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

8 DAVID HANSEN AND YUSRA NAQVI
Recall that χr is odd here, so χr(−1) = χ r/d(−1)χd(−1) = −1, and so we have that
χd(−1) = −χ r/d(−1) = ±1. Selberg inversion applies again, and the derivatives of modular
forms appearing in the definition of g(z) arise naturally from the linear form in m and n
appearing in (3.11).
In the odd case, it is not immediately clear that g(z) is in fact a modular form, since it
contains derivatives of modular forms. However, it is in fact easy to prove modularity by
employing the following useful fact.
Proposition 3.1. Let f(z) be a modular form of weight k on some subgroup of SL2(Z).
Then 1 d
2πi dz f(z) = ( ˜ f(z) + kE2(z)f(z))/12, where E2(z) is the Eisenstein series defined in
(1.1) and ˜ f(z) is a modular form of weight k + 2.
Note that E2 is not a modular form (see [6]). Using this proposition, we easily obtain
g(z) = 1
πi
χd(−1)df ′ (dz)f(rz/d)
= 1
2π 2 i 2
= 1
12πi
= 1
12πi
d|r
gcd(d,r/d)=1
d|r
gcd(d,r/d)=1
d|r
gcd(d,r/d)=1
d|r
gcd(d,r/d)=1
χd(−1)f(rz/d) ∂
∂z f(dz)
χd(−1)f(rz/d)( ˜ fd(z) + kE2(z)f(dz))
χd(−1)f(rz/d) ˜ fd(z),
where ˜ fd(z) is a modular form of weight k+2 and level dN. The sum involving E2’s vanishes
due to cancellation in characters, namely χd(−1) = −χ r/d(−1).
To complete the proofs of Theorems 1.1 and 1.2, it suffies to compute the levels of the
relevant Shimura lifts. Because g(z) lies in the space M2k(N ′ r, ψ 2 ), it is easy to see by the
general theory of twists (see [6], Sec. 2.2) that gχr(z) ∈ M2k(N ′ r 3 , χ 2 rψ 2 ). However, we can
in fact show that gχr(z) lies in the space M2k(N ′ r 2 , χ 2 rψ 2 ). To do this, we demonstrate the
invariance of gχr(z) under a complete set of representatives of right cosets of Γ0(N ′ r 3 ) in
Γ0(N ′ r 2 ). By Proposition 2.5 of [2], we have that [Γ0(N ′ r 2 ) : Γ0(N ′ r 3 )] = r, so such a set
of representatives is given by
(3.12) αj :=
1 0
jN ′ r2
1
for j = 0, 1, 2, ..., r − 1. For convenience, we define the slash operator for γ ∈ GL + 2 (Q) by
(3.13) f(z) |k γ := f(γz)(cz + d) −k (det γ) k/2 .
With this notation, we need to show gχr(z) |k αj = gχr(z) for j = 0, 1, 2, ..., r − 1. Using
Proposition 17 in Sec. 3.3 of [3] and defining τ(χr) := r−1
m=0 χr(m)e 2πim/r , we first write