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

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36 W. DUKE AND Y. LI where x j (A ′ , B) is the preimage of the Heegner point τ j (A ′ , B) on X 0 (N) and ⎛ ⎞ ∑ T f := ⎜ ⎝ c(f, −m)T m − (N + 1) ∑ c(f, −Nm ′ )T m ′W N ⎟ ⎠ , m≥1 m ′ ≥1 N∤m N∤m ′ ( U f,B,ψ = c f,1ψ(B[n]) + c f,2 ψ(B −1 [n]) + c f,3 ψ(B[n] −1 ) + c f,4 ψ(B −1 [n] −1 ) ∑ ψ 2 (A ′ ) log |u 12H(p)(N 2 A ′|) − 1) A ′ for some constants c f,j ∈ Q determined by f(z). Since f(z) ∈ S 1 , the coefficient c(f, −m) vanishes for all m ≥ 1 satisfying N 2 |m. So we have N ∤ m ′ in the summation above. Using (49) and equation (2.16) in [24, p. 240], one can show that the c f,j ’s are integral linear combinations of c(f, −m), m ≥ 1. Since f is in the integral span of S 1 , Lemma 6.4 implies that c(f, −m) ∈ Z for all m ≥ 1. Then the denominator of constant in the definition of U f,B,ψ is independent of f. Given any newform h(z) ∈ S 2 (N), we know that (h|W N )(z) = −(h|U N )(z). It is then easy to check that T f (h(z)) = 0. Since N is prime, S 2 (N) is spanned by newforms. So T f annihilates any h(z) ∈ S 2 (N). Then T f ((x 1 (A, B)) − (0)) is a trivial divisor on J 0 (N) defined over H, since the actions of the Hecke operators and Fricke involution on the Jacobian are the same as those on S 2 (N). So it is the divisor of a rational function φ f,A,B on X 0 (N), defined over H. By the axioms defining height pairings, we can rewrite Σ f,N,B,ψ as Σ f,N,B,ψ = ∑ ψ 2 (A) log |u f,B (A)| + U f,B,ψ , where A∈Cl(F ) u f,B (A) = φ f,A,B(τ 2 (f, A, B)) ∈ H. φ f,A,B (∞) Given any σ C ∈ Gal(H/F ) associated to C ∈ Cl(F ), it sends the Heegner point x 1 (A, B) to the Heegner point x 1 (AC −1 , B). Since the Galois action commutes with the Hecke action, we see that φ σ C f,A,B (z) φ f,AC −1 ,B(z) ∈ H× is a constant. So σ C (u f,B (A)) = u f,B (AC −1 ). Thus, the proposition holds for all f(z) in the integral span S 1 . □ 7. Proof of Theorem 1.1 From §5 and §6, we know that the regularized inner product between f lift,N,B (z) ∈ M !,+ 1 (p) and g ψ (z) can be put into a form quite similar to (10) and satisfies condition (iii) in Theorem 1.1, whenever f(z) ∈ M 0(N) ! has rational Fourier coefficients and N is either 1 or a prime number different from p. For a mock-modular form ˜g ψ (z) = ∑ n∈Z r+ ψ (n)qn as in Proposition 4.2, (76) gives rise to the following equation involving a linear combination of r + ψ (n)’s ∑ δ(k)r B (pm − Nk)r + ψ (k) =〈f lift,N,B , g ψ 〉 reg (86) m∈Z δ N (m)c(f, −m) ∑ k∈Z + N ∑ m ′ ≥0 c(f, −Nm ′ )ρ N,B,ψ (m ′ ).
MOCK-MODULAR FORMS OF WEIGHT ONE 37 In this section, we will show that these linear combinations of r + ψ (n)’s are sufficient to determine ˜g ψ (z) up to some element in S1 − (p). Then, one can choose an appropriate ˜g ψ (z) satisfying Theorem 1.1. The following lemma is key to the proof of Theorem 1.1. Recall that δ(n) is the same as in equation (12) and δ N (m) appears in Definition 6.3(ii). Lemma 7.1. Let N 0 = 1 and {N j : 1 ≤ j ≤ (p − 1)/2} be a set of primes congruent to 1 modulo N. Suppose {d(n)} n≥1 is a set of complex numbers such that d(n) = 0 for all n with χ p (n) = 1 and they satisfy the following equations ∑ m∈Z δ N (m)c(f, −m) ∑ n≥1 ∑ δ(n)d(n)r B (pm − n) = 0, n≥1 δ(n)d(n)r B (pm − Nn) = 0, for all classes B ∈ Cl(F ) and f(z) ∈ M !,new 0 (N j ), 1 ≤ j ≤ (p − 1)/2. Then the power series D(z) := ∑ n≥1 d(n)qn is a weight one cusp form in S − 1 (p). Proof. Define the sum s j,B (m) for m ≥ 1 and the series φ j,B (z) by s j,B (m) := ∑ n≥1 φ j,B (z) := ∑ m≥1 s j,B (m)q m . δ(n)d(n)r B (pm − N j n), First, it is not hard to see that φ j,B (z) ∈ S 2 (N j ). When j = 0, this is clear since φ j,B (z) is identically 0. When 1 ≤ j ≤ (p−1)/2, N j is a prime and a power series h(z) = ∑ m≥1 c(h, m)qm ∈ CJqK is in S 2 (N j ) if and only if ∑ δ Nj (m)c(f, −m)c(h, m) = 0, m∈Z for all f(z) = ∑ m∈Z c(f, m)qm ∈ M !,new 0 (N j ). The sufficiency is a consequence of Lemma 6.4. Next, it is easy to see that (87) φ j,B (z) = (ϑ B (z)D(N j z))|U p + ϑ B (z)(D|U p )(N j z), = 1 p k=0 p−1 ∑ ( (ϑ B ) ) z+k + ϑ p B (z) D ( Nj z+k p ) where U p acts formally on power series in CJqK and we used N j ≡ 1 (mod p) to reach the second step. Set M := (p+1)/2 ∏ j ′ =1 M j := M N j . N j ′,
Page 1 and 2: MOCK-MODULAR FORMS OF WEIGHT ONE W.
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Page 31 and 32: Similar to the proof of Theorem 5.1
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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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