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

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$Math 33B/1 S00 K. Solutions to Midterm Exam G1, Version B If you ...$

$Math 3C Homework 3 Solutions$

14 W. DUKE AND Y. LI If f(z) ∈ M 1 (p), then this is the usual Petersson inner product. Now let ĝ(z) ∈ H 1 (p) be a preimage of g(z) under ξ 1 with the following Fourier expansions where W p = ( −1 p ĝ(z) = ∑ n∈Z c + ∞(n)q n − ∑ n≥1 c(g, n)β 1 (n, y)q −n , (ĝ| 1 W p )(z) = ∑ c + 0 (n)q n − ∑ c(g| 1 W p , n)β 1 (n, y)q −n , n∈Z n≥1 ) is the Fricke involution and it acts on f(z) ∈ H ! 1 (p) by ( ) (f| 1 W p )(z) = √ 1 pz f − 1 . pz The expression for (ĝ| 1 W p )(z) follows from the commutativity between ξ 1 and the slash operator. Note that ∑ n∈Z c+ ∞(n)q n and ∑ n∈Z c+ 0 (n)q n are mock-modular forms with shadows g(z) and (g| 1 W p )(z) respectively. Suppose f(z) ∈ M 1(p) ! has the Fourier expansion ∑ n∈Z c ∞(f, n)q n and ∑ n∈Z c 0(f, n)q n at the cusp infinity and 0 respectively. Then we can express 〈f, g〉 reg in terms of these Fourier coefficients. Lemma 3.1. Let f(z) ∈ M ! 1(p) and g(z) ∈ S 1 (p). In the notations above, we have (28) 〈f, g〉 reg = ∑ n∈Z c + ∞(n)c ∞ (f, −n) + c + 0 (n)c 0 (f, −n). Proof. Since g(z) = ξ 1 (ĝ(z)) = −2iy ∂ĝ ∂z (29) 〈f, g〉 reg = − lim Y →∞ and dzdz = 2idxdy, we can rewrite equation (27) as ∫ f(z) ∂ĝ ∂z dzdz. F(Y ) As an application of Stokes’ Theorem, we have ∫ ( f(z) ∂ĝ ∂z + ∂f ) ∮ ∂z ĝ(z) dzdz = F(Y ) ∂F(Y ) f(z)ĝ(z)dz, where the line integral is in the counterclockwise direction. Using the modularity of f(z) and ĝ(z), we can cancel most of the contour integral on the right hand side, except the contributions from the horizontal segment from − 1 + iY and 1 + iY near ∞ and the arc 2 2 center at 0 connecting the points σ 0 (± 1 ∂f + iY ). Also, f(z) is holomorphic, hence = 0. 2 ∂z Thus, the equation above becomes ∫ f(z) ∂ĝ ∫ − 1 F(Y ) ∂z dzdz = 2 +iY f(z)ĝ(z) + (f| 1 W p )(z)(ĝ| 1 W p )(z)dz. 1 2 +iY Now combining this with equations (29) and the bound |β 1 (n, y)| = O(e −4πny ), we obtain equation (28) □ Remark. Notice that the right hand side of equation (28) depends on the choice of ĝ(z), whereas the left hand side only depends on g(z). So if we replace ĝ(z) with h(z) ∈ M ! 1(p), then Lemma 3.1 still holds and we obtain 0 = ∑ n∈Z c ∞ (h, n)c ∞ (f, −n) + c 0 (h, n)c 0 (f, −n), where h(z) has Fourier expansions ∑ n∈Z c ∞(h, n)q n and ∑ n∈Z c 0(h, n)q n at the cusp infinity and 0 respectively.
MOCK-MODULAR FORMS OF WEIGHT ONE 15 Now, we want to split up the space H ! 1(p) and M ! 1(p) canonically into the direct sum of two subspaces. This decomposition behaves nicely with respect to ξ 1 and regularized inner product. For ɛ = ±1, we define the following space (30) H !,ɛ 1 (p) := {f(z) = ∑ n∈Z a(n, y)q n ∈ H ! 1(p) : a(n, y) = 0 whenever χ p (n) = −ɛ}. By imposing the same condition on the Fourier expansions, we can define H1(p), ɛ M !,ɛ 1 (p), M1(p), ɛ S1(p), ɛ M !,ɛ 1 (p) and M ɛ 1(p). Notice that M !,ɛ 1 (p) is the same as the space A ɛ k (p, χ p) defined on page 51 of [8] for k = 1. By the definition of R p (n) in (5), it is clear that E 1 (z) ∈ M 1 + (p). Also, the non-holomorphic weight one Eisenstein series Ẽ1(z) belongs to M !,− 1 (p). This can be checked from the factorization (6) and the fact that E 1 (z) ∈ M 1 + (p). Clearly, H !,ɛ 1 (p) is a subspace of H 1(p) ! for ɛ = ±1. In fact, the space H 1(p) ! can be decomposed into direct sum of these subspaces using the operators defined as follows. Let U p be the standard operator on H 1(p) ! defined by (31) f(z)| 1 U p = √ 1 p−1 ∑ ( f| 1 jp ) 1 . p Its action on the Fourier expansion of f(z) = ∑ m a(m, y)qm ∈ H 1(p) ! is as follows (32) (f| 1 U p )(z) = ∑ ( ) a pm, y q m . p m It is easy to see that U p preserves the space H 1(p). ! This is also true for the Fricke involution W p , since χ p is a real character. Applying W p twice produces a negative sign for odd weight, i.e. (f| 1 Wp 2 )(z) = −f(z). Define the following operator on H 1(p) ! for ɛ ∈ {±1} j=0 ( ɛi(f| 1 U p W p ) + f (33) pr ɛ (f) := 1 2 Using these operators, we can decompose H 1(p) ! into the direct sum of H !,+ 1 (p) and H !,− 1 (p). On M ! 1(p), one can define U p and W p by considering their action on the associated harmonic Maass form in H 1(p) ! and obtain the same decomposition of spaces. Lemma 3.2. For ɛ = ±1, the operator pr ɛ is the identity operator when restricted to the subspace H !,ɛ 1 (p) and the zero operator when restricted to the subspace H !,−ɛ 1 (p). Equivalently, we have (34) (f| 1 W p )(z) = −iɛ(f| 1 U p )(z), for all f(z) ∈ H !,ɛ 1 (p). Also, we can write (35) H ! 1(p) = H !,+ 1 (p) ⊕ H !,− 1 (p). Similarly, we have this decomposition for M ! 1(p). Remark. The decomposition above clearly holds for subspaces of H ! 1(p) or M ! 1(p), such as H 1 (p), M ! 1(p), M 1 (p), S 1 (p) and M 1 (p). Proof. Let f(z) = ∑ n∈Z a(n, y)qn ∈ H 1(p) ! and h := f| 1 U p W p . First, we will show that f ∈ H !,ɛ 1 (p) if and only if pr −ɛ (f) = 0. With some matrix calculations, such as in Lemma 3 of [8], we obtain (36) h(z) = √ 1 ∑ (f| 1 W p V p )(z) + ε p χ p (n)a(n, y)q n , p n∈Z ) .
Page 1 and 2: MOCK-MODULAR FORMS OF WEIGHT ONE W.
Page 3 and 4: MOCK-MODULAR FORMS OF WEIGHT ONE 3
Page 13: MOCK-MODULAR FORMS OF WEIGHT ONE 13
Page 31 and 32: Similar to the proof of Theorem 5.1
Page 39 and 40: where T := (T kk ′) 0≤k,k ′

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

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