Mock-modular forms of weight one - UCLA Department of Mathematics

34 W. DUKE AND Y. LI
When r ≥ 1, the situation is similar. In this case, we have
( )
f lift,N,B (z) = − (N + 1)T N j lift,B
N r m
(z) + (Nj ′ N r−1 m ′(pz) + j N r+1 m ′(pz))T N(ϑ B )
+ g f (z),
〈f lift,N,B , g ψ 〉 reg = − (N + 1)(ψ([n]) + ψ([n]))〈j lift,B
m ′ , g ψ 〉 reg + Nρ N,B,ψ (N r−1 m ′ )
+ ρ N,B,ψ (N r+1 m ′ ) + 〈g f , g ψ 〉,
for some g f (z) ∈ M 1 + (p) with integral linear combination of c(f, n)’s as Fourier coefficients.
By Proposition 4.1 and Corollary 5.2, the proposition holds for f(z) ∈ S 2 .
Now, we will prove the proposition when f(z) is in the integral span of S 1 and show that
the denominator of the constant β f,B in (68) is independent of f. This will finish the proof
of Proposition 6.1. By Lemma 6.2 and Lemma 6.6, it is enough to show that Σ f,N,B,ψ can
also be written in the form of the right hand side of equation (68). To accomplish that,
we will relate Σ f,N,B,ψ to the automorphic Green’s function using the counting arguments in
Proposition (3.11) from [24]. In their notation, let R N be a subset of M 2 (Z) containing all
matrices whose lower left entry is divisible by N. For z, z ′ ∈ H and k ≥ 0, let ρ m N (z, z′ , k) be
the number of γ = ( a c d b ) ∈ R N/{±1} such that det(γ) = m and
cosh d(z, γz ′ ) = 1 + 2Nk
pm .
Define the automorphic Green’s function of level N to be
∑
G N,s (z, z ′ ) :=
g s (z, γz ′ ) +
γ∈Γ 0 (N)/{±1}
z≠γz ′
where
∑
γ∈Γ 0 (N)/{±1}
z=γz ′
g s (z) := lim
w→z
(g s (z, w) − log |2πiη(z) 4 (z − w)| 2
g s (z),
= − log |2π(z − z)η(z) 4 | 2 + 2 Γ′ Γ′
(s) − 2 (1).
Γ Γ
When gcd(N, m) = 1, the m th Hecke operator T m acts on z ′ in G N,s (z, z ′ ) and defines
G m N,s (z, z′ ). It can be written as
∑
G m N,s(z, z ′ ) :=
g s (z)
γ∈R N /{±1}
det γ=m
= −2 ∑ k>0
γ∈R N /{±1}
det γ=m
z=γz ′
z≠γz ′ g s (z, γz ′ ) + ∑
ρ m N(z, z ′ , k)Q s−1
(1 + 2Nk
pm
)
+ ρ m N(z, z ′ , 0)g s (z).
Notice that ρ m N (z, z′ , k) could be non-zero when k is not an integer. When z, z ′ ∈ H are
Heegner points of discriminant −p, then ρ m N (z, z′ , k) is necessarily supported on integral k’s.
For j = 1, 2, let A j ∈ Cl(F ) and τ j ∈ Γ 0 (N)\H be the image of the Heegner points
x j = (O F , A j , [n]). Let a j be integral ideals in the class A j such that n|a j , N(a j ) = A j . Then
by Proposition (3.11) in [24], the counting function ρ m N (τ 2, τ 1 , k) satisfies
ρ m N(τ 2 , τ 1 , k) =#{(α, β) ∈ (a −1
1 a −1
2 × a −1
1 a −1
2 n)/{±1} | N F/Q (α) = Nk+pm
A 1 A 2
,
N F/Q (β) = Nk
A 1 A 2
, A 1 A 2 (α − β) ≡ 0 (mod d)},
=r A1 A −1(Nk
+ pm)r A
2 1 A 2 [n] −1(k)δ(k)