18 W. DUKE AND Y. LI Pro<strong>of</strong>. The necessity part follows from Proposition 3.4. Since h(z) is a cusp form, the summation in (44) is only over n < 0. When ˜g(z) ∈ M −ɛ 1 (p) is <strong>modular</strong>, the sufficiency follows from an application <strong>of</strong> Serre duality [5, §3] (also see [8, Theorem 6]). When ˜g(z) is mock-<strong>modular</strong>, let ∑ n
MOCK-MODULAR FORMS OF WEIGHT ONE 19 For convenience, we write E(τ, s) for E 1 (τ, s). Kr<strong>one</strong>cker’s first limit formula states that (49) ζ(2s)E(τ, s) = π s − 1 + 2π ( γ − log(2) − log( √ y|η(τ)| 2 ) ) + O(s − 1), where γ is the Euler constant. Using (49) and Rankin-Selberg unfolding trick, we can relate the inner product between dihedral new<strong>forms</strong> to logarithm <strong>of</strong> u A as follows. Proposition 4.1. Let ψ, ψ ′ be characters <strong>of</strong> Cl(F ), with ψ non-trivial. When ψ ′ = ψ or ψ, we have (50) 〈g ψ , g ψ ′〉 = 1 ∑ ψ 2 (A) log |u A |. 12 Otherwise, 〈g ψ , g ψ ′〉 = 0. A∈Cl(F ) Pro<strong>of</strong>. Since p is prime, the Eisenstein series E p (z, s) has a simple pole at s = 1 with residue , which is independent <strong>of</strong> z. So we have the relationship 3 π(p+1) 3 π(p + 1) · 〈g ψ, g ψ ′〉 = Res s=1 g ψ (z)g ψ ′(z)E p (z, s)y ∫Γ dxdy . 0 (p)\H y 2 Now, we can use the Rankin-Selberg method to unfold the right hand side and obtain ∫ g ψ (z)g ψ ′(z)E p (z, s)y dxdy = Γ(s) ∑ r ψ (n)r ψ ′(n) y 2 (4π) s n s Γ 0 (p)\H Let ρ ψ : Gal(Q/Q) → GL 2 (C) be the representation induced from ψ. Then it is also the <strong>one</strong> attached to g ψ (z) via Deligne-Serre’s theorem. Up to Euler factors at p, the right hand side is L(s, ρ ψ ⊗ ρ ψ ′), the L-function <strong>of</strong> the tensor product <strong>of</strong> the representations ρ ψ and ρ ψ ′. From the character table <strong>of</strong> the dihedral group D 2H(p) , we see that ρ ψ ⊗ ρ ψ ′ = ρ ψψ ′ ⊕ ρ ψψ ′. So when ψ ′ ≠ ψ or ψ ′ , the L-function L(s, ρ ψ ⊗ρ ψ ′) is holomorphic at s = 1 and 〈g ψ , g ψ ′〉 = 0. Otherwise, we have ∑ r ψ (n)r ψ (n) = ζ(s)L(s, χ p)L(s, ρ ψ 2) , n s ζ(2s)(1 + p −s ) n≥1 Putting these together, we obtain (51) 〈g ψ , g ψ 〉 = p 2π L(1, χ p)L(1, ρ 2 ψ 2). From the theory <strong>of</strong> quadratic <strong>forms</strong>, we have √ ∑ pL(s, ρψ 2) = ζ(2s) ψ 2 (A)E(τ A , s). A∈Cl(F ) Since ψ is non-trivial and H(p) is odd, ψ 2 is non-trivial and (49) implies that L(1, ρ ψ 2) = −√ 2π ∑ ψ 2 (A) log( √ y A |η(τ A )| 2 ), p A∈Cl(F ) Along with the class number formula for p > 3 L(1, χ p ) = 2πH(p) w p √ p n≥1 = πH(p) √ p ,