Differential operators related to harmonic Maass forms

This shows that e^{−2πizκρ/tρ}×f|_kγρhas a Fourier expansion at i∞with period t_ρas a function of x. Using Lemma 8, this can be written by

n−∞

∑

c⁺_f(n)q^n/tρ+c⁻_f(0)y^1−k+

∑

n∞n6=0

c⁻_f (n)Γ(1−k,−4πny/t_ρ)q^n/tρ.

Multiplying e^{−2πizκρ/tρ} =q^κρ/tρ both sides, we have a desired result.

Lemma 10 Let f ∈H_k^!(N,χ)andα∈GL⁺(2,Q). Then f|_kα satisfies similar growth condition as f at every cusp.

Proof 20 Letρ∈Q∪ {i∞}be a cusp ofΓ₀(N)andβ ∈SL(2,Z)such thatβ(i∞) =ρ. Without loss of generality, we assume thatα β= p q

r s

!

has integer entries; if not, multiply a suitable integer toα β without affecting f|_kα β. Let p=ap⁰and r=ar⁰where a= (p,r). Then since(p⁰,r⁰) =1, there are two integers x,y such that xp⁰+yr⁰=1and we obtain x y

−r⁰ p⁰

! p q r s

!

= a b

0 d

!

where b=xq+ys and d=−r⁰q+p⁰s are both integers. Since xp⁰+yr⁰=1, we observe that x y

−r⁰ p⁰

!

∈SL(2,Z).

Denote the inverse of this matrix byγ. Then we can write α β=γ

a b 0 d

! .

Now using(51)with width t_ρand cusp parameterκρ, we obtain f|_kα β(z) =f|_kγ

a b 0 d

!

(z) = (f|_kγ) k

a b 0 d

! (z)

=







∑

n−∞

c⁺_f(n)q

n+κρ

tρ +c⁻_f(0)y^1−kq

κρ

tρ +

∑

n∞n6=0

c⁻_f(n)Γ(1−k,−4πny/t_ρ)q

n+κρ tρ





 k

a b 0 d

! (z)

=e^2πi

bκρ dtρ

a d

k/2







∑

n−∞

d⁺_f (n)q

a(n+κρ)

dtρ +

a d

1−k

c⁻_f(0)y^1−kq

aκρ

dtρ +

∑

n∞n6=0

d⁻_f (n)Γ(1−k,−4πany/d)q

a(n+κρ) dtρ





,

where d^±_f (n) =e^2πibn/dtρc^±_f (n). Since a,d>0, the lemma follows.

(2) The weight k hyperbolic Laplacian operator can be express as

−∆k=L_k+2◦Rk+k=Rk−2◦Lk. (3) If f is an eigenfunction of∆kwith eigenvalueλ, then

∆_k+2(Rk(f)) = (λ+k)Rk(f), ∆k−2(Lk(f)) = (λ−k+2)Lk(f).

Proof 21

(1) Letα= a b c d

!

∈SL(2,Z), and denote f⁰=∂f

∂z. Then we have R_k(f|_kα)(z) =2i∂

∂z((cz+d)^−kf(αz)) +k

y((cz+d)^−kf(αz))

=2i

(cz+d)^−k−2f⁰(αz)−ck(cz+d)^−k−1f(αz)

+k

y((cz+d)^−kf(αz))

=2i(cz+d)^−(k+2)f⁰(αz) + (cz+d)^−(k+2)f(αz)

−2ick(cz+d) +k

y(cz+d)²

=2i(cz+d)^−(k+2)f⁰(αz) + (cz+d)^−(k+2)f(αz)k

y(cz+d)(cz+d)

=2i(cz+d)^−(k+2)f⁰(αz) + (cz+d)^−(k+2)f(αz)k

y|cz+d|². On the other hand,

(Rk(f)|_k+2α)(z) =2i(f⁰|_k+2α)(z) + k

Im(z)f k+2

α

(z)

=2i(cz+d)^−(k+2)f⁰(αz) + (cz+d)^−(k+2) k

Im(αz)f(αz)

=2i(cz+d)^−(k+2)f⁰(αz) + (cz+d)^−(k+2) k

Im(z)|cz+d|²f(αz).

Comparing two equations, we see that R_k(f|_kα) =R_k(f)|_k+2α.

Next, note that f is also dependent on z and so we will abuse the notation of f(z) as f(z,z).

Then we have

L_k(f|_kα)(z) =−2iy² ∂

∂z((cz+d)^−kf(αz))

=−2iy²

f(αz)∂

∂z(cz+d)^−k+ (cz+d)^−k ∂

∂zf(αz)

=−2iy²(cz+d)^−k ∂

∂zf(αz,αz)

=−2iy²(cz+d)^−k∂f

∂z(αz)(cz+d)⁻².

On the other hand,

(Lk(f)|_k−2α)(z) =−2i(cz+d)^−(k−2)(Im(αz))²∂f

∂z(αz)

=−2i(cz+d)^−(k−2)

Im(z)

|cz+d|² 2

∂f

∂z(αz)

=−2i(cz+d)^−k(Im(z))² (cz+d)²

∂f

∂z(αz).

Comparing two equations, we see that Lk(f|_kα) =Lk(f)|_k−2α.

Finally, by using (2) which we will prove below, we have

∆k(f|kα) = (Rk−2◦Lk)(f|kα) =Rk−2(Lk(f)|k−2α) = (Rk−2◦Lk)(f)|kα=∆k(f)|kα. (2) By definition, we obtain

(L_k+2◦R_k+k)(f) = (L_k+2◦R_k)(f) +k f =

−2iy² ∂

∂z 2i∂f

∂z +k yf

+k f

=4y²∂

∂z

∂

∂zf−2iy² ∂

∂z k

yf

+k f

=4y²∂

∂z

∂

∂zf−2iy²

−1 2ik

y²f+k y

∂f

∂z

+k f

=4y²∂

∂z

∂

∂zf−2kiy∂f

∂z

=−∆_kf.

(Rk−2◦Lk)(f) =

2i∂

∂z+k−2

y −2iy²∂f

∂z

=4∂

∂z

y²∂f

∂z

−2(k−2)iy²∂f

∂z

=

4y²∂

∂z

∂

∂zf−4iy∂f

∂z

−2(k−2)iy²∂f

∂z

=4y²∂

∂z

∂

∂zf−2kiy∂f

∂z

=−∆kf.

(3) Using∆kf =λf and (1), we have−(Rk−2◦Lk)(f) =λf . Then−(Rk−2◦Lk)(Lk(f)) =λLk(f) and so

−(R_k−2◦Lk+k−2)(Lk(f)) = (λ−k+2)Lk(f).

Similarly, we have−(L_k+2◦R_k+k)(Rk(f)) =λR_k(f)and so

−(L_k+2◦R_k)(Rk(f)) = (λ+k)Rk(f).

Now we introduce two differential operators acting on the spaceH_k^!(N,χ), which are associated with Maass rasing and lowering operators and play an important role in studying the Fourier expansion of

harmonic Maass forms.

In order to understand the motivation for which these operators were designed, define firstD-operator asD:= 1

2πi

∂

∂z. In general, the derivative of a modular form is not modular, but the Maass raising op- eratorRk=−4πD+k

y does. On the other hand,Rkdoes not preserve meromorphicity in general. From these observations, we want to get operators which preserves both modularity and meromorphicity. For weight 0,R₀=−4πDdoes. But for general weights, we need to modifyD-operator and raising operator as follows.

For any positive integern, we define thek-iteration of Maass raising operatorsas Rⁿ_k:=R_k+2(n−1)◦ · · · ◦R_k+2◦R_k,

and forn=0, defineR⁰_k:=1.

Lemma 12 For k∈Zand n∈N∪ {0}, we have Rⁿ_k=

n

∑

m=0

(−1)^m n

m

(k+m)^(n−m)y^m−n(4πD)^m, where x^(l)=x(x+1)· · ·(x+l−1)is a rising factorial.

Proof 22 We will use induction on n. For n=0, it is trivial. Suppose that the above equation holds for natural n. By the chain rule and properties of binomial coefficient, we have

Rⁿ⁺¹_k =R_k+2n◦Rⁿ_k=−4πD◦Rⁿ_k+ (k+2n)y⁻¹◦Rⁿ_k

=

n m=0

∑

(−1)^m n

m

(k+m)^(n−m){(n−m)y^m−n−1(4πD)^m−y^m−n(4πD)^m+1} +

n

∑

m=0

(−1)^m n

m

(k+m)^(n−m)(k+2n)y^m−n−1(4πD)^m

=

n

∑

m=0

(−1)^m n

m

(k+m)^(n−m+1)y^m−n−1(4πD)^m +

n+1

∑

m=1

(−1)^m n

m−1

(k+m−1)^(n−m+1)y^m−n−1(4πD)^m

=k⁽ⁿ⁺¹⁾y⁻ⁿ⁻¹+ (−1)ⁿ⁺¹(4πD)ⁿ⁺¹ +

n

∑

m=1

(−1)^m n

m

(k+m)^(n−m+1)+ n

m−1

(k+m−1)^(n−m+1)

y^m−n−1(4πD)^m

=

n+1

∑

m=0

(−1)^m n+1

m

(k+m)^(n+1−m)y^m−(n+1)(4πD)^m. This proves the lemma.

Now we can achieve our goals by relating the iteration of raising operatorRⁿ_kwhich preserves modu- larity and the(1−k)-th order differential operatorD^1−kwhich preserves meromorphicity. This relation is called ‘Bol’s identity’, and the operatorD^1−k is often calledBol’s operator.

Lemma 13 For k≤0, we have

D^1−k= 1

(−4π)^1−kR^1−k_k . In particular, we have that

D^1−k:M_k^!(N,χ)→M_2−k^! (N,χ).

Proof 23 Putting n=1−k≥1in Lemma 12, we observe that the rising factorial(k+m)^(1−k−m)is0 unless m=1−k. Thus we have R^1−k_k = (−1)^1−k(4πD)^1−k.

Now, using Theorem 11 (1), we see that if f satisfies modularity of weight k then R^1−k_k satisfies modular- ity of weight k+2(1−k) =2−k. Meromorphicity at the cusps follows from the fact that D^1−kpreserves meromorphicity.

Proposition 7 (1) If f ∈H_k^!(N,χ)with Fourier series as in(48), then D^1−k(f(z)) =−(4π)^k−1(1−k)!c⁻_f (0) +

∑

n−∞

c⁺_f (n)n^1−kqⁿ. (52) (2) We have

D^1−k:H_k^!(N,χ)→M_2−k^! (N,χ).

Proof 24

(1) By the definition of incomplete gamma function, Γ(s,z)e^z=

Z _∞

z

e^−t+zt^s−1dt= Z _∞

0

e^−t(t+z)^s−1dt.

So we have

Γ(1−k,−4πny)e^−4πny= Z _∞

0

e^−t(t−4πny)^−kdt, and hence

D^1−k(Γ(1−k,−4πny)e^−4πny) = 1

2πi

1−kZ _∞

0

∂^1−k

∂z^1−ke^−t(t−4πny)^−kdt=0.

On the other hand, note that

qⁿ=e^−4πnye^4πnye^2πin(x+iy)=e^−4πnye^{2πin(x−iy)}=e^−4πnye^2πinz. Hence we have

D^1−k(Γ(1−k,−4πny)qⁿ) =D^1−k(Γ(1−k,−4πny)e^−4πny)e^2πinz=0.

Therefore we have

D^1−k(f(z)) =D^1−k(c⁻_f(0)y^1−k) +D^1−k(f⁺(z))

=c⁻_f (0) 1

2πi 1−k

∂

∂z 1−k

y^1−k+ 1

2πi 1−k

n−∞

∑

c⁺_f(n) ∂

∂z 1−k

qⁿ

=c⁻_f (0) 1

2πi 1−k

1 2

1−k

(−i)^1−k ∂

∂y 1−k

y^1−k

!

+ 1

2πi 1−k

n−∞

∑

c⁺_f(n)(2πin)^1−kqⁿ

=−(4π)^k−1(1−k)!c⁻_f(0) +

∑

n−∞

c⁺_f(n)n^1−kqⁿ. This proves (1).

(2) To show that D^1−k(f(z))∈M_2−k^! (N,χ), we need to prove that it is meromorphic at the cusps of Γ₀(N). By the proof of (1), we know that D^1−k(f(z))is holomorphic onH and meromorphic at i∞. For any cusp ρ ofΓ0(N), the Fourier expansion of f at the cuspρ is given in Lemma 9. Through a process similar to the proof of (1), we can calculate that the second series in(51) annihilates, and thus meromorphicity at other cusps follows directly.

For a lowering operators, an analogous modifications can be made to have properties similar toD^1−k. Definition 20 Theshadow operatorξk is defined by

ξk=2iy^k ∂

∂z.

The shadow operator can be express by using lowering operator and hyperbolic Laplacian operator as

ξk=y^k−2Lk, and

−ξ_2−k◦ξk=−2iy^2−k ∂

∂z 2iy^k ∂

∂z

!

=−2iy^2−k ∂

∂z

−2iy^k ∂

∂z

=−4y² ∂

∂z

∂

∂z+2kiy∂

∂z

=∆k. (53)

Proposition 8 Let k be an integer not equal to1.

(1) If f ∈H_k^!(N,χ)with Fourier series as in(48), then

ξk(f(z)) = (1−k)c⁻_f(0)−(4π)^1−k

∑

n−∞

c⁻_f (−n)n^1−kqⁿ. (54)

(2) We have

ξk:H_k^!(N,χ)→M_2−k^! (N,χ).

Moreover, this map is surjective.

Proof 25 (1) First, we have ξk(f⁺) =0 since f⁺ is holomorphic and so ∂

∂zf⁺(z) =0. Next, we observe that

∂

∂zΓ(1−k,−4πny) =1 2

∂

∂x+i∂

∂y

Γ(1−k,−4πny)

= i 2

∂

∂y Z _∞

−4πny

e^−tt^−kdt

= i 2

∂

∂y

−4πn Z _∞

y

e^4πnt(−4πnt)^−kdt

=−2πin∂

∂y

_{Z ∞}

0

− Z y

0

e^4πnt(−4πnt)^−kdt

=2πine^4πny(−4πny)^−k. Then we have

ξk







∑

n∞n6=0

c⁻_f(n)Γ(1−k,−4πny)qⁿ





=2iy^k

∑

n∞n6=0

∂

∂z

c⁻_f(n)Γ(1−k,−4πny)qⁿ

=2iy^k

∑

n∞n6=0

c⁻_f(n)2πin(−4πny)^−ke^4πnye^2πinz

=−(4π)^1−k

∑

n∞n6=0

c⁻_f (n)(−n)^1−ke^4πnye^−2πinz

=−(4π)^1−k

∑

n−∞

c⁻_f(−n)n^1−kqⁿ, and

ξk(c⁻_f(0)y^1−k) =2iy^k ∂

∂zc⁻_f (0)y^1−k=2iy^kc⁻_f(0)i 2

∂

∂yy^1−k= (1−k)c⁻_f(0).

This proves (1).

(2) Let f ∈H_k^!(N,χ). By the automorphy of f , we have f|_kγ=χ(d)f for allγ= a b c d

!

∈γ₀(N).

By definition of shadow operator, observe that

(ξkf|_2−kγ)(z) = ((y^k−2L_k)|_2−kγ)(z) = y^k−2

|cz+d|^2(k−2)L_k(f)(γz)(cz+d)^−(2−k). By Lemma 11, we have

L_k(f)|_k−2γ=L_k(f|_kγ) =L_k(χ(d)f) =χ(d)Lk(f).

Taking complex conjugate on both sides, we obtain

χ(d)Lk(f)(z) =χ(d)Lk(f)(z) = (Lk(f)|_k−2γ)(z) = (cz+d)^−(k−2)L_k(f)(γz)

⇔L_k(f)(γz) =χ(d)(cz+d)^k−2L_k(f)(z).

Thus we get

(ξkf|_2−kγ)(z) = y^k−2

|cz+d|^2(k−2)

χ(d)(cz+d)^k−2L_k(f)(z)

(cz+d)^−(2−k)

=χ(d)y^k−2Lk(f)(z)

=χ(d)(ξkf)(z).

This provesξk(f)∈M_2−k^! (N,χ).

To showξk(f(z))∈M_2−k^! (N,χ), we need to prove thatξk(f(z))is meromorphic at the cusps of Γ0(N). By the proof of (1), we know thatξk(f(z))is holomorphic onH and meromorphic at i∞.

For any cuspρ ofΓ₀(N), the Fourier expansion of f at the cuspρis given in Lemma 9. Through a process similar to the proof of (1), we can calculate that the second series in(51)changes to the holomorphic part, and thus meromorphicity at other cusps follows directly.

Proving surjectivity uses the theory of algebraic geometry, especially the Serre duality. We omit this proof here, and we refer [36] for a detailed proof and [38] for more informations of Serre duality.