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

10 DAVID HANSEN AND YUSRA NAQVI
Let δ ′ = gcd(c, δ). We have that there exists an integer y such that (δ/δ ′ )|(cy + d), and so
we get
f(δz)
k
a b
c d
= δ −k/2
f(z)
Inserting this into the definition of g(z) gives
g(z)
2k
a b
=
c d
=
δ|r
gcd(δ,r/δ)=1
δ|r
gcd(δ,r/δ)=1
aδ/δ ′ δ
k
′ (ay + b)
c/δ ′ δ ′ (cy + d)/δ
= (δ/δ ′ ) −k
δ ′2z − δ ′ y
f
χδ(−1)f(δz)f(rz/δ)
δ
2k
.
a b
c d
δ
−k r/δ
−k χδ(−1)
f
(δ, c) (r/δ, c)
δ ′ −y
0 δ/δ ′
δ ′2 z − y ′
δ
f
δ ′′2 z − y ′′
where δ ′ is as before, δ ′′ = (r/δ, c) and y ′ and y ′′ are integers that depend on δ. This
transforms into
(3.17) gγ(z) =
δ|r
gcd(δ,r/δ)=1
r
−k χδ(−1) f
(r, c)
δ ′2 z − y ′
δ
f
δ ′′2 z − y ′′
We now consider
(3.18)
a
g(2z)
2k c
b
= 2
d
−k
2
g(z)
2k 0
0 a
1 c
b
,
d
which gives us that g(2z) |2k
a b
c d = gγ(2z) if c is even or gγ((z − x)/2) if c is odd, where
x is some integer that depends on d. This yields
h(z)
2k
a b
= gγ(z) − 2
c d
k−1 gγ(2z) or
= gγ(z) − 2 −k−1 gγ((z − x)/2).
Thus, in all cases, the constant term of the Fourier expansion is a constant multiple of
( r
(r,c) )−k a(0) 2 χδ(−1), and hence this term vanishes if and only if f is a cusp form or
(3.19)
δ|r
gcd(δ,r/δ)=1
χδ(−1) = 0.
In particular, this sum vanishes if and only if χr decomposes into a product of Dirichlet
characters to prime power moduli which are not all even. Note that by [7], this is equivalent
to θ(χr; z) being a cusp form. This same method can be applied to modular forms of higher
level; however, the computations are more complicated.
δ
.
δ
,