Math. Ann. DOI 10.1007/s00208-017-1542-2

Mathematische Annalen

Connected components of Prym eigenform loci in genus three Erwan Lanneau1

· Duc-Manh Nguyen2

Received: 19 October 2015 / Revised: 18 December 2016 © Springer-Verlag Berlin Heidelberg 2017

Abstract This paper is devoted to the classification of connected components of Prym eigenform loci in the strata H(2, 2)odd and H(1, 1, 2) of the bundle of Abelian differentials M3 over M3 . These loci, discovered by McMullen (Duke Math J 133:569–590, 2006), are GL+ (2, R)-invariant submanifolds of complex dimension 3 of Mg that project to the locus of Riemann surfaces whose Jacobian variety has a factor admitting real multiplication by some quadratic order O D . These algebraic varieties are not necessarily irreducible. The main result we show is that for each discriminant D the corresponding locus has one component if D ≡ 0, 4 mod 8, two components if D ≡ 1 mod 8, and is empty if D ≡ 5 mod 8. Our result contrasts with the case of Prym eigenform loci in the strata H(1, 1) in genus 2 [studied by McMullen (Ann Math (2) 165(2):397–456, 2007)] which is connected for every discriminant D.

1 Introduction Since the work of McMullen [12] it has been known that the properties of SL2 (R)orbit closure of translation surfaces are strongly related to the endomorphisms rings

Communicated by Ngaiming Mok.

B

Erwan Lanneau [email protected] Duc-Manh Nguyen [email protected]

1

Institut Fourier, UMR CNRS 5582, Université Grenoble Alpes, BP 74, 38402 Saint-Martin-d’Hères, France

2

IMB Bordeaux-Université de Bordeaux, UMR CNRS 5251, 351, Cours de la Libération, 33405 Talence Cedex, France

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of the Jacobian of the underlying Riemann surfaces (see also [3,17]). The algebrogeometric approach emphasized by McMullen is to relate affine homeomorphisms of the flat metric (on the level of the first homology group) to self-adjoint endomorphisms of the Jacobian variety. Recall that an abelian variety A ∈ Ag admits real multiplication by a totally real number field K of degree g over Q if there exists an inclusion K → End(A) ⊗ Q such that for any k ∈ K , the action of k is self-adjoint with respect to the polarization of A. Equivalently, End(A) contains a copy of an order O ⊂ K acting by self-adjoint endomorphisms.

1.1 Brief facts summary in the genus 2 case In this situation, the locus E2 = {(X, ω) ∈ M2 : Jac(X ) admits real multiplication with ω as an eigenform}, plays an important role in the classification of SL(2, R)-orbit closures in M2 . Here A = Jac(X ) ∈ A2 , K is a real quadratic field, and the endomorphism ring is canonically isomorphic to the ring of homomorphisms of H1 (X, Z) that preserve the Hodge decomposition. The locus E2 is actually a (disjoint) union of subvarieties indexed by the discriminants of the orders O ⊂ End(Jac(X )). Since orders in quadratic fields (quadratic orders) are classified by their discriminant, the unique quadratic order with discriminant D is denoted by O D . We then define E D = {(X, ω) ∈ E2 : ω is as an eigenform for real multiplication by O D }. The subvarieties E D are of interest since they are GL+ (2, R)-invariant submanifolds of M2 (see [14,15]). We can further stratify E D by defining E D (κ) = E D ∩ H(κ) for κ = (2) or κ = (1, 1). This defines complex submanifolds of dimension 2 and 3, respectively. Hence E D (2) projects to a union of algebraic curves (Teichmüller curves) in the moduli space M2 . 1.2 Components of E D (1, 1) and E D (2) It is well known that the set of abelian varieties A ∈ A2 admitting real multiplication by O D with a specified faithful representation i : O D → End(A) is parametrized by the Hilbert modular surface X D := (H × −H)/SL(O D ). In [15], it has been shown that each E D can be viewed as a C∗ -bundle over a Zariski open subset of X D , and we have E2 =

 D≥4,D≡0,1 mod 4

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E D

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In particular E D is a connected, complex suborbifold of M2 of dimension 3. The fact that there is only one (connected) eigenform locus for each D follows from the fact that there is only one faithful, proper, self-adjoint representation in the set of all 4 × 4 matrices over Z: i : O D → Mn×n (Z) up to conjugation by Sp(4, Z) (see [15] Theorem 4.4). It follows that E D (1, 1) is a Zariski open set in E D . In particular E D (1, 1) is connected for any quadratic discriminant D. The classification of components of E D (2) has also been obtained by McMullen [13]. 1.3 Higher genera In [14] it is shown that analogues of E D exist in higher genus (up to 5). These subvarieties of Mg are called Prym eigenform loci. Surfaces in a Prym eigenform locus are pairs (X, ω) such that there exists a holomorphic involution τ : X → X such that g(X ) − g(Y ) = 2, where Y = X/ τ , τ ∗ ω = −ω, and the Prym variety Prym(X, τ ) admits a real multiplication with ω an eigenform (see Sect. 2 for precise definitions). Note that the condition g(X ) − g(Y ) = 2 is needed for our discussion. For any genus, the set of Prym eigenforms whose Prym variety admits a multiplication by O D will be denoted by E D , and the intersection of E D with a stratum H(κ) is denoted by E D (κ). The goal of this paper is to investigate the topology of the Prym eigenform loci in the strata H(2, 2) and H(1, 1, 2) of M3 . Our methods are useful to tackle other strata, but all the main difficulties already appear for H(2, 2) and H(1, 1, 2), the other cases being roughly the same. It is well known that the stratum H(2, 2) also has two components H(2, 2)odd and H(2, 2)hyp . It is not difficult to see that Prym eigenforms in H(2, 2)hyp are double covers of surfaces in E D (2) (see Proposition 2.3). Thus dim E D (2, 2)hyp = 2, and E D (2, 2)hyp is a (finite) union of GL+ (2, R) closed orbits. On the other hand, we have dim E D (2, 2)odd = 3. The stratum H(1, 1, 2) is connected and we also have dim E D (1, 1, 2) = 3 (see Proposition 2.6). Our main result reveals that the situation in genus three is quite different from the one in genus two. More precisely: Theorem A Let κ ∈ {(2, 2)odd , (1, 1, 2)}. For any discriminant D ≥ 8, with D ≡ 0, 1 mod 4, the loci E D (κ) are non empty if and only if D ≡ 0, 1, 4 mod 8, and in this case they are pairwise disjoint. Moreover the following dichotomy holds: (1) If D is even then E D (κ) is connected, (2) If D is odd then E D (κ) has exactly two connected components. Remark 1.1 One of the main differences between the cases of genus two and genus  0 , three is that the polarization of the Prym variety in genus three has the form 0J 2J which is the reason why E D (κ) is empty if D ≡ 5 mod 8. In genus two we have Prym(X, τ ) = Jac(X ) and the Prym involution τ must be the hyperelliptic involution which is unique. In genus three Prym(X, τ ) is only a factor of Jac(X ), and there may be more than one Prym involution as we will see in Sect. 3.

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τ (σ1 ) σ2

(X2 ,ω2 ) (X2 ,ω2 )

(X1 ,ω1 )

τ (σ0 )

σ0

(X0 ,ω0 )

σ1 (X0 ,ω0 )

σ0 (X1 ,ω1 )

σ2 σ1

Fig. 1 Decomposition of (X, ω) ∈ Prym(2, 2)odd (left) and (X, ω) ∈ Prym(1, 1, 2) (right) into three tori

Thus it is not obvious that one can use the discriminant to distinguish different Prym eigenform loci. It is also worth noticing that while E 9 (4) and E 16 (4) are empty (see [7]), the loci E 9 (κ) and E 16 (κ) do exist for κ ∈ {(2, 2)odd , (1, 1, 2)}. 1.4 Triple tori An important step in the proof is a description of the geometry of the surfaces, and of the boundary of these loci. Since this description is of independent interest we have described several specific decompositions into the appendix. An important tool is the use of triple tori: (1) We say that (X, ω) ∈ E D (2, 2)odd admits a three tori decomposition if there exists a triple of homologous saddle connections {σ0 , σ1 , σ2 } on X joining the two distinct zeros of ω. (2) We say that (X, ω) ∈ E D (1, 1, 2) admits a three tori decomposition if there exist two pairs of homologous saddle connections {σ0 , σ1 } and {σ0 , σ1 } on X joining the double zero to the simple zeros such that {σ0 , σ1 } = τ ({σ0 , σ1 }). If (X, ω) admits a three tori decomposition then it can be viewed as a connected sum of three slit tori (X j , ω j ), j = 0, 1, 2, (see Fig. 1). We will always assume that X 0 is preserved by the Prym involution τ and X 1 , X 2 are exchanged by τ . As a corollary of our main result, we prove the following theorem, which is used in the paper [9]: Theorem B For any discriminant D such that E D (κ) = ∅, there exist in any component of E D (κ) surfaces which admit three-tori decompositions. Theorem B is proved in Sect. 7. In “Appendix” we provide a partial compactification of E D (κ), κ ∈ {(2, 2), (1, 1, 2)} that is the first step in order to compute the Siegel–Veech constants.

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1.5 Strategy of the proof The important ingredients of the proof of the main theorem is the use of surgeries (see Sect. 5). The core of Theorem A are Theorem 6.1 on admissible saddle connection and Theorem 4.1 on non-connectedness. The proofs of Theorem 6.1 and Theorem 4.1 appear in Sects. 6 and 4 respectively. (1) An elementary way to get Prym eigenforms in H(2, 2)odd and H(1, 1, 2) is given by Lemma 2.4. Another way is the use of the surgery “Breaking up a zero” on a Prym eigenforms in H(4) (see [5]). We deduce that E D (2, 2)odd and E D (1, 1, 2) are non-empty whenever E D (4) is non-empty. (2) In Sect. 3 we prove that the loci E D (κ) are pairwise disjoint (Lemma 3.3 and Theorem 3.1). As we have noticed in Remark 1.1 a surface (X, ω) ∈ H(2, 2)odd may have more than one Prym involution. However we then show that all Prym varieties admit real multiplication by O9 . This proves in particular that E 9 (2, 2)odd is non-empty despite the fact that E 9 (4) = ∅. (3) To get an upper bound of the number of components of E D (κ) our strategy is to find in each component C of E D (κ) a surface (X, ω) such that we can collapse the zeros of ω along some saddle connections to get a surface in E D (4). Such saddle connections are called admissible (see Sect. 5). Surprisingly, it turns out that there are components that contain no surface with an admissible saddle connection, moreover all these components belong to the loci E 9 (κ) and E 16 (κ). This fact is proved in Theorem 6.1. (4) Finally, to get the exact count of the number of components we will show in Sect. 4 that if E D (4) is not connected then E D (κ) is not connected either (in contrast with the situation in genus two). This difference comes from the invariant defined in [7].

2 Background We review the necessary tools and results involved in the proof of our main result. For an introduction to translation surfaces in general, and a nice survey on this topic, see e.g. [10,18]. 2.1 Prym eigenform Let X be a Riemann surface and τ an involution of X . We define the Prym variety of (X, τ ) by Prym(X, τ ) = ((X, τ )− )∗ /H1 (X, Z)− where (X, τ )− = ker(τ +id) ⊂ (X ), (X ) is the space of holomorphic one forms on X , and H1 (X, Z)− is the anti-invariant homology of X with respect to τ . Remark that Prym(X, τ ) has naturally a polarization: the lattice H1 (X, Z)− is equipped with the restriction of the intersection form on H1 (X, Z).

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Following [14] we will call a translation surface (X, ω) a Prym form if there exists an involution τ of X such that dimC (X, τ )− = 2, and ω ∈ (X, τ )− . Note that the condition dimC − (X, τ ) = 2 is equivalent to g(X ) − g(Y ) = 2, where Y := X/ τ . In this situation, we will call τ a Prym involution of X . Note that a Riemann surface may have more than one Prym involution (see Theorem 3.1). Recall that a (real) quadratic order is a ring isomorphic to Z[x]/(x 2 + bx + c), the discriminant of the order is defined by D = b2 − 4c. Orders with the same discriminants are isomorphic. Thus for any D ∈ N, D ≡ 0, 1 mod 4, we will write square, O D to designate the unique quadratic order of discriminant D. When D is not a √ O D is a finite index subring of the integer ring in the quadratic field K := Q( D). Let A be an abelian variety of (complex) dimension 2. We say that A admits a real multiplication by O D if there exists an injective ring morphism i : O D → End(A) such that i(O D ) is a self-adjoint proper subring of End(A) (properness means that if f ∈ End(A) and if there exists n ∈ Z, n > 0, such that n f ∈ i(O D ), then f ∈ i(O D )). Definition 2.1 We will call a translation surface (X, ω) a Prym eigenform, if there exists a Prym involution τ of X such that • Prym(X, τ ) admits a real multiplication by some quadratic order O D , • ω ∈ (X, τ )− is an eigenvector of O D . The set of Prym eigenforms admitting real multiplication by O D will be denoted by E D . In [13] it is shown that E D are closed, GL+ (2, R)-invariant submanifolds of the bundle Mg . Up to now, these are the only known GL+ (2, R)-invariant submanifolds of Mg which are geometrically primitive and are not closed orbits (recall that a flat surface (X, ω) is geometrically primitive if there is no translation covering π : (X, ω) → (Y, η) with g(Y ) < g(X )). The intersection of E D with a stratum H(κ) will be denoted by E D (κ). Clearly, E D (κ) are GL+ (2, R)-invariant submanifolds of H(κ). Any translation surface in genus two is a Prym form (the Prym involution being the hyperelliptic involution). It turns out that the locus E2 of Prym eigenforms in genus two is a disjoint union of E D for D ≡ 0, 1 mod 4 and D ≥ 5. It is a fact that E D (2) is connected if D ≡ 0, 4, 5 mod 8, and has two components otherwise (D ≡ 1 mod 8). On the other hand E D (1, 1) is connected for all D (see [13,15]). McMullen [14] proved the existence of Prym eigenforms in genera 3, 4 and 5, and in particular that E D (4) and E D (6) are non-empty for infinitely many D. It is well known that the minimal stratum H(4) of M3 has two components Hhyp (4) and Hodd (4) (see [5] for precise definitions) and the whole locus Prym(4) is contained in Hodd (4). In [7] the authors gave a complete classification of E D (4), namely: Theorem 2.2 (Lanneau–Nguyen [7]) For D ≥ 17, E D (4) is non empty if and only if D ≡ 0, 1, 4 mod 8. All the loci E D (4) are pairwise disjoint. Moreover, for the values 0, 1, 4 of discriminants, the following dichotomy holds. Either (1) D is odd and then E D (4) has exactly two connected components, or (2) D is even and E D (4) is connected. For D < 17, only E 12 (4) and E 8 (4) are non-empty, each of which consists of a unique closed GL+ (2, R)-orbit.

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It follows from [14] that each connected component of E D (4) is a closed GL+ (2, R)orbit. The first striking fact about Prym eigenforms in H(4) is that E D (4) = ∅ if D ≡ 5 mod 8. This difference is actually due to the signature of the polarization of the Prym variety. The second remarkable fact is that E 9 (4) = E 16 (4) = ∅ even though 9 and 16 are not “forbidden values” of D. It is worth noticing that even though we have the same statement in the case D ≡ 1 mod 8 as McMullen’s result in E D (2) (namely, E D (4) has two components), the reason for this disconnectedness is different. Roughly speaking, the two components of E D (4) correspond to two distinct complex lines in the space (X, τ )−  H 1 (X, R)− , whereas in the case E D (2), the two components correspond to the same complex line (this is actually a consequence of the fact that E D (1, 1) is connected), they can only be distinguished by the spin invariant (see [13, Section 5]). 2.2 Prym eigenforms in H(2, 2) and H(1, 1, 2) As explained above we will concentrate on these two particular strata since all the main difficulties appear in this context. The stratum H(1, 1, 2) is connected while the stratum H(2, 2) has two connected components: H(2, 2)hyp and H(2, 2)odd (see [5]). We will not use this classification in the sequel. Proposition 2.3 If (X, ω) ∈ E D (2, 2)hyp then there exists a Prym eigenform (X , ω ) ∈ E D (2) (for some D dividing D) and an unramified double cover ρ : X → X such that ρ ∗ ω = ω. In particular, since E D (2) is a finite union of GL+ (2, R) closed orbits by [13], E D (2, 2)hyp is also a finite union of GL+ (2, R) closed orbits. Proof of Proposition 2.3 By definition X is a hyperelliptic Riemann surface, and the hyperelliptic involution ι exchanges the zeros of ω. Since ι commutes with all automorphisms of X , we have τ := τ ◦ ι is also an involution of X (where τ is the Prym involution of X ). Let X := X/ τ  and let ρ : X → X be the associated double cover. Since any element of (X ) is anti-invariant under ι we have dim ker(τ − id) = dim ker(τ + id). Note that ker(τ − id) ⊂ (X ) is canonically isomorphic to (X/ τ ) and thus has complex dimension 1. Therefore ker(τ − id) ⊂ (X ) has complex dimension 3 − 1 = 2 as well as (X ). Thus X is a Riemann surface of genus two. By the Riemann–Hurwitz formula ρ is unramified. Since τ ∗ ω = ω, there exists a holomorphic one-form ω on X such that ρ ∗ ω = ω. Since (X, ω) ∈ H(2, 2) and ρ is unramified, we conclude that (X , ω ) ∈ H(2). Remark that ρ ∗ is an isomorphism between ker(τ − id) = ker(τ + id) and (X ), and ρ maps H1 (X, Z)− to a sublattice of index two in H1 (X , Z), therefore ρ induces a two-to-one covering from Prym(X, τ ) to Jac(X ). By assumption Prym(X, τ ) admits a real multiplication by the order O D for which ω is an eigenvector. It follows that Jac(X ) also admits a real multiplication by O D ⊗ Q for which ω is an eigenvector. Thus there exists a discriminant D satisfying D |D such that (X , ω ) ∈ E D (2). This shows the first part of the proposition.

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But we know from [15] that E D (2) is a union of GL+ (2, R) closed orbits, and since the map (X, ω) → (X , ω ) is clearly GL+ (2, R)-equivariant, it follows that (X, ω) belongs to a GL+ (2, R)-closed orbit. Since any Riemann surface X in Mg admits only finitely many unramified double covers, we derive that there are only   finitely many closed orbits in E D (2, 2)hyp . In view of Proposition 2.3 we will focus on the cases E D (2, 2)odd and E D (1, 1, 2). Observe that if (X, ω) ∈ Prym(2, 2)odd then by definition the Prym involution τ exchanges the two zeros of ω, and if (X, ω) ∈ Prym(1, 1, 2) then τ exchanges the simple zeros ω while fixing the double zero. The next lemma provides us with examples of Prym eigenforms in Prym(2, 2)odd and Prym(1, 1, 2). Lemma 2.4 (Real multiplication by O D ) Let (w, h, e) ∈ Z3 be an integral vector satisfying 

w > 0, h > 0, gcd(w, √h, e) = 1, 2 D = e + 8wh, e + D > 0.

√

Let λ := e+2 D . Note that λ2 = eλ + 2wh. We denote by Sκ (w, h, e) the surface defined in Fig. 2. Then Sκ (w, h, e) ∈ E D (κ). Proof of Lemma 2.4 Each surface in Fig. 2 is a connected some of three slit tori, and admits an involution τ which fixes one torus and exchanges the other two (see also Sect. 1.4). It is not difficult to see that τ is a Prym involution, and that Sκ (w, h, e) ∈ Prym(κ). Let (X, ω) be one of the surfaces in Fig. 2. Let X 0 be the torus invariant under τ , and X 1 , X 2 be the other two tori. By construction, there are bases (ai , bi ) of H1 (X i , Z), i = 0, 1, 2, such that • τ (a0 ) = −a0 , τ (b0 ) = −b0 , τ (a1 ) = −a2 , τ (b1 ) = −b2 , • ω(a0 ) = λ, ω(b0 ) = ıλ, • For i = 1, 2 one has ω(ai ) = w and ω(bi ) = ı h. w

h h

h

w λ

λ

λ

λ h

S(2,2) (w,h,e)∈ΩED (2,2)odd Fig. 2 Real multiplication by O D

123

w

w

S(1,1,2) (w,h,e)∈ΩED (1,1,2)

Connected components of Prym eigenform loci in genus three

Set a = a1 +a2 , b = b1 +b2 . Then {a0 , b0 , a, b} isa symplectic basis of H1 (X, Z)− in  J 0 which the intersection form is given by the matrix 0 2J . Let T be the endomorphism of H1 (X, Z) which is given in the basis (a0 , b0 , a, b) by the matrix  2w 0    eId 0 2h . T =  h 20  0 0 w  0  Since the restriction of the intersection form on H1− (X, Z) is given by 0J 2J , it is easy to check that T is self-adjoint with respect this form. Note that in this basis ω is given by the vector (λ, ıλ, 2w, 2hı), therefore we have T ∗ ω = λω. It follows that T ∈ End(Prym(X, τ )). Since T satisfies T 2 = eT + 2whId4 , T generates a selfadjoint subring of End(Prym(X, τ )) isomorphic to O D for which ω is an eigenvector. The condition on the gcd shows that the ring is a proper subring. Thus Sκ (w, h, e) ∈   E D (κ). The lemma is proved. Corollary 2.5 For any D ≥ 8, D ≡ 0, 1, 4 mod 8, the loci E D (2, 2)odd and E D (1, 1, 2) are non-empty. Proof of Corollary 2.5 Given any D ≥ 8, D ≡ 0, 1, 4 mod 8 it is straightforward to construct a solution (w, h, e) ∈ Z with D = e2 + 8wh satisfying the assumptions of Lemma 2.4. Indeed: • if D ≡ 0 mod 8, the tuple (D/8, 1, 0) is a solution. • if D ≡ 1 mod 8, the tuple ((D − 1)/8, 1, −1) is a solution. • if D ≡ 4 mod 8, the tuple ((D − 4)/8, 1, −2) is a solution. Corollary 2.5 is proved.   We end this section with the following interesting problem. Question 1 For each D, compute the number of connected components of E D (2, 2)hyp . The present paper is devoted to the classification of connected components of E D (2, 2)odd . 2.3 Kernel foliation To investigate the topology of these loci we first recall the notion of the kernel foliation. Let (X, ω) ∈ H(κ) be a translation surface. In a neighborhood of (X, ω) the kernel foliation leaf of (X, ω) consists of surfaces having the same absolute periods as (X, ω). This foliation has already appeared in several papers (see for example [1,2,4,11,16]. For κ ∈ {(2, 2)odd , (1, 1, 2)}, the intersection with the kernel foliation leaves gives rise to a foliation of the Prym eigenform loci E D (κ) (see [8,9] for more details). The leaves of this foliation have complex dimension one. Since the leaves of this foliation have dimension one, for any (X, ω) ∈ E D (κ), we can use the notation (X, ω) + w, with w lies in a small disk, to denote surfaces in the same leaf and close to (X, ω) (see [9], Section 3). Moreover up to the action of GL+ (2, R), a neighborhood of (X, ω) in E D (κ) consists of surfaces in the same kernel foliation leaf as (X, ω). Namely, we have

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Proposition 2.6 ([8], Corollary 3.2) Let (X , ω ) ∈ E D (κ) close enough to (X, ω) ∈ E D (κ). Then there exists a unique pair (g, w), where g ∈ GL+ (2, R) close to id, and w ∈ C with |w| small, such that (X , ω ) = g · ((X, ω) + w). In particular, we have dim E D (κ) = 3.

3 Uniqueness The Prym involution is unique in the minimal stratum in all genera (this is a consequence of the fact that surfaces in the minimal stratum never have automorphisms). For other strata, surfaces may have more than one Prym involution; they differ by a translation (see e.g. the “Appendix”). Theorem 3.1 Let (X, ω) ∈ H(2, 2)odd be a surface having two Prym involutions τ1 = τ2 such that τ1∗ ω = τ2∗ ω = −ω. Then there exist i1 : O9 → End(Prym(M, τ1 )) and i2 : O9 → End(Prym(M, τ2 )) such that ii (O9 ) is a self-adjoint proper subring of End(Prym(X, τi )), and ω is an eigenform for both subrings i1 (O9 ) and i2 (O9 ). In particular (X, ω) ∈ E 9 (2, 2)odd . If (X, ω) ∈ H(1, 1, 2), then there exists at most one Prym involution τ such that τ ∗ ω = −ω. Corollary 3.2 For κ ∈ {(2, 2)odd , (1, 1, 2)}, if D1 = D2 then E D1 (κ)∩E D2 (κ) = ∅. We will need the following two elementary lemmas. Lemma 3.3 Let A be an abelian variety of dimension two. We regard A as a quotient C2 /L, where L is a lattice isomorphic to Z4 equipped with a non-degenerate skew-symmetric inner product ,  : L × L → Z which is compatible with the complex structure. Let v = 0 be a vector in C2 . Assume that there exists a self-adjoint endomorphism ϕ of A such that ϕ(v) = λv, with λ ∈ R, ϕ = λ · Id. Then there exists a unique discriminant D and a unique self-adjoint proper subring O of End(A) isomorphic to O D for which v is an eigenvector. Lemma 3.4 Let (X, ω) ∈ H(2, 2), and τ1 = τ2 be two Prym involutions of X satisfying τ1∗ ω = τ2∗ ω = −ω. There exists a degree three translation cover p : (X, ω) → (Y, ξ ), ramified only at the zeros of ω, where (Y, ξ ) is a flat torus. Moreover, the involutions τ1 , τ2 descend to the unique involution of Y which acts by −id on the homology and exchanges the images by p of the zeros of ω. We first give a proof of the corollary. Proof of Corollary 3.2 Let (X, ω) ∈ E D1 (κ) ∩ E D2 (κ). Let τ1 and τ2 be the corresponding Prym involutions of X . If τ1 = τ2 then by Theorem 3.1 one has κ = (2, 2)odd and D1 = D2 = 9. If τ1 = τ2 then Lemma 3.3, applied to A = Prym(X, τ1 ) = Prym(X, τ2 ), gives the uniqueness of the self-adjoint proper   subring O ⊂ End(A), which implies D1 = D2 . The corollary is then proved.

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Proof of Lemma 3.3 The proof is standard (see [7, Section 5] for related results). Let S = C · v be the complex line generated by v, and let S denote the orthogonal complement of S with respect to ,  in C2 . Note that S is also a complex line in C2 . Set w = ıv. Since ϕ is an endomorphism of A, we have ϕ(w) = iϕ(v) = λw. In other words ϕ|S = λ · id S . Since ϕ is self-adjoint, it also preserves the complex line S . Thus ϕ|S = λ · id S where λ = λ. Now the minimal polynomial χϕ ∈ Z[X ] of ϕ has degree 2. By definition λ is a real root of χϕ . Hence λ , that is a root, is also real. Moreover, since v is an eigenvector of ϕ, up to a real and imaginary scalar, all the coordinates of v in a basis of L belong to K = Q(λ). Remark that either K = Q, or K ⊂ R and [K : Q] = 2. Let K v be the subring of End(A) ⊗ Q consisting of self-adjoint endomorphisms of A for which v is an eigenvector. For any f ∈ K v , the matrix of f in the decomposition  λ( f ) 0 2 C = S ⊕ S has the form 0 λ ( f ) . We claim that K v is either isomorphic to K or to Q2 . To see this, it suffices to notice that each element of K v is uniquely determined by its eigenvalues on S and S . If λ ∈ Q then we can assume that all the coordinates of v belong to Q, hence both λ( f ) and λ ( f ) belong to Q as f is defined of over Q. → (λ( f ), λ ( f )) is an isomorphism of Q vector spaces. If Thus  : K v → Q2 , f √ λ∈ / Q, then λ ∈ K = Q( d), with d ∈ N, d is not a square. It follows that λ ( f ) is the Galois conjugate of λ( f ) in K . Consequently,  : K v → K , f → λ( f ) is an isomorphism of Q-vector spaces. Set O = K v ∩ End(A). By definition, O is the unique self-adjoint proper subring of End(A) for which v is an eigenvector. It remains to show that O  Z[X ]/(X 2 +bX +c) for some b, c ∈ Z, such that b2 − 4c > 0. We have dimQ K v = 2. For any f ∈ O such that ( f, id) is a basis of K v as a Q-vector space, we denote by ( f ) the discriminant of the minimal polynomial of f . Note that we have ( f ) > 0, and ( f ) = (λ( f ) − λ ( f ))2 . Set D = min{( f ) : f ∈ O, ( f, id) is a basis of K v }. Let ψ be an element of O such that (ψ) = D. Let us show that O = Zψ + Zid. Indeed, let f be an element of O, then we can write f = xψ + yid, with (x, y) ∈ Q2 , / N, by replacing ψ by f − [x]ψ, we can find ψ ∈ O such and ( f ) = x 2 D. If x ∈ that 0 < (ψ ) < D, therefore we must have x ∈ N. It follows that y ∈ N. Finally, since ψ ∈ End(A), the minimal polynomial of ψ has the the form ψ 2 + bψ + cid,   with b, c ∈ Z such that D = b2 − 4c. Proof of Lemma 3.4 The restriction of τ = τ1 ◦ τ2 around the local chart of the double zero of ω can be written as τ (z) = ζ z. Since τ ∗ ω = ω, it follows that ζ 3 = 1. Obviously ζ = 1: otherwise τ is the identity map in a neighborhood of P and hence τ1 = τ2 . Let p : X → Y = X/ τ  be the quotient map. Since τ has order three, p is a ramified covering of degree 3. Clearly, the two zeros of ω are branch points of p of order 3. Moreover since dim ker(τ − id) ≥ 1, one has genus(Y ) ≥ 1. These two facts, combined with the Riemann–Hurwitz formula −4 = 2 − 2 · genus(X ) = 3 · (2 − 2 · genus(Y )) − ≤−





(e p (x) − 1)

x∈X

(e p (x) − 1) ≤ −4

x∈X

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implies that genus(Y ) = 1 and the two zeros of ω are the only branch points of p. Since τ ∗ ω = ω, the form ω descends to a holomorphic 1-form ξ on Y i.e. ω = p ∗ ξ . Now the subgroup of Aut(X ) generated by τ , namely {id, τ1 ◦ τ2 , τ2 ◦ τ1 }, is clearly invariant under the conjugations by τ1 and τ2 . Therefore τ1 and τ2 induces two involutions, say ι1 and ι2 , on Y . Since τi permutes the zeros of ω, the equality τi∗ ω = −ω reads ιi∗ ξ = −ξ , and ιi exchanges the images of the zeros of ω by p. Now   Y is a torus: there exists only one such involution. Hence ι1 = ι2 . Proof of Theorem 3.1 Writing τ = τ1 ◦ τ2 , one has τ ∗ ω = ω and τ fixes all the zeros of ω. If (X, ω) ∈ H(1, 1, 2) then τ is the identity map (it is the identity map around the double zero of ω) thus τ1 = τ2 . Now assume (X, ω) ∈ H(2, 2)odd . Let P, Q ∈ X denote the zeros of ω. Let {α, β} be a pair of simple closed geodesics passing through a fixed point of ι and missing p(P) and p(Q) = ι( p(P)). Clearly this pair forms a basis of H1 (Y, Z) invariant under ι that does not contain p(P) nor p(Q) (see Lemma 3.4 for definition of p). We will construct a symplectic basis of X and a self-adjoint endomorphism of Prym(X, τ1 ). We lift the curves α, β to X in the following manner. Let R ∈ Y be the (unique) intersection point of α with β. Hence ι(R) = R and R is a regular point of the covering p. Since p ◦τ1 = ι◦ p, the involution τ1 induces a transposition of p −1 (R) = {R0 , R1 , R2 }. Therefore τ1 fixes some point, say R0 . Thus τ2 (R0 ) = R0 . Let α0 (respectively, β0 ) be the pre-image of α (respectively, β) passing through R0 . For j = 1, 2 let α j = τ j (α0 ), β j = τ j (β0 ), where τ = τ1 ◦ τ2 . Clearly

αi , α j  = βi , β j  = 0 for any i = j ∈ {0, 1, 2}. On the other hand α0 , β0  = 1 and

α0 , β1  = τ1 (α0 ), τ2 (β0 ) = α0 , τ2 (β0 ) = 0 since τ2 (β0 ) is the pre-image of β passing through τ2 (R0 ) = R0 . The same argument shows that for any i, j ∈ {0, 1, 2},

αi , β j  = δi j . Hence (α0 , β0 , α1 , β1 , α2 , β2 ) is a symplectic basis of H1 (X, Z). This allows us to construct a symplectic basis (a0 , b0 , a1 , b1 ) of H1 (X, Z)− = ker(τ1 + id) as usual: 

a0 = α0 a1 = α1 + α2

b0 = β0 b1 = β1 + β2

 0  . One can normalize by using The intersection form is given by the matrix 0J 2J GL+ (2, R) so that ξ(α) = 1 and ξ(β) = ı ∈ C. Then ω (viewed as an element of H 1 (X, C)− ) is the vector (in the basis dual to (a0 , b0 , a1 , b1 )) (ω(a0 ), ω(b0 ), ω(a1 ), ω(b1 )) = (1, ı, 2, 2ı). By straightforward computation, the Teichmüller discs of (X, ω) and S(2,2) (1, 1, −1) ∈  E 9 (2, 2) coincide. Now the matrix T =

−id2 2·id2 id2 0

(in the basis (a0 , b0 , a1 , b1 )) is

self-adjoint with respect to the restriction of the intersection form on H1 (M, Z)− . Moreover (1, ı, 2, 2ı) · T = (1, ı, 2, 2ı) thus ω is an eigenform for T : hence (X, ω) ∈ E D (2, 2)odd . Since T 2 = −T + 2, the endomorphism T generates a proper subring of End(Prym(M, τ1 )) isomorphic to O D where D = 1 + 4 · 2 = 9. By Lemma 3.3, this subring is unique.

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The same argument shows that End(Prym(M, τ2 )) also contains a unique selfadjoint proper subring isomorphic to O9 , for which ω is an eigenform. The proof of the theorem is now complete.  

4 Non connectedness of E D (κ) In this section we will show that when D ≡ 1 mod 8, the number of components of E D (2, 2)odd and E D (1, 1, 2) is at least two. It is worth noticing that this is not true in genus two, namely, E D (2) has two connected components, while E D (1, 1) is connected (see [13]). Theorem 4.1 For any D ≥ 9 satisfying D ≡ 1 mod 8, the loci E D (2, 2)odd and E D (1, 1, 2) are not connected. Proof of Theorem 4.1 First of all by Corollary 2.5, E D (κ) is non-empty. Before going into the details, we first explain why E D (4) is not connected when D ≡ 1 mod 8 (see [7, Theorem 6.1]). For every surface (X, ω) ∈ E D (4) we denote by S the subspace of H 1 (X, R)− = ker(τ + Id) ⊂ H 1 (X, R) generated by {Re(ω), Im(ω)} and by S the orthogonal complement of S with respect to the intersection form in H 1 (X, R)− . By definition there is an injective ring morphism i : O D → End(Prym(X, τ )) such that i(O D ) is a self-adjoint proper subring of End(Prym(X, τ )), and for any T ∈ i(O D ), S is an eigenspace of T . It turns out that an element T ∈ Im(i) is uniquely determined by its minimal polynomial and by the eigenvalue of its restriction to S. Indeed, the minimal polynomial of T has degree two; thus if T|S = λid S then T|S = λ id S , where λ and λ are the roots of the minimal polynomial of T . Therefore T is given by the matrix λ0 λ0 in the decomposition H 1 (X, R)− = S ⊕ S . In [7, Section 6] for each D ≡ 1 mod 8, with D ≥ 17, we constructed two surfaces (X i , ωi ) ∈ E D (4), i = 0, 1, where the corresponding generators of the order T0 ∈ Im(i0 ) and T1 ∈ Im(i1 ) satisfies: • T0 and T1 have the same minimal polynomial, • T0|S0 = λid S0 and T1|S1 = λid S1 , • , |Im(T0 ) = 0 mod 2 and , |Im(T1 ) = 0 mod 2. where Im is the image. Now if (X 0 , ω0 ) and (X 1 , ω1 ) belong to the same connected component (i.e. A · (X 0 , ω0 ) = (X 1 , ω1 ) where A ∈ GL+ (2, R)) then by construction, there exists an isomorphism f : H1 (X 0 , Z)− → H1 (X 1 , Z)− such that T1 := f −1 ◦ T1 ◦ f defines an endomorphism of Prym(X 0 , τ0 ). By uniqueness of the Prym involution and the map i0 : O D → End(X 0 , τ0 ), it follows that both T0 and T1 belong to i0 (O D ). By construction T0 and T1 have the same minimal polynomial and T0|S0 = (T1 )|S0 = λid S0 . Thus in view of the above remark, T0 = T1 . This is a contradiction since , |Im(T0 ) = , |Im(T1 ) mod 2. We now go back to the proof of Theorem 4.1 and apply similar ideas. Let (w, h, e) ∈ Z3 be as in Lemma 2.4 where D = e2 + 8wh ≡ 1 mod 8. Note that e is odd. We will show that the two surfaces (X 0 , ω0 ) := Sκ (w, h, −e) ∈ E D (κ) and (X 1 , ω1 ) := Sκ (w, h, e) ∈ E D (κ)

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do not belong to the same component. Recall that by construction, for j = 0, 1, we have associated to (X j , ω j ) a generator T j of the order i j (O D ) ⊂ End(Prym(X j , τ j )). Recall that, in the symplectic basis (a0 , b0 , a, b) of H1 (X j , Z) given in Lemma 2.4, the endomorphism is given by the matrix  2w 0    2w 0     −eId eId 0 2h 0 2h and T1 =  h 20  respectively. T0 =  h 02 0 0 0 w 0 w Let us assume that there is a continuous path γ : [0, 1] → E D (κ) such that γ (i) = (X i , ωi ) for i = 0, 1, we will draw a contradiction. Let γ˜ be a lift of γ to the vector bundle T3 over the Teichmüller space T3 . We will denote by (X s , ωs ) the image of s ∈ [0, 1] by γ˜ . Let  be the base surface of the Teichmüller space. By construction the path γ˜ induces a continuous map which sends every s ∈ [0, 1] to a tuple (Js , τs , L s , is , Ss ), where • Js is the complex structure of H 1 (, R), induced by the complex structure of X s , • τs ∈ Sp(6, Z) is the matrix which gives the action of the Prym involution of X s on H 1 (, R), • L s is the lattice H 1 (, Z) ∩ ker(τs + id), 1 − 1 • is : O D → End(H 1 (, R)− s ), where H (, R)s = ker(τs + id) ⊂ H (, R), is an injective ring morphism where is (O D ) is a self-adjoint proper subring of End(H 1 (, R)− s ) that preserves L s . • Ss is the subspace of H 1 (X, R)− = ker(τs + Id) generated by {Re(ωs ), Im(ωs )}. Remark that since ω is holomorphic, Ss is invariant under Js . The action of GL+ (2, R) preserves the subspace Ss ⊂ H 1 (, R)− , and the kernel foliation leaves invariant [Re(ω)] and [Im(ω)]. Therefore Ss is invariant along the path γ˜ . Clearly, the matrix τs is also invariant along the deformation γ˜ . This implies that L s and is are also invariant along γ˜ . In particular S0 = S1 = S and i0 = i1 . Thus T0 , T1 belong to the same ring i0 (O D ) ⊂ End(Prym(X 0 , τ0 )). Remark that T1 and T0 + e · Id have the same minimal polynomial X 2 − eX − 2wh. In addition the eigenvalues of T1 on √S1 = C · ω1 , and the eigenvalue of T0 + e · Id on C · ω0 are both equal to λ = (e + D)/2. Hence T1 = T0 + e · Id. Now Im(T0 + e · Id) mod 2 is generated by {a, b}. The restriction of the intersection form ,  to this subspace is equal to 0 mod 2. On the other hand the restriction of the intersection form to Im(T1 ) does not vanish modulo 2:

T1 (a0 ), T1 (b0 ) ≡ a0 , b0  ≡ 1 mod 2. This is a contradiction, and the theorem follows.

 

In Sect. 5.4 we will give a topological argument for the non connectedness of E 9 (2, 2)odd .

5 Collapsing zeros along a saddle connection In this section we describe a surgery of collapsing several zeros of Prym eigenforms together such that the resulting surface is still a non degenerate Prym eigenform. This

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can be thought as the converse of the surgery “breaking up a zero”. We emphasize that the point is to remain an eigenform, otherwise this is standard (see [5]). In what follows, all the zeros are labelled and all the saddle connections are oriented: if (X, ω) ∈ E D (2, 2)odd , we label the zeros by P and Q and if (X, ω) ∈ E D (1, 1, 2), we label the simple zeros by R1 , R2 and the double zero by Q. Let σ0 be a saddle connection on X . Convention 1 We will always assume that: (1) If (X, ω) ∈ E D (2, 2)odd then σ0 is a saddle connection from P to Q that is invariant under τ . (2) If (X, ω) ∈ E D (1, 1, 2) then σ0 is a saddle connection from R1 to Q. Observe that such saddle connections always exist on any (X, ω) ∈ E D (κ): for κ = (2, 2)odd , take σ0 to be the union of a path of minimal length from a regular fixed point of τ to a zero of ω and its image by τ , for κ = (1, 1, 2), take a path of minimal length from the set {R1 , R2 } to {Q}. 5.1 Admissible saddle connections We begin with the following definition. Definition 5.1 Let (X, ω) ∈ Prym(κ) be a Prym form. (1) κ = (2, 2)odd : we say that σ0 is admissible if for any saddle connection σ = σ0 from P to Q satisfying ω(σ ) = λω(σ0 ), with λ ∈ R+ , one has λ > 1. (2) κ = (1, 1, 2): we say that σ0 is admissible if, for any saddle connection σ = σ0 starting from R1 and satisfying ω(σ ) = λω(σ0 ), with λ ∈ R+ , either λ > 1 if σ ends at Q, or λ > 2 if σ ends at R2 . Observe that by definition the subset consisting of surfaces having an admissible saddle connection is an open GL+ (2, R)-invariant subset. Lemma 5.2 Let κ ∈ {(2, 2)odd , (1, 1, 2)}. For any (X, ω) ∈ E D (κ) and any σ0 satisfying Convention 1, there exists a sufficiently small neighborhood of (X, ω) a surface (X , ω ) ∈ E D (κ) with a saddle connection σ0 (corresponding to σ0 and also satisfying Convention 1) such that any saddle connection σ on X in the same direction as σ0 , if it exists, satisfies (1) Case κ = (2, 2)odd : ω (σ ) = ω (σ0 ). (2) Case κ = (1, 1, 2): (a) If σ connects a simple zero to the double zero then ω (σ ) = ω (σ0 ). (b) If σ connects two simple zeros then ω (σ ) = 2ω (σ0 ). Proof of Lemma 5.2 For any vector v ∈ R2 small enough we denote by σ0 the saddle connection on (X , ω ) = (X, ω) + v corresponding to σ0 . Observe that the set Slope(X, ω) = {s ∈ R ∪ {∞} : s is the slope of ω(γ ) = 0, with [γ ] ∈ H1 (X, Z)}

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is countable. Hence there exists a vector v ∈ R2 small enough so that the slope of ω (σ0 ) does not belong to the set Slope(X, ω) = Slope(X , ω ). Case κ = (2, 2)odd . Let σ be a saddle connection starting from P in the same direction as σ0 . If σ ends at P then [σ ] ∈ H1 (X , Z) and 0 = ω (σ ) has the same slope as ω (σ0 ). This is a contradiction. Thus σ ends at Q and [γ ] = [σ0 ∗ (−σ )] ∈ H1 (X , ω ). If ω (γ ) = 0 then we get again a contradiction. Therefore ω (γ ) = 0 i.e. ω (σ ) = ω (σ0 ). Case κ = (1, 1, 2). Let σ be a saddle connection starting from R1 in the same direction as σ0 . If σ ends at R1 we run into the same contradiction. If σ ends at Q then we also run into the same conclusion namely ω (σ ) = ω (σ0 ). Hence let us assume that σ ends at R2 . Thus σ0 ∗ τ (σ0 ) ∗ (−σ ) is a closed path from R1 to R1 (through Q and R2 ). Therefore [γ ] = [σ0 ∗ τ (σ0 ) ∗ (−σ )] ∈ H1 (X , ω ). The same contradiction   shows that ω (γ ) = 0, i.e. ω (σ ) = 2ω (σ0 ). The lemma is proved. 5.2 Non-admissible saddle connection and twin/double-twin Lemma 5.2 leads to the following natural definition. Definition 5.3 Let (X, ω) ∈ E D (κ) and let σ0 be a saddle connection on X satisfying Convention 1. If κ = (2, 2)odd : a saddle connection σ is a twin of σ0 if σ joins P to Q and ω(σ ) = ω(σ0 ). If κ = (1, 1, 2): (1) a saddle connection σ is a twin of σ0 if it has the same endpoints and ω(σ ) = ω(σ0 ). (2) a saddle connection σ is a double-twin of σ0 if σ joins R1 to R2 and ω(σ ) = 2ω(σ0 ). When (X, ω) ∈ E D (2, 2)odd , since the angle between two twin saddle connections is a multiple of 2π and the angle at P is 6π , we see that each saddle connection σ0 has at most two twins. When (X, ω) ∈ E D (1, 1, 2), the same remark shows that σ0 has at most one twin or one double-twin. Moreover the midpoint of any double twin saddle connection is fixed by the Prym involution. As an immediate corollary of Lemma 5.2, we draw Corollary 5.4 Let κ ∈ {(2, 2)odd , (1, 1, 2)}. For any connected component C of E D (κ), there exist (X, ω) ∈ C and a saddle connection σ0 on (X, ω) satisfying Convention 1 such that either σ0 is admissible, or σ0 has a twin or a double twin. 5.3 Collapsing admissible saddle connections We have the following proposition that is a converse to the surgery “breaking up a zero” (see [5]). We prove the proposition only in the setting of Corollary 5.4. A more general statement holds but it is not needed in this paper.

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Proposition 5.5 Let (X, ω) ∈ E D (κ) be a Prym form and σ0 a saddle connection satisfying Convention 1. We assume that σ0 is admissible. Then one can collapse the zeros of ω along σ0 by using the kernel foliation so that the resulting surface belongs to E D (4). In particular, if there is no saddle connection in the same direction as σ0 connecting two different zeros, then one can collapse σ0 to get a surface in E D (4). We detail the proof since the arguments are not standard for real multiplication. Proof We first consider the case when (X, ω) ∈ E D (2, 2)odd . The proof we describe is constructive. Set  = |σ0 |. As usual we assume that σ0 is horizontal. By definition of admissible saddle connection, any other horizontal saddle connection from P to Q has length > . For any horizontal geodesic ray emanating from a zero of ω we say that the ray is positive if it has direction (1, 0) and negative if it has direction (−1, 0). For instance by convention σ0 is a positive ray for P and a negative ray for Q. Since the conical angle at P and Q is 6π , there are two other positive horizontal rays emanating from + + − − and σ P,2 , as well as two other negative rays for Q, say σ Q,1 and σ Q,2 . P, say σ P,1 We parametrize each ray by its length to the zero where it emanates. Obviously, if a positive ray intersects a negative ray then it corresponds to a (horizontal) saddle connection. We will first prove the proposition under a slightly stronger assumption (C) any horizontal saddle connection other than σ0 has length > 2. We will construct a set T which is a union of horizontal rays as follows: the first + + − − element of T is σ0 . Now if a ray σ P,1 or σ P,2 intersects σ Q,1 or σ Q,2 by assumption, the associated saddle connection has length λ > 2. Hence we can choose ε > 0 + + − − and σ P,2 at time  + ε do not intersect σ Q,1 and σ Q,2 at such that positive rays σ P,1 time  + ε in their interior. These are the next elements of T . Finally we consider the negative rays from P and positive rays from Q at time ε. By the assumption, the union T of all of these rays is an embedded tree in X (see Fig. 3). We now consider a neighborhood T (δ) = { p ∈ X ; h( p, T ) < δ} of T , where h is the distance measured in the vertical direction. For δ > 0 small enough, T is a deformation retract of T (δ). We can easily construct T (δ) from 10 Euclidian rectangles whose heights are equal to δ and widths are equal to  + 2ε. We will now change the flat metric of T (δ) without changing the metric outside of this neighborhood. Given any  ∈]0, [, by varying the points where the rectangles are sewn (see the figure), we can obtain a new surface (X , ω ) in E D (2, 2)odd with a saddle connection invariant under the involution whose length is equal to  . Note +ε

+ε ε

ε

ε

ε

ε

+ε

ε

ε

ε

+ε

ε

ε

+ε

ε

+ε

ε

ε

+ε

2+ε

Fig. 3 Collapsing two zeros along a saddle connection invariant under τ

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that the surfaces obtained from this construction belong to the same leaf of the kernel foliation as (X, ω). When  = 0, we get a new closed surface (X 0 , ω0 ) ∈ H(4) sharing the same absolute periods as (X, ω). Moreover there exists an involution τ0 on X 0 such that τ0∗ ω0 = −ω0 . Hence (X 0 , ω0 ) ∈ E D (4).  us now give the proof of the lemma without the additional condition (C). Using  1 Let 0 t 0 e , t > 0, we can assume that any non-horizontal saddle connection has length > 4. Set 0 = min{|σ |, σ is a simple horizontal closed geodesic passing through P or Q}. Choose any δ ≤ 1/2 min{, 0 }, and consider B(P, δ) := {x ∈ X, d(x, P) < δ} and B(Q, δ) := {x ∈ X, d(x, Q) < δ}, where d is the distance defined by flat metric. By assumption, B(P, δ) and B(Q, δ) are two embedded topological disks in X which are disjoint. Therefore, the surface (X, ω) + (−δ, 0), which is obtained by moving P by δ/2 to the right, and Q by δ/2 to the left, is well defined. In the new surface, any horizontal saddle connection from P to Q has length reduced by δ, but the lengths of all horizontal geodesic loops are unchanged since they are absolute periods of ω. Note also that if σ is another horizontal saddle connection joining P to Q, then |σ | − |σ0 | is also unchanged. It follows that after finitely many steps, we can find a surface in the horizontal leaf of (X, ω) such that |σ0 | < 1/20 and |σ0 | < 1/2|σ | for any other horizontal saddle connection σ from P to Q, so that (C) holds. We can now apply the above arguments to conclude. We now turn to the case when (X, ω) ∈ E D (1, 1, 2). The construction is similar, we keep the same convention: σ0 is a positive ray for R1 and a negative ray for Q. Note that τ (σ0 ) is a horizontal saddle connection from Q to R2 , it is a positive ray for Q ± , i = 1, 2, the two other positive/negative and negative ray for R2 . We denote by σ Q,i + − rays from Q, and σ R1 (resp. σ R2 ) the other positive (resp. negative) ray from R1 (resp. from R2 ). We again prove the proposition with a slightly stronger assumption that for any other horizontal saddle connection σ , one has |σ | > 4|σ0 |. We will construct a set T which is a union of positive/negative rays parametrized by the length to its origin. The first elements of T are σ0 and τ (σ0 ). We then add to T • the rays σ R+1 and σ R−2 of length 2 + ε • the negative rays from R1 and positive rays from R2 of length ε. • the positive and negative rays from Q other than σ0 and τ (σ0 ) of length  + ε. with ε > 0 small. Now if the ray σ R+1 intersects any negative horizontal ray then, by assumption, the associated saddle connection has length > 4. Hence we can choose ε > 0 such that positive ray σ R+1 at time 2 + ε does not intersect any negative ray from Q at time  + ε, nor any negative ray from R1 or R2 at time 2 + ε. Similar arguments apply for other positive rays. It follows that T is a tree. We now consider a neighborhood T (δ) = { p ∈ X ; h( p, T ) < δ} of T as before.   The rest of the proof follows the same lines as the case κ = (2, 2)odd .

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Connected components of Prym eigenform loci in genus three Fig. 4 On the left: (X, ω) = S(2,2) (1, 1, 1), on the right: (X, ω) + (0, ε). In (X, ω), σ0 has no twin, but in (X, ω) + (0, ε) it has one

1

1 1

2

σ0 1

2

σ0

σ0

5.4 Twins and non connectedness of Prym eigenform loci when D = 9 In this section we give another elementary proof of the non connectedness of the loci E 9 (2, 2)odd (Theorem 4.1). Another proof of Theorem 4.1 , case E 9 (2, 2) Set X 0 := S(2,2) (1, 1, −1) and X 1 := S(2,2) (1, 1, 1) (see Lemma 2.4). For i = 0, 1, let Ci be the connected component of E 9 (2, 2)odd containing (X i , ωi ). We claim that on any surface in C0 , any saddle connection which connects the two zeros of ω0 has exactly two twins. Since this property is not satisfied by (X 1 , ω1 ) (the longest horizontal saddle connection on (X 1 , ω1 ) connects the zeros of ω1 and has no other twins) this will prove the theorem for E 9 (2, 2)odd . By construction the surface (X 0 , ω0 ) has three distinct Prym involutions, each of which preserves exactly one of the tori in the decomposition shown in Fig. 2. By Lemma 3.4 there exists a ramified covering p : X 0 → N0 of degree three (ramified at the zeros of ω0 ) where N0 is a torus. Hence any saddle connection on X 0 which connects the zeros of ω0 has two other twins. For any surface in the kernel foliation leaf or in the GL+ (2, R)-orbit of (X 0 , ω0 ), this property is clearly preserved (since surfaces still have three Prym involutions satisfying the above property). Thus (X 1 , ω1 ) does   not belong to the component of (X 0 , ω0 ). Remark 5.6 On the surface S(2,2) (1, 1, 1), the longest horizontal saddle connection (the one that is contained in the boundary components of the bottom cylinder) satisfies Convention 1, and has no twins. It is not admissible since there are two other saddle connections in the same direction with smaller length. If we move this surface slightly in the kernel foliation leaf to break this parallelism, we will find a twin of this saddle connection (see Fig. 4). This example shows that having a saddle connection satisfying Convention 1 which has no twins does not imply the existence of an admissible one. On the other hand, we know that S(2,2) (1, 1, 1) belongs to E 9 (2, 2)odd and E 9 (4) = ∅, so there exists no admissible saddle connection on S(2,2) (1, 1, 1).

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5.5 Twins and triple tori The next lemma provides a useful criterion to have triple tori from twin saddle connections (see Sect. 1.4 for the definition of triple tori). Lemma 5.7 Let (X, ω) be a translation surface and let σ0 be a saddle connection on X satisfying Convention 1. We assume that σ1 is a twin of σ0 that is not invariant under τ . (1) If (X, ω) ∈ Prym(2, 2)odd and σ1 ∪ τ (σ1 ) is separating then the triple of saddle connections σ0 , σ1 , τ (σ1 ) decomposes (X, ω) into a triple of flat tori. (2) If (X, ω) ∈ Prym(1, 1, 2) and σ0 ∪ σ1 is separating then the pairs (σ0 , σ1 ) and (τ (σ0 ), τ (σ1 )) decomposes (X, ω) into a triple of flat tori. Proof of Lemma 5.7 As usual we assume first that (X, ω) ∈ Prym(2, 2)odd . Let σ2 = τ (σ1 ). We first begin by observing that (X, ω) is the connected sum of a flat torus (X 0 , ω0 ) with a surface (X , ω ) ∈ H(1, 1), along σ1 ∪ σ2 . Indeed the saddle connections σ1 and σ2 determine a pair of angles (2π, 4π ) at P and Q. Since τ permutes P and Q, and preserves the orientation of X , it turns out that the angles 2π at P and the angle 2π at Q belong to the same side of σ1 ∪ σ2 . As subsurfaces of X , X 0 and X have a boundary that consists of the saddle connections σ1 and σ2 . We can glue σ1 and σ2 together to obtain two closed surfaces that we continue to denote by X 0 and X . We now have on X 0 (respectively, on X ) a marked geodesic segment σ (respectively, a saddle connection σ ) that corresponds to the identification of σ1 and σ2 . Note also that σ0 is contained in X . The involution τ induces two involutions: τ0 on X 0 and τ on X . The involution τ0 is uniquely determined by the properties τ0 (ω0 ) = −ω0 and τ0 permutes the endpoints of σ (namely, P and Q). Hence τ0 is the elliptic involution and has in particular 4 fixed points: the midpoint of σ and three fixed points of τ . Since τ has 4 fixed points, τ has exactly 2 fixed points: the midpoint of σ0 (σ0 is invariant under τ ) and the midpoint of σ . Let ι be the hyperelliptic involution of X . Since ι has 6 fixed points, we derive τ = ι. We claim that ι(σ0 ) = −σ . Indeed, ι(σ0 ) is a saddle connection such that ω (ι(σ0 )) = −ω (σ0 ). Hence ι(σ0 ) = −σ0 or ι(σ0 ) = −σ . If ι(σ0 ) = −σ0 then τ ◦ι(σ0 ) = σ0 , hence τ ◦ι is the identity map in the neighborhoods of P and Q. Therefore τ ◦ ι = id X : this is a contradiction since we know that τ = ι. Now the closed curve σ0 ∗ (−σ ) is preserved by ι, hence it is separating. Cutting X along this closed curve we obtain two flat tori (X 1 , ω1 ) and (X 2 , ω2 ). It is not difficult to see that X 1 and X 2 are exchanged by τ . By construction, X is the connected sum of X 0 , X 1 , and X 2 which are glued together along the slits corresponding to σ0 , σ1 , σ2 . The proof for the case (X, ω) ∈ Prym(1, 1, 2) is similar, it follows the same lines as the above discussion.  

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6 Collapsing surfaces to E D (4) The goal of this section is to establish the following theorem, which is a key step in the proof of Theorem A. Theorem 6.1 Let κ ∈ {(2, 2)odd , (1, 1, 2)}. Let C be a connected component of E D (κ). If for every surface (X, ω) ∈ C there is no admissible saddle connection on (X, ω) then D ∈ {9, 16}. More precisely, under this assumption, C contains one of the following surfaces (see Lemma 2.4): Sκ (1, 1, −1), Sκ (1, 1, 1) ∈ E 9 (κ), or Sκ (1, 2, 0), S(2,2) (2, 1, 0) ∈ E 16 (κ). Recall that as an immediate consequence we draw an upper bound on the number of connected components of E D (2, 2)odd and E D (1, 1, 2) (see Sect. 7).

6.1 Strategy of the proof of Theorem 6.1 Let (X, ω) be a Prym eigenform. We will first prove the theorem under the assumption that (X, ω) admits a three tori decomposition: X 0 is preserved and X 1 , X 2 are exchanged by the Prym involution τ (given by Lemma 5.7). The strategy is to find suitable directions on the tori so that either the conclusion of Theorem 6.1 holds or we get some restrictions on the tori. Applying this idea many times, we eventually obtain to desired result. To prove the theorem in full generality, we show that if there is no three tori decomposition then D is a square. In this situation, we again apply the same idea to find suitable directions in which the surface decomposes nicely. In the sequel let σ0 be a saddle connection satisfying Convention 1. In view of Corollary 5.4 we can assume that σ0 has a twin or a double twin, say σ1 , otherwise the theorem is proved. Depending the strata, we will distinguish three cases as follows: • κ = (2, 2)odd : Case A τ (σ1 ) = −σ1 and σ1 ∪ τ (σ1 ) is separating. Case B τ (σ1 ) = −σ1 and σ1 ∪ τ (σ1 ) is non-separating. Case C τ (σ1 ) = −σ1 . • κ = (1, 1, 2): Case A σ1 is a twin and σ0 ∪ σ1 is separating. Case B σ1 is a twin and σ0 ∪ σ1 is non-separating. Case C σ1 is a double twin (τ (σ1 ) = −σ1 ). We will first prove Theorem 6.1 under the assumption of Case A. This case is simpler since Lemma 5.7 applies: (X, ω) admits a three tori decomposition. In the other two cases, by Lemma 6.2 D is a square. We then prove that Case B and Case C can be reduced to Case A: this corresponds to Sects. 6.3 and 6.4, respectively.

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6.2 Proof of Theorem 6.1 under the assumption of Case A From now on we will assume that for any (X, ω) ∈ C ⊂ E D (κ), there exists no admissible saddle connection. Let σ2 be the image of σ1 by the Prym involution τ . Thanks to Lemma 5.7 (X, ω) admits a three tori decomposition: X 0 is preserved and X 1 , X 2 are exchanged by the Prym involution τ . Claim 1 There exists (X, ω) ∈ C such that the horizontal direction is periodic on the tori (X 0 , X 1 , X 2 ). Proof of Claim 1 By moving in the kernel foliation leaf and using GL+ (2, R) action, we can assume that the slits σi are parallel to a simple closed geodesic in (X 0 , ω0 ) which is horizontal. Since (X, ω) is completely periodic in the sense of Calta (see [8]), the claim follows.   In the sequel we assume that (X, ω) is decomposed into three horizontal cylinders, say C0 , C1 , C2 , along the saddle connections σ0 , σ1 , σ2 , where C0 is fixed, and C1 , C2 are exchanged by the Prym involution. We let s = |σi |. We denote by i , h i the width and the height of the cylinder Ci . Obviously 1 = 2 and h 1 = h 2 (see Fig. 1 for the notations). 6.2.1 Case κ = (2, 2)odd Claim 2 One of the following two equalities holds: h 0 = h 1 or h 0 = 2h 1 . Proof of Claim 2 Let δ be a saddle connection in C0 joining P to Q which crosses the  curve of C0 only once. Note that δ is anti-invariant under τ . Using U =  core { 01 1t , t ∈ R}, we can assume that δ is vertical. By assumption, δ is not admissible. Changing the length of the slits (the length of σi ) if necessary and using the argument in Lemma 5.2, we can assume that δ has a twin δ . In particular δ intersects (at least) one of the core curve of C1 or C2 . Therefore we have |δ | = mh 1 + nh 0 with m ∈ Z, m ≥ 1. The condition |δ | = |δ| implies n = 0. Thus h 0 = mh 1 . Assume that h 0 > 2h 1 . Let η1 be a geodesic segment in C1 joining P to the midpoint of σ0 . Set η2 = τ (η1 ) and η = η1 ∪ η2 . Observe that η is a saddle connection invariant under τ . Again, by using the subgroup U we assume that η is vertical. Hence |η| = 2h 1 < h 0 . Clearly any other vertical (upward) geodesic ray starting from P must intersect C0 . Thus, if there exists another vertical saddle connection η joining P to Q, we must have |η | ≥ h 0 > 2h 1 = |η|. Hence η is admissible and the claim is proved.   Claim 3 If h 0 = h 1 then, either: (1) 0 = 1 and (X, ω) is contained in the same component as Sκ (1, 1, −1) ∈ E 9 (κ), or (2) 0 = 21 and (X, ω) is contained in the same component as Sκ (1, 2, 0) ∈ E 16 (κ). Proof of Claim 3 Let h := h 0 = h 1 . Up to the action of the horocycle flow we assume that δ0 is a saddle connection in C0 joining P to Q such that ω(δ0 ) = (0, h). Since

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Connected components of Prym eigenform loci in genus three Fig. 5 Claim 3: h 1 = h 0 : the surfaces Sκ (1, 1, −1) and diag(1, 21 ) · Sκ (1, 2, 0)

Q σ2 P C2

γ

Q σ2 P C2

σ0

γ σ0

C1

C1 σ1

h0 = h1

σ1

γ

C0

C0

σ2 0

=

1

γ

γ

h0 = h1

σ2

0

=2

1

δ0 is not admissible, there exists a vertical saddle connection δ1 joining P and Q such that ω(δ1 ) = (0, λ), where 0 < λ ≤ h. Actually ω(δ1 ) = (0, h). Since δ1 cannot be contained in C0 , δ1 is contained in C1 . Thus δ2 = τ (δ1 ) is contained in C2 . Let γ be the saddle connection contained in C 1 ∪ C 2 joining P to Q and passing through the midpoint of σ0 as shown in Fig. 5. By assumption there exists another saddle connection γ joining P to Q parallel to γ such that |γ | ≤ |γ |. We claim that either γ or τ (γ ) starts in C0 . This is clear if γ is not invariant under τ . If γ is invariant under τ and starts from C2 then it must end in C1 = τ (C2 ). The topology of the surfaces (C1 , C2 and C0 are cylinders glued together in a necklace, in this order) implies that γ must cross C0 at least once. In particular γ crosses twice the core curves of cylinders. Hence the vertical coordinate ω(γ ) is greater than 2h. Therefore |γ | > |γ | that is a contradiction. We can now suppose that γ starts in C0 . Observe that ω(γ ) = (s − 21 , 2h) (recall that s = |σi |). If γ is not contained in C 0 = X 0 (Fig. 5, left), γ must end up in C1 . Since |γ | ≤ |γ | elementary calculation shows that ω(γ ) = (s − 0 − 1 , 2h). Now γ and γ are parallel, thus 0 = 1 and (X, ω) = Sκ (1, 1, −1). If γ is contained in C 0 (Fig. 5, right), γ must intersect twice the core curve of C0 . Thus ω(γ ) = (s − 0 , 2h), from which we deduce 0 = 21 and (X, ω) =   diag(1, 21 ) · Sκ (1, 2, 0). The claim is proved. Claim 4 If h 0 = 2h 1 then, either: (1) 0 = 1 and (X, ω) is contained in the same component as Sκ (2, 1, 0) ∈ E 16 (κ), or (2) 0 = 21 and (X, ω) is contained in the same component as Sκ (1, 1, 1) ∈ E 9 (κ). Proof of Claim 4 Let h := h 1 = h 0 /2. Let δ1 be a geodesic segment, contained in C1 , joining the midpoint of σ0 to P. Using the horocycle flow we assume δ1 to be vertical. Set δ2 = τ (δ1 ) and δ = δ1 ∪ δ2 . By construction δ is a saddle connection which is invariant under τ . By assumption, there exists another vertical saddle connection δ joining P to Q such that |δ | ≤ |δ|. It is easy to see that any other vertical geodesic ray emanating from P intersects the core curve of C0 . Since ω(δ) = (0, 2h) and h 0 = 2h, δ is contained in C0 .

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Q

P

C2 C1

P

C2

γ

γ

C1

γ C0

h0 = 2h1

C0 γ 0

=

h0 = 2h1

1

γ

0

=2

1

Fig. 6 Claim 4: h 0 = 2h 1 : the surface on the left belongs to component of Sκ (2, 1, 0), and on the right belongs to the component of Sκ (1, 1, 1)

Now let γ be the saddle connection in C 1 ∪ C 2 passing through the midpoint of σ0 , joining P to Q such that ω(γ ) = (21 , 2h) (see Fig. 6). By assumption, there exists a saddle connection γ in the same direction as γ such that |γ | ≤ |γ |. As in Claim 3 either γ or τ (γ ) starts in C0 . The proof follows the same lines. Up to permutation, γ starts in C0 . Since |γ | ≤ |γ | and γ is parallel to γ , γ is actually contained in C0 . In particular ω(γ ) = (k0 , 2h) for some k ∈ Z, k ≥ 1 and ω(γ ) = ω(γ ). We draw 21 = k0 . Now we claim that the inequality 0 ≥ 1 holds. Indeed there exists a horizontal saddle connection σ0 in X 0 such that σ1 ∪ σ0 and σ2 ∪ σ0 are the two boundaries components of the cylinder C0 . Similarly, there exists a pair of horizontal saddle connections σ1 , σ2 where σi is contained in X i such that σi ∪ σ0 is a boundary component of Ci . By construction we have τ (σ0 ) = −σ0 , τ (σ1 ) = −σ2 , and 0 = |σ1 | + |σ0 |

and

1 = |σ0 | + |σ1 | = |σ0 | + |σ2 |.

If 0 < 1 then |σ0 | < |σ1 | = |σ2 |. Hence σ0 is admissible, contradicting our assumption. In conclusion 21 = k0 ≥ 1 implies 0 = 1 , or 0 = 21 . The corresponding surfaces are represented in Fig. 6. It is not hard to check that those two surfaces belong to the same connected component that Sκ (2, 1, 0) and Sκ (1, 1, 1), respectively. The claim is proved.   6.2.2 Case κ = (1, 1, 2) Claim 5 One of the following two equalities holds: h 0 = h 1 , or h 0 = 2h 1 . Proof of Claim 5 The proof of this claim follows the same lines as the proof of Claim 2.   Claim 6 If h 0 = h 1 then, either (1) 0 = 1 , and (X, ω) is contained in the same component as Sκ (1, 1, −1) ∈ E 9 (κ), or

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Connected components of Prym eigenform loci in genus three R2 C2

R2

τ (γ)

C2

τ (γ)

Q C0

γ

Q C0

δ

γ

δ

Q C1

γ

Q γ

C1

δ1 R1

0

=

1

δ1 R1

0

=2

1

Fig. 7 Claim 6: h 0 = h 1 : the surfaces Sκ (1, 1, −1) and diag(1, 21 ) · Sκ (1, 2, 0)

(2) 0 = 21 , and (X, ω) is contained in the same component as Sκ (1, 2, 0) ∈ E 16 (κ). Proof of Claim 6 Set h := h 0 = h 1 . We first consider a saddle connection δ contained in C0 , joining R1 to Q and intersecting the core curve of C0 only once. As usual we assume δ to be vertical (hence ω(δ) = (0, h)). By assumption δ has a twin or a doubletwin δ1 (necessarily δ1 starts in C1 ). Clearly δ1 is a twin: otherwise it must end in C2 , hence it must cross the core curve of C0 at least once. In particular its length satisfies |δ1 | ≥ 3h > 2|δ| that is a contradiction. Let γ be the saddle connection contained in C1 and joining R1 to Q, as shown in Fig. 7. By assumption ω(γ ) = (1 , h). Now there exists another saddle connection γ starting from R1 in the same direction as γ . Observe that γ must start in C0 . Using Lemma 5.2, we can assume that γ is either a twin or a double-twin of γ . If γ is a twin of γ then ω(γ ) = (1 , h). Hence 0 = 1 : (X, ω) = Sκ (1, 1, −1). If γ is a double-twin of γ then ω(γ ) = (21 , 2h). Hence γ crosses twice the core curves of C0 implying that 0 = 21 : (X, ω) = diag(1, 21 ) · Sκ (1, 2, 0). This proves the claim.   Claim 7 If h 0 = 2h 1 then 0 = 21 . In addition (X, ω) belongs to the same connected component as Sκ (1, 1, 1) ∈ E 9 (κ). Proof Set h = h 1 . Let δ1 be a saddle connection contained in C1 joining R1 to Q and intersecting the core curve of C1 only once. We can suppose that δ1 is vertical. By assumption, there exists another vertical saddle connection δ starting from R1 . Observe that δ must intersect the core curve of C0 , thus we have |δ| ≥ h 0 = 2|δ1 |. By assumption, δ must be a double-twin of δ1 , which means that δ joins R1 to R2 and is contained in C0 . Now, let γ be the saddle connection in C1 joining R1 to Q as shown in Fig. 8. By assumption, there exists a saddle connection γ starting from R1 and parallel to γ . The same argument as above shows that γ is a double-twin of γ and is contained

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τ (γ)

C2

R2

Q

C0

R2 γ

R1 γ

C1

Q

R1

in C0 . It follows that 0 = 21 . Thus (X, ω) belongs to the component of Sκ (1, 1, 1). The claim is proved.   6.3 Reduction from Case B to Case A Let (X, ω) ∈ E D (κ) and let σ0 be a saddle connection in X satisfying Convention 1. We suppose that σ0 has a twin σ1 . Moreover, if κ = (2, 2)odd we assume that σ2 := −τ (σ1 ) = σ1 and σ1 ∪ σ2 is non-separating, and if κ = (1, 1, 2) we assume that σ0 , σ1 is non-separating. Our aim is to show that there exists in the component of (X, ω) a surface having a family of homologous saddle connections satisfying Case A (this is Lemma 6.3). We first show Lemma 6.2 Let (X, ω) ∈ E D (κ), where κ ∈ {(2, 2)odd , (1, 1, 2)}. If there exists c ∈ H1 (X, Z)− satisfying c = 0 and ω(c) = 0 then D is a square. In particular, up to rescaling by GL+ (2, R), all the absolute periods of ω belong to Q + ıQ. Proof of Lemma 6.2 We can assume that c is primitive in H1 (X, Z), that is for any 1 / H1 (X, Z). Pick a symplectic basis (α1 , β1 , α2 , β2 ) of H1 (X, Z)− n ∈ N, n > 1, c ∈ n with β2 = c. Set μi = αi , βi , where ,  is the intersection form of H1 (X, Z), and ω(α1 ) = x1 + ı y1 , ω(β1 ) = z 1 + ıt1 , ω(α2 ) = x2 + ı y2 . Since  Area(X, ω) = μ1 det

x1 y1

z1 t1



 + μ2 det

x2 y2

0 0



 = μ1 det

x1 y1

z1 t1

 >0

it follows that (x1 + ı y1 , z 1 + ıt1 ) is a basis of R2 . Using GL+ (2, R) we can assume that (x1 , y1 ) = (1, 0) and (z 1 , t1 ) = (0, 1). By [7, Proposition 4.2] there exists a unique generator T of O D which is given in the basis (α1 , β1 , α2 , β2 ) by a matrix of the form a b   eId 2 c d T = μ1  d −b  0 a μ2 −c

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with (a, b, c, d, e) ∈ Z5 such that T ∗ ω = λω and λ > 0. Observe that the discriminant μ1 (bc − ad). Since Re(ω) = (1, 0, x2 , 0) and Im(ω) = D satisfies D = e2 − 4 μ 2 (0, 1, y2 , 0) in the above coordinates, direct computations show that b = d = 0.   Hence D = e2 and (x2 , y2 ) ∈ Q2 . The lemma is proved. Lemma 6.3 Assume that κ = (2, 2)odd . Then there exists in the component of (X, ω) a surface having a triple of homologous saddle connections γ0 , γ1 , γ2 , where γ0 is invariant, γ1 and γ2 are exchanged by the involution. Proof Let c0 , c1 , and c2 denote the simple closed curves σ1 ∗ (−σ2 ), σ0 ∗ (−σ1 ) and σ0 ∗ (−σ2 ) respectively. Note that we have c0 = c2 − c1 and τ (c1 ) = −c2 . By assumption 0 = c0 ∈ H1 (X, Z). If 0 = c1 ∈ H1 (X, Z) then c2 = −τ (c1 ) = 0 ∈ H1 (X, Z), which implies that c0 = 0 ∈ H1 (X, Z). Thus we can conclude that all of the curves c0 , c1 , c2 are non-separating. Cutting X along σ0 , σ1 , σ2 , we obtain a connected surface whose boundary has three components corresponding to c0 , c1 , c2 . Gluing the pair of geodesic segments in each boundary component together, we get a closed translation surface (X , ω ) with three marked geodesic segments. Since the angle between two consecutive twin saddle connections is 2π , we derive that ω has no zeros, thus (X , ω ) must be a torus. We denote the geodesic segments in X corresponding to c0 , c1 , c2 by c0 , c1 , c2 respectively. The involution τ of X induces an involution τ on X , which leaves c0 invariant and exchanges c1 and c2 . Let Pi and Q i , i = 0, 1, 2, denote the endpoints of ci , where Pi (respectively, Q i ) corresponds to P (respectively, to Q). Observe that as (X, ω) moves in the leaf of the kernel foliation, the surface (X , ω ) is the same, only the segments ci vary. Therefore we can assume that (X , ω ) is the standard torus C/(Z ⊕ ıZ), and the length of c0 is small. Let δ1 (respectively, η1 ) denote the geodesic segment of minimal length from P0 to P1 (respectively, from Q 0 to Q 1 ). Note that as (X, ω) moves in the kernel foliation leaf, ω (δ1 ) and ω (η1 ) are invariant. Therefore, we can assume that δ1 and c0 are not parallel. Since c0 and c1 are parallel and have the same length, we see that c0 ∪ δ1 ∪ c1 ∪ η1 is the boundary of an embedded parallelogram in X . It follows in particular that δ1 and η1 are parallel and have the same length. Let δ2 = τ (η1 ) and η2 = τ (δ1 ). We have δ = δ1 ∪ δ2 is a geodesic segment joining P1 to P2 , and η = η1 ∪ η2 is a geodesic segment joining Q 1 to Q 2 . Since ω(c0 ) = 0, by Lemma 6.2, for any c ∈ H1 (X, Z), we have ω(c) ∈ Q + ıQ. Therefore ω (δ) = ω (η) ∈ Q + ıQ. Since X is the standard torus, there exists a pair of parallel simple closed geodesics α + , α − of X such that δ ⊂ α + , and η ⊂ α − (see Fig. 9). When |ci | is small enough and non-parallel to α ± , the geodesics α + and α − cut X into two cylinders, one of which contains all the segments c0 , c1 , c2 . Recall that (X, ω) is obtained from (X , ω ) by slitting along c0 , c1 , c2 , and regluing the geodesic segments in the boundary. By construction, we see that (X, ω) admits a decomposition into four cylinders in the direction of α ± as shown in Fig. 9. Let C0 denote the largest cylinder in this decomposition, then it is easy to see that there exist in C0 three homologous saddle connections γ0 , γ1 , γ2 such that τ (γ0 ) = −γ0 ,   τ (γ1 ) = −γ2 , and γ1 ∪ γ2 is a separating curve as desired.

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σ1

α+ α−

P1

P0

P2

Q1

Q0

Q2

γ1

γ2

σ0

σ0

σ1

γ0

γ1

σ2

σ2

(X, ω)

(X , ω ) Fig. 9 Cylinders decomposition in direction of α ±

σ1

Fig. 10 Configurations of the saddle connections σ1 , σ1 , σ2 , σ2 on the torus (X , ω )

σ1

σ2

σ2 R2

Q1

R2

Q

Q2

σ2

σ2

R1

R1 σ1

σ1

Lemma 6.4 Assume that κ = (1, 1, 2) and σ0 has a twin σ1 such that the curve σ0 ∗ (−σ1 ) is non-separating. Then there exists in the component of (X, ω) a surface having two pairs of homologous saddle connections (σ1 , σ1 ) and (σ2 , σ2 ), where σi and σi join the simple zero Ri to the double zero Q, and {σ2 , σ2 } = τ ({σ1 , σ1 }). Proof We will use similar ideas to the proof of Lemma 6.3. Let c1 = σ0 ∗ (−σ1 ) and c2 = τ (c1 ). By the cutting-gluing construction along c1 and c2 (using the assumption that c1 is non-separating), we get a flat torus (X , ω ) with three marked geodesic segments c1 , c2 , c such that ω (c1 ) = ω (c2 ) = 1/2ω (c ) (see Fig. 10). For i = 1, 2, we denote the endpoints of ci by Ri and Q i so that Ri corresponds to Ri and Q i corresponds to Q. We denote the endpoints of c by R1 and R2 such that Ri corresponds to Ri . The midpoint of c corresponds to Q, we denote it by Q . We denote the subsegment of c between Q and Ri by ci . The Prym involution of X gives rise to an involution τ of X which satisfies τ (c1 ) = −c2 , τ (c1 ) = −c2 . As (X, ω) moves in the leaf of the kernel foliation, the surface (X , ω ) remains the same, only the segments c1 , c2 , c vary. Therefore we can assume that c1 , c2 , c are contained in three distinct parallel simple closed geodesics of X . Changing the direction of c slightly, we see that there exist geodesic segments σi from Q i to Ri , and σi from Q to Ri , i = 1, 2, (see Fig. 10) such that • τ (σ1 ) = −σ2 and τ (σ1 ) = −σ2 , • σi ∗ ci and σi ∗ ci are homologous.

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Reconstruct (X, ω) from (X , ω ) we see that σi and σi are homologous saddle con  nections, and the pairs (σ1 , σ1 ) and (σ2 , σ2 ) have the desired properties. 6.4 Reduction from Case C to Case A Let (X, ω) be a Prym eigenform in Prym(2, 2)odd  Prym(1, 1, 2), and let σ0 be a saddle connection on X satisfying Convention 1. If (X, ω) ∈ Prym(2, 2)odd we suppose that σ0 has a twin σ1 which is also invariant under the Prym involution τ , and if (X, ω) ∈ Prym(1, 1, 2) we suppose σ0 has a double twin σ1 (which is invariant under τ ). Our aim is to show that there exists in the component of (X, ω) a surface having a family of saddle connections satisfying Case A for both κ = (2, 2)odd and κ = (1, 1, 2). We first show Lemma 6.5 Define • c = σ0 ∗ (−σ1 ), if (X, ω) ∈ Prym(2, 2)odd , • c = σ0 ∗ τ (s0 ) ∗ (−σ1 ), if (X, ω) ∈ Prym(1, 1, 2). Then in both cases, we have c = 0 ∈ H1 (X, Z)− , and ω(c) = 0. Proof From the definition of c, we have τ (c) = −c, hence c ∈ H1 (X, Z)− . It is also clear that ω(c) = 0. All we need to show is that c = 0 ∈ H1 (X, Z). We first consider the case (X, ω) ∈ Prym(2, 2)odd . Remark that the pair of angles at P and Q determined σ0 and σ1 is (2π, 4π ). Since τ (σ0 ) = −σ0 and τ (σ1 ) = −σ1 we see that the angle 2π at P and the angle 4π at Q belong to the same side of c, and vice versa. Cutting X along c we get a surface whose boundary has two components, each of which is a union of two geodesic segments corresponding to σ0 and σ1 . Since σ0 and σ1 are twins, the two segments in each component has the same length, therefore we can glue them together to get a closed (possibly disconnected) translation surface (X , ω ) with two marked geodesic segments η1 , η2 . If the new surface is disconnected, then each component is a translation surface with only one singularity of angle 4π . Since such a surface does not exist, we conclude that X is connected, and hence c = 0 in H1 (X, Z). For the case (X, ω) ∈ Prym(1, 1, 2), by a similar construction, that is cutting along c, then closing the boundary components of the new surface (by gluing the path corresponding to σ0 ∪ τ (σ0 ) and the segment corresponding to σ1 ), we also get a translation surface (X , ω ) having two singularities with cone angle 4π . The same argument as above shows that this surface belongs to H(1, 1), therefore c = 0 ∈   H1 (X, Z). Lemma 6.6 Let (X, ω) ∈ Prym(2, 2)odd be a Prym eigenform having a twin σ1 of σ0 that is invariant under τ . Then one can find in the connected component of (X, ω) another surface having a triple of homologous saddle connections. Proof of Lemma 6.6 Cutting X along c = σ0 ∪σ1 and gluing the two segments of each boundary component together, we get a closed translation surface (X , ω ) ∈ H(1, 1) (see the proof of Lemma 6.5). By construction there exist on X a pair of disjoint geodesic segments η1 , η2 such that ω (η1 ) = ω (η2 ) = ω(σ0 ) and ηi joins a zero of ω

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to a regular point. Let (Pi , Q i )i=1,2 denote the endpoints of ηi , where Pi (respectively, Q i ) corresponds to P (respectively, to Q). The numbering is chosen so that P1 and Q 2 are the zeros of ω . The involution τ of X descends to an involution of X exchanging η1 and η2 . We denote this involution by τ . Remark that τ has two fixed points in X , none of which are contained in the segments η1 , η2 . Note also that as (X, ω) moves in its leaf of the kernel foliation, (X , ω ) also moves in its leaf of the kernel foliation in H(1, 1) (only the relative periods change). Let ι be the hyperelliptic involution of X . Since ι has six fixed points but τ has two, we have ι = τ . Remark that ι◦τ is also an involution of X satisfying (ι◦τ )∗ ω = ω . The surface X = X / ι ◦ τ  is an elliptic curve. Let π is the branch covering π : X → X , which is is ramified at P1 and Q 2 . Then ω descends to a holomorphic 1-form ω on X so that ω = π ∗ ω . For i = 1, 2 let Pi , Q i , ηi denote the images of Pi , Q i , ηi in X . Note that we have ω (η1 ) = ω (η2 ) = ω(σ0 ). We consider the tuple (X , ω , P1 , P2 ) as an element in H(0, 0), that is the moduli space of flat tori with two marked points. We first observe that as (X, ω) moves in its leaf of the kernel foliation, the corresponding surfaces (X , ω , P1 , P2 ) are the same in H(0, 0) (only ω (η1 ) = ω (η2 ) change). Indeed all the coordinates of (X , ω , P1 , P2 ) are determined by the absolute periods of (X, ω). Let α1 be a simple closed geodesic of (X , ω ) which passes through P1 and does not contain P2 . Using GL(2, R), we can assume that α1 is horizontal. By moving in the kernel foliation leaf of (X, ω), we can also assume that ηi are parallel to α1 (ω (ηi ) = λω (α1 ), with 0 < λ < 1). By construction, the surface (X , ω ) admits a decomposition into cylinders in the horizontal direction. Note that X must have three horizontal cylinders, otherwise there would be horizontal saddle connection joining P1 to Q 2 , which is excluded since α1 does not contain η2 (see Fig. 11).

P2 Q 2 P1 Q 1

P1 Q 1

Q P σ1 P2 Q 2

P2 Q 2

(X , ω )

P

Q σ0

P1 Q 1

σ0

(X, ω)

(X , ω )

Q

σ0

σ0

˜ ω (X, ˜)

P

˜ ω (X, ˜)

Fig. 11 σ0 and σ1 are invariant under τ , X admits two involutions: the hyperelliptic one ι which fixes each of the cylinders, and the involution τ induced by τ which exchanges the pair simple cylinders and fixes the larger one. Observe that τ exchanges η1 = P1 Q 1 and η2 = P2 Q 2

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We can reconstruct (X, ω) from (X , ω ): one sees that (X, ω) also admits a decomposition into three horizontal cylinders (see Fig. 11). Consider the surface ( X˜ , ω) ˜ = (X, ω) + (0, −), with  > 0 small as shown in Fig. 11. We see that ( X˜ , ω) ˜ admits a decomposition into four horizontal cylinders, three of which are simple. It is easy to check that there exists a triple of twin saddle connections γ0 , γ1 , γ2 in the largest horizontal cylinder of X˜ (which is preserved by the Prym involution) which satisfy τ (γ0 ) = −γ0 , τ (γ1 ) = −γ2 , and γ1 ∪ γ2 is a separating curve. This proves the lemma.   Lemma 6.7 Let (X, ω) ∈ Prym(1, 1, 2) be a Prym eigenform having a double twin σ1 of σ0 . Then one can find in the connected component of (X, ω) a surface having two pairs of homologous saddle connections (σ1 , σ1 ) and (σ2 , σ2 ) that are exchanged by the Prym involution. Proof Let (X , ω ) ∈ H(1, 1) be the surface obtained by the “cutting-gluing” construction along c = σ0 ∗ τ (σ0 ) ∗ (−σ1 ) (see Lemma 6.5). Note that we have on X two marked and disjoint geodesic segments η1 and η2 (corresponding to c) such that the midpoint of ηi is a zero of ω , and ω (ηi ) = ω(σ1 ) = 2ω(σ0 ). We will consider ηi as a slit with two sides ηi and ηi , where ηi corresponds to σ1 , and ηi corresponds to the union σ0 ∪ τ (σ0 ). Remark that the Prym involution τ induces an involution τ on X which is not the hyperelliptic involution. It follows that X is a double cover of a torus. By construction, as (X, ω) moves in its kernel foliation leaf (X , ω ) is fixed, only the marked geodesic segments (slits) ηi are changed. Using GL+ (2, R), we can assume that X is horizontally periodic, and it has three horizontal cylinders (one may use the fact that X is a double cover of a torus to show that a periodic direction with three cylinder does exist). We can arrange so that the slits are also horizontal, and ηi are contained in the boundary of the largest cylinder (see Fig. 12). Reconstruct (X, ω) from (X , ω ), we see that (X, ω) has two pairs of homologous saddle connections (σ1 , σ1 ) and (σ2 , σ2 ), such that τ (σ1 ) = −σ2 , τ (σ1 ) = σ2 , and σ1 ∗ σ1 ∗ σ2 is homologous to the core curve of the largest horizontal cylinder in X . The lemma is then proved.   τ (σ0 ) σ1

σ2

σ1

σ1

σ2

η1 σ1

η2

σ2

(X , ω )

σ0

σ1

σ1

σ2

(X, ω)

Fig. 12 Prym(1, 1, 2) case C: σ0 has a double-twin

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(Xε ,ωε )=S(2,2) (2,1,0)+(0 )

(Xε ,ωε )

(Y,η)

Fig. 13 Connecting S(2,2) (2, 1, 0) to S(2,2) (1, 2, 0)

7 Proof of the main result Let us now give the proof of our main theorem. Proof of Proof of Theorem A For all D ∈ N, D ≡ 0, 1, 4 mod 8, the loci E D (κ) are non empty: this is Corollary 2.5. We now consider the cases D = 0, 1, 4 mod 8, D ≥ 9, and D ∈ / {9, 16}. By Theorem 6.1 and Corollary 5.4, any component of E D (κ) contains a surface with an admissible saddle connection that collapses to a point in E D (4). Recall that E D (4) is a finite collection of Teichmüller discs. By Proposition 2.6 for any connected component C of E D (4), there exists at most one component of E D (κ) adjacent to C i.e. whose closure contains C. Therefore, the number of connected components of E D (κ) is bounded from above by the number of components of E D (4). In particular, when D ≡ 0, 4 mod 8, since E D (4) is connected (see Theorem 2.2), so is E D (κ). For D ≡ 1 mod 8, by Theorem 2.2 we know that E D (4) has two components, so E D (κ) has at most two components. On the other hand, Theorem 4.1 tells us that E D (κ) cannot be connected. Thus we can conclude that E D (κ) has exactly two connected components. For D ≡ 5 mod 8, if E D (κ) is non-empty then again Theorem 6.1, Corollary 5.4 and Proposition 5.5 implies that E D (4) is also non-empty, which contradicts Theorem 2.2. We now consider the cases D ∈ {9, 16}. Since E D (4) = ∅, by Proposition 5.5, there exists no admissible saddle connection on any surface in E 9 (κ)  E 16 (κ). By Theorem 6.1, we see that any surface in E 9 (κ)  E 16 (κ) belongs to the same component as one of the surfaces Sκ (1, 1, −1), Sκ (1, 1, 1), Sκ (1, 2, 0) or S(2,2) (2, 1, 0) (see Lemma 2.4 for the definition). It follows immediately that E 16 (2, 2) and E 9 (κ) have at most two connected components and E 16 (1, 1, 2) is connected. The fact that E 9 (κ) is not connected is proved in Theorem 4.1. Hence E 9 (κ) has exactly two connected components. It remains to prove that E 16 (2, 2) is connected. It is sufficient to show that S(2,2) (1, 2, 0), S(2,2) (2, 1, 0) ∈ E 16 (2, 2)odd belong to the same component. We consider (X ε , ωε ) = S(2,2) (2, 1, 0) + (0, ε), with ε > 0 small enough (see Fig. 13).

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Observe that (X ε , ωε ) admits a decomposition into four horizontal cylinders. Moving horizontally in the kernel foliation leaf of (X ε , ωε ) we get a surface (Y, η) = (X ε , ωε ) + v, with v ∈ R × {0}, which admits a decomposition into three vertical cylinders. It is not difficult to see that (Y, η) can be connected to S(2,2) (1, 2, 0) by using the action of GL+ (2, R) and moving in the kernel foliation leaves. The proof of Theorem A is now complete.   As a direct corollary we prove Theorem B i.e. the existence in any component of E D (κ) of surfaces which admit three-tori decompositions. Proof of Theorem B Let (w, h, e) ∈ Z be as in Lemma 2.4 where D = e2 + 8wh. We consider the corresponding surfaces (X ± , ω± ) := Sκ (w, h, ±e). By Lemma 2.4 (X ± , ω± ) ∈ E D (κ). If D ≡ 1 mod 8 then by Theorem A, E D (κ) is connected and (X ± , ω± ) admits a three-tori decomposition. If D ≡ 1 mod 8 then by Theorem A, E D (κ) has two connected components and from the proof of Theorem 4.1, (X + , ω+ ) and (X − , ω− ) do not belong to the same connect component. This ends the proof of Theorem B.   Acknowledgements We would like to thank Alex Eskin, Martin Möller, and Barak Weiss for useful discussions. We would also thank Université de Bordeaux and Institut Fourier in Grenoble for the excellent working condition during the preparation of this work. Some of the research visits which made this collaboration possible were supported by the ANR Project GeoDyM. The authors are partially supported by the ANR Project GeoDyM. We thank the anonymous referee for his thorough review and highly appreciate the comments and suggestions.

Appendix: A partial compactification of E D (κ), κ ∈ {(2, 2), (1, 1, 2)} The subvarieties E D (2, 2)odd and E D (1, 1, 2) of M3 are not compact. In order to compute their topological and dynamical invariants (e.g. Euler characteristic, Siegel– Veech constants) it is important to embed these varieties into compact ones such that the complement are divisors. The aim of this section is to provide a partial compactification of E D (κ) having good properties. Degeneration of surfaces We first introduce some notations. (1) We denote by Prymtriple (0, 0, 0) the set of triple tori {(X j , ω j , P j ), j = 0, 1, 2}, We where (X j , ω j , P j ) ∈ H(0), and (X 1 , ω1 , P1 ) and (X 2 , ω2 , P2 ) are isometric.  denote by  j the lattice of C corresponding to (X j , ω j ), that is  j := { c ω j , c ∈ H1 (X j , Z)}. Given a natural number D ∈ N satisfying D ≡ 0, 1, 4 mod 8, we define E D (0, 0, 0) to be the subset of Prym(0, 0, 0) consisting of triples that satisfy the following condition: given any basis (u 0 , v0 ) of 0 , there exists a tuple (w, h, t, e) ∈ Z4 satisfying

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 P D,triple (0, 0, 0)

w > 0, h > 0, 0 ≤ t < gcd(w, h),, gcd(w, h, t, e) = 1 D = e2 + 8wh

by and a basis (u 1 , v1 ) of 1 such that u 1 is parallel to u 0 and up to a rescaling √ GL+ (2, R), we have u 0 = λ, v0 = ıλ, u 1 = w, v1 = t + ı h, where λ = e+2 D . (2) We denote by E D (2)∗ the space of triples (X, ω, W ), where (X, ω) is a Prym eigenform in E D (2) (in particular X is a Riemann surface of genus 2), and W is a Weierstrass point of X which is not the zero of ω. (3) We denote by E D (2)∗∗ the space of tuples (X, ω, W1 , W2 ), where (X, ω) is a Prym eigenform in E D (2), and W1 , W2 are two Weierstrass points, not a zero of ω. (4) Let Prym(03 ) denote the set of tuples (X, ω, P0 , P1 , P2 ), where (X, ω, P0 ) ∈ H(0), and P1 , P2 ∈ X are permuted by the (elliptic) involution fixing P0 . Given D = e2 ∈ N, with e ≥ 3, we denote by E D (03 ) the subset of Prym(03 ) consisting of elements that satisfy the following condition: there exists a ∈ Z such that  P D (03 ) : 0 < a < e/2, gcd(a, e) = 1 and up to a rescaling by GL+ (2, R), we have (X, ω)  (C/(Z ⊕ ıZ), dz), and P0 , P1 , P2 are the projections of 0, − ae , ae respectively. (5) For any D = e2 , with e ≥ 3, we denote by  E˜ D (02 ) the set of surfaces ( X˜ , ω) ˜ ∈ H(1, 1) that are double covers of surfaces (X, ω, P1 , P2 ) ∈ H(0, 0) ramified over exactly the two marked points P1 , P2 , where (X, ω, P1 , P2 ) satisfies the following condition: there exists a ∈ Z such that  P D (02 ) : 0 < a < e/2, gcd(a, e) = 1 and up to a rescaling by GL+ (2, R), we have (X, ω)  (C/(Z ⊕ ıZ), dz), and P1 , P2 are the projections of − ae , ae respectively. Partial compactification of E D (2, 2) Theorem 7.1 Let (X, ω) ∈ E D (2, 2)odd be a Prym eigenform and τ be the Prym involution of X . Let S denote the set of horizontal saddle connections joining the two zeros P and Q of ω. We assume that S is not empty. Let σ0 be a saddle connection of minimal length in S, and t0 = |σ0 |. Define (X t , ωt ), t ∈ [0, t0 ) to be the surface in the leaf of the kernel foliation through (X, ω), which is obtained by contracting σ0 by the amount t, that is (X t , ωt ) = (X, ω)−(t, 0). Let M be the limit surface as t tends to t0 . We have (I) If σ0 has no twin then M ∈ E D (4). (II) If σ0 has only one other twin σ1 then either (II.a): σ0 and σ1 are permuted by the Prym involution and M ∈ E D (2)∗ , with D ∈ {D, D/4}, or (II.b): both σ0 , σ1 are invariant under the Prym involution and M ∈  E˜ D (02 ).

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(III) If σ0 has two other twins σ1 , σ2 then we can always assume that σ0 is invariant under τ . Set η = σ1 ∗(−σ2 ), then either (III.a): η is a separating curve and M ∈ E D,triple (0, 0, 0), or (III.b): η is a non-separating curve and M ∈ E D (03 ). Note that σ0 cannot have more than two twins, and Cases (II.b) and (III.b) can only occur when D is a square. Remark 7.2 • Since τ permutes the zeros of ω, τ (σ0 ) is also an element of S which has the same length as σ0 . Thus if σ0 has no twin then σ0 must be invariant under τ . • If σ0 and σ1 are two twins saddle connections in S both invariant by τ , then the simple closed curve σ0 ∗ (−σ1 ) must be non-separating (see Lemma 6.5), and consequently D must be a square (see [9, Lem. 9.2(2)]). • Since τ induces a permutation of order two on any family of twins saddle connections, if S contains three twins saddle connections, then one of them must be invariant under τ . Proof Cases (I) follows from Proposition 5.5. Cases (II.a), and (III.a) were actually proved in [9, Theorem 9.1]. We will only give the proofs of the remaining cases. Case (II.b) σ0 and σ1 are twins and both are invariant under τ . Applying the cuttinggluing procedure described in Lemma 6.5 to η := σ0 ∪ σ1 , we get a surface (X , ω ) ∈ H(1, 1) which is a double cover of a torus (X , ω ). Along this process, P (resp. Q) gives rise to two points P1 , P2 (resp. Q 1 , Q 2 ) in X , and η gives rise to two geodesic segments η1 , η2 , where ηi joins Pi to Q i . By convention, the double cover from X to X is ramified over P1 and Q 2 . As (X, ω) moves in the leaf of the kernel foliation, (X , ω ) and P1 , P2 are fixed, only ηi vary. Thus, as t tends to t0 , ηi shrinks to a point, and (X , ω ) converges in H(1, 1) to the double cover of (X , ω ) ramified over P1 and P2 . All we need to show is that (X , ω , P1 , P2 ) satisfies the conditions in (P D (02 )). We first seek a symplectic basis of H1 (X, Z)− . Let γ be a geodesic segment of minimal length in X joining P1 to P2 . The pre-image of γ in X consists of two segments, one of them joins a zero of ω to a regular point both of which correspond to the zero P of ω. Let γ denote this segment (see Fig. 14). It follows that γ corresponds to a simple closed curve γ on X which contains P. By construction, we have γ , η = ±1 in H1 (X, Z). Note that τ (γ ) is a saddle connection joining Q to itself, and τ (γ ) is homologous to −γ . Thus γ ∈ H1 (X, Z)− as an element of the homology group. Let (α , β ) be a basis of H1 (X , Z) which is represented by a pair of simple closed curves disjoint from γ ∪η1 ∪η2 . The pre-image of α (resp. of β ) in X corresponds to two simple closed curves α1 , α2 (resp. β1 , β2 ) in X which satisfy τ (α1 ) = −α2 (resp. τ (β1 ) = −β2 ). Set α = α1 +α2 , β = β1 +β2 , then {α, β, γ , η} is a symplectic basis of H1 (X, Z)− . Since ω(η) = 0, the arguments in Lemma 6.2 show that up to a rescaling by GL+ (2, R), we can assume that ω(α) = 2, ω(β) = 2ı, and ω(γ ) ∈ Q + ıQ. Now, observe that we have ω(α) = 2ω (α ), ω(β) = 2ω (β ), and ω(γ ) = ω (γ ). Thus (X , ω ) is the standard torus (C/(Z⊕ıZ), dz). Since ω (γ ) ∈ Q⊕ıQ, we can find a basis (αˆ , βˆ ) of H1 (X , Z) such that αˆ and γ are parallel. Applying again a matrix in SL(2, Z), we can assume that ω (αˆ ) = 1, ω (βˆ ) = ı, and ˆ γ , η) be the symplectic basis of H1 (X, Z)− with ˆ β, ω (γ ) = x, with x ∈ R>0 . Let (α, ˆ ˆ restriction of the intersection (α, ˆ β) arising from  (α0ˆ , β ). Note that in this basis, the − ∗ form is given by 2J 0 J . As an element of (H1 (X, Z) ) , ω corresponds to the vector

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Q1

Q2

Q

Q

α2

β

η1 Q1

P1

P2

γ η2

Q2

P1

P2

γ η2

Q2

P1 β1

β2 Q2

α1

α2 P1

P

η1 Q1

P1

γ

P2

P1

(X , ω ) ∈ H(1, 1)

P

β2

σ0 Q2

Q

β1

P σ0

Q

Q

α1

α (X , ω , P1 , P2 ) ∈ H(0, 0)

Q

P Q

γ σ1

P

P

Q (X, ω) ∈ Prym(2, 2)odd

Fig. 14 Finding a symplectic basis adapted to γ , (X , ω ) is the double cover of (X , ω ) ramified over P1 and Q 2 , (X, ω) is obtained from (X , ω ) by slitting along η1 , η2 , then identifying the left side of η1 with the right side of η2 , and the right side of η1 with the left side of η2

(2, 2ı, x, 0). By [7, Prop.  4.2], thereexists a unique generator T of O D which is given by a matrix of the form

e 0 2d −2c

0 e −2b 2a

a c 0 0

b d 0 0

with (a, b, c, d, e) ∈ Z5 , gcd(a, b, c, d, e) = 1,

such that ω·T = λω and λ > 0. The condition ω·T = λω implies that b = c = d = 0, λ = e, and x = 2a/e. It follows that T satisfies the equation T 2 = eT . Since T is a generator of O D , we must have D = e2 . Note that γ is an embedded segment, therefore we must have 0 < x < 1, which implies that 0 < a < e/2. All the conditions in (P D (02 )) are then fulfilled. Case (III.b) σ0 is invariant under τ and has two other twins σ1 , σ2 satisfying η := σ1 ∗ (−σ2 ) is a non-separating curve. We first show that σ1 , σ2 are exchanged by τ . Suppose that it is not the case, then both σ1 , σ2 are invariant under τ . Applying the cuttinggluing procedure described in Lemma 6.6 to σ0 and σ1 , we then get a surface (X , ω ) ∈ H(1, 1) together with an involution τ induced by τ . By construction, σ2 becomes a saddle connection in X joining the two zeros of ω . The arguments in Lemma 6.6 actually show that the hyperelliptic involution ι of X sends σ2 into another saddle connection σ2 . Since σ2 must correspond to a saddle connection of X , we derive that σ0 has three other twins, which is impossible. Thus we can conclude that σ1 , σ2 are exchanged by τ . Apply now the cutting-gluing procedure described Lemma 6.3 (see also Fig. 9) to the family {σ0 , σ1 , σ2 }, we obtain a flat torus (X , ω ) with three geodesic segments η0 , η1 , η2 which are parallel and have the same length. Let Pi , Q i be the endpoints of ηi , where Pi (resp. Q i ) corresponds to P (resp. to Q). By construction, τ induces an involution τ of X that leaves η0 invariant and exchanges η1 , η2 . Let Pˆi be the midpoint of ηi . Note that τ fixes Pˆ0 and exchanges Pˆ1 , Pˆ2 . As (X, ω) moves in the leaf of the kernel foliation, (X , ω ) and τ are fixed, only ηi vary. In particular, as t tends to t0 , ηi shrinks to the point Pˆi . Thus the limit surface belongs to Prym(03 ). All we need to show is that (X , ω , Pˆ0 , Pˆ1 , Pˆ2 ) satisfies the conditions in (P D (03 )). For this purpose, we first need to specify a symplectic basis of H1 (X, Z)− . Set ηi := σ0 ∗ (−σi ), i = 1, 2. Observe that τ (η1 ) = −η2 and τ (η2 ) = −η1 . Thus ηˆ := η1 + η2 ∈ H1 (X, Z)− . Let γ1 be a geodesic segment of minimal length

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from Pˆ1 to Pˆ0 . Let γ2 = τ (γ1 ). Reconstruct (X, ω) from (X , ω ) (see Fig. 9), we see that γ1 and γ2 correspond respectively to two simple closed geodesics γ1 , γ2 that are ˆ = 2 (up exchanged by τ . Let γˆ = γ1 + γ2 , we then have γˆ ∈ H1 (X, Z)− , and γˆ , η to a convenient choice of orientations). Let α, β be a basis of H1 (X , Z) which can be represented by simple closed curves disjoint from η0 ∪ η1 ∪ η2 ∪ γ1 ∪ γ2 . Since τ is an elliptic involution of X , we have τ (α) = −α and τ (β) = −β in H1 (X , Z). It follows that α and β correspond to two ˆ is a symplectic elements of H1 (X, Z)− which satisfy α, β = 1. Thus {α, β, γˆ , η} − , in which the restriction of the intersection form is given by the Z) basis of H1 (X,  0 . matrix 0J 2J Since ω(η) ˆ = 0, the arguments in Lemma 6.2 show that up to a rescaling by GL+ (2, R), we can assume that ω(α) = 1, ω(β) = ı, and ω(γˆ ) ∈ Q+ıQ. We can then ˆ γˆ , η} ˆ of H1 (X , Z) such that αˆ and γ are parallel. Note that {α, ˆ β, ˆ choose a basis (α, ˆ β) i − is also a symplectic basis of H1 (X, Z) . Applying an appropriate element of SL(2, Z), we can assume that ω corresponds to a vector (1, ı, x, 0), with x ∈ R>0 , with respect to this basis. The rest of the proof then follows from the same arguments as Case (II.b).   Partial compactification of E D (1, 1, 2) Theorem 7.3 Let (X, ω) ∈ E D (1, 1, 2) be a Prym eigenform in H(1, 1, 2), and τ be the Prym involution of X . We denote by Q the double zero of ω, and by R1 , R2 the simple zeros. Let S denote the set of horizontal saddle connections with distinct endpoints in X . We assume that S is not empty. Let σ0 be a saddle connection of minimal length in S, and t0 = |s0 |. Note that we can always assume that R1 is an endpoint of σ0 . For any t ∈ [0, t0 ), let (X t , ωt ) be the surface in the kernel foliation leaf through (X, ω) which is obtained by reducing the length of σ0 by the amount t, that is (X t , ωt ) = (X, ω) − (t, 0). Let M be the limit surface as t tends to t0 . We have (I) If σ0 joins R1 to Q and has no twin nor double-twin then M ∈ E D (4). (II) If σ0 joins R1 to Q and has a double-twin then M ∈  E˜ D (02 ). (III) If σ0 joins R1 to Q and has a twin σ1 , set η := σ0 ∗ (−σ1 ), then either (III.a): η is separating and M ∈ E D,triple (0, 0, 0), or (III.b): η is non-separating and M ∈ E D (03 ) (IV) If σ0 joins R1 to R2 , then either (IV.a): σ0 has no twin and M ∈ E D (2, 2)hyp , or (IV.b): σ0 has a twin and M ∈ E D (2)∗∗ with D ∈ {D, D/4}. Note that Cases (II) and (III.b) can only occur when D is a square. Proof Case (I) is actually proved in Proposition 5.5. We will give a sketch of proof for the remaining cases. Case (II) σ0 has a double-twin σ1 . Recall that σ1 joins R1 to R2 and ω(σ1 ) = 2ω(σ0 ). Note also that σ1 is unique. Set η = σ0 ∗τ (σ0 )∗(−σ1 ). By Lemma 6.5, we know that η ∈ H1 (X, Z)− is non-separating and ω(η) = 0. Applying the cutting-gluing procedure described in Lemma 6.7 to the curve η, we obtain a surface (X , ω ) ∈ H(1, 1) together with two geodesic segments η1 , η2 corresponding to η. The midpoints of η1 and η2

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are the zeros of ω . The arguments in Lemma 6.7 show that as (X, ω) moves in a leaf of the kernel foliation, (X , ω ) is fixed, only η1 , η2 vary. Thus, as t tends to t0 , ηi shrinks to a zero of ω . Hence the limit surface is (X , ω ). By construction, τ induces an involution τ of X which has two fixed points. Therefore τ is not the hyperelliptic involution of X . It follows that (X , ω ) is a double cover of a flat torus (X , ω ) ramified over two points P1 , P2 . We can now use the same arguments as in Theorem 7.1 Case (II.b) to conclude that (X , ω ) ∈  E˜ D (02 ). Case (III.a) σ0 has a twin σ1 and η = σ0 ∗ (−σ1 ) is separating. Applying the cuttinggluing procedure in Lemma 5.7 to η and τ (η), we get a triple of tori {(X j , ω j ), j = 0, 1, 2}. The Prym involution τ induces an involution on this family of tori which leaves (X 0 , ω0 ) invariant and exchanges (X 1 , ω1 ) and (X 2 , ω2 ). By this construction, η and τ (η) give rise to a slit η j on X j such that |η0 | = 2|η1 | = 2|η2 | = 2|σ0 |. As (X, ω) moves in a leaf of the kernel foliation, the family {(X j , ω j ), j = 0, 1, 2} is fixed, only the slits η j vary. Thus as t tends to t0 , the slit η j shrinks to a point P j ∈ X j . Clearly, the triple of marked flat tori {(X j , ω j , P j ), j = 0, 1, 2} belongs to Prymtriple (0, 0, 0). We can now use the arguments in [9, Lem. 11.2] to conclude that {(X j , ω j , P j ), j = 0, 1, 2} ∈ E D,triple (0, 0, 0). Case (III.b) σ0 has a twin σ1 and η = σ0 ∗ (−σ1 ) is non-separating. Applying the cutting-gluing procedure in Lemma 6.4 to the curves η and τ (η), we obtain a flat torus (X , ω ) together with three slits η0 , η1 , η2 such that |η0 | = 2|η1 | = 2|η2 |. In Fig. 10, η0 , η1 , η2 are respectively the segments R1 R2 , R1 Q 1 , R2 Q 2 . Note that τ induces an elliptic involution of X that leaves η0 invariant and exchanges η1 , η2 . Let Q denote the midpoint of η0 , and Q 1 , Q 2 be respectively the endpoint of η1 and η2 that correspond to the double zero Q of ω. Note that Q also corresponds to Q. As (X, ω) moves in a leaf of the kernel foliation, (X , ω ), Q , Q 1 , Q 2 are fixed, only the slits ηi vary. When t tends to t0 , η0 shrinks to Q , η1 and η2 shrink to Q 1 and Q 2 respectively. Thus the limit surface is (X , ω , Q , Q 1 , Q 2 ) ∈ Prym(03 ). Using similar arguments as the proof of Theorem 7.1 Case (III.b), we get that (X , ω , Q , Q 1 , Q 2 ) ∈ E D (03 ). Case (IV.a) σ0 joins R1 to R2 and has no twin. The arguments in Proposition 5.5 show that we can collide R1 and R2 by collapsing σ0 . Therefore, as t tends to t0 , (X t , ωt ) ˆ in H(2, 2). Since the set of Prym eigenforms converges (in M3 ) to a surface ( Xˆ , ω) ˆ ∈ E D (2, 2). is closed in Mg (see [14]), we have ( Xˆ , ω) Let τˆ be the Prym involution of Xˆ . Recall that the Prym involution τ on X exchanges ˆ R1 , R2 . Since R1 and R2 collide in the limit, τˆ fixes the resulting double zero of ω. Thus τˆ fixes both zeros of ω. ˆ It follows that ( Xˆ , ω) ˆ is the standard orienting double cover of a (meromorphic) quadratic differential in Q(12 , −12 ). But it is well-known that such double covers belong to the component Hhyp (2, 2) (see [6]). Therefore, we can conclude that ( Xˆ , ω) ˆ belongs to E D (2, 2)hyp . By Proposition 2.3, we also know that ( Xˆ , ω) ˆ is a unramified double cover of an eigenform in H(2). In particular, the ˆ is closed. GL+ (2, R)-orbit of ( Xˆ , ω) Case (IV.b) σ0 joins R1 to R2 and has a twin σ1 (which also joins R1 to R2 ). Let   η := σ0 ∗ (−σ1 ).

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Claim 8 η is non-separating. Proof Assume that η is separating. By cutting along η, then closing up the boundaries of the resulting components, we obtain a slit torus (X 1 , ω1 ) with a slit η1 , and a surface (X 2 , ω2 ) ∈ H(2) with a marked geodesic segment η2 . The Prym involution τ induces the involutions τ1 on X 1 and τ2 on X 2 . Observe that τ1 leaves η1 invariant. Since X 1 is an elliptic curve, τ1 must have four fixed points, three of which are not contained in η1 . Since X 1 \η1 is a subsurface of X , we derive that those three fixed points of τ1 are also fixed points of τ . Recall that by definition, τ has four fixed points, and one of which must be the double zero of ω. Therefore, the double zero of ω (which is also the double zero of ω2 ) is the unique fixed point of τ in X 2 \η2 . It follows that τ2 has exactly two fixed points: the unique zero of ω2 and the midpoint of η2 . In particular, τ2 is not the hyperelliptic involution ι2 of X 2 . On the other hand, the unique zero of ω2 is fixed by both of τ2 and ι2 , and the differentials of τ2 and ι2 at this point are both equal to −Id. Therefore, we must have τ2 = ι2 , which is a contradiction. We can now conclude that η is non-separating.   Claim 9 σ0 and σ1 are exchanged by τ . Proof Since η is non-separating, cutting X along η, then closing up the boundary of the resulting surface, we obtain a surface (X , ω ) ∈ H(2) with to marked geodesic segments η1 , η2 arising from η. Note that η1 and η2 are parallel and have the same length. By construction, τ induces an involution τ on X . Assume that both σ0 and σ1 are invariant under τ . A careful inspection on the action of τ in a neighborhood of η shows that τ maps η1 to η2 . Recall that τ has four fixed points in X , two of which are contained in η (i.e. the midpoints of σ0 and σ1 ). It follows that τ has only two fixed points, which correspond to the fixed points of τ not contained in η. We derive in particular that τ is not the hyperelliptic involution of X . But since we have τ ∗ ω = −ω , the same argument as in the previous Claim shows that we also get a contradiction.   Let (X , ω ), η1 , η2 be as in Claim 9. Let P1 , P2 denote respectively the midpoints of η1 , η2 . A direct consequence of Claim 9 is that both η1 , η2 are invariant under the involution τ on X which is induced by τ . It follows that τ has exactly 6 fixed points: the four fixed points of τ together with P1 and P2 . Thus τ is the hyperelliptic involution of X . We can now use the arguments in [9, Lem. 8.7] to prove that (X , ω ) belongs to E D (2), with D ∈ {D, D/4}. Let (α1 , β1 , α2 , β2 ) be the symplectic basis of H1 (X, Z)− , where β2 = β21 + β22 , as shown in Fig. 15. We can view ω as a vector in H 1 (X, C)−  C4 . Let (x, y, u, v) be the row vector corresponding to ω with respect to this basis.  0  . Note that the intersection form in H1 (X, Z)− is given by the matrix 0J 2J There exists a generator T of O D ⊂ End(Prym(X, τ )) whose matrix in the basis {α1 , β1 , α2 , β2 } has the form  T =

e 0 d −c

0 e −b a

2a 2c 0 0

2b  2d 0 0

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β21

α1

η2

η1 α2 β1

β22

(X , ω ) Fig. 15 Finding a symplectic basis for H1 (X, Z)− : (X, ω) is obtained from (X , ω ) by identifying the left side of η1 with the right side of η2 , and the right side of η1 with the left side of η2 . Note that β21 and β22 are homologous in X but not in X . The Prym involution of X is induced by the hyperelliptic involution of X . A symplectic basis of H1 (X, Z)− is given by {α1 , β1 , α2 , β21 + β22 }

with (a, b, c, d, e) ∈ Z5 and gcd(a, b, c, d, e) = 1, such that ω · T = λω, with λ ∈ R>0 . Observe that by construction, we have ω(β2 ) = ω(β21 )+ω(β22 ) = 2ω(β21 ) (since we have β22 = −τ (β21 ) and τ ∗ ω = −ω). Consider now α1 , β1 , α2 , β21 as 1cycles on X . Observe that {α1 , β1 , α2 , β21 } is a symplectic basis of H1 (X , Z). Again, by construction, we have ⎧ ω (α1 ) = ω(α1 ) = x, ⎪ ⎪ ⎨ ω (β1 ) = ω(β1 ) = y, ⎪ ω (α2 ) = ω(α2 ) = u, ⎪ ⎩ ω (β21 ) = ω(β21 ) = 1/2ω(β2 ) = v/2, which means that ω corresponds to the row vector (x, y, u, v/2) in H 1 (X , C). Let T be the following matrix T :=



e 0 d −2c

0 e −b 2a

2a 2c 0 0

b d 0 . 0

Note that T is self-adjoint with respect to the intersection form of H1 (X , Z). It is straightforward to check that the condition ω · T = λω is equivalent to ω · T = λω . Hence (X , ω ) is an eigenform in H(2). Thus (X , ω ) ∈ E D (2), for some discriminant D . Note that O D must be generated by a matrix T ∈ End(Jac(X )) such that T = kT , with k ∈ Z. Thus we must have k = gcd(2a, 2c, b, d, e). Since we have gcd(a, b, c, d, e) = 1, it follows that k ∈ {1, 2}, which implies that D ∈ {D, D/4}. To conclude, we finally remark that as (X, ω) moves in a leaf of the kernel foliation, (X , ω ), and P1 , P2 are fixed. Hence, as t tends to t0 , η1 and η2 shrink to P1 and P2 , and the limit surface belongs to E D (2)∗∗ . The proof of Theorem 7.3 is now complete.

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