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

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8 W. DUKE AND Y. LI<br />

2. Existence <strong>of</strong> Harmonic Maass Forms <strong>of</strong> Weight One<br />

Given a cusp form <strong>of</strong> <strong>weight</strong> <strong>one</strong>, we will show the existence <strong>of</strong> a <strong>weight</strong> <strong>one</strong> mock-<strong>modular</strong><br />

form with it as shadow in this section. There are several different approaches to proving this<br />

result, such as a geometric approach in [9]. For completeness and since it might have some<br />

independent interest, in this section we prove the existence by analytically continuing <strong>weight</strong><br />

<strong>one</strong> Poincaré series via the spectral expansion.<br />

We begin with some basic definitions. Let k ∈ Z. For any function f : H → C and<br />

γ ∈ GL + 2 (R), define the <strong>weight</strong> k slash operator | k γ by<br />

(f| k γ)(z) := (det(γ))k/2 f(γz),<br />

(cz+d) k<br />

where γz is the linear fractional transformation <strong>of</strong> z under γ. We will write f|γ for f| k γ<br />

when the <strong>weight</strong> <strong>of</strong> f is understood. For M ∈ Z + let Γ 0 (M) denote the usual congruence<br />

subgroup <strong>of</strong> SL 2 (Z) <strong>of</strong> level M, namely<br />

Γ 0 (M) = {( a c d b ) ∈ SL 2(Z) ( a c d b ) ≡ ( ∗ 0 ∗<br />

∗ ) mod M}.<br />

Let ν : Z/MZ → C × be a Dirichlet character such that ν(−1) = (−1) k and ν(γ) := ν(d) for<br />

γ = ( a c d b ) ∈ Γ 0(M). Let F k (M, ν) be the space <strong>of</strong> smooth functions f : H → C such that for<br />

all γ ∈ Γ 0 (M)<br />

(f | k γ)(z) = ν(γ)f(z).<br />

Recall from (1) and (2) the differential operator ξ k and the <strong>weight</strong> k hyperbolic Laplacian<br />

∆ k . Let z = x + iy. Then ∆ k can be written as<br />

−∆ k = −y 2 (<br />

∂ 2<br />

+ ∂2<br />

∂x 2<br />

) (<br />

+ iky<br />

∂y 2<br />

∂<br />

+ i ∂<br />

∂x ∂y<br />

Then f(z) ∈ F k (M, ν) is a <strong>weight</strong> k harmonic weak Maass form <strong>of</strong> level M and character ν<br />

(or more briefly, a weakly harmonic form) if it satisfies the following properties.<br />

(i) f(z) is real-analytic.<br />

(ii) ∆ k (f) = 0.<br />

(iii) The function f(z) has at most linear exp<strong>one</strong>ntial growth at all cusps <strong>of</strong> Γ 0 (M).<br />

From now on we fix k = 1 and write | for the slash operator | 1 . Let H ! 1(M, ν) be the<br />

space <strong>of</strong> weakly harmonic <strong>modular</strong> <strong>forms</strong> <strong>of</strong> <strong>weight</strong> <strong>one</strong>, level M and character ν. Denote<br />

by M ! 1(M, ν) , M 1 (M, ν) and S 1 (M, ν) the usual subspaces <strong>of</strong> weakly holomorphic <strong>modular</strong><br />

<strong>forms</strong>, holomorphic <strong>modular</strong> <strong>forms</strong> and cusp <strong>forms</strong>, respectively. Let M ! 1(M, ν) be the space<br />

<strong>of</strong> mock-<strong>modular</strong> <strong>forms</strong> with shadows in M 1 (M, ν) and M 1 (M, ν) ⊂ M ! 1(M, ν) the subspace<br />

<strong>of</strong> mock-<strong>modular</strong> <strong>forms</strong> whose shadows are in S 1 (M, ν).<br />

From (2), it is clear that the image <strong>of</strong> H ! 1(M, ν) under ξ 1 lies in M ! 1(M, ν). Let H 1 (M, ν)<br />

be the preimage <strong>of</strong> S 1 (M, ν) under ξ 1 . Then H 1 (M, ν) is canonically isomorphic to M 1 (M, ν).<br />

We want to show that the map ξ 1 is in fact a surjection onto S 1 (M, ν).<br />

To proceed, we will construct two families <strong>of</strong> Poincaré series P m (z, s), Q m (z, s), where the<br />

first family is similar to the <strong>one</strong> used in [6]. They are a priori defined for Re(s) > 1 and will<br />

be analytically continued to Re(s) > 0 through their spectral expansions. Unlike the cases<br />

k ≥ 2, this will only be a statement about existence, and not a formula that could be used<br />

to calculate the holomorphic coefficients <strong>of</strong> the preimage explicitly. To prove the analytic<br />

continuation, we will refer to results in [31] and [35].<br />

Given any positive integer m, define<br />

(16)<br />

(17)<br />

)<br />

.<br />

φ ∗ m(z, s) := e 2πimx (4π|m|y) −1/2 M sgn(m)<br />

,s− 1 (4π|m|y),<br />

2 2<br />

ϕ ∗ m(z, s) := e 2πimx (4π|m|y) s−1/2 e −2π|m|y .

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