Shimura lifts of half-integral weight modular forms - Department of ...

6 DAVID HANSEN AND YUSRA NAQVI
3. Proofs of Theorems 1.1 and 1.2
We begin by presenting the proof of the formula for the lift in Theorem 1.1. From the
definition of F (z), we have F (z) = ∞ n=0 b(n)qn with
(3.1) b(n) =
n − m2
χr(m)a ,
4r
m∈Z
where f(z) = ∞ n=0 a(n)qn is as in the statement of Theorem 1.1. As above, the Shimura
lift is given by
∞
(3.2) S1(F ) = A(n)q n
with the coefficients A(n) defined by
(3.3)
∞
n=1
n=1
A(n)n −s = L(s − k + 1, χrψχ 2k
4 )
We also need the coefficients defined by
∞
(3.4)
cd(n)q n := f(dz)f(rz/d) =
n=0
n=0 m∈Z
∞
b(n 2 )n −s .
n=1
∞
n − dm
a(m)a q
r/d
n .
Throughout the proof, we use the convention that a modular form coefficient is zero if its
argument is negative or not integral. From (3.1) it is easy to see that
(3.5) b(n 2 ) =
(n − m)(n + m)
χr(m)a
.
4r
m∈Z
This is a finite sum with non-zero coefficients whenever (n − m)(n + m)/(4r) ∈ N. Note
that n + m and n − m must both be even for n and m to be integers with 4|(n 2 − m 2 ). Let
gcd((n − m)/2, r) = d. Thus, m ≡ n (mod 2d) and m ≡ −n (mod 2r/d). Now suppose
gcd(d, r/d) = d ′ > 1. This implies that m ≡ n ≡ −n ≡ 0 (mod 2d ′ ), so d| gcd(m, r) and
so χr(m) = 0. Therefore, we only consider the cases in which gcd(d, r/d) = 1. We have
m = n + 2dm ′ for some m ′ ∈ Z, so n − m = −2dm ′ and n + m = 2n + 2dm ′ . Thus,
(3.6)
(n − m)(n + m)
4r
= −m′ (n + dm ′ )
.
r/d
Also, since m ≡ n (mod d) and m ≡ −n (mod r/d), we have that
= χd(m)χ r/d(m)
= χ r/d(−1)χd(n)χ r/d(n)
χr(m)
= χd(n)χ r/d(−n)
= χ r/d(−1)χr(n),