Recently, Neurerer-Oliver [28] proved a converse theorem for Maass form of arbitrary levelN, which is analogous to Theorem 7. As in the case of modular form, we need several functional equations of the twistedL-functions. They use the same twisting moduli as Weil’s. To include the case of non-cuspidal form, they weaken the condition of the twistedL-function to be meromorphic. Their poles are arised from the 0-th coefficient of Fourier-Whittaker series (25).
Theorem 13 (Theorem 3.1 in [28]) Let N be a positive integer,χ be a Dirichlet character modulo N, ε ∈ {0,1},ν∈Cbe such that 1
4−ν2>0, and let an, bn be sequences of complex numbers such that
|an|,|bn|=O(nσ)for someσ∈R. LetPbe a set given in Definition 11. Assume that
(1) (i) Ifε =0 andν6=0, then Λν(f,s) andΛν(g,s) continue to meromorphic functions at most simple poles in the set{±ν,1±ν}.
(ii) Ifε =0 andν=0, then Λν(f,s) andΛν(g,s) continue to meromorphic functions at most double poles in the set{0,1}.
(iii) Ifε=1, thenΛν(f,s)andΛν(g,s)continue to entire functions.
(2) For primitive charactersψ of conductor mψ ∈P, the functionsΛν(f,ψ,s)andΛν(g,ψ,s)con- tinue to entire functions.
(3) For all primitive charactersψ of conductor m=mψ ∈P∪ {1}, the functions Λν(f,ψ,s)and Λν(g,ψ,s)are uniformly bounded on every vertical strip outside of a small neighborhood around each pole, and satisfy the functional equation
Λν(f,ψ,s) = (−1)εCψ(m2N)12−sΛν(g,ψ,1−s), (32) where
Cψ =CN,ψ=χ(m)ψ(N)τ(ψ)/τ(ψ) =χ(m)ψ(N)τ(ψ)2/m.
For n<0, define an= (−1)εa−n, and bn= (−1)εb−n. Define a0(z)and b0(z)as follows
• Forν6=0,
a0(y) =−Ress=−νΛν(f,s)y12+ν−Ress=νΛν(f,s)y12−ν,
b0(y) =−Ress=1+νΛν(g,s)(Ny)12+ν−Ress=1−νΛν(g,s)(Ny)12−ν. (33)
• Forν=0,
a0(y) =−Ress=0Λν(f,s)y12−Ress=0sΛν(f,s)y12logy,
b0(y) =−Ress=0Λν(g,s)y12−Ress=0sΛν(g,s)y12logy. (34) Putting these together, we define f and g as
f(z) =a0(y) +2
∑
n6=0
an
√yKν(2π|n|y)e2πinx, g(z) =b0(y) +2
∑
n6=0
bn
√yKν(2π|n|y)e2πinx. (35) Then f and g are weight 0Maass forms on Γ0(N) of parity ε, character χ (resp. χ) and eigenvalue
1
4−ν2. Furthermore f(z) =g(−1/Nz)for all z∈H.
We will survey their proof with 3 steps. We will omit some complicated computations.
Proof 16 (proof (sketch))
(Step 1) First, we will prove f(z) =g(−1/Nz). This can be done by additional computations to Hecke- Maass’s method in Lemma 12.
For a convenience, we rewrite a Fourier-Whittaker expansion of f given in(35)as f(z) =a0(y) +4(−i)ε
∞
∑
n=1
an√
yKν(2πny)cos(ε)(2πnx),
where cos(ε) is a ε-th derivative of cos. In fact, this represents even and odd form at the same time. Denote f˜(z) = f(z)−a0(y). Denoting g andg analogously.˜
Next, let m∈P be a prime such that (m,N) =1 and let α = ma ∈Qfor some a∈Z. We de- fine the ‘additive twists’ of L(f,s)as
L(f,s,α,cos(k)) =
∞
∑
n=1
cos(k)(2πnα)ann−s. Let
γ(−)
k
f (s) =π−s−[ε+k]Γ
s+ [ε+k] +ν 2
Γ
s+ [ε+k]−ν 2
,
where(−)k denotes+if k is even and−if k is odd, and[k+ε]is1if k+ε is even and0if k+ε is odd. Using this, we define the completion of the additive twists by
Λν(f,s,α,cos(k)) =γ(−)
k
f (s)L(f,s,α,cos(k)).
In order to continue the proof, we introduce the following formula related to the hypergeometric function2F1.
Lemma 6 (6.699(3-4) in [29]) Let w∈R, k≥0, andν∈C. ForRe(−s±ν)<ε, 4
Z ∞
0
Kν(2y)cos(k)(2wy)ysdy
y =ik(2w)[k]πsγ(−)
k+ε
f (s)2F1
s+[k]+ν 2 ,s+[k]−2 ν
1 2+[k]
−w2
! .
Proposition 4 Let f˜(z) =4(−i)ε∑∞n=1an
√yKν(2πny)cos(ε)(2πnx) with polynomially bounded an,ε∈ {0,1}, w∈Randα∈Q. The Mellin transform of f is given by˜
Z ∞
0
f˜(wy+iy+α)ys−12dy y
=
∑
j∈{0,1}
i−j(2w)[ε+j]Λν(f,s,α,cos(j))2F1
s+[ε+j]+ν
2 ,s+[ε+2j]−ν
1 2+[ε+j]
−w2
! .
Proof 17 Recall the sum formulae for sine and cosine functions
cos(a+b) =cosacosb−sinasinb,sin(a+b) =sinacosb+cosasinb.
Sincecos(1)=sin, forε∈ {0,1}we can combine two formulas as cos(ε)(a+b) =
∑
j∈{0,1}
(−1)jcos(j)acos(ε+j)b.
Compute using Lemma 6 and above formula, the result follows directly.
Z ∞
0
f˜(wy+iy+α)ys−12dy y
=4(−i)ε
∞
∑
n=1
an
Z ∞
0
Kν(2πny)cos(ε)(2πn(wy+α))ysdy y
=4(−i)ε
∞
∑
n=1
an
∑
j∈{0,1}
(−1)jcos(j)(2πnα) Z ∞
0
Kν(2πny)cos(ε+j)(2πnwy)ysdy y
=4(−i)ε
∞
∑
n=1
an
∑
j∈{0,1}
(−1)jcos(j)(2πnα)(πn)−s Z ∞
0
Kν(2πny)cos(ε+j)(2wy)ysdy y
=
∞ n=1
∑
ann−s
∑
j∈{0,1}
i−jcos(j)(2πnα)(2w)[ε+j]γ(−)
j
f (s)2F1
s+[ε+j]+ν 2 ,s+[ε+j]−2 ν
1 2+[ε+j]
−w2
!
=
∑
j∈{0,1}
i−j(2w)[ε+j]Λν(f,s,α,cos(j))2F1
s+[ε+j]+ν
2 ,s+[ε+2j]−ν
1 2+[ε+j]
−w2
! .
In the case ofα=0, we can skip using (17), and so the Mellin transform of f is given by˜ Z ∞
0
f˜(wy+iy)ys−12dy
y = (2π)εΛν(f,s)2F1
s+ε+ν 2 ,s+ε2−ν
1 2+ε
−w2
! ,
and its inverse transform
f˜(wy+iy) =(2π)ε 2πi
Z c+i∞
c−i∞ Λν(f,s)2F1
s+ε+ν 2 ,s+ε−ν2
1 2+ε
−w2
!
y12−sds, (36) for wy+iy∈H and c>1+|Re(ν)|.
For the next step, we need a following asymptotic behavior of2F1.
Lemma 7 (Section 7.1 in [30]) IfIm(s)→∞, the function2F1has the following asymptotic.
2F1 a+λ,b−λ
c
1−z 2
!
∼2a+b−1Γ(1−b+λ)Γ(c)(1+e−iν)c−a−b−1/2 (λ π)1/2Γ(c−b+λ)(1−e−iν)c−−1/2
e(λ−b)iν+e±iπ(c−1/2)−(λ+a)iν+O(1/λ) .
For s=σ+it, putting a=σ+ε+ν
2 , b=1−σ+ε+ν
2 , c=1+2ε
2 , andλ=it
2. Then we have
2F1
s+ε+ν 2 ,s+ε2−ν
1+2ε 2
−w2
!
=O(evt/2).
Finally, using Stirling’s formula, both gamma factors is O(e−πt/4) so we have the following asymptotic behavior.
Γ
s+ε+ν 2
Γ
s+ε−ν 2
2F1
s+ε+ν 2 ,s+ε−ν2
1 2+ε
−w2
!
decays exponentially as|Im(s)| →∞for all w∈R.
Hence we can shift the path of integration in (36) by the Phragmèn-Lindelöf principle, and by the functional equation forΛν(f,s), we have f˜(wy+iy) =g(−1/N(wy˜ +iy)) +H(wy+iy), where
H(wy+iy) =
0, ε =1,
∑x∈{±ν,1±ν}Ress=x2F1
s+ν 2 ,s+−2ν
1 2
−w2
Λν(f,s)y12−s, ε =0.
Forε=1, we have f(z) =g(−1/Nz)directly. Forε=0, calculate residues in H(wy+iy)using some formulae for hypergeometric function, we have a0(y)−b0(y)given in(33)ifν6=0, and(34) ifν=0. Finally, by the analytic continuation to the wholeH, we conclude f(z) =g(−1/Nz).
(Step 2) Next, we will use the functional equation(32)to obtain the twisted modular relation.
For a primitive Dirichlet characterψmodulo m, define a twisted version of f , L(f,s), andΛν(f,s) as follows
fψ(z) =
∑
n6=0
ψ(n)an
√yKν(2π|n|y)e2πinx,
L(f,ψ,s) =
∞ n=1
∑
ψ(n)ann−s, Λν(f,ψ,s) =γ(−)
k
f (s)L(f,ψ,s), and define gψ analogously. Note thatΛν(f,ψ,s) =Λν(fψ,s).
We need some formulas related to the Gauss sum. Let p be a prime such that(p,N) =1. Then we have
χ(n)τ(χ) =
p
∑
u=1
χ(u)e nu
p
.
Summing overχ modulo p, we have e
n p
=1− p
p−1χ0(n) + 1
p−1
∑
χ(modp) χ6=χ0
τ(χ)χ(n), (37)
where the sums are over Dirichlet characters modulo p andχ0is the trivial Dirichlet character modulo p. Separating the real and imaginary part of(37)gives the following two equations.
cos 2πna
p
=1− p
p−1ψ0(n) + 1
p−1
∑
ψ(modp) ψ6=ψ0
ψ(−1)=1
τ(ψ)ψ(an),
sin 2πna
p
=− i
p−1
∑
ψ(modp) ψ(−1)=−1
τ(ψ)ψ(an).
Using these formulas, forα =ma ∈Qwe can represent L(f,s,α,cos(k))as a linear combination of L(f,ψ,s)
L(f,s,α,cos(k)) = ik
m−1
∑
ψ(modm) ψ6=ψ0
ψ(−1)=(−1)k
τ(ψ)ψ(a)L(f,ψ,s)
+
(−1)k/2
L(f,s)− m
m−1L(f,ψ0,s)
, if k is even,
0, if k is odd.
We can also representΛν(f,s,α,cos(k))as follows Λν(f,s,α,cos(k)) = ik
m−1
∑
ψ(modm) ψ6=ψ0 ψ(−1)=(−1)k
τ(ψ)ψ(a)Λν(f,ψ,s)
+
(−1)k/2
Λν(f,s)− m
m−1Λν(f,ψ0,s)
, if k is even,
0, if k is odd.
(38)
Then by(32), we have the following functional equations Λν(f,s,α,cos(k)) = (−1)εik(m2N)12−sχ(m)
m−1
∑
ψ(modm) ψ6=ψ0
ψ(−1)=(−1)k
ψ(aN)τ(ψ)Λν(g,ψ,1−s)
+
(−1)k/2
Λν(f,s)− m
m−1Λν(f,ψ0,s)
, if k is even,
0, if k is odd.
(39)
Next, letα = ma,β =mb ∈Qand let z=wy+iy∈H. We will consider a function f(z+α)− f(z+β) = f˜(z+α)−f˜(z+β). Applying the inverse Mellin transform to Proposition 4, we have
f(z+α)−f(z+β) =
∑
j∈{0,1}
i−j(2w)[ε+j]
× 1 2πi
Z c+i∞
c−i∞ (Λν(f,s,α,cos(j))−Λν(f,s,β,cos(j)))2F1
s+[ε+j]+ν 2 ,s+[ε+j]−ν2
1 2+[ε+j]
−w2
!
y12−sds.
(40) By(38),Λν(f,s,α,cos(j))−Λν(f,s,β,cos(j))is just a linear combination of twistedΛν(f,s)by characters of conductor m. By assumption (2) in Theorem 13, it is entire function. Shifting the path of integration and using the functional equation(39), we have
Z c+i∞
c−i∞ (Λν(f,s,α,cos(j))−Λν(f,s,β,cos(j)))2F1
s+[ε+j]+ν 2 ,s+[ε+j]−2 ν
1 2+[ε+j]
−w2
! y12−sds
=
Z 1−c+i∞
1−c−i∞ (Λν(f,s,α,cos(j))−Λν(f,s,β,cos(j)))2F1
s+[ε+j]+ν 2 ,s+[ε+j]−ν2
1 2+[ε+j]
−w2
! y12−sds
= Z c+i∞
c−i∞
(Λν(f,1−s,α,cos(j))−Λν(f,1−s,β,cos(j)))2F1
1−s+[ε+j]+ν
2 ,1−s+[ε+2 j]−ν
1 2+[ε+j]
−w2
! ys−12ds
= (−1)εijχ(m)
m−1
∑
ψ(modm) ψ6=ψ0 ψ(−1)=(−1)j
ψ(N)(ψ(a)−ψ(b))τ(ψ)
× Z c+i∞
c−i∞
Λν(g,ψ,s)2F1
1−s+[ε+j]+ν
2 ,1−s+[ε+j]−ν 2 1 2+[ε+j]
−w2
!
(Nm2y)s−12ds. (41) Now we state the Euler identity for2F1as follows.
2F1 a,b
c
z
!
= (1−z)c−a−b2F1 c−a,c−b
c
z
!
Applying this and the inverse Mellin transform(36)for gψ, the last integration equals Z c+i∞
c−i∞
Λν(g,ψ,s)2F1
1−s+[ε+j]+ν
2 ,1−s+[ε+j]−ν 2 1 2+[ε+j]
−w2
!
(Nm2y)s−12ds
= Z c+i∞
c−i∞ Λν(gψ,s)2F1
s+[ε+j]+ν
2 ,s+[ε+2j]−ν
1 2+[ε+j]
−w2
!
1 (1+w2)Nm2y
12−s
ds
=2πi(2w)−[ε+j]gψ
− 1 Nm2z
. (42)
Substituting(41)and(42)into(40), we have f(z+α)−f(z+β) = χ(m)
m−1
∑
ψ(modm) ψ6=ψ0
ψ(−N)(ψ(a)−ψ(b))τ(ψ)gψ
− 1 Nm2z
. (43)
On the other hand, using(37), we obtain f(z+α)−f(z+β) =
∑
n6=0
an
√yKν(2π|n|y)e2πinx(e2πinma −e2πinbm)
=
∑
n6=0
an
√yKν(2π|n|y)e2πinx
1
m−1
∑
ψ(modm) ψ6=ψ0
τ(ψ)ψ(n)(ψ(a)−ψ(b))
= 1
m−1
∑
ψ(modm) ψ6=ψ0
τ(ψ)(ψ(a)−ψ(b))fψ(z). (44)
Comparing(43)and(44), we have
∑
ψ(modm) ψ6=ψ0
ψ(a)
τ(ψ)fψ(z)−χ(m)ψ(−N)τ(ψ)gψ
− 1 Nm2z
=
∑
ψ(modm) ψ6=ψ0
ψ(b)
τ(ψ)fψ(z)−χ(m)ψ(−N)τ(ψ)gψ
− 1 Nm2z
.
This shows that the expression on the left-hand side is independent of the choice of a satisfying (a,m) =1. Since the set of all characters modulo m is linearly independent, the coefficients must be zero. This can be concluded with the following proposition.
Proposition 5 Letψ be a primitive Dirichlet character of conductor m and(m,N) =1. If an,bn
andΛν(f,ψ,s),Λν(g,ψ,s)satisfy the assumption of Theorem 13, then fψ(z) =χ(m)ψ(−N)τ(ψ)
τ(ψ)gψ
− 1 Nm2z
=Cψ(gψ|0ωNm2)(z).
(Step 3) The most crucial difference appears when we prove modularity. To see this, we are going to review the results obtained through techniques used in Step 2 in the proof of Theorem 7. Note that the proof of these results do not depend much on the properties of the modular form, but rather on Dirichlet characters and matrix calculations.
For a primitive Dirichlet characterψ modulo m∈Pwe have fψ =τ(ψ)−1
m u=1
∑
ψ(u)f|0Tu/m
where Tr= 1 r 0 1
!
. We recall the matrix in(14) as follows. For two integers m,v such that (m,vN) =1, there exist two integers n,u such that mn−uvN =1. Then we get the following matrix.
γ(v/m) = m −v
−uN n
!
∈Γ0(N).
Following the proof of Lemma 4, and Lemma 3, we obtain the same result to Proposition 3 with respect to the assumptions in 13.
Proposition 6 Let m,n be distinct elements ofP. Then forγ= m −v
−uN n
!
andγ0= m v uN n
! , the matrix
β =γ−1T−2v/mγ0T−2v/m= 1 −2v/m 2uN/n −3+4/mn
!
is an elliptic matrix of infinite order. Let g be a holomorphic function onH satisfy assumptions in Theorem 13, and let h:=g|0(χ(m)−γ). Then we have
h|0β=h. (45)
In the proof of Lemma 3 (3), we used the fact that a holomorphic function onH which is invari- ant under an infinite order elliptic matrix is constant. Recall that an elliptic matrix has an unique fixed point inH. From this, geometrically speaking, we can say that a holomorphic function on H which is radially symmetric must be constant. However, this is not true in general for the non- holomorphic case. For example, there are spherical functions (formed by radially-symmetrizing Im(z)s).
Neurerer and Oliver resolve this problem using the following ‘two (hyperbolic) circles method’.
(They noted on their paper that this method was proposed by David Farmer.)
Theorem 14 (Theorem 3.11 in [28]) If f is a continuous function onH that is invariant under two infinite order elliptic matrices with distinct fixed points inH, then it is constant.
From Theorem 14, we can complete the proof of Theorem 13 by finding two infinite order elliptic matrices such that for allγ= a b
cN d
!
∈Γ0(N), g|0(χ(m)−γ)is invariant to these matrices.
Since a and d are prime to cN, there exist two integers s and t such that p1 =a+t1cN∈P and q=d+scN∈P. By the definition of P, there are infinitely many such p1 and q. Let r1=s(a+t1cN) +b+t1d. Then
g|0 a b cN d
!
=g|0 1 −t1
0 1
! p1 r1 cN q
! 1 −s
0 1
!
=g|0 p1 r1 cN q
! 1 −s
0 1
! .
By(45), the function h1:=g|0 χ(m)− p1 r1 cN q
!!
is invariant under the elliptic matrix
β1:= 1 −2r1/p1
−2cN/q −3+4/p1q
! .
Let p2=a+t2cN∈Pwhich is different to p1and q, and let r2=s(a+t2cN) +b+t2d. Observe that
g|0 p2 r2 cN q
!
=g|0 1 t2 0 1
! a b cN d
! 1 s 0 1
!
=g|0 a b cN d
! 1 s 0 1
!
=g|0 p1 r1 cN q
! .
Again by(45), h2:=g|0 χ(m)− p2 r2 cN q
!!
=h1is invariant under the elliptic matrix
β2:= 1 −2r2/p2
−2cN/q −3+4/p2q
! .
We can compute the fixed points ofβ1andβ2as follows.
z1=− q cN
1 p1q−1
+i
s
− q2 c2N2
1 p1q−1
2
+ 4r1q p1cN, z2=− q
cN 1
p2q−1
+i s
− q2 c2N2
1 p2q−1
2
+ 4r2q p2cN.
Comparing real parts, we find z16=z2 if p1 6=p2. Hence by Theorem 14, h1 is constant. On the other hand, we can check that both g|0 p1 r1
cN q
!
and g are eigenfunctions of the hyperbolic Laplacian ∆with the same positive eigenvalue. Since constant functions have eigenvalue0, h1 must be identically zero. Through this, we conclude that
g|0 a b cN d
!
=g|0 p1 r1 cN q
! 1 −s
0 1
!
=χ(m)g|0 1 −s
0 1
!
=χ(m)g.
This completes the proof.
In the same paper, Neurerer and Oliver proved a slightly different version of converse theorem (Theorem 1.1 in [28]). Moreover, they applied this to show that the quotient of the symmetric square L-function of a Maass newform and the Riemann zeta function has infinitely many poles (Corollary 1.2 in [28]).