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

MOCK-MODULAR FORMS OF WEIGHT ONE 25
Using the Fourier expansion of Φ B (ĝ ψ , z) and Rankin-Selberg unfolding, we can rewrite equation
(61) as
( )
k
(62) a B (ĝ ψ , m) + 24σ 1 (m)a B (ĝ ψ , 0) = lim δ(k)r B (pm + k)r ψ (k)φ s ,
where
φ s (µ) :=
∫ ∞
1
s→0
∑
k≥1
du
(1 + µu) 1+s u .
Comparing Q s−1 (t) and φ s−1 (µ), we see that φ 0 (µ) = 2Q 0 (1 + 2µ) and
φ s−1 (µ) − 2Γ(2s) Q
sΓ(s) 2 s−1 (1 + 2µ) = O(µ −1−s ).
( ) ( )
2
That means we can substitute Q
sΓ(s) 2 s−1 1 + 2k
k
for φ
pm s−1 and in the limit as s approaches
1. Together with (8) and (60), equation (61)
pm
becomes
(63)
〈j lift,B
m , g ψ 〉 reg +24σ 1 (m)〈ϑ B , g ψ 〉 =
− lim
s→1
∑
A
ψ(A) ∑ k≥1
pm
( ))
δ(k)r B (pm + k)r A (k)
(−2Q s−1 1 + 2k .
pm
Now the counting argument in Proposition (3.11) in [23] tells us that
δ(k)r B (pm + k)r A (k) = ρ m (A, B, k),
where ρ m (A, B, k) counts the number of γ ∈ M 2 (Z)/{±1} such that det(γ) = m and
cosh d(τ √ AB −1 , γτ √ AB ) = 1 + 2k
pm .
In particular when k = 0, the number of γ ∈ M 2 (Z)/{±1} such that det(γ) = m and
γτ √ AB = τ√ AB
is exactly r −1 B (pm) = r B (m). Since k ≥ 1 in the summation, equation (63)
becomes
(64)
〈jm
lift,B ,g ψ 〉 reg + 24σ 1 (m)〈ϑ B , g ψ 〉 = − lim
= − lim
ɛ→0
lim
s→1
∑
A
s→1
∑
A
ψ(A)
∑
(
g s τ
√
AB −1, γτ √ AB
γ∈M 2 (Z)/{±1}
det(γ)=m,
γτ √ AB≠τ√ AB −1
ψ(A) ( G m s (τ √ AB −1 + ɛ, τ √ AB ) − r B(m)g s (τ √ AB −1 + ɛ, τ √ AB −1 ) )
where g s is defined in (57). From this expression, it is clear that the choice of these CM
points do not matter.
)
Since g 1 (τ + ɛ, τ) = − log
(1 + 4Im(τ)2 and ψ is non-trivial, we see that
ɛ 2
lim lim
ɛ→0
s→1
∑
A
ψ(A)g s (τ √ AB −1 + ɛ, τ √ AB −1 ) = − ∑ A
Also by equation (59), we have
∑
lim ψ(A)G m s (τ √
s→1
AB
+ ɛ,τ √ −1 AB ) = ∑
A
A
4πσ 1 (m) ∑ A
ψ(A) log(y 2 √
AB −1).
ψ(A) log |Ψ m (τ √ AB −1 + ɛ, τ √ AB )|2 −
ψ(A) lim
s→1
(E(τ √ AB −1 + ɛ, s) + E(τ √ AB , s)).
)