactly two points q0 , q1 ∈ Yg /H lying above R0 and there exists a unique point q∞ ∈ Yg /H lying above R∞ . To finish the analysis of the combinatorics of these coverings let us denote by s1 , . . . , s the ramification indices of Yg /K → Yg /H. By the Hurwitz–Riemann formula,

2g(Yg /K) − 2 = [H : K] 2g(Yg /H) − 2 + (1 − 1/si ) i=1

and this leads to = 2. Necessarily q∞ is a branching point of Yg /K → Yg /H, because its ramification index as a branching point of Yg /K → Yg /G is 4. Thus, only one among {q0 , q1 } is a branching point of Yg /K → Yg /H, and we interchange their labels if necessary so that Yg /K → Yg /H ramifies over q0 but not over q1 .

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(5.4.3) Formulae for the automorphisms a and b. We are now in a position to calculate an explicit formula for the automorphism b of Yg or, equivalently, of the field extension C(x, y)|C. To that end we begin with b2 . Let b2 K : Yg /K → Yg /K be the automorphism of the sphere Yg /K = Σ induced by b2 . Since p0 is a ramification point of the covering Yg /K → Yg /H, while both p−1 and p1 lie over the same point q1 , the automorphism b2 K of Yg /K or, equivalently, of the field extension C(x)|C, fixes x = 0 and switches x = 1 and x = −1. On the other hand, as Yg /K is a sphere, b2 K is a linear fractional transformation. Hence, for some α1 , α2 , α3 ∈ C, b2 K(x) =

α1 x α2 x + α3

and

α1 = −1, α2 + α3

α1 = 1. α2 − α3

Therefore α2 = 0 and α3 = −α1 , that is, b2 K(x) = −x. In other words, b2 (x) = −x when we see b as an automorphism of the function field C(x, y) of Yg . The function field of the orbit space Yg /H is C(z), with z = x2 , and let bH be the automorphism of Yg /H or, equivalently, of the field extension C(z)|C, induced by b. Since the points q0 and q1 lie over the same point R0 of Yg /G, the automorphism bH switches these points. Moreover, since q∞ is the only point in the fiber of R∞ under the projection Yg /H → Yg /G, the automorphism bH fixes q∞ . Since x = ∞ is the only fixed point of b apart from x = 0, we deduce that q∞ ≡ z = ∞, while q0 ≡ z = 0 and q1 ≡ z = 1. Thus, with coordinates, bH(0) = 1,

bH(1) = 0

and bH(∞) = ∞.

It follows easily that bH(z) = 1−z, that is, bH(x2 ) = 1−x2 . Hence b(x2 ) = 1−x2 , and to calculate b(x) we must find a square root of 1 − x2 in C(x, y). Note that in this field, y 2g+2 = (x − 1)xg−1 (x + 1)g+2 =⇒ y 2g+2 x2 = (x2 − 1)xg+1 (x + 1)g+1 . (5.4.4) Computation of b(x). In what follows we write √ g0 = (g + 1)/4, which is an integer because g ≡ 3 (mod 4). Taking i = −1, we have 1 − x2 =

−x2 y 2g+2 −x2 y 8g0 = . xg+1 (x + 1)g+1 x4g0 (x + 1)4g0

In other words, the function v ∈ C(x, y) defined by v(x, y) =

y 2g0 + 1)g0

xg0 (x


131

satisfies the equality (ixv 2 )2 = −x2 v 4 = 1 − x2 . Hence, without loss of generality, we choose b(x) = −ixv 2 . Observe that v 4 = (x2 − 1)/x2 . (5.4.5) Computation of b(y). It remains to determine b(y). Note that b(x2 ) = 1 − x2 and so b

x2 − 1 x2

=

x2 b(x2 ) − 1 = 2 . 2 b(x ) x −1

We can write y 8g0 = x4g0 (x + 1)4g0

x2 − 1 x2

and so

x2 . x2 − 1 After multiplying the respective sides of both equalities we get b(y)8g0 = b(x)4g0 (b(x) + 1)4g0

y 8g0 b(y)8g0 = x4g0 (x + 1)4g0 b(x)4g0 (b(x) + 1)4g0 = x4g0 (x + 1)4g0 (−ixv 2 )4g0 (1 − ixv 2 )4g0 . Therefore the quotient y 2 b(y)2 x(x + 1)(−ixv 2 )(1 − ixv 2 ) is a 4g0 -th root of unity. Among the different possibilities we choose the simplest one, namely, y 2 b(y)2 = x(x + 1)(−ixv 2 )(1 − ixv 2 ). Thus, we are led to find a square root u ∈ C(x, y) of x(x + 1)(−ixv 2 )(1 − ixv 2 ) = −ix2 (x + 1)v 2 − x3 (x + 1)v 4 . 2 Once this is done, yb(y) = u2 and we choose b(y) = u/y. Since v 4 = (x2 − 1)/x2 we have [C(x, v 2 , u) : C(x, v 2 )] = 2 = [C(x, v) : C(x, v 2 )] and [C(x, v 2 ) : C(x)] = 2. Therefore both C(x, v 2 , u)|C(x) and C(x, v)|C(x) are degree 2 subextensions of the cyclic Galois extension C(x, y)|C(x) and so they coincide. Hence, u ∈ C(x, v), that is, there exist Ai ∈ C(x) with 0 ≤ i ≤ 3 such that u = A0 (x) + A1 (x)v + A2 (x)v 2 + A3 (x)v 3 .

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After squaring we get u2 = (A20 + A22 v 4 + 2A1 A3 v 4 ) + 2(A0 A1 + A2 A3 v 4 )v + (A21 + A23 v 4 + 2A0 A2 )v 2 + 2(A0 A3 + A1 A2 )v 3 . On the other hand, we already observed that u2 = h + v 2 , where h = −x3 (x + 1)v 4

and

= −ix2 (x + 1).

Consequently, the functions Ai are a solution of the system of equations ⎧ ⎪ ⎪ A3 A0 ⎨ A1 A0 2 ⎪ ⎪ A0 ⎩ A21

+ A1 A2 + A3 v 4 A2 + A22 v 4 + 2A1 A3 v 4 + A23 v 4 + 2A0 A2

= = = =

0, 0, h, .

We are going to prove that (A0 , A2 ) = (0, 0). Otherwise we deduce, from the two first equations above, that A21 = A23 v 4 , and so A1 = εA3 v 2 for some ε = ±1. After substituting these values in the third and fourth equations, and multiplying the first one by v 2 , we get ⎧ = 0, ⎨ A1 (εA0 + A2 v 2 ) 2 2 4 2 2 A0 + A2 v + 2εA1 v = h, ⎩ = /2. A21 + A0 A2 The first equation leads us to distinguish two cases. First, if A1 = 0 then

A20 + A22 v 4 = h, = /2. A0 A2

Consequently, h=

2 + A22 v 4 =⇒ 4v 4 A42 − 4hA22 + 2 = 0. 4A22

The discriminant 16(h2 − 2 v 4 ) of this polynomial (in the variable A2 ) must be a square in the field C(x). But this is false, because h2 − 2 v 4 = x6 (x + 1)2 v 8 + x4 (x + 1)2 v 4 = x4 (x + 1)2 (x2 v 8 + v 4 ), and the factor x2 v 8 + v 4 =

(x2 − 1)2 x2 − 1 + = x2 − 1 x2 x2


133

is not a square in C(x). Thus A1 = 0, which implies A0 = −εA2 v 2 . Hence,

A22 v 4 + εA21 v 2 = h/2, A21 − εA22 v 2 = /2.

Substitute A21 by εA22 v 2 + /2 in the first equation to obtain A22 v 4 + εv 2 (εA22 v 2 + /2) = h/2 =⇒ h − εv 2 = 4A22 v 4 . Therefore, −x3 (x + 1)v 4 − εv 2 = 4A22 v 4 and so, v2 =

4A22

−ε ∈ C(x), + x3 (x + 1)

a contradiction. This way A0 = A2 = 0 and this implies

2A1 A3 v 4 = h, + A23 v 4 = . A21

The first equation reads 2A1 A3 v 4 = h = −x3 (x+1)v 4 =⇒ 2A1 A3 = −x3 (x+1) =⇒ A3 =

−x3 (x + 1) . 2A1

Then, the second one says = A21 +

x6 (x + 1)2 v 4 =⇒ 4A41 − 4A21 + x6 (x + 1)2 v 4 = 0. 4A21

After solving this equation one gets A21 =

ix2 (x + 1)(−1 ± x) ± x3 (x + 1) = . 2 2

This forces the sign to be “−” and so A21 = −ix2 (x + 1)2 /2. Since c = (1 + i)/2 satisfies c2 = i/2 we get, up to sign, A1 (x) = cix(x + 1). Therefore, A3 =

−x3 (x + 1) −x2 = cx2 . = 2A1 2ci

This way u = cix(x + 1)v + cx2 v 3 ∈ C(x, y) and finally b(y) =

y 2g0 cix(x + 1)v + cx2 v 3 , where v(x, y) = g0 and 4g0 = g + 1. y x (x + 1)g0

We are ready to state the following result.

134


Theorem 5.4.6. Let g be a positive integer with g ≡ 3 (mod 4) and let C(x, y), where y 2g+2 = (x − 1)xg−1 (x + 1)g+2 , be the field of rational functions of the Kulkarni surface Yg of genus g. Let ξ be a primitive (2g + 2)-th root of unity, c = (1 + i)/2 and g0 = (g + 1)/4. Consider the rational functions v(x, y) =

y 2g0 xg0 (x + 1)g0

and u = cix(x + 1)v + cx2 v 3 .

Then the automorphism group of the field extension C(x, y)|C is generated by the automorphisms a and b defined by a(x) = x,

a(y) = ξy;

b(x) = −ixv 2 ,

b(y) =

u . y

Moreover, these automorphisms satisfy the equalities a2g+2 = b4 = (ab)2 = b2 ab2 ag = 1. Proof. The groups Aut(C(x, y)|C) and a, b coincide because they both have 8(g + 1) elements. The other claims are checked straightforwardly. (5.4.7) The symmetries of the Kulkarni surface. Let C(x, y) denote the field of rational functions of the Kulkarni surface Yg of genus g, where y 2g+2 = (x − 1)xg−1 (x + 1)g+2 . Recall that the symmetries of Yg are nothing else but the automorphisms of order 2 of the field extension C(x, y)|R that do not fix i. We already observed that the automorphism σ of C(x, y) over R induced by the assignment σ(x) = x;

σ(y) = y;

σ(i) = −i

is a symmetry of Yg . As we announced at the beginning of this section, we are going to show that {σ, a ◦ σ, σ ◦ b} are representatives of the three conjugacy classes of symmetries of Yg having fixed points. It is clear that both τ = a ◦ σ and γ = σ ◦ b are automorphisms of the field extension C(x, y)|R and they do not fix i. Let us prove now that they have order 2. We keep the notations from Theorem 5.4.6. Since σ(x) = x = a(x) it follows that τ 2 fixes x. Moreover, τ 2 (y) = τ (a(y)) = τ (ξ r y) = a(ξ¯r y) = ξ¯r a(y) = ξ¯r ξ r y = y. Hence τ has order 2. As to the automorphism γ, note that

σ(v) = σ

y 2g0 =v xg0 (x + 1)g0

and so

γ(x) = σ(−ixv 2 ) = ixv 2 .


135

On the other hand,

b(v) = b

u2g0 xg0 (x + 1)g0 1 y 2g0 = = = . xg0 (x + 1)g0 y 2g0 (−ixv 2 )g0 (1 − ixv 2 )g0 y 2g0 v

Henceforth, γ 2 (x) = γ(ixv 2 ) = σ(b(ix)b(v)2 ) =

xv 2 v2

= x.

Thus γ 2 fixes x and it also fixes y. In fact, γ(y) =

σ(u) cix(x + 1)v + c¯x2 v 3 = σ(y) y

and this implies y ci(−ixv)(1 − ixv 2 ) − c¯x2 v . b(γ(y)) = u Therefore

y ci(ixv)(1 + ixv 2 ) − cx2 v γ (y) = cix(x + 1)v + c¯x2 v 3 2

and so it remains to check the equality ci(ixv)(1 + ixv 2 ) − cx2 v = cix(x + 1)v + c¯x2 v 3 , which follows automatically from ci = −c. We have proved that σ, τ = a ◦ σ and γ = σ ◦ b are symmetries of Yg and now we shall show that they are pairwise analytically non-conjugate. (1) Suppose first that σ and τ are conjugate. Then there exists φ = ak bj ∈ Aut+ (Yg ) such that aσ = φ−1 σφ = b−j a−k σak bj =⇒ ak bj aσ = σak bj =⇒ ak bj aσ(v) = σak bj (v). Recall that σ(v) = v, b(v) = 1/v and observe that

a(v) = a

y 2g0 ξ 2g0 y 2g0 = = iv. g g + 1) 0 x 0 (x + 1)g0

xg0 (x

Thus, if j is even then bj (v) = v and so ik+1 v = ak bj aσ(v) = σak bj (v) = σ(ik v) = (−i)k v,

136


which is false. If j is odd then bj (v) = 1/v and we have 1/ik−1 v = ak (i/v) = ak bj (iv) = ak bj aσ(v) = σak bj (v) = σak (1/v) = σ(1/ik v) = 1/(−i)k iv, a contradiction again. (2) Suppose now that σ and γ are conjugate. Then there exists an automorphism φ = ak bj ∈ Aut+ (Yg ) such that σb = φ−1 σφ = b−j a−k σak bj . So bj σb1−j = a−k σak and, in particular, bj σb1−j (x) = a−k σak (x). But σ(x) = a(x) = x and so the last equality reads bj σb1−j (x) = x. Recall that b(x) = −ixv 2 . Therefore, (2.1) For j = 0 we get x = σb(x) = −σ(ixv 2 ) = ixv 2 , an absurdity. (2.2) For j = 1 we get x = bσ(x) = b(x) = −ixv 2 , an absurdity again. (2.3) For j = 2 we get x = b2 σb−1 (x) = b2 σb3 (x). Notice that b2 (x) = b(−ixv 2 ) = −i(−ixv 2 )/v 2 = −x ⇒ b3 (x) = −b(x) = ixv 2 . Consequently, x = b2 σb3 (x) = b2 σ(ixv 2 ) = b2 (−ixv 2 ) = (−i)(−x)v 2 = ixv 2 , a contradiction. (2.4) For j = 3 we get x = b3 σb−2 (x) = b3 σb2 (x) = b3 σ(−x) = b3 (−x) = −ixv 2 , a contradiction again. (3) Suppose finally that τ and γ are conjugate. Let φ = ak bj ∈ Aut+ (Yg ) such that σb = φ−1 aσφ = b−j a−k aσak bj =⇒ bj σb1−j = a1−k σak . Evaluating at x we get bj σb1−j (x) = a1−k σak (x) = x ; however, we have just proved that this is false. This way we have shown that the symmetries σ, τ = a◦σ and γ = σ ◦b are pairwise analytically non-conjugate. To finish this section we must prove the following. Proposition 5.4.8. With the above notations, σ, τ = a ◦ σ and γ = σ ◦ b are symmetries with fixed points of the Kulkarni surface Yg . Proof. In case g ≡ 7 (mod 8) the surface Yg admits no fixed point free symmetry. So in what follows we suppose that g ≡ 3 (mod 8). The symmetry σ fixes points, because the equation y 2g+2 = (x − 1)xg−1 (x + 1)g+2 admits a real solution for y whenever x ≥ 1. As to the symmetry τ, note that τ (x) = x and τ (y) = a(y) = ξy where ξ = eπi/(g+1) . Therefore, since ξ is unitary,


137

¯ τ ((1 + ξ)y) = (1 + ξ)ξy = (1 + ξ −1 )ξy = (1 + ξ)y. Hence, after multiplying by κ = (1 + ξ)2g+2 and introducing the linear change of coordinates x = x, ω = (1 + ξ)y, the equation of the function field of Yg becomes ω g+2 = κ(x − 1)xg−1 (x + 1)g+2 and κ is a negative real number. Indeed, as the argument of ξ equals π/(g + 1), we have that arg(1 + ξ) = π/2(g + 1) and so arg(1 + ξ)2g+2 = π. Notice that ¯ τ (ω) = τ (1 + ξ)τ (y) = (1 + ξ)ξy = (1 + ξ −1 )ξy = (1 + ξ)y = ω. Thus, τ fixes both x and ω and τ (i) = −i, that is, τ is complex conjugation with respect to the new coordinates. Moreover, it fixes points because the equation ω g+2 = κ(x−1)xg−1 (x+1)g+2 admits a real solution for ω whenever −1 ≤ x ≤ 1. To prove that γ also fixes points we will repeat the strategy above, that is, we will find a new equation of the function field of Yg so that γ acts as complex conjugation with respect to the new coordinates. However, the process is now much more involved and it requires many computations. Recall that we defined g0 = (g + 1)/4 and consider the auxiliary functions q=

y2 ; x(x + 1)

v = q g0 ;

1 t=q+ ; q

r=v+

1 v

and s =

v q + . q v

As we proved in part (9) in Proposition 5.3.4, the subgroup M = ag+1 , b of Aut+ (Yg ) generated by ag+1 and b is isomorphic to Z2 ⊕ Z4 . (5.4.9) We claim that Fix(M ) = C(t), where Fix(M ) is the subfield of C(x, y) which is fixed by M . Indeed, b(q) = =

b(y 2 ) u2 = 2 2 b(x)b(x + 1) y (−ixv )(1 − ixv 2 ) 1 x(x + 1)(−ixv 2 )(1 − ixv 2 ) = 2 2 2 y (−ixv )(1 − ixv ) q

and so b(t) = t. Moreover, a(q) =

ξ2 y2 a(y)2 = = ξ2 q a(x)(1 + a(x)) x(1 + x)

which implies ag+1 (q) = ξ 2g+2 q = q and so a(t) = t. Therefore C(t) ⊂ Fix(M ). Thus, to prove the equality it is enough to check that [C(x, y) : C(t)] = [Fix(M ) : C(t)] = |M | = 8.

138


Clearly, [C(q) : C(t)] = 2 because the irreducible polynomial Irr(q, C(t)) of q over C(t) is T 2 − tT + 1. Also, [C(q, x) : C(q)] = 2 since x2 = 1/(1 − q g+1 ); finally [C(x, y) : C(x, q)] = 2 because y 2 = x(x + 1)q. Hence, [C(x, y) : C(t)] = [C(x, y) : C(x, q)] · [C(x, q) : C(q)] · [C(q) : C(t)] = 8, as desired. (5.4.10)

Let us prove now that C(x, q) = Fix(ag+1 ). Indeed, a(x) = x and ag+1 (q) =

ξ 2g+2 y 2 y2 (ag+1 (y))2 = = = q. x(x + 1) x(x + 1) x(x + 1)

Thus C(x, q) ⊂ Fix(ag+1 ) and the equality follows because [C(x, y) : C(x, q)] = 2 = |ag+1 | = [C(x, y) : Fix(ag+1 )]. (5.4.11) Now we find an element f ∈ C(x, y) fixed by γ. Note that, as b has order 4, we may assume that there exists f ∈ C(x, y) such that b(f ) = if . There are many such elements, and we choose √ 2(i − 1)(v 4 − 1)x f= . 2v Recall that b(v) = 1/v and b(x) = −ixv 2 . Hence, a direct computation gives

b(f ) =

√ 2(i − 1)v v14 − 1 (−ixv 2 ) 2

= if.

Consequently, γ(f ) = f because σ(v) = v and σ(x) = x, and so, √ γ(f ) = σ(b(f )) = σ(if ) = (−i)σ(f ) = (5.4.12)

2(−i)(−i − 1)(v 4 − 1)x = f. 2v

For later purposes let us prove that f 4 = −r2 (r2 − 4).

Of course, f4 =

(i − 1)4 (v 4 − 1)4 x4 −(v 4 − 1)4 x4 = 4v 4 v4

and (v 4 − 1)x2 = −1, that is, 2

−(v 4 − 1)2 (v 4 − 1)x2 −(v 4 − 1)2 1 2 2 = = − v − f = v4 v4 v2 2

1 1 1 = − v 2 + 2 + 4 = − v 2 + 2 + 2 v 2 + 2 − 2 = −r2 (r2 − 4). v v v 4


(5.4.13)

C(t, f ) = Fix(ag+1 ),

139

[C(t, f ) : C(t)] = 4 and

Irr(f, C(t)) = T 4 + r2 (r2 − 4). Indeed, recall that a(v) = iv and a(x) = x. Thus, √ a(f ) =

2(i − 1)(a(v)4 − 1)a(x) = f, 2a(v)

while b2 (f ) = b(if ) = ib(f ) = −f . Therefore C(t, f ) = Fix(ag+1 ) and so [C(t, f ) : C(t)] =

8 8 [C(x, y) : C(t)] = = = 4. [C(x, y) : C(t, f )] [C(x, y) : Fix(ag+1 )] |ag+1 |

All we need to check now is that −f 4 = r2 (r2 − 4) ∈ C(t) = Fix(M ). But a(v) = iv and b(v) = 1/v. Thus a4 (v) = v and since g+1 ∈ 4Z, also ag+1 (v) = v. Therefore a fixes r2 (r2 − 4). Moreover,

b(r2 (r2 − 4)) = b(v)2 −

2 1 2 1 = − v 2 = r2 (r2 − 4). 2 2 b(v) v

Hence r2 (r2 − 4) is fixed by both ag+1 and b, that is, r2 (r2 − 4) ∈ C(t). 3 (5.4.14) The sum ψ0 = i=0 bi (y) is fixed by b and σ but ag+1 (ψ0 ) = −ψ0 . The first assertion is evident since b has order 4. As to the third, recall that b and ag+1 commute, by part (9) in Proposition 5.3.4. Moreover, a(y) = ξy, where ξ = eπi/g+1 , and so ag+1 (y) = −y. Hence, ag+1 (ψ0 ) = ag+1

3

i=0

3 3 bi (y) = bi (ag+1 (y)) = bi (−y) = −ψ0 . i=0

i=0

As to the second, a straightforward but cumbersome calculation leads us to the expression ψ0 =

y(v 2 − 1)(v 2 x + x − 1)(v − q) . qv 2

Since σ fixes y, q and v it follows that σ(ψ0 ) = ψ0 , and so also γ(ψ0 ) = ψ0 . √ (5.4.15) The quotient ψ = rψ0 / 2 is also fixed by b and σ, while ag+1 (ψ) = −ψ. Moreover, ψ 2 = r(s − 2). Indeed b(r)√= b(v) + 1/b(v) = 1/v + v = r and we already know that σ fixes v and 2. This proves the first part. For the second, a(v) = iv and, as g + 1 is a multiple of 4, we have ag+1 (v) = v. Thus ag+1 fixes r, and therefore ag+1 (ψ) = −ψ.

140


For the last assertion, observe that 1 − v 4 = 1/x2 and y 2 = qx(x + 1). This way, ψ02 y 2 (1 − v 2 )2 (x(1 + v 2 ) − 1)2 = (v − q)2 q qv 4 x(x + 1)(1 − v 2 )2 (x2 (1 + v 2 )2 + 1 − 2x(1 + v 2 )) = qv 4 2 4 2 x(x + 1)(x (1 − v ) + (1 − v 2 )2 − 2x(1 − v 4 )(1 − v 2 )) = qv 4 4 x(x + 1)(1−v ) + x(x + 1)(1−v 2 )2 − 2(1 + x)x2 (1−v 4 )(1−v 2 ) = qv 4 2 2x(x + 1)(1 − v ) − 2(1 + x)(1 − v 2 ) 2(x2 − 1)(1 − v 2 ) = = qv 4 qv 4 2 2 2 2 2 2x (1 − v )(1 + v ) 2 2x (1 − v ) = = . = q q(1 + v 2 ) q(1 + v 2 ) This way we get (1 + v 2 )2 (v − q)2 (1 + v 2 )(v − q)2 r2 ψ02 = = 2 2 2 v q(1 + v ) qv 2 v

1 + v 2 v 2 + q 2 − 2qv 1 q = +v + − 2 = r(s − 2). = v qv v q v

ψ2 =

(5.4.16) Let ζ = f + ψ. Then C(x, y) = C(t, ζ). Moreover, the symmetry γ = σ ◦ b satisfies γ(i) = −i;

γ(t) = t;

γ(ζ) = ζ.

Hence γ acts on Yg as complex conjugation with respect to the coordinates t and ζ. Indeed, as proved in (5.4.9), we have C(t) = Fix(M ), where M = ag+1 , b Z2 ⊕ Z4 . Recall that ag+1 , b2 and ag+1 b2 are the only elements of order 2 of M , see (9) in Proposition 5.3.4, that is, the maximal subfields of C(x, y) containing C(t) are Fix(ag+1 ), Fix(b2 ) and Fix(ag+1 b2 ). Thus, to check the equality C(x, y) = C(t, ζ) it suffices to prove that neither ag+1 nor b2 nor ag+1 b2 fixes ζ. Indeed, by (5.4.13) and (5.4.15) ag+1 (ζ) = ag+1 (f ) + ag+1 (ψ) = f − ψ = ζ, b2 (ζ) = b2 (f ) + b2 (ψ) = −f + ψ = ζ, ag+1 b2 (ζ) = ag+1 (−f + ψ) = −ζ = ζ. As to the action of γ on i, t and ζ, observe that γ(i) = σ(i) = −i while γ(t) = t because σ fixes q and b(q) = 1/q, by (5.4.9). Moreover, with the notations used in (5.4.15), γ(ψ) = ψ and, by (5.4.11), γ(f ) = f . Thus, γ fixes ζ.


141

Therefore C(t, ζ) is the field of rational functions of Yg and so the irreducible polynomial of ζ over C(t) is a defining equation for the Kulkarni surface. Since C(x, y)|C(t) is a Galois extension with Galois group M = ag+1 , b, we have (5.4.17) Irr(ζ, C(t)) = φ∈M (T − φ(ζ)). In order to prove that the symmetry γ fixes points we are going to calculate explicitly this last polynomial. First we will prove the following equality.

2 2 (5.4.18) Irr(ζ, C(t)) = T 2 − r(s − 2) − f 4 − 16f 4 r(s − 2)T 2 . To calculate Irr(ζ, C(t)) consider first the factor F1 (T ) = (T − ζ)(T − ag+1 (ζ)) = (T − f − ψ)(T − f + ψ) = (T − f )2 − ψ 2 = T 2 − 2f T + f 2 − ψ 2 = T 2 − 2f T + f 2 − r(s − 2). 2

Secondly, we calculate the polynomial F2 = F1b whose coefficients are the images of the coefficients of F1 under b2 . We observed in (5.4.9) that b(q) = 1/q and so b(v) = 1/v. Hence b fixes r and s while b(f ) = if by (5.4.11). Hence b2 fixes r and s but b2 (f ) = −f . Consequently, F2 (T ) = T 2 + 2f T + f 2 − r(s − 2). After multiplying we get F3 (T ) = F1 (T )F2 (T ) = (T 2 + f 2 − r(s − 2))2 − 4f 2 T 2 . Now we calculate, using the equality b(f 2 ) = b(f )2 = −f 2 , the polynomial F3b (T ) = (T 2 − f 2 − r(s − 2))2 + 4f 2 T 2 . Finally, we get Irr(ζ, C(t)) = F3 (T )F3b (T ) = (T 2 +f 2 −r(s −2))2 −4f 2 T 2 (T 2 −f 2 − r(s − 2))2 +4f 2 T 2 2 2 = T 2 + f 2 − r(s − 2) T 2 − f 2 − r(s − 2) − 16f 4 T 4

2 2 + 4f 2 T 2 T 2 + f 2 − r(s − 2) − T 2 − f 2 − r(s − 2)

2 2 = T 2 −r(s − 2) −f 4 −16f 4T 4 +16f 4T 2 T 2 +f 2 − r(s−2) 2

2 = T 2 − r(s − 2) − f 4 − 16f 4 r(s − 2)T 2

2 2 = T 2 − r(s − 2) + r2 (r2 − 4) + 16r3 (r2 − 4)(s − 2)T 2 .

142


We should express the coefficients of Irr(ζ, C(t)) as rational functions of the variable t. Notice that r=v+

1 1 = q g0 + g0 v q

and s =

q v 1 + = q g0 −1 + g0 −1 . q v q

Thus it is enough to prove the following formula: (5.4.19) For each non-negative integer n the function fn (t) = q n + 1/q n satisfies

fn (t) =

[n/2]

1 2n−1

k=0

n n−2k 2 t (t − 4)k , 2k

where [n/2] is the greatest integer less than or equal to n/2. Notice that f0 (t) = 2 and f1 (t) = t. Moreover, 1 1 fn+1 (t) + fn−1 (t) = q n+1 + n+1 + q n−1 + n−1 q q

1 1 = q n+1 + n−1 + q n−1 + n+1 q q 1

1 1 qn + n = q qn + n + q q q = tfn (t). Thus the sequence {fn (t)}n satisfies the second order recurrence relation fn+1 (t) − tfn (t) + fn−1 (t) = 0, whose characteristic polynomial is χ(U ) = U 2 − tU + 1, with roots U = (t ± √ t2 − 4)/2. Hence there exist α, β ∈ C such that

t + √t2 − 4 n

t − √t2 − 4 n +β . fn (t) = α 2 2 α+β=2 Since f0 (t) = 2 and f1 (t) = t we get , that is, α = β = 1 and so α−β=0 fn (t) =

(t +

√ √ t2 − 4)n + (t − t2 − 4)n , 2n

which yields, by Newton binomial, the formula in the statement. As we pointed out, r = fg0 (t) and s = fg0 −1 (t). Moreover, g0 is odd because g = 4g0 − 1 ≡ 3 (mod 8), and so


r = r(t) = fg0 (t) =

g0 g0 −2k 2 t (t − 4)k ∈ R[t], and 2k

(g0 −1)/2

1 2g0 −1

s = s(t) = fg0 −1 (t) =

143

1 2g0 −2

k=0

g0 − 1 g0 −1−2k 2 t (t − 4)k ∈ R[t]. 2k

(g0 −1)/2

k=0

Then we get the irreducible polynomial of ζ over C(t) or, equivalently, a new equation of the function field of Yg , after substituting these values of r(t) and s(t) in the formula of (5.4.18), that is, P (ζ, t) = 0, where P (ζ, t) =

2 2 ζ 2 − r(s − 2) + r2 (r2 − 4) + 16r3 (r2 − 4)(s − 2)ζ 2 ∈ R[ζ, t].

Hence, to show that the symmetry γ = σ ◦ b fixes points, it suffices to see that the polynomial P (ζ, t) has infinitely many real zeros. For each θ ∈ (π/2, π) define v(θ) = e4θi/g+1 ∈ R. Then q(θ) = v(θ)(g+1)/4 = θi e and this implies 1 = eθi + e−θi and q(θ) v(θ) q(θ) + = e(g−3)θi/(g+1) + e(3−g)θi/(g+1) . s(θ) = q(θ) v(θ)

r(θ) = q(θ) +

√ Thus −2 < r(θ) < 0 and −2 < s(θ) < 2 because g ≥ 7. Then r(θ)(s(θ)−2) > 0 and so there exists a real number ζ1 (θ) such that ζ1 (θ)2 = r(θ)(s(θ) − 2). Hence, P (ζ1 (θ), t(θ)) = r4 (θ)(r2 (θ) − 4) r2 (θ) − 4 + 16(s(θ) − 2)2 < 0, while P (ζ2 , t(θ)) > 0 for big enough values of ζ2 . Therefore, for each value of θ ∈ (π/2, π) there exist real numbers t(θ), ζ(θ) such that P (ζ1 (θ), t(θ)) = 0, and we are done.

Chapter 6

Appendix

Let Mg be the moduli space of complex isomorphism classes of complex algebraic curves of genus g ≥ 2. Let us denote by MR g the complex moduli space of real algebraic curves of genus g, which consists of complex isomorphism classes of complex algebraic curves of genus g ≥ 2 that are defined by real polynomials. Since Mg is a quasiprojective variety defined in some projective space Pn (C) by means of polynomials with real (in fact rational) coefficients, complex conjugation induces an anticonformal involution σg∗ : Mg → Mg . It is clear that the set Fix(σg∗ ) of fixed points of σg∗ contains MR g but, as observed by Clifford Earle [40] at the beginning ∗ of the 1970’s, the inclusion MR g ⊂ Fix(σg ) is proper if g > 1. The curves whose isomorphism classes occur in this difference Fix(σg∗ ) \ MR g are called asymmetric. In this appendix we present an example of a family of hyperelliptic asymmetric curves of given genus. It is worthwhile mentioning that Seppälä showed in [110] that every asymmetric curve is, in fact, a covering of a real algebraic curve.

6.1 Compact Riemann Surfaces Without Symmetries It is a classical result that for g ≥ 3, a “sufficiently general” (smooth, irreducible) complex projective curve of genus g has no birational automorphism, except the identity, as proved by Baily [6]. In the same vein, most complex algebraic curves admit no symmetry. However, Earle in [40] was the first one who found asymmetric Riemann surfaces admitting antianalytic automorphisms. Here we describe an infinite family of hyperelliptic Riemann surfaces admitting no symmetry. The group of analytic automorphisms of the surfaces in this family is a direct product of two cyclic groups. The construction we present is a particular case of the one in [26], where the authors obtain a complete characterization of those hyperelliptic surfaces which possess an anticonformal automorphism but are not symmetric. A Family of Asymmetric Surfaces Let n, be two positive integers, where n is even, and denote g = n − 1. Let ξ = eπi/n and let B = {b1 , . . . , b2 } ⊂ C be a finite subset of complex numbers such that all roots of the polynomial E. Bujalance et al., Symmetries of Compact Riemann Surfaces, Lecture Notes in Mathematics 2007, DOI 10.1007/978-3-642-14828-6 6, c Springer-Verlag Berlin Heidelberg 2010

145

146

6 Appendix

f (x) = (xn − bn1 ) · · · (xn − bn2 ) are simple. The Riemann surface SB with function field C(x, y), where y 2 = f (x), is a hyperelliptic surface of genus (deg f −2)/2 = g. The automorphism of the field extension C(x)|C defined by x → ξ 2 x induces a bijection on the set of roots of f , because (ξ 2k bi )n = bni . Thus, there exists an automorphism a of the field extension C(x, y)|C satisfying a(x) = ξ 2 x. (6.1.1) We look for a suitable choice of the set B such that the surface SB admits an anticonformal automorphism τ with τ 2 = a. This implies, in particular, that τ is not a symmetry. Since the covering SB → Σ = SB /a ; (x, y) → x ramifies over x = 0 and x = ∞, the automorphism τ must switch these two points. Hence, there should exist a complex number α such that τ (x) = 1/(αx). In fact, ξ 2 x = a(x) = τ 2 (x) = τ

1 αx

=

αx . α ¯

Therefore, as ξ is unitary, we can write ξ/ξ¯ = ξ 2 = α/α ¯ and we choose α = −ξ. This way, for each index i we have 1 −1 ¯ −b¯i τ (x − bi ) = − bi = x+ ¯ . ξx x bi ξ Consequently, −1/b¯i ξ should be a root of f for each root bi of f . Since ξ n = −1 n and (−1)n = 1, we get (−1/b¯i ξ)n = −1/b¯i and so, for each factor xn − bni also n n x + 1/b¯i is a factor of the polynomial f . The equalities τ (x − n

τ

bni )

=

−1 ξx

n

n −1 ¯ n −b¯i 1 n n ¯ x + n − bi = n − bi = n x x b¯i

n −1 1 1 1 1 1 xn + n = + n = n − n = n n (xn − bni ) ¯ ξx bi bi x bi x bi

show that the polynomial f can be assumed to have the form

f (x) =

i=1

(x − n

bni )

1 n x + n . b¯i

and

6.1 Compact Riemann Surfaces Without Symmetries

147

Moreover, let c ∈ C be a square root of the product

n ¯ i=1 (−bi /bi ) .

Then

1 n n n (x − bi ) x + n τ (y) = τ (f (x)) = τ b¯i i=1 ¯ n f (x) cy 2 −b i = = . bi x2n xg+1 i=1

2

Henceforth, the field extension C(x, y)|R associated to the surface SB admits the anticonformal automorphism τ defined by τ (i) = −i;

τ (x) =

−1 ; ξx

τ (y) =

cy . xg+1

As to the automorphism a of C(x, y)|C, the equality a = τ 2 implies a(y) = ξ g+1 y. Indeed, a(y) = τ

cy c¯cyξ g+1 xg+1 = = ξ g+1 y xg+1 (−1)g+1 xg+1

because g + 1 = n is even and c is unitary. (6.1.2) The automorphism a has order n and the cyclic group a does not contain the hyperelliptic involution ρ of SB , defined by ρ(x) = x and ρ(y) = −y. Thus the group Aut(SB ) of conformal automorphisms of SB contains the group a ⊕ ρ = Zn ⊕ Z2 . Moreover, the automorphisms τ and aρ have orders 2n and n respectively. For the first statement, observe that ak (y) = ξ k(g+1) y and ak (x) = ξ 2k x. Hence k a is the identity if and only if 2n divides both k(g + 1) and 2k or, equivalently, n divides k. On the other hand, suppose that ρ = aj for some exponent j. This implies ξ j(g+1) y = aj (y) = ρ(y) = −y

and ξ 2j x = aj (x) = ρ(x) = x

or, equivalently, j(g + 1) must be divisible by n but not by 2n, and 2j must be a multiple of 2n. This is impossible because g + 1 is even. The third statement is evident because ρ commutes with a. Finally, τ 2n = an = 1 and, as n is even, n = |a| = |τ 2 | =

|τ | =⇒ |τ | = 2n, gcd(2, |τ |)

and |aρ| = lcm(|a|, |ρ|) = n, where lcm stands for the least common multiple. (6.1.3) Let H = ρ be the subgroup generated by the hyperelliptic involution and let G+ be the subgroup of the group G = τ ⊕ H = Z2n ⊕ Z2 consisting of the conformal elements of G. We claim that G = Aut(SB ) unless the factor group

148

6 Appendix

Aut+ (SB )/H contains a subgroup containing aH and which is isomorphic either to the dihedral group Dn or to the cyclic group Zkn for some integer k ≥ 2. Indeed, suppose that G Aut(SB ). Then, the difference Aut(SB ) \ G contains some conformal automorphism, and so also Zn = G+ /H Aut+ (SB )/H. But this last factor group is a finite group of automorphisms of the Riemann sphere SB /H and so Aut+ (SB )/H is isomorphic either to the symmetric group S4 , or to the alternating groups A4 , A5 , or to the dihedral group Dm or to the cyclic group Zm , for some multiple m of n. Assume there exists a multiple m of n such that either Zm = Aut+ (SB )/H or Dm = Aut+ (SB )/H. In the first case Zn = G+ /H Aut+ (SB )/H Zm and so Aut+ (SB )/H, which contains aH, is isomorphic to Zkn for some integer k ≥ 2. In the second case also Dn ≤ Dm = Aut+ (SB )/H, and this proves our claim in these two cases. Thus we may suppose that Aut+ (SB )/H A4 , S4 or A5 , and so n = 2, 4, because S4 , A4 and A5 contain no element of order ≥ 6. Assume first that n = 2. The Sylow 2-subgroups of A4 and A5 are isomorphic to the Klein group Z2 ⊕ Z2 , while the Sylow 2-subgroups of S4 are isomorphic to the dihedral group D4 . In either case the Sylow 2-subgroup of Aut+ (SB )/H containing aH is isomorphic to Z2 and it is contained either in Z4 or in D2 . Finally, if n = 4 then Aut+ (SB )/H = A4 or A5 , because these last groups have no element of order 4. Thus Aut+ (SB )/H S4 and so it contains the dihedral group D4 . This completes the proof of our claim. (6.1.4) We are ready to determine explicit conditions on the set B = {b1 , . . . , b2 } which imply that Aut(SB ) = τ ⊕ H = Z2n ⊕ Z2 . Notice that if this is the case then the surface SB admits no symmetry. Indeed, the anticonformal automorphisms of SB are of the form τ j ρ for some odd integer j. If one of them were a symmetry, its square aj = τ 2j would be the identity, that is, n would divide j, a contradiction. Hence, our task now is to find conditions on the set B such that the factor group Aut+ (SB )/H contains no subgroup containing aH and which is isomorphic either to the dihedral group Dn or to the cyclic group Zkn for some integer k ≥ 2. First assume that Aut+ (SB )/H Zkn . In this case x = 0 and x = ∞ are fixed by a conformal automorphism of SB of order kn. Thus, if ω = e2πi/kn , the map x → ωx is an automorphism of order kn of the field extension C(x)|C which lifts to C(x, y). Therefore, by (6.1.1), for each factor xn − bni of f, the polynomials kn kn xkn − bkn − (−1)k /b¯i are also factors of f . Hence, for some divisor k i and x of , the surface SB should admit an equation of the form /k

(−1)k+1 kn y2 = (xkn − bkn + i ) x kn b¯i i=1

.

(6.1)

6.1 Compact Riemann Surfaces Without Symmetries

149

Assume now that Aut+ (SB )/H Dn . Then SB admits a conformal automorphism of order 2 that interchanges x = 0 and x = ∞. Therefore the automorphism x → 1/x of C(x)|C induces a bijection of the set B. Hence, for each factor xn − bni of f there exists an index j such that anj = 1/ani or −1/a¯j n = a¯i n . Consequently, must be even and the surface SB should admit an equation of the form 2

y =

/2 i=1

(x − n

bni )

1 1 n n n x + n (xn + b¯i ). x − n ¯ bi bi

(6.2)

This completes the proof of the following theorem. Theorem 6.1.5. Let n, be positive integers with n even, and denote g = n − 1. Let B = {b1 , . . . , b2 } ⊂ C be a finite set such that all roots of the polynomial f (x) = (xn − bn1 ) · · · (xn − bn2 ) are simple. Suppose also that bn+j = −1/¯bnj for j = 1, . . . , . (1) The compact Riemann surface SB with function field C(x, y), where y 2 = f (x), is a hyperelliptic surface of genus g. (2) Let ρ be the hyperelliptic involution of SB , let ξ = eπi/n and let c ∈ C be a square root of the product i=1 (−b¯i /bi )n . The surface SB admits the group τ ⊕ ρ = Z2n ⊕ Z2 as a group of automorphisms, where τ is the anticonformal automorphism of SB induced by the automorphism of the field extension C(x, y)|R defined as τ (i) = −i;

τ (x) =

−1 ; ξx

τ (y) =

cy . xg+1

(3) Suppose that a defining equation for SB cannot be expressed in the form (6.1) or (6.2). Then τ ⊕ ρ = Z2n ⊕ Z2 is the full group of automorphisms of SB . In particular, this surface admits no symmetry.

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Index

Symbols H -number, 95 H, 1 Aut(S), 16 dim(Λ), 13 L(ω), 71 || S ||, xv || σ ||, 33 μf (g), 26, 27 μi (g), 29 μw i (g), 29 ν(g), xv sp(σ), xii ϕ-conjugacy, 91 ci ∼ cj , 10 (M−q )-symmetry, 61 C(Λ, c), 9 A abstractly orientable group, 21 orientation in an, 21 abstractly oriented group, 21 Accola-Maclachlan surface, xviii, 110 area of a signature, 5 of an NEC group, 5 asymmetric surface, xix automorphism group of a Riemann surface, 16 of a Riemann surface, 16 of the hyperbolic plane, 2

canonical surface symbol, 2 contribution of a reflection, 34 D dianalytic structure, 11 F Fuchsian group, 2 triangle, 3 fundamental parallelogram of a lattice, 69 fundamental region of an NEC group, 2 G genus of a Klein surface, 11 H Hurwitz automorphism group, 95 Hurwitz group, 95 Hurwitz-Riemann formula, 6 hyperelliptic involution, 82 hyperelliptic Riemann surface, 82 I isotropy group, 91 K Klein quartic, xvii, 95 Klein surface, xii, 11 Kulkarni surface, xviii, 110

B bordered surface NEC group, 11 C canonical Fuchsian group, 2

L lattice, 69 hexagonal, 71 square, 71

157

158

Index

P period cycle, 3 presentation of an NEC group, 3 proper periods, 3 pseudosymmetry, 87

S set of canonical generators of an NEC group, 3 signature abstract, 3 Fuchsian, 6 maximal Fuchsian, 13 maximal NEC, 13 NEC, 6 of a Fuchsian group, 3 of an NEC group, 3 triangle Fuchsian, 3 triangle NEC, 4 Singerman symmetry, xvii smooth epimorphism, 12 smooth factor of an NEC group, 12 species of a symmetry, xii, 16 surface NEC group, 11 symmetric Riemann surface, xi, 15 of the first type, 92 of the second type, 92 symmetry non-separating, xii, 15 of a Riemann surface, xi, 15 separating, xii, 15 symmetry type, 16

R real algebraic curve, xi purely imaginary, xii real form, xii real structure of a Riemann surface, 15

T Teichmüller space, 14 topological type of a symmetry, xii, 15 topologically equivalent actions, 12

liftings of a Möbius transformation, 83 link periods, 3

M Macbeath-Singerman surface, 96 Macbeath-Singerman symmetry, xvii modular group, 14 moduli space, 14

N NEC group, 2 maximal, 13 triangle , 4 normal pair, 13

O orbit genus, 3 oval of a symmetry, 15

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Symmetries of compact Riemann surfaces

Compact Riemann surfaces

Symmetries of Compact Riemann Surfaces (Lecture Notes in Mathematics 2007)

Geometry and spectra of compact riemann surfaces

Function theory on compact Riemann surfaces

Riemann surfaces

Riemann surfaces

Riemann Surfaces

Riemann Surfaces

Geometry and Spectra of Compact Riemann Surfaces (Progress in Mathematics)

Families of Meromorphic Functions on Compact Riemann Surfaces

Riemann Surfaces of Infinite Genus

Lectures on Riemann surfaces

Compact Riemann Surfaces: An Introduction to Contemporary Mathematics

Introduction to Compact Riemann Surfaces and Dessins d'Enfants

Compact Riemann Surfaces (Lectures in Mathematics. ETH Zurich)

Compact Riemann Surfaces: An Introduction to Contemporary Mathematics

Kuramochi Boundaries of Riemann Surfaces

Riemann Surfaces of Infinite Genus

Lectures on Riemann Surfaces

Riemann Surfaces (Mathematics Series)

Conformal mapping on Riemann surfaces

Computational Approach to Riemann Surfaces

Conformal mapping on Riemann surfaces

Extremal Polynomials and Riemann Surfaces

Ernst Equation and Riemann Surfaces

Theta functions on Riemann surfaces

Theta Functions on Riemann Surfaces

Hardy classes on Riemann surfaces

Riemann surfaces, dynamics, and geometry

Theta Functions On Riemann Surfaces

Symmetries of compact Riemann surfaces

Compact Riemann surfaces