Abelian Actions on Pseudo-real Riemann Surfaces

A compact Riemann surface is called pseudo-real if it admits orientation-reversing automorphisms but none of them has order two. In this paper, we find necessary and sufficient conditions for the existence of an action on a pseudo-real surface of genus g⩾2\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$g\geqslant 2$$\end{document} of an abelian group containing orientation-reversing automorphisms. Several consequences are obtained, such as the solution of the minimum genus problem for such abelian actions.


Introduction
A compact Riemann surface is called pseudo-real if it admits anti-conformal (orientation-reversing) automorphisms, but no anti-conformal automorphism of order 2, or equivalently, if the surface is reflexible but not definable over the reals. An anti-conformal involution is usually called symmetry, so another term used for such surfaces is asymmetric. Their importance stems from the fact that in the moduli space of compact Riemann surfaces of given genus, pseudo-real surfaces represent the points that have real moduli but are not definable over the reals.
The first results on the existence of pseudo-real surfaces are due to Earle [12] and Shimura [25]. Seppälä [24] showed that the complex algebraic curves of real moduli are coverings of algebraic curves defined over real numbers. This was generalized by Bagiński and Gromadzki [2] who proved, using the language of Riemann surfaces, that a pseudo-real Riemann surface is a cyclic unbranched covering of degree a power of 2 of a compact Riemann surface having a purely imaginary real form. Other examples of pseudo-real surfaces in [14] for non-orientable surfaces, and by McCullough in [21] for bordered surfaces.
Cyclic groups acting on Riemann surfaces and containing elements which reverse the orientation of the surface were considered by Etayo in [13]. He found the necessary and sufficient conditions on the signature of a proper NEC group for which there exists a smooth epimorphism onto a cyclic group. An application of his results led to finding the largest order of a cyclic group acting on a pseudo-real Riemann surface. Also, the minimum genus problem for cyclic actions on pseudo-real surfaces was considered by Bagiński and Gromadzki in [2], with the right solution given by Conder and Lo in [11]. Non-cyclic abelian actions on pseudo-real surfaces were considered in [5,Section 4], with the focus on actions of the largest possible order.
Following [2], we say that a group A acts essentially on a pseudo-real surface if A contains elements which reverse the orientation of the surface. In this paper, we consider essential abelian actions on pseudo-real surfaces. Given a proper NEC group Λ and a finite abelian group A we find necessary and sufficient conditions for the existence of a smooth epimorphism θ : Λ → A with symmetry-free image. They are given in terms of the signature of Λ and the invariant factors of A. If these conditions are satisfied, then H/ ker θ is a pseudo-real Riemann surface on which the abelian group A acts essentially with the prescribed signature of Λ. This is the main result of this paper, see Theorems 2.1 and 2.2. Several consequences are obtained. The minimum genus problem for essential abelian actions is solved in Sect. 3. When the abelian group has just two invariant factors we find an explicit expression for the minimum genus, Theorem 3.3. In the general case, the solution is given as a minimum to be found among several quantities, Theorem 3.4. In Sect. 4 we fix the order N and let A vary among the abelian groups of order N. We find the minimum genus of a pseudo-real surface on which an abelian group of order N acts essentially. Further, the unique abelian group among those of order N attaining the minimum genus is also determined, Theorem 4.1. Finally, in Sect. 5 we obtain a short proof of three of the bounds computed in [5], Theorem 5.1.

Preliminaries
We begin this section with some background about Riemann surfaces and their automorphism groups which can be found in [9]. We then provide some further information about group actions on pseudo-real surfaces, with special emphasis on abelian groups.

Riemann Surfaces, NEC Groups and Their Signatures
Any compact Riemann surface S of genus g > 1 can be represented as the orbit space H/Γ of the upper half-plane H under the action of some surface Fuchsian group Γ, (that is, a torsion-free discrete cocompact subgroup of Aut + (H) = PSL(2, R), the group of all orientation-preserving isometries of H). A non-Euclidean crystallographic (NEC, for short) group Λ is a discrete cocompact subgroup of Aut(H) = PGL(2, R), the group of all isometries of We will refer to the last one as the long relation. The integers m 1 , . . . , m r are called proper periods. The hyperbolic area of a fundamental region for an NEC group Λ with this signature is 2πμ(Λ) where We call μ(Λ) the reduced area of Λ. If Γ is a surface Fuchsian group and the surface H/Γ has genus g then its reduced area is μ(Γ) = 2g − 2. If Δ is any subgroup of finite index in Λ, then Δ is also an NEC group, and the hyperbolic areas of fundamental regions for Δ and Λ satisfy the Riemann-Hurwitz formula μ(Δ) = |Λ : Δ| μ(Λ). (1.2)

Group Actions on Pseudo-real Surfaces
Now suppose the surface S = H/Γ is pseudo-real. Then, the full group Aut(S) = Λ/Γ contains orientation-reversing automorphisms but none of them has order two. Therefore, Λ contains no hyperbolic reflection but it contains some glide reflection. This means that the signature of Λ is of the form (1.1). Orientation-reversing automorphisms of order two are usually called symmetries, so we will say that Aut(S) is symmetry-free. Observe that, as S admits no symmetry, it also admits no orientation-reversing automorphism of order 2n with n odd. Otherwise, the n-th power of such an automorphism would be a symmetry. So, the order of every element of Aut(S) Aut + (S) is divisible by 4. In particular, every involution in Aut(S) preserves orientation, and |Aut(S)| is divisible by 4. Now, let G be a group that acts on a surface S containing some automorphism that reverses orientation. Following [2], we call such an action essential. The group G may contain no symmetry, but this does not guarantee that the surface S is pseudo-real. In fact, it may happen that Aut(S) strictly contains G with a symmetry in Aut(S) G. This is a key point in the study of pseudo-real surfaces. A detailed discussion of this topic can be Lemma 1.2. Let q be a prime number and R, α 1 , . . . , α t , β 1 , . . . , β r be nonnegative integers with α i α i+1 and β i β i+1 . There is an epimorphism if and only if the following condition holds: if R < t, then for each i t − R, the elementary divisor q αi divides at least t − R + 1 − i elementary divisors q βj .
Clearly, the existence of an epimorphism Λ ab → A from the abelianization Λ ab of an NEC group Λ onto an abelian finite group A is equivalent to the fulfillment of the conditions in Lemma 1.2 for each prime q dividing the order of A. Our goal now is to describe the structure of Λ ab in terms of the signature of Λ.
We write m i = p μi1 1 · · · p μis s with prime numbers p 1 < · · · < p s and nonnegative integers μ ij such that μ 1j + · · · + μ rj > 0. For each prime p j , we rearrange the integers μ 1j , . . . , μ rj to obtain increasing sequences of integers μ 1j · · · μ rj and define m i = p μi1 1 · · · p μis s . Then m i | m i+1 . Some m i may be equal to 1. We can obtain the abelianization Λ ab by computing the Smith normal form of the relation matrix of the abelianized canonical presentation of Λ.
Proof. See [23, Lemma 3.4] for instance.  In fact, if 2 μr is the largest 2-power divisor of any proper period, then the largest even elementary divisor of Λ ab is 2 μr+1 .
We will also need the following inequality, a proof of which can be found in [22]: A finite abelian group A has a unique invariant factor decomposition are the elementary divisors of A.

Main Results
Let Λ be an NEC group with signature (γ; −; In this section, we find the necessary and sufficient conditions on the integers γ, m 1 , . . . , m r , v 1 , . . . , v t for which there exists a smooth epimorphism θ : Λ → A with symmetry-free image. If these conditions are satisfied, then H/ ker θ is a pseudo-real Riemann surface on which the abelian group A acts essentially with signature (γ; − ; [m 1 , . . . , m r ]; {−}). Observe that the condition γ + r > 3 is imposed by Proposition 1.1.
We first consider the case when just one invariant factor of A is divisible by four, Theorem 2.1. The general case will be analyzed in Theorem 2.2. We say that an even integer 2k is singly even if k is odd.
Assume that A has n cyclic factors of singly even order and let Then, there exists a smooth epimorphism Λ → A with symmetry-free image if and only if the following conditions hold: Proof. Assume first that there exists a smooth epimorphism θ : Λ → A with symmetry-free image.
(i) As ker θ is torsion-free, each θ(x i ) has order m i . This order has to divide the exponent v t of A. So v t /m i is an integer, and we now show that it is even for all i. Let us write C vt = C 2 α t × C v t with v t odd, and let u 1 , . . . , u t be generators of C v1 , . . . , C vt−1 , C v t , respectively, which are orientation-preserving since v 1 , . . . , v t−1 , v t are either odd or singly even. Suppose there exists some proper period m i0 divisible by 2 αt . Then, A would be generated by u 1 , . . . , u t , θ(x i0 ), and it would not contain orientation-reversing elements, a contradiction. Hence v t /m i is even for all i.
(ii) Lemmas 1.2 and 1.3 applied to the epimorphism Λ ab → A yields that if γ − 1 < t then for each i t − γ + 1 every elementary divisor of C vi divides at least t − γ + 2 − i elementary divisors of Λ ab . Since each odd elementary divisor of Λ ab is a prime power divisor of a proper period of Λ, see Remark 1.4, it follows that condition (ii) is true for odd elementary divisors of C vi . The situation with the even ones is the same except that the largest even elementary divisor of Λ ab is not a divisor of some m i but a divisor of 2m i for some i, see Remark 1.4. The application of Lemma 1.2 to i = t − γ + 1 yields that the elementary divisor 2 αt−γ+1 of C vt−γ+1 divides at least one elementary divisor of Λ ab . Hence, 2 αt−γ+1 divides 2m i for some i.
(iii) Let ω t be a generator of C 2 α t . To generate C 2 α t , the exponent of ω t in θ(d j ) must be odd for some j since v t /m i is even for all i. We claim that, in fact, the exponent of ω t in θ(d j ) is odd for all j. This is clear if γ = 1. So assume γ > 1 and that the exponent of ω t in, say θ(d 1 ), is an odd integer a. Suppose, to get a contradiction, that the exponent of ω t in, say θ(d 2 ), is an even integer b. Then, there exists an even integer c such that ac + b ≡ 2 αt−1 (mod 2 αt ). The orientation-reversing element θ(d c 1 d 2 ) satisfies that its square has odd order. This is impossible in a pseudo-real surface. This shows the claim.
Let e it be the exponent of ω t in θ(x i ). It follows from the above claim and the long relation that 1 2 i e it + γ is even. We now compare the parities of 1 2 e it and 1 2 (v t /m i ). Let 2 μi be the largest power of 2 dividing m i , so that v t /m i = 2 αt−μi v i with v i odd. Since the 2 μi -th power of ω eit t vanishes, we get e it = 2 αt−μi e it for some e it . Hence, if v t /m i is a multiple of 4 then also e it is a multiple of 4. We claim that the converse also holds if 8 | 2 αt or n = 0. In fact, assume 4 | e it and suppose, to get a contradiction, that v t /m i is singly even. Then m i is even. If m i is singly even then α t = 2 (so 8 2 αt ) and e it ≡ 0 (mod 2 αt ). Thus the non-zero components of θ(x i ) in the Sylow 2-subgroup (C 2 ) n × C 2 α t lie in (C 2 ) n . But n = 0 by hypothesis (since 8 2 αt ). This gives a contradiction. If m i is divisible by four then the (m i /2)-th power of θ(x i ) has null components in (C 2 ) n (if any) but not in C 2 α t . It follows that e it = 2 αt−μi e it with e it odd. So if 4 | e it then 4 | (v t /m i ). This shows our claim. Consequently, if 8 | 2 αt or n = 0 then 1 2 v t /m i and 1 2 e it have the same parity and thus 1 2 i v t /m i + γ is even. This shows the first part in (iii). Finally, if γ + 1 2 v t /m i is odd (so α t = 2 and n > 0) and γ is even then the component of θ( This prevents the number of elements θ(x i ) of (singly) even order from being equal to one. In addition, since the sum 1 2 v t /m i is odd, there is an odd number of odd terms v t /(2m i ). These correspond to even proper periods. This shows condition (iii). We prove the sufficiency of the conditions by defining epimorphisms θ q : Λ → A q for each prime q in the set {q 1 , . . . , q λ } of prime numbers dividing the order of A, and a smooth epimorphism θ : Λ → A as the direct product epimorphism For readability, we write μ i = μ i (q) in the definition of each homomorphism θ q . Recall that q μi divides m i but q μi+1 does not. Also, we assume that μ i μ i+1 ; otherwise, there is a permutation τ q of {1, . . . , r} such that μ τq(i) μ τq(i+1) and we replace x i by x τq(i) and μ i by μ τq(i) in the definition of θ q (x i ) below. The order of θ q (x i ) has to be q μi , so that the order of θ(x i ) is m i . In addition, each θ q has to transform the long relation into the identity. We write ω i = ω i (q) for a generator of C q α i , admitting the possibility ω i = 1 when α i = 0 (to lighten notation).
We first define θ 2 . Let Σ = i 2 αt−μi , which is even by (i), and satisfies that 1 2 Σ has the same parity as 1 2 i v t /m i . Note that the following definition for θ 2 prevents θ(Λ) from containing symmetries since the exponent of ω t is even in θ 2 (x i ), i = 1, . . . , r, and odd in θ 2 (d i ), i = 1, . . . , γ. For the same reason, ker θ has no orientation-reversing elements.
If γ + 1 2 i v t /m i is even and γ < t then we define It is easy to check that θ 2 is onto and that it preserves the long relation. The order of ω 2 α t −μ i t is 2 μi . So, θ 2 (x i ) has order 2 μi for i = 1, . . . , r − t + γ − 1. If μ i 1 then the same happens to Suppose this element is not trivial. Then 2 is an elementary divisor of C 2 α i for some i ∈ {1, . . . , t−γ}. Since γ − 1 < t, we may apply condition (ii) to get that 2 divides at least t − γ + 2 − i 2 proper periods. This contradicts that m r is odd. The proof that if μ i = 0 for i = r then θ 2 (x i ) = 1 is similar.
If γ + 1 2 i v t /m i is even and γ t then we define (2. 2) It is straightforward to check that θ 2 is onto, preserves the long relation and that θ 2 (x i ) has order 2 μi for all i. If γ + 1 2 i v t /m i is odd, then 8 v t (so α t = 2) and n > 0 by (iii ). Observe that each proper period is odd or singly even by (i). In addition, if γ is odd then 1 2 i v t /m i is even and so there is an even number of even proper periods. Moreover, there are some even proper periods (at least two) since otherwise θ(d 1 · · · d γ ) 2 = θ(x 1 · · · x r ) −1 would have odd order, which is impossible in a pseudo-real surface because θ(d 1 · · · d γ ) reverses the orientation. On the other hand, if γ is even, then 1 2 i v t /m i is odd and there is an odd number of even proper periods-at least three by (iii ). In particular, r 2 for any value of γ. If γ < t then we define θ 2 as in (2.1) but we redefine the images of x r−1 , x r and d 1 in the following way: By definition of n, the elements ω 1 , . . . , ω t−n−1 are trivial whilst ω t−n , . . . , ω t−1 have order two. If γ t, then we define θ 2 as in (2.2) but we redefine the images of x r−1 , x r and d 1 as It is easy to see that in both cases θ 2 is onto, preserves the long relation and θ 2 (x i ) has order 2 μi . Now, let q = 2 be a prime number dividing the order of A, C q α 1 × · · · × C q α t the Sylow q-subgroup of A, α i α i+1 , ω i a generator of C q α i and Σ ≡ i q αt−μi (mod q αt ), 0 Σ < q αt . We also write s = −Σ/2 − 1 if Σ is even and s = (q αt − Σ)/2 − 1 otherwise, so that ω 2 s+2 t = ω −Σ t . If γ t then we define: If γ > t, then we define: For either value of γ, the proof that θ q is onto, preserves the long relation and that θ q (x i ) has order q μi , can be done in a similar way as in the case q = 2.
We now consider the case when more than one invariant factor of A is divisible by four. Some parts of the proof of Theorem 2.2 are similar to the proof of Theorem 2.1 and will be omitted. Proof. Assume first that θ : Λ → A is a smooth epimorphism with symmetryfree image.
(i) The order of θ(x i ) is m i and so it divides the exponent v t of A. If r = 1 and v t /m 1 is odd, then the long relation is not preserved.
(ii) These conditions follow from Lemma 1.2 applied to the epimorphism Λ ab → A.
(iv) The long relation is not preserved if either 2 αt divides only one proper period or α t−1 < α t and 2 αt divides an odd number of proper periods. If m = 2, α t−1 = α t and 2 αt divides an odd number of proper periods, then the long relation would not be preserved, some orientation-reversing element would belong to ker θ or θ(Λ) would contain a symmetry. We prove the sufficiency of the conditions (i)-(iv) by defining epimorphisms θ q as in Theorem 2.1 for q = 2, and θ 2 as follows. Let ω i be a generator of C 2 α i and Σ = i 2 αt−μi , so that Σ is odd if and only if i v t /m i is odd. Note that, if Σ is odd, then 2 αt divides an odd number of proper periods; recall condition (iv) in that case.
(a) 2 αt m i . If m < γ t: if γ is even, If γ > t: if γ is even, (Here, we replace ω i by ω i−1 in the definition of θ 2 (d i ) if also n = 0.) If γ m: (b) 2 αt |m i for some i and Σ is even.

Minimum Genus
As an application of Theorems 2.1 and 2.2, in this section, we find the minimum genus > 1 of a pseudo-real Riemann surface on which a given abelian group A acts essentially as a group of automorphisms. We denote this value by g * (A). Recall that the order of A is divisible by 4. The case when A is cyclic was solved by Conder and Lo in [11]. Here, we provide a shorter proof based on our previous results.

Theorem 3.1.
Let v 1 be an integer such that 4 | v 1 . The minimum genus of a pseudo-real surface on which the cyclic group C v1 acts essentially is Proof. An easy application of Theorem 2.1 shows that the signature . Theorem 2.1 shows that C 4n does act with the first three signatures if n = 3, n = 6 and n = 15, respectively. Since their reduced area is also 1 − 1 2n , they provide additional actions of the same group C 4n on genus 2n for these particular values of n. Theorem 2.1 shows that C 4n does not act with the last signature unless m 3 = 2n. In fact, it has to be m 3 = 2n/k, by condition (i), and k odd, by condition (iii). If k > 1 and p is a prime divisor of k then the elementary divisor p α of C 4n does not divide, at least, one proper period, contradicting condition (ii). Therefore, C 4n does not act with signature of reduced area smaller than 1 − 1 2n . Remark 3.2. Observe that if n = 3, 6, 15 then the signature with which C 4n acts attaining the bound is unique, namely, (1; −; [2, 2, 2n]; {−}). The epimorphism is also unique, and the surfaces are hyperelliptic, see [4].
Proof. The following signatures fulfill the conditions of Theorem 2.1 or Theorem 2.2 assuring the existence of an essential action of C v1 × C v2 on a pseudo-real surface of genus 1 + 3 4 v 1 v 2 − R with R as above: We have to prove that the reduced area of any other signature fulfilling the conditions of Theorem 2.1 or Theorem 2.2 is greater than or equal to that of the above signatures. The reduced area of each of these is smaller than 3/2. So we focus on signatures of the form (1.1) with γ + r > 3 for which the reduced area is smaller than 3/2. They satisfy (γ, r) = (2, 2), (1,3) or (1,4), as is easy to see. We deal just with the first case, 8 | v 1 or v 1 = 4, the other cases can be proven likewise. The reduced area of (2; −; , so we will prove that any signature satisfying the conditions of Theorem 2.2 (the one applicable for these values of v 1 ) has reduced area   For i ∈ {t − m + 1, . . . , t}, we consider the following signatures: otherwise.

Least Minimum Genus of Abelian Groups of the Same Order
For a fixed value of N there are several abelian groups of the same order N.
In this section we find the minimum genus g * (N ) of a pseudo-real surface on which an abelian group of order N acts essentially. The minimum is attained by a unique group, and we determine such a group. We again recall that N is divisible by 4.
The minimum genus is attained uniquely by C 3 × C 12 , C N and either C 2 × C N/2 if 16 N or C 2 × C 2 × C N/4 if 16 | N , respectively.