Hecke eigenforms and representation numbers of quadratic forms LYNNE H. WALLING

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1 ecke eigenforms and representation numbers of quadratic forms YE. WAIG 1. Introduction. When looking for multiplicative relations satisfied by representation numbers of quadratic forms, it is natural to study the effect of the ecke operators on theta series attached to quadratic forms; in this paper we use such theta series to construct ecke eigenforms and thereby obtain relations on weighted averages of representation numbers of quadratic forms. When working over the rationals we obtain the relations r(gen, mm = λ (m r(gen, m a (m,m a>1 χ (aa k 1 r (gen, mm where is an even rank lattice equipped with a positive definite quadratic form, r(gen, n is the average number of times the lattices in the genus of represent n, λ (m is an easily computed constant, χ is a quadratic character associated to, and is a particular sublattice of scaled by 1/m (see Theorem 3.9. We assume herein that we are working with a totally positive quadratic form Q on a vector space V of even dimension 2k over a totally real number field K; for a lattice on V we define the theta series attached to to be the ilbert modular form θ(, τ = x e πit r(q(xτ. a 2 We observe that there is a family of lattices related to which is partitioned into nuclear families such that the ecke operators essentially permute the weighted averages of theta series attached to lattices within a nuclear family. Analyzing these permutations allows us to construct ecke eigenforms, and analyzing the behavior of these eigenforms allows us to infer relations on the weighted averages of representation numbers of lattices within a nuclear family (as described above for K = Q. ote that when the character associated to the lattice is nontrivial, the eigenforms we construct are only eigenforms for the subalgebra of the ecke algebra which is known to map theta series to linear combinations of theta series with the same weight, level, and character. 2. Definitions. et K be a totally real number field of degree n over Q; let O denote its ring of integers and its different. et be an integral ideal, I a fractional 1

2 ideal, k Z +, and χ a numerical character modulo ; we let {( ( a b O I Γ 0 (, I = 1 1 c d I O } : ad bc O, ad bc 0 and we let M k (Γ 0 (, I, χ denote the space of ilbert modular forms of (uniform weight k and character χ for the group Γ 0 (, I. As defined in [8] (see also (2.21 of [4], we have ecke operators T (P : M k (Γ 0 (, I, χ M k (Γ 0 (, IP, χ where P is a prime not dividing 2 ; we also have operators S(J : M k (Γ 0 (, I, χ M k (Γ 0 (, IJ 2, χ where J is any fractional ideal relatively prime to 2 (i.e. ord P J ord P 2 = 0 for all finite primes P. The collection of these operators T (P and S(Q generate a commutative algebra T which we call the ecke algebra. Since the mapping from M k (Γ 0 (, I, χ onto M k (Γ 0 (, αi, χ defined by f f ( α 1 0 (where α 0 is an isomorphism 0 1 which commutes with the operators of T, we set M k (, χ = M k (Γ 0 (, I, χ I where the sum runs over a complete set of strict ideal class representatives I, and we consider T as an algebra of operators on M k (, χ. (otice that f ( α 0 = f ( α whenever α 0, α 0 and αi = α I. We let T 0 be the subalgebra of T which acts on each component space M k (, χ ; thus T 0 is generated as a vector space by all operators of the form T (M 1 T (M s S(J where J is a fractional ideal relatively prime to 2, the M i are integral ideals (not necessarily distinct and αm 1 M s J 2 = O for some α 0. As shown in Theorem 3.1 of [8], a T 0 -eigenform f M k (Γ 0 (, I, χ can be lifted to several linearly independent T -eigenforms F M k (, χ. Remark. The ecke operators T (P defined here differ slightly from those defined in [2] and in [7]; letting T (P denote the latter operator, we have (up to identification of isomorphic spaces via the map f f ( α K/Q (P k/2 T (PS(P 1 = T (P. 2

3 We also have S(P 1 = V (P where V (P is the operator defined in [2] and [7]. et V be a quadratic space of dimension 2k over K with Q a totally positive quadratic form and associated bilinear form B such that B(x, x = Q(x. For a lattice on V we define the theta series attached to by θ(, τ = x πit r(q(xτ e where τ n (and denotes the complex upper half-plane. As shown in [7] (see also [2], θ(, τ is a modular form of (uniform weight k with quadratic character χ for the group Γ 1 0(S(, n = {A Γ 0 (S(, n : det A = 1} where n, the norm of, is the fractional ideal of O generated by 1 2Q(, and S(, the level or stufe of, is the product of (n 1 (n # 1 and perhaps some dyadic primes (see [7]. ere # denotes the dual of ; note that S( is an integral ideal and χ is a quadratic character modulo S(. Also note that a nondyadic prime P divides S( if and only if P is not modular (as defined in [3]. (.B.: The norm n defined here is not the same as the norm defined in [3]. et P be a prime not dividing 2S(; as shown in [7], when /P is hyperbolic we can realize θ(, τ T (P as a linear combination of theta series attached to P-sublattices of, when /P is not hyperbolic we can realize θ(, τ T (P 2 as a linear combination of θ(p, τ and theta series attached to P 2 -sublattices of, and θ(, τ S(P is a constant multiple of θ(p, τ. (A sublattice of is a P-sublattice of if P and /P is a maximal totally isotropic subspace of /P; a P-sublattice of a P- sublattice of is a P 2 -sublattice of if dim /( P =dim /P. Thus the subspace of M k (S(, χ generated by forms whose components are theta series is invariant under the action of the subalgebra T of T where T is generated as an algebra by all operators S(I, T (P 2, and by T (P when /P is hyperbolic. Thus the subspace of M k (Γ 0 (S(, n(, χ M k (Γ 1 0(S(, n(, χ spanned by theta series is invariant under the subalgebra T 0 = T T Constructing the eigenforms. From now on we set = Snd we require that for all dyadic primes P, the localization P is an even unimodular lattice. As remarked in [8], χ = 1 if and only if /P is hyperbolic for all primes P/ 2. The space /P is hyperbolic if and only if ( 1 k disc P = π 2 ɛ where π is some nonzero element of K P and 3

4 ɛ O P such that ɛ is a square modulo P (and disc P denotes the discriminant of P ; since disc P = discv (in O P for all primes P and by 65:19 of [3] discv is a square at an infinite number of primes, /P is hyperbolic for an infinite number of primes P. Definition. A lattice K is in the family of, denoted fam, if K is a lattice on V α for some α 0, α relatively prime to 2, and for every prime P there exists some u P O P such that K u P P P (where denotes isometry; here V α denotes the quadratic space V scaled by α and K u P denotes the lattice K P scaled by u P. We say K fam is in the nuclear family of, denoted fam +, if there exists some u O such that K u is in the genus of. Remark. In the case that χ = 1, the requirement that α be relatively prime to 2 is superfluous: For any β 0, we can find some α 0 such that α is relatively prime to 2 and the ilbert symbols ( α, ( 1 k discv and ( β, ( 1 k discv are equal at all primes P P P dividing 2 (see 63B of [3]. (ote that if P is nondyadic then specifying the square class of β in O P allows us to control (β, ( 1 k discv P ; if P is dyadic then, by the ocal Square Theorem, taking β α (mod 4P we get β = αw 2 for some w 1 + 2P and hence (β, ( 1 k discv P = (α, ( 1 k discv P. For P a prime not dividing 2, discv P = disc P, and since χ = 1, 3 of [7] shows that ( 1 k disc P is a square. Thus for every prime P we have ( α, ( 1 k discv P = ( β, ( 1 k discv P, so by 66:5 of [3] we have V α V β. χ It was remarked in [8] that the number of isometry classes in fam is finite when = 1; we now prove emma 3.1. The number of nuclear families in fam is 2 r where r Z. Proof: As shown in [7], P is modular at all primes P/ 2, so it follows from 92:1a of [3] that u P P P for all u P O P. In fact, for P/ 2 and u P O P, we must have P unless P has an odd rank Jordan component and (u P P = 1 (see 92:2 of u P P [3]. For P 2 and u P O P, the norm group of P is equal to the norm group of u P P (since P is even unimodular, so by 93:16 of [3], u P P P if and only if V u P P V P, and by 63 of [3], V u P P V P if and only if the ilbert symbol (u P, ( 1 k disc P P = (u P, ( 1 k discv P P = 1. Consider P to be a bad prime for if P/ 2 and P has an odd rank Jordan component, or if P 2 and there exists some u P O P such that (u, P ( 1k disc P P = 1. et Q 1,..., Q t denote the bad primes for. Thus there are at most 2 t genera in fam. We associate to K fam the vector ((u 1 : Q 1,..., (u t : Q t where K u i Q i Qi and 4

5 ( : Q is the egendre symbol if Q/ 2, and ( : Q is the ilbert symbol (, ( 1 k disc Q Q if Q 2; thus each genus within fam is associated to one such vector. We claim that these vectors associated to the genera in fam form a multiplicative subgroup of {±1} t. Since gen is associated to (1,..., 1 and each vector has order 1 or 2, we only need to verify that the set is closed. Take J, K fam. Thus K is on V α where α 0, α is relatively prime to 2, and genk is associated to the vector ((α : Q 1,..., (α : Q t. For any prime P/ 2, we have K P K u P P P, so V u P P V P. From 63 of [3], we see this means 1 = (α, ( 1 k discv P = (α, ( 1 k disc P P. Since P is modular and of even rank, disc P = β 2 ɛ for some β K P and ɛ O P. If ord P α is odd, then ( 1 k ɛ must be a square, which implies that /P is hyperbolic (see 3 of [7]. Thus we have αp 1 P s I 2 = O where the P i are primes not dividing 2 such that /P (and hence J/PJ is hyperbolic, and I is some fractional ideal. et J 0 = J, and let J i be a P-sublattice of J i 1. Then IJk α must be in fam, and for any bad prime Q, (IJk α Q = JQ α. Thus if genj is associated to the vector (u 1,..., u t, the genus gen(ijk α is associated to the vector (u 1 (α : Q 1,..., u t (α : Q t. Thus the set of vectors associated to the genera within fam forms a subgroup of the group {±1} t. A genus genk lies within fam + if and only if there is some totally positive unit u such that genk u = gen; thus each nuclear family within fam contains the same number of genera, which is the number of distinct vectors in the set U = {(..., (u : Q i... : u O, u 0 }. So each nuclear family is associated to a unique coset of {±1} t /U, and thus there are 2 r nuclear families within fam where r is some nonnegative integer. q.e.d. Remark. If K is a P 2 -sublattice of then P 1 K gen fam + since at all primes Q = P we have K Q = Q = P Q, and K P P P by construction. (For a more detailed discussion of P- and P 2 -sublattices, see 6 of [7]. et 1,..., 2 r represent the distinct nuclear families within fam, and define θ(fam + i, τ = K 1 θ(k, τ o(k where the sum is taken over a complete set of isometry class representatives K fam + i and o(k denotes the order of the orthogonal group O(K of K. otice that θ(fam + i, τ M k (Γ 0 (, (, χ. To help describe the action of T 0 on θ(fam + i, τ, we first prove 5

6 emma 3.2. et P be a prime ideal such that P/ 2, and consider T (P as a map from M k (Γ 0 (, n, χ into M k (Γ 0 (, Pn, χ. 1 If /P is hyperbolic then θ(fam +, τ T (P = K/Q (P k/2 ( K/Q (P k θ(fam + K, τ where K is any P-sublattice of. 2 If /P is not hyperbolic then Proof: θ(fam +, τ T (P 2 = K/Q (P k ( K/Q (P 2k 2 K/Q (P k θ(fam + P, τ. The first assertion follows from emma 5.2 of [8] and the observation that θ(fam +, τ = u θ(genu, τ where the sum runs over a finite set of totally positive units u (which represent the distinct vectors in the set U defined in the proof of the preceding lemma. To prove the second assertion, we let 1,..., m represent the distinct isometry classes in fam + ; as remarked above, any P 2 -sublattice of lies in fam + P. We set g ij = the number of isometries σ of V which map P j to a P2 -sublattice of i. Then we have that 1 o( j g ij is the number of P 2 -sublattices of i which are isometric to P j, 1 o( i g ij is the number of P 2 -sublattices of j which are isometric to P i, and by Proposition 7.3 of [7], 1 o( j j g ij = i 1 o( i g ij = (P k (P k 2 (P 0 ((P k + 1 ((P ow Proposition 6.1 and Theorem 7.4 of [7] yield the desired result. q.e.d. et genst operators be the commutative monoid consisting of all (finite products of the {T (P : /P is hyperbolic } {T (P 2 : /P is not hyperbolic } {S(I} where it is understood that P is a prime ideal, P/ 2, and I is a fractional ideal relatively prime to 2. Then as vector spaces, T is generated by genst and T 0 is generated by genst 0 = T 0 genst. ow we prove emma 3.3. et T genst0. Then θ(fam + i, τ T = λ T θ(fam + σt (i, τ where λ T is nonzero and dependent only on T, and σ T is a permutation of {1,..., 2 r }. Furthermore, 6

7 σ T 2 = 1 and σ T has a fixed point only if σ T = 1; the set {σ T : T genst 0 } forms a commutative group of order 2 r which acts transitively on {1,..., 2 r }. Proof: et Q 1,..., Q t be the bad primes for as in the proof of emma 3.1, and associate as before each genus within fam with a unique element of {±1} t, and each nuclear family within fam with a unique element of the quotient group {±1} t /U. Take T genst 0. Thus T = T (P 1 T (P l T (P 2 l+1 T (P 2 l+ss(i where the P i are primes (not necessarily distinct such that /P i is hyperbolic for 1 i l and not hyperbolic for i > l, and αp 1 P l P 2 l+1 P2 l+s I2 = O for some α 0. et K = K l+s K 1 K 0 = be lattices such that K i is a P i -sublattice of K i 1 for 1 i l, and K i is a P 2 i -sublattice of K i 1 for i > l. Then repeated use of the preceding lemma and Proposition 6.1 of [7] shows that θ(fam +, τ T = λ T θ(fam + IK α, τ (where λ 0 depends only on T ; the techniques used in emma 3.1 show that fam + IK α is associated to the coset (..., (α Q i,...u. Similarly, if fam + i is associated to (..., ɛ i,...u then θ(fam + i, τ T = λ T θ(fam + j, τ where fam + j is associated to (..., ɛ i (α Q i,...u. ence we may associate to T a permutation σ T of {1,..., 2 r } where σ T 2 = 1 and σ T has a fixed point only when σ T = 1. Clearly {σ T : T genst 0 } is an abelian group of order 2 r ; that this group is transitive follows from emma 3.2 and the proof of emma 3.1. q.e.d. This shows us that for T genst 0 we have ai θ(fam + i, τ T = λ T ai θ(fam + σt (i, τ = a σt (i θ(fam + i, τ (since σ T 2 = 1 so to find the eigenspaces of T 0 on θ(fam + i, τ i we merely need to find the eigenspaces of {σ T : T genst 0 } acting on C 2r by (a 1, a 2,... (a σt (1, a σt (2,.... Since σ T 2 = 1 and {σ T : T genst 0 } is transitive, a vector (a 1, a 2,... is an eigenvector for {σ T : T genst0 } only if there is some a C such that a i = ±a for each i. Clearly (1, 1,..., 1 is an eigenvector; this corresponds to θ(fam, τ = i θ(fam+ i, τ, which was shown to be a T 0 -eigenform in [8] in the case that χ = 1. 7

8 emma 3.4. For 1 i r, let ρ i1 be the permutation ρ i1 = (1 2 i (2 2 i 1(3 2 i 2 (2 i 1 2 i and for 1 < j 2 r i, define ρ ij inductively by ρ ij = (2 i + a 1 2 i + a 2 (2 i + a l 1 2 i + a l where ρ ij 1 = (a 1 a 2 (a l 1 a l. Set σ i = 2 r i j=1 ρ ij. Then up to a reordering of fam + 1,..., fam + 2 r, the permutations σ 1,..., σ r generate {σ T : T genst 0 }. Proof: Clearly the σ i commute, σ 2 i = 1, and for i 1 < i 2 < < i l, we have σ il σ i2 σ i1 (1 1. ence if we can demonstrate that σ 1,..., σ r {σ T : T genst0 } then it must be the case that these σ i generate {σ T : T genst0 }. We begin by choosing σ T1 1 and then ordering the nuclear families within fam such that σ T1 = (1 2(3 4 (2 r 1 2 r = σ 1. ext, using the transitivity of the group {σ T : T genst 0 }, we choose σ T2 such that σ T2 (1 = 4. ow, σ T1 σ T2 = σ T2 σ T1, so σ T2 (2 = 3. Thus σ T2 = (1 4(2 3. ext we observe that σ T2 (5 6 else σ T1 σ T2 is a nontrivial permutation with a fixed point; thus we can reorder the nuclear families such that σ T2 (5 = 8. otice that this we can choose this reordering to preserve the equality σ T1 = σ 1. Since we have σ T1 σ T2 = σ T2 σ T1, we get σ T2 = (1 4(2 3(5 8(6 7. Reasoning as above, σ T2 (9 10, thus a reordering of {fam + i } gives us σ T2 = (1 4(2 3(5 8(6 7(9 12(10 11 and we still have σ T1 = σ 1. Continuing this process of reordering {fam + i } gives us σ T2 = σ 2 and σ T1 = σ 1. 8

9 ow we choose σ T3 such that σ T3 (1 = 8; the subgroup σ T1, σ T2, σ T3 is commutative and has no nontrivial elements with fixed points, so arguing as before we can reorder the nuclear families {fam + i } so that σ T3 = σ 3, σ T2 = σ 2, and σ T1 = σ 1. Continuing this process, we find σ T4,..., σ T r {σ T : T genst 0 } and a reordering of the nuclear families {fam + i } such that σ T r = σ r,..., σ T1 = σ 1. q.e.d. This shows that the T 0 -eigenforms of the form i a i θ(fam + i, τ correspond to the vectors (..., a i,... which are eigenforms for {σ 1,..., σ r }. ow we show emma 3.5. et σ 1,..., σ r be as in emma 3.4. For i = 1,..., r, let v i = (v i1,..., v i2 r where v ij = ( 1 a with a Z such that a 2 i 1 < j (a i 1. Then v 1,..., v r generate an abelian group of order 2 r under componentwise multiplication, and every element in this group is an eigenvector for {σ T : T genst0 } (where the permutation σ maps the vector (a 1, a 2,... (C 2r to (a σ(1, a σ(2,.... Furthermore, if v and v are distinct elements of this group, then v and v have distinct eigenvalues for some σ T. Proof: otice that the entries of the vector v i occur in blocks of 2 i 1, so σ j (v i = { vi if j < i v i if j i. For 1 j r and v = v i1 v is with i 1 < < i s, take l s such that j i l and j < i l+1 if l + 1 s; then { v if l is even σ j (v = v if l is odd. Since σ(vv = σ(vσ(v, two vectors v and v have the same eigenvalues for all the σ i if and only if v = v. q.e.d. Remark. For any vector v = (a 1, a 2,... in this group v 1,..., v r we have i a i = 0 unless v = (1, 1,..., 1. To see this, write v = v i1 v is where i 1 < < i s, and observe that for i < i s and j l (mod 2 is 1, we have v ij = { vil if i < i s or j l (mod 2 i s v il if i = i s and j l (mod 2 i s. Thus a j = a l if j l (mod 2 i s, and a j = a l if j l (mod 2 i s 1 but j l (mod 2 i s. 9

10 et v r+1,..., v 2 r denote the other vectors in the group generated by v 1,..., v r, and write v i = (v i1,..., v i2 r. With the nuclear families of fam ordered as in emma 3.4, set E i (τ = j v ij θ(fam + j, τ. Then we have Theorem 3.6. Those E i (τ which are nonzero are linearly independent T 0 -eigenforms. We also have i θ(fam + i, τ = Ei (τ, i so the number of forms θ(fam + 1, τ,..., θ(fam + 2 r, τ which are linearly independent is equal to the number of E i (τ which are nonzero; here 2 r is the number of nuclear families within fam. Proof: emmas 3.3 and 3.5 show that for T genst 0, E i (τ { λt E T = i (τ if σ T (v i = v i λ T E i (τ if σ T (v i = v i ; since v i and v j have the same eigenvalues for σ 1,..., σ r only when i = j, the nonzero E i (τ must be linearly independent. To finish proving the theorem, it suffices to show that the matrix A = (v ij is nonsingular. ow, the i, j-entry of AA t is the dot product of v i with v j, but this is just the sum of the entries of the vector v i v j. As remarked above, the sum of the entries of any vector v in the group generated by v 1,..., v r is 0 if v (1, 1,..., 1; thus AA t = 2 r I where I denotes the 2 r 2 r identity matrix. q.e.d. ext we use Theorem 3.1 of [8] to lift the T 0 -eigenforms E i (τ to T -eigenforms Ẽ i (τ M k (, χ where χ is a ecke character extending χ χ with χ (ã = sgn(ã k for an adele ã K A. (Thus χ is a ecke character such that the finite part of its conductor divides the conductor of χ, χ(ã = χ (a when ã, the -part of the adele ã, is a unit at all primes dividing and a K such that a ã (mod ; c.f. [8] or [4]. ote that for u O, χ (u θ(fam +, τ = θ(fam +, τ ( u 1 0 = sgn(u 0 u k θ(fam +, τ so χ (u = sgn(u k as required in [8]. The lift of E i (τ involves theta series attached to lattices in the extended family of, which is defined as follows. 10

11 Definition. A lattice K is connected to by a prime-sublattice chain if there exist lattices K 0 =, K 1,..., K s such that K i is a P i - or a P 2 i -sublattice of K i 1 (depending on whether /P i is hyperbolic and K = IKs α for some fractional ideal I and some α 0. ere it is understood that P 1,..., P s are prime ideals (not necessarily distinct which do not divide 2, and I and αo are relatively prime to 2 ; also, we take K i to be a P i -sublattice of K i 1 if /P i is hyperbolic, and we take K i to be a P 2 i -sublattice of K i 1 otherwise. If K is connected by a prime-sublattice chain to a lattice in fam then we say K is in the extended family of, denoted xfam. We can lift any E i (τ regardless of whether E i (τ is nonzero (although if E i (τ = 0 then the lift of E i (τ is also zero. Theorem 3.1 of [8] shows us that each nonzero T 0 - eigenform E i (τ can be lifted to c i h linearly independent T -eigenforms Ẽi(τ where h is the class number of K/Q and c i is the cardinality of {[Q] : Q = αp 1 P l I 2 with α 0 and /P i hyperbolic }; here [Q] denotes the complex of the strict ideal class of Q, and it is understood that I is a fractional ideal relatively prime to 2. Thus if χ = 1, each nonzero E i (τ can be lifted to h linearly independent T -eigenforms where h is the strict class number of K/Q. Furthermore, an examination of this lift gives us Corollary 3.7. For each i = 1,..., 2 r, each component of Ẽi(τ is a linear combination of theta series attached to lattices in xfam, and Ẽ i (τ T (P = δi (P( K/Q (P k Ẽi(τ where δ i (P = ±1 so that σ T (P (v i = δ i (Pv i. To obtain nice relations on the Fourier coefficients of the lifts Ẽi(τ, we follow [4] and define the Fourier coefficients for an integral ideal M by a i (Q = a µ (ξ K/Q (I µ k/2 where ξ 0, ξi 1 µ = M, and the form ζ a µ(ζe 2πiTr(ζτ M k (Γ 0 (, I µ, χ is a component of Ẽi(τ. Thus for ξ n, ξ 0, and M = ξ(n 1 we have a i (M = K/Q (n k/2 j v ij r(fam + j, ξ 11

12 where r(fam +, ξ = 1 o( r(, ξ with the sum running over isometry class representatives fam +, and where we take n to be one of the strict ideal class representatives used to define M k (, χ = I M k (Γ 0 (, I, χ. Since Ẽi(τ is an eigenform we have κ(ma i (M = M+M A χ (A K/Q (A k 1 a i (MM A 2 where T (M T and κ(m is some constant such that Ẽi(τ T (M = κ(mẽi(τ (see (2.23 of [4]. These relations and a scrutiny of the lift will yield relations on the numbers r(fam +, ξ; as a step toward that goal we prove emma 3.8. For a prime P/ 2, define { 1 if /P is hyperbolic, ε (P = 1 otherwise; and define ε (A = P A ε (P ord P A. Then for any a O such that a is relatively prime to, χ (a = sgn(a k ε (ao. Moreover, if we let I 1,..., I h be strict ideal class representatives such that Ẽ i (τ M k (, χ η M k (Γ 0 (, I η n, χ (where χ is a ecke character extending χ χ see the discussion following emma 3.6, then for M = ξ(n 1 with ξ n, ξ 0, M relatively prime to 2, and A 2 M, a i (A 2 M = χ (Aε (A K/Q (n k/2 v ij r(fam + A j, 2ξ where χ (A = χ(ã for an adele ã such that ã P ã = 1 for all primes P with ord P A = 0 and ão = A. ( a b Proof: etting Γ c d 0 (, n such that ad bc = 1, we see that the restrictions on d in the transformation formula (2 of [7] are unnecessary, but one needs to replace (d m 2 by d m 2. Then the arguments used to prove Theorem 3.7 of [7] show that χ (a = sgn(a k ε (a for any a O with (ao, = j

13 ow, take µ and a K such that A = ai µ and a 0. Then from the construction of Ẽi(τ we find that a i (A 2 M = χ (I µ K/Q (I µ k v ij r(fam + I µ j, 2α 2 ξ K/Q (n(i µ k/2 j = χ (I µ ε (I µ K/Q (n k/2 j v ij r(fam + αi µ j, 2ξ. Since χ is quadratic, we have χ (Aε (A = χ (I µ ε (I µ χ (aε (ao = χ (I µ ε (I µ, and so the lemma follows. q.e.d. For a prime ideal P/ 2 we now define and λ(p = K/Q (P k/2 ( K/Q (P k when ε (P = 1, λ(p 2 = K/Q (P k ( K/Q (P 2k 2 K/Q (P k when ε (P = 1. We inductively define λ(m where T (M T by defining λ(p a λ(p b = min{a,b} c=0 K/Q (P c(2k 1 λ(p a+b 2c and λ(mλ(m = λ(mm whenever M and M are relatively prime. Theorem 3.9. Take ξ n, ξ 0. Write ξ(n 1 = MM where M and M are integral ideals and T (M T. Then r(fam +, 2ξ = λ(m K/Q (M k/2 r(fam +, 2ξ ε (A K/Q (A k 1 r(fam + A, 2ξ M+M A A O where n = Mn and is connected to by a prime-sublattice chain. In the case that K = Q, we can scale to assume n = Z; then for m, m Z such that T (m T we have r(fam +, 2mm = λ(m m k/2 r(fam +, 2m χ (a a k 1 r (fam +, 2mm a (m,m a>1 13 a 2

14 where fam with fam + determined by m. Proof: Take ξ n, ξ 0, and write ξ(n 1 = MM with T (M T (i.e. M is relatively prime to 2 and ord P M is even whenever ε (P = 1 for P/ 2. To lift E i (τ, we may as well assume that Mn is one of the strict ideal class representatives used to define M k (, χ; then we observe that E i (τ T (M 2 S(M 1 = K/Q (M k λ(m 2 E i (τ so we can fix a square root of χ (M and define the Mn-component of Ẽi(τ to be χ (M K/Q (Pk/2 λ(m E i T (M = χ (M K/Q (M k/2 j v ij θ( j, τ where n j = M n and j is connected to j by a prime-sublattice chain. (otice that if M = I 2 then we can interpret χ (M (Mk/2 K/Q v ij θ( j, τ j as ε (I χ (I K/Q (I k j v ijθ(i j, τ since I j fam + j in this case. Then Ẽ i (τ T (M = λ(m χ (M K/Q (Mk/2 Ẽ i (τ. Since M = ξm 1 (n 1, we have a i (M = χ (M K/Q (M k/2 j v ij r(fam + j, 2ξ K/Q (M k/2 K/Q (n k/2 so κ i (Ma i (M = λ(m K/Q (M k/2 K/Q (n k/2 j v ij r(fam + j, 2ξ (where E i (τ T (M = κ i (M E i (τ. Therefore we have a i (MM = κ i (Ma i (M χ (A K/Q (A k 1 a i (M A 2 M+M A A O 14

15 which, together with Corollary 3.7 and emma 3.8, gives us v ij r(fam + j, 2ξ = λ(m K/Q (M k/2 j M+M A A O j v ij r(fam + j, 2ξ ε (A K/Q (A k 1 j v ij r(fam + A j, 2ξ. Summing on i and normalizing yields the desired result. q.e.d. 4. inear independence of theta series when K = Q, χ = 1, and rank > 2. Suppose now that K = Q, χ = 1, and rank = 2k with k > 1. Since the object of this section is to show that the E i (τ constructed in 3 are nonzero (and thus by Thoerem 3.6 the forms {θ(fam + i, τ} are linearly independent, we may assume that the lattice has been scaled so that n = Z. et q 1,..., q t denote the bad primes for (as defined in the proof of emma 3.1. Since is even unimodular when localized at 2, an analysis of the ilbert symbol (, ( 1 k disc 2 shows that 2 is not a bad prime for. Furthermore, since χ = 1, there are 2 t nuclear families within fam. (Using the Chinese Remainder Theorem we can realize each vector in {±1} t as ((a : q 1,..., (a : q t where a = p 1 p s with the p i primes (not necessarily distinct and p i / 2. hyperbolic, K 1 a Since /p i is fam where K = K s, K 0 =, and K i is a p i -sublattice of K i 1. Then fam + K 1 a = genk 1 a is associated to the vector ((a : q 1,..., (a : q t. For any prime p the localized lattice p has the Jordan decomposition p J 1 J l ; letting e s = ord p nj s and m s = rankj s = dimj s Q, for b > 0 with ord p b = e we define rank p (b, = (e e s m s. e s <e otice that rank p (b, depends on fam, p and b; since fam is fixed, we simply write rank p (b for rank p (b,. emma 4.1. Fix a prime q and suppose fam + such that for all primes p we have p p except when p = q. (Thus q must be a bad prime for. Then θ(fam +, τ θ(fam +, τ = 2 mass 15 b>0, (a,b=1 rank q b odd b, (bτ a k

16 and θ(fam +, τ + θ(fam +, τ = 2 mass 1 + b>0, (a,b=1 rank q b even b, (bτ a k where and mass = fam + b, = i k (disc 1 2 b k 1 o( is an invariant for fam. x /b e πi a b Q(x Proof: First we verify that mass is an invariant of fam. From our construction we know that the zero-fourier coefficient of θ(fam +, τ is mass; from the definition of ecke operators we know that the zero-fourier coefficient of θ(fam +, τ T (p must be (p k times the zero-fourier coefficient of θ(fam +, τ. identifying isomorphic spaces of modular forms via the map f f ( 1 0, 0 p we see that for p a prime such that χ (p = 1 we have θ(fam +, τ T (p = (p k θ(fam + K 1 p, τ where K is a p-sublattice of and hence K 1 p Using emma 3.2 and fam. Since the zero-fourier coefficient of θ(fam + K 1 p, τ is massk 1 p, we have that mass = massk 1 p. ow, as discussed at the beginning of this section, any nuclear family fam + fam has a representative of the form (K s 1 a where a = p 1 p s with p 1,..., p s primes (not necessarily distinct such that χ (p i = 1 and K 0 =, K 1,..., K s are lattices with K i a p i -sublattice of K i 1 ; thus it follows that mass = mass for any fam. From equation (82 of [5] we have θ(fam +, τ = mass 1 + b, (bτ a k a,b where the sum is over all integers a and b with b > 0 and (a, b = 1, and b, ( = i k (disc 1 2 b k a e b Q(x 16 x /b

17 (here we write e(α for e πiα. Using the results of 3 of [7] we have x /b e b Q(x = (b k χ (b p prime p e b x p e b/b e b Q(x where b = b b with (b, b = 1 = (b,. Fix a prime p b and let e = ord p b; by assumption p 2 (since is even and unimodular when localized at 2 and so 2/ ; see 3 of [7] hence we can write p J 1 J l where rankj s = m s and J s p s e 1,..., 1, ɛs with ɛs Z and e s < e s+1. Since we are assuming Z = n we have e 1 = 0. We know /p e p /p e p (where the isomorphism is as Z/p e Z-modules, and if x and x p such that x x p e p then Q(x Q(x (mod p e Z. Thus, choosing µ 2Z such that µ pe b (mod pe, the techniques of 3 of [7] give us x p e b/b where e b Q(x = l s=1 = p 2ke d 1,...,d m s Z/pe Z e s <e e ( µap es e (d d 2 m s 1 + ɛ s d 2 m s p m s(e s e/2 c s b,, p 1 ( if 2 (e e s, ms ( 1 c s b,, p 2 p = ( if 2/ (e e s but 2 m s, ( 1 m s 1 2 aɛ s p p 1 2 e(µd 2 if 2/ (e e s m s. d Z/pZ (Recall that e 1 = 0 since n = Z, so e > e 1. For fam such that p u p with u Z p, we see that c s b,, p = (u p c s b,, p if e s < e and 2/ (e s s m s, and c s b,, p = c s b,, p otherwise. Since (u p = 1 only when p = q, the lemma now follows. q.e.d. ow we can prove 17

18 emma 4.2. For each i = 1,..., 2 t we have t E i (τ = mass + ( 1 j=1(1 δij + 2 t a,b b, (bτ a k where the sum runs over all integers a, b with b > 0, (a, b = 1, and for each j, δ ij is chosed to be 0 or 1 such that rank qj b δ ij (mod 2. Also, t j=1 (1 + ( 1δ ij = 0 unless E i (τ = θ(fam, τ. Proof: Choose T 1,..., T t genst such that θ(fam +, τ Tj = λ Tj θ(fam +, τ for some fam with qs q s except when s = j. Then we can arrange the nuclear families within fam such that σ j = σ Tj, 1 j t, where σ j is as defined in emma 3.4. et { 0 if σj (v δ ij = i = v i, 1 if σ j (v i = v i (where v i is defined in emma 3.5 and E i (τ = t j=1 v ij θ(fam + j, τ. Then the argument used to prove emma 4.1 now gives us 1 mass E i(τ 2 t 1 = (1 + ( 1 δ it + 2 j=1 rank q1 b δ i1 (2 2 t 2 = (1 + ( 1 δ it (1 + ( 1 δ it j=1 b, j (bτ a k rank q1 b δ i1 (2 rank q2 b δ i2 (2 b, j (bτ a k. t = (1 + ( 1 δ ij + 2 t j=1 rank qj b δ ij (2 (1 j t b, 1 (bτ a k. q.e.d. Finally we have 18

19 Theorem 4.3. For each i = 1,..., 2 t, the form E i (τ is nonzero, and hence the set {θ(fam + 1, τ..., θ(fam + 2 t, τ} is linearly independent. Proof: If E i (τ = θ(fam, τ then the zero-fourier coefficient of E i (τ is nonzero. So suppose E i (τ θ(fam, τ; we will show that the first Fourier coefficient of E i (τ, a i (1, is nonzero. As before, let q 1,..., q t denote the bad primes for and set Q = q δ i1 1 q δ it t where the δ ij are as defined in the preceding proof. From the preceding lemma we have (2 t mass 1 E i (τ = = b = b b 1 rank qi b δ ij (2 ( 1 b, a Z (a,b=1 a (Z/bZ (a Q b, (bτ a k (bτ a + bm k m Z ( 1 b, (a Q ( 1k 1 2kζ(k B k a (Z/bZ = ( 1k 1 2kζ(k B k b k b ( 1 b, a (Z/bZ m k 1 e 2πim(τ a b m=1 ( ( a Q m k 1 e m=1 ma 2πi b e 2πimτ (ere ζ denotes the Riemann-zeta function, and B k is the kth Bernoulli number. ow, take Q Z such that 0 if p/ b, ord p Q = 1 if ord p Q = 1, 2 otherwise; then for b 1 with rank qi b δ ij (mod 2, we have ( a Q e 2πi ma b a (Z/bZ = a Z/bZ = q e b = q e b ( a Q ( b ordq Q q q e ( b ordq Q q q e ma 2πi e b a Z/q e Z c Z/qZ (a q ord q Q e 2πi ma (c q ord q Q mc 2πi e q e 19 q e d Z/q e 1 Z md 2πi e q e 1.

20 where it is understood that in the above product q is a prime dividing b. Unless q e 1 m, the above sum over d is zero; also, the sum over c is 1 if q e / m and ord q Q is even. Thus in the Fourier expansion of E i (τ, the coefficient for e 2πiτ is ( 1 k 1 2kζ(k B k n Z +, (n,q=1 n square free (Qn k ( = ( 1k 1 2kζ(k 1 Q k B k Q, n Z +, (n,q=1 n square free ote that ( 1k 1 2kζ(k Q k B k ( 1 q Q ( ( 1 Qn, Qn q n k χ (n(n Q. Q, 0. We know that c Z/qZ q (c q e 2πi c q = q ; ( a Q e 2πi a b a (Z/nZ c Z/qZ (c q e 2πi c q thus to show that the coefficient of e 2πiτ is nonzero, we only need to show that the last sum in the above expression is nonzero. We have n Z +, (n,q=1 n square free n k χ (n(n Q = 1 + n>1 n square free n k χ (n (n Q and n k χ (n(n Q < n 2 < 1. q.e.d. n>1 n>1 5. Concluding remarks. Since the techniques used in this paper are in some ways rather general, one can imagine there are many extensions of the results presented here. First let us note that equation (82 of [5] (which we used in 4 to show E i (τ 0 in the case that K = Q, χ = 1 and k > 1 is extended in [6] to allow K Q; however, in [6] Siegel only concerns himself with free lattices. Thus the techniques of 4 can be used to show that the E i (τ are all nonzero in the case that the strict class number of K is 1; this author suspects that Siegel s results hold regardless of the strict class number of K, and thus in general all the E i (τ are nonzero. ext, let us observe that Theorem of 20

21 [1] implies that the results of 3 can be extended to Siegel modular forms when K = Q and χ = 1. In view of the techniques used to prove Theorem 7.4 of [7], this author suspects that Theorem of [1] can be extended to allow K Q and χ 1. If this is indeed the case, then the results of 3 herein may be extended to general Siegel modular forms. Furthermore, Siegel s papers [5] and [6] extend equation (82 of [5] to include Siegel modular forms attached to free lattices; thus the results of 4 could probably be extended to Siegel modular forms. Finally, we mention that these results should extend to include lattices of odd rank; this, in fact, is the author s current concern. References [1] A.. Andrianov, Quadratic Forms and ecke Operators, Springer-Verlag, ew York, [2] M. Eichler, On theta functions of real algebraic number fields, Acta Arith. 33 (1977, [3] O.T. O Meara, Introduction to Quadratic Forms, Springer-Verlag, ew York, [4] G. Shimura, The special values of the zeta functions associated with ilbert modular forms, Duke Math. J. (1978, [5] C.. Siegel, Über die analytische Theorie der quadratischen Formen, in Gesammelte Abhandlungen, Springer-Verlag, ew York, 1966, pp [6], Über die analytische Theorie der quadratischen Formen III, in Gesammelte Abhandlungen, Springer-Verlag, ew York, 1966, pp [7].. Walling, ecke operators on theta series attached to lattices of arbitrary rank, Acta Artih. 54 (to appear [8],On lifting ecke eigenforms, Trans. AMS (to appear. 21