CYCLE INTEGRALS OF MEROMORPHIC MODULAR FORMS AND

CYCLE INTEGRALS OF MEROMORPHIC MODULAR FORMS AND CM-VALUES OF AUTOMORPHIC FORMS

arXiv:1409.0793v1 [math.NT] 2 Sep 2014

KATHRIN BRINGMANN AND BEN KANE

1. Introduction and statement of results While investigating the Doi-Naganuma lift, Zagier [16] encountered interesting weight 2k cusp forms (τ ∈ H, k ∈ 2N) X Q(τ, 1)−k , (1.1) fk,δ (τ ) := Q∈Qδ

where Qδ is the set of integral binary quadratic forms of discriminant δ > 0. Zagier’s cusp forms were then used by Kohnen and Zagier [11] to construct a kernel function for the Shimura and Shintani lifts between integral and half-integral weight cusp forms. An interesting application is the proof of non-negativity of twisted central L-values [11]. These cusp forms may also be viewed as a sum of Poincar´e series which appeared in previous work of Petersson [13]. To give another example of their importance, Kohnen and Zagier [12] showed that the even periods of the forms fk,δ are rational. This question was more recently revisited by Duke, Imamoglu, and T´oth [7] in the framework of weakly holomorphic modular forms. Similarly to (1.1), one can define cusp forms by restricting the sum defining fk,δ to those Q in a fixed SL2 (Z)-equivalence class [Q0 ] of binary quadratic forms. These functions span the space of weight 2k cusp forms [10]. Choosing the discriminant negative instead, one constructs meromorphic modular forms with poles at CM-points of a given discriminant. To describe these, for D > 0 such that −D is a discriminant, let A be an equivalence class of positive definite integral binary quadratic forms with discriminant −D. For an even positive integer k, define X k (1.2) fk,−D,A (τ ) := D 2 Q(τ, 1)−k . Q∈A

Following Zagier’s proof of modularity for fk,δ and noting that Q(τ, 1) has a unique root zQ ∈ H, it follows that the functions fk,−D,A are meromorphic modular forms. Using the absolute and uniform convergence proven in Proposition 6.1 of [1], one further obtains that they decay like cusp forms towards infinity. Following Petersson, who constructed elliptic Poincar´e series [13, 14, 15] related to (1.2) (see (21) of [15]), we call such meromorphic modular forms which vanish towards infinity like cusp forms meromorphic cusp forms. Date: September 3, 2014. The research of the first author was supported by the Alfried Krupp Prize for Young University Teachers of the Krupp foundation and the research leading to these results has received funding from the European Research Council under the European Union’s Seventh Framework Programme (FP/2007-2013) / ERC Grant agreement n. 335220 - AQSER. The research of the second author was supported by grant project number 27300314 of the Research Grants Council. 1

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KATHRIN BRINGMANN AND BEN KANE

Returning to the classical functions fk,δ , Kohnen and Zagier [11] showed that the Petersson inner product of fk,δ against a weight 2k cusp form f equals the δth coefficient of the first Shintani lift of f , which is given as a trace of cycle integrals of f . In this paper, we compute inner products of meromorphic cusp forms against fk,−D,[Q] , where we assume throughout that Q = [a, b, c] is positive definite with discriminant −D < 0. The situation is much more involved than in the case of cusp forms. Since the naive inner product diverges, it must be regularized. Petersson [15] defined a regularization for inner products of meromorphic modular forms. However, his regularization sometimes diverges so we consider here a generalization hf, gimer (see Section 2). One can show that our regularization exists for all weight k meromorphic cusp forms and agrees with Petersson’s whenever his is finite. Since our regularization always exists, we expect that it is useful in other applications where Petersson’s diverges.

 To give the evaluation of the inner product f, fk,−D,[Q] mer , we let   1 if D 6= 3, 4, 1 ωD := #ΓQ = 2 if D = 4,  2 3 if D = 3,

where ΓQ is the stabilizer of the action of SL2 (Z) on Q, and for Q = [a, b, c] and τ = u+iv ∈ H a|τ |2 + bu + c . v Theorem 1.1. If f is a weight 2k meromorphic cusp form and k > 3, then Z arctanh √D  1−k X Qz


 2 D Resτ =z f (τ )Q(τ, 1)k−1 sinh2k−2 (θ)dθ. (1.3) f, fk,−D,[Q] mer = ωD z∈H 0 Qτ :=

z6=zQ

In particular, fk,−D,[Q] is orthogonal to cusp forms. Remarks. (1) The left-hand side of (1.3) is independent of the choice of representative in [Q] by construction. One can use modularity of f and Q to directly show that the righthand side is also independent of the representative. (2) Petersson already showed that fk,−D,[Q] is orthogonal to cusp forms using his regularization (see Satz 8 of [15] with r = 2k and ν = −k). However, his proof does not yield the value of the inner product in general and the method used in this paper differs from Petersson’s proof. (3) Using a new regularization, one can further prove that fk,−D,[Q] is orthogonal to all weakly holomorphic modular forms, i.e., those meromorphic modular forms whose poles only occur at cusps (see Section 9). (4) Theorem 1.1 is a natural generalization of cycle integrals for negative discriminants. If the discriminant is positive, then there are two roots of the binary quadratic form on the real line and the cycle integral is the integral along the intersection with H with the circle centered on the real line which intersects both of these points, modulo the action of ΓQ . For −D < 0, only one root of the binary quadratic form is in the upper half plane. Since the roots no longer come in pairs, a reasonable definition for the generalization of a cycle integral is an integral along a circle centered at this unique

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−1 root, factored out by the action of ΓQ . Since ΓQ is finite, one obtains ωD times the integral along the entire circle, which gives the residue.

Our next goal is to view the right-hand side of (1.3) in the framework of automorphic forms. For this, we rewrite the occuring residue, using the elliptic expansion of f around τ =z X af (z; n)wz (τ )n , (1.4) f (τ ) = (τ − z)−2k n≫−∞

where

τ −z . τ −z We omit f and z in af (z; n) whenever the dependence is clear from the context. Note that for z = x + iy ∈ H, the functions y −2k−n af (z; n) satisfy weight 2k + 2n modularity (see Lemma 8.1). For ℓ ∈ N0 , let cℓ (̺, k, n; z) be the ℓth coefficient of the Taylor expansion around τ = z of Gz,̺,n (τ ) := (2Im(̺))1−k (τ − z)n−2k (τ − ̺)k−1 (τ − ̺)k−1 . We then define the Poincar´e series X gn (z, M̺), (1.5) Fn (z, ̺) = Fk,n (z, ̺) := y 2k−n wz (τ ) :=

M ∈SL2 (Z)/Γ̺

where Γ̺ is the stabilizer of ̺ in SL2 (Z) and 8 Z 2 arcosh(1− u(z,̺)u(z,̺) ) gn (z, ̺) := cn−1 (̺, k, n; z) sinh2k−2 (θ)dθ 0

with

|z − ̺|2 =: 2 cosh(d(z, ̺)) − 2. Im(̺)Im(z) Note that d(z, ̺) is the hyperbolic distance between z and ̺. The functions Fn converge absolutely for k > 3 (see Lemmas 8.2 and 8.3, noting that the CM-points zQ are dense in H) and are SL2 (Z)-invariant as functions of ̺ by construction. We next represent the inner product as a linear combination of the functions Fn evaluated at the poles of f and the CM-point ̺ = zQ . u(z, ̺) :=

Theorem 1.2. If k > 3 and the poles of f are at z1 , . . . , zr modulo SL2 (Z), then r 2k−1 X X

 f, fk,−D,[Q] mer = Im (zℓ )−2k+n af (zℓ ; −n) Fn (zℓ , zQ ) . ℓ=1 n=1

Remarks. (1) The images of lifts between integral and half-integral weight weak Maass forms have Fourier expansions which can be written as CM-traces for negative discriminants and cycle integrals for positive discriminants [4, 6, 8]. Thus the appearance of CM-values of SL2 (Z)-invariant functions in Theorem 1.2 is very natural. Since the generating function of Zagier’s cusp forms for positive discriminants yields the kernel function for the first Shintani lift, an interesting question is whether the generating function for negative discriminants also has some connection to a lift. However, the naive

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KATHRIN BRINGMANN AND BEN KANE

generating function does not converge and furthermore would have a dense set of poles in the upper half-plane. This will be investigated further in upcoming work [5]. (2) Note that Fn (z, ̺) has a singularity at z = ̺. If zℓ = zQ , then Fn (zQ , zQ ) in Theorem 1.2 should be interpreted as the sum defining Fn without the term corresponding to M ∈ ΓQ = ΓzQ , which is the unique term contributing to the singularity. This is further explained in the proof of Theorem 1.2 and (8.2). As mentioned above, Petersson’s original regularization does not suffice for our purposes. In particular, his associated Petersson norm never exists for meromorphic forms with

poles in 2upper half-plane. The special case f = fk,−D,[Q] in Theorem 1.1 yields the value of fk,−D,[Q] mer .

Corollary 1.3. If k > 3, then  

fk,−D,[Q] 2 = 2k Fk (zQ , zQ ) − Im (zQ )k gk (zQ , zQ ) . (1.6) mer

We next investigate the automorphic properties of the functions Fn . For a formal definition of meromorphic harmonic Maass forms see Section 8. Theorem 1.4. (1) If k > 3, then the functions Fn (z, ̺) are SL2 (Z)-invariant as functions of ̺ and they satisfy weight −2k + 2n modularity as functions of z. (2) For k > 3 the function F1 (z, ̺) is a weight 2 − 2k meromorphic harmonic Maass form as a function of z. Remarks. (1) The evaluations of the functions Fn (z, ̺) at ̺ = zQ are given more explicitly in (8.8), rewritten using Lemma 8.3. (2) A new class of weight 2 − 2k modular objects, locally harmonic Maass forms, was investigated in [3] and, for k = 1, in [9]. These transform like modular forms and are annihilated by the hyperbolic Laplacian away from an exceptional set of measure zero along which they exhibit discontinuities. For a fixed class A of quadratic forms of discriminant δ > 0, these functions have the form (see (3.2) and (3.8) of [3]) Z arctan √δ X Qz F1−k,δ,A(z) := sgn (Qz ) Q (z, 1)k−1 sin2k−2 (θ)dθ. 0

Q∈A

This closely resembles the weight 2−2k meromorphic harmonic Maass form, rewritten using Lemma 8.3, (8.8), and (4.1), Z arctanh √D  X Qz 1−k sinh2k−2 (θ)dθ. F1 (z, zQ ) = (2i)1−2k D 2 Q(z, 1)k−1 Q∈[Q]

0

The meromorphic harmonic Maass form F1 (z, zQ ) satisfies interesting properties under the
∂ 1 ∂ 2k−1 natural operators ξ2−2k := 2iy 2−2k ∂z and D 2k−1 := 2πi . ∂z Theorem 1.5. (1) We have

ξ2−2k (F1 (z, zQ )) =

1 fk,−D,[Q](z). (−2i)2k−1

CYCLE INTEGRALS OF MEROMORPHIC FORMS AND CM-VALUES

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(2) We have D 2k−1 (F1 (z, zQ )) = −

4

(2k − 2)!

k− 21

(4πi)2k−1

fk,−D,[Q](z).

Remarks. (1) Theorem 1.5 (2) implies that the function F1 (z, zQ ) has poles in the meromorphic part. By Theorem 1.5 (1), it also has singularities in its non-meromorphic part. (2) It is rather unusual for a harmonic Maass form to map to a constant multiple of the same function under ξ2−2k and D 2k−1 . For example, if the image under the ξ-operator is a cusp form, then this cannot occur. However, the locally harmonic Maass forms investigated in [3] have a similar property. (3) The non-meromorphic part of F1 (z, zQ ) is annihilated by D 2k−1 . However, if one directly differentiates the meromorphic part, then the pole order is at least 2k because differentiation raises the order of the pole. There is hence interesting cancellation between the singularities in the meromorphic and non-meromorphic parts. The paper is organized as follows. In Section 2, we introduce our new regularization. In Section 3, it is shown that the integral of the -kth coefficient of the elliptic expansion (1.4) around zQ vanishes. We then introduce some tools in

Section 4which are used in Section 5 to make a change of variables for the contribution to f, fk,−D,[Q] mer from the remaining terms from the elliptic expansion. We then compute the integral of the remaining terms in Section 6. We show Theorem 1.1 in Section 7. Theorems 1.2, 1.4, and 1.5 are proven in Section 8. Finally, we describe how to obtain orthogonality of the fk,−D,[Q] against weakly holomorphic modular forms in Section 9. 2. A new regularized Petersson inner product and unfolding For weight 2k meromorphic cusp forms f and g whose poles occur at distinct points z1 , . . . , zr modulo SL2 (Z), we define the inner product hf, gimer as the constant term of the Laurent expansion in the variables s1 , . . . , sr around s1 = 0, s2 = 0, . . . , sr = 0 of the meromorphic continuation (if existent) of Z dudv (2.1) hf Hs , gi = f (τ )Hs (τ )g(τ )v 2k 2 . v SL2 (Z)\H

Here s = (s1 , . . . , sr ) and

Hs (τ ) = Hs1 ,...,sr ,z1 ,...,zr (τ ) :=

r Y

hsℓ ,zℓ (τ )

ℓ=1

with

hsℓ ,z (τ ) := |wz (Mτ )|sℓ , where M ∈ SL2 (Z) is chosen such that Mτ is contained in a fixed fundamental domain F ∗ . The fundamental domain is such that each zℓ lies in the interior of F ∗ if zℓ 6≡ i, ρ πi (mod SL2 (Z)), i lies in the interior of F ∗ ∪ SF ∗ , and ρ := e 3 is in the interior of F ∗ ∪ UF ∗ ∪ 1 −1 U 2 F ∗ , where S := ( 01 −1 0 ) and U := ( 1 0 ) . Although the choice of fundamental domain depends on f and g, one can show that the regularized inner product is independent of the choice of fundamental domain. We throughout denote the constant term of the Laurent expansion in the variables s1 , . . . , sr simply as coeff [s0 ] .

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KATHRIN BRINGMANN AND BEN KANE

To compute hf, fk,−D,Aimer , note that we can write X k Q(τ, 1)−k 2k γ fk,−D,A(τ ) = D 2 γ∈ΓQ \ SL2 (Z)

with Q ∈ A. Here |κ is the usual weight κ slash-operator, which, if there is possible ambiguity for the variable in the slash, we sometimes write as |κ,τ . Unfolding gives ! Z k −k (2.2) hf Hs , fk,−D,Aimer = D 2 coeff [s0 ] f (τ )Hs (τ )Q (τ , 1) v 2k−2 dudv . ΓQ \H

k

(2.3)

D2 ωD

times Z  −k 2k−2 coeff [s0 ] f (τ )Hs (τ )Q (τ , 1) v dudv .

We can then rewrite (2.2) as

H

The following lemma, which follows by direct calculation, is useful for evaluating (2.3). Lemma 2.1. If ε > 0 is sufficiently small and |τ − zQ | < ε, then s hs,zQ (τ ) = wzQ (τ ) .

3. Contribution from the −kth coefficient to (2.3)

In this section, we compute the contribution from the −kth coefficient of the elliptic expansion (1.4) around τ = zQ to (2.3), namely af (−k) (τ − zQ )−2k wzQ (τ )−k .

(3.1)

This special term arises when computing the Petersson norm of fk,−D,[Q], since for Q = [a, b, c], we have (3.2)

Q(τ, 1) = a (τ − zQ ) (τ − zQ ) = a (τ − zQ )2 wzQ (τ ).

It is not a surprise that special treatment is required for the term (3.1), since Petersson’s regularized inner product does not exist whenever af (−k) 6= 0. Since the only pole of (3.1) for the corresponding term in (2.3) is at τ = zQ , we may set sℓ = 0 for zℓ 6= zQ . For η ∈ C, we then let hη := hη,zQ . We are now ready to state the main result of this section. Proposition 3.1. We have  Z −2k −k 2k−2 −k (3.3) coeff [η0 ] wzQ (τ ) hη (τ )Q (τ , 1) v dudv = 0. (τ − zQ ) H

Proof. For ε > 0 sufficiently small so that Lemma 2.1 holds, we split the integral in (3.3) into a range with |τ − zQ | < ε and η one for which |τ − zQ | ≥ ε. For the range |τ − zQ | < ε, we may plug in hη (τ ) = wzQ (τ ) , while for |τ − zQ | ≥ ε we show that the integral converges for η = 0. By (3.2), we may then rewrite (3.3) as  Z η−2k 2k−2 −4k −k v dudv . wzQ (τ ) (3.4) a coeff [η0 ] |τ − zQ | H

CYCLE INTEGRALS OF MEROMORPHIC FORMS AND CM-VALUES

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To show that the integral in (3.3) converges absolutely for η = 0 and |τ − zQ | ≥ ε, we rewrite, using (3.1) and (3.2), the integral over the range |τ − zQ | ≥ ε for η = 0 as Z −k (3.5) a |τ − zQ |−2k |τ − zQ |−2k v 2k−2 dudv. τ ∈H |τ −zQ |≥ε Writing zQ = xQ + iyQ , one easily sees that (3.6)

v ≤ |τ − zQ | + yQ ,

|τ − zQ | ≥ |τ − zQ | .

Hence (3.5) may be bounded from above against Z −k a |τ − zQ |−4k (|τ − zQ | + yQ )2k−2 dudv |τ −zQ |≥ε Z  yQ 2k−2 −k |τ − zQ |−2−2k dudv. ≤a 1+ ε |τ −zQ |≥ε

We then write τ − zQ in polar coordinates and note that the resulting integral evaluates as π −2k ε , and hence converges. k To compute (3.4), we make the change of variables wzQ (τ ) = Reiϑ with 0 ≤ R < 1 and 0 ≤ ϑ < 2π. A straightforward calculation shows that 2k  v 2k dudv 1 − R2 4R dRdϑ. , = = 4k 4yQ v2 (1 − R2 )2 |τ − zQ | We therefore obtain that Z Z η−2k 2k−2 −4k 1−2k −2k |τ − zQ | wzQ (τ ) v dudv = 4 yQ

0

H

1

1−R


2 2k−2

R

1−2k+η

The Binomial Theorem then gives Z 1 2k−2 X 2k − 2
(−1)ℓ 2 2k−2 1−2k+η . 1−R R dR = 2 − 2k + 2ℓ + η ℓ 0 ℓ=0

dR

Z

2π

dϑ.

0

The constant term of the Laurent expansion around η = 0 is thus the sum over all terms ℓ 6= k − 1 with η = 0. Making the change of variables ℓ 7→ 2k − 2 − ℓ, one obtains that X 2k − 2 (−1)ℓ = 0, ℓ 2 − 2k + 2ℓ 0≤ℓ≤2k−2 ℓ6=k−1

yielding the proposition.

 4. Location of the poles

In this section, we begin to lay the groundwork needed to determine the contribution to (2.3) coming from the terms X af (zQ ; n) wzQ (τ )n . h(τ ) := (τ − zQ )−2k n≫−∞ n6=−k

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KATHRIN BRINGMANN AND BEN KANE

To make a two-variable change of variables in Section 5, we require some preliminaries. Specifically, we change u and v to τ and √ ! D (4.1) θτ = θQ,τ := arctanh . Qτ For θ > 0, we let Sθ = Sθ,Q be the counter-clockwise oriented curve consisting of τ ∈ H satisfying √ D (4.2) = tanh (θ) . Qτ The curves Sθ are circles which collapse to the point zQ as θ → ∞ and whose radii increase unbounded as θ → 0. y

Q Lemma 4.1. If θ > 0, then Sθ is a counter-clockwise oriented circle centered at xQ + i tanh(θ) yQ which does not intersect the real line. of radius sinh(θ)

Proof. Since tanh(θ) > 0, the condition in (4.2) becomes !2 √ 2  D D b2 b − +a v− +c− (4.3) a u+ = 0. 2a 2a tanh(θ) 4a 4a tanh2 (θ)

√

D b + i 2a tanh(θ) We use D = 4ac − b2 to rewrite (4.3) as the equation for the circle centered at − 2a

of radius

√

D . 2a sinh(θ)

Since

√ b D (4.4) xQ = − , yQ = , 2a 2a we see that the radius and center are precisely as given in the lemma.



It turns out to be useful to determine whether τ ∈ Sθ , τ is in the interior     y y Q Q < Dθ := τ ∈ H τ − xQ + i tanh(θ) sinh(θ)

of the convex hull of Sθ , or τ is in the exterior of Dθ .

Lemma 4.2. (1) If τ 6= zQ , then θQ,τ is the unique θ > 0 such that τ ∈ Sθ . Furthermore, we have τ ∈ Dθ if and only if 0 < θ < θQ,τ . (2) For every 0 < ε < 1, there exists a constant θε such that for all τ ∈ H with |τ − zQ | > ε and v < 2y1Q ε , we have θQ,τ < θε . (3) For every θ > 0, we have zQ ∈ Dθ . Proof. (1) Defining (4.5)

 hτ (θ) := (u − xQ ) + v − 2

yQ tanh(θ)

2

−

2 yQ

sinh2 (θ)

,

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we see by Lemma 4.1 that (4.6)

τ ∈ Sθ ⇔ hτ (θ) = 0,

τ ∈ Dθ ⇔ hτ (θ) < 0.

Since τ ∈ SθQ,τ by construction (and hence hτ (θQ,τ ) = 0), (1) is proved once we show that hτ (θ) < 0 for 0 < θ < θQ,τ and hτ (θ) > 0 for θ > θQ,τ . It thus suffices to confirm that hτ is strictly monotonically increasing. Expanding and simplifying, we have 2vyQ 2 + yQ . (4.7) hτ (θ) = (u − xQ )2 + v 2 − tanh(θ) We then obtain (1) since tanh(θ) is strictly monotonically increasing. (2) Since limθ→∞ tanh(θ) = 1, there exists for every δ > 0 a constant ϑδ such that θ > ϑδ 1 implies that tanh(θ) ≥ 1+δ . Hence, by (4.7), |τ − zQ | > ε yields that hτ (θ) ≥ (u − xQ )2 + (v − yQ )2 − 2vyQ δ ≥ ε2 − 2vyQ δ.

The restriction v < 2y1Q ε then implies that hτ (θ) > ε2 − δε . Choosing δ = ε3 , we have hτ (θ) > 0 and hence θε := ϑε3 suffices for the claim. (3) Taking the limit θ → ∞ in (4.7) and recalling that limθ→∞ tanh(θ) = 1, we have limθ→∞ hzQ (θ) = 0. By (4.7), hzQ is strictly increasing, and hence (4.6) implies that zQ ∈ Dθ for all θ > 0, concluding the proof.  Since we later integrate over poles of f , we next determine some properties of the set of θ for which a pole of f lies on Sθ . We denote these by n o Bf := θ > 0 ∃τ ∈ Sθ such that f has a pole at τ .

Lemma 4.3. The set Bf is discrete.

Proof. By Lemma 4.2 (1), there exists for each pole τ 6= zQ of f precisely one θ > 0 for which τ ∈ Sθ . Furthermore, zQ ∈ / Sθ for all θ > 0 by Lemma 4.2 (2). Since the poles of f are discrete, the lemma follows.  5. A Change of variables In Section 6, we compute the contribution from h to (2.3). For this we make a change of variables from u and v to τ and θ. The main result of this section is the following. Proposition 5.1. The terms coming from h in (2.3) equal Z ∞  Z 1 2k−2 −k k−1 (5.1) D 2 coeff [s0 ] sinh (θ) h(τ )Hs (τ )Q(τ, 1) dτ dθ . 0

Sθ

Before proving Proposition 5.1, we first show some useful identities.

Lemma 5.2. (1) For τ satisfying (4.2), we have sinh2 (θ) = (2) We have

wz (τ ) = Q

Dv 2 . |Q(τ, 1)|2

1 . cosh(θ) + sinh(θ)

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KATHRIN BRINGMANN AND BEN KANE

Proof. (1) From (4.2), we obtain sinh2 (θ) = The result now follows since

Q2τ

D . −D

|Q(τ, 1)|2 (5.2) −D = . v2 (2) Expanding (3.2) yields c = a |zQ |2 . Combining this with (4.4), one easily obtains √ 2  2 Qτ v − Dv 1 1 − tanh(θ) wz (τ ) = √ = . = Q 1 + tanh(θ) cosh(θ) + sinh(θ) Qτ v + Dv Q2τ



We next write dudv in terms of dτ dθ. Lemma 5.3. We have

√

D dudv = dτ dθ. Q (τ , 1) Proof. We first write dθ in terms of dτ and dτ , namely √ !! √ !! D D ∂ ∂ dθ = d arctanh = arctanh dτ + Qτ ∂τ Qτ ∂τ

arctanh

√ !! D dτ . Qτ

Since dτ ∧ dτ = 0, we only need to compute the contribution of dτ . Note that iQ(τ, 1) ∂ Qτ = − . ∂τ 2v 2 Using (5.2), we obtain √ √ √ !   i DQ(τ, 1) i D 1 ∂ D = = arctanh . (5.3) ∂τ Qτ 2v 2 Q2τ − D 2Q (τ , 1) Expanding dτ in terms of du and dv yields the statement.



Proof of Proposition 5.1. The claim follows directly by Lemma 5.2 (1) and Lemma 5.3.



6. Contribution to the inner product from h Define (6.1)

RQ,f (θ) := 2πi

X

z∈Dθ z6=zQ


Resτ =z f (τ )Q(τ, 1)k−1 .

The goal of this section is to determine the contribution from the terms h in (2.3). Proposition 6.1. One has  Z Z 1 −k 2k−2 −k coeff [s0 ] h(τ )Hs (τ )Q (τ , 1) v dudv = D 2 H

∞

sinh2k−2 (θ)RQ,f (θ)dθ.

0

Before proving Proposition 6.1, we show a number of auxilary results.

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6.1. Splitting the integral. Lemma 6.2. For any ε > 0, we have  Z −k 2k−2 k− 21 coeff [s0 ] dudv h(τ )Hs (τ )Q (τ , 1) v (6.2) D H Z ε  Z ∞ Z 2k−2 2k−2 k−1 = sinh (θ)RQ,f (θ)dθ + coeff [s0 ] sinh (θ) f (τ )Hs (τ )Q(τ, 1) dτ dθ . ε

0

Sθ

Remark. The contribution over θ < ε to the integral on the left-hand side is unchanged on the right-hand side, while the integral over θ > ε on the right-hand side is the integral on the left-hand side for s = 0. The purpose of Lemma 6.2 is to obtain Proposition 6.1 by taking ε → 0. Before proving Lemma 6.2, we first determine certain residues. Lemma 6.3. We have Hence


Resτ =zQ h(τ )Q(τ, 1)k−1 = 0.

RQ,f (θ) = 2πi

X

z∈Dθ


Resτ =z h(τ )Q(τ, 1)k−1 .

Proof. Using (3.2) to rewrite Q(τ, 1) and expanding h around zQ with (1.4), we obtain X (6.3) h(τ )Q(τ, 1)k−1 = ak−1 ah (n) (τ − zQ )n+k−1 (τ − zQ )−1−k−n . n≫−∞

The terms from n > −k do not contribute to the residue at τ = zQ and ah (−k) = 0. To determine the Taylor expansion of (τ − zQ )−1−k−n around zQ for n < −k, we obtain by the Binomial Theorem −1−k−n X −1 − k − n −1−k−n (τ − zQ ) (2iyQ )−1−k−n−ℓ (τ − zQ )ℓ . = ℓ ℓ=0 Hence the terms in (6.3) with n < −k equal   −2 X X −1 − k − n k−1 (2iyQ )−2−m (τ − zQ )m . a ah (n) m−n−k+1 n 0. Moreover, the poles z of f are discrete and their corresponding imaginary parts are bounded since i∞ is not a pole of f . Hence Lemma 4.2 (2) implies that for θ sufficiently large and z 6= zQ , we have z ∈ / Dθ . For such θ, we thus can plug in sℓ = 0 for zℓ 6= zQ . In this range, we may hence assume without loss of generality that s = s1 ∈ C. We next claim that for θ sufficiently large s (6.4) hs (τ ) = wz (τ ) . Q

12

KATHRIN BRINGMANN AND BEN KANE

Assuming that (6.4) is true, choose θ0 so that for θ > θ0 , (6.4) holds and zQ is the only pole of f in Dθ . Then, by (6.4) and Lemma 5.2 (2), the contribution of θ > θ0 to (5.1) equals Z ∞ Z sinh2k−2 (θ) 1 −k 2 (6.5) D h(τ )Q(τ, 1)k−1 dτ dθ. s (cosh(θ) + sinh(θ)) θ0 Sθ

Using the Residue Theorem and the fact that zQ is the only pole inside Dθ , Lemma 6.3 implies that for θ > θ0 Z
(6.6) h(τ )Q(τ, 1)k−1 dτ = RQ,f (θ) = 2πi Resτ =zQ h(τ )Q(τ, 1)k−1 = 0, Sθ

and hence (6.5) vanishes. To show (6.4), we separate into the cases D 6= 3, 4 and D = 3, 4. For D 6= 3, 4, we claim that for θ sufficiently large, we have Dθ ⊆ F ∗ . Since F ∗ is chosen so that there exists a ball B centered at zQ of some radius δ for which B ⊆ F ∗ , it suffices to prove that for θ sufficiently large Dθ√ ⊆ B. We have shown that zQ ∈ Dθ , so for every τ ∈ Dθ , |zQ − τ | ≤ 2rθ with D rθ := 2a sinh(θ) . Furthermore, rθ → 0 as θ → ∞, and hence |zQ − τ | → 0 as θ → ∞. Lemma 2.1 hence yields (6.4) for D 6= 3, 4. For D = 4, we also have |τ − i| ≤ 2rθ for each τ ∈ Dθ . Since i is in the interior of F ∗ ∪ SF ∗ and rθ → 0 as θ → ∞, for θ sufficiently large we have Dθ ⊆ F ∗ ∪ SF ∗. If τ ∈ F ∗ , then hs (τ ) = |wi (τ )|s by definition. If τ ∈ SF , then we obtain (6.4) using |wi (Sτ )| = |wi (τ )| . The proof for D = 3 is similar and uses 3 fundamental domains around the point ρ. We next rewrite the contribution to (6.2) from small θ. The integral over θ < ε yields the second summand in (6.2). We claim that for ε < θ < θ0 we can plug in s = 0. Since Bf is countable by Lemma 4.3, the contribution of ε < θ < θ0 to (6.2) equals Z Z 2k−2 (6.7) sinh (θ) h(τ )hs (τ )Q(τ, 1)k−1 dτ dθ. ε