Resolvent Trace Formula and Determinants of Laplacians on Orbifold Riemann Surfaces
For a nonnegative integer, we consider the -Laplacian Δₙ acting on the space of -differentials on a confinite Riemann surface X which has ramification points. The trace formula for the resolvent kernel is developed along the line à la Selberg. Using the trace formula, we compute the regularized...
Saved in:
| Published in: | Symmetry, Integrability and Geometry: Methods and Applications |
|---|---|
| Date: | 2021 |
| Main Author: | |
| Format: | Article |
| Language: | English |
| Published: |
Інститут математики НАН України
2021
|
| Online Access: | https://nasplib.isofts.kiev.ua/handle/123456789/211444 |
| Tags: |
Add Tag
No Tags, Be the first to tag this record!
|
| Journal Title: | Digital Library of Periodicals of National Academy of Sciences of Ukraine |
| Cite this: | Resolvent Trace Formula and Determinants of Laplacians on Orbifold Riemann Surfaces. Lee-Peng Teo. SIGMA 17 (2021), 083, 40 pages |
Institution
Digital Library of Periodicals of National Academy of Sciences of Ukraine| _version_ | 1859723033553403904 |
|---|---|
| author | Teo, Lee-Peng |
| author_facet | Teo, Lee-Peng |
| citation_txt | Resolvent Trace Formula and Determinants of Laplacians on Orbifold Riemann Surfaces. Lee-Peng Teo. SIGMA 17 (2021), 083, 40 pages |
| collection | DSpace DC |
| container_title | Symmetry, Integrability and Geometry: Methods and Applications |
| description | For a nonnegative integer, we consider the -Laplacian Δₙ acting on the space of -differentials on a confinite Riemann surface X which has ramification points. The trace formula for the resolvent kernel is developed along the line à la Selberg. Using the trace formula, we compute the regularized determinant of Δₙ + ( + 2 − 1), from which we deduce the regularized determinant of Δₙ, denoted by det′Δₙ. Taking into account the contribution from the absolutely continuous spectrum, det′Δₙ is equal to a constant Cₙ times ( ) when ≥ 2. Here ( ) is the Selberg zeta function of . When = 0 or = 1, ( ) is replaced by the leading coefficient of the Taylor expansion of ( ) around = 0 and = 1, respectively. The constants Cn are calculated explicitly. They depend on the genus, the number of cusps, as well as the ramification indices, but are independent of the moduli parameters.
|
| first_indexed | 2026-03-15T10:21:32Z |
| format | Article |
| fulltext |
Symmetry, Integrability and Geometry: Methods and Applications SIGMA 17 (2021), 083, 40 pages
Resolvent Trace Formula and Determinants
of n Laplacians on Orbifold Riemann Surfaces
Lee-Peng TEO
Department of Mathematics, Xiamen University Malaysia,
Jalan Sunsuria, Bandar Sunsuria, 43900, Sepang, Selangor, Malaysia
E-mail: lpteo@xmu.edu.my
Received April 07, 2021, in final form September 05, 2021; Published online September 13, 2021
https://doi.org/10.3842/SIGMA.2021.083
Abstract. For n a nonnegative integer, we consider the n-Laplacian ∆n acting on the space
of n-differentials on a confinite Riemann surface X which has ramification points. The trace
formula for the resolvent kernel is developed along the line à la Selberg. Using the trace
formula, we compute the regularized determinant of ∆n + s(s + 2n − 1), from which we
deduce the regularized determinant of ∆n, denoted by det′∆n. Taking into account the
contribution from the absolutely continuous spectrum, det′∆n is equal to a constant Cn
times Z(n) when n ≥ 2. Here Z(s) is the Selberg zeta function of X. When n = 0 or n = 1,
Z(n) is replaced by the leading coefficient of the Taylor expansion of Z(s) around s = 0
and s = 1 respectively. The constants Cn are calculated explicitly. They depend on the
genus, the number of cusps, as well as the ramification indices, but is independent of the
moduli parameters.
Key words: determinant of Laplacian; n-differentials; cocompact Riemann surfaces; Selberg
trace formula
2020 Mathematics Subject Classification: 14H15; 11F72; 11M36
Dedicated to Professor Leon Takhtajan
on the occasion of his 70th birthday
1 Introduction
Let H be the upper half plane, and let Γ be a Fuchsian group, a discrete subgroup of PSL(2,R).
The group Γ acts discontinuously on H and the quotient X = Γ\H is a Riemann surface. In this
work, we consider the case where X is a cofinite Riemann surface. In other words, X has finite
hyperbolic volume. This includes compact Riemann surfaces, as well as surfaces with finitely
many cusps and ramification points.
Let
∆0 = −y2
(
∂2
∂x2
+
∂2
∂y2
)
be the Laplacian operator acting on functions. It is an invariant operator under the action of
PSL(2,R).
In the seminal paper [18], Selberg developed a theory that can be used to study the spectrum
of ∆0 on Riemann surfaces, making full use of the invariant property of the Laplacian operator.
The theory was subsequently developed and elaborated in [7, 11, 12, 19, 23, 24, 25]. One of
the extensions of Selberg’s theory is to consider the spectrum of Dn for n ≥ 1, where Dn =
−Kn−1Ln−n(n− 1), and Kn and Ln are the Maass operators1 (see Section 2 for more details).
This paper is a contribution to the Special Issue on Mathematics of Integrable Systems: Classical and Quan-
tum in honor of Leon Takhtajan.
The full collection is available at https://www.emis.de/journals/SIGMA/Takhtajan.html
1Our Dn has a negative sign compared to that used in [6], while Fischer [7] denote it by −∆n.
mailto:lpteo@xmu.edu.my
https://doi.org/10.3842/SIGMA.2021.083
https://www.emis.de/journals/SIGMA/Takhtajan.html
2 L.-P. Teo
The significance of this theory is an identity that relates the spectral trace to the geometric
trace.
In mathematical physics, the regularized determinant of the Laplacian is of special interest.
This has been considered in [4, 17] for compact Riemann surfaces. Subsequently, Efrat [5] exten-
ded the result to smooth Riemann surfaces with cusps. In the special case of arithmetic surfaces,
this has also been considered in [14, 15]. In [9], Gong derived the regularized determinant in full
generality using the trace formula developed in [7] directly, which can be applied to cofinite
Riemann surfaces and for the operators Dn with unitary twists.
The goal of this paper is to study the explicit expression for the resolvent trace formula and
the regularized determinant of n-Laplacians ∆n on Riemann surfaces that have finite volumes.
∆n has the same spectrum as Dn + n(n − 1). In principle, one can extract the expression
for the trace formula and the determinant of ∆n from the formulas obtained in [7, 9, 12, 23].
However, we find that a direct approach from a theory for ∆n would be more appealing for
further application. This is the tasks undertaken in this paper.
2 Laplacians of n-differentials
Let H be the upper half of the complex plane equipped with the hyperbolic metric
ds2 =
dx2 + dy2
y2
.
The corresponding metric density is ρ(z) = (Im z)−2 and the area form is
dµ(z) =
dx dy
y2
= ρ(z) d2z.
The group PSL(2,R) acts on H transitively by Möbius transformations
γ =
(
a b
c d
)
: z 7→ γ(z) =
az + b
cz + d
.
The hyperbolic metric is an invariant metric under this group action.
The distance between two points z and w on H, denoted by d(z, w), is given by
cosh d(z, w) = 1 + 2u(z, w),
where
u(z, w) =
|z − w|2
4 Im z Imw
is a point-pair invariant, i.e.,
u(γz, γw) = u(z, w) for all γ ∈ PSL(2,R).
Let Γ be a cofinite Fuchsian group, namely, it is a discrete subgroup of PSL(2,R) such that
X = Γ\H is a Riemann surface with finite hyperbolic volume. Such Γ is finitely generated.
More precisely, if X is a genus g Riemann surface with q-punctures and v ramification points,
then Γ is generated by 2g hyperbolic elements α1, β1, . . . , αg, βg, q parabolic elements κ1, . . . , κq,
as well as v elliptic elements τ1, . . . , τv of orders m1, . . . ,mv respectively, with
2 ≤ m1 ≤ m2 ≤ · · · ≤ mv.
Resolvent Trace Formula and Determinants of n Laplacians 3
For each 1 ≤ j ≤ v, τ
mj
j = I. The generators of Γ satisfy the additional relation
α1β1α
−1
1 β−1
1 · · ·αgβgα
−1
g β−1
g κ1 · · ·κqτ1 · · · τv = I,
where I is the identity element. We say that the Riemann surface X and the group Γ are of
type (g; q;m1,m2, . . . ,mv).
Let K be the canonical bundle of X. For a nonnegative integer n, let S(n) be the space of
sections of Kn/2 ⊗ K̄−n/2. A function in S(n) can be realized as a function f : H → C satisfying
f(γz)
(
cz̄ + d
cz + d
)n
= f(z), for all γ =
(
a b
c d
)
∈ Γ.
In the context of analytic number theory, it is natural to consider the Maass operators Kn:
S(n) → S(n+ 1), Ln : S(n) → S(n− 1) and Dn : S(n) → S(n) given by [6]
Kn = (z − z̄)
∂
∂z
+ n,
Ln = − (z − z̄)
∂
∂z̄
− n,
Dn = −y2
(
∂2
∂x2
+
∂2
∂y2
)
+ 2iny
∂
∂x
.
These operators are invariant with respect to the PSL(2,R) action and
Dn+1Kn = KnDn, Dn−1Ln = LnDn,
Dn = −Ln+1Kn − n(n+ 1) = −Kn−1Ln − n(n− 1).
For applications of the theory of Riemann surfaces in physics, it is more natural to consider the
space of tensors f(z)(dz)n, or called n-differentials, on X. These are sections of Kn, which are
functions f : H → C satisfying
f(γz)γ′(z)n = f(z), for all γ ∈ Γ.
We denote this space by H2
n(Γ). It is a Hilbert space with the inner product
⟨f, g⟩ =
∫∫
X
f(z)g(z)ρ(z)−ndµ(z).
The n-Laplacian operator ∆n = 4∂̄∗n∂̄n is an operator on H2
n(Γ) with explicit formula given by
∆n = −4y2−2n ∂
∂z
y2n
∂
∂z̄
.
This is a positive operator. We put a factor 4 in front of ∂̄∗n∂̄n so that when n = 0, we get the
usual Laplacian on functions ∆0.
Throughout this paper, the nonnegative integer n is fixed. Our goal is to study the trace
formula for the resolvent of ∆n and the regularized determinant of ∆n. The isometry I:
H2
n(Γ) → S(n) defined by
f(z) 7→ ynf(z)
conjugates ∆n with Dn + n(n− 1). Hence, ∆n and Dn + n(n− 1) have the same spectrum.
For a Riemann surface with cusps, the spectrum of ∆n consists of a discrete part and a con-
tinuous part. To study the spectrum of the continuous part, we need to consider Eisenstein
series, which we discuss in next section.
4 L.-P. Teo
A powerful tool to study the spectrum of ∆n is the Selberg trace formula. The Selberg trace
formula for the operator Dn has been developed extensively in [7, 12, 23]. It can be adapted
to ∆n. To find the determinants of Laplacian, we will follow Fischer [7] and derive the trace
formula for the resolvent kernel of the Laplacian operator first.
Let Ψ: R → C be a function that vanishes at infinity and satisfies other regularity conditions
to be specified when needed. Define k : H×H → C by
k(z, w) = Ψ(u(z, w))
(−4)n
(z − w̄)2n
= Ψ(u(z, w))Hn(z, w), (2.1)
where
Hn(z, w) =
(−4)n
(z − w̄)2n
.
Notice that Hn(z, z) = ρ(z)n. Since Hn(γz, γw)γ
′(z)nγ′(w)
n
= Hn(z, w), we find that
k(γz, γw)γ′(z)nγ′(w)
n
= k(z, w).
In other words, k(z, w) is a point-pair invariant kernel on H.
Given s ∈ C, the function f(z) = (Im(z))s−n is an “eigenfunction” of ∆n with eigenvalue
−(s− n)(s+ n− 1). The following proposition shows that f(z) is also an eigenfunction for the
operator defined by the point-pair invariant kernel k(z, w).
Proposition 2.1. Let Ψ: R → C be a continuous function with compact support and let
k(z, w) = Ψ(u(z, w))Hn(z, w)
be the corresponding point-pair invariant kernel. If s ∈ C, then∫∫
H
k(z, w)(Imw)s−n(Imw)2ndµ(w) = Λs(Im z)s−n,
where
Λs =
∫∫
H
k(i, w)(Imw)s−n(Imw)2ndµ(w).
Proof. Given z ∈ H, notice that γ =
(
a b
0 d
)
, with 1
d = a =
√
Im z and ab = Re z maps i to z.
For such γ, γ′(z) = a2 = Im z. Let
g(z) =
∫∫
H
k(z, w)(Imw)s−n(Imw)2ndµ(w).
Using k(γi, γw)γ′(i)nγ′(w)
n
= k(i, w) and Im(γw) = Im(w)|γ′(w)|, a change of variables w 7→ γw
gives
g(z) =
∫∫
H
k(γi, γw)(Im γw)s+ndµ(w) = a2s−2n
∫∫
H
k(i, w)(Imw)s+ndµ(w)
= Λs(Im z)s−n,
where
Λs =
∫∫
H
k(i, w)(Imw)s+ndµ(w). ■
Resolvent Trace Formula and Determinants of n Laplacians 5
Given the point-pair invariant kernel k(z, w) on H, we can define a point-pair invariant ker-
nel K(z, w) for the Riemann surface X by
K(z, w) =
∑
γ∈Γ
k(γz, w)γ′(z)n, z, w ∈ H. (2.2)
Notice that if f ∈ H2
n(Γ), then∫∫
X
K(z, w)f(w)ρ(w)−ndµ(w) =
∫∫
H
k(z, w)f(w)ρ(w)−ndµ(w).
We want to find a function Ψn,s(u) = Ψ(u) so that the corresponding K(z, w) is the resolvent
kernel of the operator ∆n+s(s+2n−1). The reason to use s(s+2n−1) instead of (s−n)(s+n−1)
is so that when s = 0, we get ∆n. This function Ψ(u) must satisfy the differential equation
∆n(Ψ(u(z, w))Hn(z, w)) = −s(s+ 2n− 1)Ψ(u(z, w))Hn(z, w).
Using the fact that
∆n(Ψ(u(z, w))Hn(z, w)) =
(
−u(u+ 1)
∂2Ψ
∂u2
+ [(2n− 2)u− 1]
∂Ψ
∂u
)
Hn(z, w),
we find that
u(u+ 1)
∂2Ψ
∂u2
− [(2n− 2)u− 1]
∂Ψ
∂u
− s(s+ 2n− 1)Ψ = 0.
A solution is given by
Ψn,s(u) =
(u+ 1)−s
4π
Γ(s)Γ(s+ 2n)
Γ(2s+ 2n)
2F1
(
s, s+ 2n
2s+ 2n
;
1
u+ 1
)
, (2.3)
where
2F1
(
a, b
c
; z
)
=
Γ(c)
Γ(a)Γ(b)
∞∑
k=0
Γ(a+ k)Γ(b+ k)
Γ(c+ k)
zk
k!
is the hypergeometric function. The normalization constant 1/4π is chosen so that when u→ 0+,
Ψ(u) ∼ 1
4π
log
1
u
.
More explicitly, one can show that as u→ 0+,
Ψn,s(u) =
1
4π
{
log
1
u
+ 2ψ(1)− ψ(s+ 2n)− ψ(s)
}
+O(u). (2.4)
Here
ψ(s) =
Γ′(s)
Γ(s)
is the logarithmic derivative of the gamma function Γ(s), and ψ(1) = −γ, where γ is the Euler
constant.
6 L.-P. Teo
3 The Eisenstein series
For Riemann surfaces that are not compact, it is well-known that the spectrum of the Laplacians
contain a continuous part, which are related to the Eisenstein series.
The Riemann surface X has q cusps corresponding to the q parabolic elements κ1, . . . , κq.
For 1 ≤ i ≤ q, let xi ∈ R∪ {∞} be the fixed point of κi. Then xi is a representative of the cusp
associated to κi. Let σi ∈ PSL(2,R) be an element that conjugates κi to
(
1 ±1
0 1
)
, namely,
σ−1
i κiσi =
(
1 ±1
0 1
)
.
Then σi(∞) = xi. If B is the parabolic subgroup generated by
(
1 1
0 1
)
, then Γi = σiBσ
−1
i is the
stabilizer of the cusp xi in Γ.
Define the Eisenstein series associated to the cusp xi by
Ei(z, s;n) =
∑
γ∈Γi\Γ
[
Im
(
σ−1
i γz
)]s−n[(
σ−1
i γ
)′
(z)
]n
,
when Re s > 1. Here to simplify notation, we write σ−1 ◦ γ as σ−1γ, and we write
(
σ−1 ◦ γ
)
(z)
as σ−1γz. One can check that the definition of the Eisenstein series here differs from the one used
in [7, 12, 23] by the factor yn. This makes good sense in view of the isometry between H2
n(Γ)
and S(n) that we discussed earlier.
Using the fact that ∆ny
s−n = −(s− n)(s+ n− 1)ys−n, and ∆n is invariant with respect to
the action of PSL(2,R), we find that
∆nEi(z, s;n) = −(s− n)(s+ n− 1)Ei(z, s;n).
The theory of Eisenstein series has been developed extensively in the books [7, 12, 13]. In the
following, we follow [13] to give a brief exposition of the facts needed in this work.
Given 1 ≤ i, j ≤ q, there is a double coset decomposition of the group σ−1
i Γσj into disjoint
double cosets given by
σ−1
i Γσj = δijΩ∞ ∪
⋃
c>0
⋃
dmod c
Ωd/c.
Here Ω∞ contains all the upper triangular matrices in σ−1
i Γσj , and Ωd/c is the double coset
Bωd/cB, where ωd/c is an element of σ−1
i Γσj of the form
ωd/c =
(
∗ ∗
c d
)
.
Using this, one can show that when y → ∞,
Ei(σjz, s;n)σ
′
j(z)
n = δijy
s−n + φij(s;n)y
1−s−n + exponentially decaying terms,
where
φij(s;n) =
√
π
Γ(s)Γ
(
s− 1
2
)
Γ(s+ n)Γ(s− n)
∑
c>0
∑
dmod c
1
c2s
is a Dirichlet series. From this explicit expression, one finds that
φij(s;n) =
Γ(s)2
Γ(s+ n)Γ(s− n)
φij(s; 0). (3.1)
Namely, there is a simple relation between φij(s;n) and φij(s; 0).
Resolvent Trace Formula and Determinants of n Laplacians 7
The q × q matrix Φ(s) = [φij(s)] is called the scattering matrix and it plays an important
role in the spectral theory. It has the following properties:
Φ(s) = Φ(s̄), Φ(s)T = Φ(s).
Moreover, if we denote by Ξ(z, s;n) the column matrix with components Ei(z, s;n), then
Ξ(z, s;n) = Φ(s;n) Ξ(z, 1− s;n).
It follows that
Φ(s)Φ(1− s) = Iq.
Denote by φ(s) the determinant of Φ(s), namely, φ(s) = detΦ(s). Then we find that
φ(s) = φ(s̄), φ(s)φ(1− s) = 1.
4 The trace of the resolvent kernel
The spectral theory for the Riemann surface X = Γ\H states that there is a countable orthonor-
mal system {uk}k≥0 of eigenfunctions of ∆n with eigenvalues 0 = λ0 ≤ λ1 ≤ λ2 ≤ · · · , and
eigenpackets given by the Eisenstein series so that for any f ∈ H2
n(Γ),
f(z) =
∞∑
k=0
⟨f, uk⟩uk(z) +
1
4π
q∑
j=1
∫ ∞
−∞
〈
f,Ej
(
·, 1
2
+ ir;n
)〉
Ej
(
z,
1
2
+ ir;n
)
dr.
We want to use this formula to find the trace of the resolvent (∆n + s(s+2n− 1))−1. However,
the kernel function K(z, w) for this resolvent defined by Ψn,s(u) in (2.3) has singularity along
z = w. To circumvent this problem, we consider the kernel function K(z, w) with Ψ(u) =
Ψn,s(u)−Ψn,a(u), for some fixed a. For this kernel function,
⟨K(·, w), uk⟩ = Λ(λk)uk(w),〈
K(·, w), Ej
(
·, 1
2
+ ir;n
)〉
= Λ̃(r)Ej
(
w,
1
2
+ ir;n
)
,
where
Λ(λk) =
1
λk + s(s+ 2n− 1)
− 1
λk + a(a+ 2n− 1)
,
Λ̃(r) =
1(
s+ n− 1
2
)2
+ r2
− 1(
a+ n− 1
2
)2
+ r2
.
Therefore, the spectral decomposition of K(z, w) is
K(z, w) =
∞∑
k=0
Λ(λk)uk(z)uk(w) +
1
4π
q∑
j=1
∫ ∞
−∞
Λ̃(r)Ej
(
z,
1
2
+ ir;n
)
Ej
(
w,
1
2
+ ir;n
)
dr.
Setting z = w, we have
K(z, z) =
∞∑
k=0
Λ(λk)|uk(z)|2 +
1
4π
q∑
j=1
∫ ∞
−∞
Λ̃(r)
∣∣∣∣Ej
(
z,
1
2
+ ir;n
)∣∣∣∣2dr.
8 L.-P. Teo
Integrating over X, one would have
∞∑
k=0
Λ(λk) =
∫∫
X
(
K(z, z)− 1
4π
q∑
j=1
∫ ∞
−∞
Λ̃(r)
∣∣∣∣Ej
(
z,
1
2
+ ir;n
)∣∣∣∣2dr
)
y2ndµ(z).
As in [13], some regularizations are needed to make this integral well-defined. This will be
explained in the following.
By definition (2.1) and (2.2), K(z, w) can be written as a series
K(z, w) =
∑
γ∈Γ
k(γz, w)γ′(z)n,
where
k(z, w) = (Ψn,s(u(z, w))−Ψn,a(u(z, w)))
(−4)n
(z − w̄)2n
,
with Φn,s(u) defined in (2.3). Decompose the elements of the group Γ into the set that con-
tains only the identity, and the three sets that contain respectively hyperbolic elements, elliptic
elements and parabolic elements, we have
K(z, w) = k(z, w) +
∑
γ∈Γ
γ is hyperbolic
k(γz, w)γ′(z)n +
∑
γ∈Γ
γ is elliptic
k(γz, w)γ′(z)n
+
∑
γ∈Γ
γ is parabolic
k(γz, w)γ′(z)n.
Therefore,
∞∑
k=0
Λ(λk) = Ξ0 + ΞH + ΞE + ΞP , (4.1)
where
Ξ0 =
∫∫
X
k(z, z)y2ndµ(z),
ΞH =
∫∫
X
∑
γ∈Γ
γ is hyperbolic
k(γz, z)γ′(z)ny2ndµ(z),
ΞE =
∫∫
X
∑
γ∈Γ
γ is elliptic
k(γz, z)γ′(z)ny2ndµ(z).
The term ΞP contains the parabolic contribution as well as the absolutely continuous spectrum.
We need to do some regularization to make it finite. Let F be a fundamental domain of X on H.
Given Y > 0, let
F Y = F \
q⋃
j=1
F Y
j ,
where
F Y
j = F ∩ σj
(
{x+ iy | y > Y }
)
.
Resolvent Trace Formula and Determinants of n Laplacians 9
Then we define
ΞP = lim
Y→∞
{∫∫
FY
∑
γ∈Γ
γ is parabolic
k(γz, w)γ′(z)ny2ndµ(z)
− 1
4π
q∑
j=1
∫ ∞
−∞
Λ̃(r)
∫∫
FY
∣∣∣∣Ej
(
z,
1
2
+ ir;n
)∣∣∣∣2 y2ndµ(z) dr
}
.
In Appendix D, we show that this limit indeed exist.
By (2.4), we have
k(z, z)y2n = − 1
4π
{
ψ(s+ 2n) + ψ(s)− ψ(a+ 2n)− ψ(a)
}
.
Therefore,
Ξ0 = −|X|
4π
{
ψ(s+ 2n) + ψ(s)− ψ(a+ 2n)− ψ(a)
}
,
where |X| is the hyperbolic area of X given by
|X| = 2π
{
2g − 2 + q +
v∑
j=1
(
1− 1
mj
)}
.
The computations of ΞH , ΞE and ΞP are much more complicated. We leave them to Appen-
dices B, C and D, and quote the results here.
For the hyperbolic contributions, let P be the primitive hyperbolic conjugacy classes in Γ.
For a representative γ of a primitive hyperbolic class, let N(γ) > 1 be the multiplier of γ. Then
ΞH = EH(s)− EH(a), where
EH(s) =
1
2s+ 2n− 1
∑
[γ]∈P
∞∑
k=0
logN(γ)
N(γ)s+n+k − 1
.
The elliptic contribution is given by ΞE = EE(s)− EE(a), where
EE(s) =
1
2s+ 2n− 1
v∑
j=1
mj−1∑
r=0
[
2αmj (r − n) + 1−mj
2m2
j
ψ
(
s+ r
mj
)
+
2αmj (r + n) + 1−mj
2m2
j
ψ
(
s+ 2n+ r
mj
)]
.
If m is a positive integer greater than 1, and k is an integer, αm(k) is defined to be the least
positive residue modulo m. Namely, it is the smallest nonnegative integer congruent to k mo-
dulo m.
Finally, we have ΞP = EP (s)− EP (a), where
EP (s) =
1
(2s+ 2n− 1)2
[
q − TrΦ
(
1
2
)]
+
1
2
Σ(s)
+
q
2(2s+ 2n− 1)
{
ψ(s) + ψ(s+ 2n)− 2 log 2− 2ψ
(
s+ n+
1
2
)
− 2ψ(s+ n)
}
,
where
Σ(s) =
1
2π
∫ ∞
−∞
1
r2 +
(
s+ n− 1
2
)2 φ′
φ
(
1
2
+ ir
)
dr.
Gathering all the results, we obtain the resolvent trace formula for ∆n.
10 L.-P. Teo
Theorem 4.1 (the resolvent trace formula). The resolvent trace formula of the n-Laplacian ∆n
on the Riemann surface X is given by
TS(s)− TS(a) = TG(s)− TG(a),
where TS(s) and TG(s) are respectively the spectral trace and the geometric trace given by
TS(s) =
∞∑
k=0
1
λk + s(s+ 2n− 1)
− 1
2
Σ(s),
TG(s) =− |X|
4π
[
ψ(s+ 2n) + ψ(s)
]
+
1
2s+ 2n− 1
∑
[γ]∈P
∞∑
k=0
logN(γ)
N(γ)s+n+k − 1
+
1
2s+ 2n− 1
v∑
j=1
mj−1∑
r=0
[
2αmj (r − n) + 1−mj
2m2
j
ψ
(
s+ r
mj
)
+
2αmj (r + n) + 1−mj
2m2
j
ψ
(
s+ 2n+ r
mj
)]
+
q
2(2s+ 2n− 1)
[
ψ(s) + ψ(s+ 2n)− 2 log 2− 2ψ
(
s+ n+
1
2
)
− 2ψ(s+ n)
]
+
A
(2s+ 2n− 1)2
,
and A is the constant
A = Tr
[
I − Φ
(
1
2
)]
.
In [22], we have explained that A is an even nonnegative integer. In applying the resolvent
trace formula, we always take a to be a positive constant that is large enough.
As mentioned in the introduction, ∆n has the same spectrum as Dn + n(n − 1). Therefore,
∆n + s(s+ 2n− 1) has the same spectrum as
Dn + s(s+ 2n− 1) + n(n− 1) = Dn + (s+ n)(s+ n− 1).
Hence, if we replace the s in the resolvent trace formula for Dn that were derived in [7, 12]
by s+ n, one should get the resolvent trace formula for ∆n obtained in Theorem 4.1.
In principle, we can directly quote the formulas in [7, 12] and proceed directly to derive the
regularized determinants of n-Laplacians. However, we have found some inconsistencies in the
results in these two references. One of the goals of this work is to resolve this inconsistency, giving
a concise reference to the computations that lead to the resolvent trace formula, supplementing
the n = 0 case done in the book [13]. When doing so, we find that it is more natural to consi-
der ∆n which is a matter of taste. The use of ∆n+s(s+2n−1) instead of ∆n+(s−n)(s+n−1)
is so that setting s = 0 in ∆n + s(s+ 2n− 1) give ∆n.
5 Dimensions of spaces holomorphic differentials
For n ≥ 1, the space of holomorphic n-differentials is the subspace of H2
n(Γ) which are holomor-
phic. Hence, they are eigenvectors of ∆n with eigenvalue 0. We also call them zero modes or
holomorphic cusp forms of weight 2n.
Let dn be the dimension of the space of holomorphic n-differentials of X. Although the
formula for dn is well-known, we would like to show how this can be derived from the resolvent
trace formula, similar to what have been done in [12].
Resolvent Trace Formula and Determinants of n Laplacians 11
When n = 0, d0 is the dimension of space of holomorphic functions, which is equal to 1,
corresponding to the constant functions.
Notice that the hyperbolic contribution to the resolvent trace formula can be written as
ΞH =
1
2s+ 2n− 1
d
ds
logZ(s+ n)− 1
2a+ 2n− 1
d
ds
logZ(a+ n),
where Z(s) is the Selberg zeta function of X defined as
Z(s) =
∏
[γ]∈P
∞∏
k=0
(
1−N(γ)−s−k
)
.
It is well known that Z(s) has a zero of order 1 at s = 1 and it is absolutely convergent
when Re s > 1. Therefore, when n = 1, d
ds logZ(s + n) has residue 1 at s = 0. When n ≥ 2,
d
ds logZ(s+ n) has residue 0 at s = 0.
As we explained above, the dimension of the space of holomorphic n-differentials dn is equal
to the multiplicity of 0 as a discrete eigenvalue of ∆n. According to Theorem D.1, the residue
of Σ(s) at s = 0 is 0. Hence, dn is equal to (2n−1) times the residue at s = 0 of the function on
the left hand side of the resolvent trace formula. To determine the value of dn from the resolvent
trace formula, we need to find the residue at s = 0 of the function at the right-hand side of
the trace formula. It then amounts to understanding the zeros and poles of Z(s) at positive
integers, as well as the residues of ψ(s) at rational numbers.
Since
ψ(s) = −γ +
∞∑
k=0
(
1
k + 1
− 1
s+ k
)
,
we find that the residues of ψ(s) at s = 0,−1,−2, . . . are all equal to −1. ψ(s) does not have
poles at other points.
Using the fact that αmj (−1) = mj − 1, we find that
d1 =
1
2
[
2g − 2 + q +
v∑
j=1
(
1− 1
mj
)]
+ 1− 1
2
v∑
j=1
(
1− 1
mj
)
− q
2
= g,
which is a well-known result, since the space of holomorphic one-differentials is exactly the space
of abelian differentials.
When n ≥ 2,
dn =
2n− 1
2
[
2g − 2 + q +
v∑
j=1
(
1− 1
mj
)]
+
1
2
v∑
j=1
mj − 1− 2αmj (−n)
mj
− q
2
.
Since
αmj (−n)
mj
= − n
mj
−
⌊
− n
mj
⌋
,
we find that when n ≥ 2,
dn = (2n− 1)(g − 1) + (n− 1)q +
v∑
j=1
(⌊
n− n
mj
⌋)
.
12 L.-P. Teo
6 The determinant of n-Laplacian
In this section, we want to derive our main result – the formula for the determinant of n-
Laplacian ∆n for a cofinite Riemann surface X in terms of the Selberg zeta function for the
Riemann surface. This extends our result in [22] to the case where n ≥ 1. The results are
not entirely new. For compact hyperbolic surfaces, the relation between the determinant of ∆n
and the Selberg zeta function have been obtained by D’Hoker and Phong [4] and Sarnak [17].
In [5], Efrat considered the determinant of ∆0+ s(s− 1) for cofinite hyperbolic surfaces without
elliptic points. For congruence subgroups Γ0(N), Γ1(N) and Γ(N), Koyama has obtained the
relation in his work [14, 15]. Gong has considered the more general case of Laplacian operators
on automorphic forms of nonzero weights in [9], but this work has not attracted much attention.
We would first derive the determinant of ∆n + s(s + 2n − 1). Since ∆n contains absolutely
continuous spectrum, we need to be careful with defining the determinant. Following [5, 15, 24],
let
ζ(w, s) =
∞∑
k=0
1
(λk + s(s+ 2n− 1))w
− 1
4π
∫ ∞
−∞
1[(
s+ n− 1
2
)2
+ r2
]w φ′
φ
(
1
2
+ ir
)
dr
be the spectral zeta function of X correspond to ∆n+s(s+2n−1). This expression is well-defi-
ned when Rew is large enough. It can be analytically continued to a neighbourhood of w = 0.
The zeta regularized determinant det(∆n + s(s+ 2n− 1)) is defined as
det((∆n + s(s+ 2n− 1)) = exp (−ζw(0, s)) .
By uniqueness of analytic continuation, we find that
d
ds
1
2s+ 2n− 1
d
ds
log det(∆n + s(s+ 2n− 1)) =
d
ds
1
2s+ 2n− 1
d
ds
(−ζw(0, s))
=
d
ds
{ ∞∑
k=0
[
1
λk + s(s+ 2n− 1)
− 1
λk + a(a+ 2n− 1)
]
− 1
4π
∫ ∞
−∞
[
1(
s+ n− 1
2
)2
+ r2
− 1(
a+ n− 1
2
)2
+ r2
]
φ′
φ
(
1
2
+ ir
)
dr
}
.
To compute this using the resolvent trace formula, we first recall that the Alekseevskii–Barnes
double gamma function Γ2(s) is defined as [1, 3]:
Γ2(s+ 1) =
1
(2π)
s
2
e
s
2
+ γ+1
2
s2
∞∏
k=1
(
1 +
s
k
)−k
es−
s2
2k .
It satisfies the equation
Γ2(s+ 1) =
Γ2(s)
Γ(s)
.
Using
d
ds
log Γ2(s+ 1) = −1
2
log(2π) +
1
2
+ (γ + 1)s− s
∞∑
k=1
(
1
k
− 1
s+ k
)
,
and
ψ(s) =
d
ds
log Γ(s) = −γ − 1
s
−
∞∑
k=1
(
1
s+ k
− 1
k
)
,
Resolvent Trace Formula and Determinants of n Laplacians 13
we find that
d
ds
log
(2π)2s+2n−1Γ2(s+ 2n)2Γ2(s)
2Γ(s+ 2n)2n−1
Γ(s)2n+1
= −(2s+ 2n− 1)(ψ(s+ 2n) + ψ(s)− 2).
Hence, the resolvent trace formula says that
d
ds
1
2s+ 2n− 1
d
ds
log det(∆n + s(s+ 2n− 1))
=
d
ds
1
2s+ 2n− 1
d
ds
log
[
Z∞(s)Z(s+ n)Zell(s)Zpar(s)
]
,
where
Z∞(s) =
[
(2π)2s+2n−1Γ2(s+ 2n)2Γ2(s)
2Γ(s+ 2n)2n−1
Γ(s)2n+1
] |X|
4π
,
Zell(s) =
v∏
j=1
mj−1∏
r=0
Γ
(
s+ r
mj
) 2αmj (r−n)+1−mj
2mj
Γ
(
s+ 2n+ r
mj
) 2αmj (r+n)+1−mj
2mj
=
v∏
j=1
∏mj−1
r=0 Γ
(
s+r
mj
)αmj (r−n)
mj Γ
(
s+2n+r
mj
)αmj (r+n)
mj[
(2π)mj−1m
−(2s+2n−1)
j Γ(s+ 2n)Γ(s)
] 1
2
(
1− 1
mj
) ,
Zpar(s) =
[
Γ(s)Γ(s+ 2n)
22s+2n−1Γ(s+ n)2Γ
(
s+ n+ 1
2
)2]q/2(s+ n− 1
2
)A
2
. (6.1)
It follows that
det(∆n + s(s+ 2n− 1)) = Z∞(s)Z(s+ n)Zell(s)Zpar(s)e
B(s+n− 1
2)
2
+D
for some constants B and D. To determine B and D, we need to study the behavior of both
sides when s→ ∞.
First we define the heat kernel
θ(t) =
∞∑
k=0
e−tλk − 1
4π
∫ ∞
−∞
e
−t
(
r2+[n− 1
2 ]
2
)
φ′
φ
(
1
2
+ ir
)
dr
for t > 0. Notice that
ζ(w, s) =
1
Γ(w)
∫ ∞
0
tw−1θ(t)e−t(s(s+2n−1))dt.
Although we can derive the explicit formula for θ(t) using the general trace formula, but we do
not need it. We only need the asymptotic expansion of θ(t) as t → 0+. By comparison to the
n = 0 or the general theory about heat kernel, we know that the asymptotic expansion of θ(t)
has the form
θ(t) =
(
b
t
+
c log t√
t
+
d√
t
+ h
)
e−t(n− 1
2)
2
+O
(√
t
)
for some constants b, c, d and h. Let
u = s+ n− 1
2
.
14 L.-P. Teo
Then as u→ ∞,
log det(∆n + s(s+ 2n− 1)) ∼ − ∂
∂w
∣∣∣∣
w=0
1
Γ(w)
∫ ∞
0
tw−1
(
b
t
+
c log t√
t
+
d√
t
+ h
)
e−tu2
dt.
Using
∂
∂w
∣∣∣∣
w=0
1
Γ(w)
∫ ∞
0
tw−1e−tu2
dt = −2 log u,
∂
∂w
∣∣∣∣
w=0
1
Γ(w)
∫ ∞
0
tw−2e−tu2
dt = −u2 + 2u2 log u,
∂
∂w
∣∣∣∣
w=0
1
Γ(w)
∫ ∞
0
tw− 3
2 e−tu2
dt = −2
√
πu,
∂
∂w
∣∣∣∣
w=0
1
Γ(w)
∫ ∞
0
tw− 3
2 log t e−tu2
dt = −2
√
πu(2− 2 log 2− γ − 2 log u),
we find that as u→ ∞,
log det(∆n + s(s+ 2n− 1)) = bu2 − 2bu2 log u+ 2c
√
πu(2− 2 log 2− γ − 2 log u)
+ 2d
√
πu+ 2h log u+ o(1).
On the other hand, it is obvious from definition that logZ(s + n) is o(1) as s → ∞. From the
asymptotic behaviors
log Γ(s) =
(
s− 1
2
)
log s− s+
1
2
log 2π + o(1),
log Γ2(s+ 1) = −1
2
s2 log s+
3
4
s2 − s
2
log(2π) +
1
12
log s− ζ ′(−1) + o(1), (6.2)
we can deduce the following. As s→ ∞,
logZ∞(s) =
|X|
4π
{(
−2u2 + 2n2 − 1
6
)
log u+ 3u2 − 4ζ ′(−1)
}
+ o(1),
logZpar(s) =
q
2
{−(2u+ 1) log u+ 2u− log(2π)− 2u log 2}+ A
2
log u+ o(1).
The asymptotic behavior of Zell(s) is the most complicated one. We leave the computation to
Appendix F and quote the result here. We find that as s→ ∞,
logZell(s) = B log u+ D + o(1),
where
B =
v∑
j=1
(
m2
j − 1
6mj
−
αmj (n)(mj − αmj (n))
mj
)
,
D = −
v∑
j=1
(
m2
j − 1
6mj
−
αmj (n)(mj − αmj (n))
mj
)
logmj .
Resolvent Trace Formula and Determinants of n Laplacians 15
Hence, we find that as s→ ∞,
log det(∆n + s(2 + 2n− 1))
=
|X|
4π
{(
−2u2 + 2n2 − 1
6
)
log u+ 3u2 − 4ζ ′(−1)
}
+ B log u+ D
+
q
2
{
−(2u+ 1) log u+ 2u− log(2π)− 2u log 2
}
+
A
2
log u+Bu2 +D + o(1)
= bu2 − 2bu2 log u+ 2c
√
πu(2− 2 log 2− γ − 2 log u) + 2d
√
πu+ 2h log u+ o(1).
Comparing the two asymptotic expansions give
B = −|X|
2π
,
D =
|X|
π
ζ ′(−1) +
q
2
log(2π) +
v∑
j=1
(
m2
j − 1
6mj
−
αmj (n)(mj − αmj (n))
mj
)
logmj . (6.3)
This gives the following result.
Theorem 6.1. Let n be a nonnegative integer and let ∆n be the n-Laplacian of the Riemann
surface X. Then
det (∆n + s(s+ 2n− 1)) = Z∞(s)Z(s+ n)Zell(s)Zpar(s)e
B(s+n− 1
2)
2
+D,
where Z(s) is the Selberg zeta function, Z∞(s), Zell(s) and Zpar(s) are defined in (6.1), and the
constants B and D are given by (6.3).
Since ∆n has zero eigenvalues, we need to remove these zero modes when we define the
regularized determininant of ∆n. When n ≥ 2, φ(s+ n) is regular when s = 0. Hence, a reaso-
nable definition is
det′∆n = lim
s→0
det(∆n + s(s+ 2n− 1))
[s(s+ 2n− 1)]dn
=
1
(2n− 1)dn
lim
s→0
det(∆n + s(s+ 2n− 1))
sdn
, (6.4)
where dn is the dimension of holomorphic n-differentials. When n = 1, Proposition D.1 shows
that we can also use (6.4) to define det′∆1.
The n = 0 case is more complicated. As discussed in [23], the possible zero of φ(s) at s = 0
would give some extra contribution. If n0 is the order of zero of φ(s) at s = 0, then we should
define
det′∆0 = lim
s→0
det(∆0 + s(s− 1))
[s(s− 1)]1−n0
= (−1)n0−1 lim
s→0
det(∆0 + s(s− 1))
s1−n0
.
Now we want to derive the formula for det′∆n from Theorem 6.1. We discuss the case n = 0
and n ≥ 1 separately. When n = 0, we find that when s→ 0,
Z∞(s) ∼
(
1
2π
) |X|
4π
s−
|X|
2π ,
Zell(s) ∼
v∏
j=1
s
mj−1
mj m
1−mj
mj
j
mj−1∏
r=1
Γ
(
r
mj
) 2r+1−mj
mj
,
Zpar(s) ∼
(
−1
2
)A
2
(
2
π
) q
2
.
16 L.-P. Teo
These imply that
det′∆0 = C0 lim
s→∞
Z(s)
s2g−1+q−n0
,
where
C0 = (−1)
A
2
+1−n02q−
A
2 (2π)−
q
2
− |X|
4π e
B
4
+D
v∏
j=1
m
1−mj
mj
j
mj−1∏
r=1
Γ
(
r
mj
) 2r+1−mj
mj
. (6.5)
When n ≥ 1, we find that as s→ 0,
Z∞(s) ∼ s(2n−1)
|X|
4π
[
(2π)2n−1Γ2(2n)
2Γ(2n)2n−1
] |X|
4π ,
Zell(s) ∼
v∏
j=1
{
s
mj−1−2αmj (−n)
2mj m
2αmj (−n)+1−mj
2mj
j
×
mj−1∏
r=1
Γ
(
r
mj
) 2αmj (r−n)+1−mj
2mj
mj−1∏
r=0
Γ
(
2n+ r
mj
) 2αmj (r+n)+1−mj
2mj
}
,
Zpar(s) ∼ s−
q
2
[
Γ(2n)
22n−1Γ(n)2Γ
(
n+ 1
2
)2]q/2(n− 1
2
)A
2
.
Using also the fact that Z(s) has a zero of order 1 at s = 1, and Z(n) is nonzero, we conclude
that when n = 1,
det′∆1 = C1Z ′(1),
while when n ≥ 2,
det′∆n = CnZ(n).
The constant Cn is given by
Cn =
[
(2π)2n−1Γ2(2n)
2Γ(2n)2n−1
] |X|
4π
×
v∏
j=1
{
m
2αmj (−n)+1−mj
2mj
j
mj−1∏
r=1
Γ
(
r
mj
) 2αmj (r−n)+1−mj
2mj
mj−1∏
r=0
Γ
(
2n+ r
mj
) 2αmj (r+n)+1−mj
2mj
}
×
[
22n−1
πΓ(2n)
]q/2
(2n− 1)−dn
(
n− 1
2
)A
2
eB(n−
1
2)
2
+D. (6.6)
Finally, we obtain the main result of our paper.
Theorem 6.2. When n ≥ 0, the regularized determinant of the n-Laplacian ∆n of X is given by
det′∆n =
C0Z0, n = 0,
C1Z ′(1), n = 1,
CnZ(n), n ≥ 2,
where
Z0 = lim
s→0
Z(s)
s2g−1+q−n0
,
C0 is given by (6.5) and for n ≥ 1, Cn is given by (6.6).
Resolvent Trace Formula and Determinants of n Laplacians 17
This establishes the exact relation between the regularized determinant det′∆n and the Sel-
berg zeta function. Notice that the constant Cn only depends on the type of the Riemann surface
and is independent of the moduli. In [20] and [21], Takhtajan and Zograf established the local in-
dex theorem for Riemann surfaces with cusps and with ramification points, using Z ′(1) and Z(n),
n ≥ 2, as defintions for det′∆1 and det′∆n, n ≥ 2 respectively. Since they only considered the sec-
ond variation of log det′∆n on the moduli space, Theorem 6.2 justifies their approach. However,
if one wants to consider the holomorphic factorization of the determinant of Laplacian, as con-
sidered in [16] for compact Riemann surfaces, the precise value of Cn becomes important. They
would also be important for the extension of the work [8] by Freixas i Montplet and von Pippich.
We would like to point out that our definition of the regularized determinant of Laplacian
has taken into account the contribution from the absolutely continuous spectrum. For a generic
cofinite Riemann surface with cusps, it is not known whether it has an infinite discrete spectrum.
In fact, it has been conjectured to be the opposite. The inclusion of the absolutely continuous
spectrum is essential to render the determinant to be equal to a moduli independent constant
times Z(n) (for n ≥ 2).
In the resolvent trace formula, the contribution from the absolutely continuous spectrum
depends on φ(s), the determinant of the scattering matrix, which is moduli dependent. It would
be interesting if one can compute the contribution of this term in the local index theorem
explicitly.
A The inversion formulas
In this appendix, we want to derive a useful formula that is needed in the computation of the
elliptic contribution to the trace equation (4.1) in Appendix C.
When
s =
1
2
+ ir,
we find that
−(s− n)(s+ n− 1) = r2 +
(
n− 1
2
)2
.
Given a function H (λ), define
h(r) = H
(
r2 +
(
n− 1
2
)2)
.
In this work, we let H (λ) be a function such that the function h(r) satisfies the following
conditions:
� h(r) is holomorphic in the strip | Im(r)| ≤ 1
2 + δ for some δ > 0,
� there exist positive constants M and ε such that |h(r)| ≤ M
(1+|r|)2+ε in the strip.
Using Proposition 2.1 with k(z, w) the kernel of the operator H (∆n), we find that
H (−(s− n)(s+ n− 1))
=
∫ ∞
0
∫ ∞
−∞
Ψ
(
x2 + (y − 1)2
4y
)
(−4)n
(x− i(y + 1))2n
ys+ndx dy
y2
= 2(−1)n
∫ ∞
0
∫ ∞
−∞
Ψ
(
x2 +
(y − 1)2
4y
)
1(
x− i
(y + 1)
2
√
y
)2n y
s− 3
2dx dy.
18 L.-P. Teo
Let
Q(v) = 2(−1)n
∫ ∞
−∞
Ψ(x2 + v)
1(
x− i
√
v + 1
)2ndx
= 2(−1)n
∫ ∞
0
Ψ(x2 + v)
(
x+ i
√
v + 1
)2n
+
(
x− i
√
v + 1
)2n(
x2 + 1 + v
)2n dx.
Then
h(r) =
∫ ∞
0
Q
(
(y − 1)2
4y
)
yir−1dy =
∫ ∞
−∞
Q
(
sinh2
t
2
)
eitrdt.
Define
g(t) = Q
(
sinh2
t
2
)
.
Then
h(r) =
∫ ∞
−∞
g(t)eitrdt.
Namely, h(r) is the Fourier transform of g(t). The theory of Fourier transform implies that
g(t) =
1
2π
∫ ∞
−∞
h(r)e−irtdr.
When
Ψ(u) = Ψn,s(u)−Ψn,a(u),
where Ψn,s(u) is defined in (2.3), we have
h(r) =
1
r2 +
(
n− 1
2
)2
+ s(s+ 2n− 1)
− 1
r2 +
(
n− 1
2
)2
+ a(a+ 2n− 1)
=
1
r2 +
(
s+ n− 1
2
)2 − 1
r2 +
(
a+ n− 1
2
)2 .
It is well-known that h(r) is the Fourier transform of the function
g(t) =
1
2s+ 2n− 1
e−t(s+n− 1
2) − 1
2a+ 2n− 1
e−t(a+n− 1
2).
Next we want to express Ψ(u) in terms of Q(v) in the particular case where Ψ(u) = Ψn,s(u) −
Ψn,a(u). Inspired by the formula in [11], we claim that
Ψ(x) = − 1
2π
∫ ∞
−∞
Q′(x+ t2
)(√
x+ 1 + t2 − t
)2n
dt
= − 1
2π
∫ ∞
0
Q′(x+ t2
) [(√
x+ 1 + t2 − t
)2n
+
(√
x+ 1 + t2 + t
)2n]
dt.
Since Ψn,s(u) is a linear combination of terms of the form
1
(u+ 1)α
,
it suffices to consider the case where
Ψ(x) =
1
(x+ 1)α
.
Resolvent Trace Formula and Determinants of n Laplacians 19
In this case,
Q(v) = 4(−1)n
∫ ∞
0
1
(x2 + v + 1)α+2n
n∑
r=0
(
2n
2r
)
(v + 1)n−r(−1)n−rx2rdx
= 2
n∑
r=0
(
2n
2r
)
(v + 1)n−r(−1)r
∫ ∞
0
xr−
1
2
(x+ v + 1)α+2n
dx
=
2
(v + 1)α+n− 1
2
n∑
r=0
(
2n
2r
)
(−1)r
Γ
(
r + 1
2
)
Γ
(
α+ 2n− r − 1
2
)
Γ(α+ 2n)
.
Some manipulations give
Q(v) =
2
(v + 1)α+n− 1
2
Γ
(
α+ 2n− 1
2
)
Γ(α+ 2n)
Γ
(
1
2
) n∑
r=0
(−n)r
(
−n+ 1
2
)
r(
1
2
)
r
r!
(
1
2
)
r(
−α− 2n+ 3
2
)
r
=
2
√
π
(v + 1)α+n− 1
2
Γ
(
α+ 2n− 1
2
)
Γ(α+ 2n)
2F1
(
−n, −n+ 1
2
−α− 2n+ 3
2
; 1
)
=
2
√
π
(v + 1)α+n− 1
2
Γ
(
α+ 2n− 1
2
)
Γ(α+ 2n)
Γ
(
−α− 2n+ 3
2
)
Γ (−α+ 1)
Γ (−α− n+ 1)Γ
(
−α− n+ 3
2
)
=
2
√
π
(v + 1)α+n− 1
2
Γ(α+ n)Γ
(
α+ n− 1
2
)
Γ(α)Γ(α+ 2n)
.
It follows that
Q′(v) = − 2
√
π
(v + 1)α+n+ 1
2
Γ(α+ n)Γ
(
α+ n+ 1
2
)
Γ(α)Γ(α+ 2n)
.
Next, we compute
J(x) =
∫ ∞
−∞
1(
x+ 1 + t2
)α+n+ 1
2
(√
x+ 1 + t2 − t
)2n
dt
= 2
∫ ∞
0
1(
x+ 1 + t2
)α+n+ 1
2
n∑
r=0
(
2n
2r
)
(x+ 1 + t2)n−rt2rdt
= 2
n∑
r=0
(
2n
2r
)∫ ∞
0
t2r(
x+ 1 + t2
)α+r+ 1
2
dt
=
1
(x+ 1)α
n∑
r=0
(−n)r
(
−n+ 1
2
)
r(
1
2
)
r
r!
Γ
(
r + 1
2
)
Γ (α)
Γ
(
α+ r + 1
2
) .
It follows that
J(x) =
Γ(α)
Γ
(
α+ 1
2
) √
π
(x+ 1)α
n∑
r=0
(−n)r
(
−n+ 1
2
)
r
r!
1(
α+ 1
2
)
r
=
Γ(α)
Γ
(
α+ 1
2
) √
π
(x+ 1)α
2F1
(
−n, −n+ 1
2
α+ 1
2
; 1
)
=
Γ(α)
Γ
(
α+ 1
2
) √
π
(x+ 1)α
Γ
(
α+ 1
2
)
Γ(α+ 2n)
Γ(α+ n)Γ
(
α+ n+ 1
2
) .
20 L.-P. Teo
This proves that for
Ψ(x) =
1
(x+ 1)α
,
the inversion formula holds. Namely, if
Q(v) = 2(−1)n
∫ ∞
−∞
Ψ
(
x2 + v
) 1(
x− i
√
v + 1
)2ndx,
then
Ψ(x) = − 1
2π
∫ ∞
−∞
Q′(x+ t2
)(√
x+ 1 + t2 − t
)2n
dt.
Thus, this formula works for Ψ(u) = Ψn,s(u)−Ψn,a(u).
B The hyperbolic contribution
In this section, we want to compute the hyperbolic contribution
ΞH =
∑
γ∈Γ
γ is hyperbolic
∫∫
Γ\H
k(γz, z)γ′(z)ny2ndµ(z)
to the trace equation (4.1).
For any γ ∈ Γ, let
Γγ =
{
α ∈ Γ | αγα−1 = γ
}
.
It is a cyclic subgroup of Γ generated by some γ0 ∈ Γ such that γ = γℓ0 for some nonzero integer ℓ.
The elements in Γ are partitioned into conjugacy classes. The conjugacy class of γ consists
of all elements in Γ of the form αγα−1, when α runs through all elements in Γ. Two elements α
and β in Γ define the same element in the class containing γ, namely,
αγα−1 = βγβ−1,
if and only if α−1β ∈ Γγ .
Let γ ∈ Γ be a hyperbolic element. There exists σ ∈ PSL(2,R) and a real number λγ > 1
such that
σ−1γσ = Dγ =
(
λ
1/2
γ 0
0 λ
−1/2
γ
)
.
λγ is called the multiplier of γ. Each hyperbolic element γ in Γ belongs to a conjugacy class.
All elements in the same conjugacy class has the same multiplier. There exists γ0 ∈ Γ such that
γ = γℓ0 for some nonzero integer ℓ, and Γγ is generated by γ0. γ0 is called a primitive hyperbolic
element.
Let P be the set of conjugacy classes of primitive hyperbolic elements in Γ. Then the set of
hyperbolic elements in Γ can be written as
⋃
[γ0]∈P
∞⋃
ℓ=1
{
αγℓ0α
−1 | α ∈ Γγ0\Γ
}
.
Resolvent Trace Formula and Determinants of n Laplacians 21
Given γ0 a representative of a primmitive hyperbolic conjugacy class, and ℓ a positive integer,
let γ = γℓ0. We first consider the integral
IH =
∑
α∈Γγ\Γ
∫∫
F
k
(
αγα−1z, z
) [
(αγα)′(z)
]n
y2ndµ(z).
Using the invariant property of the kernel k(z, w), we find that
IH =
∫∫
σ−1(Γγ\H)
k(λγz, z)λ
n
γy
2ndµ(z).
In the following, we abbreviate λγ as λ. Let λ = λℓ0. Then
σ−1 (Γγ\H) = {x+ iy | 1 ≤ y ≤ λ0} .
It follows that
IH = λn
∫ λ0
1
∫ ∞
−∞
k(λz, z)y2ndx
dy
y2
= λn
∫ λ0
1
∫ ∞
−∞
Ψ
(
(λ− 1)2
4λ
(
x2 + 1
)) (−4)n
[(λ− 1)x+ i(λ+ 1)]2n
dx
dy
y
=
2(−1)nλ
1
2
λ− 1
lnλ0
∫ ∞
−∞
Ψ
(
x2 +
(λ− 1)2
4λ
)
1[
x+ iλ+1
2
√
λ
]2ndx
=
√
λ
λ− 1
lnλ0 g(lnλ) =
1
2s+ 2n− 1
λ−s−n
1− λ−1
lnλ0 −
1
2a+ 2n− 1
λ−a−n
1− λ−1
lnλ0.
Now we can sum up the contributions from γ = γℓ0, where ℓ ranges from 1 to ∞. To avoid
confusion with the eigenvalues of Laplacian, we denote λγ by N(γ). Since
∞∑
ℓ=1
λ−ℓ(s+n)
1− λ−ℓ
=
∞∑
k=0
∞∑
ℓ=1
λ−ℓ(s+n+k) =
∞∑
k=0
1
λs+k+n − 1
,
we find that ΞH = EH(s)− EH(a), where
EH(s) =
1
2s+ 2n− 1
∑
[γ]∈P
∞∑
k=0
N(γ)−(s+n+k)
1−N(γ)−(s+n+k)
lnN(γ).
C Elliptic contribution
In this appendix, we compute the elliptic contribution
ΞE =
∑
γ∈Γ
γ is elliptic
∫∫
Γ\H
k(γz, z)γ′(z)ny2ndµ(z)
to the trace equation (4.1).
For simplicity, let us just do the calculation with
Ψ(u) = Ψn,s(u),
22 L.-P. Teo
understanding that the singularity at u = 0 is eliminated by subtracting Ψn,a(u). At the end,
we will restore the result with
Ψ(u) = Ψn,s(u)−Ψn,a(u).
Recall that the Riemann surface X has v ramification points corresponding to the v elliptic
elements τ1, . . . , τv of orders m1, . . . ,mv respectively. For each 1 ≤ j ≤ v, there is a σj ∈
PSL(2,R) such that
σ−1
j τjσj =
(
cos θj − sin θj
sin θj cos θj
)
,
where θj =
π
mj
. If γ is an elliptic element of Γ, then there exists α ∈ Γ, 1 ≤ j ≤ v and an integer
1 ≤ ℓ ≤ mj − 1 so that
α−1γα = τ ℓj .
This implies that
ΞE =
v∑
j=1
mj−1∑
ℓ=1
∑
α∈Γτj \Γ
∫∫
Γ\H
k
(
ατ ℓjα
−1z, z
)[(
ατ ℓjα
−1
)′
(z)
]n
y2ndµ(z)
=
v∑
j=1
mj−1∑
ℓ=1
∫∫
Γτj \H
k
(
τ ℓj z, z
)(
τ ℓj
)′
(z)ny2ndµ(z) =
v∑
j=1
mj−1∑
ℓ=1
IE,j(θj,ℓ),
where θj,ℓ =
πℓ
mj
, and
IE(θ) =
1
mj
∫∫
H
k(Rθz, z)R
′
θ(z)
ny2ndµ(z),
Rθ =
(
cos θ − sin θ
sin θ cos θ
)
.
In the following, we omit the subscript j in our calculation of IE,j . Now,
u(Rθz, z) + 1 =
sin2 θ
(
1 + x2 + y2
)2
+ 4 cos2 θy2
4y2
,
(−4)n
R′
θ(z)
n
(Rθz − z̄)2n
=
(−4)n(
sin θ
(
1 + x2 + y2
)
− 2iy cos θ
)2n .
With
t =
1 + x2 + y2
2y
,
we have
x =
√
2yt− 1− y2, dx =
y dt√
2yt− 1− y2
.
Then
IE(θ) = (−1)n
2
m
∫ ∞
0
∫ ∞
1+y2
2y
Ψ
(
t2 sin2 θ − sin2 θ
)
(t sin θ − i cos θ)2n
dt√
2yt− 1− y2
dy
y
Resolvent Trace Formula and Determinants of n Laplacians 23
= (−1)n
2
m
∫ ∞
1
∫ t+
√
t2−1
t−
√
t2−1
dy
y
√
2yt− 1− y2
Ψ
(
t2 sin2 θ − sin2 θ
)
(t sin θ − i cos θ)2n
dt.
It has been computed that∫ t+
√
t2−1
t−
√
t2−1
dy
y
√
2yt− 1− y2
= π.
Therefore,
IE(θ) = (−1)n
2π
m
∫ ∞
1
Ψ
(
t2 sin2 θ − sin2 θ
)(
t sin θ − i cos θ
)2n dt.
This integral is difficult to compute. Fischer [7] used the power series expansion of Ψn,s(u)
followed by a lot of tedious calculations. The generalization from n = 0 to general positive n is
not easy, and Hejhal was unable to proceed in his first volume [11]; only succeeded to circumvent
the problem in his second volume [12]. We will use the method of Hejhal [12].
First of all, using the substitution u = t2 sin2 θ − sin2 θ, we find that
IE(θ) = (−1)n
π
m sin θ
∫ ∞
0
Ψ(u)(√
u+ sin2 θ − i cos θ
)2n du√
u+ sin2 θ
.
Using the inversion formula (see Appendix A)
Ψ(u) = − 1
2π
∫ ∞
−∞
Q′(u+ t2
)(√
u+ 1 + t2 − t
)2n
dt,
we find that
IE(θ) =
(−1)n+1
2m sin θ
∫ ∞
0
∫ ∞
−∞
Q′(u+ t2
) (√
u+ 1 + t2 − t
)2n(√
u+ sin2 θ − i cos θ
)2ndt du√
u+ sin2 θ
=
(−1)n+1
2m sin θ
∫ ∞
−∞
∫ ∞
t2
Q′(v)
(√
v + 1− t
)2n(√
v − t2 + sin2 θ − i cos θ
)2n dv√
v − t2 + sin2 θ
dt
=
(−1)n+1
2m sin θ
∫ ∞
0
Q′(v)H (v) dv,
where
H (v) =
∫ √
v
−
√
v
(√
v + 1− t
)2n(√
v − t2 + sin2 θ − i cos θ
)2n dt√
v − t2 + sin2 θ
.
Making a change of variables
t =
√
v + sin2 θ sinφ,
we find that
H (v) =
∫ φ0(v)
−φ0(v)
(√
v + 1−
√
v + sin2 θ sinφ
)2n(√
v + sin2 θ cosφ− i cos θ
)2n dφ,
where
φ0(v) = sin−1
√
v√
v + sin2 θ
.
24 L.-P. Teo
By definition, H (0) = 0, and hence,
IE(θ) =
(−1)n
2m sin θ
∫ ∞
0
Q(v)H ′(v) dv.
Now, we calculate H ′(v). It is straightforward to find that
H ′(v) = φ′
0(v)
[(√
v + 1−
√
v + sin2 θ sinφ0
)2n(√
v + sin2 θ cosφ0 − i cos θ
)2n +
(√
v + 1 +
√
v + sin2 θ sinφ0
)2n(√
v + sin2 θ cosφ0 − i cos θ
)2n
]
+
∫ φ0(v)
−φ0(v)
d
dv
[√
v + 1−
√
v + sin2 θ sinφ√
v + sin2 θ cosφ− i cos θ
]2n
dφ.
If we define ω so that
sinhω =
cos θ√
v + sin2 θ
,
we find that
√
v + 1√
v + sin2 θ
= coshω.
Observe that
d
dv
√
v + 1−
√
v + sin2 θ sinφ√
v + sin2 θ cosφ− i cos θ
=
d
dω
coshω − sinφ
cosφ− i sinhω
dω
dv
= −i
d
dφ
coshω − sinφ
cosφ− i sinhω
dω
dv
= − i cos θ
2(v + sin2 θ)
√
v + 1
d
dφ
coshω − sinφ
cosφ− i sinhω
.
These imply that
H ′(v) =
sin θ
2
√
v(v + sin2 θ)
[(√
v + 1−
√
v
)2n
(sin θ − i cos θ)2n
+
(√
v + 1 +
√
v
)2n
(sin θ − i cos θ)2n
]
− i cos θ
2(v + sin2 θ)
√
v + 1
∫ φ0(v)
−φ0(v)
d
dφ
[√
v + 1−
√
v + sin2 θ sinφ√
v + sin2 θ cosφ− i cos θ
]2n
dφ
=
(−1)n sin θ e2inθ
2
√
v(v + sin2 θ)
[(√
v + 1−
√
v
)2n
+
(√
v + 1 +
√
v
)2n]
− (−1)ni cos θ e2inθ
2(v + sin2 θ)
√
v + 1
[(√
v + 1−
√
v
)2n −
(√
v + 1 +
√
v
)2n]
.
It follows that
H ′
(
sinh2
t
2
)
d
dt
sinh2
t
2
=
(−1)ne2inθ
2
(
sinh2 t
2 + sin2 θ
){ sin θ cosh
t
2
(etn + e−tn) + i cos θ sinh
t
2
(etn − e−tn)
}
=
(−1)ne2inθ
cosh t− cos 2θ
{
sin θ
(
e
t
2 + e−
t
2
)
(etn + e−tn)
2
+ i cos θ
(
e
t
2 − e−
t
2
)
(etn − e−tn)
2
}
=
(−1)ne2inθ
cosh t− cos 2θ
{
sin θ
e(n+
1
2)t + e−(n+
1
2)t + e(n−
1
2)t + e−(n−
1
2)t
2
Resolvent Trace Formula and Determinants of n Laplacians 25
+ i cos θ
e(n+
1
2)t + e−(n+
1
2)t − e(n−
1
2)t − e−(n−
1
2)t
2
}
.
This implies that
IE(θ) =
(−1)n
2m sin θ
∫ ∞
0
Q
(
sinh2
t
2
)
H ′
(
sinh2
t
2
)
d
dt
sinh2
t
2
dt
=
e2inθ
2m sin θ
∫ ∞
0
g(t)
cosh t− cos 2θ
{
sin θ
e(n+
1
2)t + e−(n+
1
2)t + e(n−
1
2)t + e−(n−
1
2)t
2
+ i cos θ
e(n+
1
2)t + e−(n+
1
2)t − e(n−
1
2)t − e−(n−
1
2)t
2
}
.
Using
g(t) =
1
2s+ 2n− 1
e−t(s+n− 1
2),
and ∫ ∞
0
e−µt
cosh t− cos 2θ
dt =
2
sin 2θ
∞∑
k=1
sin 2kθ
µ+ k
,
we find that
IE(θ) =
1
(2s+ 2n− 1)
e2inθ
2m sin θ
1
sin 2θ
×
{
sin θ
[ ∞∑
k=1
sin 2kθ
s+ k − 1
+
∞∑
k=1
sin 2kθ
s+ k
+
∞∑
k=1
sin 2kθ
s+ k + 2n
+
∞∑
k=1
sin 2kθ
s+ k + 2n− 1
]
+ i cos θ
[ ∞∑
k=1
sin 2kθ
s+ k − 1
−
∞∑
k=1
sin 2kθ
s+ k
+
∞∑
k=1
sin 2kθ
s+ k + 2n
−
∞∑
k=1
sin 2kθ
s+ k + 2n− 1
]}
=
1
(2s+ 2n− 1)
e2inθ
2m sin θ
1
sin 2θ
×
{
sin θ
[ ∞∑
k=0
sin(2k + 2)θ
s+ k
+
∞∑
k=0
sin 2kθ
s+ k
+
∞∑
k=0
sin 2kθ
s+ k + 2n
+
∞∑
k=0
sin(2k + 2)θ
s+ k + 2n
]
+ i cos θ
[ ∞∑
k=0
sin(2k+2)θ
s+ k
−
∞∑
k=0
sin 2kθ
s+ k
+
∞∑
k=0
sin 2kθ
s+ k + 2n
−
∞∑
k=0
sin(2k+2)θ
s+ k + 2n
]}
=
1
2s+ 2n− 1
ie2inθ
2m sin θ
( ∞∑
k=0
e−i(2k+1)θ
s+ k
−
∞∑
k=0
ei(2k+1)θ
s+ 2n+ k
)
.
Now we find the sum
m−1∑
ℓ=1
IE
(
πℓ
m
)
.
Notice that is k is an integer,
Sk =
m−1∑
ℓ=1
i
sin πℓ
m
exp
(
−πiℓ(2k + 1)
m
)
26 L.-P. Teo
=
1
2
[
m−1∑
ℓ=1
i
sin πℓ
m
exp
(
−πiℓ(2k + 1)
m
)
+
m−1∑
ℓ=1
i
sin π(m−ℓ)
m
exp
(
−πi(m− ℓ)(2k + 1)
m
)]
=
m−1∑
ℓ=1
exp
(
πiℓ(2k+1)
m
)
− exp
(
−πiℓ(2k+1)
m
)
exp
(
πiℓ
m
)
− exp
(
−πiℓ
m
) =
m−1∑
ℓ=1
x2k+1
ℓ − x−2k−1
ℓ
xℓ − x−1
ℓ
,
where
xℓ = exp
(
πiℓ
m
)
.
Using the fact that
m−1∑
ℓ=1
x2hℓ =
{
m− 1 if m divides h,
−1 if m does not divide h,
we find that if k is a nonnegative integer
Sk =
m−1∑
ℓ=1
(
x2kℓ + x2k−2
ℓ + · · ·+ x−2k+2
ℓ + x−2k
ℓ
)
= −2k − 1 +m
(
2
⌊
k
m
⌋
+ 1
)
= −2m
{
k
m
}
+m− 1.
Here {x} = x−⌊x⌋. The case when k is a negative integer is more complicated. First we notice
that for any integer x, if q =
⌊
x
m
⌋
, then x = qm+ r, where 0 ≤ r ≤ m− 1. It follows that
−x− 1 = −qm− r − 1 = (−q − 1)m+m− r − 1,
where
0 ≤ m− r − 1 ≤ m− 1.
This shows that⌊
−x− 1
m
⌋
= −q − 1 = −
⌊
x
m
⌋
− 1.
If k is a negative integer, let k′ = −k. Then k′ is a positive integer and
Sk = −
m−1∑
ℓ=1
x2k
′−1
ℓ − x−2k′+1
ℓ
xℓ − x−1
ℓ
= −
m−1∑
ℓ=1
(
x2k
′−2
ℓ + x2k
′−4
ℓ + · · ·+ x−2k′+4
ℓ + x−2k′+2
ℓ
)
= −
{
−2k′ + 1 +m
(
2
⌊
k′ − 1
m
⌋
+ 1
)}
= −2k − 1−m
(
−2
⌊
k
m
⌋
− 1
)
= −2m
{
k
m
}
+m− 1.
This shows that for any integer k,
Sk = −2αm(k) +m− 1,
Resolvent Trace Formula and Determinants of n Laplacians 27
where
αm(k) = m
{
k
m
}
is the least positive residue of k modulo m. We have established that
αm(−k) = m− 1− αm(k).
Hence,
S−k = −Sk.
It follows that
m−1∑
ℓ=1
IE
(
πℓ
m
)
=
1
2m
1
2s+ 2n− 1
{ ∞∑
k=0
1
s+ k
(−2αm(k − n) +m− 1)
+
∞∑
k=0
1
s+ 2n+ k
(−2αm(k + n) +m− 1)
}
.
Writing k = mq + r, where 0 ≤ r ≤ m− 1, we have
m−1∑
ℓ=1
IE
(
πℓ
m
)
=
1
2m
1
2s+ 2n− 1
{ ∞∑
q=0
m−1∑
r=0
1
s+mq + r
(−2αm(r − n) +m− 1)
+
∞∑
q=0
m−1∑
r=0
1
s+ 2n+mq + r
(−2αm(r + n) +m− 1)
}
.
Notice that when r runs from 0 to m−1, r−n and r+n respectively runs through a complete
residue system modulo m. Hence,
m−1∑
r=0
αm(r − n) =
m−1∑
r=0
[m− 1− αm(r − n)]
and
m−1∑
r=0
αm(r + n) =
m−1∑
r=0
[m− 1− αm(r + n)].
Therefore,
m−1∑
r=0
(−2αm(r − n) +m− 1) =
m−1∑
r=0
(−2αm(r + n) +m− 1) = 0. (C.1)
Thus
m−1∑
ℓ=1
IE
(
πℓ
m
)
=
1
2m
1
2s+ 2n− 1
{ ∞∑
q=0
m−1∑
r=0
[
1
s+mq+ r
− 1
m(q + 1)
]
(−2αm(r − n) +m− 1)
+
∞∑
q=0
m−1∑
r=0
[
1
s+ 2n+mq + r
− 1
m(q + 1)
]
(−2αm(r + n) +m− 1)
}
28 L.-P. Teo
=
1
2s+ 2n− 1
m−1∑
r=0
[
2αm(r − n) + 1−m
2m2
ψ
(
s+ r
m
)
+
2αm(r + n) + 1−m
2m2
ψ
(
s+ 2n+ r
m
)]
.
Finally, we find that the elliptic contribution is given by ΞE = EE(s)− EE(a), where
EE(s) =
1
2s+ 2n− 1
v∑
j=1
mj−1∑
r=0
[
2αmj (r − n) + 1−mj
2m2
j
ψ
(
s+ r
mj
)
+
2αmj (r + n) + 1−mj
2m2
j
ψ
(
s+ 2n+ r
mj
)]
.
D Contribution from parabolic elements
and absolutely continuous spectrum
In this appendix, we compute the parabolic contribution ΞP to the trace equation (4.1).
D.1 Contribution from parabolic elements
In this section, we compute the term
ΞP,1 =
∫∫
FY
∑
γ∈Γ
γ is parabolic
k(γz, z)γ′(z)ny2ndµ(z).
Again we first consider the case where Ψ(u) = Ψn,s(u). Recall that the Riemann surface X
has q cusps corresponding to the q parabolic elements κ1, . . . , κq. For each 1 ≤ i ≤ q, there is
a σi ∈ PSL(2,R) such that
σ−1
i κiσi =
(
1 ±1
0 1
)
.
If γ is a parabolic element of Γ, then there exists α ∈ Γ, 1 ≤ j ≤ q and a nonzero integer ℓ so
that
α−1γα = κℓj .
For an integer ℓ, define
Tℓ =
(
1 ℓ
0 1
)
.
Then
ΞP,1 =
∫∫
FY
q∑
j=1
∑
ℓ ̸=0
∑
α∈Γκj \Γ
k
(
ακℓjα
−1z, z
)[(
ακℓjα
−1
)′
(z)
]n
y2ndµ(z)
=
q∑
j=1
∑
ℓ ̸=0
∑
α∈σ−1
j Γκiσj\σ−1
j Γσj
∫∫
σ−1
j (FY )
k
(
αTℓα
−1z, z
)[(
αTℓα
−1
)′
(z)
]n
y2ndµ(z)
=
q∑
j=1
∑
ℓ ̸=0
∫∫
HY
j
k(Tℓz, z)T
′
ℓ(z)
ny2ndµ(z).
Resolvent Trace Formula and Determinants of n Laplacians 29
Here
HY
j = HY \
⋃
k ̸=j
⋃
α∈Γκj \Γ
σ−1
j
(
α
(
F Y
k
))
,
where
HY =
{
x+ iy | 0 ≤ x ≤ 1, 0 < y ≤ Y
}
.
Since the hyperbolic area of H \HY is O
(
Y −1
)
, we find that
ΞP,1 = qIP +O
(
Y −1
)
,
where
IP =
∑
ℓ̸=0
∫∫
HY
k(Tℓz, z)T
′
ℓ(z)
ny2ndµ(z).
Now
u(Tℓz, z) =
ℓ2
4y2
,
(−4)n
T ′
ℓ(z)
n
(Tℓz − z̄)2n
=
(−4)n
(ℓ+ 2iy)2n
.
Hence, we have
IP = (−1)n
∞∑
ℓ=1
∫ Y
0
Ψ
(
ℓ2
4y2
)[
(2y)2n
(l + 2iy)2n
+
(2y)2n
(l − 2iy)2n
]
dy
y2
= (−1)n
∞∑
ℓ=1
2
ℓ
∫ 2Y/ℓ
0
Ψ
(
1
y2
)[
y2n
(1 + iy)2n
+
y2n
(1− iy)2n
]
dy
y2
.
As in [13], if F (y) is a well-behaved function, standard techniques in analytic number theory
give
∞∑
ℓ=1
2
ℓ
∫ 2Y
ℓ
0
F (y) dy =
∫ ∞
1−
∫ 2Y
v
0
F (y) dy d
(∑
ℓ≤v
2
ℓ
)
=
∫ 2Y
0
F (y)
[∫ 2Y/y
1−
d
(∑
ℓ≤v
2
ℓ
)]
dy
=
∫ 2Y
0
F (y)
[
2 log
2Y
y
+ 2γ +O
(
yY −1
)]
dy.
Hence,
IP = (−1)n
∫ 2Y
0
Ψ
(
1
y2
)[
y2n
(1 + iy)2n
+
y2n
(1− iy)2n
][
2 log
2Y
y
+ 2γ +O
(
yY −1
)]dy
y2
= (log(2Y ) + γ)IP,0 + IP,1 +O
(
Y −1 log Y
)
,
where
IP,0 = 2(−1)n
∫ ∞
0
Ψ
(
1
y2
)[
y2n
(1 + iy)2n
+
y2n
(1− iy)2n
]
dy
y2
,
IP,1 = 2(−1)n
∫ ∞
0
Ψ
(
1
y2
)[
y2n
(1 + iy)2n
+
y2n
(1− iy)2n
]
log
1
y
dy
y2
.
30 L.-P. Teo
A straightforward computation gives
IP,0 = 2(−1)n
∫ ∞
0
Ψ(x2)
[
1
(x+ i)2n
+
1
(x− i)2n
]
dx = Q(0) =
1
2s+ 2n− 1
.
The computation of IP,1 is more complicated:
IP,1 = 2(−1)n
∫ ∞
0
Ψ(u2)
[
1
(u+ i)2n
+
1
(u− i)2n
]
log udu
=
(−1)n
π
∫ ∞
0
∞∑
k=0
Γ(s+ k)Γ(s+ 2n+ k)
Γ(2s+ 2n+ k)
1
k!
1
(u2 + 1)k+s+2n
×
n∑
m=0
(
2n
2m
)
u2m(−1)n−m log udu
=
1
4π
n∑
m=0
(
2n
2m
)
(−1)m
∞∑
k=0
Γ(s+ k)Γ(s+ 2n+ k)
Γ(2s+ 2n+ k)
1
k!
∫ ∞
0
um
(u+ 1)k+s+2n
log u
1√
u
du.
This implies that
IP,1 = G′(0),
where
G(α) =
1
4π
n∑
m=0
(
2n
2m
)
(−1)m
∞∑
k=0
Γ(s+ k)Γ(s+ 2n+ k)
Γ(2s+ 2n+ k)
1
k!
∫ ∞
0
um+α− 1
2
(u+ 1)k+s+2n
du.
From [10], we find that∫ ∞
0
um+α− 1
2
(u+ 1)k+s+2n
du =
Γ
(
m+ α+ 1
2
)
Γ
(
s+ 2n+ k −m− α− 1
2
)
Γ(s+ 2n+ k)
.
Hence,
G(α) =
1
4π
n∑
m=0
(
2n
2m
)
(−1)mΓ
(
m+ α+
1
2
) ∞∑
k=0
Γ(s+ k)Γ
(
s+ 2n+ k −m− α− 1
2
)
Γ(2s+ 2n+ k)
1
k!
.
Using the definition and identity of hypergeometric functions (see [2]), we have
∞∑
k=0
Γ(s+ k)Γ
(
s+ 2n+ k −m− α− 1
2
)
Γ(2s+ 2n+ k)
1
k!
=
Γ(s)Γ
(
s+ 2n−m− α− 1
2
)
Γ (2s+ 2n)
2F1
(
s, s+ 2n−m− α− 1
2
2s+ 2n
; 1
)
=
Γ(s)Γ
(
s+ 2n−m− α− 1
2
)
Γ
(
m+ α+ 1
2
)
Γ (s+ 2n) Γ
(
s+m+ α+ 1
2
) .
Now, we notice that(
2n
2m
)
=
(2n)(2n− 1) · · · (2n− 2m+ 1)
1 · 2 · · · (2m− 1)(2m)
=
(−n)m
(
−n+ 1
2
)
m(
1
2
)
m
m!
,
(−1)mΓ
(
s+ 2n−m− α− 1
2
)
= (−1)m
Γ
(
s+ 2n− α− 1
2
)(
s+ 2n− α− 1
2 − 1
)
· · ·
(
s+ 2n− α− 1
2 −m
)
Resolvent Trace Formula and Determinants of n Laplacians 31
=
Γ
(
s+ 2n− α− 1
2
)(
−s− 2n+ α+ 1
2 + 1
)
· · ·
(
−s− 2n+ α+ 1
2 +m
)
=
Γ
(
s+ 2n− α− 1
2
)(
−s− 2n+ α+ 3
2
)
m
.
This gives
G(α) =
1
4π
Γ
(
α+ 1
2
)2
Γ(s)Γ
(
s+ 2n− α− 1
2
)
Γ(s+ 2n)Γ
(
s+ α+ 1
2
) H(α),
where
H(α) =
n∑
m=0
(−n)m
(
−n+ 1
2
)
m(
1
2
)
m
m!
(
α+ 1
2
)
m
(
α+ 1
2
)
m(
−s− 2n+ α+ 3
2
)
m
(
s+ α+ 1
2
)
m
= 4F3
(
−n, −n+ 1
2 , α+ 1
2 , α+ 1
2
1
2 , −s− 2n+ α+ 3
2 , s+ α+ 1
2
; 1
)
.
This is a balanced hypergeometric series. According to [2, Theorem 3.3.3], we find that
H(α) =
Γ(α+ 1− s)Γ(α+ s+ 2n)Γ
(
α− s− 2n+ 3
2
)
Γ
(
α+ s+ 1
2
)
Γ(α+ 1− s− n)Γ(α+ s+ n)Γ
(
α− s− n+ 3
2
)
Γ
(
α+ s+ n+ 1
2
)
× 4F3
(
−n, −n+ 1
2 , −α, −α
1
2 , s− α, 1− s− 2n− α
; 1
)
.
As a function of α,
4F3
(
−n, −n+ 1
2 , −α, −α
1
2 , s− α, 1− s− 2n− α
; 1
)
= 1 +O
(
α2
)
as α→ 0.
Hence, IP,1 = G′(0) = L′(0), where L(α) is given by
L′(α) =
1
4π
Γ
(
α+ 1
2
)2
Γ(s)Γ
(
s+ 2n− α− 1
2
)
Γ(α+ 1− s)Γ(α+ s+ 2n)Γ
(
α− s− 2n+ 3
2
)
Γ(s+ 2n)Γ(α+ 1− s− n)Γ(α+ s+ n)Γ
(
α− s− n+ 3
2
)
Γ
(
α+ s+ n+ 1
2
) .
It is then straightforward to obtain
IP,1 =
1
4π
Γ
(
1
2
)2
Γ(s)Γ
(
s+ 2n− 1
2
)
Γ(1− s)Γ(s+ 2n)Γ
(
−s− 2n+ 3
2
)
Γ (s+ 2n) Γ(1− s− n)Γ(s+ n)Γ
(
−s− n+ 3
2
)
Γ
(
s+ n+ 1
2
)
×
{
2ψ
(
1
2
)
− ψ
(
s+ 2n− 1
2
)
+ ψ(1− s) + ψ(s+ 2n) + ψ
(
−s− 2n+
3
2
)
− ψ(1− s− n)− ψ(s+ n)− ψ
(
−s− n+
3
2
)
− ψ
(
s+ n+
1
2
)}
.
Using the fact that Γ(z + 1) = zΓ(z), we find that
Γ
(
s+ n+
1
2
)
=
(
s+ n− 1
2
)
Γ
(
s+ n− 1
2
)
,
and
ψ
(
s+ n+
1
2
)
=
1
s+ n− 1
2
+ ψ
(
s+ n− 1
2
)
.
32 L.-P. Teo
On the other hand, using
Γ(z)Γ(1− z) =
π
sinπz
,
we find that
Γ(s)Γ(1− s)
Γ(1− s− n)Γ(s+ n)
=
sinπ(s+ n)
sinπs
= (−1)n,
Γ(s+ 2n− 1
2)Γ
(
−s− 2n+ 3
2
)
Γ
(
−s− n+ 3
2
)
Γ
(
s+ n− 1
2
) =
sinπ
(
s+ n− 1
2
)
sinπ(s+ 2n− 1
2)
= (−1)n,
ψ(s)− ψ(1− s) + ψ(1− s− n)− ψ(s+ n) = 0,
ψ
(
s+ n− 1
2
)
− ψ
(
−s− 2n+
3
2
)
+ ψ
(
−s− n+
3
2
)
− ψ
(
s+ n− 1
2
)
= 0.
Together with
Γ
(
1
2
)
=
√
π, ψ ≤
(
1
2
)
= −γ − 2 log 2,
we have
1
4π
Γ
(
1
2
)2
Γ(s)Γ
(
s+ 2n− 1
2
)
Γ(1− s)Γ
(
−s− 2n+ 3
2
)
Γ(1− s− n)Γ(s+ n)Γ
(
−s− n+ 3
2
)
Γ
(
s+ n+ 1
2
) =
1
2(2s+ 2n− 1)
,
and
IP,1 =
1
2(2s+ 2n− 1)
{
ψ(s) + ψ(s+ 2n)− 4 log 2− 2ψ
(
s+ n+
1
2
)
− 2ψ(s+ n)
}
− γ
2s+ 2n− 1
+
1
(2s+ 2n− 1)2
.
Collecting the various contributions, we find that
IP =
1
2s+ 2n− 1
log Y +
1
(2s+ 2n− 1)2
+
1
2(2s+ 2n− 1)
{
ψ(s) + ψ(s+ 2n)− 2 log 2− 2ψ
(
s+ n+
1
2
)
− 2ψ(s+ n)
}
.
Therefore, ΞP,1 = EP,1(s)− EP,1(a), where
EP,1(s) =
q
2s+ 2n− 1
log Y +
q
(2s+ 2n− 1)2
+
q
2(2s+ 2n− 1)
×
{
ψ(s) + ψ(s+ 2n)− 2 log 2− 2ψ
(
s+ n+
1
2
)
− 2ψ(s+ n)
}
+O
(
Y −1
)
.
D.2 Contribution of the absolutely continuous spectrum
In this section, we want to compute
J =
1
4π
q∑
j=1
∫ ∞
−∞
h(r)
∫∫
FY
∣∣∣∣Ej
(
z,
1
2
+ ir;n
)∣∣∣∣2y2ndµ(z) dr,
Resolvent Trace Formula and Determinants of n Laplacians 33
where
h(r) = Λ̃
(
1
4
+ r2
)
=
1
r2 +
(
s+ n− 1
2
)2 − 1
r2 +
(
a+ n− 1
2
)2 .
From Appendix E, we find that
q∑
j=1
∫∫
FY
∣∣∣∣Ej
(
z,
1
2
+ ir;n
)∣∣∣∣2y2ndµ(z)
=
1
2ir
Tr
[
Φ
(
1
2
− ir
)
Y 2ir − Φ
(
1
2
+ ir
)
Y −2ir
]
+ 2q log Y − φ′
φ
(
1
2
+ ir
)
.
Therefore,
J = J1 + J2 + J3,
where
J1 =
q log Y
2π
∫ ∞
−∞
h(r) dr,
J2 =
1
8πi
∫ ∞
−∞
h(r)
r
Tr
[
Φ
(
1
2
− ir
)
Y 2ir − Φ
(
1
2
+ ir
)
Y −2ir
]
dr,
J3 = − 1
4π
∫ ∞
−∞
h(r)
φ′
φ
(
1
2
+ ir
)
dr.
Since
1
2π
∫ ∞
−∞
h(r) dr = g(0),
we find that
J1 = q log Y
(
1
2s+ 2n− 1
− 1
2a+ 2n− 1
)
.
As in [13],
J2 =
1
8πi
∫ ∞
−∞
h(r)
r
Tr
[
Φ
(
1
2
− ir
)
Y 2ir − Φ
(
1
2
+ ir
)
Y −2ir
]
dr
=
1
4
TrΦ
(
1
2
)
h(0) +O
(
Y −2ε
)
=
[
1
(2s+ 2n− 1)2
− 1
(2a+ 2n− 1)2
]
TrΦ
(
1
2
)
+O
(
Y −2ε
)
,
where ε is an arbitrary positive number. Collecting everything together, we find that
J = q log Y
(
1
2s+ 2n− 1
− 1
2a+ 2n− 1
)
+
[
1
(2s+ 2n− 1)2
− 1
(2a+ 2n− 1)2
]
TrΦ
(
1
2
)
− 1
4π
∫ ∞
−∞
h(r)
φ′
φ
(
1
2
+ ir
)
dr +O
(
Y −2ε
)
.
When h(r) = 1
r2+α2 for some α with Reα > 0, we can follow the method of [23] to compute the
integral
J3 = − 1
4π
∫ ∞
−∞
h(r)
φ′
φ
(
1
2
+ ir
)
dr.
34 L.-P. Teo
From (3.1) and the results for the case n = 0, we can deduce that φ(s) is holomorphic on the
half plane Re s ≥ 1
2 except for for a finite number of poles on
(
1
2 , 1
]
, with order not larger than q.
From the relation φ(s)φ(1 − s) = 1, we then deduce that on the half plane Re s < 1
2 , φ(s) can
only have zeros on
[
0, 12
)
. Using these and the residue theorem, we find that
1
2π
∫ ∞
−∞
1
r2 + α2
φ′
φ
(
1
2
+ ir
)
dr =
1
2s+ 2n− 1
φ′(s+ n)
φ(s+ n)
−
∑
ρ is a pole of φ(s)
Re ρ< 1
2
order (ρ)
α2 −
(
ρ− 1
2
)2
+
∑
σ is a pole of φ(s)
σ∈( 1
2
,1]
order (σ)
α2 −
(
σ − 1
2
)2 + c. (D.1)
Here order(z) is the order of pole of φ(s) at s = z, c is a constant independent of α.
We are interested in the particular case where α = s+ n− 1
2 . The following result is needed
when we want to discuss the dimension of the space of holomorphic n-differentials.
Proposition D.1. Given a positive integer n, let Σ(s) be the function
Σ(s) =
1
2π
∫ ∞
−∞
1
r2 +
(
s+ n− 1
2
)2 φ′
φ
(
1
2
+ ir
)
dr.
Then the residue of Σ(s) at s = 0 is 0.
Proof. According to (D.1), Σ(s) is a sum of three terms. We denote these three terms by Σ1(s),
Σ2(s) and Σ3(s) respectively.
We need to discuss the case where n = 1 and n ≥ 2 separately.
When n = 1, the residue of Σ1(s) at s = 0 is equal to −n0, where n0 is the order of pole
of φ(s) at s = 1, which is equal to the order of zero of φ(s) at s = 0. Since φ(s) does not have
pole at s = 0, the residue of Σ2(s) at s = 0 is 0. Since φ(s) has a pole of order n0 at s = 1, the
residue of Σ3(s) at s = 0 is n0. This shows that the residue of Σ(s) at s = 0 is 0.
When n ≥ 2, the residue of Σ1(s) is equal to
1
2n−1 times the order of zero of φ(s) at s = n.
Since φ(s) does not have pole when Re s > 1, the residue of Σ3(s) at s = 0 is 0. For Σ2(s), the
order of zero of φ(s) at s = n is equal to the order of pole of φ(s) at 1−n. Therefore, the residue
of Σ2(s) at s = 0 is the negative of the residue of Σ1(s) at s = 0. This proves the assertion that
the residues of Σ(s) at s = 0 is 0. ■
D.3 The term ΞP
Collecting the results from Appendices D.1 and D.2, we find that the parabolic contribution
ΞP = lim
Y→∞
{∫∫
FY
∑
γ∈Γ
γis parabolic
k(γz, z)γ′(z)ny2ndµ(z)
− 1
4π
q∑
j=1
∫ ∞
−∞
Λ̃(r)
∫∫
FY
∣∣∣∣Ej
(
z,
1
2
+ ir;n
)∣∣∣∣2y2ndµ(z) dr
}
is given by ΞP = EP (s)− EP (a), where
EP (s) =
1
(2s+ 2n− 1)2
[
q − TrΦ
(
1
2
)]
+
1
4π
∫ ∞
−∞
1
r2 +
(
s+ n− 1
2
)2 φ′
φ
(
1
2
+ ir
)
dr
+
q
2(2s+ 2n− 1)
{
ψ(s) + ψ(s+ 2n)− 2 log 2− 2ψ
(
s+ n+
1
2
)
− 2ψ(s+ n)
}
.
Resolvent Trace Formula and Determinants of n Laplacians 35
E Maass–Selberg relation
The Maass–Selberg relation is very important in the computation of the contribution from the
absolutely continuous spectrum. The results in this part follows almost the same as the case
n = 0 as presented in [13].
Given a function f in H2
n(Γ) such that ∆nf = λf , it has the following Fourier expansion
around the cusp associated to κj .
f(σjz)σ
′
j(z)
n =
∞∑
k=−∞
f
(j)
k (y)e2πikx.
We define
fY (z) =
{
f(z), z ∈ F Y ,
f(z)− f
(j)
0
(
σ−1
j z
)
, z ∈ F Y
j .
Theorem E.1. If f and g are functions in H2
n(Γ) such that
∆nf = λ1f, ∆ng = λ2g.
Then
(λ1 − λ2)
∫∫
F
fY gY y2n−2dx dy = −Y 2n
q∑
j=1
(
f
(j)
0 (Y )g
(j)′
0 (Y )− f
(j)′
0 (Y )g
(j)
0 (Y )
)
.
Proof.
(λ1 − λ2)
∫∫
FY
fḡy2n−2dx dy =
∫∫
FY
∆nfḡy
2n−2dx dy −
∫∫
FY
f∆ngy
2n−2dx dy
= − 2i
∫∫
FY
{(
∂
∂z
y2n
∂
∂z̄
f
)
ḡ − f
(
∂
∂z̄
y2n
∂
∂z
ḡ
)}
dz ∧ dz̄
= − 2i
∫
∂FY
(
y2n
∂f
∂z̄
ḡdz̄ + fy2n
∂ḡ
∂z
dz
)
=
q∑
j=1
−2i
∫
σ−1
j ∂FY
j
(
y2n
∂f
∂z̄
(σjz)σ
′
j(z)
nσ′j(z) g(σjz)σ
′(z)n dz̄
+ f(σjz)σ
′
j(z)
ny2n
∂g
∂z̄
(σjz)σ′j(z)
nσ′j(z) dz
)
= Y 2n
q∑
j=1
2i
∫ 1
0
(
∂f
∂z̄
(σjz)σ
′
j(z)
nσ′j(z) g(σjz)σ
′(z)n dx
+ f(σjz)σ
′
j(z)
n∂g
∂z̄
(σjz)σ′j(z)
nσ′j(z) dx
)
.
Notice that
∂f
∂z̄
(σjz)σ
′
j(z)
nσ′j(z) =
∞∑
k=−∞
πikf
(j)
k (y)e2πikx +
i
2
∞∑
k=−∞
∂f
(j)
k
∂y
(y)e2πikx.
Hence,
(λ1 − λ2)
∫∫
FY
fḡy2n−2dx dy = −Y 2n
q∑
j=1
∞∑
k=−∞
(
∂f
(j)
k
∂y
(Y )g
(j)
k (Y )− f
(j)
k (Y )
∂g
(j)
k
∂y
(Y )
)
.
36 L.-P. Teo
Applying ∆nf = λf , where λ = −(s− n)(s+ n− 1), to the Fourier expansion, we find that
y2
∞∑
k=−∞
(2πk)2f
(j)
k (y)e2πikx − y2
∞∑
k=−∞
∂2f
(j)
k
∂y2
(y)e2πikx − 2ny
∞∑
k=−∞
2πkf
(j)
k (y)e2πikx
− 2ny
∞∑
k=−∞
∂f
(j)
k
∂y
(y)e2πikx = −(s− n)(s+ n− 1)
∞∑
k=−∞
f
(j)
k (y)e2πikx.
Hence,
y2
∂2f
(j)
k
∂y2
(y) + 2ny
∂f
(j)
k
∂y
(y)−
(
4πk2y2 − 4πnky + (s− n)(s+ n− 1)
)
f
(j)
k = 0.
This implies that
f
(j)
0 (y) = α0y
s−n + β0y
1−s−n.
On the other hand,
∂
∂y
y2n
(
∂f
(j)
k
∂y
g
(j)
k − f
(j)
k
∂g
(j)
k
∂y
)
= y2n−2[(s1− n)(s1 + n− 1)− (s2− n)(s2 + n− 1)]f
(j)
k g
(j)
k .
Hence, if k ̸= 0,
−Y 2n
q∑
j=1
∑
k ̸=0
(
∂f
(j)
k
∂y
(Y )g
(j)
k (Y )− f
(j)
k (Y )
∂g
(j)
k
∂y
(Y )
)
= −(λ1 − λ2)
q∑
j=1
∑
k ̸=0
∫ ∞
Y
y2n−2f
(j)
k g
(j)
k dy = −(λ1 − λ2)
q∑
j=1
∫∫
FY
j
fY ḡY y2n−2dx dy.
It follows that
(λ1 − λ2)
∫∫
FY
fḡy2n−2dx dy = − Y 2n
q∑
j=1
(
∂f
(j)
0
∂y
(Y )g
(j)
0 (Y )− f
(j)
0 (Y )
∂g
(j)
0
∂y
(Y )
)
− (λ1 − λ2)
q∑
j=1
∫∫
FY
j
fY ḡY y2n−2dx dy.
Hence,
(λ1 − λ2)
∫∫
F
fY gY y2n−2dx dy = −Y 2n
q∑
j=1
(
f
(j)
0 (Y )g
(j)′
0 (Y )− f
(j)′
0 (Y )g
(j)
0 (Y )
)
. ■
Theorem E.2 (Maass–Selberg relation). If s1 and s2 are regular points of the Eisenstein series
Ei(z, s1;n) and Ei(z, s2;n), s1 ̸= s̄2 and s1 + s̄2 ̸= 1, then∫∫
F
EY
i (z, s1;n)EY
j (z, s2;n)y
2n−2dx dy
= δij
1
s1 + s2 − 1
Y s1+s2−1 +
1
s2 − s1
φij(s1)Y
s2−s1 +
1
s1 − s2
φji(s2)Y
s1−s2
− 1
s1 + s2 − 1
q∑
k=1
φik(s1)φjk(s2)Y
1−s1−s2 .
Resolvent Trace Formula and Determinants of n Laplacians 37
Proof. By Theorem E.1, we have∫∫
F
EY
i (z, s1;n)EY
j (z, s2;n)y
2n−2dx dy =
Y 2n
−(s1 − n)(s1 + n− 1) + (s2 − n)(s2 + n− 1)
×
q∑
k=1
(
[δikY
s1−n+φik(s1)Y
1−s1−n][δjk(s2− n)Y s2n−1 + (1− s2− n)φjk(s2)Y −s2−n]
− [δik(s1 − n)Y s1−n−1 + (1− s1 − n)φik(s1)Y
−s1−n][δjkY s2−n + φjk(s2)Y 1−s2−n]
)
= δij
1
s1 + s2 − 1
Y s1+s2−1 +
1
s2 − s1
φij(s1)Y
s2−s1 +
1
s1 − s2
φji(s2)Y
s1−s2
− 1
s1 + s2 − 1
q∑
k=1
φik(s1)φjk(s2)Y
1−s1−s2 . ■
Theorem E.3. As Y → ∞,
q∑
j=1
∫∫
FY
∣∣∣∣Ej
(
z,
1
2
+ ir;n
)∣∣∣∣2y2ndµ(z)
=
1
2ir
Tr
[
Φ
(
1
2
− ir
)
Y 2ir − Φ
(
1
2
+ ir
)
Y −2ir
]
+ 2q log Y − φ′
φ
(
1
2
+ ir
)
.
Proof. From the proof of the previous theorem, we find that∫∫
FY
∣∣∣∣Ej
(
z,
1
2
+ ir;n
)∣∣∣∣2y2ndµ(z) = ∫∫
F
∣∣∣∣EY
j
(
z,
1
2
+ ir;n
)∣∣∣∣2y2ndµ(z)
+ exponentially decaying terms.
Setting s1 = s2 = σ + ir in the previous theorem, we find that∫∫
F
∣∣∣∣EY
j
(
z,
1
2
+ ir;n
)∣∣∣∣2y2ndµ(z) = 1
2σ − 1
Y 2σ−1 − 1
2ir
φjj(σ + ir)Y −2ir
+
1
2ir
φjj(σ + ir)Y 2ir
− 1
2σ − 1
q∑
k=1
φjk(σ + ir)φjk(σ + ir)Y 1−2σ.
We want to take the limit when σ → 1/2. We find that as σ → 1/2,
Y 2σ−1 = 1 + (2σ − 1) log Y +O
(
(2σ − 1)2
)
,
φjk(σ + ir) = φjk
(
1
2
+ ir
)
+
(
σ − 1
2
)
φ′
jk
(
1
2
+ ir
)
+O
(
(2σ − 1)2
)
.
Hence,
q∑
j=1
∫∫
F
∣∣∣∣EY
j
(
z,
1
2
+ ir;n
)∣∣∣∣2y2ndµ(z)
=
1
2ir
Tr
[
Φ
(
1
2
− ir
)
Y 2ir − Φ
(
1
2
+ ir
)
Y −2ir
]
+
q
2σ − 1
(
1 + (2σ − 1) log Y
)
− 1
2σ − 1
q∑
j=1
q∑
k=1
[
φjk
(
1
2
+ ir
)
+
(
σ − 1
2
)
φ′
jk
(
1
2
+ ir
)]
38 L.-P. Teo
×
[
φkj
(
1
2
− ir
)
+
(
σ − 1
2
)
φ′
kj
(
1
2
− ir
)](
1− (2σ − 1) log Y
)
+O(2σ − 1).
Now the matrix Φ(s) =
[
φij(s)
]
satisfies
Φ(s)Φ(1− s) = I.
Hence, for each 1 ≤ j ≤ q,
q∑
k=1
φjk
(
1
2
+ ir
)
φkj
(
1
2
− ir
)
= 1.
Moreover,
Φ(1− s) = Φ(s)−1.
Since
d
ds
log detΦ(s) = Tr
[
Φ′(s)Φ−1(s)
]
,
and
d
ds
log detΦ(s) = − d
ds
log detΦ(1− s)
we find that
Tr
[
Φ′
(
1
2
+ ir
)
Φ−1
(
1
2
+ ir
)]
= Tr
[
Φ′
(
1
2
− ir
)
Φ−1
(
1
2
− ir
)]
.
This gives
q∑
j=1
∫∫
F
∣∣∣∣EY
j
(
z,
1
2
+ ir;n
)∣∣∣∣2y2ndµ(z) = 1
2ir
Tr
[
Φ
(
1
2
− ir
)
Y 2ir − Φ
(
1
2
+ ir
)
Y −2ir
]
+ 2q log Y − Tr
[
Φ′
(
1
2
+ ir
)
Φ−1
(
1
2
+ ir
)]
.
The assertion of the theorem follows. ■
F Asymptotic behavior of the zeta function Zell(s)
In this section, we want to compute the asymptotic behavior of logZell(s), where
Zell(s) =
v∏
j=1
mj−1∏
r=0
Γ
(
s+ r
mj
) 2αmj (r−n)+1−mj
2mj
Γ
(
s+ 2n+ r
mj
) 2αmj (r+n)+1−mj
2mj
,
when u = s+ n− 1
2 is large. From the asymptotic behavior of log Γ(s) (6.2), we have
logZell(s) = A u log u+ B log u+ Cu+ D + o(1),
where
A =
v∑
j=1
αj
mj
, B =
v∑
j=1
βj , C =
v∑
j=1
1
mj
(−αj − αj logmj),
Resolvent Trace Formula and Determinants of n Laplacians 39
D = −
v∑
j=1
βj logmj +
1
2
log(2π)
v∑
j=1
αj ,
and
αj =
mj−1∑
r=0
(
2αmj (r − n) + 1−mj
2mj
+
2αmj (r + n) + 1−mj
2mj
)
,
βj =
mj−1∑
r=0
{
2αmj (r − n) + 1−mj
2mj
(−n+ 1
2 + r
mj
− 1
2
)
+
2αmj (r + n) + 1−mj
2mj
(
n+ 1
2 + r
mj
− 1
2
)}
.
From (C.1), we find that αj = 0 for 1 ≤ j ≤ 0. Hence, A = C = 0, and
D = −
v∑
j=1
βj logmj .
It remains to calculate βj . First we notice that
βj =
1
mj
mj−1∑
r=0
{
r
2αmj (r − n) + 1−mj
2mj
+ r
2αmj (r + n) + 1−mj
2mj
}
because of (C.1). If αmj (n) = ℓ, then
αmj (r + n) =
{
ℓ+ r, 0 ≤ r ≤ mj − ℓ− 1,
ℓ+ r −mj , mj − ℓ ≤ r ≤ mj − 1,
αmj (r − n) =
{
mj + r − ℓ, 0 ≤ r ≤ ℓ− 1,
r − ℓ, ℓ ≤ r ≤ mj − 1.
With the help of a standard computer algebra, we find that
βj =
m2
j − 1
6mj
−
αmj (n)(mj − αmj (n))
mj
.
Hence,
B =
v∑
j=1
(
m2
j − 1
6mj
−
αmj (n)(mj − αmj (n))
mj
)
,
D = −
v∑
j=1
(
m2
j − 1
6mj
−
αmj (n)(mj − αmj (n))
mj
)
logmj .
Acknowledgements
This research is supported by the Ministry of Higher Education Malaysia through the Fundamen-
tal Research Grant Scheme (FRGS) FRGS/1/2018/STG06/XMU/01/1. We would like to thank
L. Takhtajan and J. Friedman who have given helpful comments and suggestions. We would
also like to thank the referees for reading the paper carefully and giving valuable comments.
40 L.-P. Teo
References
[1] Alekseevskii V.P., On functions similar to the gamma function, Comm. Proc. Kharkov Math. Soc. 1 (1889),
169–238.
[2] Andrews G.E., Askey R., Roy R., Special functions, Encyclopedia of Mathematics and its Applications,
Vol. 71, Cambridge University Press, Cambridge, 1999.
[3] Barnes E.W., The theory of the G-function, Q. J. Math. 31 (1900), 264–314.
[4] D’Hoker E., Phong D.H., On determinants of Laplacians on Riemann surfaces, Comm. Math. Phys. 104
(1986), 537–545.
[5] Efrat I., Determinants of Laplacians on surfaces of finite volume, Comm. Math. Phys. 119 (1988), 443–451.
[6] Fay J.D., Fourier coefficients of the resolvent for a Fuchsian group, J. Reine Angew. Math. 293–294 (1977),
143–203.
[7] Fischer J., An approach to the Selberg trace formula via the Selberg zeta-function, Lecture Notes in Math.,
Vol. 1253, Springer-Verlag, Berlin, 1987.
[8] Freixas i Montplet G., von Pippich A.-M., Riemann–Roch isometries in the non-compact orbifold setting,
J. Eur. Math. Soc. (JEMS) 22 (2020), 3491–3564, arXiv:1604.00284.
[9] Gong D.G., Zeta-determinant and torsion functions on Riemann surfaces of finite volume, Manuscripta
Math. 86 (1995), 435–454.
[10] Gradshteyn I.S., Ryzhik I.M., Table of integrals, series, and products, 6th ed., Academic Press Inc., San
Diego, CA, 2000.
[11] Hejhal D.A., The Selberg trace formula for PSL(2,R), Vol. 1, Lecture Notes in Math., Vol. 548, Springer-
Verlag, Berlin, 1976.
[12] Hejhal D.A., The Selberg trace formula for PSL(2,R), Vol. 2, Lecture Notes in Math., Vol. 1001, Springer-
Verlag, Berlin, 1983.
[13] Iwaniec H., Spectral methods of automorphic forms, 2nd ed., Graduate Studies in Mathematics, Vol. 53,
Amer. Math. Soc., Providence, RI, 2002.
[14] Koyama S., Determinant expression of Selberg zeta functions. I, Trans. Amer. Math. Soc. 324 (1991),
149–168.
[15] Koyama S., Determinant expression of Selberg zeta functions. III, Proc. Amer. Math. Soc. 113 (1991),
303–311.
[16] McIntyre A., Takhtajan L.A., Holomorphic factorization of determinants of Laplacians on Riemann surfaces
and a higher genus generalization of Kronecker’s first limit formula, Geom. Funct. Anal. 16 (2006), 1291–
1323, arXiv:math.CV/0410294.
[17] Sarnak P., Determinants of Laplacians, Comm. Math. Phys. 110 (1987), 113–120.
[18] Selberg A., Harmonic analysis and discontinuous groups in weakly symmetric Riemannian spaces with
applications to Dirichlet series, J. Indian Math. Soc. (N.S.) 20 (1956), 47–87.
[19] Selberg A., Göttingen lectures on harmonic analysis, in Alte Selberg Collected Papers, Springer Collected
Works in Mathematics, Springer-Verlag, Berlin, 1989, 626–675.
[20] Takhtajan L.A., Zograf P.G., A local index theorem for families of ∂-operators on punctured Riemann
surfaces and a new Kähler metric on their moduli spaces, Comm. Math. Phys. 137 (1991), 399–426.
[21] Takhtajan L.A., Zograf P.G., Local index theorem for orbifold Riemann surfaces, Lett. Math. Phys. 109
(2019), 1119–1143, arXiv:1701.00771.
[22] Teo L.-P., Ruelle zeta function for cofinite hyperbolic Riemann surfaces with ramification points, Lett. Math.
Phys. 110 (2020), 61–82, arXiv:1901.07898.
[23] Venkov A.B., Spectral theory of automorphic functions, Proc. Steklov Inst. Math. 153 (1982), 1–163.
[24] Venkov A.B., Kalinin V.L., Faddeev L.D., A nonarithmetic derivation of the Selberg trace formula, J. Soviet
Math. 8 (1977), 177–199.
[25] Voros A., Spectral functions, special functions and the Selberg zeta function, Comm. Math. Phys. 110
(1987), 439–465.
https://doi.org/10.1017/CBO9781107325937
https://doi.org/10.1007/BF01211063
https://doi.org/10.1007/BF01218082
https://doi.org/10.1515/crll.1977.293-294.143
https://doi.org/10.1007/BFb0077696
https://doi.org/10.4171/jems/992
https://arxiv.org/abs/1604.00284
https://doi.org/10.1007/BF02568004
https://doi.org/10.1007/BF02568004
https://doi.org/10.1007/BFb0079608
https://doi.org/10.1007/BFb0079608
https://doi.org/10.1007/BFb0061302
https://doi.org/10.1007/BFb0061302
https://doi.org/10.1090/gsm/053
https://doi.org/10.2307/2001500
https://doi.org/10.2307/2048513
https://doi.org/10.1007/s00039-006-0582-7
https://arxiv.org/abs/math.CV/0410294
https://doi.org/10.1007/BF01209019
https://doi.org/10.1007/BF02431886
https://doi.org/10.1007/s11005-018-01144-w
https://arxiv.org/abs/1701.00771
https://doi.org/10.1007/s11005-019-01222-7
https://doi.org/10.1007/s11005-019-01222-7
https://arxiv.org/abs/1901.07898
https://doi.org/10.1007/BF01212422
1 Introduction
2 Laplacians of n-differential
3 The Eisenstein series
4 The trace of the resolvent kernel
5 Dimensions of spaces holomorphic differentials
6 The determinant of n-Laplacian
A The inversion formulas
B The hyperbolic contribution
C Elliptic contribution
D Contribution from parabolic elements and absolutely continuous spectrum
D.1 Contribution from parabolic elements
D.2 Contribution of the absolutely continuous spectrum
D.3 The term Xi P
E Maass–Selberg relation
F Asymptotic behavior of the zeta function Z ell(s)
References
|
| id | nasplib_isofts_kiev_ua-123456789-211444 |
| institution | Digital Library of Periodicals of National Academy of Sciences of Ukraine |
| issn | 1815-0659 |
| language | English |
| last_indexed | 2026-03-15T10:21:32Z |
| publishDate | 2021 |
| publisher | Інститут математики НАН України |
| record_format | dspace |
| spelling | Teo, Lee-Peng 2026-01-02T08:35:30Z 2021 Resolvent Trace Formula and Determinants of Laplacians on Orbifold Riemann Surfaces. Lee-Peng Teo. SIGMA 17 (2021), 083, 40 pages 1815-0659 2020 Mathematics Subject Classification: 14H15; 11F72; 11M36 arXiv:2104.00895 https://nasplib.isofts.kiev.ua/handle/123456789/211444 https://doi.org/10.3842/SIGMA.2021.083 For a nonnegative integer, we consider the -Laplacian Δₙ acting on the space of -differentials on a confinite Riemann surface X which has ramification points. The trace formula for the resolvent kernel is developed along the line à la Selberg. Using the trace formula, we compute the regularized determinant of Δₙ + ( + 2 − 1), from which we deduce the regularized determinant of Δₙ, denoted by det′Δₙ. Taking into account the contribution from the absolutely continuous spectrum, det′Δₙ is equal to a constant Cₙ times ( ) when ≥ 2. Here ( ) is the Selberg zeta function of . When = 0 or = 1, ( ) is replaced by the leading coefficient of the Taylor expansion of ( ) around = 0 and = 1, respectively. The constants Cn are calculated explicitly. They depend on the genus, the number of cusps, as well as the ramification indices, but are independent of the moduli parameters. This research is supported by the Ministry of Higher Education Malaysia through the Fundamental Research Grant Scheme (FRGS) FRGS/1/2018/STG06/XMU/01/1. We would like to thank L. Takhtajan and J. Friedman, who have given helpful comments and suggestions. We would also like to thank the referees for carefully reviewing the paper and providing valuable comments. en Інститут математики НАН України Symmetry, Integrability and Geometry: Methods and Applications Resolvent Trace Formula and Determinants of Laplacians on Orbifold Riemann Surfaces Article published earlier |
| spellingShingle | Resolvent Trace Formula and Determinants of Laplacians on Orbifold Riemann Surfaces Teo, Lee-Peng |
| title | Resolvent Trace Formula and Determinants of Laplacians on Orbifold Riemann Surfaces |
| title_full | Resolvent Trace Formula and Determinants of Laplacians on Orbifold Riemann Surfaces |
| title_fullStr | Resolvent Trace Formula and Determinants of Laplacians on Orbifold Riemann Surfaces |
| title_full_unstemmed | Resolvent Trace Formula and Determinants of Laplacians on Orbifold Riemann Surfaces |
| title_short | Resolvent Trace Formula and Determinants of Laplacians on Orbifold Riemann Surfaces |
| title_sort | resolvent trace formula and determinants of laplacians on orbifold riemann surfaces |
| url | https://nasplib.isofts.kiev.ua/handle/123456789/211444 |
| work_keys_str_mv | AT teoleepeng resolventtraceformulaanddeterminantsoflaplaciansonorbifoldriemannsurfaces |