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

MOCK-MODULAR FORMS OF WEIGHT ONE 11
c n,u = 0 for u ≤ 0. Applying the Rankin-Selberg unfolding trick to 〈 ˜Q m (·, s), e n (·)〉 gives us
∫ ∞
〈 ˜Q m (·, s), e n (·)〉 = (4π|m|) 1/2 e −2π|m|y (4π|m|y) s−1 c n,m W sgn(m)
0
= (4π|m|) 1/2 Γ ( s − 1
c − it ) (
2 n Γ s −
1
+ it )
2 n
(21)
n,m ( )
Γ s − sgn(m)
2
2 ,s n− 1 2
(4π|m|y) dy
y
The last step uses the Mellin transform of the W -Whittaker function [2, eq. (8b)] and
the substitution s n = 1/2 + it n . When Re(s) > 1, the sum defining D(z, s) is absolutely
convergent [35, Satz 8.1 ], since y 1/2 Q m (z, s) ∈ ˜D 1 (M, ν) for Re(s) > 1. When 0 < Re(s) ≤ 1,
we can write D(z, s) as
D(z, s) = (4π|m|) ∑ 1/2 Γ ( s + 1 − 1
c − it ) (
2 n Γ s + 1 −
1
+ it )
2 n s − sgn(m)
n,m (
)
2
e n (z) ·
n∈N
Γ s + 1 − sgn(m)
(s − 1
2
2 )2 + t 2 n
Since t 2 n = λ n − 1/4 and ∑ n>M λ−2 n converges [35, Satz 8.1], we can apply Cauchy-Schwarz
inequality to see that the sum on the right hand side above converges absolutely on compact
subsets of {s ∈ C : Re(s) > 0, s ≠ 1/2}. At s = 1/2, the first R terms in the sum produce a
simple pole since t n = 0 for all 1 ≤ n ≤ R. The rest of the sum still converges absolutely. So
the right hand side above gives the analytic continuation of D(z, s) to Re(s) > 0.
For the continuous spectrum, the contribution C(z, s) can be treated similarly. For any
cusp ι, we can write the Fourier expansion of E ι (z, s) at infinity in the following form (for
ι = ∞, see [20, eq. (76)’])
E ι (z, s) = y s + ψ ι (s)y 1−s + ∑ ψ ι,m (s)W sgn(m)
,s− 1 (4π|m|y)e 2πimx ,
2 2
m≠0
where ψ ι (s) and ψ ι,m (s) are products of gamma factors and Selberg-Kloosterman zeta functions.
It is well-known that The Eisenstein series can be analytically continued in s to the
whole complex plane. When Re(s) > 1/2, the poles of E ι (z, s) are in the interval s ∈ (1/2, 1]
[35, Satz 10.3 ]. On the line Re(s) = 1/2, E ι (z 0 , s) is holomorphic in s for any fixed z 0 ∈ H
[35, Satz 10.4]. So both ψ ι (s) and ψ ι,m (s) admit analytic continuation to Re(s) > 0 and are
holomorphic on Re(s) = 1/2. Using the same unfolding trick above, we can evaluate
〈 ˜Q m (·, s), E ι (·, 1 + ir)〉 = ψ ( 1 2 ι,m + ir) (4π|m|) 1/2 Γ ( s − 1 − ir) Γ ( s − 1 2 ( ) + ir) 2 .
2
Γ s − sgn(m)
2
Then C(z, s) can be written as
C(z, s) = ∑ ι
(22)
= ∑ ι
∫
1 ∞
4π −∞
∫
1 ∞
4π −∞
(
ψ 1
ι,m + ir) (4π|m|) 1/2 Γ ( s − 1 − ir) Γ ( s − 1 2
( ) + ir) 2
E
2 ι (z, 1 + ir)dr
Γ s − sgn(m)
2
2
(
ψ 1
ι,m + ir) (4π|m|) 1/2 Γ ( s + 1 − 1 − ir) Γ ( s + 1 − 1
(
2
)
+ ir) 2
2
Γ s + 1 − sgn(m)
2
(s − sgn(m) )
2
·
(s − 1 − ir)(s − 1 + ir) · E ι(z, 1 + ir)dr.
2
2 2