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

32 W. DUKE AND Y. LI
From the proof of Theorem 5.1, we know that Φ 1,B[n] (ĝ ψ , z) + Φ 1,B[n] −1(ĝ ψ , z) has a constant
term ρ N,B,ψ (0) and no pole. So Φ ∗ N,B (ĝ ψ, z) is also O(1/y) at the cusp 0.
For m ≥ 1, let P m,N (z, s) be the Poincaré series of level one, weight two defined by
∑
P m,N (z, s) := (y s e 2πimz )| 2 γ.
It is characterized by the property that
γ∈Γ ∞\Γ 0 (N)
lim 〈h(z), P m,N(z, s)〉 = mc(h, m)
s→0
for any h(z) = ∑ m≥1 c(h, m)qm ∈ S 2 (N). Since {c(f, −m) : m ≥ 1} ∈ Λ N , we know that
〈
(78) lim h(z), ∑ 〉
δ N (m)
c(f, −m)P
s→0 m m,N(z, s) = 0,
m≥1
δ N (m)
m c(f, −m)P m,N(z, s) ∈ S 2 (N) is 0.
for any h(z) ∈ S 2 (N). So lim s→0
∑m≥1
Now we can consider the inner product between Φ ∗ N,B (ĝ ψ, z) and this linear combination
of Poincaré series and obtain
∑
δ
4π lim N (m)
c(f,
s→0 m −m)〈Φ∗ N,B(ĝ ψ , z), P m,N (z, s)〉 = 0.
m≥1
Since Φ ∗ N,B (ĝ ψ, z) is O(1/y) at both cusps, we can apply Rankin-Selberg unfolding as in the
proof of Theorem 5.1. The limit on the left hand side above then breaks up into two parts.
The first part is
⎛
a N,B (ĝ ψ , m) − ∑ a N,B (ĝ ψ , −n)c(P n,N , m)
∑
⎜
n≥0
⎟
(79)
m≥1
δ N (m)c(f, −m) ⎜
⎝
− 24Nρ N,B,ψ(0)
N 2 − 1
(
σ1
( m
N
⎞
)
− σ1 (m) ) ⎟
⎠
Since (f|W N )(z) = −N(f|U N )(z) and c(P n,N , 0) = 0 for n ≥ 1, Lemma 6.5 tells us that for
n ≥ 0
∑
δ N (m)c(f, −m)c(P n,N , m) = −δ N (n)c(f, n).
m≥1
So expression (79) becomes
Σ ′ f,N,B,ψ − 24Nρ N,B,ψ(0)
N 2 − 1
∑
δ N (m)c(f, −m) ( (
σ m
)
1 N − σ1 (m) ) .
m≥1
The second part involves the limit of an infinite sum, which can be evaluated as
Σ f,N,B,ψ − (N + 1) ∑ m ′ ≥1
c(f, −Nm ′ )(ρ N,B,ψ (m ′ ) + 24σ 1 (m ′ )ρ N,B,ψ (0)).
Adding the last two expressions together, which yields 0, and substituting into (76), we
obtain equation (74). The constant C f appears since
∑ ( (
(80) c(f, 0) = 24
m
)
N 2 Nσ1
−1
N − σ1 (m) ) δ N (m)c(f, −m)
m≥1
from applying Lemma 6.5 to P 0,N (z) ∈ M 2(N) ! and f(z) ∈ M !,new
0 (N). Notice if c(f, −m) ∈ Z
for all m ≥ 1, then the denominator of c(f, 0) is independent of f.
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