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

MOCK-MODULAR FORMS OF WEIGHT ONE 

W. DUKE AND Y. LI 

Abstract. The object of this paper is to initiate a study of the Fourier coefficients of 

a weight one mock-modular form and relate them to the complex Galois representations 

associated to the form’s shadow, which is a weight one newform. In this paper, our focus 

will be on weight one dihedral newforms of prime level p ≡ 3 (mod 4). In this case we give 

properties of the Fourier coefficients themselves that are similar to (and sometimes reduce 

to) cases of Stark’s conjectures on derivatives of L-functions. We also give a new modular 

interpretation of certain products of differences of singular moduli studied by Gross and 

Zagier. Finally, we provide some numerical evidence that the Fourier coefficients of a mockmodular 

form whose shadow is exotic are similarly related to the associated complex Galois 

representation. 

1. Introduction 

Roughly speaking, a mock-modular form is the holomorphic part of a harmonic modular 

form. For our purposes a harmonic modular form of weight k ∈ 1 2 Z is a Maass form for Γ 0(M) 

that is killed by the weight k Laplacian and that is allowed to have polar-type singularities 

in the cusps. Associated to such a form f is the weight 2 − k weakly holomorphic form 

(1) ξ k f(z) = 2iy k ∂¯z f(z). 

The operator ξ k is related to the weight k Laplacian ∆ k through the identity 

(2) ∆ k = ξ 2−k ξ k . 

A special class of harmonic forms have ξ k f holomorphic in the cusps and a Fourier expansions 

at ∞ of the shape 

(3) f(z) = ∑ n≥n 0 

c + (n)q n − ∑ n≥0 

c(n)β k (n, y)q −n . 

This expansion is unique and absolutely uniformly convergent on compact subsets of H, the 

upper half-plane. Here q = e 2πiz with z = x + iy ∈ H and β k (n, y) is given for n > 0 by 

β k (n, y) = 

∫ ∞ 

y 

e −4πnt t −k dt 

while for k ≠ 1 we have β k (0, y) = y 1−k /(k − 1) and β 1 (0, y) = − log y. For such f the Fourier 

expansion of ξ k f is simply 

ξ k f(z) = ∑ c(n)q n . 

n≥0 

After Zagier, the function ∑ n c+ (n)q n is said to be mock-modular with shadow ∑ n≥0 c(n)qn . 

It is important to observe that a mock-modular form is only determined by its shadow up to 

the addition of a holomorphic form. 

Date: July 3, 2012. 

Supported by NSF grant DMS-10-01527. 

Supported by NSF. 

1

2 W. DUKE AND Y. LI 

Some (non-modular) mock-modular forms have Fourier coefficients that are well-known 

arithmetic functions. Let σ(n) = ∑ m|n 

m and H(n) be the Hurwitz class number. Then 

−8π ∑ ( ) 

σ(n)q n σ(0) = − 

1 

24 

n≥0 

is mock-modular of weight k = 2 for the full modular group with shadow 1 and 

−16π ∑ ( ) 

H(n)q n H(0) = − 

1 

12 

n≥0 

is mock-modular of weight 3/2 for Γ 0 (4) with shadow the Jacobi theta series θ(z) = ∑ n∈Z qn2 . 

(See [45]). Actually, these two examples will play a fundamental role in our work on the weight 

one case. 

In general, the Fourier coefficients of mock-modular forms are not well understood. Especially 

in the case of weight 1/2, which includes the mock-theta functions of Ramanujan, there 

has been considerable progress made by Zwegers in his thesis [47], by Bringmann-Ono [6], 

and others (see [46] for a good exposition). The Fourier coefficients of mock-modular forms 

of weight 1/2 whose shadows Shimura–lift to cusp forms attached to elliptic curves have also 

been shown to be quite interesting by Bruinier and Ono [10] and Bruinier [13]. We remark 

that mock-modular forms whose shadows are only weakly holomorphic are also of interest 

(see [17],[18]) but in this paper we only consider those with holomorphic shadows. 

The self-dual case k = 1 presents special features and difficulties. The Riemann-Roch 

theorem is without content when k = 2 − k, and the existence of cusp forms is a subtle 

issue. Furthermore, the infinite series representing the Fourier series of weight one harmonic 

Poincaré series are difficult to handle. 

The fact that interesting non-modular mock-modular forms of weight 1 exist follows from 

work of Kudla, Rapoport and Yang [30]. Suppose that M = p > 3 is a prime with p ≡ 3 

(mod 4) and that χ p (·) = ( · ) is the Legendre symbol. Let 

p 

(4) E 1 (z) = 1 2 H(p) + ∑ n≥1 

R p (n)q n 

be Hecke’s Eisenstein series of weight one for Γ 0 (p) with character χ p , where for n > 0 

(5) R p (n) = ∑ m|n 

χ p (m). 

It follows from [30] that Ẽ1(z) := ∑ n≥0 R+ p (n)q n is mock-modular of weight k = 1 with 

shadow E 1 (z), where for n > 0 we have 

(6) R p + (n) = −(log p)ord p (n)R p (n) − ∑ 

log q(ord q (n) + 1)R p (n/q), 

χ p(q)=−1 

and where R p + (0) is a constant. 1 The associated harmonic form is constructed using the 

s-derivative of the non-holomorphic Hecke-Eisenstein series of weight 1. An arithmetic interpretation 

of its coefficients is given in [30]; it is shown that (−2R + (n) + 2 log pR p (n)) is the 

degree of a certain 0-cycle on an arithmetic curve made from elliptic curves with CM by the 

ring of integers in F = Q( √ −p). 

1 Explicitly, R + p (0) = −H(p)( L′ (0,χ p) 

L(0,χ p) + ζ′ (0) 

ζ(0) − c), where c = 1 2 γ + log(4π3/2 ) and γ is Euler’s constant.

MOCK-MODULAR FORMS OF WEIGHT ONE 3 

In this paper we will study the Fourier coefficients of mock-modular forms whose shadows 

are newforms. For simplicity we will continue to assume that M = p > 3 is a prime. To each 

A ∈ Cl(F ), the class group of F , one can associate a theta series ϑ A (z) defined by 

(7) ϑ A (z) := 1 + ∑ 

q N(a) = ∑ r 

2 A (n)q n . 

n≥0 

a⊂O F 

[a]∈A 

Hecke showed that ϑ A (z) ∈ M 1 (p, χ p ), the space of weight one holomorphic modular forms 

for Γ 0 (p) with character χ p . Let ψ be a character of Cl(F ) and consider g ψ (z) ∈ M 1 (p, χ p ) 

defined by 

(8) g ψ (z) := ∑ 

r ψ (n)q n . 

A∈Cl(F ) 

ψ(A)ϑ A (z) = ∑ n≥0 

When ψ = ψ 0 is the trivial character, the form g ψ0 (z) is just E 1 (z) from (4), as a consequence 

of Dirichlet’s fundamental formula 

(9) R p (n) = ∑ 

r A (n). 

A∈Cl(F ) 

Otherwise, g ψ (z) is a newform in S 1 (p, χ p ), the subspace of M 1 (p, χ p ) consisting of cusp forms. 

Let H be the Hilbert class field of F with ring of integers O H . The following result shows 

that the Fourier coefficients of certain mock-modular forms of weight one with shadow g ψ (z) 

can be expressed in terms of logarithms of algebraic numbers in H. 

Theorem 1.1. Let p ≡ 3 (mod 4) be a prime with p > 3. Let ψ be a non-trivial character 

of Cl(F ), where F = Q( √ −p). Then there exists a weight one mock-modular form 

˜g ψ (z) = ∑ n≥n 0 

r + ψ (n)qn 

with shadow g ψ (z) such that the following hold. 

(i) When χ p (n) = 1 or n < − p+1 24 r+ ψ 

(n) equals to zero. 

(ii) The coefficients r + ψ 

(n) are of the form 

(10) 

∑ 

r + ψ 

(n) = −β 

A∈Cl(F ) 

ψ 2 (A) log |u(n, A)|, 

where β ∈ Q, u(n, A) are units in O H when n ≤ 0 and algebraic numbers in H when 

n > 0. In addition, β depends only on p and u(n, A) depends only on n and A. 

(iii) Let σ C ∈ Gal(H/F ) be the element associated to the class C ∈ Cl(F ) via Artin’s isomorphism. 

Then it acts on u(n, A) by 

(iv) N H/Q (u(n, A)) is an integer and satisfies 

for all non-zero integers n. 

σ C (u(n, A)) = u(n, AC −1 ). 

− β 2 log(|N H/Q(u(n, A))|) = R + p (n) 

Observe that parts (ii) and (iii) are very similar in form to Stark’s Conjectures for special 

values of derivatives of L-functions. In fact, they are consequences of known cases of his 

conjectures when n ≤ 0. Part (iv) can be viewed as a mock-modular version of Dirichlet’s 

identity (9). It is important to observe that ˜g ψ (z) is not necessarily unique.


The proof of this result relies heavily on the Rankin-Selberg method for computing heights 

of Heegner divisors as developed in [24], but entails the use of weight one harmonic forms 

with polar singularities in cusps in place of weight one Eisenstein series and hence requires 

regularized inner products. In order to get to the individual mock-modular coefficients, it 

is necessary to consider modular curves of large prime level N and their Heegner divisors of 

height zero. The relation of this part of the proof to [24] has some independent interest. 

Zagier pointed out that his identity with Gross for the norms of differences of singular 

values of the modular j-function can be neatly expressed in terms of the coefficients R + p (n) 

given in (6). For simplicity, let −d < 0 be a fundamental discriminant not equal to −p. and 

set F ′ = Q( √ −d). As is well-known, the modular j-function is well-defined on ideal classes 

of F and F ′ and takes values in the rings of integers of their respective Hilbert class fields. 

Also, values of the j-function at different ideal classes are Galois conjugates of each other. 

For any A ∈ Cl(F ) define the quantity 

(11) a d,A := ∏ 

A ′ ∈Cl(F ′ ) 

(j(A) − j(A ′ )). 

The norm of a d,A to F is thus ∏ A∈Cl(F ) a d,A and is an ordinary integer. The result of Gross 

and Zagier [23, Theorem 1.3] is that this integer can be expressed in terms of R p + (n) as 

follows: 

(12) log ∏ 

A∈Cl(F ) 

|a d,A | 2/w d 

= − 1 4 

∑ 

k∈Z 

δ(k) R + p ( pd−k2 

4 

), 

where w d is the number of roots of unity in F ′ and δ(k) = 2 if p|k and 1 otherwise. 

There are two proofs of this factorization in [23]: one analytic and one algebraic. In fact, 

the algebraic approach, which is based on Deuring’s theory of supersingular elliptic curves 

over finite fields, gives the factorization of the ideal (a d,A ) in O H for each class A ∈ Cl(F ). 

To state it, suppose l is a rational prime such that χ p (l) ≠ 1. Then the ideal (l) factors in 

O H as 

(13) l = ∏ 

l δ(l) 

A . 

A∈Cl(F ) 

The l A ’s are primes in H above l uniquely labeled so that σ C (l A ) = l AC −1 for all C ∈ Cl(F ) 

and complex conjugation sends l A to l A −1. Let A 0 be the principal class. It is shown in [23] 

that the order of a d,A0 at the place associated to the prime l A is given by 

∑ 

(14) ord lA (a d,A0 ) = 1 δ(k) ∑ ( ) 

pd−k 

r 2 

2 

A 2 . 

4l m 

m≥1 

k∈Z 

The Galois action then yields the prime factorization of the ideal (a d,A ) for any A. 

Our second main result gives a mock-modular interpretation of the individual values |a d,A |. 

It is convenient to give it as a twisted version of (12). 

Theorem 1.2. For any ˜g ψ (z) = ∑ n≥n 0 

r + ψ (n)qn given by Theorem 1.1 and −d < 0 any 

fundamental discriminant different from −p we have 

∑ 

∑ 

(15) 

ψ 2 (A) log |a d,A | 2/w d 

= − 1 δ(k) r + 4 

ψ ( pd−k2 ) 

4 

A∈Cl(F ) 

where a d,A is defined in (11). 

k∈Z


Since p > 3 is a prime number, Cl(F ) has odd order H(p) and ψ 2 is only trivial when ψ is 

trivial. Thus we can express each logarithmic term in the sum on the left hand side of the 

following identity in terms of the coefficients of a mock-modular form with a theta series as 

shadow (see Corollary 9.3 below). 

As with Theorem 1.1, this is proved using the methods of [24] and not the analytic technique 

of [23], which uses the restriction to the diagonal of an Eisenstein series for a Hilbert modular 

group. In particular, we will define a real-analytic function Φ(z), which transforms with 

weight 3/2 and level 4, and use holomorphic projection to obtain an equation between a 

finite linear combination of r + ψ 

(n)’s and an infinite sum, similar to the one in [24]. We 

also make use of machinery from [29]. One interesting new feature is an elementary counting 

argument needed to construct a Green’s function evaluated at CM points. Actually, equation 

(15) is a particular example of a more general identity involving values of certain Borcherds 

lifts. Although we will not carry this out here, it will become clear that similar methods can 

be used to prove a level N version in this form. 

Let us illustrate our Theorems in the first non-trivial case, which occurs when p = 23. 

The class group Cl(F ) has size 3 and two non-trivial characters ψ and ψ. The Hilbert class 

field H is generated by X 3 − X − 1 over F . Let α = 1.32472 . . . be the unique real root 

of X 3 − X − 1, which is also the absolute value of the square of a unit in H. The space 

S 1 (23, χ 23 ) is one dimensional and spanned by the cusp form 

g ψ (z) = η(z)η(23z), 

where η(z) = q ∏ 1/24 n≥1 (1 − qn ) is the eta function. According to Theorem 1.1, there exists 

a mock-modular form ˜g ψ (z) having a simple pole and the following Fourier expansion 

˜g ψ (z) = 

∑ 

r + ψ (n)qn . 

n≥−1 

χ 23 (n)≠1 

The condition on the principal part determines ˜g ψ (z) uniquely in this case, though this is 

not true in general. Using Stokes’ Theorem and Stark’s calculation at the end of [40, II], one 

can show that 

r + ψ (−1) = 〈g ψ, g ψ 〉 = 3 log(α) and r + ψ 

(0) = − log(α). 

From the numerical calculations of r + ψ 

(n), we can predict the values of β and u(n, A) in 

Theorem 1.1. Let ˜β and ũ(n, A) denote the guessed values for β and u(n, A) respectively. 

The following table lists r + ψ 

(n), which are calculated numerically, and the guessed values 

˜β, ũ(n, A 0 ) for the principal class A 0 ∈ Cl(F ) for 1 ≤ n ≤ 23 and a few other values of n, all 

with χ 23 (n) ≠ 1. It also contains the norms of ũ(n, A 0 ), which agree with condition (iv) in 

Theorem 1.1. 

n r + ψ (n) ˜β ũ(n, A0 ) N H/Q (ũ(n, A 0 )) ˜β/2 

5 1.1001149692823391 2 π 5 5 2 

7 1.7161505040673007 2 π 7 7 2 

10 3.9614773685309742 2 5α −6 π5 −1 5 4 

11 0.052996471463740862 2 π 11 11 2 

14 1.6582443878082415 2 7α −4 π7 −1 7 4


15 -1.9437136922512246 2 5απ5 −1 

5 4 

17 -4.2163115309750479 2 π 17 17 2 

19 -2.7119255841404505 2 π 19 19 2 

20 5.9051910607821988 2 5α −7 5 6 

21 6.7198367256215547 2 7α −10 π −1 

7 7 4 

22 -4.2709900863081686 2 11α 5 π −1 

11 11 4 

23 -3.8460181706191355 2 π 23 23 

28 4.2179936148444277 2 7α −5 7 6 

34 -2.532478252776036 2 17α 8 π −1 

17 17 4 

38 -9.942055260392833 2 19α 15 π −1 

19 19 4 

40 -14.92826076712649 2 5α 19 π 5 5 8 

Table 1. Coefficients of ˜g ψ (z) for g ψ (z) = η(z)η(23z) ∈ S 23 (p) 

Here, π l is a generator of the prime ideal l A0 in O H and is given below in terms of α. 

l 

π l 

5 2α 2 − α − 1 

7 α 2 + α − 2 

11 2α 2 − α 

17 2α 2 + 3α + 3 

19 3α 2 + α 

1 

23 √ −23 

(10α 2 + 8α + 1) 

Table 2. Generators of l A0 . 

To illustrate how Theorem 1.2 supports these predictions, consider the classical example 

−d = −7, which also appeared in [43]. First, we can combine equation (15) and Theorem 

1.1 to write (see Corollary 9.3) 

4 log |a d,A | 2/w d 

= β ∑ ( ) 

∣ pd−k 

δ(k) log ∣u 

2 

, A 

4 

∣ . 

k∈Z 

For d = 7, p = 23 and A = A 0 , we can use the predicted values in Table 1 to rewrite the 

equation above as 

( ) 

log |u(40, A0 )| + log |u(38, A 0 )| + log |u(34, A 0 )|+ 

4 log |a 7,A0 | = 4 

log |u(28, A 0 )| + log |u(20, A 0 )| + log |u(10, A 0 )| 

= 4 log |5 3 · 7 · 17 · π17 −1 · 19 · π19 −1 · α 24 |. 

This agrees with the exact of value of a 7,A0 , which is 5 3 · 7 · 17 · π17 −1 · 19 · π19 −1 · α 24 . Applying 

the Galois action to these predicted u(n, A 0 ) shows that they also agree with Theorem 1.2. 

This and several other numerical examples, together with equation (14), motivate us to 

make a conjecture about the factorization of the fractional ideal generated by u(n, A) in 

Theorem 1.1. It is not hard to see that the following conjecture implies (14). 

Conjecture. In (ii) of Theorem 1.1 we have the following. 

(i) The number 2/β is an integer dividing 24H(p)h H , where h H is the class number of H. 

(ii) For B ∈ Cl(F ), let l B be a prime ideal above the rational prime l as in (13). Then the 

order of u(n, A) at the place of H corresponding to l B is 

∑ ( n 

) 

ord lB (u(n, A)) = 2 r 

β (A −1 B) 2 . 

l m 

m≥1


At all other places of H, u(n, A) is a unit. In particular, u(n, A) ∈ O H for all n, A. 

In general (for prime level p ≡ 3 (mod 4)) , the Deligne-Serre Theorem [15] (see also [36]) 

identifies the L-function of a newform f ∈ S 1 (p, χ p ) with the Artin L-function of an irreducible, 

odd, two-dimensional, complex representation ρ f of the Galois group of a normal 

extension K/Q. Such a ρ f gives rise to a projective representation ˜ρ f : Gal(K/Q) → PGL 2 (C) 

whose image is isomorphic to D 2n , S 4 , or A 5 , in which case we say that f is dihedral, octahedral, 

or icosahedral, respectively. The forms g ψ (z) are precisely the dihedral forms and for 

them K = H. Since g ψ1 (z) = g ψ2 (z) if and only if ψ 1 = ψ 2 or ψ 1 = ψ 2 , there are exactly 

(H(p) − 1)/2 such forms. Theorem 1.1 relates the Fourier coefficients of a mock-modular 

form with dihedral shadow g ψ (z) to linear combinations of logarithms of algebraic numbers 

in the number field K = H determined by the Galois representation. Furthermore, the coefficients 

in these linear combinations are algebraic numbers in the field generated by the 

Fourier coefficients of the shadow. 

Non-dihedral newforms are often referred to as being “exotic” since their occurrence is 

rare and unpredictable. We have carried out some numerical calculations for mock-modular 

forms whose shadows are certain exotic newforms and have observed that they also seem to 

be related to linear combinations of logarithms of algebraic numbers in the number field K 

determined by the associated Galois representation. For instance, when p = 283 the space 

S 1 (283, χ 283 ) contains a pair of octahedral newforms associated to a Galois representation of 

Gal(K/Q), where K is the degree 2 extension of the normal closure of Q[X]/(X 4 − X − 1) 

with Gal(K/Q) ∼ = GL 2 (F 3 ). Similar to the dihedral case, the Petersson norms of these 

newforms come from the residue of a certain degree four L-function. As a consequence of 

proven cases of Stark’s conjecture, this residue is the logarithm of a unit inside a subfield F 6 

of K. This determines the principal part of a mock modular form whose shadow is one of 

these octahedral newforms. Using this hint and some numerical calculations, we noticed that 

the other coefficients also seem to be linear combinations of logarithms of algebraic numbers 

in the same subfield F 6 . Furthermore, these algebraic numbers generate ideals with nice 

factorizations. Thus, it is natural to expect a statement analogous to Theorem 1.1 and the 

Conjecture should hold for exotic newforms. In the final section of the paper we will give 

computational details when p = 283. 

We end the introduction with a brief outline of the paper. The first two sections concern 

general properties of weight one mock-modular forms: in §2 we deal with the existence of 

weight one mock modular forms, while in §3 we will decompose the space of weight one 

harmonic Maass forms into a plus space and minus space. In §4 we study the principal part 

at infinity of a mock-modular form with shadow g ψ (z). The proof of Theorem 1.1 is in §5 − 

§7. In §5 and §6 , we will prove Theorem 5.1 and Proposition 6.1, which imply that different 

types of linear combinations of r + ψ 

(n)’s can be put into the form of the right hand side of 

equation (10). In §7 we verify that each individual r + ψ 

(n) can be solved from these linear 

combinations. Theorem 1.2 is proved in §8 and §9. Finally, an analysis of the specific case 

p = 283 is given in §10. 

Acknowledgements. We thank Ö. Imamoḡlu for a remark made to one of us some time ago 

that the self-dual case of weight one mock modular forms should be especially interesting. 

Also, we thank J. Bruinier, S. Ehlen, A. Venkatesh, M. Viazovska, D. Zagier and S. Zwegers 

for their comments. After we obtained the results of this paper, we learned that there is 

some intersection with [42] and [19].


2. Existence of Harmonic Maass Forms of Weight One 

Given a cusp form of weight one, we will show the existence of a weight one mock-modular 

form with it as shadow in this section. There are several different approaches to proving this 

result, such as a geometric approach in [9]. For completeness and since it might have some 

independent interest, in this section we prove the existence by analytically continuing weight 

one Poincaré series via the spectral expansion. 

We begin with some basic definitions. Let k ∈ Z. For any function f : H → C and 

γ ∈ GL + 2 (R), define the weight k slash operator | k γ by 

(f| k γ)(z) := (det(γ))k/2 f(γz), 

(cz+d) k 

where γz is the linear fractional transformation of z under γ. We will write f|γ for f| k γ 

when the weight of f is understood. For M ∈ Z + let Γ 0 (M) denote the usual congruence 

subgroup of SL 2 (Z) of level M, namely 

Γ 0 (M) = {( a c d b ) ∈ SL 2(Z) ( a c d b ) ≡ ( ∗ 0 ∗ 

∗ ) mod M}. 

Let ν : Z/MZ → C × be a Dirichlet character such that ν(−1) = (−1) k and ν(γ) := ν(d) for 

γ = ( a c d b ) ∈ Γ 0(M). Let F k (M, ν) be the space of smooth functions f : H → C such that for 

all γ ∈ Γ 0 (M) 

(f | k γ)(z) = ν(γ)f(z). 

Recall from (1) and (2) the differential operator ξ k and the weight k hyperbolic Laplacian 

∆ k . Let z = x + iy. Then ∆ k can be written as 

−∆ k = −y 2 ( 

∂ 2 

+ ∂2 

∂x 2 

) ( 

+ iky 

∂y 2 

∂ 

+ i ∂ 

∂x ∂y 

Then f(z) ∈ F k (M, ν) is a weight k harmonic weak Maass form of level M and character ν 

(or more briefly, a weakly harmonic form) if it satisfies the following properties. 

(i) f(z) is real-analytic. 

(ii) ∆ k (f) = 0. 

(iii) The function f(z) has at most linear exponential growth at all cusps of Γ 0 (M). 

From now on we fix k = 1 and write | for the slash operator | 1 . Let H ! 1(M, ν) be the 

space of weakly harmonic modular forms of weight one, level M and character ν. Denote 

by M ! 1(M, ν) , M 1 (M, ν) and S 1 (M, ν) the usual subspaces of weakly holomorphic modular 

forms, holomorphic modular forms and cusp forms, respectively. Let M ! 1(M, ν) be the space 

of mock-modular forms with shadows in M 1 (M, ν) and M 1 (M, ν) ⊂ M ! 1(M, ν) the subspace 

of mock-modular forms whose shadows are in S 1 (M, ν). 

From (2), it is clear that the image of H ! 1(M, ν) under ξ 1 lies in M ! 1(M, ν). Let H 1 (M, ν) 

be the preimage of S 1 (M, ν) under ξ 1 . Then H 1 (M, ν) is canonically isomorphic to M 1 (M, ν). 

We want to show that the map ξ 1 is in fact a surjection onto S 1 (M, ν). 

To proceed, we will construct two families of Poincaré series P m (z, s), Q m (z, s), where the 

first family is similar to the one used in [6]. They are a priori defined for Re(s) > 1 and will 

be analytically continued to Re(s) > 0 through their spectral expansions. Unlike the cases 

k ≥ 2, this will only be a statement about existence, and not a formula that could be used 

to calculate the holomorphic coefficients of the preimage explicitly. To prove the analytic 

continuation, we will refer to results in [31] and [35]. 

Given any positive integer m, define 

(16) 

(17) 

) 

. 

φ ∗ m(z, s) := e 2πimx (4π|m|y) −1/2 M sgn(m) 

,s− 1 (4π|m|y), 

2 2 

ϕ ∗ m(z, s) := e 2πimx (4π|m|y) s−1/2 e −2π|m|y .


Here M µ,ν (y) is the M-Whittaker function defined by 

(18) M µ,ν (z) := z ν+1/2 e z/2 1F 1 

( 1 

2 + µ + ν; 1 + 2ν; −z) , 

with 1 F 1 (α; β; z) being the generalized hypergeometric function. Averaging them over the 

coset representatives of Γ ∞ \Γ 0 (M), we can define the following Poincaré series 

∑ 

P m (z, s, ν) := 

ν(γ)(φ ∗ m | γ)(z, s), 

Q m (z, s, ν) := 

γ∈Γ ∞\Γ 0 (M) 

∑ 

γ∈Γ ∞\Γ 0 (M) 

ν(γ)(ϕ ∗ m | γ)(z, s). 

For convenience, we shall omit ν and write P m (z, s) and Q m (z, s) instead. When Re(s) > 

1, both families have Fourier expansions in z that converges absolutely and uniformly on 

compact subset of H (see [31, Satz 6.5] for the weight zero case). In that region, both 

families are absolutely convergent and define a holomorphic function in s. Also, P m (z, s) is 

an eigenfunction of −∆ 1 with eigenvalue (s − 1/2)((1 − s) − 1/2) and belongs to F 1 (M, ν). 

Let ∆ ′ k be the differential operator defined by 

( ) 

−∆ ′ k := −y 2 ∂ 2 

+ ∂2 + iky ∂ . 

∂x 2 ∂y 2 ∂x 

It is related to ∆ k by the following equation 

Define the space D 1 (M, ν) by 

∆ ′ k + k 2 

( 

1 − 

k 

2) 

= y k/2 ∆ k y −k/2 . 

D 1 (M, ν) := {y 1/2 f(z) : f(z) ∈ F 1 (M, ν) is smooth with compact support on H} 

Let ˜D 1 (M, ν) be the completion of D 1 (M, ν) with respect to the Petersson norm 

∫ 

||g|| 2 = 〈g, g〉 := |g(z)| 2 dxdy . 

y 2 

Γ 0 (M)\H 

Satz 3.2 in [35] implies that −∆ ′ k has a self-adjoint extension − ˜∆ ′ k from D 1(M, ν) to ˜D 1 (M, ν). 

Also, − ˜∆ ′ k has a countable system of Maass cusp forms {e n(z)} n∈N ⊂ ˜D 1 (M, ν) forming the 

discrete spectrum. Each Maass cusp form e n (z) has eigenvalue λ n and each eigenvalue has 

finite multiplicity. For any f ∈ ˜D 1 (M, ν), the discrete spectrum contributes the following 

sum to its spectral expansion 

∞∑ 

〈f, e n 〉e n (z), 

n=1 

which converges absolutely and uniformly for all z ∈ H [35, Satz 8.1]. 

Let ι be a cusp of Γ 0 (M), σ ι ∈ GL 2 (R) the scaling matrix sending ∞ to ι, and Γ ι ⊂ Γ 0 (M) 

the subgroup fixing ι. One can define the weight one real analytic Eisenstein series E ι (z, s) 

by 

E ι (z, s) := y1/2 

2 

∑ 

γ∈Γ ι\Γ 0 (M) 

ν(γ) 

cz+d (Im(σ−1 ι γz)) s−1/2 . 

Selberg showed that the Eisenstein series has meromorphic continuation to the whole complex 

plane in s. For ι ranging over the cusps of Γ 0 (M), the Eisenstein series {E ι (z, s)} ι make up


the continuous spectrum of − ˜∆ ′ 1. So for any f ∈ ˜D 1 (M, ν), the contribution of the Eisenstein 

series to the spectral expansion of f is 

1 

4π 

∑ 

ι 

∫ ∞ 

−∞ 

〈f(·), E ι (·, 1 2 + ir)〉E ι(z, 1 2 + ir)dr. 

If f is smooth, then the integral in r converges absolutely and uniformly for z in any fixed 

compact subset of H [35, Satz 12.3]. Applying the completeness theorem [35, Satz 7.2 ], we 

have the spectral expansion for any smooth f ∈ ˜D 1 (M, ν) in the form 

∞∑ 

(19) f(z) = e n 〉e n (z) + 

n=1〈f, ∑ ι 

1 

4π 

∫ ∞ 

−∞ 

〈f(·), E ι (·, 1 2 + ir)〉E ι(z, 1 2 + ir)dr, 

where the sum over n ∈ N and the integral in r both converge uniformly and absolutely for 

z in any compact set of H. 

By setting f(z) = y 1/2 Q m (z, s) and comparing P m (z, s) to Q m (z, s) as in [31], we can 

deduce the following proposition. 

Proposition 2.1. For any positive integer m, the Poincaré series Q m (z, s) and P m (z, s) 

have analytic continuation to Re(s) > 0. At s = 1/2, P m (z, s) has at most a simple pole. 

Furthermore, the residues of P m (z, s) at s = 1/2 generate S 1 (M, ν). 

Proof. First, we will prove the analytic continuation Q m (z, s). For any m > 0, the function 

˜Q m (z, s) := y 1/2 Q m (z, s) is square integrable for Re(s) > 1, hence contained in ˜D 1 (M, ν). So 

we can write out its spectral expansion 

(20) 

˜Q m (z, s) = D(z, s) + C(z, s), 

D(z, s) := 

∞∑ 

〈 ˜Q m (·, s), e n (·)〉e n (z), 

n=1 

C(z, s) := 1 

4π 

∑ 

ι 

∫ ∞ 

−∞ 

〈 ˜Q m (·, s), E ι (·, 1 2 + ir)〉E ι(z, 1 2 + ir)dr, 

where D(z, s) and C(z, s) are the contributions from the discrete and continuous spectrum 

of − ˜∆ ′ 1 respectively. 

By Satz 5.2 and Satz 5.5 in [35], e n (z) has eigenvalue λ n ∈ [1/4, ∞) under − ˜∆ ′ 1. If 

λ n = 1/4, then y −1/2 e n (z) is in S 1 (M, ν). Since each eigenvalue has finite multiplicity, we 

can let R ≥ 0 such that {y −1/2 e n (z) : 1 ≤ n ≤ R} is an orthonormal basis of S 1 (M, ν). Note 

S 1 (M, ν) could be empty, in which case R = 0. 

For n ∈ N, let t n = √ λ n − 1/4, s n = 1/2 + it n . We can use equation (66) in [20], the 

bound M µ,ν (y) = O µ,ν (e y ) as y → ∞ and the vanishing property of cusp forms at all cusps 

to write 

e n (z) = 

∞∑ 

u=−∞,u≠0 

c n,u W sgn(u) 

,s 1 (4π|u|y)e 2πiux , 

2 n− 

2 

where W µ,ν (z) is the W -Whittaker function and c n,u are constants. Note if t n = 0, i.e. 

y −1/2 e n (z) is a holomorphic cusp form, then W sgn(u)/2,sn−1/2(4π|u|y) = (4π|u|y) 1/2 e −2π|u|y and


c n,u = 0 for u ≤ 0. Applying the Rankin-Selberg unfolding trick to 〈 ˜Q m (·, s), e n (·)〉 gives us 

∫ ∞ 

〈 ˜Q m (·, s), e n (·)〉 = (4π|m|) 1/2 e −2π|m|y (4π|m|y) s−1 c n,m W sgn(m) 

0 

= (4π|m|) 1/2 Γ ( s − 1 

c − it ) ( 

2 n Γ s − 

1 

+ it ) 

2 n 

(21) 

n,m ( ) 

Γ s − sgn(m) 

2 

2 ,s n− 1 2 

(4π|m|y) dy 

y 

The last step uses the Mellin transform of the W -Whittaker function [2, eq. (8b)] and 

the substitution s n = 1/2 + it n . When Re(s) > 1, the sum defining D(z, s) is absolutely 

convergent [35, Satz 8.1 ], since y 1/2 Q m (z, s) ∈ ˜D 1 (M, ν) for Re(s) > 1. When 0 < Re(s) ≤ 1, 

we can write D(z, s) as 

D(z, s) = (4π|m|) ∑ 1/2 Γ ( s + 1 − 1 

c − it ) ( 

2 n Γ s + 1 − 

1 

+ it ) 

2 n s − sgn(m) 

n,m ( 

) 

2 

e n (z) · 

n∈N 

Γ s + 1 − sgn(m) 

(s − 1 

2 

2 )2 + t 2 n 

Since t 2 n = λ n − 1/4 and ∑ n>M λ−2 n converges [35, Satz 8.1], we can apply Cauchy-Schwarz 

inequality to see that the sum on the right hand side above converges absolutely on compact 

subsets of {s ∈ C : Re(s) > 0, s ≠ 1/2}. At s = 1/2, the first R terms in the sum produce a 

simple pole since t n = 0 for all 1 ≤ n ≤ R. The rest of the sum still converges absolutely. So 

the right hand side above gives the analytic continuation of D(z, s) to Re(s) > 0. 

For the continuous spectrum, the contribution C(z, s) can be treated similarly. For any 

cusp ι, we can write the Fourier expansion of E ι (z, s) at infinity in the following form (for 

ι = ∞, see [20, eq. (76)’]) 

E ι (z, s) = y s + ψ ι (s)y 1−s + ∑ ψ ι,m (s)W sgn(m) 

,s− 1 (4π|m|y)e 2πimx , 

2 2 

m≠0 

where ψ ι (s) and ψ ι,m (s) are products of gamma factors and Selberg-Kloosterman zeta functions. 

It is well-known that The Eisenstein series can be analytically continued in s to the 

whole complex plane. When Re(s) > 1/2, the poles of E ι (z, s) are in the interval s ∈ (1/2, 1] 

[35, Satz 10.3 ]. On the line Re(s) = 1/2, E ι (z 0 , s) is holomorphic in s for any fixed z 0 ∈ H 

[35, Satz 10.4]. So both ψ ι (s) and ψ ι,m (s) admit analytic continuation to Re(s) > 0 and are 

holomorphic on Re(s) = 1/2. Using the same unfolding trick above, we can evaluate 

〈 ˜Q m (·, s), E ι (·, 1 + ir)〉 = ψ ( 1 2 ι,m + ir) (4π|m|) 1/2 Γ ( s − 1 − ir) Γ ( s − 1 2 ( ) + ir) 2 . 

2 


2 

Then C(z, s) can be written as 

C(z, s) = ∑ ι 

(22) 

= ∑ ι 

∫ 

1 ∞ 

4π −∞ 

∫ 

1 ∞ 

4π −∞ 

( 

ψ 1 

ι,m + ir) (4π|m|) 1/2 Γ ( s − 1 − ir) Γ ( s − 1 2 

( ) + ir) 2 

E 

2 ι (z, 1 + ir)dr 


2 

2 

( 

ψ 1 

ι,m + ir) (4π|m|) 1/2 Γ ( s + 1 − 1 − ir) Γ ( s + 1 − 1 

( 

2 

) 

+ ir) 2 

2 

Γ s + 1 − sgn(m) 

2 

(s − sgn(m) ) 

2 

· 

(s − 1 − ir)(s − 1 + ir) · E ι(z, 1 + ir)dr. 

2 

2 2


When Re(s) > 0, C(z, s + 1) is absolutely convergent. So when Re(s) ≠ 1/2, we can apply 

Cauchy-Schwarz to bound expression (22) by 

∫ ∣ 

∞ 

s − sgn(m) ∣∣∣∣ 

2 

(23) 

∣(s − 1 dr. 

2 )2 + r 2 

−∞ 

When Re(s) > 1/2, let ˜C(z, s) be the expression (22), which gives the analytic continuation 

of C(z, s) in this region. When Re(s) < 1/2, we can define ˜C(z, s) in a similar fashion as in 

§6 of [31]. We start with (22) and Re(s) ∈ (1/2, 1/2 + ɛ) for some small ɛ > 0, then deform 

the contour so that it goes above 1/2−s and below s−1/2 . In the process, the following residue 

i 

i 

is picked up 

(24) 

(π|m|) 1/2 Γ(2s − 1) 

( ) ∑ 


2 

ι 

ψ ι,m (s)E ι (z, s) + ψ ι,m (1 − s)E ι (z, 1 − s). 

Finally we can reduce the real part of s to less than 1/2 and change the contour back to 

the real line. So when Re(s) < 1/2, let ˜C(z, s) be the sum of (24) and (22), which are 

holomorphic. The bound (23) and m ≥ 1 guarantees the existence of the limits of ˜C(z, s) as 

Re(s) approaches 1/2 from the left and from the right. The procedure defining ˜C(z, s) shows 

that the limits agree. So we define ˜C(z, s) on Re(s) = 1/2 to be this limit. Then ˜C(z, s) gives 

the analytic continuation of C(z, s) to Re(s) > 0. Putting these together with the analytic 

continuation of D(z, s), we have the analytic continuation of Q m (z, s) to Re(s) > 0. 

Now to analytically continue P m (z, s), we can simply compare it to Q m (z, s). Applying 

the power series expansions of 1 F 1 (α; β; z) and the exponential map to (16) and (17), we see 

that for y small and 0 < Re(s) < 2, 

|(4π|m|y) −s+1/2 (φ ∗ m(z, s) − ϕ ∗ m(z, s))| = O m (y). 

So the difference P m (z, s) − Q m (z, s) defined by term-wise subtraction is a holomorphic function 

in s for Re(s) > 0. That means the analytic continuation of P m (z, s) to Re(s) > 0 

follows from that of Q m (z, s). Furthermore, they have the same poles and residues in the 

region Re(s) > 0. Thus, to prove the second half of the proposition, it suffices to analyze the 

poles and residues of Q m (z, s) at s = 1/2. 

Since m > 0, expression (23) is bounded by an absolute constant for all s ∈ (1/2, 2]. So 

˜C(z, s) does not contribute to the pole at s = 1/2. For a Maass cusp form e n (z), the right 

hand side of (21) vanishes at s = 1/2 if t n ≠ 0. Otherwise, y −1/2 e n (z) ∈ S 1 (M, ν) and 

〈 ˜Q m (·, s), e n (·)〉 has a simple pole of residue (4π|m|) 1/2 c n,m . Thus, we have 

∑ R 

(25) Res s=1/2 P m (z, s) = Res s=1/2 Q m (z, s) = (4π|m|) 1/2 c n,m y −1/2 e n (z). 

Since {y −1/2 e n (z) : n = 1, . . . , R} is a basis of S 1 (M, ν), the matrix 

{c n,m : 1 ≤ n ≤ R, 1 ≤ m ≤ K} 

has rank equals to R for K sufficiently large. Therefore the residues at s = 1/2 of P m (z, s) 

generate S 1 (M, ν). 

□ 

The following theorem is an immediate consequence of the proposition above. 

Theorem 2.2. Using the notations above, the following map is a surjection 

ξ 1 : H 1 (M, ν) → S 1 (M, ν), 

n=1


i.e. for any cusp form h(z) ∈ S 1 (M, ν), there exists ˜h(z) ∈ M 1 (M, ν) with shadow h(z). 

Proof. When k = 1, equation (2) becomes ∆ 1 = ξ 1 ◦ ξ 1 . So the Poincaré series P m (z, s) 

satisfies 

(26) ∆ 1 (P m (z, s)) = ( s − 1 2) 2 

Pm (z, s) 

when Re(s) > 1. Since the difference between both sides is holomorphic in s and P m (z, s) 

can be analytically continued to Re(s) > 0 as in Proposition 2.1, equation (26) is valid for 

Re(s) > 0. At s = 1/2, suppose P m (z, s) has the following Taylor series expansion in s 

P m (z, s) = g −1 (z) ( s − 1 2) −1 

+ g0 (z) + g 1 (z) ( s − 1 2) 

+ Oz 

( (s 

− 

1 

2 

) 2 

) 

, 

with g j (z) ∈ F 1 (M, ν) real-analytic for j = −1, 0, 1. Since ξ 1 commutes with the slash 

operator and ∆ 1 (g 1 (z)) = ξ1(g 2 1 (z)) = g −1 (z), we have ξ 1 (g 1 (z)) ∈ F 1 (M, ν) is real-analytic 

and a preimage of g −1 (z) under ξ 1 . 

By considering the Fourier expansion of g 1 (z), we know that if g 1 (z) has at most linear 

exponential growth near the cusps, then ξ 1 (g 1 (z)) has the same property. Suppose Q m (z, s) 

has the Laurent expansion 

Q m (z, s) = g −1 (z) ( s − 1 2) −1 

+ g 

′ 

0 (z) + g ′ 1(z) ( s − 1 2) 

+ Oz 

( (s 

− 

1 

2 

near s = 1/2. Then from the spectral expansion of Q m (z, s), i.e. y −1/2 (D(z, s) + C(z, s)) 

as in (20), it is not hard to see that g 1(z) ′ has at most linear exponential growth near the 

cusps. The difference P m (z, s) − Q m (z, s) is a Poincaré series defined for Re(s) > 0. So the 

coefficient of (s − 1/2) of its Laurent series expansion around s = 1/2, say g 1(z), ′′ has at most 

linear exponential growth near the cusps. That means the sum of g 1(z) ′ and g 1(z), ′′ which is 

g 1 (z) by analytic continuation, also has this property. Thus, ξ 1 (g 1 (z)) is in H 1 (M, ν) and a 

preimage of g −1 (z) ∈ S 1 (M, ν) under ξ 1 . Since the functions {Res s=1/2 P m (z, s) : m ≥ 1} span 

the space S 1 (M, ν), the map ξ 1 : H 1 (M, ν) → S 1 (M, ν) is surjective. 

□ 

3. Regularized Inner Products and the Weight One Plus Space 

From now on, we fix M = p to be a prime number congruent to 3 modulo 4 and ν = χ p . The 

spaces H 1(p, ! χ p ), H 1 (p, χ p ), M 1(p, ! χ p ), M 1 (p, χ p ), S 1 (p, χ p ), M ! 1(p, χ p ), M 1 (p, χ p ) are the same 

as before, and we will drop χ p in these notations. In this section, we will relate the regularized 

inner products between g(z) ∈ S 1 (p) and f(z) ∈ M 1(p) ! to linear combinations of coefficients 

of a mock-modular form ˜g(z), whose shadow is g(z), via Stokes’ theorem. This technique 

has been used in [9], [11] and [18]. Then, we will split up the space M ! 1(p) canonically into 

the plus and minus spaces to facilitate further analysis. Whenever convenient, we will switch 

between M ! 1(p) and H 1(p). 

! 

Given f(z) ∈ M 1(p) ! and g(z) ∈ S 1 (p), the usual Petersson inner product 〈f, g〉 can be 

regularized as follows. Since p is prime, ( Γ 0 (p) has two inequivalent cusps, 0 and ∞. They 

−1/ 

are related by the scaling matrix σ 0 = 

√ p 

√ p 

). Take a fundamental domain of Γ 0 (p)\H, 

cut off the portion with Im(z) > Y for a large Y and intersect it with its translate under σ 0 . 

We will call this the truncated fundamental domain F(Y ). Now, define the regularized inner 

product by 

(27) 〈f, g〉 reg := lim 

Y →∞ 

∫ 

F(Y ) 

f(z)g(z)y dxdy 

y 2 . 

) 2 

)


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.


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 

) 

.


where 

V p = 

( { 

p 1, p ≡ 1 (mod 4) 

, ε 

1) 

p = 

i, p ≡ 3 (mod 4) 

Since p ≡ 3 (mod 4), we know that ε p = i. From the definition of pr −ɛ , we have 

(37) 

(pr −ɛ f)(z) = 1 (−ɛih(z) + f(z)) 

2 

( 

= 1 −ɛi 

√ (f| 1 W p V p )(z) + ∑ ) 

(1 − χ p (n)ɛ) a(n, y)q n ∈ H !,ɛ 

2 p 

1 (p). 

n∈Z 

Notice that the action V p produces an expansion with coefficients supported on multiples of 

p. So if pr −ɛ (f) = 0, then a(n, y) = 0 whenever χ p (n) = −ɛ and f ∈ H !,ɛ 

1 (p) by definition. 

Conversely if f ∈ H !,ɛ 

1 (p), then 

pr −ɛ (f) ∈ H !,+ 

1 (p) ∩ H !,− 

1 (p), 

which means the coefficients ξ 1 (pr −ɛ (f)) ∈ M ! 1(p) are supported on multiples of p. A lemma 

due to Hecke ([33, Lemma 6, p.32] ) implies that 

(38) M !,+ 

1 (p) ∩ M !,− 

1 (p) = {0}. 

So ξ 1 (pr −ɛ (f)) = 0 and pr −ɛ (f) is holomorphic and also contained in M !,+ 

1 (p) ∩ M !,− 

1 (p). By 

the same argument, pr −ɛ (f) vanishes. From this, we can deduce that H !,ɛ 

1 (p) is the eigenspace 

of iU p W p with eigenvalue ɛ, and pr ɛ restricted to H !,ɛ 

1 (p) is the identity. Using the relationship 

(f| 1 Wp 2 )(z) = −f(z), we can then deduce equation (34). It is easy to see that we have the 

following direct sum decomposition 

H 1(p) ! = H !,+ 

1 (p) ⊕ H !,− 

1 (p) 

f = pr + (f) + pr − (f). 

Applying the same procedure to the harmonic Maass form associated to any mock-modular 

form in M ! 1(p), we have the same decomposition of space 

M ! 1(p) = M !,+ 

1 (p) ⊕ M !,− 

1 (p). 

The decomposition of the space H 1(p) ! above behaves well under the action of ξ 1 . By Theorem 

2.2, we can prove a statement about surjectivity of ξ 1 : H1(p) ɛ → S1 −ɛ (p). Furthermore, 

there is a nice characterization of the space S1(p) ɛ as a consequence of facts about weight one 

modular forms. 

Lemma 3.3. For ɛ = ±1, we have a surjection 

(39) ξ 1 : H ɛ 1(p) → S −ɛ 

1 (p). 

In other words, for any g(z) ∈ S −ɛ 

1 (p), there exists a mock-modular form ˜g(z) ∈ M ɛ 1(p) whose 

shadow is g(z). 

In addition, S + 1 (p) contains all dihedral cusp forms. If there are 2d − non-dihedral forms in 

S 1 (p), then the spaces S + 1 (p) and S − 1 (p) have dimensions 1 2 (H(p)−1)+d − and d − respectively 

Proof. By Theorem 2.2, the following map is surjective 

(40) ξ 1 : H 1 (p) → S −ɛ 

1 (p). 

□


Since ξ 1 commutes with the action of U p and W p and conjugates their coefficients, it commutes 

with the operator pr ɛ as follows 

(41) ξ 1 (pr ɛ (f)) = pr −ɛ (ξ 1 (f)). 

The operator pr −ɛ is the identity when restricted to S1 −ɛ (p). So the preimage of S1 −ɛ (p) under 

ξ 1 lies in H1(p) ɛ and the map ξ 1 : H1(p) ɛ → S1 −ɛ (p) is surjective for ɛ = ±1. 

Now if f ∈ S 1 (p) is a dihedral newform, then it is a linear combination of theta series, whose 

n th coefficient is zero if χ p (n) = −1. So f ∈ S 1 + (p) by definition. If f(z) = ∑ n≥1 

c(f, n)qn 

is a octahedral or icosahedral newform, then f(z) and f(z) := f(z) = ∑ n≥1 c(f, n)qn are 

linearly independent newforms. When l ≠ p is a prime number, we have the relationship 

(42) c(f, l) = χ p (l)c(f, l). 

This is a consequence of the formula of the adjoint of the l th Hecke operator with respect to 

the Petersson inner product [34, p.21]. By the definition of S1(p), ɛ we have f + f ∈ S 1 + (p) 

and f − f ∈ S1 − (p). Since the number of octahedral and icosahedral newforms is set to be 

2d − , we obtain the formulae for the dimensions. 

□ 

Remark: The spaces S ɛ 1(p) should be compared to the spaces M + and M − in (9.1.2) in 

[36]. If all octahedral and icosahedral newforms f(z) ∈ S 1 (p) satisfy 

(43) (f| 1 W p )(z) = −if(z), 

then M + is the span of the weight one Eisenstein series and S + 1 (p) and M − = S − 1 (p). 

Besides compatibility with ξ 1 , the decomposition also behaves nicely with respect to the 

regularized inner product. As a consequence of Lemma 3.1, we have the following proposition. 

Proposition 3.4. Let f(z) ∈ M !,ɛ 

1 (p), g(z) ∈ S ɛ′ 

1 (p) and ˜g(z) ∈ M −ɛ′ 

1 (p) with shadow g(z) 

and Fourier expansion ∑ n∈Z c+ (n)q n . If ɛ = ɛ ′ , then 

(44) 〈f, g〉 reg = ∑ n∈Z 

δ(n)c(f, −n)c + (n). 

Here δ(n) is 2 if p|n and 1 otherwise. If ɛ ≠ ɛ ′ , then 〈f, g〉 reg = 0. 

Proof. Let ĝ(z) ∈ H −ɛ′ 

1 (p) be the harmonic Maass form associated to ˜g(z). Then both 

assertions are immediate consequences of Lemma 3.1 and 

(f| 1 W p )(z) = −iɛ(f| 1 U p )(z), (ĝ| 1 W p )(z) = −iɛ ′ (ĝ| 1 U p )(z). 

Since the adjoint of W p (resp. U p ) is W −1 

p 

= −W p (resp. W p U p Wp 

−1 ) with respect to the 

regularized inner product, it is easy to check that pr ɛ is self-adjoint, which also proves the 

second claim. 

□ 

When g(z) above is zero, i.e. ˜g(z) is modular, equation (44) reduces to an orthogonal 

relationship between Fourier coefficients of weakly holomorphic modular forms. In particular, 

we can take ˜g(z) to be cusp forms and obtain relations satisfied by {c(f, −n) : n ≥ 1}. These 

relations turn out to characterize the space M !,ɛ 

1 (p), which gives a nice characterization of 

the space M ɛ 1(p). 

Proposition 3.5. Let g(z) ∈ S1(p). ɛ Then there exists a mock-modular form ˜g(z) ∈ M −ɛ 

1 (p) 

with shadow g(z) and prescribed principal part ∑ n


Proof. The necessity part follows from Proposition 3.4. Since h(z) is a cusp form, the summation 

in (44) is only over n < 0. When ˜g(z) ∈ M −ɛ 

1 (p) is modular, the sufficiency follows 

from an application of Serre duality [5, §3] (also see [8, Theorem 6]). 

When ˜g(z) is mock-modular, let ∑ n


For convenience, we write E(τ, s) for E 1 (τ, s). Kronecker’s first limit formula states that 

(49) ζ(2s)E(τ, s) = π 

s − 1 + 2π ( γ − log(2) − log( √ y|η(τ)| 2 ) ) + O(s − 1), 

where γ is the Euler constant. Using (49) and Rankin-Selberg unfolding trick, we can relate 

the inner product between dihedral newforms to logarithm of u A as follows. 

Proposition 4.1. Let ψ, ψ ′ be characters of Cl(F ), with ψ non-trivial. When ψ ′ = ψ or ψ, 

we have 

(50) 〈g ψ , g ψ ′〉 = 1 ∑ 

ψ 2 (A) log |u A |. 

12 

Otherwise, 〈g ψ , g ψ ′〉 = 0. 

A∈Cl(F ) 

Proof. Since p is prime, the Eisenstein series E p (z, s) has a simple pole at s = 1 with residue 

, which is independent of z. So we have the relationship 

3 

π(p+1) 

3 

π(p + 1) · 〈g ψ, g ψ ′〉 = Res s=1 g ψ (z)g ψ ′(z)E p (z, s)y 

∫Γ dxdy . 

0 (p)\H 

y 2 

Now, we can use the Rankin-Selberg method to unfold the right hand side and obtain 

∫ 

g ψ (z)g ψ ′(z)E p (z, s)y dxdy = Γ(s) ∑ r ψ (n)r ψ ′(n) 

y 2 (4π) s n s 

Γ 0 (p)\H 

Let ρ ψ : Gal(Q/Q) → GL 2 (C) be the representation induced from ψ. Then it is also the 

one attached to g ψ (z) via Deligne-Serre’s theorem. Up to Euler factors at p, the right hand 

side is L(s, ρ ψ ⊗ ρ ψ ′), the L-function of the tensor product of the representations ρ ψ and ρ ψ ′. 

From the character table of the dihedral group D 2H(p) , we see that 

ρ ψ ⊗ ρ ψ ′ = ρ ψψ ′ ⊕ ρ ψψ ′. 

So when ψ ′ ≠ ψ or ψ ′ , the L-function L(s, ρ ψ ⊗ρ ψ ′) is holomorphic at s = 1 and 〈g ψ , g ψ ′〉 = 0. 

Otherwise, we have 

∑ r ψ (n)r ψ (n) 

= ζ(s)L(s, χ p)L(s, ρ ψ 2) 

, 

n s ζ(2s)(1 + p −s ) 

n≥1 

Putting these together, we obtain 

(51) 〈g ψ , g ψ 〉 = p 

2π L(1, χ p)L(1, ρ 2 ψ 2). 

From the theory of quadratic forms, we have 

√ ∑ 

pL(s, ρψ 2) = ζ(2s) ψ 2 (A)E(τ A , s). 

A∈Cl(F ) 

Since ψ is non-trivial and H(p) is odd, ψ 2 is non-trivial and (49) implies that 

L(1, ρ ψ 2) = −√ 2π ∑ 

ψ 2 (A) log( √ y A |η(τ A )| 2 ), 

p 

A∈Cl(F ) 

Along with the class number formula for p > 3 

L(1, χ p ) = 2πH(p) 

w p 

√ p 

n≥1 

= πH(p) √ p 

,


we arrive at 

(52) 〈g ψ , g ψ 〉 = −H(p) 

∑ 

A∈Cl(F ) 

Since ψ 2 is non-trivial, we have equation (50). 

ψ 2 (A) log( √ y A |η(τ A )| 2 ). 

Remarks. 

i) By the same procedure, one can analyze the inner product between any weight one 

newforms. In particular, if g(z), h(z) ∈ S 1 (p) arises from different types of Galois representations, 

then 〈g, h〉 = 0. 

ii) This proposition is really a proven case of Stark’s conjecture in the abelian, rank one 

case. 

Next we prove the existence of the special pre-image. 

Proposition 

∑ 

4.2. Let ψ be a non-trivial character of Cl(F ). Then there exists ˜g ψ (z) = 

n≥n 0 

r + ψ (n)qn ∈ M − 1 (p) with shadow g ψ (z) such that 

(i) When χ p (n) = 1 or n < − p+1 , the coefficient 24 r+ ψ 

(n) equals to zero. 

(ii) For n ≤ 0, the coefficients r + ψ 

(n) are of the form 

r + ψ 

(n) = −β 

∑ 

A∈Cl(F ) 

ψ 2 (A) log |u(n, A)|, 

where β ∈ Q depends only on p, and u(n, A) ∈ O × H 

depends only on n and A. 

(iii) Let σ C ∈ Gal(H/F ) be the element associated to the class C ∈ Cl(F ) via Artin’s isomorphism. 

Then it acts on the units u(n, A) by 

σ C (u(n, A)) = u(n, AC −1 ). 

Proof. From Lemma 3.3, we know that the dimension of S + 1 (p), denoted by d + , is 1 2 (H(p) − 

1)+d − . Let {g 1 , g 2 , . . . , g d+ } be a basis of S + 1 (p). Suppose we have chosen n 1 , n 2 , . . . , n d+ > 0 

such that χ p (n k ) ≠ −1 and the matrix 

P := [c(g j , n k )] 1≤j,k≤d+ 

is invertible. Then for any ψ, there exists ˜g ψ (z) ∈ M − 1 (p) with shadow g ψ (z) and principal 

part coefficients r + ψ (−n) = 0 for all positive integer n not in the set {n 1, n 2 , . . . , n d+ } by 

Proposition 3.5. Furthermore, these r + ψ 

(−n) can be uniquely determined from solving a 

d + × d + system of equations. We will show that such ˜g ψ (z) can be made to satisfy the 

proposition. 

First, we can choose {n j : 1 ≤ j ≤ d + } such that n j ≤ (p + 1)/24. Let {g 1 , g 2 , . . . , g d+ } to 

be a q-echelon basis. Then these n j ’s can be chosen to be bounded by the supremum of the 

set 

{ord ∞ h(z) : h(z) ∈ S 1 + (p)}. 

Now, the square of a cusp form in S 1 (p) is in S 2 (p), the space of weight 2 cusp form of trivial 

nebentypus on Γ 0 (p). This gives rise to a holomorphic differential form on the modular curve 

X 0 (p). Atkin showed in [1] that ∞ is not a Weiestrass point of the modular curve X 0 (p), 

i.e. ord ∞ f(z) is no more than the genus of X 0 (p). Using the Riemann-Hurwitz formula (see 

[37]), we know that the genus of X 0 (p) is bounded by (p + 1)/12. So the n j ’s can be chosen 

to be bounded by (p + 1)/24. 

□


Let {g j (z) : 1 ≤ j ≤ (H(p) − 1)/2} be the dihedral newforms, and label the class group 

characters as {ψ j : 1 ≤ j ≤ H(p) − 1} such that g ψj (z) = g j (z) and ψ j = ψ H(p)−j for 1 ≤ j ≤ 

(H(p) − 1)/2. When (H(p) + 1)/2 ≤ j ≤ d + , let g j (z) be linear combinations of non-dihedral 

newforms such that g j (z) has integral coefficients and the set {g j (z) : (H(p) + 1)/2 ≤ j ≤ d + } 

is linearly independent. Then for each non-trivial ψ, the values {r + ψ (−n k) : 1 ≤ k ≤ d + } 

satisfies the equation 

P · [δ(n k )r + ψ (−n k)] 1≤k≤d+ = R, 

where R is the d + × 1 matrix (〈g j , g ψ 〉) 1≤j≤d+ 

. By Proposition 4.1, the inner product 〈g j , g ψ 〉 

∑ 

is either 0 or 1 

12 A∈Cl(F ) ψ2 (A) log |u A |. So we have 

⎛ 

⎞ 

0. 

∑ 

0 

1 

R = 

12 A ψ2 (A) log |u A | 

⎜ 

⎟ 

⎝ 

0. ⎠ 

0 

Let A 0 be the principal class in Cl(F ) and label the non-principal ( classes as A i for 1 ≤ 

i ≤ H(p) − 1 such that A −1 = A H(p)−i . Let M ′ M1 0 

be the matrix 0 Id d− 

), where M 1 is an 

i 

1 

(H(p) − 1) × 1 (H(p) − 1) matrix defined by 

2 2 

[ψ j (A i ) + ψ j (A H(p)−i ) − 2] 1≤i,j≤(H(p)−1)/2 

and Id d− is the d − × d − identity matrix. The product M ′ P is the matrix 

⎛ 

⎞ 

H(p)(r A1 (n 1 ) − r A0 (n 1 )) · · · H(p)(r A1 (n d+ ) − r A0 (n d+ )) 

. . 

.. . 

M ′ H(p)(r 

P = 

A(H(p)−1)/2 (n 1 ) − r A0 (n 1 )) · · · H(p)(r A(H(p)−1)/2 (n d+ ) − r A0 (n d+ )) 

, 

c(g 

⎜ 

(H(p)+1)/2 , n 1 ) · · · c(g (H(p)+1)/2 , n d+ ) 

⎝ 

. ⎟ 

. 

.. . 

⎠ 

c(g d+ , n 1 ) · · · c(g d+ , n d+ ) 

which has integer coefficients. It is also non-singular since 

{ϑ Aj − ϑ A0 : 1 ≤ j ≤ (H(p) − 1)/2} ∪ {g j : (H(p) + 1)/2 ≤ j ≤ d + } 

is a basis of S 1 + (p). The product M ′ R is the d + × 1 matrix 

⎛ ∑ 

A ψ2 (A) log |u A (A 1 )| 

where 

Also by (47), we have 

M ′ R = 

⎜ 

⎝ 

1 

12 

1 

12 

⎞ 

∑ 

. 

A ψ2 (A) log |u A (A (H(p)−1)/2 )| 

, 

⎟ 

0. 

⎠ 

0 

u A (B) := u A √ B u A √ B −1 

u 2 A 

σ C (u A (B)) = u AC −1(B) 

∈ O × H .


for any class A, B, C ∈ Cl(F ). 

Since M ′ P is a non-singular matrix with integer entries, its inverse, say (α i,j ) 1≤i,j≤d+ , has 

rational entries. Thus, r + ψ (−n i) can be written as 

δ(n i )r + ψ (−n i) = ∑ A 

(H(p)−1)/2 

∑ 

ψ 2 (A) α i,j log |u A (A j )| 

j=1 

with α i,j ∈ Q for all 1 ≤ i, j ≤ d + . From this, we can choose β ∈ Q such that 

u(n i , A) = 

(H(p)−1)/2 

∏ 

j=1 

u A (A j ) −α i,j/δ(n i )β ∈ O × H . 

α i,j 

2H(p)β ∈ Z and 

Then r + ψ (−n i) satisfies conditions (ii) and (iii) in the proposition. Furthermore, the unit 

u(n i , A) is an H(p) th power in O × H . 

Finally, apply Propositions 4.1 and 3.4 to the Eisenstein series E 1 (z) and the cusp form 

g ψ (z), we get 

∑ 

δ(n)R p (n)r + ψ 

(−n) = 0. 

n≥0 

So we can write r + ψ 

(0) = −β log |u(0, A)| with 

u(0, A) = 

d + 

∏ 

i=1 

which satisfies condition (iii) as well. 

u(n i , A) −δ(n i)R p(n i )/H(p) ∈ O × H , 

□ 

5. Rankin’s method and values of Green’s function I 

From the proof of Proposition 4.2, we saw that information about r + ψ 

(n), n ≤ 0 can be 

retrieved from their various linear combinations coming from inner products. In this section, 

we will consider a different linear combination of the coefficients r + ψ 

(n). They can be viewed 

as the regularized inner product between g ψ (z) and certain weakly holomorphic forms in 

M !,+ 

1 (p) lifted from modular functions of level one. Following the same procedure as in [24], 

we will show in Theorem 5.1 that these linear combination of r + ψ 

(n)’s can be expressed as the 

twisted sum of logarithms of the modular polynomial evaluated at certain CM points. This 

result is more or less a twisted version of the Gross-Zagier formula at level one, where there 

is no weight two cusp form or L-function. Also, the information provided by the theorem, 

as formulated in Corollary 5.2, will be a stepping stone to the proof of Theorem 1.1. As a 

modular form on the degenerated Hilbert modular surface, the modular polynomial can be 

expressed in terms of a Borcherds lift of modular functions of level one. Thus, it is natural 

to phrase Theorem 5.1 in terms of a regularized inner product and this Borcherds lift. This 

prepares the way for §6, where we treat the level N case. 

For m ≥ 0, let j m (z) = q −m + O(q) be the unique modular function of level one with a 

pole of order m at infinity, e.g. j 0 (z) = 1 and j 1 (z) = j(z) − 744. For each B ∈ Cl(F ), choose 

an associated CM point τ B = x B + iy B ∈ H and define j lift,B 

m (z) ∈ M !,+ 

1 (p) by 

jm 

lift,B (z) := j m (pz)ϑ B (z).


Let M 2 (Z) be the space of 2 × 2 matrices with integer coefficients. For m ≥ 1, (τ 1 , τ 2 ) ∈ 

(PSL 2 (Z)\H) 2 , define the function Ψ m (τ 1 , τ 2 ) by 

∏ 

Ψ m (τ 1 , τ 2 ) := 

(j(τ 1 ) − j(γτ 2 )) . 

γ∈SL 2 (Z)\M 2 (Z) 

det(γ)=m 

It is the value of the modular polynomial ϕ m (X, Y ) at X = j(τ 1 ), Y = j(τ 2 ), and also the 

Borcherds lift of j m (z) to a function on the degenerated Hilbert modular surface. The main 

result of this section is as follows. 

Theorem 5.1. Let ψ be a non-trivial character of Cl(F ) and m ≥ 1. Then 

∑ 

( log |Ψm (τ 

(53) 〈jm 

lift,B , g ψ 〉 reg = −2 lim ψ 2 (A ′ A 

) 

′ B ′−1 + ɛ, τ A ′ B ′)| + ) 

ɛ→0 

r B (m) log |y A ′ B ′−1| , 

A ′ ∈Cl(F ) 

where B ′ ∈ Cl(F ) is the unique class such that B ′2 = B. 

It will be clear in the proof that equation (53) is independent of the choice of τ B . Before 

stating the proof, we need a crucial result in [23]. For two distinct points τ j = x j + iy j ∈ H, 

the invariant hyperbolic distance d(τ 1 , τ 2 ) between them is defined by 

(54) cosh d(τ 1 , τ 2 ) = (x 1 − x 2 ) 2 + y1 2 + y2 

2 . 

2y 1 y 2 

Notice for d(τ 1 , τ 2 ) = d(γτ 1 , γτ 2 ) for all γ ∈ SL 2 (R). The Legendre function of the second 

kind Q s−1 (t) is defined by 

∫ ∞ 

Q s−1 (t) = (t + √ t 2 − 1 cosh u) −s du, Re(s) > 1, t > 1, 

(55) 

0 

Q 0 (t) = 1 log ( ) 

1 + 2 

2 t−1 

For two distinct points τ 1 , τ 2 ∈ PSL 2 (Z)\H, the following convergent series defines the automorphic 

Green’s function 

(56) G s (τ 1 , τ 2 ) := ∑ 

g s (τ 1 , γτ 2 ), Re(s) > 1, 

where 

γ∈PSL 2 (Z) 

(57) g s (τ 1 , τ 2 ) := −2Q s−1 (cosh d(τ 1 , τ 2 )). 

Recall that E(τ, s) is defined in (48) and ϕ 1 (s) is the coefficient of y 1−s in the Fourier 

expansion of E(τ, s). 

Proposition 5.1 in [23] tells us that for distinct τ 1 , τ 2 ∈ PSL 2 (Z)\H, the values of the 

j-function is related to the values of the automorphic Green’s function by 

(58) log |j(τ 1 ) − j(τ 2 )| 2 = lim 

s→1 

(G s (τ 1 , τ 2 ) + 4πE(τ 1 , s) + 4πE(τ 2 , s) − 4πϕ 1 (s)) − 24. 

Apply the m th Hecke operator to τ 2 on both sides gives us [23, eq.(5.2)] 

( G 

m ) 

(59) log |Ψ m (τ 1 , τ 2 )| 2 s (τ 1 , τ 2 ) + 4πσ 1 (m)E(τ 1 , s)+ 

= lim 

s→1 4πm s σ 1−2s (m)E(τ 2 , s) − 4πσ 1 (m)ϕ 1 (s) 

where σ ν (m) = ∑ d|m dν and 

G m s (τ 1 , τ 2 ) = 

∑ 

γ∈SL 2 (Z)\M 2 (Z) 

det(γ)=m 

g s (τ 1 , γτ 2 ). 

− 24σ 1 (m),


Using this result and appropriate Rankin-Selberg unfolding trick as in [24], we can prove 

Theorem 5.1. 

Proof of Theorem 5.1. Let ˜g ψ (z) ∈ M − 1 (p) be a mock-modular form as in Proposition 4.2 

with Fourier expansion 

˜g ψ (z) = ∑ r + ψ (n)qn 

n∈Z 

and ĝ ψ (z) the associated harmonic Maass form. For a class B ∈ Cl(F ), define 

Φ B (ĝ ψ , z) := Tr p 1(−i(ĝ ψ |W p )(z)ϑ B (z)) 

= (ĝ ψ |U p )(z)ϑ B (z) + (ĝ ψ (z)ϑ B (z))|U p . 

This should be compared to ˜Φ s (z) in Proposition (1.2) of [24, §IV]. When N = 1, the 

derivative of ˜Φ s (z) at s = 1 is more or less our Φ B (ĝ ψ , z) for ψ trivial. 

Here, Φ B (ĝ ψ , z) transforms like a level one, weight two modular form and has the following 

Fourier expansion at infinity 

Φ B (ĝ ψ , z) = ∑ n∈Z(b B (ĝ ψ , n, y) + a B (ĝ ψ , n))q n , 

b B (ĝ ψ , n, y) = − ∑ k∈Z 

a B (ĝ ψ , n) = ∑ k∈Z 

δ(k)r ψ (k)β 1 

(k, y p 

δ(k)r + ψ (k) r B(pn − k). 

) 

r B (pn + k), 

By the choice of ˜g ψ (z), we have ord ∞ (˜g ψ (z)) ≥ − p+1 > −p. So the sum a 24 B(ĝ ψ , n) is zero for 

all n < 0. It is easy to see from the definition above that jm lift,B ∈ M !,+ 

1 (p). So we can apply 

Proposition 3.4 to obtain 

(60) 

〈j lift,B 

m , g ψ 〉 reg = a B (ĝ ψ , m). 

In particular when m = 0, we have j lift,B 

0 = 1 lift,B = ϑ B (z) and 

〈ϑ B , g ψ 〉 = a B (ĝ, 0). 

Now, let Ẽ2(z) the non-holomorphic Eisenstein series of level one, weight two defined by 

Ẽ 2 (z) = 1 − 24 ∑ n≥1 

σ 1 (n)q n − 3 

πy . 

Then we can use Poincaré series to apply holomorphic projection to the function 

Φ ∗ B(ĝ ψ , z) := Φ B (ĝ ψ , z) − a B (ĝ ψ , 0)Ẽ2(z), 

since it satisfies the growth condition Φ ∗ B (ĝ ψ, z) = O(1/y) at the cusp infinity. For m > 0, let 

P m,1 (z, s) be the Poincaré series of level one, weight two defined by 

∑ 

P m,1 (z, s) := 

(y s e 2πimz )| 2 γ. 

γ∈Γ ∞\PSL 2 (Z) 

The limit as s goes to zero of the inner product between P m,1 (z, s) and Φ ∗ B (ĝ ψ, z) is the m th 

Fourier coefficient of a cusp form of weight two, level one. Since the space of such forms is 

trivial, we have 

(61) lim 

s→0 

〈Φ ∗ B(ĝ ψ , z), P m,1 (z, s)〉 = 0.


Using the Fourier expansion of Φ B (ĝ ψ , z) and Rankin-Selberg unfolding, we can rewrite equation 

(61) as 

( ) 

k 

(62) a B (ĝ ψ , m) + 24σ 1 (m)a B (ĝ ψ , 0) = lim δ(k)r B (pm + k)r ψ (k)φ s , 

where 

φ s (µ) := 

∫ ∞ 

1 

s→0 

∑ 

k≥1 

du 

(1 + µu) 1+s u . 

Comparing Q s−1 (t) and φ s−1 (µ), we see that φ 0 (µ) = 2Q 0 (1 + 2µ) and 

φ s−1 (µ) − 2Γ(2s) Q 

sΓ(s) 2 s−1 (1 + 2µ) = O(µ −1−s ). 

( ) ( ) 

2 

That means we can substitute Q 

sΓ(s) 2 s−1 1 + 2k 

k 

for φ 

pm s−1 and in the limit as s approaches 

1. Together with (8) and (60), equation (61) 

pm 

becomes 

(63) 

〈j lift,B 

m , g ψ 〉 reg +24σ 1 (m)〈ϑ B , g ψ 〉 = 

− lim 

s→1 

∑ 

A 

ψ(A) ∑ k≥1 

pm 

( )) 

δ(k)r B (pm + k)r A (k) 

(−2Q s−1 1 + 2k . 

pm 

Now the counting argument in Proposition (3.11) in [23] tells us that 

δ(k)r B (pm + k)r A (k) = ρ m (A, B, k), 

where ρ m (A, B, k) counts the number of γ ∈ M 2 (Z)/{±1} such that det(γ) = m and 

cosh d(τ √ AB −1 , γτ √ AB ) = 1 + 2k 

pm . 

In particular when k = 0, the number of γ ∈ M 2 (Z)/{±1} such that det(γ) = m and 

γτ √ AB = τ√ AB 

is exactly r −1 B (pm) = r B (m). Since k ≥ 1 in the summation, equation (63) 

becomes 

(64) 

〈jm 

lift,B ,g ψ 〉 reg + 24σ 1 (m)〈ϑ B , g ψ 〉 = − lim 

= − lim 

ɛ→0 

lim 

s→1 

∑ 

A 

s→1 

∑ 

A 

ψ(A) 

∑ 

( 

g s τ 

√ 

AB −1, γτ √ AB 

γ∈M 2 (Z)/{±1} 

det(γ)=m, 

γτ √ AB≠τ√ AB −1 

ψ(A) ( G m s (τ √ AB −1 + ɛ, τ √ AB ) − r B(m)g s (τ √ AB −1 + ɛ, τ √ AB −1 ) ) 

where g s is defined in (57). From this expression, it is clear that the choice of these CM 

points do not matter. 

) 

Since g 1 (τ + ɛ, τ) = − log 

(1 + 4Im(τ)2 and ψ is non-trivial, we see that 

ɛ 2 

lim lim 

ɛ→0 

s→1 

∑ 

A 

ψ(A)g s (τ √ AB −1 + ɛ, τ √ AB −1 ) = − ∑ A 

Also by equation (59), we have 

∑ 

lim ψ(A)G m s (τ √ 

s→1 

AB 

+ ɛ,τ √ −1 AB ) = ∑ 

A 

A 

4πσ 1 (m) ∑ A 

ψ(A) log(y 2 √ 

AB −1). 

ψ(A) log |Ψ m (τ √ AB −1 + ɛ, τ √ AB )|2 − 

ψ(A) lim 

s→1 

(E(τ √ AB −1 + ɛ, s) + E(τ √ AB , s)). 

)


By (49) and the substitution A = A ′2 , we have 

lim 

s→1 4π ∑ A 

ψ(A)(E(τ √ AB −1 , s)+E(τ √ AB , s)) = −24(ψ(B)+ψ(B)) ∑ A ′ ψ 2 (A ′ ) log | √ y A ′η 2 (τ A ′)|. 

From (52) and 

H(p)ϑ B (z) = ∑ ψ ′ 

ψ ′ (B)g ψ ′(z), 

it is easy to see that 

〈ϑ B , g ψ 〉 = −(ψ(B) + ψ(B)) ∑ A ′ ψ 2 (A ′ ) log | √ y A ′η 2 (τ A ′)|. 

After substituting these into equation (64) and changing A, B to A ′2 , B ′2 respectively, we 

obtain equation (53). 

□ 

For any m ≥ 1, r ≥ 0 and τ, τ ′ ∈ H, define the level one modular function Ψ ∗ m(z; τ, τ ′ , r) 

by 

Ψ ∗ m(z; τ, τ ′ Ψ m (τ, z) 

, r) := 

(j(τ ′ ) − j(z)) r 

In terms of Ψ ∗ m(z; τ, τ ′ , r), we can re-write the right hand side of (53) as 

−2 

∑ 

ψ 2 (A ′ ) (log |Ψ ∗ m(τ A ′ B ′; τ A ′ B ′−1, τ A ′ B ′, r B(m))| + r B (m) log |y A ′ B ′j′ (τ A ′ B ′)|) 

A ′ ∈Cl(F ) 

The quantity Ψ ∗ −1 

m(τ A ′ B ′; τA ′ B 

, τ ′ A ′ B ′, r B(m)) is well-defined and non-zero, since the order of 

Ψ m (τ A ′ B ′−1, z) at z = τ A ′ B ′ is exactly r B(m) from the proof above. By equation (47) and the 

fact that 

(j ′ (z)) 6 = j(z) 4 (j(z) − 1728) 3 ∆(z), 

we have 

σ C 

⎛ 

⎝ 

∏ 

C ′ ∈Cl(F ) 

y 6 A ′ B ′j′ (τ A ′ B ′)6 

y 6 C ′ ∆(τ C ′) 

⎞ 

⎛ 

⎠ = ⎝ 

∏ 

C ′ ∈Cl(F ) 

y 6 C −1 A ′ B ′ j ′ (τ C −1 A ′ B ′)6 

y 6 C ′ ∆(τ C ′) 

⎞ 

⎠ ∈ H. 

Similarly, the modular function Ψ ∗ m(z; τ A ′ B ′−1, τ A ′ B ′, r B(m)), which is defined over H, is sent 

to Ψ ∗ m(z; τ C −1 A ′ B ′−1, τ C −1 A ′ B ′, r B(m)) under σ C . So if we let 

⎛ 

u m,B (A ′ ) := Ψ ∗ m(τ A ′ B ′; τ A ′ B ′−1, τ A ′ B ′, r B(m)) 6H(p) · ⎝ 

⎞ 

∏ 

r B (m) 

yA 6 ′ B ′j′ (τ A ′ B ′)6 ⎠ 

yC 6 ∆(τ ′ C ′) 

, 

then we can write 

with u m,B (A ′ ) ∈ H satisfying 

〈j lift,B 

m , g ψ 〉 reg = − 1 

3H(p) 

∑ 

A ′ ∈Cl(F ) 

C ′ ∈Cl(F ) 

(65) σ C (u m,B (A ′ )) = u m,B (A ′ C −1 ). 

ψ 2 (A ′ ) log |u m,B (A ′ )| 

When m = 0, we have j lift,B 

0 (z) = ϑ B (z) and 

∑ 

〈ϑ B , g ψ 〉 = − 1 ψ 2 (A ′ ) log |u 

12H(p) 

A ′ B ′u A ′ B ′−1|, 

A ′


where u C ′ is defined in (46) for any C ′ ∈ Cl(F ). So we can let u m,B (A ′ ) = u A ′ B ′u A ′ B ′−1 when 

m = 0 and equation (65) still holds by (47). Since every modular function of weight one is 

a linear combination of j m (z), m ≥ 0, we can deduce the following corollary concerning the 

algebraic property of r + ψ 

(n) from (60) and Proposition 3.4. 

Corollary 5.2. Let ˜g ψ (z) = ∑ n∈Z r+ ψ (n)qn ∈ M − 1 (p) be a mock-modular form satisfying 

condition (i) in Proposition 4.2. For any modular function f(z) = ∑ n∈Z c(f, n)qn of level 

one with rational Fourier coefficients, there exists β f,B ∈ Q and u f,B (A) ∈ H × independent 

of the character ψ such that 

(66) 

〈f lift,B , g ψ 〉 reg = ∑ c(f, n) ∑ δ(k)r B (pn − k)r + ψ (k) 

n∈Z k∈Z 

∑ 

= −β f,B ψ 2 (A) log |u f,B (A)| 

A∈Cl(F ) 

and σ C (u f,B (A)) = u f,B (AC −1 ) for any C ∈ Cl(F ). If f has integral Fourier coefficients, then 

12H(p)β f,B ∈ Z. 

6. Rankin’s method and values of Green’s Function II 

In this section, we will work out the generalization of Corollary 5.2 to modular functions of 

prime level N. From Corollary 5.2, we see that linear combinations of r + ψ 

(n) have the desired 

algebraic property. To prove Theorem 1.1, we need different linear combinations arising from 

regularized inner product between g ψ and other f ∈ M !,+ 

1 (p). By considering analogues of 

j lift,B for modular functions of level N, we can obtain these linear combinations, which will be 

sufficient to solve each r + ψ 

(n) (see §7). Other than being a step toward the proof of Theorem 

1.1, this section is interesting on its own as the analogue of the Gross-Zagier formula with 

trivial Heegner point divisor in the Jacobian. Also, the interpretation of the regularized inner 

product in terms of Borcherds lifts continues to hold for modular functions of level N. 

Let N be a prime number different from p. For a modular function f(z) of level N and 

B ∈ Cl(F ), one can define f lift,N,B (z) ∈ M 1(p) ! by 

(67) f lift,N,B (z) := Tr Np 

p ((f|W N )(pz)ϑ B (Nz)), 

−1 

where W N = ( 

N 

) is the Fricke involution and ϑ B (z) is the weight one theta series corresponding 

to the class B ∈ Cl(F ). Notice that the form f lift,N,B (z) is the same as f lift,B (z) from 

§5 when N = 1. The main result of this section is the following generalization of Corollary 

5.2. 

Proposition 6.1. Let f(z) is a modular function of prime level N ≠ p, with rational Fourier 

coefficients. There exist β f,B ∈ Q and u f,B (A) ∈ H independent of the character ψ such that 

∑ 

(68) 〈f lift,N,B , g ψ 〉 reg = −β f,B ψ 2 (A) log |u f,B (A)|. 

A∈Cl(F ) 

For any class C ∈ Cl(F ), the algebraic integer u f,B (A) satisfies 

σ C (u f,B (A)) = u f,B (AC −1 ) 

where σ C ∈ Gal(H/F ) is associated to C ∈ Cl(F ) via Artin’s isomorphism. Furthermore, if 

f has integral principal part Fourier coefficients and satisfies equation (69) below, then the 

denominator of β f,B only depends on N and p.


The proof of Proposition 6.1 follows the same steps as in the proof of Theorem 5.1 and 

Corollary 5.2. First, we will express 〈f lift,N,B , g ψ 〉 reg in terms of other regularized inner products 

and the limit of an infinite sum. Then, via Green’s function, we will relate the limit of 

the infinite sum to the height pairings between Heegner divisors on J 0 (N), the Jacobian of 

X 0 (N). Since the Heegner divisor is trivial on J 0 (N), the height pairing is the same as the 

value of some modular function evaluated at a Heegner point. From there, we can deduce 

equation (68). Before presenting the proof, we need some notations and facts about level N 

modular forms. For this part, we only require N to be a prime number. 

Let M 2(N), ! resp. M 0(N), ! be the space of weakly holomorphic weight 2 modular forms, 

resp. modular functions, of level N, and trivial nebentypus. Denote the subspaces of M 2(N) 

! 

of cusp forms and holomorphic modular forms by M 2 (N), S 2 (N) respectively. Let M !,new 

0 (N) 

be the subspace of M 0(N) ! containing weakly holomorphic modular forms f(z) satisfying 

(69) (f|W N )(z) = −N(f|U N )(z). 

The following lemma shows that M ! 0(N) can be be decomposed into the direct sum of 

M !,new 

0 (N) and level one modular functions. 

Lemma 6.2. Let f(z) be a modular function of prime level N. Then it can be written 

uniquely as 

f(z) = f 1 (Nz) + f 2 (z), 

such that f 1 (z) is a modular function of level 1 and f 2 (z) ∈ M !,new 

0 (N). 

Proof. Let a, b ∈ Z. With some matrix calculations, we have 

( ( )) 

f| UN W N + a N (z) = 

a−1 

f(z) + N TrN 1 (f)(Nz) 

( ( ) ( )) 

f| UN W N + a UN W 

N 

N + b N (z) = 

(a−1)(b−1) 

f(z) + a+b+N−1 Tr N N 2 

N 1 (f)(Nz) 

Set f j (z) as follows and we are done 

f 1 (z) = N 

N+1 TrN 1 (f)(z), 

f 2 (z) = − N N+1 NW N ) (z) − f(z)) . 

Alternatively, one can characterize the space M ! 0(N) via the space S 2 (N). 

Definition 6.3. A set of numbers {λ m : m ≥ 1} is called a relation for S 2 (N) if 

(i) λ m ∈ C is non-zero for all but finitely many m. 

(ii) For any cusp form h(z) = ∑ m≥1 c(h, m)qm ∈ S 2 (N), the numbers {λ m } satisfy 

∑ 

(70) 

δ N (m)λ m c(h, m) = 0. 

m≥1 

where δ N (m) = N + 1 if N|m and 1 otherwise. 

A relation {λ m } m≥1 is called integral if λ m ∈ Z for all m ≥ 1. Denote Λ N the set of all 

relations for S 2 (N). 

For any f(z) ∈ M !,new 

0 (N), one knows that the principal part coefficients, {c(f, −m) : 

m ≥ 1}, is a relation for S 2 (N) from works by Petersson. By considering the Fourier expansion 

of vector-valued Poincaré series, (See [7], [27], [32]), one can also show that the converse 

is true. So given λ = {λ m } ∈ Λ N , the expression 

∑ 

λ m q −m 

m≥1 

□


is necessarily the principal part of the Fourier expansion at infinity of a unique function 

f λ (z) ∈ M !,new 

0 (N). Alternatively, this statement follows essentially from an application of 

Serre duality [5, §3]. Suppose λ ∈ Λ N is integral. Then the n th Fourier coefficient of f λ (z) is 

integral when n ≠ 0 since f λ is a rational function of j(z) and j(Nz). 

Let g N be the genus of Γ 0 (N). From [1], we know that for all h(z) ∈ S 2 (N), ord ∞ (h(z)) ≤ 

dim(S 2 (N)) = g N ≤ (N + 1)/12. Using matrix computations similar to that in the proof of 

Lemma 6.2, one can show that 

(71) h| 2 U N = −h| 2 W N 

for all h(z) ∈ S 2 (N), implying that U N ◦ U N is the identity operator on S 2 (N). Combining 

these facts together gives us the following lemma about M !,new 

0 (N). 

Lemma 6.4. The space M !,new 

0 (N) is spanned by modular functions 

∞∑ 

f m (z) = q −m + c m (n)q n , 

n=−g N 

with m ≥ g N + 1. For these m, the coefficient c m (n) are integers when n ≠ 0. Furthermore, 

for all m ≥ g N + 1, 

f N 2 m(z) − N+1 

δ N (m) f m(z) = q −N 2m − N+1 

δ N (m) q−m + O(1). 

There is a similar duality statement dictating the existence of forms in M ! 2(N). 

Lemma 6.5. For every n ≥ 0, there exists P n,N (z) = ∑ m∈Z c(P n,N, m)q m ∈ M ! 2(N) such 

that 

(i) At the cusp infinity, P n,N (z) = q −n + O(q). 

(ii) (P n,N |W N )(z) = −(P n,N |U N )(z). 

Furthermore, if f(z) = ∑ m∈Z c(f, m)qn ∈ M !,new 

0 (N), then 

∑ 

δ N (m)c(f, m)c(P n,N , −m) = 0. 

m∈Z 

Now, we will recall some facts about Heegner points on the modular curve from [22] and 

their height pairings on the Jacobian. Let X 0 (N) be the natural compactification of Y 0 (N), 

the open modular curve over Q classifying pairs of elliptic curves (E, E ′ ) with an order N 

cyclic isogeny φ : E → E ′ . The complex points of X 0 (N) has a structure of a compact 

Riemann surface and can be identified with H ∪ P 1 (Q) modulo the action of Γ 0 (N). Heegner 

points on X 0 (N) corresponds to pair of elliptic curves (E, E ′ ) having complex multiplication 

by the same ring O in some imaginary quadratic field F . This occurs only when there exists 

an O-ideal n dividing N such that O/n is cyclic of order N. In this case, E(C) is isomorphic 

to C/a with a an invertible O-submodule in F . Since this a can be chosen independent of 

homothety by elements in F × , we only need to consider [a], the class of a in Pic(O). So 

a Heegner point on X 0 (N) can be expressed as the triplet (O, n, [a]). To find the image of 

such a Heegner point in H, choose an oriented basis 〈ω 1 , ω 2 〉 of a such that an −1 has basis 

〈ω 1 , ω 2 /N〉. Then τ [a],n := ω 1 /ω 2 ∈ Γ 0 (N)\H is the image of this Heegner point. 

For our purpose, F = Q( √ −p) and O = O F . Then we require χ p (N) = 1 for Heegner 

points to exist. In that case, N splits into nn in F with 

n = ZN + Z bn+√ −p 

2 

.


Let A ∈ Pic(O F ) = Cl(F ). A point τ ∈ H corresponding to the Heegner point (O, n, A) 

must satisfy an equation 

Aτ 2 + Bτ + C = 0, 

with B 2 − 4AC = −p, N|A, B ≡ b n (mod 2N) and gcd(A/N, B, NC) = 1. We will denote 

this image by τ(A, n). Notice τ(A, n) is well-defined up to the action of Γ 0 (N). 

Let J 0 (N) be the Jacobian of X 0 (N). Its complex points is a compact Riemann surface 

and can be viewed as the set of divisors of degree zero modulo the set of principal divisors 

on X 0 (N). Let 〈, 〉 C be the height symbol on X 0 (N)(C). It is a unique real-valued function 

defined on the set of divisors of degree zero and satisfies 

〈∑ 

ni (x i ), b〉 

= ∑ n i log |f(x i )| 2 

C 

i 

if b = (f) is a principal divisor. 

For n ≥ 0 and a class B ∈ Cl(F ), denote the inner product 

(72) ρ N,B,ψ (n) := 〈j lift,B[n] 

n 

+ j lift,B[n]−1 

n , g ψ 〉 reg . 

In particular for j 0 (z) = 1, the inner product ρ N,B,ψ (0) becomes 〈T N (ϑ B ), g ψ 〉, where T N is 

the N th Hecke operator on M ! 1(p). Also, notice that 

(73) T N (ϑ B )(z) = ϑ B (Nz) + (ϑ B |U N )(z) = ϑ B[n] (z) + ϑ B[n] −1(z). 

By Corollary 5.2, we know that ρ N,B,ψ (n) can be written in the form of the right hand 

side of (68) and satisfies the conditions in Proposition 6.1. The following lemma relates 

〈f lift,N,B , g ψ 〉 reg to ρ N,B,ψ (n) and an infinite sum analogous to the right hand side of (62). 

Lemma 6.6. Suppose f(z) ∈ M !,new 

0 (N). In the notations above, we have 

(74) 〈f lift,N,B , g ψ 〉 reg = ∑ m ′ ≥1 

c(f, −Nm ′ )ρ N,B,ψ (m ′ ) + C f ρ N,B,ψ (0) − Σ f,N,B,ψ , 

where 

∑ 

Σ f,N,B,ψ := − lim 

s→1 

C f = 

m≥1 

N∤m 

(N + 1) lim 

( 

c(f, −m) ∑ k≥1 

s→1 

∑ 

m ′ ≥1 

c(f, −Nm ′ ) ∑ k ′ ≥1 

Nc(f, 0) + 24 ∑ m ′ ≥1 

c(f, −Nm ′ )σ 1 (m ′ ) 

) 

δ(k)r B (pm + Nk)r ψ (k)2Q s−1 

(1 + 2kN + 

pm 

) 

δ(k ′ )r B (k ′ )r ψ (Nk ′ − pm ′ )2Q s−1 

(−1 + 2k′ N 

, 

pm ′ 

) 

. 

Proof. Let ˜g ψ (z) ∈ M − 1 (p) be a mock-modular form as in Proposition 4.2 with Fourier expansion 

˜g ψ (z) = ∑ n∈Z r+ ψ (n)qn and ĝ ψ (z) the associated harmonic Maass form. By Theorem 

5.1, we have the following identities 

) 

ρ N,B,ψ (n) = −24σ 1 (n)ρ N,B,ψ (0) + lim δ(k)c(T N (ϑ B ), pm + k)r ψ (k)2Q s−1 

(1 + 2k , 

pm 

ρ N,B,ψ (n) = ∑ k∈Z 

s→1 

∑ 

k≥1 

r + ψ (k) ( ( 

r pn−k 

) 

B N + rB (pNn − Nk) ) δ(k).

Similar to the proof of Theorem 5.1, we define 


Φ N,B (ĝ, z) := Tr Np 

N (−i(ĝ|W p)(Nz)ϑ B (z)) 

= (ĝ|U p )(Nz)ϑ B (z) + (ĝ(z)ϑ B (Nz))|U p . 

When N = 1, this is the same as Φ B (ĝ ψ , z) defined in the proof of Theorem 5.1. It transforms 

like a modular form of level N, weight two and has the following Fourier expansion at infinity 

Φ N,B (ĝ ψ , z) = ∑ m∈Z(b N,B (ĝ ψ , m, y) + a N,B (ĝ ψ , m))q m , 

b N,B (ĝ ψ , m, y) = − ∑ k∈Z 

a N,B (ĝ ψ , m) = ∑ k∈Z 

Suppose f(z) has the Fourier expansion 

at infinity. Define the sum Σ ′ f,N,B,ψ by 

δ(k)r ψ (k)β 1 

( 

k, Ny 

p 

r + ψ (k) r B(pm − Nk)δ(k). 

f(z) = ∑ m∈Z 

c(f, m)q m 

(75) Σ ′ f,N,B,ψ := ∑ m∈Z 

c(f, −m)δ N (m)a N,B (ĝ ψ , m). 

) 

r B (pm + Nk), 

Since f lift,N,B (z) ∈ M !,+ 

1 (p), we can apply Proposition 3.4 to obtain 

⎛ ∑ 

〈f lift,N,B , g ψ 〉 reg = ∑ r + ψ (k)r ⎞ 

B(pm − Nk)δ(k)− 

c(f, −m) ⎜ 

k∈Z 

⎝ 

m∈Z N ∑ r + ψ (k)r ( pm−Nk 

) 

⎟ 

⎠ 

B N δ(k) 2 

k∈Z 

= −N ∑ c(f, −Nm ′ ) ∑ ( ( ) 

) 

r + ψ (k) pm 

r ′ −k 

B + r 

N B (pNm ′ − Nk) δ(k) 

m ′ ∈Z 

k∈Z 

+ ∑ m∈Z 

δ N (m)c(f, −m)a N,B (ĝ ψ , m) 

(76) 

= −N ∑ m ′ ≥0 

c(f, −Nm ′ )ρ N,B,ψ (m ′ ) + Σ ′ f,N,B,ψ 

The sum over m ′ ∈ Z changes to be over only m ′ ≥ 0, since r + ψ 

(k) = 0 for all k ≤ −p by the 

choice of ˜g ψ (z). 

For n ≥ 0, let P n,N (z) = q −n + O(q) ∈ M 2(N) ! be the weakly holomorphic modular form 

as in Lemma 6.5. When n = 0, we can take P 0,N (z) ∈ M 2 (N) to be 

P 0,N (z) = 

Consider Φ ∗ N,B (ĝ ψ, z), defined by 

(77) 

Φ ∗ N,B(ĝ ψ , z) := Φ N,B (ĝ ψ , z) − ∑ n≥0 

(N−1)Ẽ2(Nz) N − (N−1)Ẽ2(z). 

1 

a N,B (ĝ ψ , −n)P n,N (z) + Nρ N,B,ψ(0) 

N 2 − 1 

(Ẽ2(Nz) − Ẽ2(z)). 

Now it is easy to check that Φ ∗ N,B (ĝ ψ, z) is O(1/y) at the cusp infinity. At the cusp 0, some 

calculations show that 

Φ N,B (ĝ ψ , z)|W N = −Φ N,B (ĝ ψ , z)|U N + Φ 1,B[n] (ĝ ψ , z) + Φ 1,B[n] −1(ĝ ψ , z)


From the proof of Theorem 5.1, we know that Φ 1,B[n] (ĝ ψ , z) + Φ 1,B[n] −1(ĝ ψ , z) has a constant 

term ρ N,B,ψ (0) and no pole. So Φ ∗ N,B (ĝ ψ, z) is also O(1/y) at the cusp 0. 

For m ≥ 1, let P m,N (z, s) be the Poincaré series of level one, weight two defined by 

∑ 

P m,N (z, s) := (y s e 2πimz )| 2 γ. 

It is characterized by the property that 

γ∈Γ ∞\Γ 0 (N) 

lim 〈h(z), P m,N(z, s)〉 = mc(h, m) 

s→0 

for any h(z) = ∑ m≥1 c(h, m)qm ∈ S 2 (N). Since {c(f, −m) : m ≥ 1} ∈ Λ N , we know that 

〈 

(78) lim h(z), ∑ 〉 

δ N (m) 

c(f, −m)P 

s→0 m m,N(z, s) = 0, 

m≥1 

δ N (m) 

m c(f, −m)P m,N(z, s) ∈ S 2 (N) is 0. 

for any h(z) ∈ S 2 (N). So lim s→0 

∑m≥1 

Now we can consider the inner product between Φ ∗ N,B (ĝ ψ, z) and this linear combination 

of Poincaré series and obtain 

∑ 

δ 

4π lim N (m) 

c(f, 

s→0 m −m)〈Φ∗ N,B(ĝ ψ , z), P m,N (z, s)〉 = 0. 

m≥1 

Since Φ ∗ N,B (ĝ ψ, z) is O(1/y) at both cusps, we can apply Rankin-Selberg unfolding as in the 

proof of Theorem 5.1. The limit on the left hand side above then breaks up into two parts. 

The first part is 

⎛ 

a N,B (ĝ ψ , m) − ∑ a N,B (ĝ ψ , −n)c(P n,N , m) 

∑ 

⎜ 

n≥0 

⎟ 

(79) 

m≥1 

δ N (m)c(f, −m) ⎜ 

⎝ 

− 24Nρ N,B,ψ(0) 

N 2 − 1 

( 

σ1 

( m 

N 

⎞ 

) 

− σ1 (m) ) ⎟ 

⎠ 

Since (f|W N )(z) = −N(f|U N )(z) and c(P n,N , 0) = 0 for n ≥ 1, Lemma 6.5 tells us that for 

n ≥ 0 

∑ 

δ N (m)c(f, −m)c(P n,N , m) = −δ N (n)c(f, n). 

m≥1 

So expression (79) becomes 

Σ ′ f,N,B,ψ − 24Nρ N,B,ψ(0) 

N 2 − 1 

∑ 

δ N (m)c(f, −m) ( ( 

σ m 

) 

1 N − σ1 (m) ) . 

m≥1 

The second part involves the limit of an infinite sum, which can be evaluated as 

Σ f,N,B,ψ − (N + 1) ∑ m ′ ≥1 

c(f, −Nm ′ )(ρ N,B,ψ (m ′ ) + 24σ 1 (m ′ )ρ N,B,ψ (0)). 

Adding the last two expressions together, which yields 0, and substituting into (76), we 

obtain equation (74). The constant C f appears since 

∑ ( ( 

(80) c(f, 0) = 24 

m 

) 

N 2 Nσ1 

−1 

N − σ1 (m) ) δ N (m)c(f, −m) 

m≥1 

from applying Lemma 6.5 to P 0,N (z) ∈ M 2(N) ! and f(z) ∈ M !,new 

0 (N). Notice if c(f, −m) ∈ Z 

for all m ≥ 1, then the denominator of c(f, 0) is independent of f. 

□


Proof of Proposition 6.1. If f(z) = f 1 (Nz) with f 1 (z) a modular function of level one, then 

(f|W N )(z) = (f|U N )(z) and 

f lift,N,B (z) = f lift,B 

1 (z) = f 1 (pz)ϑ B (z). 

This case of the proposition follows from Corollary 5.2. By Lemma 6.2, it is enough to prove 

the proposition for f(z) ∈ M !,new 

0 (N). In that case, some calculations show that 

(81) f lift,N,B (z) = −N(f|U N )(pz)ϑ B (Nz) + (f(pz)ϑ B (z)) |U N . 

Let χ p (N) = ɛ. It is easy to see that f lift,N,B (z) ∈ M !,ɛ 

1 (p). Since g ψ (z) ∈ S + 1 (p), Proposition 

3.4 tells us that 〈f lift,N,B , g ψ 〉 reg = 0 when ɛ = −1. So we can suppose χ p (N) = ɛ = 1. 

By Lemma 6.4, it is easy to see that M !,new 

0 (N) is generated by the set S 1 ∪ S 2 , where 

(82) 

S 1 := {f m (z) : m ≥ 1 + g N , N 2 ∤ m} 

S 2 := {f N 2 m(z) − N+1 

δ N (m) f m(z) : m ≥ 1 + g N }. 

Let f(z) = f N 2 m(z) − N+1 

δ N (m) f m(z) ∈ S 2 . We can uniquely write the integer m as N r m ′ with 

r ≥ 0, m ′ ≥ −1 and gcd(N, m ′ ) = 1. Then at the cusp infinity, we have 

f(z) = q −N r+2 m ′ − N+1 

δ N (N r m ′ ) q−N r m ′ + O(1). 

From equation (81), it follows that the −n th Fourier coefficient of f lift,N,B (z) can be written 

as 

( 

( )) 

( ) 

pN r m ′ 

c(f lift,N,B pN 

, −n) = −N r r+1 m ′ −n 

B − N+1 r N 

−n 

N 

δ N (N r m ′ ) B N 

( 

) 

+ r B (−Nn + pN r+2 m ′ ) − N+1 r δ N (N r m ′ ) B(−Nn + pN r m ′ ) . 

When r = 0, we can rewrite the equation above as 

c(f lift,N,B , −n) = −(N + 1) ( ( ) 

r B pm ′ − n N + rB (pm ′ − Nn) ) 

( ) 

) 

pNm 

+ 

(r ′ −n 

B + r 

N 

B (N(pNm ′ − n)) 

( 

) 

= −(N + 1)c T N (j lift,B 

m 

), −n + c (j ′ 

Nm ′(pz)T N (ϑ B )(z), −n) . 

Notice that j pNm ′(pz)T N (ϑ B )(z) = j lift,B[n] 

Nm 

(z) + j lift,B[n]−1 

′ 

Nm 

(z). Since this is true for all n ≥ 1 

′ 

and the principal part uniquely determines a form in M !,+ 

f lift,N,B (z) = −(N + 1)T N 

( 

j lift,B 

m ′ 

) 

(z) 

+ j lift,B[n] 

Nm ′ 

1 (p) up to forms in M + 1 (p), we have 

(z) + j lift,B[n]−1 (z) + g f (z) 

for some g f (z) ∈ M + 1 (p) with integral linear combination of c(f, n)’s as Fourier coefficients. 

Since T N is self-adjoint with respect to the regularized inner product and 

T N (g ψ )(z) = (ψ([n]) + ψ([n]))g ψ (z) 

for N prime with χ p (N) = 1, we can rewrite 〈f lift,N,B , g ψ 〉 reg as 

〈f lift,N,B , g ψ 〉 reg = −(N + 1)(ψ([n]) + ψ([n]))〈j lift,B 

m ′ , g ψ 〉 reg + ρ N,B,ψ (Nm ′ ) + 〈g f , g ψ 〉. 

Nm ′


When r ≥ 1, the situation is similar. In this case, we have 

( ) 

f lift,N,B (z) = − (N + 1)T N j lift,B 

N r m 

(z) + (Nj ′ N r−1 m ′(pz) + j N r+1 m ′(pz))T N(ϑ B ) 

+ g f (z), 

〈f lift,N,B , g ψ 〉 reg = − (N + 1)(ψ([n]) + ψ([n]))〈j lift,B 

m ′ , g ψ 〉 reg + Nρ N,B,ψ (N r−1 m ′ ) 

+ ρ N,B,ψ (N r+1 m ′ ) + 〈g f , g ψ 〉, 

for some g f (z) ∈ M 1 + (p) with integral linear combination of c(f, n)’s as Fourier coefficients. 

By Proposition 4.1 and Corollary 5.2, the proposition holds for f(z) ∈ S 2 . 

Now, we will prove the proposition when f(z) is in the integral span of S 1 and show that 

the denominator of the constant β f,B in (68) is independent of f. This will finish the proof 

of Proposition 6.1. By Lemma 6.2 and Lemma 6.6, it is enough to show that Σ f,N,B,ψ can 

also be written in the form of the right hand side of equation (68). To accomplish that, 

we will relate Σ f,N,B,ψ to the automorphic Green’s function using the counting arguments in 

Proposition (3.11) from [24]. In their notation, let R N be a subset of M 2 (Z) containing all 

matrices whose lower left entry is divisible by N. For z, z ′ ∈ H and k ≥ 0, let ρ m N (z, z′ , k) be 

the number of γ = ( a c d b ) ∈ R N/{±1} such that det(γ) = m and 

cosh d(z, γz ′ ) = 1 + 2Nk 

pm . 

Define the automorphic Green’s function of level N to be 

∑ 

G N,s (z, z ′ ) := 

g s (z, γz ′ ) + 

γ∈Γ 0 (N)/{±1} 

z≠γz ′ 

where 

∑ 

γ∈Γ 0 (N)/{±1} 

z=γz ′ 

g s (z) := lim 

w→z 

(g s (z, w) − log |2πiη(z) 4 (z − w)| 2 

g s (z), 

= − log |2π(z − z)η(z) 4 | 2 + 2 Γ′ Γ′ 

(s) − 2 (1). 

Γ Γ 

When gcd(N, m) = 1, the m th Hecke operator T m acts on z ′ in G N,s (z, z ′ ) and defines 

G m N,s (z, z′ ). It can be written as 

∑ 

G m N,s(z, z ′ ) := 

g s (z) 

γ∈R N /{±1} 

det γ=m 

= −2 ∑ k>0 

γ∈R N /{±1} 

det γ=m 

z=γz ′ 

z≠γz ′ g s (z, γz ′ ) + ∑ 

ρ m N(z, z ′ , k)Q s−1 

(1 + 2Nk 

pm 

) 

+ ρ m N(z, z ′ , 0)g s (z). 

Notice that ρ m N (z, z′ , k) could be non-zero when k is not an integer. When z, z ′ ∈ H are 

Heegner points of discriminant −p, then ρ m N (z, z′ , k) is necessarily supported on integral k’s. 

For j = 1, 2, let A j ∈ Cl(F ) and τ j ∈ Γ 0 (N)\H be the image of the Heegner points 

x j = (O F , A j , [n]). Let a j be integral ideals in the class A j such that n|a j , N(a j ) = A j . Then 

by Proposition (3.11) in [24], the counting function ρ m N (τ 2, τ 1 , k) satisfies 

ρ m N(τ 2 , τ 1 , k) =#{(α, β) ∈ (a −1 

1 a −1 

2 × a −1 

1 a −1 

2 n)/{±1} | N F/Q (α) = Nk+pm 

A 1 A 2 

, 

N F/Q (β) = Nk 

A 1 A 2 

, A 1 A 2 (α − β) ≡ 0 (mod d)}, 

=r A1 A −1(Nk 

+ pm)r A 

2 1 A 2 [n] −1(k)δ(k)


for k ≥ 0, where d = ( √ −p) is the different of F . The first equality is established by the 

bijection 

γ = ( a c d b ) ∈ R N/{±1} ↦→ α = cτ 1τ 2 + dτ 2 − aτ 1 − b, 

β = cτ 1 τ 2 + dτ 2 − aτ 1 − b. 

The Fricke involution W N sends x j to the Heegner point (O F , A j [n] −1 , [n] −1 ). Set A ′ 1 = 

A 1 [n] −1 and a ′ 1 ∈ A ′ 1 a class such that n|a ′ 1, then the same counting argument establishes 

ρ m′ 

N (τ 2 , τ ′ 1, k ′ ) =#{(α, β) ∈ ((a ′ 1) −1 ā −1 

2 n × (a ′ 1) −1 a −1 

2 )/{±1} | N F/Q (α) = Nk′ 

A 1 A 2 

, 

N F/Q (β) = Nk′ −pm ′ 

A 1 A 2 

, A 1 A 2 (α − β) ≡ 0 (mod d)}, 

=r A ′ 

1 A 2 

(Nk ′ − pm)r A ′ 

1 A −1 

2 [n]−1 (k ′ )δ(k ′ ) 

for k ′ ≥ pm ′ /N. Apply these counting arguments to G m N,s (τ 2, τ 1 ) and G m′ 

N,s (τ 2, τ 1), ′ we obtain 

G m N,s(τ 2 , τ 1 ) = − 2 ∑ ( ) 

r A1 A −1(Nk 

+ pm)r A 

2 1 A 2 [n] −1(k)δ(k)Q s−1 1 + 2Nk + 

pm 

(83) 

k≥1 

r A1 A −1(m)g 

s(τ 2 ), 

2 

∑ 

( ) 

G m′ 

N,s(τ 2 , τ 1) ′ = − 2 r A ′ 

1 A 2 

(Nk ′ − pm ′ )r A ′ 

1 A −1 (k ′ )δ(k ′ )Q 

2 [n]−1 s−1 −1 + 2Nk′ + 

pm ′ 

(84) 

k ′ >pm ′ /N 

r A ′ 

1 A −1 

2 [n]−1 ( m′ 

N )g s(τ 2 ). 

Since N ∤ m ′ in the definition of G m′ 

N,s (z, z′ ), the term r A ′ 

1 A −1 ( m′ )g 

2 [n]−1 N s(τ 2 ) above vanishes. 

By Proposition (2.23) in [24, p.242], we can relate the Green’s function to the height pairing 

between (x 2 ) − (∞), (x 1 ) − (0) and W N ((x 1 ) − (0)) on the Jacobian as follows 

( ) 

G 

m 

N,s (τ 2 , τ 1 ) + 4πσ 1 (m)E N (W N τ 2 , s) 

〈(x 2 ) − (∞), T m ((x 1 ) − (0))〉 C = lim 

s→1 + 4πm s σ 1−2s (m)E N (τ 1 , s) + σ 1(m)κ N 

s−1 

−σ 1 (m)(λ N + 2κ N ), 

( ) 

G 

m 

N,s(τ ′ 

2 , τ 1) ′ + 4πσ 1 (m ′ )E N (τ 2 , s) 

〈(x 2 ) − (∞), T m ′W N ((x 1 ) − (0))〉 C = lim 

s→1 

+ 4π(m ′ ) s σ 1−2s (m ′ )E N (W N τ 1 , s) + σ 1(m ′ )κ N 

s−1 

−σ 1 (m ′ )(λ N + 2κ N ). 

Here E N (z, s) is the Eisenstein series defined in (48) and κ N , λ N are explicit constants dependent 

on N. 

For A ′ , B ∈ Cl(F ), substitute A 1 = A ′√ B[n], A 2 = A ′√ B −1 [n] and A ′ 1 = A ′√ B[n] −1 into 

equations (83) and (84). Denote τ 1 = τ A1 ,n, τ 2 = τ A2 ,n and τ 1 ′ = τ ′ A 1 [n] −1 ,¯n by τ 1(A ′ , B), τ 2 (A ′ , B) 

and τ 1(A ′ ′ , B) respectively. Summing together equations (83), (84) over m, m ′ and A ′ with a 

character ψ 2 gives us 

Σ f,N,B,ψ = ∑ m≥1 

N∤m 

c(f, −m) 

∑ 

A ′ ∈Cl(F ) 

− (N + 1) ∑ m ′ ≥1 

c(f, −Nm ′ ) 

ψ 2 (A ′ )(G m N,s(τ 2 (A ′ , B), τ 1 (A ′ , B)) − r B (m)g s (τ 2 (A ′ , B))) 

∑ 

A ′ ∈Cl(F ) 

ψ 2 (A ′ )G m N,s(τ 2 (A ′ , B), τ ′ 1(A ′ , B)) 

We can rewrite the sum above in terms of height pairings as follows 

(85) Σ f,N,B,ψ = ∑ 

ψ 2 (A ′ )〈(x 2 (A ′ , B)) − (∞), T f ((x 1 (A ′ , B)) − (0))〉 C + U f,B,ψ , 

A ′ ∈Cl(F )


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 ′ ).


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 ′,


Then φ j,B (pM j z) is a modular form of weight two, level pM for all 0 ≤ j ≤ (p − 1)/2. Define 

the following matrices with entries in CJqK by 

( ) 

) 

L := 

(ϑ B M j z + k + ϑ 

p B (pM j z) 

( ( )) 

X := D Mz + k p 

0≤k≤p−1 

R := (pφ j,B (pM j z)) 0≤j≤(p−1)/2 

. 

0≤j≤(p−1)/2 

0≤k≤p−1. 

The dimensions of L, X and R are (p + 1)/2 × p, p × 1 and (p + 1)/2 × 1 respectively. Then 

we can rewrite equation (87) in the following matrix expression 

(88) L · X = R. 

Notice that matrix L has rank (p + 1)/2. This can be seen by considering the product 

( 

) 

( ) 

∑ 

1 /p 

L · 

p e−2πikk′ = δ(k ′ ) r B (n)q M jn 

, 

0≤k,k ′ ≤p−1 

n≡k ′ mod p 

0≤j≤(p−1)/2 

0≤k ′ ≤p−1. 

which, after permuting columns, is composed of a (p + 1)/2 × (p − 1)/2 zero submatrix and 

a (p + 1)/2 × (p + 1)/2 submatrix of the form 

( 

) 

∑ 

L ′ := δ(n ν ) r B (n)q M jn 

. 

n≡n ν mod p 

0≤j,ν≤(p−1)/2 

Here n 0 = 0 and 0 < n 1 < n 2 < · · · < n (p−1)/2 

determinant of this submatrix is a nonzero power series in q with leading term 

⎛ 

⎞ 

(p−1)/2 

∏ 

⎝ r B (n B,ν ) ⎠ q ∑ 1≤j≤(p−1)/2 M jn B,j 

, 

ν=1 

where {n B,ν } 1≤ν≤(p−1)/2 is a set of positive integers such that 

(i) 0 < n B,1 < n B,2 < · · · < n B,(p−1)/2 , 

(ii) r B (n B,ν ) ≠ 0 for all 1 ≤ ν ≤ (p − 1)/2, 

(iii) r B (n) = 0 if n ≡ n B,ν ′ (mod p) for some ν ′ and n < n B,ν ′. 

So the submatrix L ′ has full rank in M (p+1)/2,(p+1)/2 (CJqK) and the null space of L is spanned 

by the column vectors 

( 

1 

p e−2πikk′ /p 

) 

0≤k≤p−1 

with k ′ running over the (p − 1)/2 non-residues 

modulo p. This also shows that L has rank (p + 1)/2 in M (p−1)/2,(p−1) (C) after substituting 

q = e 2πiz for all z in a dense subset U ∈ H. 

Let γ = ( ) 

a b 

pMc d ∈ Γ0 (pM), then applying | 2 γ to both sides of equation (87) gives us a new 

equation with the same right hand side. Let c denote the multiplicative inverse of c modulo 

p and we have the following standard transformation of ϑ B (z) 

{ 

( 

) 

) ( 

ϑ B 

(M j z + k | a b 

) χ 

p 1 pMc d = p (a + ck)ϑ B M j z + (a+ck)dk , if a + ck ≢ 0 mod p 

p 

i √ pχ p (−c)ϑ B (pM j z), 

if a + ck ≡ 0 mod p 

( 

ϑ B (pM j z) | a b 

) 

χ √ ) 

1 pMc d = p(−c)i p 

ϑ 

p B 

(M j z + cd . 

p 

Using these properties, equation (88) transforms into 

L · T · X ′ = R,

where 

T := (T kk ′) 0≤k,k ′ ≤p−1, 

T kk ′ := χp(−c)i √ p 

(δ cd(k) ′ − gp(k) ) + 

p 

δ ′ α(β) := 

{ 

1, if α = β 

0, otherwise 


p−1 

∑ 

{ p 

g p (k) := 1 (1 + χ 

2 

p (n))e 2πink/p = 

, if p | k 

2 

i √ pχ p(k) 

, if p ∤ k 

n=0 

2 

( ( ) ) 

X ′ := D Mz + k′ | 

p 1 γ 

0≤k ′ ≤p−1 

{ 

χp (a + ck ′ )(δ ′ (a+ck ′ )dk ′(k′ ) − gp(k) ), 

p 

if k ≢ −ac mod p 

iχ p(−c) 

2 √ g p p(k), if k ≡ −ac mod p 

So when q = e 2πiz for any z ∈ H, the column vector X − T · X ′ with entries in C is in) 

the null 

space of L. If z ∈ U, the null space is spanned by the column vectors 

with k ′ ∈ Z and χ p (k ′ ) = −1. 

Now define a 1 × p row vector V by 

( 

1 

V := (g p p(k) + 1) . 

2 

)0≤k≤p−1 

It is orthogonal to the null space of L when z ∈ U. So for all z ∈ U, 

Simple calculation shows that 

V · X = V · T · X ′ . 

V · T = χ p (d)V, 

V · X = χ p (d)D(Mz), 

V · X ′ = χ p (d)(D(Mz)| 1 γ). 

( 

1 

p e−2πikk′ /p 

0≤k≤p−1 

So we know that (D(Mz)| 1 γ) = χ p (d)D(Mz) for all z ∈ U. Since U ⊂ H is dense and both 

sides are holomorphic, we have (D(Mz)| 1 γ) = χ p (d)D(Mz) for all z ∈ H. This is true for 

any γ ∈ Γ 0 (pM), or equivalently 

for all ( ) 

a Mb 

pc d ∈ Γ0 (p), b ∈ Z. 

Let Γ 0 (p, M) := {γ = ( a b 

( 

(D| a Mb 

1 pc d 

) 

)(z) = χp (d)D(z) 

c d ) : p|a, M|b}. It is not too hard to see that Γ 0(p) is generated 

by Γ 0 (p, M) and T = ( 1 1 1 ) ∈ Γ 0 (p). Since D(z) is also invariant under the action of T , it 

has level p. By the shape of the Fourier expansion of D(z), it is in S1 − (p). 

□ 

Corollary 7.2. Let N 0 = 1 and N j be distinct primes congruent to 1 modulo p for 1 ≤ j ≤ 

(p − 1)/2. Then the space M !,+ 

1 (p) is spanned by the set 

{f lift,N j,B (z) : B ∈ Cl(F ), f(z) ∈ M !,+ 

0 (N j ), 0 ≤ j ≤ (p − 1)/2}. 

Proof. Denote the subspace of M !,+ 

1 (p) spanned by the set above by M !,+,lift 

1 (p). Suppose 

there exists f(z) ∈ M !,+ 

1 (p) not in M !,+,lift 

1 (p). Then by the perfect pairing between M !,+ 

1 (p) 

and S1 − (p), there exists 

D(z) = ∑ d(n)q n ∈ CJqK 

n≥1 

such that


(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 ) 

□


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


and we have 

r + ψ (n) = −β ∑ A 

ψ 2 (A) log |u(n, A)| = −β ′ ∑ A 

ψ 2 (A) log |u ′ (n, A)|. 

By the choice of β ′ and u ′ (n, A), we also have 

(90) 

∑ 

R p + (n) = r + A (n) 

A∈Cl(F ) 

∑ 

= −β ′ log |u ′ (n, A)| 

A∈Cl(F ) 

= − β′ 

2 log |N H/Q(u ′ (n, A))|. 

This is an analogue of equation (9) for mock-modular forms as mentioned in the introduction. 

□ 

8. Counting CM points 

In preparation for the proof of Theorem 1.2, we will count the number of CM points on a 

hyperbolic circle in terms of the representation numbers of positive definite binary quadratic 

forms. As before, such a counting argument is needed to construct a Green’s function at 

special points. This construction follows ideas of [23], but in the counting argument given 

there one also sums over all classes of a given discriminant. Even when one discriminant is 

prime, which we are assuming, their proof involves quite an intricate application of algebraic 

number theory. Surprisingly, the refined version we need for a fixed class has a completely 

elementary proof using the classical theory of composition of binary quadratic forms. It 

has the further advantage that it applies without extra effort when the other discriminant 

is not fundamental. Although we will not give details here, the argument adapts to give a 

corresponding refinement of the level N case in [24]. 

Good references for the theory of binary quadratic forms are the books by Buell [3] and 

Cox [14]. Let −d < 0 be a discriminant, not necessarily fundamental, and Q −d the set of 

positive definite integral binary quadratic forms 

Q(x, y) = ax 2 + bxy + cy 2 

with −d = b 2 − 4ac, where a, b, c ∈ Z. For any Q ∈ Q −d the associated CM point is defined 

to be 

τ Q = −b + √ −d 

∈ H, 

2a 

where H is the upper half plane. Clearly Q(τ Q , 1) = 0. The modular group Γ = PSL 2 (Z) acts 

on Q ∈ Q −d by a linear change of variables, which induces linear fractional transformation 

on τ Q . Let w Q be the number of stabilizers of Q. The number of equivalence classes of 

quadratic forms is the Hurwitz class number H(d). Those classes represented by primitive 

forms (those with gcd(a, b, c) = 1) comprise a finite abelian group under composition, the 

class group. When −d is fundamental, this class group is canonically isomorphic to the ideal 

class group of Q( √ −d) by sending [Q] to the class A ∈ Cl(Q( √ −d)) containing the fractional 

ideal Z + Zτ Q . For primitive Q ∈ Q −d , let Q 2 denote a representative of the square class of 

Q. 

Associated to Q is the counting function 

r Q (k) = 1 2 #{(x, y) ∈ Z2 | Q(x, y) = k},


which is finite and can be non-zero only for non-negative integers k. Clearly r Q (k) is a class 

invariant. When −d is fundamental and A ∈ Cl(Q( √ −d)) is the class associated to [Q], the 

counting function r Q (k) is the same as r A (k), the number of integral ideals in the class A 

having norm equals to k. 

The invariant hyperbolic distance d(τ, τ ′ ) between two points τ = x + iy and τ ′ = x ′ + iy ′ 

in H is defined by (54). The goal now is to count the CM points τ Q ′ of a given discriminant 

−d that are at a fixed hyperbolic distance from a fixed CM point τ Q of a (possibly) different 

negative discriminant −D. The number we will study is given by 

(91) ρ Q (k, d) = #{Q ′ ∈ Q −d | cosh d(τ Q ′, τ Q ) = k √ 

dD 

}. 

Since the hyperbolic distance is invariant under linear fractional transformation, the quantity 

ρ Q (k, d) depends only on the class of Q. In case −D is the negative of a prime we will give 

a formula for ρ Q in terms of the counting function r Q 2 for the squared class of the quadratic 

form Q. 

Proposition 8.1. Let Q be a positive definite binary quadratic form of discriminant −D 

where D = p ≡ 3 (mod 4) is a prime. Suppose that −d < 0 is a discriminant and that k ≥ 0. 

Then 

(92) ρ Q (k, d) = δ(k) r Q 2( k2 −pd 

4 

). 

Our proof of this result is an entirely elementary exercise in the classical theory of binary 

quadratic forms. It relies on a remarkable algebraic identity given in (97) below for a particular 

quadratic form in the square class, whose existence can be coaxed out of the work of 

Dirichlet [16]. 

Lemma 8.2. Every class of primitive positive definite binary quadratic forms of discriminant 

−D contains a representative of the form 

(93) Q(x, y) = Ax 2 + Bxy + ACy 2 

with −D = B 2 − 4A 2 C, A > 0 and gcd(A, B) = 1. Furthermore, the class of the square of Q 

is represented by the form 

(94) Q 2 (x, y) = A 2 x 2 + Bxy + Cy 2 . 

Proof. It is a well–known result of Gauss ([21, §228], see also [3, Prop 4.2 p.50] ) that a 

primitive form of discriminant −D properly represents a positive integer A prime to −D and 

is thus, upon an appropriate change of variables, equivalent to a form of the shape (A, b, c) 

with b 2 − 4Ac = −D. In particular we have that gcd(A, b) = 1. 

Next, by means of a simple translation transformation x ↦→ x + ty, y ↦→ y we will arrange 

that (A, b, c) ∼ (A, B, AC) as desired. This transformation leaves A alone and changes b to 

B = b + 2At. We now choose t to force 

Since c = b2 +D 

4A 

B 2 = (b + 2At) 2 ≡ −D (mod 4A 2 ). 

this is equivalent to solving for t the congruence 

tb ≡ c 

(mod A), 

which is possible since gcd(b, A) = 1. 

The fact that (A 2 , B, C) represents the square class of (A, B, AC) is a consequence of a 

classical result of Dirichlet on the convolution of “united” forms (see [3, p.57]). □

Thus 

(An 

− (2A 2 x + By) )( An + (2A 2 x + By) ) ≡A 2 n 2 − (2A 2 x + By) 2 


Turning now to the proof of Proposition 8.1, it is easily checked using (54) that for 

Q ′ (x, y) = ax 2 + bxy + cy 2 ∈ Q −d 

and Q ∈ Q −D as in (93) above we have 

√ 

(95) 

dD cosh d(τQ ′, τ Q ) = 2Ac + 2ACa − Bb. 

It follows that the statement of Theorem 8.1 is equivalent to the equality 

(96) #{(a, b, c) ∈ Z 3 | b 2 − 4ac = −d and 2Ac + 2ACa − Bb = k} = δ(k) r Q 2( k2 −dD 

4 

), 

when D = p. Note that a > 0 for any (a, b, c) in the set since k ≥ 0. 

In order to prove this we will establish a direct bijection between the solutions to the 

relevant equations. A calculation verifies the truth of the following key identity: 

(97) 4 Q 2 (x, y) = (2Ac + 2ACa − Bb) 2 − (B 2 − 4A 2 C)(b 2 − 4ac) 

where Q 2 is as in (94) above and 

x =c − Ca 

y =Ba − Ab. 

Thus every solution (a, b, c) from (96) gives rise to a solution (x, y) of 

(98) Q 2 (x, y) = k2 − dD 

. 

4 

In fact, there is no need to assume that D = p for this part of the argument. 

This assumption simplifies our treatment of the converse, to which we now turn. Suppose 

we are given a solution (x, y) of (98). The result is trivial when d = pm 2 and k = pm for 

some integer m, so assume otherwise. Then (−x, −y) is another distinct solution. Observe 

that since p = 4A 2 C − B 2 we have 

4A 2 Q 2 (x, y) ≡ A 2 (4A 2 x 2 + 4Bxy + 4Cy 2 ) ≡ (2A 2 x + By) 2 

(mod p). 

Since Q is assumed to be primitive we know that p ∤ A. 

If p ∤ k we can find an integer a such that 

An = ±(2A 2 x + By) + pa 

≡A 2 n 2 − 4A 2 Q 2 (x, y) ≡ 0 

(mod p). 

in precisely one of the two cases of ±. Suppose as we may that An = 2A 2 x + By + pa. Thus 

we have 

a = An − 2A2 x − By 

. 

p 

Then since p = 4A 2 C − B 2 we have 

k = 2A(x + Ca) + 2ACa − B Ba − y 

A . 

Since gcd(A, B) = 1 we must have that A | (Ba − y) and we may define 

b =(Ba − y)/A 

c =x + Ca.


Then we have 2Ac + 2ACa − Bb = k. A computation also shows that b 2 − 4ac = −d. If p | k 

then we get two distinct triples (a, b, c) for ±(x, y) which both satisfy these. Obviously the 

definition of (a, b, c) inverts the map we started with to get (x, y), at least for those pairs 

that correspond to an (a, b, c). This finishes the proof of Proposition 8.1. 

To give a level N version of Proposition 8.1, we can start with the more general identity 

4N Q ∗ (x, y) = (2ANc + 2ACNa − Bb) 2 − (B 2 − 4A 2 NC)(b 2 − 4Nac) 

where Q ∗ (x, y) = A 2 Nx 2 + Bxy + Cy 2 and 

x =c − Ca 

y =Ba − Ab. 

9. Proof of Theorem 1.2 

In this section we will prove Theorem 1.2. It will be deduced as a corollary of a more 

general identity for special values of certain Borcherds lifts of weight 1/2 weakly holomorphic 

forms. We again apply the Rankin method to forms with poles. We remark that the method 

also yields another proof of (12), whose analytic proof in [23] uses Hecke-Eisenstein series of 

weight one. In fact, the method can be generalized to higher levels in order to obtain some 

refinements of the type we have been considering of certain results of [25] on height pairings 

of Heegner divisors. 

Let M 

1/2 ! be the space of weakly holomorphic modular forms of weight 1/2 and level 4 

satisfying Kohnen’s plus space condition. It has a canonical basis {f d } d≥0 with d ≡ 0, 3 

(mod 4) and Fourier expansions 

f d (z) = q −d + ∑ n≥1 

c(f d , m)q m . 

Let f(z) ∈ M 

1/2 ! be a weakly holomorphic form with integral Fourier coefficients c(f, n). In 

[4], Borcherds constructed an infinite product Ψ f (z) using c(f, n) as exponents, and showed 

that it is a modular form of weight c(f, 0) and some character. The divisors of Ψ f (z) are 

supported on cusps and imaginary quadratic irrationals. In particular, if τ is a quadratic 

irrational of discriminant −D < 0, then its multiplicity in Ψ f (z) is 

ord τ (Ψ f ) = ∑ k>0 

c(f, −Dk 2 ). 

For example, when f(z) = f d (z) with d > 0, the Borcherds product Ψ fd (z) equals to 

∏ 

(99) 

(j(z) − j(τ q )) 1/wq , 

q∈Q −d /Γ 

where Γ = PSL 2 (Z). Note that when −d is fundamental, w d , the number of roots of unity in 

Q( √ −d), is equal to 2w q for all q ∈ Q −d . 

Given f(z) ∈ M ! 1/2 , define a modular form f lift,θ (z) ∈ M ! 1(p) by 

(100) f lift,θ (z) := U 4 (f(pz)θ(z)), 

where U 4 is the standard U-operator. It is easy to verify that f lift,θ (z) ∈ M !,+ 

1 (p) from its 

Fourier expansion. 

For ψ a non-trivial character of Cl(F ), let 

˜g ψ (z) = ∑ n∈Z 

r + ψ (n)qn ∈ M − 1 (p)


be a mock-modular form with shadow g ψ (z). By Proposition 3.4, the regularized inner 

product 〈f lift,θ , g ψ 〉 reg can be expressed in terms of r + ψ 

(n) as 

(101) 〈f lift,θ , g ψ 〉 reg = ∑ ∑ 

( ) 

−pm − k 

r + 2 

ψ 

c(f, m)δ(k). 

4 

m∈Z 

k∈Z 

The following theorem relates this inner product, hence the linear combination of r + ψ 

(n), to 

the values of the Borcherds lift Ψ f (z). 

Theorem 9.1. Let f(z) ∈ M 

1/2 ! be a weakly holomorphic modular form with integral Fourier 

coefficients c(f, n), and Ψ f (z) its Borcherds lift. Suppose ψ is a non-trivial character of 

Cl(F ). Then we have 

∑ 

(102) 〈f lift,θ , g ψ 〉 reg = −2 lim ψ 2 (A) ( log |Ψ f (τ A + ɛ)| 2 + C f log |y A | ) , 

ɛ→0 

A∈Cl(F ) 

where C f = ∑ k∈Z c(f, −pk2 ) is the constant term of f lift,θ (z), and τ A is a CM point associated 

to the class A. 

Remark. Recall that jm 

lift,B (z) = j m (pz)ϑ B (z) for B ∈ Cl(F ). When B = A 0 is the principal 

class in Cl(F ), we can write 

2j lift,A 0 

m (z) = j m (pz)U 4 (θ(pz)θ(z)) 

= (j m (4z)θ(z)) lift,θ 

= ∑ t∈Z 

f lift,θ 

4m−t 

(z). 2 

t 2 ≤4m 

Suppose r A0 (m) = 0, then m is not a perfect square and the right hand side of equation (53) 

becomes 

−2 

∑ 

ψ 2 (A) log |Ψ m (τ A , τ A )| . 

A∈Cl(F ) 

By Kronecker’s identity, this is the same as 

⎛ 

−2 

∑ 

A∈Cl(F ) 

∑ 

ψ 2 (A) ⎜ 

⎝ log |Ψ f4m−t 2 

(τ A )| ⎟ 

⎠ . 

t∈Z 

t 2 ≤4m 

Thus, this case of Theorem 5.1 is a consequence of Theorem 9.1. 

The plan of the proof goes as follows. First, notice that both sides of equation (102) are 

additive with respect to f(z). So it suffices to prove the theorem when f(z) = f d (z) for all 

d ≥ 0 and d ≡ 3 or 0 modulo 4. For convenience, we denote 

(103) a(d, ψ) := 〈f lift,θ 

d 

, g ψ 〉 reg . 

Then, we define a function Φ(z) analogous to G D (z) in [29], where the inner product between 

G D (z) and a weight k+1/2 level 4 eigenform g(z) gives special values of L-function associated 

to the Shimura lift of g(z). For us, Φ(z) is non-holomorphic, transforms with weight 3/2 and 

level 4, and satisfies Kohnen’s plus space condition. Since there is no holomorphic cusp form 

of weight 3/2, level 4 satisfying the plus space condition, the inner product between the 

holomorphic projection of (almost) Φ(z) and the Poincaré series with an s-variable giving 

rise to such cusp forms is zero as s approaches zero. This gives us the identity relating finite 

⎞


linear combinations of r + ψ 

(n) to the limit of an infinite sum. The finite linear combination 

will be exactly the right hand side of equation (101). Then, we will relate the limit of the 

infinite sum to Green’s function, hence Ψ fd (z), via the counting argument in §8. In some 

sense, the proof is in the same spirit as Zagier’s proof of Borcherds theorem in [44]. 

The following lemma is analogous to Lemma 6.6 and relates a(d, ψ) with the limit of an 

infinite sum involving r ψ (n). 

Lemma 9.2. Let d ≥ 1 be an integer congruent to 0 or 3 modulo 4. Then we have 

( 

a(0, ψ)H(d) ∑ 

( ) ( ) ) 

(104) a(d, ψ) − = lim δ(k)ϱ s−1 −1 + k2 −pd + k 

2 

r ψ , 

H(0) s→1 pd 4 

k∈Z 

where the function ϱ s (µ) is defined by 

ϱ s (µ) := 

∫ ∞ 

1 

du 

(µu + 1) 1+s 

2 u , µ > 0. 

Proof. Let ˜g ψ (z) ∈ M − 1 (p) be a mock-modular form with shadow g ψ (z) and 

ord ∞˜g ψ (z) ≥ − p 4 . 

Forms satisfying these conditions exist by Proposition 4.2. The restriction on the order of 

˜g ψ (z) at infinity simplifies the proof and is not important. But the requirement that ˜g ψ (z) 

be in the minus space is crucial for the validity of the statement. 

Denote ĝ ψ (z) ∈ H − 1 (p) the associated harmonic Maass form and define the function Φ(z) 

by 

Φ(z) = Tr 4p 

4 ((ĝ ψ |U p )(4z)θ(pz)), 

where Tr 4p 

4 is the trace down operator from level 4p to level 4. The function Φ(z) transforms 

like a weight 3/2 modular form of level 4 and should be compared to G D (z) in [29], which 

was defined by applying the trace down operator Tr 4D 

4 to the product of the holomorphic 

weight k Eisenstein series and θ(|D|z). If one replaces the weight k holomorphic Eisenstein 

series by the weight one real-analytic Eisenstein series with an s variable, and consider the 

derivative with respect to s, then it is something quite similar to this Φ(z). 

Using the coset representatives 

Γ 0 (4p)\Γ 0 (4) = 

p−1 

⊔ 

µ=0 

( 

1 

4 1 

) ( 

1 µ 

1 

) ⊔ 

( 

1 

1) 

, 

and the fact that (ĝ ψ |U p )(z) = −i(ĝ ψ |W p )(z), We can calculate the Fourier expansion of 

Φ(z) = Tr 4p 

4 ((ĝ ψ |U p )(4z)θ(pz)) as 

Φ(z) = ∑ n∈Z(b(n, y) + a(n))q n , 

b(n, y) = − ∑ 

a(n) = ∑ k∈Z 

k 2 −4m=pn 

δ(k)r + ψ 

δ(k)r ψ (m)β 1 

(m, 4y p 

( pn − k 

2 

Since p ≡ 3 (mod 4), Φ(z) satisfies Kohnen’s plus space condition and n ≡ 0 or 3 (mod 4). 

Since ord ∞˜g ψ (z) ≥ −p/4, the right hand side of equation (101) equals to a(d) for f(z) = f d (z). 

4 

) 

. 

) 

,


So we have 

(105) a(d) = a(d, ψ). 

Let F(z) be the weight 3/2 Eisenstein series studied in [28], which has the following Fourier 

expansion 

∞∑ 

∑ ∞ 

F(z) = H(n)q n + y −1/2 1 

16π β 3/2(m 2 , y)q −m2 , 

n=0 

m=−∞ 

and satisfies Kohnen’s plus space condition. For all n ∈ Z, notice that b(n, y)q n is exponentially 

decaying as y → ∞. Also, a(n) vanishes for all n < 0. Thus, the function 

Φ ∗ (z) := Φ(z) − a(0)F(z) 

H(0) 

is O(y −1/2 ) as y → ∞. The same decaying property holds at the other two cusps of Γ 0 (4) as 

well, since Φ ∗ (z) satisfies Kohnen’s plus space condition. So we can consider the holomorphic 

projection of Φ ∗ (z) to the Kohnen plus space S + 3/2 (Γ 0(4)). Define the weight 3/2 Poincaré 

series by 

∑ 

(106) 

P d (z, s) := 

j(γ, z) −3 e 2dπiγz Im(γz) s/2 , 

where for γ ∈ Γ 0 (4) 

γ∈Γ ∞\Γ 0 (4) 

j(γ, z) := θ(γz) 

θ(z) . 

This series converges absolutely for Re(s) > 1 and can be analytically continued to Re(s) ≥ 

0. As s → 0, the inner product 〈Φ ∗ (z), P d (z, s)〉 is the d th coefficient of a cusp form in 

S + 3/2 (Γ 0(4)), since Φ ∗ (z) is already in the plus space. Given S + 3/2 (Γ 0(4)) = {0}, we know the 

limit is zero and obtain the following equation after applying Rankin-Selberg unfolding, 

( Γ( 

1+s 

(107) lim 

) ( 

2 

a(d) − a(0)H(d) ) ∫ ∞ 

) 

+ b(d, y)e −4πdy 1/2+s/2 dy 

y = 0. 

s→0 (4πd) 1/2+s/2 H(0) 

y 

After some manipulations, we have 

∫ ∞ 

After substituting µ = k2 

pd 

(108) − 

∫ ∞ 

Here, we used the fact 

0 

0 

β 1 (d, µy)e −4πdy 1/2+s/2 dy 

y 

y 

− 1 and µ = 

pk2 

d 

b(d, y)e −4πdy 1/2+s/2 dy 

y 

y 

0 

= 

Γ( 

1+s 

2 ) 

(4πd) 1/2+s/2 ϱ s (µ) . 

− 1, we arrive at the following expression 

∑ 

( ) −pd + k 

2 

δ(k)r ψ ϱ s 

(−1 + k2 

4 

pd 

1+s 

Γ( 

2 

= ) 

(4πd) 1+s 

2 

k∈Z 

β 1 (d, αy) = β 1 (dα, y) 

for all α, y, d ∈ R >0 . Now substituting (108) and (105) into (107) gives us equation (104) 

Proof of Theorem 9.1. When d = 0, f d (z) = θ(z) is the Jacobi theta function and θ lift,θ (z) = 

2ϑ A0 (z) is twice the weight one theta series associated to the principal class A 0 . The 

Borcherds lift of θ(z) is η 2 (z) and equation (102) becomes 

〈2ϑ A0 , g ψ 〉 = −4 ∑ 

ψ 2 (A) log ∣ √ y A η 2 (τ A ) ∣ A∈Cl(F ) 

) 

. 

□


This is justified by Proposition 4.1 and the facts that 

1 ∑ 

〈ϑ A0 , g ψ 〉 = 

〈g ψ ′, g ψ 〉, 

H(p) 

ψ ′ :Cl(F )→C × ∣ √ y A η 2 (τ A ) ∣ = ∣ √ y A η 2 (τ A ) ∣ . 

When d > 0 is congruent to 0 or 3 modulo 4, we need to study the right hand side 

of equation (104), which will yield the Green’s function. The procedure here follows part 

of §5 in [23] and is analogous to the proof of Theorem 5.1. Recall that Q s−1 (t) is the 

Legendre function of the second kind defined in (55). By comparing it with ϱ s (µ), we see 

that ϱ 0 (µ) = 2Q 0 ( √ µ + 1) and 

Q s−1 ( √ µ + 1) − 

( ) 

Substituting 22−s Γ(2s) 

Q |k| 

sΓ(s) 2 s−1 

√ pd 

equation (104) yields 

a(d, ψ) − 

a(0, ψ)H(d) 

H(0) 

⎛ 

= lim ⎝2 

s→1 

sΓ(s)2 

2 2−s Γ(2s) ϱ s−1(µ) = O(µ −1/2−s/2 ). 

( ) 

for ϱ s−1 −1 + k2 in the limit on the right hand side of 

∑ 

A∈Cl(F ) 

pd 

ψ(A) ∑ 

k> √ pd 

δ(k)r A 

( −pd + k 

2 

4 

) 

2Q s−1 

( k 

√ pd 

) ⎞ ⎠ . 

The factor of 2 next to the summation comes from the contribution k < − √ pd. 

Let Q A ∈ Q −p be the quadratic form such that Q A (τ A , 1) = 0. By Proposition 8.1 we have 

( ) ( ) 

ρ QA (k, d) = δ(k)r Q 2 

A 

= δ(k)r A 2 , 

k 2 −pd 

4 

k 2 −pd 

4 

where ρ QA (k, d) is defined in (91) and counts the number of quadratic forms Q ′ ∈ Q −d such 

that cosh d(τ QA , τ Q ′) = √ k 

pd 

and is independent of the choice of τ A . So equation (104) becomes 

a(d, ψ) − 

a(0, ψ)H(d) 

H(0) 

= (−2) lim 

= (−2) lim 

∑ 

s→1 

A∈Cl(F ) 

∑ 

s→1 

A∈Cl(F ) 

= (−2) lim ⎜ 

⎝ lim 

ɛ→0 

⎛ 

⎛ 

ψ(A 2 ) ⎝ ∑ 

ψ 2 (A) ∑ 

∑ 

s→1 

A∈Cl(F ) 

k> √ pd 

⎞ 

( ) 

k 

−2Q s−1 

√ pd 

ρ QA (k, d) ⎠ 

g s (τ A , τ Q ) 

Q∈Q −d 

τ Q ≠τ A 

ψ 2 (A) ∑ 

⎞ 

g s (τ A + ɛ, τ Q ) ⎟ 

⎠ 

Q∈Q −d 

τ Q ≠τ A 

where g s (τ 1 , τ 2 ) is defined in (57). 

As long as d is not of the form pm 2 for any integer m, the condition τ Q ≠ τ A is satisfied 

by all Q ∈ Q −d . Otherwise, there exists exactly one Q ∈ Q −d such that τ Q = τ A . Since 

c(f d , 0) = 0, the number of such Q ∈ Q −d can also be written as 

1 

C 2 f d 

= ∑ c(f d , −pm 2 ) = 1. 

m>0


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 

∑ 


∑ 

ɛ→0 

A∈Cl(F ) 

⎞ 


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. 

□


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 

∣ 

∏ 


(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 ) 


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 


⎞ 

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.


for any d. By the same argument in the proof of Theorem 9.1, one can show that equation 

(12) holds when d ≠ pm 2 for any integer m. So in that case, the equation above becomes 

∣ 

− 1 ∑ 

( ) pd − k 

r + 2 

1 ∑ 

B 

δ(k) = ψ(B −2 ) ∑ ∣∣∣∣∣ 

∏ 

ψ 2 (A) log 

4 

2 4 

H(p) 

− j(τ q )) 

k∈Z 

ψ 

A ∣q∈Q −d /Γ(j(A) 1/wq ∣ ∣∣∣∣∣ ∏ 

= log 

(j(B) − j(τ q )) 1/wq . 

∣ 


10. Case p = 283: Some numerical calculations 

Finally, we will present another numerical example to demonstrate that statements similar 

to the Conjecture in §1 should be true for octahedral newforms. These calculations were 

conducted in SAGE [41]. When p = 283, Cl(F ) has order 3 and the space S 1 (p) is 3- 

dimensional spanned by dihedral newform h(z) and octahedral newforms f ± (z) = f 1 (z) ± 

√ −2f2 (z), where 

h(z) = q + q 4 − q 7 + q 9 + O(q 10 ), 

f 1 (z) = q − q 4 + 2q 6 − q 7 − q 9 + O(q 10 ), 

f 2 (z) = q 2 − q 3 − q 5 + O(q 10 ). 

By Lemma 3.3, h(z), f 1 (z) ∈ S + 1 (p) and f 2 (z) ∈ S − 1 (p). By Deligne-Serre Theorem, f ± (z) 

arises from ρ : Gal(K/Q) → GL 2 (C), where K is the degree 2 extension of the normal closure 

of Q[X]/(X 4 − X − 1) such that Gal(K/Q) ∼ = GL 2 (F 3 ). 

Let F 6 be the subfield of K fixed by a subgroup of Gal(K/Q) isomorphic to Z/8Z, and H 3 

the cubic subfield of F 6 . Explicitly, we can write 

H 3 = Q[Y ]/(Y 3 + 4Y + 1), 

F 6 = Q[X]/(X 6 − 3X 5 + 6X 4 − 7X 3 + 10X 2 − 7X + 6), 

□ 

with the embedding 

H 3 ↩→ F 6 

Y ↦→ X 2 − X + 1. 

Fix a complex embedding of F 6 ↩→ C such that X = t, Y = θ with t, θ ∈ C having positive 

imaginary part. Since F 6 is totally complex, the unit group of F 6 has rank 2 and is generated 

by 

u 1 = t 2 − t + 1, u 2 = t 3 + t − 1. 

As in the case of dihedral newforms, one can express the Petersson norms of f + (z) and f − (z) 

as 

〈f + , f + 〉 = 〈f − , f − 〉 = p + 1 

12 

Res s=1 

ζ(s)L(s, F 6 ) 

L(s, H 3 )ζ(2s)(1 + p −s ) . 

Notice the ratio L(s, F 6 )/L(s, H 3 ) is the L-function of the quadratic Hecke character of H 3 

associated to the Hilbert quadratic norm residue symbol of F 6 /H 3 . So it is holomorphic at 

s = 1 and the right hand side can be evaluated to be 8 log |u 1 u 2 2|.


By Proposition 3.5, there exists mock modular forms ˜f 1 (z) ∈ M − 1 (p) and ˜f 2 (z) ∈ M + 1 (p) 

with shadows f 1 , f 2 respectively such that 

∑ 

˜f 1 (z) = c + 1 (−4)q −4 + c + 1 (−1)q −1 + c + 1 (0) + 

c + 1 (n)q n , 

˜f 2 (z) = c + 2 (−2)q −2 + c + 2 (0) + 

Using Proposition 3.4, we find that 

∑ 

n>0,χ 283 (n)≠−1 

n>0,χ 283 (n)≠1 

c + 2 (n)q n . 

−c + 1 (−4) = c + 1 (−1) = c + 2 (−2) = 2 log |u 1 u 2 2|. 

Since M 1 + (p) ⊂ M + 1 (p) and M1 − (p) ⊂ M − 1 (p) are both non-empty, the principal part of ˜f j (z) 

does not determine it uniquely. However, once c + 1 (2) is chosen, then ˜f 1 is fixed. Similarly, 

once c + 2 (0), c + 2 (1) and c + 2 (4) are chosen, ˜f2 is fixed. Since we expect c + j (n) to be logarithms 

of absolute values of algebraic numbers in F 6 , it is natural to choose c + 1 (2), c + 2 (0), c + 2 (1) and 

c + 2 (4) this way and study the other coefficients. So we will write 

c + j (n) = β j log |u j (n)|, 

where β j ∈ Q and u j (n) ∈ C × is some complex number. Then we fix ˜f j (z) by letting 

β 1 = 2, β 2 = u 2 (0) = u 2 (1) = 1 and 

u 2 (4) = u 2 1(2) = − 7 16 t5 + 5 4 t4 − 7 8 t3 + 47 

16 t2 − 13 

16 t + 19 

8 . 

From the numerical calculations, we can make predictions of the algebraicity of u j (n). In the 

table below, we list c + j (l) for j = 1, 2 and various primes l. Also, we list the predicted values 

˜β j and fractional ideals generated by ũ j (n) in F 6 . 

When χ p (l) ≠ 1, the ideal (l) splits into L 0 L 1 in H 3 , and L 1 splits into l l,1 l l,2 in F 6 . 

When χ p (l) = 1 and c(f 1 , l) = ±1, the ideal (l) splits into l 1 l 2 in F 6 . When χ p (l) = 1 and 

c(f 1 , l) = 0, the ideal (l) splits into L l,0 L l,1 in H 3 and L l,1 splits into l l,1 l l,2 in F 6 , where 

l l,j has order 4 in Cl(F 6 ), the class group of F 6 . Since F 6 and H 3 have class numbers 8 and 

2 respectively, the fractional ideals in Table 3 and Table 4 are all principal. For example, 

the value we chose for u 2 1(2) generates the fractional ideal (l 2,1 /l 2,2 ) 4 , where l 2,1 and l 2,2 have 

order 8 in Cl(F 6 ). So it is necessary to make ˜β 1 = 1/2. The numerical pattern also justifies 

this choice of ˜f j (z). 

Table 3. Coefficients of ˜f 1 (z) 

l c(f 2 , l) c + 1 (l) ˜β1 (ũ 1 (l)) 

2 1 1.2075349695016218 1/2 (l 2,1 /l 2,2 ) 4 

3 -1 -0.44226603950742649 1/2 (l 3,1 /l 3,2 ) 4 

5 -1 -3.9855512247433431 1/2 (l 5,1 /l 5,2 ) 4 

17 0 -3.2181607607379323 1/2 (l 17,1 /l 17,2 ) 4 

19 1 5.3481233955176073 1/2 (l 19,1 /l 19,2 ) 4 

31 1 -1.7192005338244623 1/2 (l 31,1 /l 31,2 ) 4 

37 0 0.32541651822318252 1/2 (l 37,1 /l 37,2 ) 4


43 1 -4.6200896216743352 1/2 (l 43,1 /l 43,2 ) 4 

47 -1 -1.0203031328088645 1/2 (l 47,1 /l 47,2 ) 4 

53 0 5.8419201851710110 1/2 (l 53,1 /l 53,2 ) 4 

67 0 -3.9318486618330462 1/2 (l 67,1 /l 67,2 ) 4 

79 0 7.5720154112893967 1/2 (l 79,1 /l 79,2 ) 4 

107 0 -0.81774052769784944 1/2 (l 107,1 /l 107,2 ) 4 

109 -1 4.8808523053451562 1/2 (l 109,1 /l 109,2 ) 4 

283 0 6.86483405137284 1/2 (l 283,1 /l 283,2 ) 4 

Table 4. Coefficients of ˜f 2 (z) 

l c(f 1 , l) c + 2 (l) ˜β2 (ũ 2 (l)) 

7 -1 -3.27983974462451 1 l 7,1 /l 7,2 

11 1 -2.56257986300244 1 l 11,1 /l 11,2 

13 1 -5.57196302179201 1 l 13,1 /l 13,2 

23 -1 1.01652189648251 1 l 23,1 /l 23,2 

29 -1 1.54494007675715 1 l 29,1 /l 29,2 

41 1 0.771808245755645 1 l 41,1 /l 41,2 

71 0 -4.99942007705695 1 (l 71,1 /l 71,2 ) 2 

73 0 -1.64986308549260 1 (l 73,1 /l 73,2 ) 2 

83 -2 8.97062724307569 1 1 

89 -1 -0.399183274865547 1 l 89,1 /l 89,2 

101 0 5.99108448704487 1 (l 101,1 /l 101,2 ) 2 

283 1 4.48531362153791 1 1 

643 2 2.32782185606303e-10 1 1 

773 2 5.08403073372794e-9 1 1 

859 -2 -8.97062719191082 1 1 

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UCLA Mathematics Department, Box 951555, Los Angeles, CA 90095-1555 

E-mail address: wdduke@ucla.edu 

UCLA Mathematics Department, Box 951555, Los Angeles, CA 90095-1555 

E-mail address: yingkun@ucla.edu

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

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