Hilbert Modular Forms of Weight 1/2 Citation Achimescu, Sever (2005) Hilbert Modular Forms of Weight 1/2. Dissertation (Ph.D.), California Institute of Technology. doi:10.7907/PBNR-3X07. https://resolver.caltech.edu/CaltechETD:etd-04272005-172254 Abstract NOTE: Text or symbols not renderable in plain ASCII are indicated by [...]. Abstract is included in .pdf document. Let [...] be the space of Hilbert modular forms of half integral weight of all levels N and characters X. We denote by φ N : O F → C a periodic function of period N. Let Θ be the C-linear space of the functions f : H X H → C, [...] where, for each f,t ∈ O F runs through a finite subset of totally positive integers of F. Main Theorem. H = Θ Using this theorem, for some fixed F's, an explicit basis can be found. Some examples are given in Chapter 4. Item Type: Thesis (Dissertation (Ph.D.)) Subject Keywords: Hilbert modular forms of half integral weight Degree Grantor: California Institute of Technology Division: Physics, Mathematics and Astronomy Major Option: Mathematics Thesis Availability: Public (worldwide access) Research Advisor(s): Ramakrishnan, Dinakar Thesis Committee: Ramakrishnan, Dinakar (chair) Wales, David B. Flach, Matthias Dunfield, Nathan M. Defense Date: 27 September 2004 Record Number: CaltechETD:etd-04272005-172254 Persistent URL: https://resolver.caltech.edu/CaltechETD:etd-04272005-172254 DOI: 10.7907/PBNR-3X07 Default Usage Policy: No commercial reproduction, distribution, display or performance rights in this work are provided. ID Code: 1520 Collection: CaltechTHESIS Deposited By: Imported from ETD-db Deposited On: 02 May 2005 Last Modified: 21 May 2020 22:42 Thesis Files Preview PDF (Full thesis) - Final Version See Usage Policy. 289kB Preview PDF (titlepage.pdf) - Final Version See Usage Policy. 17kB Preview PDF (copyrightpage.pdf) - Final Version See Usage Policy. 11kB Preview PDF (ackno.pdf) - Final Version See Usage Policy. 16kB Preview PDF (abstract.pdf) - Final Version See Usage Policy. 52kB Preview PDF (contents.pdf) - Final Version See Usage Policy. 16kB Preview PDF (ch1-5.pdf) - Final Version See Usage Policy. 203kB Repository Staff Only: item control page

iv Abstract Let H= X M 12 (N , χ) N ,χ be the space of Hilbert modular forms of half integral weight of all levels N and characters χ. We denote by ϕN : OF → C a periodic function of period N . Let Θ be the C-linear space of the functions f : H × H → C, f (z) = X X t 0 ϕNt (ξ)exp(tπi(ξ 2 z1 + ξ 2 z2 )) ξ∈OF where, for each f , t ∈ OF runs through a finite subset of totally positive integers of F . Main Theorem. H=Θ Using this theorem, for some fixed F ’s, an explicit basis can be found. Some examples are given in Chapter4.  lil Acknowledgement I am grateful to my advisor, Professor Dinakar Ramakrishnan, for his guidance and encouragement. I thank Caltech for its financial support. 1 Chapter 1 Introduction Classically, a modular form of weight k ∈ Z for a congruent subgroup µ ¶ a b , N |c} < SL2 (Z) is a holomorphic function Γ := Γ0 (N ) := {γ = c d f on the upper-half plane H := {z ∈ C , Imz > 0} satisfying µ f (γz) = (cz + d)k f (z) , ∀γ = a b c d ¶ ∈Γ and an additional property called “holomorphicity at the cusps.” Using the natural inner product, namely, the Petterson inner product on the C-vector space of modular forms, we get finite a dimensional Hilbert space for each pair (Γ, k). There are Hecke operators {T (p)}p6 |N prime which preserve these spaces. A generalization is when we allow k ∈ Z[ 12 ] and we get the half integral weight modular forms. Shimura introduced a family of corresponding Hecke operators {T (p2 )}p prime . In [Serre-Stark], Serre and Stark proved a theorem which gives explicit basis consisting of theta functions for some spaces of half integral weight modular forms. 2 The Hilbert modular forms (of half integral weight) are modular forms (of half integral weight) of several variables. The main result in my dissertation is an analog, for certain real quadratic fields of class number one, of the theorem of Serre-Stark which gives algorithms to compute basis of spaces of half integral weight modular forms over Q. An alternative proof of a part of the Serre-Stark theorem, which was outlined (using adellic representations) by Pierre Deligne in a letter to Jean Pierre Serre, is expanded and generalized in this thesis to some quadratic fields of class number one in Chapter 3. Some of the results of Serre and Stark over Q, which Deligne implicitly assumed, require different arguments over F . In particular, one need to use certain nontrivial, structural results of Goro Shimura on Hilbert modular forms of half integral weight. There is already a vast generalization of the Serre-Stark theorem in a slightly different form, due to Shimura, valid over all totally real fields, but this method seems to only give generators, not a basis. The result in Chapter 3 is in a more explicit form and hopefully more suitable for applications. See Chapter 4 for examples. 3 In Chapter 5 we define the congruent number problem over F and we take a few steps towards generalizing Tunnell’s results over Q. Here the √ generalization is not trivial, since already for Q( 2) there are examples of √ Q( 2)-congruent numbers α such that the associated elliptic curve Eα : √ y 2 = x3 − α2 x over Q( 2) has zero rank; α corresponds to torsion points √ which Eα acquires over Q( 2). 4 Chapter 2 Preliminaries Let F be a totally real number field with class number one and dis√ √ criminant D; F = Q( d0 ) = Q( D) with d0 square free and such that √ OF = Z[ d0 ], where OF denotes the ring of integers of F . The two √ √ real embeddings are a + b d0 7→ a ± b d0 . The discriminant (of F ) is √ DF/Q = d0 , the different is D := DF/Q = 2 d0 OF and the inverse different is D−1 = 2√1d OF . 0 For P prime ideal of OF we denote P = $OF , $ ∈ OF . µ ¶ a b Put SL2 (OF ) = { , a, b, c, d ∈ OF , ad − bc = 1}. For ξ = c d √ √ m + n d0 ∈ OF we denote ξ 0 = m − n d0 its conjugate. For N an ideal in OF we denote µ a b Γ0 (N ) = { c d ¶ ∈ SL2 (OF ), c ∈ N }. Put H = {z ∈ C, Imz > 0} and z = (z1 , z2 ) ∈ H × H. Γ0 (N ) acts on H × H as follows: first we view Γ0 (N ) ,→ GL2 (R) × 5 GL2 (R) via µ a b c d ¶ µ a b 7 ( → c d ¶ µ 0 a , c0 ¶ b0 ) d0 then GL2 (R) × GL2 (R) acts on H × H componentwise: µ a c b d ¶ µ 0 a × c0 b0 d0 ¶ az1 + b a0 z2 + b0 , (z1 , z2 ) → 7 ( , ) cz1 + d c0 z2 + d0 All square roots of numbers in C − {x < 0} are to be taken in the right half-plane. Define θ : H × H → C as follows: θ(z) := X exp(πi(ξ 2 z1 + ξ 02 z2 )) ξ∈OF Observe that θ is holomorphic on H × H. Recall that a function f on an open set in Cn is said to be holomorphic iff it has an absolutely convergent power series expansion in each compact polydisc in that open set. We have |θ(z)| = | X exp(πiξ 2 z1 )exp(πiξ 02 z2 )| ≤ ξ∈OF | X ξ∈OF 1 exp(2πiξ 2 z1 )| 2 | X ξ∈OF 1 exp(2πiξ 02 z2 )| 2 6 Because the one-dimensional theta function H → C , z 7→ X exp(2πin2 z) n∈Z is holomorphic, the series in the righthand side of the above inequality converge absolutely on each compact (on H), therefore the series defining θ converges absolutely on each compact (in H × H). For a prime ideal pOF and β ∈ OF we define a quadratic symbol ( ( β ) = +1 ifβ is a square in (OF /pOF )∗ pOF β ) = −1 ifβ is a non square in (OF /pOF )∗ pOF ( β ) = 0 if β ∈ pOF pOF We extend it by multiplicativity to all nonzero ideals of OF . For 0 6= α ∈ β β OF , write ( α ) = ( αO ). F For γ ∈ Γ0 (N ) we consider the automorphy factor of weight 12 , denoted h(γ, z), introduced in [Shi2], Proposition 1.2 pg. 285. The definition is intrinsic, though not explicit. This automorphy factor is a holomorphic 7 function on H × H satisfying h(γ, z)2 = t(cz1 + d)(c0 z2 + d0 ) where t is a root of unity. A multiplicative character modulo N is a group homomorphism χ : (OF /N )∗ → C∗ . We also view a character χ as a periodic function on OF by putting χ(ν) = 0, ∀ν ∈ N . Definition. A holomorphic Hilbert modular form of weight 12 , character χ and level N is a function f : H × H → C which is holomorphic, both on H × H and at the cusps, and satisfies µ f (γz) = χ(d)h(γ, z)f (z) , ∀γ = a b c d ¶ ∈ Γ0 (N ) , ∀z ∈ H × H Note that, since in our case [F : Q] > 1, by the Koecher principle, the condition “holomorphic at the cusps” is superfluous. The definition of Hilbert modular forms of weight k2 , with k odd integer, is similar; but with h(γ, z) replaced by h(γ, z)k . We denote the C-vector space of holomorphic Hilbert modular forms of weight 12 , character χ and level N by M 12 (N , χ). Following [Shi1], for f, g ∈ M 12 (N , χ) we define their inner product 8 < f, g > by Z < f, g >=| Γ\H × H | 1 −1 f (z)g(z)(y1 y2 ) 2 dH (z) Γ\H×H where y =Imz and dH (x + iy) = (y1 y2 )−2 dx1 dx2 dy1 dy2 . This integral converges if one of the forms is a cusp form. But in fact < f, g > makes sense, as noted by Deligne, for all f, g in M 12 (N ). Given a function f : H × H → C we denote ft : H × H → C, ft (z) = f (tz) where t totally positive, t ∈ OF . Also, for t totally positive, t ∈ OF , we define a “quadratic character” χt as follows: for d ∈ OF , d 6= 0, χt (d) = ( t0 ) if t0 ≡ 1(mod4) d χt (d) = ( 4t0 ) otherwise d χt (u) = 1 , ∀u ∈ (OF )∗ where t0 denotes the square-free part of N (t). For a character ψ modulo N , define θψ : H × H → C as follows: θψ (z) := X ξ∈OF ψ(ξ)exp(πi(ξ 2 z1 + ξ 02 z2 )) 9 Theorem. θψ ∈ M 12 (4(cond ψ)2 , ψ). Proof. This is Lemma 4.3 pg. 784 of [Shi1]. In particular, θ ∈ M 12 (4OF ). Proposition 1. Let t ∈ OF be totally positive. Then f ∈ M 12 (N , χχt ) ⇒ ft ∈ M 12 (tN , χ) In particular, since χ2t = 1, f ∈ M 12 (N , χ) ⇒ ft ∈ M 12 (tN , χχt ) Proof. Need to prove µ ¶ a b ft (γz) = χ(d)h(γ, z)ft (z) , ∀γ = ∈ Γ0 (tN ) , ∀z ∈ H × H c d µ ¶ µ ¶ a b a bt Fix γ = ∈ Γ0 (tN ) and put γt = c ∈ Γ0 (N ). By hypothc d d t esis, f ∈ M 12 (Γ0 (N ), χχt ) thus f (γt tz) = χ(d)χt (d)h(γt , tz)f (tz) , ∀z ∈ H × H The conclusion follows by noticing that γt tz = tγz and using the following 10 Lemma. χt (d)h(γt , tz) = h(γ, z) , ∀z ∈ H × H Proof: The squares of both sides are equal, by [Shi2] pg. 286. It suffices to check the sign. Equivalently, since h(γ, z) = θ(γz) θ(z) , it suffices to check the Proposition for f = θ. But this is a known fact. 11 Chapter 3 The main result Let H= X M 12 (N , χ) N ,χ be the space of Hilbert modular forms of half integral weight of all levels N and characters χ. We denote by ϕN : OF → C a periodic function of period N , i.e. ϕ(x) = ϕ(x + α) , ∀x ∈ OF , ∀α ∈ N . Let Θ be the C-linear space of the functions f : H × H → C, f (z) = X X t 0 ϕNt (ξ)exp(tπi(ξ 2 z1 + ξ 2 z2 )) ξ∈OF where, for each f , t ∈ OF runs through a finite subset of totally positive integers of F . Main Theorem. H=Θ Proof: 12 The inclusion Θ⊆H follows from Proposition 1 of the previous chapter and the following Lemma. The function z 7→ X 0 ϕN (ξ)exp(πi(ξ 2 z1 + ξ 2 z2 )) ξ∈OF is a Hilbert modular form of weight 12 possibly with a character. Proof: This is a corollary of the Theorem on pg.154 from [Garrett]. Now we prove the key inclusion H⊆Θ Let Af denote the finite adeles of F . It is known that (cf. [Del], pg. f 2 (Af ) of SL2 (Af ) acts on H, preserves 259) the metaplectic 2-covering SL the scalar product and leaves Θ stable. Under this action, H decomposes into a direct sum of irreducible representations. Let Hi be one of them. We want to prove that Hi is contained in Θ. 13 Proposition 1. For suitable (N , χ), Hi ∩ M 12 (N , χ) 6= 0. Proof: There exists a family of compact open subgroups {K0 (N )}N , which is also a fundamental system of neighborhoods of the unity of SLg 2 (Af ), such that Γ0 (N ) = K0 (N ) ∩ SL2 (Af ), ∀N . Here we have used the fact that F has class number 1. There is a character χ̃ whose restriction to Γ0 (N ) is K0 (N ),χ̃ χ. If V is a vector space on which SLg for the 2 (Af ) acts, we write V subgroup of vectors v ∈ V such that kv = χ̃(k)v. One knows H K0 (N ),χ̃ = M 21 (N , χ) We have K (N ),χ̃ H K0 (N ),χ̃ = ⊕i Hi 0 Now fix any Hi . Then the admissibility of Hi as an SLg 2 (Af )-module implies K (N ),χ̃ that there exists (N , χ̃) such that Hi 0 6= 0. Therefore there exists (N , χ) such that Hi ∩ M 21 (N , χ) 6= 0. QED One knows that for every prime ideal P of OF , including those dividing N , there is a Hecke operator T (P 2 ) acting on M 12 (N , χ). They come 14 from the action of SL2 (Af ). If P, Q are primes, then T (P 2 ) commute with T (Q2 ). The operators T (P 2 ) preserve the scalar product and have a common eigenvector in M 12 (N , χ). Proposition 2. We have : 0 f (z) = Σµ∈OF a(tµ2 )exp(tπi(µ2 z1 + µ 2 z2 )) where t ∈ OF , t is totally positive, t|N , and, ∀α ∈ OF , a(αµ2 ) = a(α)ψ(µ) if (µ, N ) = 1 a(αP 2 ) = cP a(α) if P | N where ψ is a multplicative character mod 2N and T (P 2 )f = cP f . Proof: This is a consequence of the following lemmas: Lemma 1. We have (i) There is a basis of M 12 (N , χ) consisting of forms whose coefficients belong to a number field, 15 (ii) If f (z) = P ξ∈OF a(ξ)exp(πi(ξ 2 0 z1 + ξ 2 z2 )) belongs to M 12 (N , χ) and the a(ξ) are algebraic numbers then the a(ξ) have bounded denominators (i.e., there exists a nonzero integer D such that Da(ξ) is an algebraic integer for all ξ). Proof of Lemma 1: The analogous results for integral weight forms are known. Also it is known that there is an orthogonal decomposition M 12 (N , χ) = S 12 (N , χ) ⊕ E 12 (N , χ) for the scalar product,where S 12 (N , χ) denotes the space of cusp forms and E 21 (N , χ) denotes the space of Eisenstein forms. In [Shi2] it is proved that there is a basis of E 12 (N , χ) consisting of forms whose coefficients belong to a number field and have bounded denominators. For both (i) and (ii), it suffices to prove the analogous results for S 21 (N , χ). By Prop 5.2, 5.3 from [Shi1], S 21 (N , χ) is spanned by Hecke eigenforms for T (P 2 ) for almost all P . Write S 12 (N , χ) = Θ0 ⊕ Θ⊥ 0 where Θ0 is the subspace of S 21 (N , χ) generated by theta series and Θ⊥ 0 is the subspace perpendicular to Θ under the scalar product a corollary of the result of this chapter is S 21 (N , χ) = Θ0 but we don’t know it yet. The analogous 16 statements of (i) and (ii) hold for Θ0 by the explicit knowledge of theta series. If S 12 (N , χ) = Θ0 then we are done. If not, then there exists a Hecke eigenform f in Θ⊥ 0 . As in [Shi1], pg. 816, we associate to f an integral weight form h. It may happen that different f ’s correspond to the same h. We denote by S 12 (N , h, χ) the subspace of S 12 (N , χ) generated by those f ’s corresponding to h. By prop 8.9(2) of [Shi1], S 12 (N , h, χ) has a basis of forms whose coefficients belong to a number field and have bounded denominators. QED Lemma 2. Let f (z) = P 0 2 2 ξ∈OF a(ξ)exp(πi(ξ z1 + ξ z2 )) be a nonzero element of M 12 (N , χ) and let P be a prime with P 6 | N . Put P = $OF . Assume that f | T (P 2 ) = cP f , with cP ∈ C. Let µ ∈ OF such that P 2 6 | µ. Then: (i) a(µP 2n ) = a(µ)χ($)n ( Pµ )n for every n ≥ 0 integer, 1 (ii) If a(µ) 6= 0 then P 6 | µ and cP = χ($)( Pµ )(1 + N (P ) ). Proof of Lemma 2: Cf. Prop5.4 pg. 788 in [Shi], the operator T (P 2 ) maps forms with 17 algebraic coefficients into themselves. Hence cP is algebraic and we may assume that the coefficients a(ξ) of f are algebraic numbers. Fix µ and consider the power series A(T ) = ∞ X a(µP 2n )T n n=0 where T is an indeterminate. As in the proof of Theorem 5.5, pg. 790 [Shi] we have A(T ) = a(µ) 1 − αT (1 − βT )(1 − γT ) where α = χ(p)N (P )−1 ( Pµ ) and β + γ = cP , βγ = χ($2 )N (P )−1 . If a(µ) = 0 then A(T ) = 0 thus a(µP 2n ) = 0 for all n ≥ 0 therefore (i). Assume now a(µ) 6= 0. Using Lemma 1(ii), A(T ) converges in the P -adic topology, if we view A(T ) as a nonzero P -adic rational function of T , in the P -adic unit disc U defined by | T |P < 1; hence A(T ) cannot have a pole in U . However, since βγ = χ($2 )P −1 , either β −1 or γ −1 belongs to U ; assume it is β −1 . Now A(T ) holomorphic at β −1 implies that the factors 1 − βT and 1 − αT cancel each other. Thus α = β and A(T ) = a(µ) (1 − γT ) 18 thus a(µP 2n ) = γ n a(µ) Now βγ 6= 0 ⇒ α 6= 0 ⇒ $ 6 | µ. Moreover, γ= βγ χ($2 )N (P )−1 µ = = χ($)( ) µ α χ($)N (P )−1 ( $ ) $ which implies (i). Finally, (ii) follows from cP = β + γ = α + γ. QED Lemma 3. Let f (z) = P ξ∈OF a(ξ)exp(πi(ξ 2 0 z1 + ξ 2 z2 )) be a nonzero element of M 21 (Γ0 (N , χ)) and let N 0 be a multiple of N . Assume that, for all P 6 | N 0 , we have f | T (P 2 ) = cP f , with cP ∈ C. Then there exists a unique (up to multiplication by a unit) square-free nonzero t ∈ OF such that a(ξ) = 0 if ξt is not a square in OF . Moreover: (i) t | N 0 , 1 0 (ii) cP = χ($)( $t )(1 + N (P ) ) if $ 6 | N , (iii) a(ξµ2 ) = a(ξ)χ(µ)( µt ) if (µ, N )0 = 1. Proof of Lemma 3: Let µ, µ0 ∈ OF such that a(µ) 6= 0 and a(µ0 ) 6= 0. For each P prime in 19 OF , P p 6 | N 0 µµ0 we have, by Lemma 2(ii), µ 1 µ0 1 χ($)( )(1 + ) = cp = χ($)( )(1 + ) $ N (P ) $ N (P ) thus, for all $ prime in OF , $ 6 | N 0 µµ0 , µ0 µ ( ) = ( ). $ $ This implies, for each prime q dividing µ and not dividing µ0 , denoting by vq (µ) the exponent of q in µ, that q vq (µ) ( )=1 $ for all $ prime in OF , $ 6 | N 0 µµ0 . In particular, for a $ large enough, this implies that vq (µ) is even. Therefore we may write µ = tµ21 , µ0 = tµ02 1 with t square free. The uniqueness of t follows repeating the argument for another pair µ, µ00 instead of µ, µ0 . We proved the first part of the Lemma 3. Now we prove (i) and (ii). Fix a prime p not dividing N 0 . Fix µ such that a(µ) 6= 0 and write µ = tµ21 = t$2n µ20 with $ not dividing µ0 . We apply Lemma 1 2 2(i) to tµ20 and we get a(µ) = a(tµ20 )χ($)( $t )(1 + N (P ) ) thus a(tµ0 ) 6= 0. By Lemma 2(ii), $ does not divide tµ20 hence $ does not divide t. So 20 1 far Lemma 3(i) proved. Now, by Lemma 2(ii), cP = χ($)( $t )(1 + N (P )) thus Lemma 3(ii) proved. Now we prove Lemma 3(iii). Because both sides are multiplicative in µ, it suffices to check it for µ = $ with $ prime not dividing N 0 . Writing ξ = ξ0 $2n with $2 6 | N and applying Lemma 2(i) we get qed. QED Lemma 4. We have a(µ$2 ) = cP a(µ) if $ | N . Proof of Lemma 4: It is a straightforward corollary of [Shi], p.788, Prop 5.4. Consider now g(z) = Σξ∈OF ,(ξ,N )=1 a(tξ 2 )exp(tπi(ξ 2 z1 + ξ 02 z2 )) Proposition 3. (i) g 6= 0; (ii) g ∈ Θ; (iii) g is (up to a scalar factor) the transform of f (z) = Σξ∈OF a(tξ 2 )exp(tπi(ξ 2 z1 + ξ 02 z2 )) 21 by ΠP |N LP where P = $OF with $ totally positive and LP is the operator which transforms h(z) into h(z) − cP h($2 z1 , $02 z2 ). (iv) Lp can be defined by the element µ 1 − cP $−1 0 0 $ ¶ g2 (FP )], where FP is the completion of F at P . of the group ring Z[SL (v) g ∈ Hi . Proof: (i)g 6= 0 because f 6= 0; (ii) Define ϕN (ξ) = a(tξ 2 ) if (ξ, N ) = 1 and ϕN (ξ) = 0 otherwise. ϕ is periodic of period N because ψ is so. Therefore g(z) = Σξ∈OF ϕN (ξ)exp(tπi(ξ 2 z1 + ξ 02 z2 )) ∈ Θ. (iii) Fix P = $OF | N with $ totally positive. We have (LP f )(z) = Σξ∈OF a(tξ 2 )exp(tπi(ξ 2 z1 + ξ 02 z2 ))− cp Σξ∈OF a(tξ 2 )exp(tπi(ξ 2 $2 z1 + ξ 02 $02 z2 )) = Σξ∈OF a(tξ 2 )exp(tπi(ξ 2 z1 + ξ 02 z2 ))− 22 Σξ∈OF a(t($ξ)2 )exp(tπi(ξ 2 $2 z1 + ξ 02 $02 z2 )) = Σξ∈OF ,(ξ,$)=1 a(tξ 2 )exp(tπi(ξ 2 z1 + ξ 02 z2 )) in other words LP “deletes” from the expansion of f those terms such that $ | ξ. The conclusion follows noticing that ΠP |N LP means composition of operators. (iv) f 2 (A) = SL f 2 (F∞ ) × SL f 2 (Af ) with SL2 (F ) embeds diagonally in SL f 2 (A) as (x∞ , xf ) componentwise multiplication. Write the elements of SL f 2 (F∞ ) and xf ∈ SL f 2 (Af ). If γ ∈ SL2 (F ) then γ∞ = γ with x∞ ∈ SL γf = γ and γ := (γ, γ) = (γ, 1)(1, γ). µ −1 ¶ µ −1 $ 0 $ Let γ = ∈ SL2 (F ). Also note that 0 $ 0 f 2 (FP ) ⊆ SL f 2 (Af ). SL 0 $ ¶ ∈ Since the automorphic forms are left invariant g2 (Af ) is the under the diagonal action of SL2 (F ), the action of γ in SL f 2 (Af ) sends h(z) to inverse of the action of (γ, 1). By part (iii), γ ∈ SL h($2 z1 , $02 z2 ). Since LP (h)(z) = h(z1 , z2 ) − cP h($2 z1 , $02 z2 ), the conclusion follows. (v) follows from (iii),(iv) and the fact that Hi is invariant. 23 Therefore Hi ∩ Θ 6= ∅. Since Hi is irreducible, this implies Hi ⊆ Θ. QED 24 Chapter 4 Examples √ √ Consider F = Q[ 2] and N = 8Z[ 2]. From the proofs in Chapter 4 follows that √ n 1 (2 Z[ M 2], χ) is genχ 2 P erated by {(gψ )t } where ψ is an even primitive character modulo N and t is a totally positive integer of F dividing 2. √ The coset representatives of OF /N are {a + b 2 + N } with a, b ∈ √ {0, ..., 7} so that OF /N has 64 elements. For α = a + b 2, we denote √ [α] := a + b 2 + N . Since OF /N is finite, the elements of (OF /N )∗ are √ exactly the nonzero divisors. Notice that [a + b 2] is a zero divisor in OF /N if and only if a is even. Thus (OF /N )∗ has 32 elements. We list the orders of all elements (and we conclude that (OF /N )∗ has exponent 8): √ [3], [5], [7] and [a + 4 2], a odd, have order 2; √ √ [a + 2 2], [a + 6 2], a odd, have order 4; √ [a + b 2], a, b odd, has order 8. Thus, counting the elements of order 2 in an abelian group of order 25 32 and exponent 8, we conclude that (OF /N )∗ ∼ = Z/8 × Z/2 × Z/2 since they have the same maximum number of 7 elements of order 2. Any other abelian group of order 32 and exponent 8 has less than 7 elements of order 2. It is straightforward to check that [1 + √ 2] generates a subgroup of √ order 8, that [−1] = [7] generates a subgroup of order 2, that [3 + 4 2] generates a subgroup of order 2, that [7] 6∈< [1 + √ 2] >, and that [3 + √ √ √ 4 2] 6∈< [7] > × < [1 + 2] >. It follows that (OF /N )∗ =< [1 + 2] > √ × < [7] > × < [3 + 4 2] >. The character group of (the finite abelian group) (OF /N )∗ is therefore < ϕ0 > × < ϕ1 > × < ϕ2 > where √ < ϕ0 >≡ 1 on < [7] > × < [3 + 4 2] >, hence < ϕ0 > can be viewed as a character of < [1 + √ 2] >; < ϕ1 >≡ 1 on < [1 + √ √ 2] > × < [3 + 4 2] >, hence < ϕ1 > can be viewed as a character of < [7] >; < ϕ2 >≡ 1 on < [7] > × < [1 + √ 2] >, hence < ϕ2 > can be viewed √ as a character of < [3 + 4 2] >. Any character ψ of (OF /N )∗ can be written as ψ = ϕj00 ϕj11 ϕj22 with 26 j0 ∈ {0, ..., 7}, j1 ∈ {0, 1}, j2 ∈ {0, 1}. Recall that all the units of OF are √ {±(1 + 2)k , k ∈ Z}, so that ψ is even if and only if ψ ≡ 1 on < [7] > × < [1 + √ 2] >, that is j0 = 0 and j1 = 0, therefore all the even characters of (OF /N )∗ are {1, ψ := ϕ2 }. So far we got that {θ, θψ , θ2 , (θψ )2 , θ√2 , (θψ )√2 } is a set of generators for √ 1 (8Z[ M 2], χ). Since these six generators have six distinct charχ 2 P acters, we conclude: √ {θ} is a basis for M 21 (4Z[ 2]); √ {θψ } is a basis for M 12 (4Z[ 2], ψ); √ √ {θ√2 } is a basis for M 12 (4 2Z[ 2], χ√2 ); √ √ {(θψ )√2 } is a basis for M 21 (4 2Z[ 2], ψχ√2 ); √ {θ2 } is a basis for M 12 (8Z[ 2], χ2 ); √ {(θψ )2 } is a basis for M 12 (8Z[ 2], ψχ2 ). A generalization is : √ √ OF = Z[ 2], N = 2n Z[ 2], n ∈ Z, n ≥ 4 With similar arguments we get that : 27 (OF /N )∗ has 22n−1 elements and has exponent 2n ; √ All the elements of the form [a + b 2] of order two are a, a 6= 1 odd √ (there are 2n−1 − 1 of them) and [a + 2n−1 2], a odd (there are 2n−1 of them). √ All the elements of order 2n are [a + b 2] with a, b odd. Counting the elements of order 2, we conclude (OF /N )∗ ∼ = Z/2n × Qn−1 j=1 Z/2. Its group of characters is generated by ϕ0 of order 2n and ϕ1 , ..., ϕn−1 each of order 2. Without loss of generality ϕ1 ≡ 1 on < [−1] >. j n−1 Any even characters ψ can be written as ϕj22 ....ϕn−1 with j2 , ..., jn−1 ∈ {0, 1}. Proposition. A basis for √ n 1 (2 Z[ M 2], χ) is given by χ 2 P j n−1 {(θψ )t , ψ = ϕj22 ....ϕn−1 , j2 , ..., jn−1 ∈ {0, 1} , t | 2n−2 , t > 0}. Remark. A character on OF is the pull-back of a character on Z via norm iff it √ is invariant under the Galois action. For N = 8Z[ 2], the forms θ and θψ 28 have such characters, since every element of order 2 in (OF /N )∗ is invariant under the Galois action. However, the forms {(θψ )√2 , θ√2 } are genuinely new. 29 Chapter 5 Towards the congruent number problem over F Let F be a totally real number field. Below α ∈ F and τ : F → R denotes a field embedding. Definitions: (i) x ∈ F is said to be τ -positive iff τ (x) > 0; (ii) x ∈ F is said to be totally positive iff τ (x) > 0 , ∀τ (iii) α ∈ F is said to be an F -congruent number with respect to τ iff 1 ∃τ −positive X, Y, Z such that X 2 + Y 2 = Z 2 and τ (X)τ (Y ) = τ (α) 2 Note that an F -congruent number with respect to τ is τ -positive. Also, note that by applying τ −1 the last equality becomes 1 XY = α 2 (iv) α ∈ F is said to be an F -congruent number for all τ iff α is an F -congruent number with respect to τ, ∀τ . 30 Our goal is to generalize the concept of Q-congruent number to F congruent number and give criteria for α to be an F -congruent number. Note that there are two generalizations: “F -congruent number with respect to a (fixed) τ ” and “F -congruent number ∀τ .” These are distinct as as the following example shows: Example √ √ √ Let F = Q , α = 78 + 58 2 , X = 3 + 4 2 , Y = 20 + 12 2 , √ √ √ Z = 21 + 12 2 , τ (a + b 2) = a + b 2 and observe that α is an F congruent number with respect to τ but α is not totally positive. Important: From now on we fix a τ and by “positive” we mean “τ positive” and by “F-congruent number” we mean “F-congruent number with respect to τ .” Problem: Find criteria for α to be an F -congruent number. Proposition 1. α ∈ F is an F-congruent number ⇒ ∀x ∈ F , x2 α is an F-congruent number. Proof. 31 By definition, ∃X, Y, Z > 0 such that X 2 + Y 2 = Z 2 and XY = 2α. Multiplying both equalities by x2 we get that x2 α is an F -congruent number corresponding to the triple xX, xY, xZ by choosing x > 0. Define an equivalence relation as follows: α1 ∼ α2 ⇔ ∃x ∈ F ∗ such that α1 = x2 α2 Definition. α ∈ OF is said to be square free iff 6 ∃P prime ideal in OF such that P 2 | (α). In particular, any unit of OF is square free. Proposition 2. ∀α ∈ F , α > 0 , ∃α0 ∈ OF , α0 > 0 such that α ∼ α0 Proof. Multiply α by the square of its denominator. Important: From now on when we want to find out if α ∈ F , α > 0 is an F-congruent number we may assume that α ∈ OF , α > 0 and square free. 32 Lemma 1. Let Eα be the elliptic curve y 2 = x3 − α2 x. All the points of Eα (F ) of order 2 are {(0, 0), (α, 0), (−α, 0)} Proof. Let O 6= P = (xP , yP ) ∈ Eα (F ) , 2P = O. Cf. [Kob],pg.34 x2P = x2 +α2 ( P2yP )2 thus 2P = O ⇒ yP = 0 ⇒ x3P − α2 xP = 0 ⇒ xP ∈ {0, ±α}. Conversely any (x, 0) ∈ Eα (F ) has order two. QED Proposition 3. Let Eα be the elliptic curve y 2 = x3 −α2 x. If ∃P ∈ Eα (F ) such that 2P 6= O then α is an F-congruent number. Proof. Replacing P by 2P we may assume that P = (x, y) with x a square in F , in particular x > 0. Because 2P 6= 0, we have that y 6= 0. Indeed, assuming by contradiction y = 0, we get, from y 2 = x3 − α2 x, that P = (x, y) ∈ {(0, 0), (α, 0), (−α, 0)} which are elements of order two in Eα (F ). Eventually replacing P = (x, y) by −P = (x, −y) we may assume that 33 y > 0. Let X = (x+α)(x−α) , Y = 2α xy . Note that XY = 2α and X 2 + Y 2 = y 2 2 ) 2 ( (x +α ) . We have that Y > 0. Also X > 0 from XY = 2α. Taking y 2 2 ) Z = (x +α > 0 we get a triple X, Y, Z > 0 defining α to be an F y congruent number. QED Fix α ∈ F ∗ . Proposition 4. If F/Q Galois then ϕ(| Eα (F )tors |) ≤ 2[F : Q] where ϕ denotes the totient Euler function. Proof. Denote K = Q(i). Let d be the density function as defined in [N euk]. Let Y = {(p) prime ideal of Z such that p ≡ 3(mod 4) and p splits completely in OF }. We have that F ∩ K = Q ⇒ d(Y ) = 12 [F 1:Q] using [Neuk], cor(13.6)pg547. Let m =| Eα (F )tors |. 34 Claim m | p + 1 , ∀(p) ∈ Y but finitely many Proof of Claim Here we denote points on the elliptic curve with projective coordinates, (x, y, z) satisfying y 2 z = x3 − n2 xz 2 so that we may assume Eα (F )tors = Eα (OF )tors . For any prime ideal P ∈ OF we define the “reduction mod P” map ϕP : Eα (F )tors = Eα (OF )tors → Eα (OF /P )tors (x, y, z) 7→ (x̄, ȳ, z̄) For every (p) ∈ Y fix P ideal of OF dividing pOF and denote ϕp = ϕP . Because p splits completely in OF we have that OF /P = Fp . From [Kob] page 40 we get | Eα (Fp )tors |= p + 1 (here we use p ≡ −1(mod 4)). Finally, for all but finitely many (p) ∈ Y we have that ϕp is injective, by using a similar argument to that one in [Kob] pp 44-45: first we notice that (x̄1 , ȳ1 , z̄1 ) = (x̄2 , ȳ2 , z̄2 ) iff P divides the gcd(y1 z2 −y2 z1 , x2 z1 −x1 z2 , x1 y2 − y2 x1 ) then, because only finitely many P ’s divide this gcd, we conclude that ϕp is injective for all but finitely many p’s. 35 We proved the claim, that is p ≡ −1(mod m) , ∀(p) ∈ Y but finitely many Let A = {p ∈ Z , p ≡ −1(mod m)}. We have d(A) ≥ d(Y ) = 21 [F 1:Q] . On the other hand Dirichlet density theorem (cf. [Neuk] pg.54) implies 1 that d(A) = ϕ(m) . QED. Corollary 1. If [F : Q] = 2 then | Eα (F )tors |∈ {4, 8, 12} Proof. It follows from Proposition 4 and Lemma 1. Proposition 5. If [F : Q] = 2 and α is not a square in F then | Eα (F )tors |6= 8 Proof. By contradiction; assume, via Lemma1, that ∃P = (x, y) ∈ Eα (F )tors x2 +α2 of order four. Then 2P has order two thus y2P = 0 and x2P = ( P2yP )2 ∈ {0, ±α} that is it is equal to α thus α is a square in F , contradiction. 36 QED Proposition 6. 3 does not divide | Eα (F )tors |. Proof. In our case we know that 3 is inert in K = Q[i] so the associated elliptic curve is supersingular there. Looking at the representation of GF on E[3] one has the following possibilities: (i) 3 is inert in F , in which case the representation is irreducible; (ii) 3 splits in F , in which case the representation splits as χ ⊕ χ with χ a non-trivial character of GF . Consequently, no vector (other than zero) in E[3] is fixed by GF , and so there is no nonzero point of order 3 in E(F ). QED Corollary 2. If [F : Q] = 2 and α is not a square in F then | Eα (F )tors |= 4 Proof. It follows from Cor 1, Prop 5 and Prop 6. 37 Lemma2. ∃ a bijection between S := {0 < X < Y < Z ∈ F , X 2 + Y 2 = Z 2 , 1 XY = α} 2 and T := {x ∈ F ∗ , x, x + α, x − α ∈ F 2 } Proof. The following two functions S −→ T Z (X, Y, Z) 7→ x = ( )2 2 and T −→ S x 7→ (X = √ x+α− √ x − α, Y = √ x+α+ √ √ x − α, Z = 2 x) are inverse each other. QED Proposition 7. If [F : Q] = 2 and α is not a square in OF TFAE: (i) ∃P ∈ Eα (F ) such that P has infinite order; 38 (ii)∃P ∈ Eα (F ) such that 2P 6= O; (iii)α is an F -congruent number. Proof. (i)⇒ (ii) trivial; (ii)⇒(iii) see Proposition 3; (iii)⇒(i) for this implication only we need the hypothesis [F : Q] = 2 and α is not a square in OF . If α is an F -congruent number then by Lemma2 ∃x ∈ F ∗ such that y := p x(x + α)(x − α) ∈ F that is (x, y) ∈ Eα (F ). We claim that (x, y) has infinite order in Eα (F ). Assume it doesn’t, then (x, y) ∈ Eα (F )tors which by Cor2 and Lemma1 is {O, (0, 0), (α, 0), (−α, 0)}. Because (x, y) 6= O , it follows that y = 0 and x ∈ {0, ±α} but x ∈ F ∗ ∩ F 2 thus α = x is a square, contradiction. QED 39 Bibliography [Del] P. Deligne, Sommes de Gauss Cubiques et Revetements de SL(2), d’apres S. J. Patterson, In: Lecture Notes in Mathematics, 770, pp. 244277, Springer, 1980. [Garret] P. B. Garret, Holomorphic Hilbert Modular Forms, Wadsworth & Brooks/Cole Mathematics Series, 1990. [Kob] N. Koblitz, Introduction to Elliptic Curves and Modular Forms, Springer, 1984 [Neuk] J. Neukirch, Algebraic Number Theory, Springer, 1999. [Serre-Stark] J. P. Serre & H. M. Stark, Modular Forms of weight 1/2, In: Modular functions of one variableVI. Lecture Notes in Mathematics, vol. 627, pp. 27-68.Berlin Heidelberg New York: Springer 1977. [Shi] G. Shimura, On Hilbert Modular Forms of Half-integral Weight, In: Duke Mathematical Journal, vol. 55, No. 4. December 1987. [Shi2] G. Shimura, On Eisenstein Series of Half-Integral Weight, In: Duke Mathematics Journal, vol. 52, No.2, June 1985 [Tunnel] J. B. Tunnel, A classical Diophantine Problem and Modu- 40 lar Forms of Weight 3/2, In: Inventiones mathematicae 72, pp. 323-334, Springer-Verlag 1983. [Vas] L. N. Vaserstein On the Group SL(2) over Dedekind Rings of Arithmetic Type, In: Math. Sbornik vol. 18, 1972  Contents Acknowledgement Abstract 1 Introduction 2 Preliminaries 3 The main result 4 Examples 5 Towards the congruent number problem Bibliography over F ili iv 11 24 29 39 il (©€)2005 Sever Achimescu All Rights Reserved HILBERT MODULAR FORMS OF WEIGHT 1/2 Thesis by Sever Achimescu In Partial Fulfillment of the Requirements for the Degree of Doctor of Philosophy California Institute of Technology Pasadena, California 2005 (Defended September 27, 2004) HILBERT MODULAR FORMS OF WEIGHT 1/2 Thesis by Sever Achimescu In Partial Fulfillment of the Requirements for the Degree of Doctor of Philosophy California Institute of Technology Pasadena, California 2005 (Defended September 27, 2004) ii c °2005 Sever Achimescu All Rights Reserved iii Acknowledgement I am grateful to my advisor, Professor Dinakar Ramakrishnan, for his guidance and encouragement. I thank Caltech for its financial support. iv Abstract Let H= X M 12 (N , χ) N ,χ be the space of Hilbert modular forms of half integral weight of all levels N and characters χ. We denote by ϕN : OF → C a periodic function of period N . Let Θ be the C-linear space of the functions f : H × H → C, f (z) = X X t 0 ϕNt (ξ)exp(tπi(ξ 2 z1 + ξ 2 z2 )) ξ∈OF where, for each f , t ∈ OF runs through a finite subset of totally positive integers of F . Main Theorem. H=Θ Using this theorem, for some fixed F ’s, an explicit basis can be found. Some examples are given in Chapter4. v Contents Acknowledgement iii Abstract iv 1 Introduction 1 2 Preliminaries 4 3 The main result 11 4 Examples 24 5 Towards the congruent number problem over F 29 Bibliography 39 1 Chapter 1 Introduction Classically, a modular form of weight k ∈ Z for a congruent subgroup µ ¶ a b , N |c} < SL2 (Z) is a holomorphic function Γ := Γ0 (N ) := {γ = c d f on the upper-half plane H := {z ∈ C , Imz > 0} satisfying µ f (γz) = (cz + d)k f (z) , ∀γ = a b c d ¶ ∈Γ and an additional property called “holomorphicity at the cusps.” Using the natural inner product, namely, the Petterson inner product on the C-vector space of modular forms, we get finite a dimensional Hilbert space for each pair (Γ, k). There are Hecke operators {T (p)}p6 |N prime which preserve these spaces. A generalization is when we allow k ∈ Z[ 12 ] and we get the half integral weight modular forms. Shimura introduced a family of corresponding Hecke operators {T (p2 )}p prime . In [Serre-Stark], Serre and Stark proved a theorem which gives explicit basis consisting of theta functions for some spaces of half integral weight modular forms. 2 The Hilbert modular forms (of half integral weight) are modular forms (of half integral weight) of several variables. The main result in my dissertation is an analog, for certain real quadratic fields of class number one, of the theorem of Serre-Stark which gives algorithms to compute basis of spaces of half integral weight modular forms over Q. An alternative proof of a part of the Serre-Stark theorem, which was outlined (using adellic representations) by Pierre Deligne in a letter to Jean Pierre Serre, is expanded and generalized in this thesis to some quadratic fields of class number one in Chapter 3. Some of the results of Serre and Stark over Q, which Deligne implicitly assumed, require different arguments over F . In particular, one need to use certain nontrivial, structural results of Goro Shimura on Hilbert modular forms of half integral weight. There is already a vast generalization of the Serre-Stark theorem in a slightly different form, due to Shimura, valid over all totally real fields, but this method seems to only give generators, not a basis. The result in Chapter 3 is in a more explicit form and hopefully more suitable for applications. See Chapter 4 for examples. 3 In Chapter 5 we define the congruent number problem over F and we take a few steps towards generalizing Tunnell’s results over Q. Here the √ generalization is not trivial, since already for Q( 2) there are examples of √ Q( 2)-congruent numbers α such that the associated elliptic curve Eα : √ y 2 = x3 − α2 x over Q( 2) has zero rank; α corresponds to torsion points √ which Eα acquires over Q( 2). 4 Chapter 2 Preliminaries Let F be a totally real number field with class number one and dis√ √ criminant D; F = Q( d0 ) = Q( D) with d0 square free and such that √ OF = Z[ d0 ], where OF denotes the ring of integers of F . The two √ √ real embeddings are a + b d0 7→ a ± b d0 . The discriminant (of F ) is √ DF/Q = d0 , the different is D := DF/Q = 2 d0 OF and the inverse different is D−1 = 2√1d OF . 0 For P prime ideal of OF we denote P = $OF , $ ∈ OF . µ ¶ a b Put SL2 (OF ) = { , a, b, c, d ∈ OF , ad − bc = 1}. For ξ = c d √ √ m + n d0 ∈ OF we denote ξ 0 = m − n d0 its conjugate. For N an ideal in OF we denote µ a b Γ0 (N ) = { c d ¶ ∈ SL2 (OF ), c ∈ N }. Put H = {z ∈ C, Imz > 0} and z = (z1 , z2 ) ∈ H × H. Γ0 (N ) acts on H × H as follows: first we view Γ0 (N ) ,→ GL2 (R) × 5 GL2 (R) via µ a b c d ¶ µ a b 7 ( → c d ¶ µ 0 a , c0 ¶ b0 ) d0 then GL2 (R) × GL2 (R) acts on H × H componentwise: µ a c b d ¶ µ 0 a × c0 b0 d0 ¶ az1 + b a0 z2 + b0 , (z1 , z2 ) → 7 ( , ) cz1 + d c0 z2 + d0 All square roots of numbers in C − {x < 0} are to be taken in the right half-plane. Define θ : H × H → C as follows: θ(z) := X exp(πi(ξ 2 z1 + ξ 02 z2 )) ξ∈OF Observe that θ is holomorphic on H × H. Recall that a function f on an open set in Cn is said to be holomorphic iff it has an absolutely convergent power series expansion in each compact polydisc in that open set. We have |θ(z)| = | X exp(πiξ 2 z1 )exp(πiξ 02 z2 )| ≤ ξ∈OF | X ξ∈OF 1 exp(2πiξ 2 z1 )| 2 | X ξ∈OF 1 exp(2πiξ 02 z2 )| 2 6 Because the one-dimensional theta function H → C , z 7→ X exp(2πin2 z) n∈Z is holomorphic, the series in the righthand side of the above inequality converge absolutely on each compact (on H), therefore the series defining θ converges absolutely on each compact (in H × H). For a prime ideal pOF and β ∈ OF we define a quadratic symbol ( ( β ) = +1 ifβ is a square in (OF /pOF )∗ pOF β ) = −1 ifβ is a non square in (OF /pOF )∗ pOF ( β ) = 0 if β ∈ pOF pOF We extend it by multiplicativity to all nonzero ideals of OF . For 0 6= α ∈ β β OF , write ( α ) = ( αO ). F For γ ∈ Γ0 (N ) we consider the automorphy factor of weight 12 , denoted h(γ, z), introduced in [Shi2], Proposition 1.2 pg. 285. The definition is intrinsic, though not explicit. This automorphy factor is a holomorphic 7 function on H × H satisfying h(γ, z)2 = t(cz1 + d)(c0 z2 + d0 ) where t is a root of unity. A multiplicative character modulo N is a group homomorphism χ : (OF /N )∗ → C∗ . We also view a character χ as a periodic function on OF by putting χ(ν) = 0, ∀ν ∈ N . Definition. A holomorphic Hilbert modular form of weight 12 , character χ and level N is a function f : H × H → C which is holomorphic, both on H × H and at the cusps, and satisfies µ f (γz) = χ(d)h(γ, z)f (z) , ∀γ = a b c d ¶ ∈ Γ0 (N ) , ∀z ∈ H × H Note that, since in our case [F : Q] > 1, by the Koecher principle, the condition “holomorphic at the cusps” is superfluous. The definition of Hilbert modular forms of weight k2 , with k odd integer, is similar; but with h(γ, z) replaced by h(γ, z)k . We denote the C-vector space of holomorphic Hilbert modular forms of weight 12 , character χ and level N by M 12 (N , χ). Following [Shi1], for f, g ∈ M 12 (N , χ) we define their inner product 8 < f, g > by Z < f, g >=| Γ\H × H | 1 −1 f (z)g(z)(y1 y2 ) 2 dH (z) Γ\H×H where y =Imz and dH (x + iy) = (y1 y2 )−2 dx1 dx2 dy1 dy2 . This integral converges if one of the forms is a cusp form. But in fact < f, g > makes sense, as noted by Deligne, for all f, g in M 12 (N ). Given a function f : H × H → C we denote ft : H × H → C, ft (z) = f (tz) where t totally positive, t ∈ OF . Also, for t totally positive, t ∈ OF , we define a “quadratic character” χt as follows: for d ∈ OF , d 6= 0, χt (d) = ( t0 ) if t0 ≡ 1(mod4) d χt (d) = ( 4t0 ) otherwise d χt (u) = 1 , ∀u ∈ (OF )∗ where t0 denotes the square-free part of N (t). For a character ψ modulo N , define θψ : H × H → C as follows: θψ (z) := X ξ∈OF ψ(ξ)exp(πi(ξ 2 z1 + ξ 02 z2 )) 9 Theorem. θψ ∈ M 12 (4(cond ψ)2 , ψ). Proof. This is Lemma 4.3 pg. 784 of [Shi1]. In particular, θ ∈ M 12 (4OF ). Proposition 1. Let t ∈ OF be totally positive. Then f ∈ M 12 (N , χχt ) ⇒ ft ∈ M 12 (tN , χ) In particular, since χ2t = 1, f ∈ M 12 (N , χ) ⇒ ft ∈ M 12 (tN , χχt ) Proof. Need to prove µ ¶ a b ft (γz) = χ(d)h(γ, z)ft (z) , ∀γ = ∈ Γ0 (tN ) , ∀z ∈ H × H c d µ ¶ µ ¶ a b a bt Fix γ = ∈ Γ0 (tN ) and put γt = c ∈ Γ0 (N ). By hypothc d d t esis, f ∈ M 12 (Γ0 (N ), χχt ) thus f (γt tz) = χ(d)χt (d)h(γt , tz)f (tz) , ∀z ∈ H × H The conclusion follows by noticing that γt tz = tγz and using the following 10 Lemma. χt (d)h(γt , tz) = h(γ, z) , ∀z ∈ H × H Proof: The squares of both sides are equal, by [Shi2] pg. 286. It suffices to check the sign. Equivalently, since h(γ, z) = θ(γz) θ(z) , it suffices to check the Proposition for f = θ. But this is a known fact. 11 Chapter 3 The main result Let H= X M 12 (N , χ) N ,χ be the space of Hilbert modular forms of half integral weight of all levels N and characters χ. We denote by ϕN : OF → C a periodic function of period N , i.e. ϕ(x) = ϕ(x + α) , ∀x ∈ OF , ∀α ∈ N . Let Θ be the C-linear space of the functions f : H × H → C, f (z) = X X t 0 ϕNt (ξ)exp(tπi(ξ 2 z1 + ξ 2 z2 )) ξ∈OF where, for each f , t ∈ OF runs through a finite subset of totally positive integers of F . Main Theorem. H=Θ Proof: 12 The inclusion Θ⊆H follows from Proposition 1 of the previous chapter and the following Lemma. The function z 7→ X 0 ϕN (ξ)exp(πi(ξ 2 z1 + ξ 2 z2 )) ξ∈OF is a Hilbert modular form of weight 12 possibly with a character. Proof: This is a corollary of the Theorem on pg.154 from [Garrett]. Now we prove the key inclusion H⊆Θ Let Af denote the finite adeles of F . It is known that (cf. [Del], pg. f 2 (Af ) of SL2 (Af ) acts on H, preserves 259) the metaplectic 2-covering SL the scalar product and leaves Θ stable. Under this action, H decomposes into a direct sum of irreducible representations. Let Hi be one of them. We want to prove that Hi is contained in Θ. 13 Proposition 1. For suitable (N , χ), Hi ∩ M 12 (N , χ) 6= 0. Proof: There exists a family of compact open subgroups {K0 (N )}N , which is also a fundamental system of neighborhoods of the unity of SLg 2 (Af ), such that Γ0 (N ) = K0 (N ) ∩ SL2 (Af ), ∀N . Here we have used the fact that F has class number 1. There is a character χ̃ whose restriction to Γ0 (N ) is K0 (N ),χ̃ χ. If V is a vector space on which SLg for the 2 (Af ) acts, we write V subgroup of vectors v ∈ V such that kv = χ̃(k)v. One knows H K0 (N ),χ̃ = M 21 (N , χ) We have K (N ),χ̃ H K0 (N ),χ̃ = ⊕i Hi 0 Now fix any Hi . Then the admissibility of Hi as an SLg 2 (Af )-module implies K (N ),χ̃ that there exists (N , χ̃) such that Hi 0 6= 0. Therefore there exists (N , χ) such that Hi ∩ M 21 (N , χ) 6= 0. QED One knows that for every prime ideal P of OF , including those dividing N , there is a Hecke operator T (P 2 ) acting on M 12 (N , χ). They come 14 from the action of SL2 (Af ). If P, Q are primes, then T (P 2 ) commute with T (Q2 ). The operators T (P 2 ) preserve the scalar product and have a common eigenvector in M 12 (N , χ). Proposition 2. We have : 0 f (z) = Σµ∈OF a(tµ2 )exp(tπi(µ2 z1 + µ 2 z2 )) where t ∈ OF , t is totally positive, t|N , and, ∀α ∈ OF , a(αµ2 ) = a(α)ψ(µ) if (µ, N ) = 1 a(αP 2 ) = cP a(α) if P | N where ψ is a multplicative character mod 2N and T (P 2 )f = cP f . Proof: This is a consequence of the following lemmas: Lemma 1. We have (i) There is a basis of M 12 (N , χ) consisting of forms whose coefficients belong to a number field, 15 (ii) If f (z) = P ξ∈OF a(ξ)exp(πi(ξ 2 0 z1 + ξ 2 z2 )) belongs to M 12 (N , χ) and the a(ξ) are algebraic numbers then the a(ξ) have bounded denominators (i.e., there exists a nonzero integer D such that Da(ξ) is an algebraic integer for all ξ). Proof of Lemma 1: The analogous results for integral weight forms are known. Also it is known that there is an orthogonal decomposition M 12 (N , χ) = S 12 (N , χ) ⊕ E 12 (N , χ) for the scalar product,where S 12 (N , χ) denotes the space of cusp forms and E 21 (N , χ) denotes the space of Eisenstein forms. In [Shi2] it is proved that there is a basis of E 12 (N , χ) consisting of forms whose coefficients belong to a number field and have bounded denominators. For both (i) and (ii), it suffices to prove the analogous results for S 21 (N , χ). By Prop 5.2, 5.3 from [Shi1], S 21 (N , χ) is spanned by Hecke eigenforms for T (P 2 ) for almost all P . Write S 12 (N , χ) = Θ0 ⊕ Θ⊥ 0 where Θ0 is the subspace of S 21 (N , χ) generated by theta series and Θ⊥ 0 is the subspace perpendicular to Θ under the scalar product a corollary of the result of this chapter is S 21 (N , χ) = Θ0 but we don’t know it yet. The analogous 16 statements of (i) and (ii) hold for Θ0 by the explicit knowledge of theta series. If S 12 (N , χ) = Θ0 then we are done. If not, then there exists a Hecke eigenform f in Θ⊥ 0 . As in [Shi1], pg. 816, we associate to f an integral weight form h. It may happen that different f ’s correspond to the same h. We denote by S 12 (N , h, χ) the subspace of S 12 (N , χ) generated by those f ’s corresponding to h. By prop 8.9(2) of [Shi1], S 12 (N , h, χ) has a basis of forms whose coefficients belong to a number field and have bounded denominators. QED Lemma 2. Let f (z) = P 0 2 2 ξ∈OF a(ξ)exp(πi(ξ z1 + ξ z2 )) be a nonzero element of M 12 (N , χ) and let P be a prime with P 6 | N . Put P = $OF . Assume that f | T (P 2 ) = cP f , with cP ∈ C. Let µ ∈ OF such that P 2 6 | µ. Then: (i) a(µP 2n ) = a(µ)χ($)n ( Pµ )n for every n ≥ 0 integer, 1 (ii) If a(µ) 6= 0 then P 6 | µ and cP = χ($)( Pµ )(1 + N (P ) ). Proof of Lemma 2: Cf. Prop5.4 pg. 788 in [Shi], the operator T (P 2 ) maps forms with 17 algebraic coefficients into themselves. Hence cP is algebraic and we may assume that the coefficients a(ξ) of f are algebraic numbers. Fix µ and consider the power series A(T ) = ∞ X a(µP 2n )T n n=0 where T is an indeterminate. As in the proof of Theorem 5.5, pg. 790 [Shi] we have A(T ) = a(µ) 1 − αT (1 − βT )(1 − γT ) where α = χ(p)N (P )−1 ( Pµ ) and β + γ = cP , βγ = χ($2 )N (P )−1 . If a(µ) = 0 then A(T ) = 0 thus a(µP 2n ) = 0 for all n ≥ 0 therefore (i). Assume now a(µ) 6= 0. Using Lemma 1(ii), A(T ) converges in the P -adic topology, if we view A(T ) as a nonzero P -adic rational function of T , in the P -adic unit disc U defined by | T |P < 1; hence A(T ) cannot have a pole in U . However, since βγ = χ($2 )P −1 , either β −1 or γ −1 belongs to U ; assume it is β −1 . Now A(T ) holomorphic at β −1 implies that the factors 1 − βT and 1 − αT cancel each other. Thus α = β and A(T ) = a(µ) (1 − γT ) 18 thus a(µP 2n ) = γ n a(µ) Now βγ 6= 0 ⇒ α 6= 0 ⇒ $ 6 | µ. Moreover, γ= βγ χ($2 )N (P )−1 µ = = χ($)( ) µ α χ($)N (P )−1 ( $ ) $ which implies (i). Finally, (ii) follows from cP = β + γ = α + γ. QED Lemma 3. Let f (z) = P ξ∈OF a(ξ)exp(πi(ξ 2 0 z1 + ξ 2 z2 )) be a nonzero element of M 21 (Γ0 (N , χ)) and let N 0 be a multiple of N . Assume that, for all P 6 | N 0 , we have f | T (P 2 ) = cP f , with cP ∈ C. Then there exists a unique (up to multiplication by a unit) square-free nonzero t ∈ OF such that a(ξ) = 0 if ξt is not a square in OF . Moreover: (i) t | N 0 , 1 0 (ii) cP = χ($)( $t )(1 + N (P ) ) if $ 6 | N , (iii) a(ξµ2 ) = a(ξ)χ(µ)( µt ) if (µ, N )0 = 1. Proof of Lemma 3: Let µ, µ0 ∈ OF such that a(µ) 6= 0 and a(µ0 ) 6= 0. For each P prime in 19 OF , P p 6 | N 0 µµ0 we have, by Lemma 2(ii), µ 1 µ0 1 χ($)( )(1 + ) = cp = χ($)( )(1 + ) $ N (P ) $ N (P ) thus, for all $ prime in OF , $ 6 | N 0 µµ0 , µ0 µ ( ) = ( ). $ $ This implies, for each prime q dividing µ and not dividing µ0 , denoting by vq (µ) the exponent of q in µ, that q vq (µ) ( )=1 $ for all $ prime in OF , $ 6 | N 0 µµ0 . In particular, for a $ large enough, this implies that vq (µ) is even. Therefore we may write µ = tµ21 , µ0 = tµ02 1 with t square free. The uniqueness of t follows repeating the argument for another pair µ, µ00 instead of µ, µ0 . We proved the first part of the Lemma 3. Now we prove (i) and (ii). Fix a prime p not dividing N 0 . Fix µ such that a(µ) 6= 0 and write µ = tµ21 = t$2n µ20 with $ not dividing µ0 . We apply Lemma 1 2 2(i) to tµ20 and we get a(µ) = a(tµ20 )χ($)( $t )(1 + N (P ) ) thus a(tµ0 ) 6= 0. By Lemma 2(ii), $ does not divide tµ20 hence $ does not divide t. So 20 1 far Lemma 3(i) proved. Now, by Lemma 2(ii), cP = χ($)( $t )(1 + N (P )) thus Lemma 3(ii) proved. Now we prove Lemma 3(iii). Because both sides are multiplicative in µ, it suffices to check it for µ = $ with $ prime not dividing N 0 . Writing ξ = ξ0 $2n with $2 6 | N and applying Lemma 2(i) we get qed. QED Lemma 4. We have a(µ$2 ) = cP a(µ) if $ | N . Proof of Lemma 4: It is a straightforward corollary of [Shi], p.788, Prop 5.4. Consider now g(z) = Σξ∈OF ,(ξ,N )=1 a(tξ 2 )exp(tπi(ξ 2 z1 + ξ 02 z2 )) Proposition 3. (i) g 6= 0; (ii) g ∈ Θ; (iii) g is (up to a scalar factor) the transform of f (z) = Σξ∈OF a(tξ 2 )exp(tπi(ξ 2 z1 + ξ 02 z2 )) 21 by ΠP |N LP where P = $OF with $ totally positive and LP is the operator which transforms h(z) into h(z) − cP h($2 z1 , $02 z2 ). (iv) Lp can be defined by the element µ 1 − cP $−1 0 0 $ ¶ g2 (FP )], where FP is the completion of F at P . of the group ring Z[SL (v) g ∈ Hi . Proof: (i)g 6= 0 because f 6= 0; (ii) Define ϕN (ξ) = a(tξ 2 ) if (ξ, N ) = 1 and ϕN (ξ) = 0 otherwise. ϕ is periodic of period N because ψ is so. Therefore g(z) = Σξ∈OF ϕN (ξ)exp(tπi(ξ 2 z1 + ξ 02 z2 )) ∈ Θ. (iii) Fix P = $OF | N with $ totally positive. We have (LP f )(z) = Σξ∈OF a(tξ 2 )exp(tπi(ξ 2 z1 + ξ 02 z2 ))− cp Σξ∈OF a(tξ 2 )exp(tπi(ξ 2 $2 z1 + ξ 02 $02 z2 )) = Σξ∈OF a(tξ 2 )exp(tπi(ξ 2 z1 + ξ 02 z2 ))− 22 Σξ∈OF a(t($ξ)2 )exp(tπi(ξ 2 $2 z1 + ξ 02 $02 z2 )) = Σξ∈OF ,(ξ,$)=1 a(tξ 2 )exp(tπi(ξ 2 z1 + ξ 02 z2 )) in other words LP “deletes” from the expansion of f those terms such that $ | ξ. The conclusion follows noticing that ΠP |N LP means composition of operators. (iv) f 2 (A) = SL f 2 (F∞ ) × SL f 2 (Af ) with SL2 (F ) embeds diagonally in SL f 2 (A) as (x∞ , xf ) componentwise multiplication. Write the elements of SL f 2 (F∞ ) and xf ∈ SL f 2 (Af ). If γ ∈ SL2 (F ) then γ∞ = γ with x∞ ∈ SL γf = γ and γ := (γ, γ) = (γ, 1)(1, γ). µ −1 ¶ µ −1 $ 0 $ Let γ = ∈ SL2 (F ). Also note that 0 $ 0 f 2 (FP ) ⊆ SL f 2 (Af ). SL 0 $ ¶ ∈ Since the automorphic forms are left invariant g2 (Af ) is the under the diagonal action of SL2 (F ), the action of γ in SL f 2 (Af ) sends h(z) to inverse of the action of (γ, 1). By part (iii), γ ∈ SL h($2 z1 , $02 z2 ). Since LP (h)(z) = h(z1 , z2 ) − cP h($2 z1 , $02 z2 ), the conclusion follows. (v) follows from (iii),(iv) and the fact that Hi is invariant. 23 Therefore Hi ∩ Θ 6= ∅. Since Hi is irreducible, this implies Hi ⊆ Θ. QED 24 Chapter 4 Examples √ √ Consider F = Q[ 2] and N = 8Z[ 2]. From the proofs in Chapter 4 follows that √ n 1 (2 Z[ M 2], χ) is genχ 2 P erated by {(gψ )t } where ψ is an even primitive character modulo N and t is a totally positive integer of F dividing 2. √ The coset representatives of OF /N are {a + b 2 + N } with a, b ∈ √ {0, ..., 7} so that OF /N has 64 elements. For α = a + b 2, we denote √ [α] := a + b 2 + N . Since OF /N is finite, the elements of (OF /N )∗ are √ exactly the nonzero divisors. Notice that [a + b 2] is a zero divisor in OF /N if and only if a is even. Thus (OF /N )∗ has 32 elements. We list the orders of all elements (and we conclude that (OF /N )∗ has exponent 8): √ [3], [5], [7] and [a + 4 2], a odd, have order 2; √ √ [a + 2 2], [a + 6 2], a odd, have order 4; √ [a + b 2], a, b odd, has order 8. Thus, counting the elements of order 2 in an abelian group of order 25 32 and exponent 8, we conclude that (OF /N )∗ ∼ = Z/8 × Z/2 × Z/2 since they have the same maximum number of 7 elements of order 2. Any other abelian group of order 32 and exponent 8 has less than 7 elements of order 2. It is straightforward to check that [1 + √ 2] generates a subgroup of √ order 8, that [−1] = [7] generates a subgroup of order 2, that [3 + 4 2] generates a subgroup of order 2, that [7] 6∈< [1 + √ 2] >, and that [3 + √ √ √ 4 2] 6∈< [7] > × < [1 + 2] >. It follows that (OF /N )∗ =< [1 + 2] > √ × < [7] > × < [3 + 4 2] >. The character group of (the finite abelian group) (OF /N )∗ is therefore < ϕ0 > × < ϕ1 > × < ϕ2 > where √ < ϕ0 >≡ 1 on < [7] > × < [3 + 4 2] >, hence < ϕ0 > can be viewed as a character of < [1 + √ 2] >; < ϕ1 >≡ 1 on < [1 + √ √ 2] > × < [3 + 4 2] >, hence < ϕ1 > can be viewed as a character of < [7] >; < ϕ2 >≡ 1 on < [7] > × < [1 + √ 2] >, hence < ϕ2 > can be viewed √ as a character of < [3 + 4 2] >. Any character ψ of (OF /N )∗ can be written as ψ = ϕj00 ϕj11 ϕj22 with 26 j0 ∈ {0, ..., 7}, j1 ∈ {0, 1}, j2 ∈ {0, 1}. Recall that all the units of OF are √ {±(1 + 2)k , k ∈ Z}, so that ψ is even if and only if ψ ≡ 1 on < [7] > × < [1 + √ 2] >, that is j0 = 0 and j1 = 0, therefore all the even characters of (OF /N )∗ are {1, ψ := ϕ2 }. So far we got that {θ, θψ , θ2 , (θψ )2 , θ√2 , (θψ )√2 } is a set of generators for √ 1 (8Z[ M 2], χ). Since these six generators have six distinct charχ 2 P acters, we conclude: √ {θ} is a basis for M 21 (4Z[ 2]); √ {θψ } is a basis for M 12 (4Z[ 2], ψ); √ √ {θ√2 } is a basis for M 12 (4 2Z[ 2], χ√2 ); √ √ {(θψ )√2 } is a basis for M 21 (4 2Z[ 2], ψχ√2 ); √ {θ2 } is a basis for M 12 (8Z[ 2], χ2 ); √ {(θψ )2 } is a basis for M 12 (8Z[ 2], ψχ2 ). A generalization is : √ √ OF = Z[ 2], N = 2n Z[ 2], n ∈ Z, n ≥ 4 With similar arguments we get that : 27 (OF /N )∗ has 22n−1 elements and has exponent 2n ; √ All the elements of the form [a + b 2] of order two are a, a 6= 1 odd √ (there are 2n−1 − 1 of them) and [a + 2n−1 2], a odd (there are 2n−1 of them). √ All the elements of order 2n are [a + b 2] with a, b odd. Counting the elements of order 2, we conclude (OF /N )∗ ∼ = Z/2n × Qn−1 j=1 Z/2. Its group of characters is generated by ϕ0 of order 2n and ϕ1 , ..., ϕn−1 each of order 2. Without loss of generality ϕ1 ≡ 1 on < [−1] >. j n−1 Any even characters ψ can be written as ϕj22 ....ϕn−1 with j2 , ..., jn−1 ∈ {0, 1}. Proposition. A basis for √ n 1 (2 Z[ M 2], χ) is given by χ 2 P j n−1 {(θψ )t , ψ = ϕj22 ....ϕn−1 , j2 , ..., jn−1 ∈ {0, 1} , t | 2n−2 , t > 0}. Remark. A character on OF is the pull-back of a character on Z via norm iff it √ is invariant under the Galois action. For N = 8Z[ 2], the forms θ and θψ 28 have such characters, since every element of order 2 in (OF /N )∗ is invariant under the Galois action. However, the forms {(θψ )√2 , θ√2 } are genuinely new. 29 Chapter 5 Towards the congruent number problem over F Let F be a totally real number field. Below α ∈ F and τ : F → R denotes a field embedding. Definitions: (i) x ∈ F is said to be τ -positive iff τ (x) > 0; (ii) x ∈ F is said to be totally positive iff τ (x) > 0 , ∀τ (iii) α ∈ F is said to be an F -congruent number with respect to τ iff 1 ∃τ −positive X, Y, Z such that X 2 + Y 2 = Z 2 and τ (X)τ (Y ) = τ (α) 2 Note that an F -congruent number with respect to τ is τ -positive. Also, note that by applying τ −1 the last equality becomes 1 XY = α 2 (iv) α ∈ F is said to be an F -congruent number for all τ iff α is an F -congruent number with respect to τ, ∀τ . 30 Our goal is to generalize the concept of Q-congruent number to F congruent number and give criteria for α to be an F -congruent number. Note that there are two generalizations: “F -congruent number with respect to a (fixed) τ ” and “F -congruent number ∀τ .” These are distinct as as the following example shows: Example √ √ √ Let F = Q , α = 78 + 58 2 , X = 3 + 4 2 , Y = 20 + 12 2 , √ √ √ Z = 21 + 12 2 , τ (a + b 2) = a + b 2 and observe that α is an F congruent number with respect to τ but α is not totally positive. Important: From now on we fix a τ and by “positive” we mean “τ positive” and by “F-congruent number” we mean “F-congruent number with respect to τ .” Problem: Find criteria for α to be an F -congruent number. Proposition 1. α ∈ F is an F-congruent number ⇒ ∀x ∈ F , x2 α is an F-congruent number. Proof. 31 By definition, ∃X, Y, Z > 0 such that X 2 + Y 2 = Z 2 and XY = 2α. Multiplying both equalities by x2 we get that x2 α is an F -congruent number corresponding to the triple xX, xY, xZ by choosing x > 0. Define an equivalence relation as follows: α1 ∼ α2 ⇔ ∃x ∈ F ∗ such that α1 = x2 α2 Definition. α ∈ OF is said to be square free iff 6 ∃P prime ideal in OF such that P 2 | (α). In particular, any unit of OF is square free. Proposition 2. ∀α ∈ F , α > 0 , ∃α0 ∈ OF , α0 > 0 such that α ∼ α0 Proof. Multiply α by the square of its denominator. Important: From now on when we want to find out if α ∈ F , α > 0 is an F-congruent number we may assume that α ∈ OF , α > 0 and square free. 32 Lemma 1. Let Eα be the elliptic curve y 2 = x3 − α2 x. All the points of Eα (F ) of order 2 are {(0, 0), (α, 0), (−α, 0)} Proof. Let O 6= P = (xP , yP ) ∈ Eα (F ) , 2P = O. Cf. [Kob],pg.34 x2P = x2 +α2 ( P2yP )2 thus 2P = O ⇒ yP = 0 ⇒ x3P − α2 xP = 0 ⇒ xP ∈ {0, ±α}. Conversely any (x, 0) ∈ Eα (F ) has order two. QED Proposition 3. Let Eα be the elliptic curve y 2 = x3 −α2 x. If ∃P ∈ Eα (F ) such that 2P 6= O then α is an F-congruent number. Proof. Replacing P by 2P we may assume that P = (x, y) with x a square in F , in particular x > 0. Because 2P 6= 0, we have that y 6= 0. Indeed, assuming by contradiction y = 0, we get, from y 2 = x3 − α2 x, that P = (x, y) ∈ {(0, 0), (α, 0), (−α, 0)} which are elements of order two in Eα (F ). Eventually replacing P = (x, y) by −P = (x, −y) we may assume that 33 y > 0. Let X = (x+α)(x−α) , Y = 2α xy . Note that XY = 2α and X 2 + Y 2 = y 2 2 ) 2 ( (x +α ) . We have that Y > 0. Also X > 0 from XY = 2α. Taking y 2 2 ) Z = (x +α > 0 we get a triple X, Y, Z > 0 defining α to be an F y congruent number. QED Fix α ∈ F ∗ . Proposition 4. If F/Q Galois then ϕ(| Eα (F )tors |) ≤ 2[F : Q] where ϕ denotes the totient Euler function. Proof. Denote K = Q(i). Let d be the density function as defined in [N euk]. Let Y = {(p) prime ideal of Z such that p ≡ 3(mod 4) and p splits completely in OF }. We have that F ∩ K = Q ⇒ d(Y ) = 12 [F 1:Q] using [Neuk], cor(13.6)pg547. Let m =| Eα (F )tors |. 34 Claim m | p + 1 , ∀(p) ∈ Y but finitely many Proof of Claim Here we denote points on the elliptic curve with projective coordinates, (x, y, z) satisfying y 2 z = x3 − n2 xz 2 so that we may assume Eα (F )tors = Eα (OF )tors . For any prime ideal P ∈ OF we define the “reduction mod P” map ϕP : Eα (F )tors = Eα (OF )tors → Eα (OF /P )tors (x, y, z) 7→ (x̄, ȳ, z̄) For every (p) ∈ Y fix P ideal of OF dividing pOF and denote ϕp = ϕP . Because p splits completely in OF we have that OF /P = Fp . From [Kob] page 40 we get | Eα (Fp )tors |= p + 1 (here we use p ≡ −1(mod 4)). Finally, for all but finitely many (p) ∈ Y we have that ϕp is injective, by using a similar argument to that one in [Kob] pp 44-45: first we notice that (x̄1 , ȳ1 , z̄1 ) = (x̄2 , ȳ2 , z̄2 ) iff P divides the gcd(y1 z2 −y2 z1 , x2 z1 −x1 z2 , x1 y2 − y2 x1 ) then, because only finitely many P ’s divide this gcd, we conclude that ϕp is injective for all but finitely many p’s. 35 We proved the claim, that is p ≡ −1(mod m) , ∀(p) ∈ Y but finitely many Let A = {p ∈ Z , p ≡ −1(mod m)}. We have d(A) ≥ d(Y ) = 21 [F 1:Q] . On the other hand Dirichlet density theorem (cf. [Neuk] pg.54) implies 1 that d(A) = ϕ(m) . QED. Corollary 1. If [F : Q] = 2 then | Eα (F )tors |∈ {4, 8, 12} Proof. It follows from Proposition 4 and Lemma 1. Proposition 5. If [F : Q] = 2 and α is not a square in F then | Eα (F )tors |6= 8 Proof. By contradiction; assume, via Lemma1, that ∃P = (x, y) ∈ Eα (F )tors x2 +α2 of order four. Then 2P has order two thus y2P = 0 and x2P = ( P2yP )2 ∈ {0, ±α} that is it is equal to α thus α is a square in F , contradiction. 36 QED Proposition 6. 3 does not divide | Eα (F )tors |. Proof. In our case we know that 3 is inert in K = Q[i] so the associated elliptic curve is supersingular there. Looking at the representation of GF on E[3] one has the following possibilities: (i) 3 is inert in F , in which case the representation is irreducible; (ii) 3 splits in F , in which case the representation splits as χ ⊕ χ with χ a non-trivial character of GF . Consequently, no vector (other than zero) in E[3] is fixed by GF , and so there is no nonzero point of order 3 in E(F ). QED Corollary 2. If [F : Q] = 2 and α is not a square in F then | Eα (F )tors |= 4 Proof. It follows from Cor 1, Prop 5 and Prop 6. 37 Lemma2. ∃ a bijection between S := {0 < X < Y < Z ∈ F , X 2 + Y 2 = Z 2 , 1 XY = α} 2 and T := {x ∈ F ∗ , x, x + α, x − α ∈ F 2 } Proof. The following two functions S −→ T Z (X, Y, Z) 7→ x = ( )2 2 and T −→ S x 7→ (X = √ x+α− √ x − α, Y = √ x+α+ √ √ x − α, Z = 2 x) are inverse each other. QED Proposition 7. If [F : Q] = 2 and α is not a square in OF TFAE: (i) ∃P ∈ Eα (F ) such that P has infinite order; 38 (ii)∃P ∈ Eα (F ) such that 2P 6= O; (iii)α is an F -congruent number. Proof. (i)⇒ (ii) trivial; (ii)⇒(iii) see Proposition 3; (iii)⇒(i) for this implication only we need the hypothesis [F : Q] = 2 and α is not a square in OF . If α is an F -congruent number then by Lemma2 ∃x ∈ F ∗ such that y := p x(x + α)(x − α) ∈ F that is (x, y) ∈ Eα (F ). We claim that (x, y) has infinite order in Eα (F ). Assume it doesn’t, then (x, y) ∈ Eα (F )tors which by Cor2 and Lemma1 is {O, (0, 0), (α, 0), (−α, 0)}. Because (x, y) 6= O , it follows that y = 0 and x ∈ {0, ±α} but x ∈ F ∗ ∩ F 2 thus α = x is a square, contradiction. QED 39 Bibliography [Del] P. Deligne, Sommes de Gauss Cubiques et Revetements de SL(2), d’apres S. J. Patterson, In: Lecture Notes in Mathematics, 770, pp. 244277, Springer, 1980. [Garret] P. B. Garret, Holomorphic Hilbert Modular Forms, Wadsworth & Brooks/Cole Mathematics Series, 1990. [Kob] N. Koblitz, Introduction to Elliptic Curves and Modular Forms, Springer, 1984 [Neuk] J. Neukirch, Algebraic Number Theory, Springer, 1999. [Serre-Stark] J. P. Serre & H. M. Stark, Modular Forms of weight 1/2, In: Modular functions of one variableVI. Lecture Notes in Mathematics, vol. 627, pp. 27-68.Berlin Heidelberg New York: Springer 1977. [Shi] G. Shimura, On Hilbert Modular Forms of Half-integral Weight, In: Duke Mathematical Journal, vol. 55, No. 4. December 1987. [Shi2] G. Shimura, On Eisenstein Series of Half-Integral Weight, In: Duke Mathematics Journal, vol. 52, No.2, June 1985 [Tunnel] J. B. Tunnel, A classical Diophantine Problem and Modu- 40 lar Forms of Weight 3/2, In: Inventiones mathematicae 72, pp. 323-334, Springer-Verlag 1983. [Vas] L. N. Vaserstein On the Group SL(2) over Dedekind Rings of Arithmetic Type, In: Math. Sbornik vol. 18, 1972