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

20 W. DUKE AND Y. LI
we arrive at
(52) 〈g ψ , g ψ 〉 = −H(p)
∑
A∈Cl(F )
Since ψ 2 is non-trivial, we have equation (50).
ψ 2 (A) log( √ y A |η(τ A )| 2 ).
Remarks.
i) By the same procedure, one can analyze the inner product between any weight one
newforms. In particular, if g(z), h(z) ∈ S 1 (p) arises from different types of Galois representations,
then 〈g, h〉 = 0.
ii) This proposition is really a proven case of Stark’s conjecture in the abelian, rank one
case.
Next we prove the existence of the special pre-image.
Proposition
∑
4.2. Let ψ be a non-trivial character of Cl(F ). Then there exists ˜g ψ (z) =
n≥n 0
r + ψ (n)qn ∈ M − 1 (p) with shadow g ψ (z) such that
(i) When χ p (n) = 1 or n < − p+1 , the coefficient 24 r+ ψ
(n) equals to zero.
(ii) For n ≤ 0, the coefficients r + ψ
(n) are of the form
r + ψ
(n) = −β
∑
A∈Cl(F )
ψ 2 (A) log |u(n, A)|,
where β ∈ Q depends only on p, and u(n, A) ∈ O × H
depends only on n and A.
(iii) Let σ C ∈ Gal(H/F ) be the element associated to the class C ∈ Cl(F ) via Artin’s isomorphism.
Then it acts on the units u(n, A) by
σ C (u(n, A)) = u(n, AC −1 ).
Proof. From Lemma 3.3, we know that the dimension of S + 1 (p), denoted by d + , is 1 2 (H(p) −
1)+d − . Let {g 1 , g 2 , . . . , g d+ } be a basis of S + 1 (p). Suppose we have chosen n 1 , n 2 , . . . , n d+ > 0
such that χ p (n k ) ≠ −1 and the matrix
P := [c(g j , n k )] 1≤j,k≤d+
is invertible. Then for any ψ, there exists ˜g ψ (z) ∈ M − 1 (p) with shadow g ψ (z) and principal
part coefficients r + ψ (−n) = 0 for all positive integer n not in the set {n 1, n 2 , . . . , n d+ } by
Proposition 3.5. Furthermore, these r + ψ
(−n) can be uniquely determined from solving a
d + × d + system of equations. We will show that such ˜g ψ (z) can be made to satisfy the
proposition.
First, we can choose {n j : 1 ≤ j ≤ d + } such that n j ≤ (p + 1)/24. Let {g 1 , g 2 , . . . , g d+ } to
be a q-echelon basis. Then these n j ’s can be chosen to be bounded by the supremum of the
set
{ord ∞ h(z) : h(z) ∈ S 1 + (p)}.
Now, the square of a cusp form in S 1 (p) is in S 2 (p), the space of weight 2 cusp form of trivial
nebentypus on Γ 0 (p). This gives rise to a holomorphic differential form on the modular curve
X 0 (p). Atkin showed in [1] that ∞ is not a Weiestrass point of the modular curve X 0 (p),
i.e. ord ∞ f(z) is no more than the genus of X 0 (p). Using the Riemann-Hurwitz formula (see
[37]), we know that the genus of X 0 (p) is bounded by (p + 1)/12. So the n j ’s can be chosen
to be bounded by (p + 1)/24.
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