Math. Z. (2010) 265:263–276 DOI 10.1007/s00209-009-0513-z
Mathematische Zeitschrift
Maass relations in higher genus Shunsuke Yamana
Received: 28 May 2008 / Accepted: 8 March 2009 / Published online: 27 March 2009 © Springer-Verlag 2009
Abstract For an arbitrary even genus 2n we show that the subspace of Siegel cusp forms of degree 2n generated by Ikeda lifts of elliptic cusp forms can be characterized by certain linear relations among Fourier coefficients. This generalizes the classical Maass relations in degree two to higher degrees. Keywords
Ikeda lifting · Saito–Kurokawa lifting · Maass spaces · Maass relations
Mathematics Subject Classification (2000)
11F30
0 Introduction The purpose of this paper is to give a characterization of the image of Ikeda’s lifting by certain linear relations among Fourier coefficients. Let us describe our results. We denote by Tr+ the set of positive definite symmetric half+ integral matrices of size r . Put Dh = det(2h) for h ∈ T2n with a fixed integer n. Let dh be n 1/2 1/2 . the absolute value of discriminant of Q(((−1) Dh ) )/Q. Put fh = (d−1 h Dh ) Fix a prime number p. The Siegel series attached to h at p is defined by b p (h, s) =
e p (−tr(hα))ν(α)−s ,
α
where α extends over all symmetric matrices of size 2n with entries in Q p /Z p . Here, we put √ −2π −1x 2n 2n e p (x) = e , where x is the fractional part of x ∈ Q p , and ν(α) = [α Z2n p +Z p : Z p ]. −s −s As is well-known, b p (h, s) is a product of two polynomials γ p (h; p ) and F p (h; p ), where
S. Yamana (B) Graduate School of Mathematics, Kyoto University, Kitashirakawa, Kyoto 606-8502, Japan e-mail:
[email protected]
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−1 n (−1)n dh pn X γ p (h; X ) = (1 − X ) 1 − (1 − p 2 j X 2 ) p j=1
and the constant term of F p (h; X ) is 1. Put F˜ p (h; X ) = X − ord p fh F p (h; p −n−1/2 X ). Then, from [6], the following functional equation holds: F˜ p (h; X ) = F˜ p (h; X −1 ). Ikeda constructed a lifting from S2k (SL2 (Z)) to Sk+n (Sp2n (Z)) for each positive integer k such that k ≡ n (mod 2). Theorem 1 [5] Let f ∈ S2k (SL2 (Z)) with k ≡ n (mod 2) be a normalized Hecke eigenform with Hecke L-function: k−1/2−s −1 (1 − α p p k−1/2−s )−1 (1 − α −1 ) . p p p + Let g be a corresponding cusp form in the Kohnen plus space Sk+1/2 (4) under the Shimura correspondence. We define the function F on the upper half-space H2n by √ c F (h)e2π −1tr(h Z ) , F(Z ) = + h∈T2n
where k−1/2
c F (h) = cg (dh )fh
F˜ p (h; α p ).
p
Here, we denote the mth Fourier coefficient of g by cg (m). Then F is a cuspidal Hecke eigenform in Sk+n (Sp2n (Z)). Put Dk = {m ∈ N | (−1)k m ≡ 0, 1 (mod 4)}. In [8] Kohnen defined an integer φ(d; h) + for each h ∈ T2n and each positive divisor d of fh , and showed that In,k :
c(m)e2π
√ −1mτ
→
d k−1 φ(d; h)c(d −2 Dh )e2π
√ −1tr(h Z )
+ d|fh h∈T2n
m∈Dk
+ is a linear map from Sk+1/2 (4) to Sk+n (Sp2n (Z)), which on Hecke eigenforms coincides with the Ikeda lifting. He also conjectured that if F ∈ Sk+n (Sp2n (Z)) has a Fourier expansion of the form
F(Z ) =
d k−1 φ(d; h)c(d −2 Dh )e2π
√ −1tr(h Z )
+ d|fh h∈T2n
for some function c : Dk → C, then F lies in the image of the map In,k . If n = 1, then this conjecture comes down to saying that the Maass space coincides with the image of the Saito–Kurokawa lifting, and hence it is true.
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Kohnen and Kojima [9] proved the conjecture for all n with n ≡ 0, 1 (mod 4). More precisely, they showed that √ + c(m)e2π −1mτ ∈ Sk+1/2 (4). g(τ ) = m∈Dk + with det(2S) = 2, provided that n ≡ 0, 1 Their proof is as follows. There exists S ∈ T2n−1 (mod 4). Let FS be the Sth Fourier–Jacobi coefficient of F, and {FS,µ0 , FS,µ1 } the modular forms of weight k + 21 associated to FS (cf. Sect. 3). They put
i(S)F(τ ) = FS,µ0 (4τ ) + FS,µ1 (4τ ). Then they showed the following facts: + (4); – i(S)F is an element of Sk+1/2 – the Fourier expansion of i(S)F is given by √ c(m)e2π −1mτ . i(S)F(τ ) = m∈Dk
These prove the conjecture for all n with n ≡ 0, 1 (mod 4). + There are no matrices S ∈ T2n−1 with det(2S) = 2 if n ≡ 2, 3 (mod 4) (cf. Lemma 2). Nevertheless, we shall show the following: + Theorem 2 Under the notation above, there exists g ∈ Sk+1/2 (4) with the following properties:
(1) If n ≡ 3 (mod 4), then c(m) = cg (m). (2) If n ≡ 2 (mod 4), then the following assertions holds: (a) F = In,k (g); (b) if m is not a square, then c(m) = cg (m); (c) c( f 2 ) = cg ( f 2 ) + (c(1) − cg (1)) f k for every positive integer f . In order to offer the reader some hint of our idea, we briefly describe the proof of Theorem 2 under the simplifying assumption n = 2. The first problem is to find auxiliary matrices which supplement the lack of the above matrix S. Fix a rational prime p and let H be a quaternion algebra with center Q ramified precisely at p and the archimedean place. Let ι be the main involution and τ the reduced trace on H . Put B = {x ∈ H | x ι = −x} and endow B with a symmetric bilinear form B(x, y) = pτ (x y ι )/2. When we identify B with an element of T3+ , using a basis of a maximal integral lattice of B , it is easy to see that det(2B) = 2 p. Analogously, we put i(B)F(τ ) = FB,µ (4 pτ ), µ
letting {FB,µ } be a collection of modular forms associated to the Bth Fourier–Jacobi coef+ ficient of F. Then i(B)F is an element of Sk+1/2 (4 p). Analyzing the Siegel series closely, we can show that the Fourier expansion of i(B)F is given by √ (−1)k m i(B)F(τ ) = 1− (c(m) − p k c( p −2 m))e2π −1mτ . p m∈Dk
On the other hand, the map
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P( p) : g →
m∈Dk
1−
(−1)k m p
(cg (m) − p k cg ( p −2 m))e2π
√ −1mτ
+ + defines an injection from Sk+1/2 (4) into Sk+1/2 (4 p). The most difficult part of the proof is to show that i(B)F belongs to the image of P( p). To solve this problem, we need the newform theory for modular forms of half-integral weight, which we recall in Sect. 2. Pursuing this analysis (for details see Sect. 4), we will + see that the existence of a function c gives rise to g ∈ Sk+1/2 (4) satisfying i(B)F = + g|P( p) for all primes p and B ∈ T3 subject to det(2B) = 2 p. This proves (b) and (c). p (h; p 1/2 ) = 0 and hence If Dh is a square, then we have p F
d −1 φ(d; h) = 0
d|fh
(see Lemma 1). From this, we get (a). Recall that we retrieve the Maass relations defined by Maass if n = 1, as was shown by Kohnen in [8, Proposition 3]. Our result provides further evidence that the above condition is a natural generalization of the Maass relations to higher genus.
Notation Put q = e(τ ) = e2π
√ −1τ
for a complex number τ . Let H be the upper half-plane. Put
Dn = {a ∈ N | (−1)n a ≡ 0, 1
(mod 4)}
with a fixed positive integer n. For a nonzero element N ∈ Q, we denote the absolute value 1/2 . For each rational of the discriminant of Q(((−1)n N )1/2 )/Q by d N . Put f N = (d−1 N N) √ prime p, we put ψ p (N ) = 1, −1, 0 according as Q p ( N ) is Q p , an unramified quadratic extension or a ramified quadratic extension. We denote by Sκ (Spr (Z)) the space of Siegel cusp forms of weight κ with respect to the Siegel modular group Spr (Z). Let Tr be the set of symmetric half-integral matrices of size r and Tr+ the set consisting of positive definite elements of Tr . We denote the hth Fourier coefficient by c F (h) for F ∈ Sκ (Spr (Z)) and h ∈ Tr+ . Put Dh = 4[r/2] det h, where [ · ] is the Gauss symbol. To simplify notation, we abbreviate dh = d Dh , fh = f Dh + for h ∈ T2n . Put h[A] = t Ah A and h(A, B) = t Ah B for matrices A and B if they are well-defined.
1 Main theorem We recall first of all a formula for the Siegel series. Let p be a rational prime and F p a finite field with p elements. Put T2n, p = T2n ⊗Z Z p . Fix a nondegenerate matrix h ∈ T2n, p and let q p be a quadratic form on V = F2n p obtained from the quadratic form x → h[x] by reducing modulo p. Let Rad(V ) be the radical of V , i.e., Rad(V ) = {x ∈ V | h[x] ≡ 0 and h[x + y] ≡ h[x] + h[y] (mod p) for all y ∈ V }.
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Writing V = V1 ⊕ Rad(V ), we put s p = s p (h) = dimF p Rad(V ), 1 if V1 is hyperbolic or s p = 2n, λ p (h) = −1 otherwise, ⎧ if s p = 0. ⎪ ⎨1 (s p −1)/2 2 j−1 X 2 ) Hn, p (h; X ) = (1 − p if 2 sp. j=1 ⎪ s p /2−1 ⎩ (s −1)/2 2 j−1 2 (1 + λ p (h) p p X ) j=1 (1 − p X ) otherwise. Set D p (h) = GL2n (Z p )\{G ∈ M2n (Z p ) ∩ GL2n (Q p ) | h[G −1 ] ∈ T2n, p },
where GL2n (Z p ) operates by left multiplication. Putting Dh = det(2h) and letting dh be the discriminant ideal of Q p (((−1)n Dh )1/2 )/Q p , we set f p (h) = (ord p Dh − ord p dh )/2.
Now the formula is
p (h; X ) = F
X −f p (h)+2 ord p det G Hn, p (h[G −1 ]; X )
G∈D p (h)
×
⎧ ⎨1
⎩1 −
dh
p
if f p (h[G −1 ]) = 0. p −1/2 X
if f p (h[G −1 ]) > 0.
(1.1)
This formula is due to Kitaoka [7] at least when p = 2. A complete and general proof including the case p = 2 was obtained by Feit [4]. Böcherer [1] obtained the formula globally in connection with the Fourier coefficients of Siegel Eisenstein series (cf. [8]). + Let h ∈ T2n . Following [8], we define the map ρh : N → C via the relation ρh (a)a −s = (1 − p −2s )Hn, p (h; p −s ). a∈N
p|fh
For each positive divisor a of fh , an integer φ(a; h) is defined by √ ρh[G −1 ] (d −2 a), φ(a; h) = a d∈N, d 2 |a G∈D (h), | det G|=d
where D(h) = GL2n (Z)\{G ∈ M2n (Z) ∩ GL2n (Q) | h[G −1 ] ∈ T2n }.
Note that there is a natural bijection between D(h) and p D p (h). For an integer e, we define le ∈ C[X, X −1 ] by ⎧ e+1 − X −e−1 ⎨X if e ≥ 0. le (X ) = X − X −1 ⎩ 0 if e < 0. For each prime number p, we put
p,N = lord p f N − ψ p ((−1)n N ) p −1/2 lord p f N −1
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(see Notation for the definitions of f N and ψ p ). The arguments in [8] assert that k−1/2
fh
p (h; X p ) = F
d k−1 φ(d; h)(d −1 fh )k−1/2
d|fh
p
p,d −2 Dh (X p ).
(1.2)
p
We have k−1/2
cg (m) = cg (dm )fm
p,m (α p )
p + (4) satisfies g|T˜ ( p 2 ) = p k−1/2 (α p + α −1 if g ∈ Sk+1/2 p )g for all primes p (see Sect. 2). Hence the following theorem follows from Theorem 1 and (1.2).
Theorem 3 [8] The notation being as in Theorem 1, we have d k−1 φ(d; h)cg (d −2 Dh ). c F (h) = d|fh + (4) → Sk+n (Sp2n (Z)) arises as in Introduction. Thus a linear map In,k : Sk+1/2 M (Sp (Z)) is defined in the following Definition 1 If k ≡ n (mod 2), then the space Sk+n 2n M way: F ∈ Sk+n (Sp2n (Z)) is an element of Sk+n (Sp2n (Z)) if there is a function c : Dk → C + such that all h ∈ T2n satisfy d k−1 φ(d; h)c(d −2 Dh ). (1.3) c F (h) = d|fh + with dh = 1, then we have Lemma 1 If n ≡ 2 (mod 4) and h is an element of T2n d k−1 φ(d; h)(d −1 fh )k = 0. d|fh M Remark 1 Lemma 1 implies that the parameter c is not uniquely determined by F ∈ Sk+n (Sp2n (Z)) if n ≡ 2 (mod 4). More precisely, a function c : Dk → C defined by c(m) if m ∈ / Q×2 c (m) = k/2 c(m) + a · m if m ∈ Q×2
is also a parameter of F for any complex number a. We shall show that these are all parameters of F (cf. Corollary 2). + such that D B = 1 under our Proof Whereas Dk contains 1, there are no matrices B ∈ T2n 1/2 hypothesis on n. It follows from (1.1) that F (h; p ) = 0 for some p (cf. [5, Lemma 15.2]). p This, together with (1.2) and p p,d −2 Dh ( p 1/2 ) = (d −1 fh )1/2 , proves the relation we want.
Our main result is the following: Theorem 2 Suppose that k ≡ n (mod 2) and F ∈ Sk+n (Sp2n (Z)) admits Fourier coeffi+ cients of the form (1.3). Then there is g ∈ Sk+1/2 (4) with the following properties: (1) If n ≡ 0, 1, 3 (mod 4), then c(m) = cg (m) for every m ∈ Dk . (2) If n ≡ 2 (mod 4), then the following assertions holds:
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(a) F = In,k (g); (b) if m is not a square, then c(m) = cg (m); (c) c( f 2 ) = cg ( f 2 ) + (c(1) − cg (1)) f k for every positive integer f . M (Sp (Z)) coincides with the image of I . Corollary 1 The space Sk+n n,k 2n
2 Modular forms of half-integral weight We recall certain properties of newforms for the Kohnen plus space (see [11] for details). For z ∈ C × , we denote by z 1/2 the square root of z such that −π/2 < arg z 1/2 ≤ π/2. For γ = ac db ∈ 0 (4) and τ ∈ H, we put c −1 (cτ + d)1/2 , j (γ , τ ) = d d where dc is the Kronecker symbol and 1 if d ≡ 1 (mod 4). d = √ −1 if d ≡ 3 (mod 4). For any function g : H → C and γ ∈ 0 (4), we define g|γ ∗ : H → C by g|γ ∗ (τ ) = j (γ , τ )−2k−1 g(γ τ ). Fix a multiple N of 4. We call a holomorphic function g on H a cusp form of weight k + 21 with respect to 0 (N ) if g|γ ∗ = g for every γ ∈ 0 (N ) and it vanishes at all cusps. The space of cusp forms of weight k + 21 with respect to 0 (N ) is denoted by Sk+1/2 (N ). We denote the mth Fourier coefficient of g ∈ Sk+1/2 (N ) by cg (m). The Kohnen plus space + Sk+1/2 (N ) is defined by + (N ) = {g ∈ Sk+1/2 (N ) | cg (m) = 0 unless m ∈ Dk }. Sk+1/2
An operator U (a) on formal power series is defined by c(m)q m |U (a) = c(am)q m m∈N
m∈N
for a positive integer a. An operator Uk (a 2 ) defined by c(m)q m |Uk (a 2 ) = c(a 2 m)q m m∈Dk
m∈Dk
+ maps Sk+1/2 (N ) into itself if each prime factor of a is that of 4−1 N . In what follows, we assume that k > 0 and N /4 is square-free. The space of newforms new,+ + Sk+1/2 (N ) for Sk+1/2 (N ) is by definition the orthogonal complement of + + Sk+1/2 ( p −1 N ) + Sk+1/2 ( p −1 N )|Uk ( p 2 ) , p|4−1 N + (N ) with respect to the Petersson where p extends over all prime divisors of 4−1 N , in Sk+1/2 new (4−1 N ). Let −1 inner product. The space of newforms for S2k ( 0 (4 N )) is denoted by S2k T˜ ( p 2 ) (resp. T ( p)) be the usual Hecke operator on the space of modular forms of half-integral (resp. integral) weight.
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Theorem 4 Suppose that N /4 is square-free and k is positive. new,+ + (N ) = ⊕a,d≥1, ad|4−1 N Sk+1/2 (4a)|Uk (d 2 ). (1) Sk+1/2 new,+ 2 2 (2) The operators T˜ ( p ) and Uk (q ), where ( p, 4−1 N ) = 1 and q|4−1 N , fix Sk+1/2 (N ). new,+ (N ) has an orthogonal C-basis which consists of common Hecke Moreover, Sk+1/2 eigenforms of these operators. (3) There is a bijective correspondence, up to scalar multiple, between Hecke eigenforms new (4−1 N ) and those in S new,+ (N ) in the following way. If f ∈ S new (4−1 N ) is a in S2k 2k k+1/2 primitive form, i.e.,
f |T ( p) = ω p f,
f |U (q) = ωq f
for every prime number p 4−1 N and prime divisor q of 4−1 N , then there is a nonzero new,+ Hecke eigenform g ∈ Sk+1/2 (N ) such that g|T˜ ( p 2 ) = ω p g, g|Uk (q 2 ) = ωq g for every prime number p 4−1 N and prime divisor q of 4−1 N . new,+ (N ), p a prime divisor of 4−1 N , and either 1 or −1. Then Proposition 1 Let g ∈ Sk+1/2 the following conditions are equivalent:
(i) if ψ p ((−1)k m) = , then cg (m) = 0; (ii) g|Uk ( p 2 ) = p k−1 g.
3 Jacobi forms and the Fourier–Jacobi coefficients It is useful to recall certain properties of Jacobi forms before turning to the proof of Theo+ rem 2. Let B ∈ T2n−1 . Put L = Z2n−1 and L ∗ = (2B)−1 L. Let T B+ be the set of all pairs (a, α) ∈ Z × L ∗ such that a > B[α]. Choose a complete representative (B) for L ∗ /L. For µ ∈ (B) we define the theta function ϑµB (τ, w) by ϑµB (τ, w) = e(τ B[α + µ] + 2B(α + µ, w)) (τ ∈ H, w ∈ C2n−1 ). α∈L
A Jacobi cusp form Φ of weight κ with index B is a holomorphic function on H × C2n−1 subject to the following conditions: (i) for every ac db ∈ SL2 (Z), we have cB[w] aτ + b w Φ , = (cτ + d)κ e Φ(τ, w); cτ + d cτ + d cτ + d (ii) for every ξ , η ∈ L, we have Φ(τ, w + ξ τ + η) = e(−2B(ξ, w) − B[ξ ]τ )Φ(τ, w); (iii) Φ has a Fourier expansion of the form cΦ (a, α)q a e(2B(α, w)). Φ(τ, w) = (a,α)∈T B+
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As a consequence of (ii), a Jacobi form Φ is written as a sum Φ(τ, w) = Φµ (τ )ϑµB (τ, w), µ∈(B)
where
Φµ (τ ) =
cΦ (a, µ)e((a − B[µ])τ ).
a∈N, a>B[µ]
The Bth Fourier–Jacobi coefficient FB of F ∈ Sκ (Sp2n (Z)) is defined by B Bα . c F (Ba,α )q a e(2B(α, w)), Ba,α = t FB (τ, w) = αB a
(a,α)∈T B+
Then FB is a Jacobi cusp form of weight κ with index B. The proof of the following proposition proceeds like the argument given for the relevant part of the proof of [3, Theorem 5.6]. + Proposition 2 Let B ∈ T2n−1 and F ∈ Sκ (Sp2n (Z)). Suppose that κ is even. Put k = κ − n, D B = det(2B)/2 and i(B)F(τ ) = FB,µ (4D B τ ). µ∈(B)
Then i(B)F is an element of
+ (4D B ). Sk+1/2
4 Proof of Theorem 2 Let v be a place of Q and B a nondegenerate symmetric matrix over Qv . We define the Hasse invariant h v (B) for B by h v (B) = (ai , a j )v , i< j
where ( , )v is the Hilbert symbol over Qv , if we take any orthogonal basis {xi } such that B[xi ] = ai . We modify h v (B) by putting ηv (B) = h v (B)((−1)n−1 , det B)v (−1, −1)n(n−1)/2 v so that an anisotropic kernel of B has dimension 2 − η p (B) when v is a finite place p (cf. [10, pp. 80–81]). We start with an easy lemma. + Lemma 2 Let B ∈ T2n−1 . Assume that D B is square-free. Then 1 if n ≡ 0, 1 (mod 4). η p (B) = −1 if n ≡ 2, 3 (mod 4). p|D B
Proof It follows from the definition of η∞ (B) that 1 if n ≡ 0, 1 (mod 4). η∞ (B) = −1 if n ≡ 2, 3 (mod 4).
(4.1)
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Note that Z2n−1 is a maximal Z p -integral lattice with respect to B for all primes p by assumpp tion. Taking the classification of maximal Z p -integral lattices (cf. [2, Sect. 9]) into account, we can observe that p divides D B whenever η p (B) = −1. Our assertion now follows from the product formula of the Hilbert symbol.
+ Lemma 3 For an arbitrary rational prime p, there exists B ∈ T2n−1 with D B = p.
Proof It suffices to prove that there exists a positive definite even integral lattice L of rank 2n − 1 with discriminant 2 p. We endow a lattice M = Z with the quadratic form x → 2(−1)n−1 px 2 . Let H be a quaternion algebra over Q which is ramified precisely at p and the real place. Fix a maximal order R of H . Take a lattice L such that L ⊗Z Z p = Pp −1 , where let Pp be the maximal ideal of R ⊗Z Z p , and L ⊗Z Zq = R ⊗Z Zq for every rational prime q distinct from p. We endow the lattice M = {x ∈ L | x ι = −x} with the quadratic form x → 2(−1)n px x ι , where x → x ι is the main involution of H . For each prime q, we define a lattice L q over Zq as follows: (M ⊗Z Zq ) ⊕ Hqn−1 if n ≡ 0, 1 (mod 4) Lq = (M ⊗Z Zq ) ⊕ Hqn−2 if n ≡ 2, 3 (mod 4) (cf. (4.1)). Here Hq Zq2 is a hyperbolic plane over Zq . Note that L q has discriminant 2 p. The Minkowski–Hasse theorem (see Theorem 6.10 on p. 225 of [10]) shows that there exists a positive definite quadratic space V over Q such that V ⊗Q Qq L q ⊗Zq Qq for every rational prime q. We can choose a lattice L ⊂ V such that L ⊗Z Zq L q for every rational prime q. Then L has the required properties.
+ Lemma 4 Let B ∈ T2n−1 and m ∈ N. Assume that D B equals a rational prime p. Then the following conditions are equivalent:
(i) there exists (a, α) ∈ T B+ such that D Ba,α = m; (ii) (−1)n m ≡ 0, 1 (mod 4) and (−1)n m −1 if n ≡ 0, 1 (mod 4).
= p 1 if n ≡ 2, 3 (mod 4). Proof We will consider the case when n ≡ 2, 3 (mod 4). The other case follows similarly. + For m to be of the form Dh for some h ∈ T2n , we must have m ∈ Dn . Choose an element c ∈ Z× in such a way that ψ (c) = −1. Since an anisotropic kernel of the quadratic form p p associated to B over Q p has dimension 3 by Lemma 2, B over Z p is equivalent to H pn−2 ⊕ K p ⊕ (−1)n−1 pc by the classification of maximal lattices √ (cf. [2, Sect. 9]), where K p is the norm form of the unramified quadratic extension Q p ( c). Since (−1)n D Ba,α = (−1)n 4 p(a − B[α]) ≡ cµ2
(mod 4 p)
for some µ ∈ Z p , (i) implies that (−1)n m is not a square modulo p or is divisible by p. One can easily check the other implication of our statement.
For there to be B ∈ Tr+ with det(2B) = 1, it is necessary and sufficient that r is a multiple of 8. Thus the following corollary is a consequence of Lemmas 3 and 4.
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Corollary 2 We have
{Dh | h ∈
+ T2n }
Dn
=
Dn − {1}
if n ≡ 0, 1, 3 (mod 4). if n ≡ 2 (mod 4).
+ in Lemma 5 Suppose that n ≡ 2, 3 (mod 4). Choose a rational prime p and B ∈ T2n−1 + such a way that D B = p. Letting (a, α) ∈ T B , we have if q = p. ∼ q (B ; X ) = q,D B ∼ (X ) F p,D B ∼ (X ) − p −1/2 p, p−2 D B ∼ (X ) if q = p.
to simplify notation. Here, we abbreviate B ∼ = Ba,α = tαBB Bα a
Proof As explained in the proof of Lemma 4, there is A ∈ GL2n (Z p ) such that 1 ⊕(n−2) 0 2 1 21 ∼ B [A] = 1 ⊕ 1 1−c ⊕ B , 2 0
where
B
∈ T2, p is of the form
2
B =
(−1)n−1 pc ∗ ∗ ∗
4
.
Note that λ p (B ∼ ) = −1, s p (B ∼ ) = s p (B ) and cd B d B∼ d B = =− . p p p It is easy to see that ψ p ((−1)n D B ∼ ) = −ψ p (−D B ) and f p (B ∼ ) = f p (B ). Moreover, the map 1 G → 2n−2 A−1 G
induces a bijection between D p
(B ) and D
p
(B ∼ ). Combining these facts with (1.1), we have
p (B ∼ ; X ) = (−1)f p (B ) F p (B ; −X ). F For any nondegenerate matrix h ∈ T2, p , it is well-known that p (h; X ) = F
p (h)
p −a/2 p,D p−a h (X ),
a=0
where p (h) = max{a ∈ N ∪ {0} | p −a h ∈ T2, p } (for example, see [6]). We conclude that p (B ; −X ) p (B ∼ ; X ) = (−1)f p (B ) F F
= (−1)f p (B ) ( p,D B (−X ) + p −1/2 p,D p−1 B (−X )) = p,D B ∼ (X ) − p −1/2 p, p−2 D B ∼ (X ), observing that p (B ) = 1 if and only if f p (B ) > 0. We can prove the formula for primes q distinct from p by a simpler way.
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The following corollary immediately follows from Lemmas 4 and 5. Corollary 3 Suppose that F ∈ Sk+n (Sp2n (Z)) has Fourier coefficients of the form (1.3). If D B equals a prime number p, then the Fourier expansion of i(B)F is as follows: (−1)k m i(B)F(τ ) = (c(m) − p k c( p −2 m))q m . 1− p m∈Dk
Put
(−1)k m ++ + =1 , (4 p) = g ∈ Sk+1/2 (4 p) | cg (m) = 0 whenever Sk+1/2 p (−1)k m new,++ new,+ =1 . (4 p) = g ∈ Sk+1/2 (4 p) | cg (m) = 0 whenever Sk+1/2 p
+ + We define the map P( p) : Sk+1/2 (N ) → Sk+1/2 (N p) by
g|P( p)(τ ) = g(τ ) − p 1−k g|(T˜ ( p 2 ) − Uk ( p 2 ))(τ ) (−1)k m = (cg (m) − p k cg ( p −2 m))q m , 1− p m∈Dk
provided that N and p are coprime. The following lemma is an easy consequence of Theorem 4 and [11, Propositions 2, 4, Corollary 2 (2)]. new,++ + ++ Lemma 6 The direct sum Sk+1/2 (4) ⊕ Sk+1/2 (4 p) is isomorphic to Sk+1/2 (4 p) via the C-linear map (g, h) → g|P( p) + h. + (4) such that Lemma 7 The notation being as in Corollary 3, there is g B ∈ Sk+1/2
i(B)F = g B |P( p). + (4) and Proof Proposition 2, together with Corollary 3 and Lemma 6, gives g ∈ Sk+1/2 new,++ h ∈ Sk+1/2 (4 p) such that
i(B)F = g|P( p) + h. Seeking a contradiction, we suppose h = 0. Note that ch (mp 2 f ) = p (k−1) f ch (m) by Proposition 1. Choose m ∈ N with properties: ch (m) = 0, Of course, we have
p −2 m ∈ / Dk .
(−1)k m p
= 1.
2f
Comparing the coefficients of q mp , we have c(m) = cg (m) +
1−
ch (m) k (−1) m p
and c(mp 2 f ) − p k c(mp 2 f −2 ) = cg (mp 2 f ) − p k cg (mp 2 f −2 ) + ch (mp 2 f )
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for f ∈ N. Therefore, we have
275
⎞
⎛
c(mp 2 f ) = cg (mp 2 f ) + ch (m) p (k−1) f ⎝1 + p + p 2 + · · · + p f −1 +
1−
pf (−1)k m p
⎠ .
It follows that the limit lim f →∞ p −k f c(mp 2 f ) is nonzero. On the other hand, for every prime q distinct from p, we have c(mq 2 f ) = O(q (k−1/2+) f ) for any > 0 by the Ramanujan– Petersson conjecture combined with Theorem 4. This is a contradiction since p is arbitrary. Hence h must be zero, and g B = g works.
Lemma 8 Suppose that p, B and g B are related as in Lemma 7, and similarly for p , B and g B . Then we have g B = g B . Proof Corollary 3 and Lemma 7 show that g B |P( p)P( p ) = g B |P( p)P( p ). Since P( p) and P( p ) are injective in view of Theorem 4 (1), we have g B = g B as claimed.
Now we are ready to prove Theorem 2. We may assume that n ≡ 2, 3 (mod 4) by [9]. We define g to be the function g B in Lemma 7, which is independent of the choice of B on account of Lemma 8. Fix a positive integer m such that (−1)k m ≡ 0, 1 (mod 4). If there exists a rational prime p such that (−1)k m
= 1, p then Corollary 3, coupled with Lemmas 3 and 7, shows that c(m) − p k c( p −2 m) = cg (m) − p k cg ( p −2 m). If (−1)k m is not a square, then we can find such a prime and deduce that c(m) = cg (m), which proves the case n ≡ 3 (mod 4). Assume that n ≡ 2 (mod 4). We then have c( f 2 ) = cg ( f 2 ) + (c(1) − cg (1)) f k . In view of Lemma 1 and Remark 1, the proof of Theorem 2 is now complete.
Acknowledgments The author should like to express his sincere thanks to Prof. Ikeda for suggesting the problem, constant help throughout this paper, encouragement and patience. The author is supported by the Grant-in-Aid for JSPS fellows.
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8. Kohnen, W.: modular forms of half-integral weight to Siegel modular forms of even genus. Math. Ann. 322, 787–809 (2002) 9. Kohnen, W., Kojima, H.: A Maass space in higher genus. Compositio. Math. 141, 313–322 (2005) 10. Scharlau, W.: Quadratic and Hermitian Forms. Springer, Berlin (1985) 11. Ueda, M., Yamana, S.: On newforms for Kohnen plus spaces. Math. Z. (to appear)
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