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

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40 W. DUKE AND Y. LI (i) d(n) = 0 for all n with χ p (n) = 1, (ii) ∑ n∈Z c(H, n)d(n)δ(n) = 0 for all H(z) = ∑ n∈Z c(H, n)qn ∈ M !,+,lift 1 (p), (iii) D(z) ∉ S1 − (p). In particular for H(z) = f lift,N,B (z), condition (ii) can be rewritten as (89) ∑ m∈Z c(f, −m) ∑ k∈Z Also, when H(z) = j lift,B[n] m ′ ∑ δ(k)d(k) δ(k)d(k)r B (pm − Nk) = N ∑ m ′ ∈Z k∈Z So equation (89) can be written as ∑ c(f, −Nm ′ ) ∑ k∈Z (z) + j lift,B[n]−1 m (z), we have (r ′ ( ) ) B + r B (pNm ′ − Nk) = 0. pm ′ −k N m∈Z δ N (m)c(f, −m) ∑ k∈Z δ(k)d(k)r B (pm − Nk) = 0. δ(k)d(k)r B ( By Lemma 7.1, D(z) ∈ S − 1 (p). This contradicts condition (iii) and we must have M !,+,lift 1 (p) = M !,+ 1 (p). ) pm ′ −k . N With Proposition 6.1 and Lemma 7.1, we are ready to give the proof of Theorem 1.1. Proof of Theorem 1.1. Denote the dimension and the q-echelon basis of S − 1 (p) by d − and the basis by {h t (z) ∈ S − 1 (p) : h t (z) = q nt + O(q n d − +1 ), 1 ≤ t ≤ d − }. Choose a mock-modular form ˜g ψ (z) = ∑ n≥n 0 r + ψ (n)qn such that it has a fixed principal part as in Proposition 4.2 and r + ψ (n t) satisfies conditions (ii) and (iii) in Theorem 1.1 for 1 ≤ t ≤ d − . Note that if S1 − (p) = ∅, then there is only one ˜g ψ (z) with a fixed principal part. We claim that this ˜g ψ (z) is a desired choice. Consider all f(z) ∈ M !,new 0 (N j ), 1 ≤ j ≤ (p − 1)/2 with rational Fourier coefficients and j m (z) for all m ≥ 0. They generate the complex vector space M !,new 0 (N j ) and M 0(1) ! respectively. They also give rise to equation (86) and equation (66), which form a system of equations satisfied by {r + ψ (n) : n ≥ n 0}. Suppose there is another set of solutions {c + (n) : n ≥ n 0 } with r + ψ (n) = c+ (n) whenever n ≤ 0, χ p (n) = 1, or n = n t for t = 1, . . . , d − . Then by Lemma 7.1, the power series D(z) := ∑ n≥1 (r + ψ (n) − c+ (n))q n is a cusp form in S1 − (p). However, since r + ψ (n t) = c + (n t ) for all 1 ≤ t ≤ d − , the cusp form D(z) must be zero in S1 − (p). So with {r + ψ (n) : n ≤ 0 or n = n t, 1 ≤ t ≤ d − } fixed, {r + ψ (n) : n ≥ d − + 1} is the unique solution to this system of equations. From Corollary 7.2 and Proposition 3.5, it is clear that r + ψ (n) can be solved from this system of equations, i.e. the r + ψ (n)’s are a rational linear combination of these regularized inner products. By Corollary 5.2 and Proposition 6.1, we can write r + ψ (n) = −β ∑ n ψ 2 (A) log |u(n, A)|, A∈Cl(F ) □
MOCK-MODULAR FORMS OF WEIGHT ONE 41 with β n ∈ Q, u(n, A) ∈ H independent of ψ and σ C (u(n, A)) = u(n, AC −1 ) for all C ∈ Cl(F ). Now to prove that we can choose β n independent of n, it is enough to show that there is a choice of β n whose denominators do not depend on n. When n > p · g N /N, it is not hard to see from equation (86) that we can explicitly express r + ψ (n) in terms of these regularized inner products and the r+ ψ (k)’s with k < n. If n = pn′ with n ′ ∈ Z, then substituting f = j n ′(z) into equation (66) gives us r + ψ (n) = 〈f lift,B , g ψ 〉 reg − ∑ k l such that kN + l ≡ 0 (mod p) and set m = (l + Nk)/p. Notice that the conditions on N and m forces N ∤ m and χ p (N) = 1. With such a choice of N and m, let f m ∈ S 1 as in (82). Then δ(n)δ N (m)r B (pm − Nn) = 1 and equation (86) becomes r + ψ (n) = − ∑ ( gN ) ∑ δ(k) δ N (m ′′ )c(f m , −m ′′ )r B (pm ′′ − Nk) r + ψ (k) k>0 m ′′ =1 − ∑ ( ) ∑ δ(k) δ N (m ′′ )c(f m , −m ′′ )r B (pm ′′ − Nk) r + ψ (k) k≤0 m ′′ ≤g N + 〈f lift,N,B m , g ψ 〉 reg + Nc(f m , 0)ρ N,B,ψ (0). When k > 0 in the summation on the right hand side, the coefficient of r + ψ (k) is an integer by Lemma 6.4. Similarly when k ≤ 0, the denominator of the coefficient of r + ψ (k) divides (N 2 − 1). Thus after choosing β n for n ≤ p · g N /N , we can choose the other β n such that their denominators are independent of n. Now choose some β and u(n, A) for all n, A such that r + ψ (n) is expressed in terms of β and u(n, A) for all non-trivial character ψ as in equation (10). Define r + A (n) by ( r + 1 A (n) := R + H(p) p (n) + ∑ ) ψ(A)r + ψ (n) = 1 H(p) = 1 H(p) ψ non-trivial ⎛ ⎞ ∏ ⎝R p + (n) + β log u(n, A ′ ) ⎠ − β log |u(n, √ A)| ∣A ′ ∈Cl(F ) ∣ ( R + p (n) + β log |N H/F (u(n, A))| ) − β log |u(n, √ A)|, where the last step follows from the transitive action of Gal(H/F ). Clearly, the generating series ∑ n∈Z r+ A (n)qn is a mock-modular form with shadow ϑ A (z). When n ≠ 0, there exists some b ∈ Q and c n ∈ Q independent of A such that 1 H(p) ( R + p (n) + β log |N H/F (u(n, A))| ) = b log |c n |. This follows from equation (6). Thus, we could choose a different β ′ ∈ Q and u ′ (n, A) ∈ H such that they satisfy condition (iii) and −β ′ log |u ′ (n, A)| = b log |c n | − β log |u(n, A)|. Then r + A (n) = −β′ log |u ′ (n, √ A)| and it is not hard to see that β ′ and u ′ (n, A) satisfies condition (ii) and (iii) in Theorem 1.1. When ψ is a non-trivial character, ψ 2 is non-trivial
Page 1 and 2: MOCK-MODULAR FORMS OF WEIGHT ONE W.
Page 3 and 4: MOCK-MODULAR FORMS OF WEIGHT ONE 3
Page 31 and 32: Similar to the proof of Theorem 5.1
Page 39: where T := (T kk ′) 0≤k,k ′

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

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