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

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$Math 3C Homework 3 Solutions$

50 W. DUKE AND Y. LI To treat both cases simultaneously, we can write ⎛ ⎞ ∑ g s (τ A + ɛ, τ Q ) = ⎝ ∑ g s (τ A + ɛ, τ Q ) ⎠ − 1C 2 f d g s (τ A + ɛ, τ A ) Q∈Q −d Q∈Q −d disc(τ Q )≠−p = ⎛ ⎝ ∑ Q∈Q −d /Γ ⎞ G s (τ A + ɛ, τ Q ) ⎠ − 1C 2 f d g s (τ A + ɛ, τ A ), where G s (τ 1 , τ 2 ) is the automorphic Green’s function defined in (56). After substituting this into equation (109), it becomes ⎛ ⎞ a(0, ψ)H(d) ∑ ∑ a(d, ψ) − =(−2) lim ⎝lim ψ 2 (A) G s (τ A + ɛ, τ Q ) ⎠ + H(0) ɛ→0 s→1 A∈Cl(F ) Q∈Q −d /Γ (109) ⎛ ⎞ C fd lim ⎝ ∑ ψ 2 (A)g 1 (τ A + ɛ, τ A ) ⎠ . ɛ→0 A∈Cl(F ) The first summand above can be evaluated using the identity (58). Substituting τ 1 = τ A +ɛ, τ 2 = τ Q into equation (58) and summing them together over A ∈ Cl(F ) and Q ∈ Q −d /Γ with coefficient ψ 2 (A) gives us ⎛ ⎞ ∑ ψ 2 (A) log |Ψ fd (τ A + ɛ)| 2 = lim ⎝ ∑ ∑ ψ 2 (A) G s (τ A + ɛ, τ Q ) + 4πE(τ A + ɛ, s) ⎠ . s→1 A A By (49), we have ⎛ lim ⎝ ∑ s→1 A ψ 2 (A) So the first summand is ⎛ −2 ⎝lim ∑ Q∈Q −d /Γ ∑ ɛ→0 A∈Cl(F ) ⎞ Q∈Q −d /Γ 4πE(τ A , s) ⎠ = −24H(d) ∑ A a(0, ψ)H(d) = − 2H(0) ψ 2 (A) log |Ψ fd (τ A + ɛ)| 2 ⎞ ( 1 + 4y2 ⎠ − ψ 2 (A) log ∣ ∣ √ y A η 2 (τ A ) ∣ ∣ a(0, ψ)H(d) . H(0) ) Substitute g 1 (τ + ɛ, τ) = − log with τ = τ ɛ 2 A into the second summand, we obtain ⎛ ⎞ a(0, ψ)H(d) ∑ a(d, ψ) − = − 2 ⎝lim ψ 2 (A) log |Ψ fd (τ A + ɛ)| 2 ⎠ a(0, ψ)H(d) − H(0) ɛ→0 H(0) − 2C fd A∈Cl(F ) ∑ A∈Cl(F ) ψ 2 (A) log |y A |. After canceling the term − a(0,ψ)H(d) H(0) , equation (109) becomes equation (102) for f = f d , d > 0. □
MOCK-MODULAR FORMS OF WEIGHT ONE 51 We can now easily deduce Theorem 1.2. It is convenient to give a slightly more general version stated in terms of theta series and quadratic forms. Corollary 9.3. Let p be a prime number congruent to 3 modulo 4, and −d a discriminant not of the form −pm 2 for any integer m. For a class B ∈ Cl(F ), suppose ˜ϑ B 2(z) = ∑ n∈Z r+ B (n)q n ∈ M − 2 1 (p) has shadow ϑ B 2(z) and ord ∞ ( ˜ϑ B 2) > −p/4. Then ∣ (110) log ∣ ∏ q∈Q −d /Γ (j(τ B ) − j(τ q )) 1/wq ∣ ∣∣∣∣∣ = − 1 4 ∑ Remark. When −d is fundamental, w d = 2w q for all q ∈ Q −d . k∈Z δ(k) r + B 2 ( pd−k2 4 ). Proof. Let f(z) = f d (z) with d not of the form pm 2 . Using the Borcherds lift of f d (z) in (99), the right hand side of (102) can be rewritten as ∣ −4 ∑ ∏ ∣∣∣∣∣ ψ 2 (A) log (j(τ A ) − j(τ q )) 1/wq . ∣ A∈Cl(F ) q∈Q −d /Γ For each non-trivial character ψ, choose ˜g ψ (z) ∈ M − 1 (p) having shadow g ψ (z) and r + ψ (n) = 0 for all n ≤ −p/4. For these ˜g ψ (z), equation (101) becomes 〈f lift,θ d , g ψ 〉 reg = ∑ ( ) pd − k r + 2 ψ δ(k). 4 k∈Z Combining these two together and dividing both sides by -4, we have ∣ ∑ ( ) pd − k (111) − 1 r + 2 4 ψ δ(k) = ∑ ∣∣∣∣∣ ∏ ψ 2 (A) log 4 (j(A) − j(τ q )) 1/wq , ∣ k∈Z A∈Cl(F ) which is precisely equation (15) when −d is fundamental. Consider the mock-modular form ˜F (z) ∈ M − 1 (p) defined by ⎛ ˜F (z) := 1 H(p) ⎜ ⎝ E+ 1 (z) + ∑ ψ:Cl(F )→C × ψ non-trivial q∈Q −d /Γ ⎞ ψ(B −2 )˜g ψ (z) ⎟ ⎠ − ˜ϑ B 2(z). ∑ Since ϑ B 2(z) = 1 H(p) ψ ψ(B−2 )g ψ (z), ˜F (z) has shadow F (z) = 0, hence is modular. Also, it has Fourier expansion in the form ˜F (z) = ∑ c + (n)q n . From the proof of Lemma 9.2, we know that ∑ ( c + k∈Z pd−k 2 4 n>−p/4 χ p(n)≠1 ) δ(k) = 〈f lift,θ d , F 〉 reg = 0.
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 and 40: where T := (T kk ′) 0≤k,k ′
Page 49: MOCK-MODULAR FORMS OF WEIGHT ONE 49

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

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