Dirichlet Scalar Determinants On

Two-Dimensional Constant Curvature Disks

Soumyadeep Chaudhuri and Frank Ferrari

Service de Physique Théorique et Mathématique
Université Libre de Bruxelles (ULB) and International Solvay Institutes
Campus de la Plaine, CP 231, B-1050 Bruxelles, Belgique

<EMAIL_ADDRESS><EMAIL_ADDRESS>
We compute exactly the scalar determinants $\det(\Delta+M^{2})$ on the two-
dimensional round disks of constant curvature $R=0$, $\mp 2$, for any finite
boundary length $\ell$ and mass $M$, with Dirichlet boundary conditions, using
the $\zeta$-function prescription. When $M^{2}=\pm q(q+1)$, $q\in\mathbb{N}$,
a simple expression involving only elementary functions and the Euler $\Gamma$
function is found. Applications to two-dimensional Liouville and Jackiw-
Teitelboim quantum gravity are presented in a separate paper.

###### Contents

  1. 1 Introduction
  2. 2 The case of infinite area
    1. 2.1 Euclidean space
    2. 2.2 Hyperbolic space
  3. 3 Generalities on the cases of finite area
    1. 3.1 Spectra and Sturm-Liouville operators
    2. 3.2 The large $M$ and large $\ell$ expansions
      1. 3.2.1 Flat case
      2. 3.2.2 Curved cases
  4. 4 Zero-mass determinants
  5. 5 Sturm-Liouville determinants
  6. 6 Ratio of determinants
    1. 6.1 Naive discussion
    2. 6.2 Correct discussion
  7. 7 Determinants on the flat disk
  8. 8 Determinants on the curved disks
  9. 9 The case $M^{2}=-\eta q(q+1)$ for $q\in\mathbb{N}$
    1. 9.1 General formula
    2. 9.2 Application: the large $\ell$ limit in negative curvature
  10. A Monotonicity of the eigenvalues as a function of the boundary length

##  1 Introduction

Motivated by applications in holography at finite cut-off and two-dimensional
quantum gravity models on finite geometries [1, 2], we compute the partition
function of a scalar field of mass $M$ on a round disk of finite boundary
length $\ell$, with Dirichlet boundary conditions, in zero, negative and
positive constant curvatures, that is to say, for a Ricci scalar
$R=2\eta/L^{2}$, $\eta=0$, $-1$ or $+1$. This amounts to evaluating the
determinants

$\mathscr{D}^{\eta}(M^{2},\ell,L)={\det}_{\text{D}}\bigl{(}\Delta_{\eta}+M^{2}\bigr{)}$
(1.1)

where $\Delta_{\eta}$ is the positive Laplacian for the appropriate disk
metrics. Explicitly, parameterizing the disk in polar coordinates
$(r,\theta)$, $0\leq r\leq 1$, we have

$\Delta_{\eta}f=-e^{-2\sigma_{\eta}}\biggl{(}\frac{1}{r}\partial_{r}\bigl{(}r\partial_{r}f\bigr{)}+\frac{1}{r^{2}}\partial_{\theta}^{2}f\biggr{)}$
(1.2)

for a conformal factor

$e^{\sigma_{0}}=\frac{\ell}{2\pi}\,\raise 2.0pt\hbox{,}\quad
e^{\sigma_{\pm}}=\frac{2r_{0}L}{1\pm r_{0}^{2}r^{2}}\,\cdotp$ (1.3)

The dimensionless parameter $r_{0}$ is related to the boundary length,
boundary extrinsic curvature and bulk area according to

$\ell^{\eta}(r_{0})=\frac{4\pi r_{0}L}{1+\eta r_{0}^{2}}\,\raise
2.0pt\hbox{,}\quad k^{\eta}(r_{0})=\frac{1-\eta r_{0}^{2}}{2r_{0}L}\,\raise
2.0pt\hbox{,}\quad A^{\eta}(r_{0})=\frac{4\pi r_{0}^{2}L^{2}}{1+\eta
r_{0}^{2}}=L\ell^{\eta}r_{0}\,.$ (1.4)

In negative curvature, $0\leq r_{0}<1$. In positive curvature, $r_{0}$ is an
arbitrary positive real number. In this case, the two disks associated with
$r_{0}$ and $1/r_{0}$ have the same boundary length but will of course yield
two different determinants. The notation $\mathscr{D}^{+}(M^{2},\ell,L)$ is
thus ambiguous and, when necessary, we use the more precise notation
$\mathscr{D}^{+>}$ and $\mathscr{D}^{+<}$ to distinguish between $r_{0}\geq 1$
and $r_{0}\leq 1$. Eventually, it will be more convenient to use the parameter
$r_{0}$ instead of $\ell$, see Eq. (1.16).

Let us briefly note that, instead of (1.1), one may prefer to study the
dimensionless determinant ${\det}_{\text{D}}[\mu^{-2}(\Delta_{\eta}+M^{2})]$,
where $\mu$ is an arbitrary renormalization scale. The equations (1.1), (1.2)
and (1.4) show that $\mu$ can be trivially absorbed by rescaling $\ell$, $L$
and $M$ appropriately. We thus set $\mu=1$ in the following and work with
dimensionless parameters.

Functional determinants like (1.1) needs to be renormalized and are thus
always defined modulo the addition of local counterterms. The possible
counterterms here are of the form

$\int\\!{\rm d}^{2}x\sqrt{g}\,,\quad\int\\!{\rm
d}^{2}x\sqrt{g}\,R\,,\quad\oint\\!{\rm d}s\,,\quad\oint\\!{\rm d}s\,k\,,$
(1.5)

where the integrals are taken over the disk bulk or the disk boundary. In
constant curvature, the two bulk terms are equivalent and yield the area
(cosmological constant) counterterm. Due to the Gauss-Bonnet formula, the
integral of the extrinsic curvature $k$ over the boundary also yields the area
and a $\ell$-independent constant. Overall, $\ln\mathscr{D}_{\eta}$ is thus
defined modulo the addition of terms of the form

$c_{0}(M,L)+c_{1}(M,L)\ell+c_{2}(M,L)A\,.$ (1.6)

When one uses a specific regularization and renormalization scheme, these
terms take a specific form, but they are not physical.

We shall use the $\zeta$-function scheme [3, 4] in the following. If
$(\lambda_{p}^{\eta})_{p\geq 0}$ denotes the set of eigenvalues of the
Laplacian, $\lambda^{\eta}_{0}\leq\lambda^{\eta}_{1}\leq\cdots$, the
$\zeta$-function is defined by

$\zeta^{\eta}(s;M^{2},\ell,L)=\sum_{p\geq
0}\frac{1}{\bigl{(}\lambda_{p}^{\eta}+M^{2}\bigr{)}^{s}}\,\cdotp$ (1.7)

The series converges for $\mathop{\text{Re}}\nolimits s>1$ and defines a
meromorphic function on the complex $s$-plane, that is regular at $s=0$. The
determinant is then defined as

$\mathscr{D}^{\eta}(M^{2},\ell,L)=e^{-(\zeta^{\eta})^{\prime}(0;M^{2},\ell,L)}\,,$
(1.8)

where the prime denotes the derivative with respect to $s$. This procedure may
seem abstract, but one can show explicitly that it is equivalent to
subtracting local counterterms from the effective action
$\ln\det\bigl{(}\Delta+M^{2}\bigr{)}$, see e.g. [4] for a pedagogical
discussion.

Let us note that the flat case can be obtained as the limit
$L\rightarrow\infty$ of either the positive (with the choice $r_{0}<1$) or the
negative curvature cases,

$\mathscr{D}^{0}(M^{2},\ell)=\lim_{L\rightarrow+\infty}\mathscr{D}^{-}(M^{2},\ell,L)=\lim_{L\rightarrow+\infty}\mathscr{D}^{+<}(M^{2},\ell,L)\,.$
(1.9)

Nevertheless, it will be useful to treat the flat case separately.

The definition (1.8) allows to immediately eliminate one parameter from our
problem. In zero curvature, we use

$\zeta^{0}\bigl{(}s;M^{2},\ell\bigr{)}=\Bigl{(}\frac{\ell}{2\pi}\Bigr{)}^{2s}\zeta^{0}\biggl{(}s;\Bigl{(}\frac{\ell
M}{2\pi}\Bigr{)}^{2},2\pi\biggr{)}$ (1.10)

to get

$\mathscr{D}^{0}\bigl{(}M^{2},\ell\bigr{)}=\Bigl{(}\frac{\ell}{2\pi}\Bigr{)}^{-2\zeta^{0}(0;(\ell
M/(2\pi))^{2},2\pi)}\mathscr{D}^{0}\biggl{(}\Bigl{(}\frac{\ell
M}{2\pi}\Bigr{)}^{2},2\pi\biggr{)}\\\
=\Bigl{(}\frac{\ell}{2\pi}\Bigr{)}^{-\frac{1}{3}+\frac{AM^{2}}{2\pi}}\mathscr{D}^{0}\biggl{(}\Bigl{(}\frac{\ell
M}{2\pi}\Bigr{)}^{2},2\pi\biggr{)}\,.$ (1.11)

We have used the general formula

$\zeta(0)=\frac{\chi}{6}-\frac{AM^{2}}{4\pi}=\frac{1}{6}-\frac{AM^{2}}{4\pi}$
(1.12)

for the value of the $\zeta$ function at $s=0$ in terms of the Euler
characteristics $\chi$, which is one for the disk, and the area $A$ of the
surface on which the Laplacian is defined. In the flat case, we thus work from
now on with $\ell=2\pi$ and note

$D^{0}(M^{2})=\mathscr{D}^{0}(M^{2},2\pi)\,.$ (1.13)

In non-zero curvature, we use

$\zeta^{\eta}\bigl{(}s;M^{2},\ell,L\bigr{)}=L^{2s}\zeta^{\eta}\bigl{(}s;(LM)^{2},\ell/L,1\bigr{)}$
(1.14)

to get

$\mathscr{D}^{\eta}\bigl{(}M^{2},\ell,L\bigr{)}=L^{-2\zeta^{\eta}(0;(LM)^{2},\ell/L,1)}\mathscr{D}^{\eta}\bigl{(}(LM)^{2},\ell/L,1\bigr{)}=L^{-\frac{1}{3}+\frac{A^{\eta}M^{2}}{2\pi}}\mathscr{D}^{\eta}\bigl{(}(LM)^{2},\ell/L,1\bigr{)}\,.$
(1.15)

From now on, we thus work with $L=1$ and note

$D^{\eta}(M^{2},r_{0})=\mathscr{D}^{\eta}(M^{2},\ell,1)\,.$ (1.16)

We shall use the parameter $r_{0}$ instead of $\ell$ to waive the ambiguity
associated with the fact that $\ell(r_{0})=\ell(1/r_{0})$ in the positive
curvature case.

Our main goal in this work will be to derive the following.

######  Theorem 1.1.

The functions $D^{0}(M^{2})$ and $D^{\eta}(M^{2},r_{0})$ are given by

$\displaystyle\begin{split}&\ln D^{0}(M^{2})=\frac{1}{3}\ln
2-\frac{1}{2}\ln(2\pi)-\frac{5}{12}-2\zeta_{\text{R}}^{\prime}(1)+\frac{1}{2}\bigl{(}\gamma-1-\ln
2\bigr{)}M^{2}\\\ &\hskip
227.62204pt+\sum_{n\in\mathbb{Z}}\ln\biggl{[}\frac{2^{|n|}|n|!}{M^{|n|}}I_{|n|}(M)e^{-\frac{M^{2}}{4(|n|+1)}}\biggr{]}\,,\end{split}$
(1.17) $\displaystyle\begin{split}&\ln
D^{\eta}(M^{2},r_{0})=-\frac{1}{2}\ln(2\pi)-\frac{5}{12}-2\zeta_{\text{R}}^{\prime}(1)+\frac{\eta
A^{\eta}(r_{0})}{3\pi}-\frac{1}{3}\ln r_{0}\\\
&+\frac{1}{2\pi}\Bigl{(}\gamma-1+\ln
r_{0}\Bigr{)}A^{\eta}(r_{0})M^{2}+\sum_{n\in\mathbb{Z}}\ln\biggl{[}f_{n}^{\eta}(1;-M^{2})e^{-\frac{A^{\eta}M^{2}}{4\pi(|n|+1)}F(1,1,|n|+2,\frac{\eta
A^{\eta}}{4\pi})}\biggr{]}\,,\end{split}$ (1.18)

where $\zeta_{\text{R}}$ is the Riemann $\zeta$-function, $\gamma$ is Euler’s
constant, $I_{n}$ is the modified Bessel function of the first kind, the
functions $f_{n}^{\eta}$ are hypergeometric functions defined in Eq. (3.5) and
(3.6), and $A^{\eta}$ is the area, defined in (1.4).

The above convergent series representation can be used to evaluate the
determinants numerically with very good accuracy. For instance, in flat space,
one can use the integral representation of the modified Bessel function,

$I_{n}(z)=\frac{(z/2)^{n}}{\sqrt{\pi}\Gamma(n+1/2)}\int_{0}^{\pi}\\!{\rm
d}t\,e^{z\cos t}(\sin t)^{2n}\,,$ (1.19)

to evaluate numerically $n!I_{n}$.111This is much more efficient than
evaluating directly $n!I_{n}$ in Mathematica, probably because it eliminates
the huge factorials by dividing explicitly by $\Gamma(n+1/2)$.

We do not know how compute the infinite sums in terms of known functions for
arbitrary values of $M^{2}$ but, interestingly, this can be done for special
values of $M^{2}$, which turn out to be important in specific quantum gravity
applications [1]. The simplifications occur when

$M^{2}=M^{2}_{\eta,q}=-\eta q(q+1)\,,\quad q\in\mathbb{N}\,.$ (1.20)

Note that in negative curvature, these special values of the mass are the
values for which the conformal dimension of the boundary operator associated
with the massive scalar field in hyperbolic space are integers,

$\Delta=\frac{1}{2}\Bigl{(}1+\sqrt{1+4M^{2}}\Bigr{)}=q+1\,.$ (1.21)

For these special values, the functions $f_{n}^{\pm}$ are polynomials of
degree $q$ of their argument $\eta A^{\eta}/(4\pi)$. For $q=0$, the infinite
sums are actually zero, term by term. For $q=1$, which is relevant in quantum
gravity, we find

$\ln
D^{\eta}\bigl{(}M^{2}=-2\eta,r_{0}\bigr{)}=-\frac{1}{2}\ln(2\pi)-\frac{5}{12}-2\zeta_{\text{R}}^{\prime}(1)+\frac{\eta}{\pi}\Bigl{(}\frac{4}{3}-\ln
r_{0}\Bigr{)}A^{\eta}\\\ -\frac{1}{3}\ln r_{0}+\ln\bigl{(}1-\eta
r_{0}^{2}\bigr{)}-\frac{3-\eta r_{0}^{2}}{1+\eta r_{0}^{2}}\ln\bigl{(}1+\eta
r_{0}^{2}\bigr{)}-2\ln\Gamma\Bigl{(}\frac{2}{1+\eta r_{0}^{2}}\Bigr{)}\,.$
(1.22)

The explicit formula for any integer $q$ is given in Section 9, Eq. (9.3).

When the mass parameter vanishes, the determinants are of course well-known,
because the calculation is then reduced to a straightforward application of
the conformal anomaly. The determinants have also been studied in infinite
volume, especially in negative curvature, motivated by applications in
holography [5] (see also [6] and the references therein). This will be
reviewed below. However, to the best of our knowledge, it is the first time
that explicit formulas are derived for massive scalar field determinants, or
determinants of the same kind, in finite volume. These calculations, and
possible generalizations, are relevant for studies of finite-size quantum
field theory and finite cut-off holography. Our main motivation came from
specific applications to two-dimensional Liouville and Jackiw-Teitelboim
quantum gravities [1].

The plan of the paper is as follows. We begin by briefly reviewing in Section
2 the case of infinite area in zero and negative curvatures, in order to put
our finite size results in perspective. In particular, it is emphasized that
the large area limit of the finite size determinants does not match the known
strictly infinite size formulas in negative curvature, an important and subtle
effect associated with hyperbolic space. In Section 3, we explain the general
set-up in finite area. The rotational invariance of the problem allows to
decompose the spectrum in terms of Fourier modes, with associated Sturm-
Liouville operators and determinants. We also study the large mass expansion
of the determinants by using the heat kernel. In flat space, the large mass
expansion is equivalent to the large area expansion, but this is not so in
curved space. In Section 4, we compute the zero mass determinants by using the
conformal anomaly. In Section 5, we compute the Sturm-Liouville determinants
for any value of the mass parameter. In Section 6, it is explained how to
compute the ratio between the massive and the massless determinants. This is
then used in Sections 7 and 8 to derive Eqs. (1.17) and (1.18). Section 9 is
devoted to the study of the special case $M^{2}=\mp q(q+1)$. We are able to
derive very explicit expressions for the determinants in these cases, in terms
of elementary functions and the Euler $\Gamma$ function, Eq. (9.3). Applying
these formulas, we obtain the large area limit of the determinants in negative
curvature.

##  2 The case of infinite area

###  2.1 Euclidean space

On the flat infinite Euclidean plane, the $\zeta$ function per unit area is
given by

$\zeta^{0}_{\infty}(s;M^{2})=\frac{1}{2\pi}\int_{0}^{\infty}\frac{\nu{\rm
d}\nu}{(\nu^{2}+M^{2})^{s}}=\frac{M^{2(1-s)}}{4\pi(s-1)}\,\cdotp$ (2.1)

The measure $\frac{1}{2\pi}\nu{\rm d}\nu$ is a Euclidean “Plancherel” measure
straightforwardly obtained by diagonalizing the Laplacian with plane waves and
going to polar coordinates in momentum space. We get the determinant

$\ln D^{0}_{\infty}=\frac{M^{2}}{4\pi}\bigl{(}1-\ln M^{2}\bigr{)}\,.$ (2.2)

In order to understand the negative curvature case below, it is interesting to
rederive this result by using a basis of eigenfunctions diagonalizing the
rotation generator around the origin. One thus expands the plane waves in
Fourier components

$e^{i\vec{p}\cdot\vec{x}}=e^{i\nu
r\cos(\theta-\phi)}=\sum_{n\in\mathbb{Z}}i^{n}J_{|n|}(\nu
r)e^{in(\theta-\phi)}\,,$ (2.3)

where $\theta$ and $\phi$ are the polar angles of $\vec{x}$ and $\vec{p}$ and
$J$ the Bessel functions. This yields an orthonormal basis of eigenfunctions

$\mathscr{J}_{\nu,n}(r,\theta)=\sqrt{\frac{\nu}{2\pi}}\,e^{in\theta}J_{|n|}(\nu
r)\,.$ (2.4)

The normalization is chosen in such a way that the usual normalization
condition for the plane waves implies

$\int_{0}^{2\pi}\\!{\rm d}\theta\,\int_{0}^{1}\\!{\rm
d}r\,r\mathscr{J}^{*}_{\nu_{1},n_{1}}(r,\theta)\mathscr{J}_{\nu_{2},n_{2}}(r,\theta)=\delta(\nu_{1}-\nu_{2})\delta_{n_{1},n_{2}}\,.$
(2.5)

The Fourier expansion of a good function $f$ in the basis
$(\mathscr{J}_{\nu,n})_{\nu\geq 0,\,n\in\mathbb{Z}}$ then reads

$f(r,\theta)=\sum_{n\in\mathbb{Z}}\int_{0}^{\infty}\\!{\rm
d}\nu\,c_{n}(\nu)\mathscr{J}_{\nu,n}(r,\theta)$ (2.6)

with coefficients

$c_{n}(\nu)=\int_{0}^{2\pi}\\!{\rm d}\theta\,\int_{0}^{1}\\!{\rm
d}r\,r\mathscr{J}^{*}_{\nu,n}(r,\theta)f(r,\theta)\,.$ (2.7)

In this formalism, the position-space $\zeta$ function reads

$\zeta^{0}_{\infty}(s;M^{2};\vec{x}_{1},\vec{x}_{2})=\sum_{n\in\mathbb{Z}}\int_{0}^{\infty}\\!{\rm
d}\nu\,\frac{\mathscr{J}^{*}_{\nu,n}(r_{1},\theta_{1})\mathscr{J}_{\nu,n}(r_{2},\theta_{2})}{(\nu^{2}+M^{2})^{s}}\,\cdotp$
(2.8)

Because Euclidean space is homogeneous, the $\zeta$ function at coinciding
points $\vec{x}_{1}=\vec{x}_{2}$ is actually space-independent. Integrating
over space then yields an infinite area factor. Factorizing this factor yields
the zeta function per unit area

$\zeta^{0}_{\infty}(s;M^{2})=\int_{0}^{\infty}\frac{\mathfrak{m}_{\text{E}}(\nu){\rm
d}\nu}{(\nu^{2}+M^{2})^{s}}\,\raise 2.0pt\hbox{,}$ (2.9)

with an integration measure, the “Euclidean Plancherel measure,” given by

$\mathfrak{m}_{\text{E}}(\nu)=\sum_{n\in\mathbb{Z}}\bigl{|}\mathscr{J}_{\nu,n}(0)\bigr{|}^{2}=\frac{\nu}{2\pi}\,\cdotp$
(2.10)

Of course, we find again the formula (2.1).

###  2.2 Hyperbolic space

There is a strong formal similarity between the spectral problems for the
Laplacian on the non-compact hyperbolic space and on the Euclidean space
reviewed above. As in the Euclidean case, the spectrum is continuous and a
basis of eigenfunctions, analogous to the plane waves, exists. Working in
units for which $L=1$ and thus the curvature is $R=-2$, the eigenvalues of the
Laplacian are parameterized as $\frac{1}{4}+\nu^{2}$ with $\nu\geq 0$. The
fact that the eigenvalues are always greater than $1/4$ is the famous
Breitenlohner-Freedman bound. The usual plane waves are replaced by non-
Euclidean plane waves which are given explicitly by an appropriate power of
the Poisson kernel. They can be expanded in Fourier components as

$\biggl{(}\frac{1-r^{2}}{1+r^{2}-2r\cos(\theta-\phi)}\biggr{)}^{\frac{1}{2}+i\nu}=\sum_{n\in\mathbb{Z}}F_{\nu,|n|}(r)e^{in(\theta-\phi)}$
(2.11)

where

$F_{\nu,n}(r)=\frac{\Gamma(1/2+i\nu+n)}{n!\Gamma(1/2+i\nu)}r^{n}F\Bigl{(}\frac{1}{2}+i\nu,\frac{1}{2}-i\nu,n+1,-\frac{r^{2}}{1-r^{2}}\Bigr{)}$
(2.12)

is given in terms of a usual hypergeometric function. This yields an
orthonormal basis

$\mathscr{F}_{\nu,n}(r,\theta)=\mathscr{N}_{n}(\nu)e^{in\theta}F_{\nu,|n|}(r)\,,$
(2.13)

with a normalization factor $\mathscr{N}_{n}(\nu)$ chosen in such a way that

$\int_{0}^{2\pi}\\!{\rm d}\theta\,\int_{0}^{1}\\!{\rm
d}r\,\frac{4r}{(1-r^{2})^{2}}\mathscr{F}^{*}_{\nu_{1},n_{1}}(r,\theta)\mathscr{F}_{\nu_{2},n_{2}}(r,\theta)=\delta(\nu_{1}-\nu_{2})\delta_{n_{1},n_{2}}\,.$
(2.14)

The non-Euclidean Fourier expansion reads

$f(r,\theta)=\sum_{n\in\mathbb{Z}}\int_{0}^{\infty}\\!{\rm
d}\nu\,c_{n}(\nu)\mathscr{F}_{\nu,n}(r,\theta)$ (2.15)

with non-Euclidean Fourier coefficients

$c_{n}(\nu)=\int_{0}^{2\pi}\\!{\rm d}\theta\,\int_{0}^{1}\\!{\rm
d}r\,\frac{4r}{(1-r^{2})^{2}}\mathscr{F}^{*}_{\nu,n}(r,\theta)f(r,\theta)\,.$
(2.16)

The $\zeta$ function per unit area is then given by

$\zeta^{-}_{\infty}(s;M)=\int_{0}^{\infty}\frac{{\rm
d}\nu\,\mathfrak{m}(\nu)}{(\nu^{2}+1/4+M^{2})^{s}}\,\raise 2.0pt\hbox{,}$
(2.17)

with an integration measure, called the Plancherel measure, given by

$\mathfrak{m}(\nu)=\sum_{n\in\mathbb{Z}}\bigl{|}\mathscr{F}_{\nu,n}(0)\bigr{|}^{2}=\bigl{|}\mathscr{N}_{0}(\nu)\bigr{|}^{2}\,.$
(2.18)

There remains the problem of determining the normalization factor, so that
(2.14) is satisfied. This looks a priori non-trivial, because the integral
involving the product of hypergeometric functions does not seem to be
standard. However, the $F_{\nu,n}$ are eigenfunctions for the Sturm-Liouville
operators $L_{n}^{-}$ defined in Eq. (3.1) below, in the case $r_{0}=1$. A
standard trick in Sturm-Liouville theory allows to compute the normalization
integrals of the eigenfunctions in terms of their asymptotics. These
asymptotics are well-known in the case of the hypergeometric function (2.12),
and thus the calculation can be done straightforwardly. Since this is not so
important for our purposes, we will refrain from giving more details here. The
result is the famous Plancherel measure on the hyperbolic space

$\mathfrak{m}(\nu)=\frac{\nu\tanh(\pi\nu)}{2\pi}\,\cdotp$ (2.19)

From this, one may obtain a rather explicit formula for the determinant. In
order to perform the analytic continuation of the $\zeta$ function to $s=0$,
it is useful to write

$\tanh(\pi\nu)=1-\frac{2e^{-2\pi\nu}}{1+e^{-2\pi\nu}}\,\cdotp$ (2.20)

This allows to isolate the non-holomorphic piece, which is entirely given by
the simple pole at $s=1$,

$\zeta^{-}_{\infty}(s;M^{2})=\frac{(M^{2}+1/4)^{1-s}}{4\pi(s-1)}-\frac{1}{\pi}\int_{0}^{\infty}\\!{\rm
d}\nu\,\frac{\nu}{e^{2\pi\nu}+1}\frac{1}{(M^{2}+\nu^{2}+1/4)^{s}}\,\cdotp$
(2.21)

This yields the logarithm of the infinite volume determinant, per unit area,

$\ln
D^{-}_{\infty}=\frac{M^{2}+1/4}{4\pi}\Bigl{(}1-\ln(M^{2}+1/4)\Bigr{)}-\frac{1}{\pi}\int_{0}^{\infty}\\!{\rm
d}\nu\,\frac{\nu}{e^{2\pi\nu}+1}\ln\bigl{(}M^{2}+\nu^{2}+1/4\bigr{)}\,.$
(2.22)

_Comments:_

i) The integral in (2.22) might be given a more explicit expression, see e.g.
[7]. The general result is that in even dimensions, the formulas are
complicated, but in odd dimensions simple closed-form formulas can be found,
because the Plancherel measures are then much simpler.

ii) Note that the expression (2.22) does not have any direct physical
significance, since it corresponds to an area counterterm. The same can of
course be said in the flat space case.

iii) The Plancherel measure has an obvious interpretation as a density of
eigenvalues per unit area of the Laplacian. More precisely, any naive discrete
finite area expression involving a sum over the eigenvalues is replaced by an
integral with the Plancherel measure,

$\frac{1}{A}\sum_{p\geq
0}f(\lambda_{p})\longrightarrow\int_{0}^{\infty}\\!{\rm
d}\nu\,\mathfrak{m}(\nu)f(\nu^{2}+1/4)\,.$ (2.23)

iv) However, the interpretation of the Plancherel measure as the infinite area
density of eigenvalues must be taken with extreme caution due to the following
subtlety associated with hyperbolic space [8].222We would like to thank Steve
Zelditch for emphasizing this crucial property.

If one considers the problem for the disk of finite area and finite
circumference $\ell$, as we are going to do below, the spectrum of the
Laplacian is discrete. At this level, the situation is very similar to the
case of flat space. One can define a density of eigenvalues per unit area

$\mathfrak{r}(\lambda)=\frac{1}{A}\sum_{p\geq 0}\delta(\lambda-\lambda_{p})$
(2.24)

such that

$\frac{1}{A}\sum_{p\geq 0}f(\lambda_{p})=\int\\!{\rm
d}\lambda\,\mathfrak{r}(\lambda)f(\lambda)=2\int_{0}^{\infty}\\!{\rm
d}\nu\,\nu\,\mathfrak{r}(\nu^{2}+1/4)f(\nu^{2}+1/4)\,.$ (2.25)

Note that the finite area density $\mathfrak{r}$ depends, of course, on the
type of boundary conditions we use; we focus in the present paper on Dirichlet
boundary conditions.

In the case of flat space, it is easy to check that the large $\ell$ limit of
the density is independent of the boundary conditions and yields the limit
measure $\frac{1}{2\pi}\nu{\rm d}\nu$, Eq. (2.10). But things are much more
complicated in hyperbolic space. The qualitative origin of the difficulty is
the well-known geometric property of hyperbolic space that, at large $\ell$,
the area scales as the circumference, $A\sim\ell$, and, moreover, the area of
an annulus of upper circumference $\ell$ and fixed geodesic width $a$ divided
by the total area goes to a constant $1-e^{-a}$ instead of zero. As a
consequence, the large $\ell$ limit is keeping track of the boundary condition
used in finite area and the limit of the density of eigenvalues (2.25) does
not coincide with the Plancherel measure (2.19) [8]. The actual form of the
limit of the density $\mathfrak{r}$ is not known, even for simple classes of
boundary conditions like Dirichlet or Neumann! This is an outstanding open
problem. Our exact finite area results below will allow us to illustrate this
subtle effect in a non-trivial way, see the end of Section 4 and especially
Section 9.2.

##  3 Generalities on the cases of finite area

As explained above, we set $\ell=2\pi$ in flat space and $L=1$ in curved
space.

###  3.1 Spectra and Sturm-Liouville operators

The spectral problem for the Laplacian is studied by decomposing the
eigenfunctions in Fourier components. This reduces the problem to computing
the spectra of purely radial Sturm-Liouville operators,

$L_{n}^{\eta}=-\frac{1}{\omega_{\eta}(r)^{2}}\Biggl{[}\frac{1}{r}\frac{{\rm
d}}{{\rm d}r}\Bigl{(}r\frac{{\rm d}}{{\rm
d}r}\,\cdot\Bigr{)}-\frac{n^{2}}{r^{2}}\Biggr{]}\,,$ (3.1)

with

$\omega_{0}(r)=1\,,\quad\omega_{\pm}(r)=\frac{2r_{0}}{1\pm
r_{0}^{2}r^{2}}\,\cdotp$ (3.2)

With Dirichlet boundary conditions at $r=1$, the operators $L_{n}^{\eta}$ are
symmetric with respect to the scalar product

$\langle
f_{1}|f_{2}\rangle=\int_{0}^{1}\omega_{\eta}(r)^{2}rf_{1}(r)f_{2}(r)\,{\rm
d}r\,.$ (3.3)

Note that $\smash{L_{n}^{\eta}=L_{-n}^{\eta}}$, it is thus enough to consider
$n\geq 0$. The spectra of the operators $\smash{L_{n}^{\eta}}$ are discrete,
and we denote their eigenvalues by $\lambda^{\eta}_{n,k}$, $n\geq 0$, ordered
such that $\smash{\lambda^{\eta}_{n,k+1}>\lambda^{\eta}_{n,k}}$. The
eigenvalues are strictly positive, because the $\smash{L_{n}^{\eta}}$ are
positive operators and the constant mode is projected out by the Dirichlet
boundary conditions. In negative curvature, we actually know that all the
eigenvalues are strictly greater than $1/4$ by the Breitenlohner-Freedman
bound, see below.

A complete basis of orthogonal eigenfunctions is straightforward to find. One
first seeks the solution of $L_{n}^{\eta}\cdot f=\lambda f$, for $\lambda>0$,
that is regular at $r=0$. One finds

$f_{n}^{0}(r;\lambda)=J_{|n|}\bigl{(}\sqrt{\lambda}\,r\bigr{)}\,,$ (3.4)

where $J_{|n|}$ is the Bessel function of the first kind, in the flat case,
and

$\displaystyle f_{n}^{-}(r;\lambda)$
$\displaystyle=r^{|n|}F\Bigl{(}\frac{1}{2}+\frac{i}{2}\sqrt{4\lambda-1},\frac{1}{2}-\frac{i}{2}\sqrt{4\lambda-1},|n|+1,-\frac{r_{0}^{2}r^{2}}{1-r_{0}^{2}r^{2}}\Bigr{)},$
(3.5) $\displaystyle f_{n}^{+}(r;\lambda)$
$\displaystyle=r^{|n|}F\Bigl{(}\frac{1}{2}+\frac{1}{2}\sqrt{4\lambda+1},\frac{1}{2}-\frac{1}{2}\sqrt{4\lambda+1},|n|+1,\frac{r_{0}^{2}r^{2}}{1+r_{0}^{2}r^{2}}\Bigr{)},$
(3.6)

where $F=_{2}\\!\\!F_{1}$ is the standard hypergeometric function, in the
negative and positive curvature cases.333These eigenfunctions are not
normalized. One then imposes the Dirichlet condition,

$f_{n}^{\eta}(1;\lambda)=0\,,$ (3.7)

whose solutions yield the full spectrum.

One can be a little bit more precise on the structure of the spectrum, by
using the fact that the eigenvalues $\lambda^{\eta}_{n,k}(r_{0})$ are strictly
decreasing functions of $r_{0}$. This intuitive result is derived explicitly
in App. A. In positive curvature, the limit $r_{0}\rightarrow\infty$ must
yield the spectrum $\\{p(p+1),p\in\mathbb{N}\\}$ of the Laplacian on the round
sphere.444One may worry that the limit yields the round sphere minus a point,
but this is immaterial at the level of $L^{2}$-spaces. In particular, it is
straightforward to check that the ground state wave function of the Dirichlet
problem, which is the ground state of the operator $L_{0}^{+}$, converges with
respect to the norm associated with the scalar product (3.3) to the s-wave
constant mode ground state of the round sphere problem. In particular, all the
eigenvalues must be strictly greater than 2, except the lowest
$\lambda_{0,0}^{+}$.

The $\zeta$ functions 1.7 we wish to study can be conveniently written as

$\zeta^{\eta}(s;M^{2})=\sum_{p\geq
0}\frac{1}{\bigl{(}\lambda_{p}^{\eta}+M^{2}\bigr{)}^{s}}=\zeta^{\eta}_{0}(s;M^{2})+2\sum_{n\geq
1}\zeta^{\eta}_{n}(s;M^{2})\,,$ (3.8)

where

$\zeta^{\eta}_{n}(s;M^{2})=\sum_{k\geq
0}\frac{1}{\bigl{(}\lambda_{n,k}^{\eta}+M^{2}\bigr{)}^{s}}$ (3.9)

is the $\zeta$-function for the operator $L^{\eta}_{n}+M^{2}$, with associated
Sturm-Liouville determinants

$d_{n}^{\eta}=\det\bigl{(}L_{n}^{\eta}+M^{2}\bigr{)}=e^{-(\zeta^{\eta}_{n})^{\prime}(0)}\,.$
(3.10)

Note that one and two-dimensional Weyl’s law, see e.g. [9], yield the
asymptotics of the eigenvalues as

$\lambda^{\eta}_{n,k}\underset{k\rightarrow\infty}{\sim}\Bigl{(}\frac{\pi
k}{a^{\eta}}\Bigr{)}^{2}\,,\quad\lambda^{\eta}_{p}\underset{p\rightarrow\infty}{\sim}\frac{4\pi
p}{A^{\eta}}\,\raise 2.0pt\hbox{,}$ (3.11)

where

$a^{\eta}=\int_{0}^{1}\\!\omega_{\eta}(r)\,{\rm d}r=\begin{cases}1&\text{if
$\eta=0$},\\\ 2\mathop{\text{arctanh}}\nolimits
r_{0}=\ln\frac{1+r_{0}}{1-r_{0}}&\text{if $\eta=-1$},\\\ 2\arctan
r_{0}&\text{if $\eta=+1$}\end{cases}$ (3.12)

and $A^{\eta}$ is the area, see Eq. (1.4). This shows that the series
representations for $\zeta_{n}^{\eta}(s)$ and $\zeta^{\eta}(s)$ converge for
$\mathop{\text{Re}}\nolimits s>1/2$ and $\mathop{\text{Re}}\nolimits s>1$,
respectively.

Our strategy will be to discuss first the Sturm-Liouville determinants, in
Section 5, before dealing with the full problem.

_Remark:_

The $\zeta$ functions are unambiguously defined when $M^{2}$ is chosen in such
a way that $\lambda_{n,k}^{\eta}+M^{2}>0$. If this is not the case, the result
for the determinant is obtained by analytically continuing in $M^{2}$. This
analytic continuation is easy to perform by using the following observation.
If one considers the $\zeta$ function defined in (1.7) and if one sets
$\zeta^{\eta,p_{0}}(s)=\sum_{p\geq p_{0}}(\lambda_{p}^{\eta}+M^{2})^{-s}$,
then the determinant is such that

$D_{\eta}=e^{-(\zeta^{\eta})^{\prime}(0)}=\Biggl{(}\prod_{p=0}^{p_{0}-1}\bigl{(}\lambda_{p}^{\eta}+M^{2}\bigr{)}\Biggr{)}e^{-(\zeta^{\eta,p_{0}})^{\prime}(0)}\,.$
(3.13)

This formula makes the analytic continuation obvious by choosing $p_{0}$ such
that $\lambda_{p}^{\eta}+M^{2}>0$ for all $p\geq p_{0}$.

###  3.2 The large $M$ and large $\ell$ expansions

####  3.2.1 Flat case

In the flat case, Eq. (1.10), (1.11) show that the large $\ell$ limit is
equivalent to the large $M$ limit. But it is well-known that the large $M$
limit is simple and given in terms of the heat kernel expansion.

Indeed, we can write555The dependence in $\ell$ is kept implicit in our
notation for the $\zeta$ function.

$\zeta^{0}(s)=\frac{1}{\Gamma(s)}\int_{0}^{\infty}\\!{\rm
d}t\,t^{s-1}e^{-M^{2}t}K^{0}(t)\,,$ (3.14)

where $K^{0}$ is the heat kernel for the flat Laplacian. At large $M$, the
integral (3.14) is dominated by the small $t$ region in which we can use the
expansion

$K^{0}(t)=\frac{1}{4\pi t}\sum_{k\in\mathbb{N}/2}a_{k}t^{k}\,.$ (3.15)

This immediately yields the large $M$ asymptotic expansion of the $\zeta$
function

$\zeta^{0}(s)\underset{M\rightarrow\infty}{=}\frac{1}{4\pi\Gamma(s)}\sum_{k\in\mathbb{N}/2}a_{k}\Gamma(s-1+k)M^{2-2k-2s}$
(3.16)

and thus of

$\ln\mathscr{D}^{0}(M^{2},\ell)\underset{M\rightarrow\infty}{=}\frac{M^{2}a_{0}}{4\pi}\bigl{(}1-\ln
M^{2}\bigr{)}+\frac{Ma_{1/2}}{2\sqrt{\pi}}+\frac{a_{1}}{2\pi}\ln M\\\
-\frac{1}{4\pi}\sum_{\begin{subarray}{c}k\in\mathbb{N}/2\\\ k\geq
3/2\end{subarray}}a_{k}\Gamma(k-1)M^{2-2k}\,.$ (3.17)

The heat kernel coefficients for Dirichlet boundary conditions are well-known,
see e.g. [10, 11]. The first coefficients for the flat disk666The first three
coefficients are universal: $a_{0}$ is Weyl’s law, $a_{1/2}$ gives the
correction to Weyl’s law in the presence of a boundary and $a_{1}$ is related
to the Euler characteristics. are

$a_{0}=A=\frac{\ell^{2}}{4\pi}\,\raise 2.0pt\hbox{,}\quad
a_{1/2}=-\frac{\sqrt{\pi}}{2}\ell\,,\quad a_{1}=\frac{2\pi}{3}\,\raise
2.0pt\hbox{,}\quad a_{3/2}=\frac{\pi^{5/2}}{16\ell}\,\raise 2.0pt\hbox{,}$
(3.18)

which yields

$\ln D^{0}(M^{2})=\frac{1}{4}M^{2}\bigl{(}1-\ln M^{2}\bigr{)}-\frac{\pi
M}{2}+\frac{1}{3}\ln M-\frac{\pi}{128M}+O\bigl{(}M^{-2}\bigr{)}\,.$ (3.19)

Using Eq. (1.11), we get

$\ln\mathscr{D}^{0}(M^{2},\ell)=\Bigl{(}\frac{\ell
M}{4\pi}\Bigr{)}^{2}\bigl{(}1-\ln M^{2}\bigr{)}-\frac{\ell
M}{4}+\frac{1}{3}\ln M-\frac{\pi^{2}}{64\ell M}+O\bigl{(}(\ell
M)^{-2}\bigr{)}\,.$ (3.20)

Per unit area, and at leading order, this matches with the infinite area
result (2.2).

####  3.2.2 Curved cases

In positive or negative curvature, the only change compared to the discussion
in the previous subsection concerns the evaluation of the heat kernel
coefficient $a_{3/2}$. The general formula yields [11]

$a_{3/2}=-\frac{\sqrt{\pi}}{64}\oint\\!{\rm
d}s\,\bigl{(}4R-k^{2}\bigr{)}=\frac{\sqrt{\pi}}{64}\,\ell\biggl{[}-9\eta+\Bigl{(}\frac{2\pi}{\ell}\Bigr{)}^{2}\biggr{]}\,.$
(3.21)

Using (1.15), we get

$\ln\mathscr{D}^{\eta}(M^{2},\ell,L)=\frac{A^{\eta}M^{2}}{4\pi}\bigl{(}1-\ln
M^{2}\bigr{)}-\frac{\ell M}{4}+\frac{1}{3}\ln M\\\ -\frac{\pi^{2}}{64\ell
M}\Bigl{(}1-\frac{9\eta\ell^{2}}{4\pi^{2}L^{2}}\Bigr{)}+O\bigl{(}1/(LM)^{2}\bigr{)}\,.$
(3.22)

Curvature effects have generated the new term
$-9\eta\ell^{2}/(4\pi^{2}L^{2})$. In negative curvature, this term shows
explicitly that the large $\ell$ expansion is distinct from the large $M$
expansion. As already mentioned at the end of Section 2, the large $\ell$
limit will be much more difficult to study; see the end of the next Section
and Section 9.2 for further discussion.

##  4 Zero-mass determinants

For $M=0$, the determinants can be straightforwardly computed using the
conformal anomaly.

The conformal anomaly relates the determinants of the Laplacian for metrics in
the same conformal class. In general, if $g$ and $g_{0}$ are two metrics such
that $g=e^{2\sigma}g_{0}$, then

$\ln\det\Delta_{g}=\ln\det\Delta_{g_{0}}-\frac{1}{12\pi}\tilde{S}_{\text{L}}(g_{0},g)\,.$
(4.1)

The “total” action $\tilde{S}_{\text{L}}$ is the sum
$\tilde{S}_{\text{L}}=S_{\text{L}}+\Delta S_{\text{L}}$ of the usual Liouville
action

$S_{\text{L}}(g_{0},g)=\int\\!d^{2}x\sqrt{g_{0}}\,\bigl{(}g_{0}^{\mu\nu}\partial_{\mu}\sigma\partial_{\nu}\sigma+R_{0}\sigma\bigr{)}+2\oint\\!{\rm
d}s_{0}\,k_{0}\sigma$ (4.2)

and of a correction term

$\Delta S_{\text{L}}(g_{0},g)=3\oint\\!{\rm
d}s_{0}\,n_{0}^{\mu}\partial_{\mu}\sigma=3\oint\Bigl{(}{\rm d}s\,k-{\rm
d}s_{0}k_{0}\Bigr{)}\,.$ (4.3)

We note $R$, $k$ and $R_{0}$, $k_{0}$ the Ricci scalar and extrinsic curvature
of the boundary for the metrics $g$ and $g_{0}$ respectively. The correction
term $\Delta S_{\text{L}}$ is a counterterm contribution and is thus
irrelevant for physics. We include it because we want to provide exact
formulas in the $\zeta$-function scheme and this counterterm is precisely
generated in the $\zeta$-function computation of the determinant with
Dirichlet boundary conditions.

In the flat case, applying the above general formulas for $g_{0}$ and $g$
being the flat disk metrics of circumferences $2\pi$ and $\ell$ respectively,
we get

$\ln\mathscr{D}^{0}(0,\ell)=\ln\mathscr{D}^{0}(0,2\pi)-\frac{1}{3}\ln\frac{\ell}{2\pi}=\ln
D^{0}(0)-\frac{1}{3}\ln\frac{\ell}{2\pi}\,\cdotp$ (4.4)

The determinant $\mathscr{D}^{0}(0,2\pi)=D^{0}(0)$ was computed in [12],

$\ln D^{0}(0)=\frac{1}{3}\ln
2-\frac{1}{2}\ln(2\pi)-\frac{5}{12}-2\zeta_{\text{R}}^{\prime}(-1)\,.$ (4.5)

This is a physically irrelevant and scheme-dependent constant, but it
corresponds to the precise value in the $\zeta$-function scheme that we are
using.

Note that the $\ell$-dependence in (4.4) also follows from Eq. (1.11) at
$M=0$. As a result, the conformal anomaly doesn’t really provide any new
information in the flat case. In this respect, the situation is more
interesting in the curved cases. Evaluating the action $\tilde{S}_{\text{L}}$
for the conformal factors $\sigma_{\pm}$ given in (1.3), using in particular
(1.4), we get

$\ln\mathscr{D}^{\pm}(0,\ell,L)=\ln D^{0}(0)\pm\frac{A^{\pm}}{3\pi
L^{2}}-\frac{1}{3}\ln(2r_{0}L)\,,$ (4.6)

where $r_{0}$ is related to $\ell$ by the relation given in (1.4). The
$L$-dependence is as predicted by (1.15) and we also obtain

$\ln D^{\pm}(0,r_{0})=\ln
D^{0}(0)\pm\frac{A^{\pm}}{3\pi}-\frac{1}{3}\ln(2r_{0})\,.$ (4.7)

_Remark:_

This formula predicts a nice large $\ell$ expansion in negative curvature, of
the form

$\ln
D^{-}\bigl{(}0,r_{0}(\ell)\bigr{)}=-\frac{\ell}{3\pi}+\frac{1}{4}-\frac{1}{2}\ln(2\pi)-2\zeta_{\text{R}}^{\prime}(-1)+\frac{2\pi^{3}}{9\ell^{3}}-\frac{8\pi^{5}}{15\ell^{5}}+O\bigl{(}1/\ell^{7}\bigr{)}\,.$
(4.8)

As expected, the leading term is proportional to the area, which is
$A^{-}=\ell+O(1)$ at leading order. However, the coefficient $-\frac{1}{3\pi}$
does not match with the coefficient predicted by the Plancherel measure, Eq.
(2.22), which is approximately $0.0538$. This is a direct illustration of the
fact, explained at the end of Section 2, that the large $\ell$ limit of the
determinants do not match with the $\ell=\infty$ Plancherel description.

##  5 Sturm-Liouville determinants

Sturm-Liouville determinants have been considered by many authors in the
literature [13].

######  Proposition 5.1.

We have, for $n\geq 0$,

$\displaystyle\frac{d^{0}_{n}(M^{2})}{d^{0}_{n}(0)}=\frac{\det(L_{n}^{0}+M^{2})}{\det
L_{n}^{0}}=\prod_{k\geq
0}\biggl{[}1+\frac{M^{2}}{\lambda^{0}_{n,k}}\biggr{]}=\frac{2^{n}n!}{M^{n}}I_{n}(M)\,,$
(5.1)
$\displaystyle\frac{d^{\pm}_{n}(M^{2})}{d^{\pm}_{n}(0)}=\frac{\det(L_{n}^{\pm}+M^{2})}{\det
L_{n}^{\pm}}=\prod_{k\geq
0}\biggl{[}1+\frac{M^{2}}{\lambda^{\pm}_{n,k}}\biggr{]}=f_{n}^{\pm}(1;-M^{2})\,,$
(5.2)

where the functions $f_{n}^{\eta}$ are defined in (3.4) and (3.5).

###### Proof.

Weyl’s law, Eq. (3.11), implies that the series representation for the
difference of the $\zeta$ functions

$\zeta_{n}^{\eta}(s;M^{2})-\zeta_{n}^{\eta}(s;0)=\sum_{k\geq
0}\biggl{[}\frac{1}{\bigl{(}\lambda^{\eta}_{n,k}+M^{2}\bigr{)}^{s}}-\frac{1}{\bigl{(}\lambda^{\eta}_{n,k}\bigr{)}^{s}}\biggr{]}$
(5.3)

converges for $\mathop{\text{Re}}\nolimits s>-1/2$. The derivative with
respect to $s$ at $s=0$ can thus be computed by taking the derivative of each
term in the sum. Using (3.10), this yields the infinite product
representations in (5.1) and (5.2). These infinite products can be computed
using standard ideas from complex analysis, as we now explain.

In the zero curvature case, Eq. (3.4) and (3.7) show that the eigenvalues
$\lambda_{n,k}^{0}$ are the squares of the zeros of the Bessel function
$J_{n}$. The result then follows from the standard infinite product
representation of the Bessel function in terms of its zeros and from the
identity $(iz)^{-n}J_{n}(iz)=z^{-n}I_{n}(z)$.

The cases of non-zero curvature are conceptually similar. Let us first note
that, by Weyl’s law (3.11), the infinite products $\prod_{k\geq
0}(1-z/\lambda^{\pm}_{n,k})$ define entire functions of the complex variable
$z$, with simple zeros at $z=\lambda^{\pm}_{n,k}$. The functions
$f_{n}^{\pm}(1;z)$ are also entire. As a consequence of the Dirichlet
condition (3.7), the zeros of these functions coincide with the eigenvalues
$\lambda_{n,k}^{\pm}$; in particular, they are all real and positive. One can
also check that these zeros are all simple. The ratio

$\frac{f_{n}^{\pm}(1;z)}{\prod_{k\geq 0}(1-z/\lambda^{\pm}_{n,k})}=e^{h(z)}$
(5.4)

is thus a non-vanishing entire function and can be written as the exponential
of an entire function $h$. Consider then

$h^{\prime}(z)=\frac{\partial_{z}f_{n}^{\pm}}{f_{n}^{\pm}}-\sum_{k\geq
0}\frac{1}{z-\lambda^{\pm}_{n,k}}\,\cdotp$ (5.5)

If we could prove that this derivative vanishes, we would conclude that $h$ is
a constant and, by evaluating the left-hand side of (5.4) at $z=0$ and using
$f_{n}^{\pm}(1;0)=1$, we would find that this constant is zero, proving Eq.
(5.2).

To derive that $h^{\prime}=0$, we consider a closed contour
$\mathcal{C}^{\pm}_{k}$ such that: i) it encircles the first $k$ eigenvalues
$\smash{\lambda^{\pm}_{n,0},\ldots,\lambda^{\pm}_{n,k-1}}$; ii) it never comes
close to any eigenvalue, i.e. the distance between $\mathcal{C}^{\pm}_{k}$ and
any eigenvalue is always greater than a strictly positive, $k$-independent
constant (this can always be achieved because the distance between successive
eigenvalues increases with $k$ at large $k$ according to Weyl’s law, Eq.
(3.11)); iii) $|z|\rightarrow\infty$ along the contour when
$k\rightarrow\infty$. Concretely, we can choose $\mathcal{C}^{\pm}_{k}$ to be
a circle centered at $z=0$ and of radius
$\smash{R^{\pm}_{k}=\frac{1}{2}(\lambda^{\pm}_{n,k-1}+\lambda^{\pm}_{n,k})}$.
Using Cauchy theorem, we note that

$\frac{1}{2i\pi}\oint_{\mathcal{C}_{k}}\frac{{\rm
d}z^{\prime}}{z^{\prime}-z}\frac{\partial_{z}f_{n}^{\pm}}{f_{n}^{\pm}}(1;z^{\prime})=\frac{\partial_{z}f_{n}^{\pm}}{f_{n}^{\pm}}\bigl{(}1;z\bigr{)}-\sum_{k^{\prime}=0}^{k-1}\frac{1}{z-\lambda^{\pm}_{n,k^{\prime}}}\,\raise
2.0pt\hbox{,}$ (5.6)

for all $z$ in the interior of the circle $\mathcal{C}_{k}$. When
$k\rightarrow\infty$, the right-hand side of this equation converges to
$h^{\prime}(z)$. To evaluate the left-hand side, we use an asymptotic analysis
of $f_{n}^{\pm}$. The result we need is discussed in [14] and can be cast in
the form

$f_{n}^{\pm}(1;z)\underset{|z|\rightarrow\infty}{\sim}c^{\pm}_{n}z^{-n/2}J_{n}(a^{\pm}\sqrt{z})\,,$
(5.7)

where the $a^{\pm}$ are defined in (3.12) and the $c_{n}^{\pm}$ depend on
$r_{0}$ and $n$ but not on $z$. Using the standard asymptotics for the Bessel
function $J_{n}$, we get in this way

$\frac{\partial_{z}f_{n}^{\pm}}{f_{n}^{\pm}}(z)=-\frac{n}{2z}-\frac{a^{\pm}}{2\sqrt{z}}\Biggl{[}\frac{1}{2a^{\pm}\sqrt{z}}+\Bigl{(}1+O\bigl{(}1/z\bigr{)}\Bigr{)}\tan\Bigl{(}a^{\pm}\sqrt{z}-\frac{n\pi}{2}-\frac{\pi}{4}\Bigr{)}\\\
+\frac{4n^{2}-1}{8a^{\pm}\sqrt{z}}\Bigl{(}1+O\bigl{(}1/z\bigr{)}\Bigr{)}\cos^{-2}\Bigl{(}a^{\pm}\sqrt{z}-\frac{n\pi}{2}-\frac{\pi}{4}\Bigr{)}\Biggr{]}\,.$
(5.8)

On the contours $\mathcal{C}^{\pm}_{k}$, we have
$\smash{z=R_{k}^{\pm}e^{i\alpha}}$. When $|\alpha|>\varepsilon$, where
$\varepsilon$ is a small, $k$-independent strictly positive constant, one
deduces from (5.8) and the elementary properties of the tangent and cosine
functions for complex arguments with a non-zero imaginary part, that
$\smash{|\partial_{z}f_{n}^{\pm}/f_{n}^{\pm}|}$ is bounded above by
$\smash{C/\sqrt{R_{k}^{\pm}}}$, for some $k$-independent constant $C$. When
$|\alpha|\leq\varepsilon$, this is also true, thanks to the fact that, by
construction, the distance between the contour and the eigenvalues
$\lambda_{n,k^{\prime}}^{\pm}$, or equivalently the poles of the tangent and
inverse cosine functions appearing in the right-hand side of Eq. (5.8), is
always strictly greater than a fixed, $k$-independent constant. When
$k\rightarrow\infty$, the integrand in the left-hand side of Eq. (5.6) thus
goes to zero as $\smash{1/(R_{k}^{\pm})^{3/2}}$ and the integral itself as
$\smash{1/\sqrt{R_{k}^{\pm}}}$. ∎

##  6 Ratio of determinants

###  6.1 Naive discussion

Naively, following the same logic as in Section 5, and using (5.1), we could
be tempted to write, e.g. in the flat case,

$\frac{D^{0}(M^{2})}{D^{0}(0)}=\prod_{n\in\mathbb{Z}}\prod_{k\geq
0}\biggl{[}1+\frac{M^{2}}{\lambda^{0}_{n,k}}\biggr{]}=I_{0}(M)\prod_{n\geq
0}\biggl{[}\frac{2^{n}n!}{M^{n}}I_{n}(M)\biggr{]}^{2}\,.$ (6.1)

However, using the large $n$ asymptotics of $I_{n}(z)$,777This expansion is
valid uniformly in $z$, as long as $|\arg z|<\frac{\pi}{2}-\epsilon$ for any
small $\epsilon>0$ and $z$ is not greater than $O(n)$.

$I_{n}(z)=\frac{z^{n}}{\sqrt{2\pi}}\frac{e^{\sqrt{n^{2}+z^{2}}}}{(n^{2}+z^{2})^{1/4}(n+\sqrt{n^{2}+z^{2}})^{n}}\biggl{[}1-\frac{1}{12\sqrt{n^{2}+z^{2}}}+O\bigl{(}1/n^{2}\bigr{)}\biggr{]}\,,$
(6.2)

we realize that the infinite product is divergent, since

$\ln\Bigl{(}\frac{2^{n}n!}{M^{n}}I_{n}(M)\Bigr{)}=\frac{M^{2}}{4n}+O\bigl{(}1/n^{2}\bigr{)}\,.$
(6.3)

In contrast to one-dimensional Sturm-Liouville operators, taking the ratio of
two-dimensional operators does not yield a finite answer. In particular, the
product of the regularized Sturm-Liouville determinants does not yield the
regularized two-dimensional determinant.

###  6.2 Correct discussion

The difficulty with the ratio of two-dimensional determinants comes from the
fact that, when we consider the subtraction

$\zeta^{\eta}(s;M^{2})-\zeta^{\eta}(s;0)=\sum_{p\geq
0}\biggl{[}\frac{1}{\bigl{(}\lambda_{p}^{\eta}+M^{2}\bigr{)}^{s}}-\frac{1}{\bigl{(}\lambda_{p}^{\eta}\bigr{)}^{s}}\biggr{]}\,,$
(6.4)

Weyl’s law, Eq. (3.11), shows that the resulting series representation
converges only for $\mathop{\text{Re}}\nolimits s>0$. Since $s=0$ is not in
the interior of the convergence region, we cannot take the derivative under
the summation and then set $s=0$; the resulting series would diverge. This is
in contrast to what happens for one-dimensional determinants, see Eq. (5.3)
and below.

To cure the problem, we thus consider

$\sum_{p\geq
0}\biggl{[}\frac{1}{\bigl{(}\lambda_{p}^{\eta}+M^{2}\bigr{)}^{s}}-\frac{1}{\bigl{(}\lambda_{p}^{\eta}\bigr{)}^{s}}+\frac{sM^{2}}{\bigl{(}\lambda_{p}^{\eta}\bigr{)}^{s+1}}\biggr{]}=\zeta^{\eta}(s;M^{2})-\zeta^{\eta}(s;0)+sM^{2}\zeta^{\eta}(s+1;0)\,,$
(6.5)

which converges for $\mathop{\text{Re}}\nolimits s>-1$. We can thus take the
derivative of both sides of this equation with respect to $s$ and then take
the limit $s\rightarrow 0$. To obtain the correct limit of the right-hand
side, one must recall that the $\zeta$-functions $\zeta^{\eta}(s;M^{2})$ have
a single pole at $s=1$, with residue given by Weyl’s law,

$\mathop{\text{res}}\nolimits_{s=1}\zeta^{\eta}=\frac{A^{\eta}}{4\pi}\,\cdotp$
(6.6)

This is a standard result that can be derived by using the heat kernel
expansion, see [10, 4]. Introducing the so-called “integrated two-point
function at coinciding points”

$C^{\eta}=\lim_{s\rightarrow
1}\Bigl{(}\zeta^{\eta}(s;0)-\frac{A^{\eta}}{4\pi}\frac{1}{s-1}\Bigr{)}$ (6.7)

we get the correct infinite product formula for the ratio of two-dimensional
determinants,

$\frac{D^{\eta}(M^{2})}{D^{\eta}(0)}=e^{C^{\eta}M^{2}}\prod_{p\geq
0}\biggl{[}\frac{\lambda_{p}^{\eta}+M^{2}}{\lambda_{p}^{\eta}}e^{-M^{2}/\lambda_{p}^{\eta}}\biggr{]}\,.$
(6.8)

Note that the right-hand side of this formula takes the form of a Weierstrass
canonical product associated with the sequence $(\lambda_{p}^{\eta})_{p\geq
0}$.

##  7 Determinants on the flat disk

In the flat case, Eq. (6.8) reads

$\frac{D^{0}(M^{2})}{D^{0}(0)}=e^{C^{0}M^{2}}\prod_{n\in\mathbb{Z}}\prod_{k\geq
0}\Biggl{[}\biggl{(}1+\frac{M^{2}}{\lambda^{0}_{n,k}}\biggr{)}e^{-M^{2}/\lambda^{0}_{n,k}}\Biggr{]}\,.$
(7.1)

The product over $k$ is evaluated as

$\prod_{k\geq
0}\Biggl{[}\biggl{(}1+\frac{M^{2}}{\lambda^{0}_{n,k}}\biggr{)}e^{-M^{2}/\lambda^{0}_{n,k}}\Biggr{]}=\Biggl{[}\prod_{k\geq
0}\biggl{(}1+\frac{M^{2}}{\lambda^{0}_{n,k}}\biggr{)}\Biggl{]}e^{-M^{2}\sum_{k\geq
0}\frac{1}{\lambda^{0}_{n,k}}}=\frac{2^{n}n!}{M^{n}}I_{n}(M)e^{-M^{2}\zeta^{0}_{n}(1;0)}\,,$
(7.2)

where we have used Eq. (5.1) and the definition of the Sturm-Liouville $\zeta$
function given in Eq. (3.9). Overall we thus obtain

$\frac{D^{0}(M^{2})}{D^{0}(0)}=e^{C^{0}M^{2}}I_{0}(M)e^{-M^{2}\zeta^{0}_{0}(1;0)}\prod_{n\geq
1}\biggl{[}\frac{2^{n}n!}{M^{n}}I_{n}(M)e^{-M^{2}\zeta^{0}_{n}(1;0)}\biggr{]}^{2}\,.$
(7.3)

The constants $\zeta^{0}_{n}(1;0)$ can be easily computed by expanding the
second equality in Eq. (5.1) in power of $M^{2}$ and identifying the
coefficients of $M^{2}$, using

$I_{n}(z)=(z/2)^{n}\sum_{k=0}^{\infty}\frac{(z/2)^{2k}}{k!(n+k)!}\,\cdotp$
(7.4)

One finds

$\zeta^{0}_{n}(1;0)=\frac{1}{4(n+1)}\,\cdotp$ (7.5)

To compute $C^{0}$ requires more work. This constant is related to the
divergence of the series representation

$\zeta^{0}(s;M^{2})=\zeta^{0}_{0}(s;M^{2})+2\sum_{n\geq
1}\zeta^{0}_{n}(s;M^{2})$ (7.6)

when $s\rightarrow 1^{+}$ and thus to the large $n$ behaviour of
$\zeta^{0}_{n}(s;M^{2})$. We are going to show that

$\zeta^{0}_{n}(s;0)\underset{n\rightarrow\infty}{\sim}\frac{1}{4\sqrt{\pi}}\frac{\Gamma(s-1/2)}{s\Gamma(s)}\frac{1}{n^{2s-1}}\,\cdotp$
(7.7)

Assuming for the moment that this is correct, consider

$\sum_{n\geq
1}\biggl{(}\zeta^{0}_{n}(s;0)-\frac{1}{4\sqrt{\pi}}\frac{\Gamma(s-1/2)}{s\Gamma(s)}\frac{1}{n^{2s-1}}\biggr{)}=\\\
\frac{1}{2}\zeta^{0}(s;0)-\frac{1}{2}\zeta^{0}_{0}(s;0)-\frac{1}{4\sqrt{\pi}}\frac{\Gamma(s-1/2)}{s\Gamma(s)}\zeta_{\text{R}}(2s-1)\,.$
(7.8)

The series on the left-hand side converges for $\mathop{\text{Re}}\nolimits
s>0$. Taking the $s\rightarrow 1$ limit on both sides of the equality yields

$C^{0}=\frac{1}{2}\Bigl{(}\gamma-1-\ln 2\Bigr{)}\,.$ (7.9)

To prove (7.7), we use the following contour integral representation

$\zeta^{0}_{n}(s;0)=\int_{\Gamma}\frac{{\rm
d}\lambda}{2i\pi}\frac{r_{n}(\lambda)}{\lambda^{s}}\,\raise 2.0pt\hbox{,}$
(7.10)

which is valid when $\mathop{\text{Re}}\nolimits s>1/2$, where the resolvent
function is

$r_{n}(\lambda)=\sum_{k\geq
0}\frac{1}{\lambda-\lambda^{0}_{n,k}}=-\biggl{[}\frac{1}{2z}\frac{{\rm
d}}{{\rm
d}z}\ln\Bigl{(}\bigl{(}z/2\bigr{)}^{-n}I_{n}(z)\Bigr{)}\biggr{]}\bigl{(}z=i\sqrt{\lambda}\bigr{)}\,,$
(7.11)

and the contour $\Gamma$ encircles counterclockwise all the
$\lambda^{0}_{n,k}$ (we take it to be close to the positive real
$\lambda$-axis, going from $+\infty$ to $\lambda^{0}_{n,0}-\epsilon$ with a
small positive imaginary part and going back from
$\smash{\lambda^{0}_{n,0}-\epsilon}$ to $+\infty$ with a small negative
imaginary part). Using the large $|\lambda|$ behaviour of
$I_{n}(i\sqrt{\lambda})=i^{n}J_{n}(\sqrt{\lambda})$ for
$\epsilon<|\arg\lambda|<\pi-\epsilon$, it is straightforward to check that the
contour can be deformed to go around the branch cut on the negative real
$\lambda$-axis counterclockwise. With $v=-\lambda$, this yields

$\zeta^{0}_{n}(s;0)=-\frac{\sin(\pi s)}{\pi}\int_{0}^{\infty}\\!{\rm
d}v\,\frac{r_{n}(-v)}{v^{s}}$ (7.12)

when $1/2<\mathop{\text{Re}}\nolimits s<1$.888Here we have ignored the
contribution of the small arc around the branch point which connects the two
legs of the deformed contour. This is valid as long as
$\mathop{\text{Re}}\nolimits s<1$. Using the uniform asymptotics (6.2), we
find, with $z=\sqrt{v}$,

$\biggl{[}\frac{1}{2z}\frac{{\rm d}}{{\rm
d}z}\ln\Bigl{(}\bigl{(}z/2\bigr{)}^{-n}I_{n}(z)\Bigr{)}\biggr{]}\bigl{(}z=\sqrt{v}\bigr{)}\underset{n\rightarrow\infty}{\sim}\frac{1}{2n}\frac{1}{1+\sqrt{1+v/n^{2}}}\,\raise
2.0pt\hbox{,}$ (7.13)

which, together with (7.11), yields

$\zeta^{0}_{n}(s;0)\underset{n\rightarrow\infty}{\sim}\frac{\sin(\pi
s)}{2\pi}\frac{1}{n^{2s-1}}\int_{0}^{\infty}\frac{{\rm
d}x}{x^{s}(1+\sqrt{1+x})}\,\cdotp$ (7.14)

Performing the integral yields (7.7), a priori for
$1/2<\mathop{\text{Re}}\nolimits s<1$, but then for any $s$ by analytic
continuation.

Let us note that Eq. (7.9) will be rederived in the next section from a more
general method.

Putting everything together, Eqs. (4.5), (7.3) and (7.9), we get the explicit
exact formula (1.17) for the determinant in flat space.

_Remark:_

Combining (1.17) and (3.19) we get the large $z$ asymptotic expansion

$\ln
I_{0}(z)+2\sum_{n=1}^{\infty}\ln\Bigl{[}\bigl{(}z/2\bigr{)}^{-n}n!I_{n}(z)e^{-\frac{z^{2}}{4(n+1)}}\Bigr{]}=\\\
\frac{z^{2}}{2}\bigl{(}2+\ln 2-\gamma-\ln z\bigr{)}-\frac{\pi
z}{2}+\frac{1}{3}\ln z\\\ -\frac{1}{3}\ln
2+\frac{1}{2}\ln(2\pi)+\frac{5}{12}+2\zeta_{\text{R}}^{\prime}(-1)-\frac{\pi}{128z}+O\bigl{(}1/z^{2}\bigr{)}\,.$
(7.15)

Higher order terms could be obtained straightforwardly from the heat kernel
expansion. It seems rather non-trivial to derive this expansion directly from
the series representation on the left-hand side of the above equation, because
we do not know an approximation of $I_{n}(z)$ at large $z$ uniformly in $n$
(whereas (6.2) is an approximation of $I_{n}(z)$ at large $n$ uniformly in
$z$).

In the case of negative curvature, for which a similar formula for the
determinant, Eq. (1.18), will be derived below, the heat kernel method is
useless to derive the interesting large $\ell$ asymptotics. It seems that this
can be done in full generality only if a method to treat directly the
asymptotics of the infinite series is devised. See Section 9.2 for a further
discussion in the special cases $M^{2}=q(q+1)$, $q\in\mathbb{N}$.

##  8 Determinants on the curved disks

Starting from Eq. (6.8) and using (5.2), we get

$\frac{D^{\pm}(M^{2},r_{0})}{D^{\pm}(0,r_{0})}=e^{C^{\pm}M^{2}}f^{\pm}_{0}(1,-M^{2})e^{-M^{2}\zeta^{\pm}_{0}(1;0)}\prod_{n\geq
1}\biggl{[}f^{\pm}_{n}(1;-M^{2})e^{-M^{2}\zeta^{\pm}_{n}(1;0)}\biggr{]}^{2}\,,$
(8.1)

which is the curved space version of (7.3).

The constants $\zeta^{\pm}_{n}(1;0)$ are computed by expanding the second
equality in Eq. (5.2) in power of $M^{2}$ and identifying the coefficients of
$M^{2}$. This yields

$\zeta_{n}^{\pm}(1;0)=-\partial_{\lambda}f_{n}^{\pm}(1;\lambda=0)=\mp(\partial_{1}-\partial_{2})F\bigl{(}1,0,n+1,\pm
A^{\pm}/(4\pi)\bigr{)}\,,$ (8.2)

where $\partial_{1}$ and $\partial_{2}$ are the partial derivatives with
respect to the first and second argument of the hypergeometric function,
respectively. From the usual series representation of the hypergeometric
function, we get $\partial_{1}F(1,0,n+1,z)=0$ and

$\partial_{2}F(1,0,n+1,z)=\frac{z}{n+1}F(1,1,n+2,z)\,,$ (8.3)

from which we obtain

$\zeta^{\pm}_{n}(1;0)=\frac{A^{\pm}}{4\pi}\frac{1}{n+1}F\bigl{(}1,1,n+2,\pm
A^{\pm}/(4\pi)\bigr{)}\,.$ (8.4)

As in flat space, the computation of the constant $C^{\pm}$, defined in Eq.
(6.7), is the trickier part. We could try to follow the direct approach used
in Section 7. This would require to start by generalizing Eq. (7.7). Instead,
we prefer to use a more general approach, which will also allow to find again
the result (7.9).

We are going to use the fact that a nice transformation law exists for $C$ as
a function of the metric, when one performs a Weyl rescaling. This is similar
to the conformal anomaly that we have used in Section 4 to compute the
determinants at zero mass. We follow some of the reasonings in [15]; the case
of the disk topology we are interested in is actually simpler than the closed
compact case studied in this reference.999The reason is that on a closed
compact surface there is a zero mode that yields non-trivial contributions.
This zero mode is absent on a disk with Dirichlet boundary condition.

We consider a disk with an arbitrary metric $g$. We note as usual the
eigenvalues of the Laplacian with Dirichlet boundary conditions $\lambda_{p}$,
with $0<\lambda_{0}<\lambda_{1}<\cdots$, and the corresponding eigenfunctions
$\psi_{p}$. The space-dependent $\zeta$ function is defined by

$\zeta(s,x,y;g)=\sum_{p\geq
0}\frac{\psi_{p}(x)\psi_{p}(y)}{\lambda_{p}^{s}}\,\raise 2.0pt\hbox{,}$ (8.5)

generalizing (1.7). The two-point function is

$G(x,y;g)=\zeta(1,x,y;g)\,.$ (8.6)

Note that these formulas have obvious generalizations to $M\not=0$, but, for
the present discussion, whose aim is to compute $C$, we need only the case of
zero mass.

Using standard perturbation theory to compute the infinitesimal variations of
the eigenvalues and of the eigenfunctions under an infinitesimal Weyl
rescaling of the metric, $\delta g=2\delta\sigma g$, under which the Laplacian
varies as $\delta\Delta=-2\delta\sigma\Delta$, one finds that the two-point
function is invariant,

$G\bigl{(}x,y;e^{2\sigma}g_{0}\bigr{)}=G\bigl{(}x,y;g_{0}\bigr{)}\,.$ (8.7)

The two-point function at coinciding point may be defined in two natural ways.
One is to subtract directly the logarithmic divergence,

$G_{\text{R}}(x;g)=\lim_{y\rightarrow x}\Bigl{(}G(x,y;g)+\frac{1}{2\pi}\ln
d(x,y;g)\Bigr{)}\,,$ (8.8)

where $d(x,y)$ denotes the geodesic distance, and another is to subtract
instead the pole at $s=1$ in the zeta function,

$G_{\text{R}}^{(\zeta)}(x;g)=\lim_{s\rightarrow
1}\Bigl{(}\zeta(s,x,x;g)-\frac{1}{4\pi}\frac{1}{s-1}\Bigr{)}\,.$ (8.9)

One can show (see e.g. [15]) that

$G_{\text{R}}^{(\zeta)}(x;g)=G_{\text{R}}(x;g)+\frac{\gamma-\ln
2}{2\pi}\,\cdotp$ (8.10)

The function $G_{\text{R}}^{(\zeta)}$ (or equivalently $G_{\text{R}}$) thus
transforms non-trivially under a Weyl rescaling, but all the dependence in the
conformal factor $\sigma$ comes from the geodesic distance,

$G_{\text{R}}^{(\zeta)}\bigl{(}x;e^{2\sigma}g_{0}\bigr{)}=G_{\text{R}}^{(\zeta)}\bigl{(}x;g_{0}\bigr{)}+\frac{\sigma(x)}{2\pi}\,\cdotp$
(8.11)

The constant $C[g]$ defined by (6.7) is expressed as

$C[g]=\int\\!{\rm d}^{2}x\sqrt{g}\,G_{\text{R}}^{(\zeta)}(x;g)$ (8.12)

and thus, using (8.11), is also given by

$C[g]=\int\\!{\rm
d}^{2}x\sqrt{g_{0}}\,e^{2\sigma}\Bigl{[}G_{\text{R}}^{(\zeta)}(x;g_{0})+\frac{\sigma(x)}{2\pi}\Bigr{]}\,.$
(8.13)

For instance, if $g_{0}=\delta$ is the metric for the flat disk of
circumference $2\pi$ and $g=(\frac{\ell}{2\pi})^{2}\delta$ is the metric for
the flat disk of circumference $\ell$, (8.13) yields

$C\Bigl{[}\Bigl{(}\frac{\ell}{2\pi}\Bigr{)}^{2}\delta\Bigr{]}=\frac{A^{0}}{\pi}\biggl{(}C\bigl{[}\delta\bigr{]}+\frac{1}{2}\ln\frac{\ell}{2\pi}\biggr{)}\,,$
(8.14)

which agrees with the formula (7.9) for

$C[\delta]=\frac{1}{2}\bigl{(}\gamma-1-\ln 2\bigr{)}\,.$ (8.15)

We want to use (8.13) to compute $C[g]$ for the hyperbolic and spherical
disks. We thus choose $g_{0}=\delta$ and the conformal factors as in Eq.
(1.3). To proceed, we need $G_{\text{R}}^{(\zeta)}(x;\delta)$. The two-point
function on the unit disk can be computed by the method of images and is given
by

$G(\vec{x},\vec{y};\delta)=-\frac{1}{2\pi}\ln\frac{|\vec{x}-\vec{y}|}{\sqrt{|\vec{x}|^{2}|\vec{y}|^{2}-2\vec{x}\cdot\vec{y}+1}}\,\cdotp$
(8.16)

It is straightforward to check that this formula satisfies all the
requirements characterizing the two-point function:
$G(\vec{x},\vec{y})=G(\vec{y},\vec{x})$,
$\Delta_{\vec{x}}G(\vec{x},\vec{y})=\delta(\vec{x}-\vec{y})$ for all $\vec{x}$
and $\vec{y}$ in the interior of the disk and $G(\vec{x},\vec{y})=0$ is
$\vec{x}$ or $\vec{y}$ is on the boundary, i.e. $|\vec{x}|=1$ or
$|\vec{y}|=1$. From (8.8) and (8.10) we thus get

$G_{\text{R}}^{(\zeta)}(\vec{x};\delta)=\frac{1}{2\pi}\Bigl{[}\ln\bigl{(}1-r^{2}\bigr{)}+\gamma-\ln
2\Bigr{]}\,.$ (8.17)

As a check, using (8.12), we get

$C[\delta]=\int_{0}^{1}\\!{\rm
d}r\,r\Bigl{[}\ln\bigl{(}1-r^{2}\bigr{)}+\gamma-\ln
2\Bigr{]}=\frac{1}{2}\bigl{(}\gamma-1-\ln 2\bigr{)}\,,$ (8.18)

which is consistent with (8.15). Finally, Eq. (8.13) yields the constants
$C=C^{\pm}$ for the spherical and hyperbolic disks,

$\displaystyle C^{\pm}$ $\displaystyle=\int_{0}^{1}\\!{\rm
d}r\,\frac{4r_{0}^{2}r}{(1\pm
r_{0}^{2}r^{2})^{2}}\Bigl{[}\ln\bigl{(}1-r^{2}\bigr{)}+\gamma-\ln
2+\ln\frac{2r_{0}}{1\pm r_{0}^{2}r^{2}}\Bigr{]}$ (8.19)
$\displaystyle=\frac{A^{\pm}}{2\pi}\Bigl{(}\gamma-1+\ln r_{0}\Bigr{)}\,.$
(8.20)

Putting together (4.5), (4.7), (8.1), (8.4) and (8.20), we find Eq. (1.18).

##  9 The case $M^{2}=-\eta q(q+1)$ for $q\in\mathbb{N}$

We now focus on the special case where the mass parameter is given either by
$M^{2}=M^{2}_{-,q}=+q(q+1)$, for negative curvature, or by
$M^{2}=M^{2}_{+,q}=-q(q+1)$, for positive curvature, as in Eq. (1.20), for any
positive integer $q$. In these cases, we shall be able to evaluate much more
explicitly the infinite sum appearing in our formula (1.18) for the
determinants. The case $q=1$ is crucially relevant for the quantum gravity
calculations presented in [1] and the result in this case is given in Eq.
(1.22).

###  9.1 General formula

We note $P_{q}$ the Legendre polynomial of degree $q$ and
$\omega_{q,k}^{\eta}(r_{0})$, $1\leq k\leq q$, the roots of the Meixner
polynomial $M_{q}(x;-2q,-\eta r_{0}^{2})$. We recall that the Meixner
polynomial is a degree $q$ polynomial that can be expressed in terms of the
hypergeometric function as

$M_{q}(x;b,c)=(b)_{q}F(-q,-x,b,1-1/c)\,,$ (9.1)

where $(b)_{q}=b(b+1)\cdots(b+q-1)$ is the Pochhammer symbol. In our case, we
have

$\begin{split}M_{q}(x;-2q,-\eta
r_{0}^{2})&=(-1)^{q}\frac{(2q)!}{q!}F\bigl{(}-q,-x,-2q,1+\eta/r_{0}^{2}\bigr{)}\\\
&=\Bigl{(}\frac{\eta r_{0}^{2}}{1+\eta
r_{0}^{2}}\Bigr{)}^{-q}(x+1-q)_{q}F\Bigl{(}-q,1+q,x+1-q,\frac{\eta
r_{0}^{2}}{1+\eta r_{0}^{2}}\Bigr{)}\\\ &=\Bigl{(}\frac{\eta r_{0}^{2}}{1+\eta
r_{0}^{2}}\Bigr{)}^{-q}\prod_{k=1}^{q}\bigl{(}x-\omega_{q,k}^{\eta}(r_{0})\bigr{)}\,,\end{split}$
(9.2)

where the second equality can be easily checked by using the explicit series
representation of the hypergeometric function. The main result of this section
is as follows.

######  Proposition 9.1.

The determinants given by Eq. (1.18) are expressed as

$\ln D^{\eta}\bigl{(}-\eta
q(q+1),r_{0}\bigr{)}=-\frac{1}{2}\ln(2\pi)-\frac{5}{12}-2\zeta_{\text{R}}^{\prime}(1)+\frac{\eta
A^{\eta}(r_{0})}{3\pi}-\frac{1}{3}\ln r_{0}\\\ -\frac{\eta
q(q+1)}{2\pi}\bigl{(}\ln r_{0}-1\bigr{)}A^{\eta}(r_{0})-q(q+1)\frac{1-\eta
r_{0}^{2}}{1+\eta r_{0}^{2}}\ln\bigl{(}1+\eta r_{0}^{2}\bigr{)}\\\ +\ln
P_{q}\Bigl{(}\frac{1-\eta r_{0}^{2}}{1+\eta
r_{0}^{2}}\Bigr{)}-2\ln\prod_{k=1}^{q}\frac{\Gamma\bigl{(}1+q-\omega_{q,k}^{\eta}(r_{0})\bigr{)}}{k!}\,.$
(9.3)

For $q=1$, one can check straightforwardly that we find the result announced
in Eq. (1.22). Note that, for negative curvature, the Legendre polynomial and
the product of the $\Gamma$ functions in the argument of the logarithm are
always strictly positive real numbers, ensuring that $\ln D^{-}$ is always
real. This is actually more generally valid for any positive value of $M^{2}$
due to the positivity of the Laplacian $\Delta_{-}$ (see appendix A). For
$M^{2}=M_{-,q}^{2}$ we will make this explicit below. In positive curvature,
the determinant may be a negative real number. This is related to the fact
that the operator whose determinant we compute may then have negative
eigenvalues101010Note that $M_{+,q}^{2}=-q(q+1)<0$. So, although the Laplacian
$\Delta_{+}$ is a positive operator (see appendix A), the operator
$(\Delta_{+}+M_{+,q}^{2})$ need not be positive., a case which is dealt with
as mentioned at the end of Section 3.1.

###### Proof.

In order to simplify some formulas, we introduce the notations

$z_{\eta}=\frac{\eta A^{\eta}}{4\pi}=\frac{\eta r_{0}^{2}}{1+\eta
r_{0}^{2}}\,\raise
2.0pt\hbox{,}\quad\tilde{z}_{\eta}=\frac{z_{\eta}-1}{z_{\eta}}=-\frac{\eta}{r_{0}^{2}}$
(9.4)

for $\eta=\pm$. For later reference, we note that $z_{+}\in[0,1[$, or
equivalently $\tilde{z}_{+}<0$; and that $z_{-}\leq 0$, or equivalently
$\tilde{z}_{-}>1$.

Let us start by evaluating the term $n=0$ in the infinite sum over
$n\in\mathbb{Z}$ in Eq. (1.18). We have

$f_{0}^{\eta}(1;-M^{2}_{\eta,q})=F(-q,q+1,1,z_{\eta})=\frac{z_{\eta}^{q}}{q!}M_{q}\Bigl{(}q;-2q,\frac{z_{\eta}}{z_{\eta}-1}\Bigr{)}=P_{q}(1-2z_{\eta})\,,$
(9.5)

where $P_{q}$ is the Legendre polynomial of degree $q$. Note that this is
always strictly positive in negative curvature, as can be checked
straightforwardly from the expansion of the hypergeometric function and the
fact that $z_{-}\leq 0$. This produces the term $\ln P_{q}$ in Eq. (9.3).
Moreover,

$F(1,1,2,z)=-\frac{\ln(1-z)}{z}\,\cdotp$ (9.6)

The $n=0$ term is thus

$\ln P_{q}\Bigl{(}\frac{1-\eta r_{0}^{2}}{1+\eta
r_{0}^{2}}\Bigr{)}+q(q+1)\ln(1+\eta r_{0}^{2})=\ln
P_{q}(1-2z_{\eta})-q(q+1)\ln(1-z_{\eta})\,.$ (9.7)

Using this, noting that the contribution of the modes is symmetric under
$n\rightarrow-n$ and comparing Eqs. (1.18) and (9.3), we see that the result
we want to prove is equivalent to the identity

$\sum_{n\geq
1}\ln\biggl{[}f_{n}^{\eta}(1;-M^{2}_{\eta,q})e^{-\frac{A^{\eta}M^{2}_{\eta,q}}{4\pi(n+1)}F(1,1,n+2,\frac{\eta
A^{\eta}}{4\pi})}\biggr{]}=\\\ q(q+1)\bigl{(}\gamma
z_{\eta}+(1-z_{\eta})\ln(1-z_{\eta})\bigr{)}-\ln\prod_{k=1}^{q}\frac{\Gamma\bigl{(}1+q-\omega_{q,k}^{\eta}(r_{0})\bigr{)}}{k!}\,\cdotp$
(9.8)

Our strategy to evaluate the sum on the left-hand side of this equation is to
write it as the limit of finite sums,

$\sum_{n\geq
1}\ln\biggl{[}f_{n}^{\eta}(1;-M^{2}_{\eta,q})e^{-\frac{A^{\eta}M^{2}_{\eta,q}}{4\pi(n+1)}F(1,1,n+2,\frac{\eta
A^{\eta}}{4\pi})}\biggr{]}=\lim_{N\rightarrow\infty}\Bigl{(}q(q+1)\Sigma_{N}^{(1)}+\Sigma_{N}^{(2)}\Bigr{)}\,,$
(9.9)

where, using, in particular, Eq. (9.4),

$\displaystyle\Sigma_{N}^{(1)}$
$\displaystyle=\sum_{n=1}^{N}\frac{z_{\eta}}{n+1}F(1,1,n+2,z_{\eta})$ (9.10)
$\displaystyle\Sigma_{N}^{(2)}$ $\displaystyle=\sum_{n=1}^{N}\ln
f_{n}^{\eta}(1;-M^{2}_{\eta,q})\,.$ (9.11)

######  Lemma 9.1.

We have, when $N\rightarrow\infty$,

$\Sigma_{N}^{(1)}=z_{\eta}\ln N+\gamma
z_{\eta}+(1-z_{\eta})\ln(1-z_{\eta})+O(1/N)\,.$ (9.12)

###### Proof.

To prove the lemma, we start by noting that, for all $z<1$,

$F(1,1,n+2,z)=\frac{1+n}{z}\Biggl{[}\sum_{s=1}^{n}\frac{1}{s}\Big{(}\frac{z-1}{z}\Big{)}^{n-s}-\Big{(}\frac{z-1}{z}\Big{)}^{n}\ln(1-z)\Biggr{]},$
(9.13)

generalizing (9.6). We can use this formula for our purposes, because the
variable $z_{\eta}$ defined in Eq. (9.4) is always strictly less than one.
Performing explicitly the trivial geometric sum, we get

$\Sigma_{N}^{(1)}=\sum_{n=1}^{N}\sum_{s=1}^{n}\frac{\tilde{z}_{\eta}^{n-s}}{s}+(1-\tilde{z}_{\eta}^{N})(1-z_{\eta})\ln(1-z_{\eta})\,.$
(9.14)

By reordering the terms and evaluating one more geometric sum, the first sum
in the above equation can be written as

$\sum_{n=1}^{N}\sum_{s=1}^{n}\frac{\tilde{z}_{\eta}^{n-s}}{s}=\sum_{s=1}^{N}\frac{1}{s}\sum_{p=0}^{N-s}\tilde{z}_{\eta}^{p}=z_{\eta}H_{N}+(1-z_{\eta})\tilde{z}_{\eta}^{N}\sum_{s=1}^{N}\frac{\tilde{z}_{\eta}^{-s}}{s}\,\raise
2.0pt\hbox{,}$ (9.15)

where the

$H_{N}=\sum_{s=1}^{N}\frac{1}{s}=\ln N+\gamma+O(1/N)$ (9.16)

are the harmonic numbers. Putting Eqs. (9.14) and (9.15) together yields

$\Sigma_{N}^{(1)}=z_{\eta}H_{N}+(1-z_{\eta})\ln(1-z_{\eta})+(1-z_{\eta})R_{N}$
(9.17)

for a remainder

$R_{N}=\tilde{z}_{\eta}^{N}\biggl{(}\sum_{s=1}^{N}\frac{\tilde{z}_{\eta}^{-s}}{s}-\ln(1-z_{\eta})\biggr{)}\,.$
(9.18)

Taking into account the form of the large $N$ expansion of the harmonic
numbers given in Eq. (9.16), the equation (9.12) we want to prove is
equivalent to the statement that

$R_{N}=O\bigl{(}1/N\bigr{)}$ (9.19)

at large $N$. To prove this asymptotic behaviour, we distinguish three cases.

##### Case one: $|\tilde{z}_{\eta}|>1$.

Note that this condition is always true in negative curvature, but not
necessarily true in positive curvature. When this condition is satisfied, we
can use the convergent series expansion of
$\ln(1-z_{\eta})=-\ln(1-1/\tilde{z}_{\eta})$ to write

$|R_{N}|=\Biggl{|}\tilde{z}_{\eta}^{N}\biggl{(}\sum_{s=1}^{N}\frac{\tilde{z}_{\eta}^{-s}}{s}+\ln(1-1/\tilde{z}_{\eta})\biggr{)}\Biggr{|}=\Biggl{|}\sum_{s=N+1}^{\infty}\frac{\tilde{z}_{\eta}^{N-s}}{s}\Biggr{|}\\\
\leq\frac{1}{N+1}\sum_{s=0}^{\infty}|\tilde{z}_{\eta}|^{-s-1}=\frac{1}{|\tilde{z}_{\eta}|-1}\frac{1}{N+1}\,\raise
2.0pt\hbox{,}$ (9.20)

which allows us to conclude.

##### Case two: $\tilde{z}_{\eta}=-1$.

Note that this can only occur in positive curvature. The reasoning used above
when $|\tilde{z}_{\eta}|>1$ must be slightly refined, replacing the inequality
in Eq. (9.20) by

$|R_{N}|=\Biggl{|}\sum_{s=1}^{N}\frac{(-1)^{s}}{s}+\ln
2\Biggr{|}=\Biggl{|}\sum_{s=N+1}^{\infty}\frac{(-1)^{s}}{s}\Biggr{|}\underset{N\rightarrow\infty}{\sim}\frac{1}{2N}\,\raise
2.0pt\hbox{,}$ (9.21)

the large $N$ estimate being obtained by using the standard trick of grouping
the terms of the alternating series two by two.

##### Case three: $|\tilde{z}_{\eta}|<1$.

Note that this can only occur in positive curvature. The term in
$\tilde{z}_{\eta}^{N}\ln(1-z_{\eta})$ in $R_{N}$ is then exponentially small
when $N\rightarrow\infty$. The result we want to prove will then follow from
the estimate

$\sigma_{N}(z)=z^{N}\sum_{s=1}^{N}\frac{z^{-s}}{s}\underset{N\rightarrow\infty}{\sim}\frac{1}{1-z}\frac{1}{N}\quad\text{when
$|z|<1$.}$ (9.22)

A nice way to prove this is to use the following integral representation for
$\sigma_{N}(z)$,

$\sigma_{N}(z)=z^{N}\int_{0}^{1/z}\frac{u^{N}-1}{u-1}\,{\rm d}u\,.$ (9.23)

For $|z|<1$, only the region very near the upper bound in the integration
range can contribute at large $N$. Setting $u=\frac{1}{z}(1-v/N)$ and taking
the large $N$ limit then yields

$\sigma_{N}(z)\underset{N\rightarrow\infty}{\sim}\frac{1}{zN}\int_{0}^{\infty}\frac{e^{-v}{\rm
d}v}{1/z-1}=\frac{1}{1-z}\frac{1}{N}\,\cdotp$ (9.24)

∎

######  Lemma 9.2.

We have, when $N\rightarrow\infty$,

$\Sigma_{N}^{(2)}=-q(q+1)z_{\eta}\ln
N-\ln\prod_{k=1}^{q}\frac{\Gamma\bigl{(}1+q-\omega_{q,k}^{\eta}(r_{0})\bigr{)}}{k!}+O(1/N)\,.$
(9.25)

###### Proof.

We are actually going to prove the identity

$\Sigma_{N}^{(2)}=\ln\prod_{k=1}^{q}\frac{k!\,\Gamma\bigl{(}N+1+q-\omega_{q,k}^{\eta}(r_{0})\bigr{)}}{(N+k)!\,\Gamma\bigl{(}1+q-\omega_{q,k}^{\eta}(r_{0})\bigr{)}}\,\raise
2.0pt\hbox{,}$ (9.26)

which is valid at finite $N$. Eq. (9.25) follows straightforwardly by using
Stirling’s formula

$\ln\Gamma(N+1+a)=N\ln N-N+\Bigl{(}a+\frac{1}{2}\Bigr{)}\ln
N+\frac{1}{2}\ln(2\pi)+O(1/N)$ (9.27)

and

$\sum_{k=1}^{q}\omega_{q,k}^{\eta}=q(q+1)z_{\eta}+\frac{1}{2}q(q-1)\,.$ (9.28)

This sum of the roots of the Meixner polynomial can be derived easily from Eq.
(9.2) and the series representation of the hypergeometric functions.

To prove Eq. (9.26), we start by noting that

$f_{n}^{\eta}(1;-M^{2}_{\eta,q})=F(-q,q+1,n+1,z_{\eta})\\\
=\frac{n!}{(n+q)!}z_{\eta}^{q}M_{q}\Bigl{(}n+q;-2q,\frac{z_{\eta}}{z_{\eta}-1}\Bigr{)}=\frac{n!q!}{(n+q)!}P_{q}^{(n,-n)}(1-2z_{\eta})=\\\
\frac{n!}{(n+q)!}\prod_{k=1}^{q}\bigl{(}n+q-\omega^{\eta}_{q,k}(r_{0})\bigr{)}\,,$
(9.29)

where $P_{q}^{(n,-n)}$ is the Jacobi polynomial, generalizing (9.5), and we
have used the last equation in (9.2) to derive the last equality. This last
equality is actually all we need, but we have indicated the expression in
terms of the Jacobi polynomials for completeness. Note that this is always
strictly positive in negative curvature, as can be checked straightforwardly
from the expansion of the hypergeometric function in the first line of the
above equation and the fact that $z_{-}\leq 0$. We thus get, from the
definition (9.11),

$\begin{split}\Sigma_{N}^{(2)}&=\sum_{n=1}^{N}\biggl{[}\ln\frac{n!}{(n+q)!}+\ln\prod_{k=1}^{q}\bigl{(}n+q-\omega^{\eta}_{q,k}(r_{0})\bigr{)}\biggr{]}\\\
&=\sum_{n=1}^{N}\ln\prod_{k=1}^{q}\frac{n+q-\omega^{\eta}_{q,k}(r_{0})}{n+k}=\ln\prod_{k=1}^{q}\prod_{n=1}^{N}\frac{n+q-\omega^{\eta}_{q,k}(r_{0})}{n+k}\,\cdotp\end{split}$
(9.30)

Using $\Gamma(z+1)=z\Gamma(z)$, one can straightforwardly check that this is
the same as Eq. (9.26). ∎

Combining Eqs. (9.9), (9.12) and (9.25), we get Eq. (9.8). ∎

###  9.2 Application: the large $\ell$ limit in negative curvature

As a simple application of the explicit formula (9.3) for the determinant, let
us study the large $\ell$ limit in negative curvature. As explained in Section
2, this limit is notoriously subtle. The logarithm $\ln D^{-}_{\infty}$ of the
strictly infinite area determinant, per unit area, is given exactly, for any
mass $M$, by Eq. (2.22). However, the infinite area limit of the finite area
determinants does not coincide with the strictly infinite area result,

$\lim_{\ell\rightarrow\infty}\frac{1}{\ell}\ln D^{-}\not=D^{-}_{\infty}\,.$
(9.31)

This was explicitly checked when the mass is zero at the end of Section 4. For
masses of the form $M^{2}=q(q+1)$, $q\in\mathbb{N}$, the goal of the present
subsection is to show the following.

######  Proposition 9.2.

Let $a_{1},\ldots,a_{q}$ be the roots of the Laguerre polynomial
$L_{q}^{(-2q-1)}$. Then the large $\ell$ asymptotics of the determinants
computed for the special masses $M^{2}=q(q+1)$, $q\in\mathbb{N}$, is given by

$\ln
D^{-}\bigl{(}M^{2}=q(q+1),r_{0}(\ell)\bigr{)}=\biggl{(}-\frac{2}{3}+\sum_{k=1}^{q}a_{k}\ln(-a_{k})\biggr{)}\frac{\ell}{2\pi}+O(1)\,.$
(9.32)

This generalizes the leading term in the expansion at zero mass given in Eq.
(4.8). Note that the roots $a_{k}$ either have an imaginary part and then come
in complex conjugate pairs (because the Laguerre polynomial $L_{q}^{(-2q-1)}$
has real coefficients) or are real and negative (because we know that $\ln
D^{-}$ must be real, as stated just below Eq. (9.3)). The logarithm of the
complex roots in (9.32) is evaluated with the determination
$\ln(-a_{k})=\ln|a_{k}|+i\arg(-a_{k})$, with $-\pi<\arg(-a_{k})<\pi$. For
instance,

$\begin{split}\ln
D^{-}\bigl{(}M^{2}=2,r_{0}(\ell)\bigr{)}&=-\Bigl{(}\frac{1}{3}+\ln
2\Bigr{)}\frac{\ell}{\pi}\,\raise 2.0pt\hbox{,}\\\ \ln
D^{-}\bigl{(}M^{2}=6,r_{0}(\ell)\bigr{)}&=-\Bigl{(}\frac{2}{3}+3\ln
12-\frac{\pi}{\sqrt{3}}\Bigr{)}\frac{\ell}{2\pi}\,\raise
2.0pt\hbox{,}\quad\text{etc.}\end{split}$ (9.33)

###### Proof.

Using (1.4) for $\ell^{-}=\ell\rightarrow+\infty$ and the fact that the
Legendre polynomial $P_{q}$ is of degree $q$, it is straightforward, starting
from (9.3), to show that

$\ln
D^{-}\bigl{(}M^{2}=q(q+1),r_{0}(\ell)\bigr{)}=\frac{q(q+1)}{2\pi}\ell\ln\frac{\ell}{4\pi}-\Bigl{(}\frac{1}{3\pi}+\frac{q(q+1)}{2\pi}\Bigr{)}\ell+q\ln\ell\\\
-2\ln\prod_{k=1}^{q}\Gamma\bigl{(}1+q-\omega^{-}_{q,k}(r_{0})\bigr{)}+O(1)\,.$
(9.34)

The non-trivial part of the calculation is to evaluate the asymptotic
behaviour of the terms involving the $\Gamma$ functions. To do this, we need
to find the asymptotic behaviour of the roots $\omega^{-}_{q,k}(r_{0})$ of the
Meixner polynomials $M_{q}(x;-2q,r_{0}^{2})$. By examining the explicit
expression of these polynomials given by the second equality in (9.2), and
noting that, in the limit we are considering,

$z_{-}=-\frac{r_{0}^{2}}{1-r_{0}^{2}}=-\frac{\ell}{4\pi}+\frac{1}{2}+O\bigl{(}1/\ell\bigr{)}\,,$
(9.35)

one finds

$\omega_{k,q}^{-}=-a_{k}z_{-}+b_{k}+O\bigl{(}1/z_{-}\bigr{)}=\frac{a_{k}\ell}{4\pi}+b_{k}-\frac{1}{2}a_{k}+O\bigl{(}1/\ell\bigr{)}$
(9.36)

where the $a_{k}$ are the roots of the polynomial

$\sum_{s=0}^{q}\frac{(q+s)!}{s!(q-s)!}x^{q-s}\,,$

which is proportional to the Laguerre polynomial $L_{q}^{(-2q-1)}$. The
$b_{k}$ (as well as the higher order terms in the large $\ell$ expansion) can
be straightforwardly expressed in terms of the $a_{k}$, but we won’t need the
(rather complicated) explicit expressions. It is sufficient to know the sums

$\sum_{k=1}^{q}a_{k}=-q(q+1)\,,\quad\sum_{k=1}^{q}b_{k}=\frac{1}{2}q(q-1)\,,$
(9.37)

which are obtained directly from Eq. (9.28). Using Eqs. (9.36) and (9.37)
together with the Stirling asymptotic expansion of the $\Gamma$ function,
already indicated in Eq. (9.27), we find Eq.(9.32). ∎

It is interesting to note that the terms in $\ell\ln\ell$ and $\ln\ell$
precisely cancel. The cancellation of the $\ell\ln\ell$ terms was actually
expected, since it is crucial for the existence of an effective action per
unit area, which is a consequence of extensivity.

### Acknowledgments

We would like to dedicate this paper to Steve Zelditch. F.F. shared with him
illuminating discussions on many subjects, including in relation with a large
part of the research presented in this paper. Steve was a magnificent
mathematician with a deep interest in physics, a warm and caring human being,
and a friend.

This work is partially supported by the International Solvay Institutes and
the Belgian Fonds National de la Recherche Scientifique FNRS (convention IISN
4.4503.15).

##  Appendix A Monotonicity of the eigenvalues as a function of the boundary
length

In this Appendix, we show that the eigenvalues $\lambda^{\pm}_{n,k}(r_{0})$ of
the Laplacians $\Delta_{\pm}$, defined by Eqs. (1.2) and (1.3), are strictly
decreasing functions of $r_{0}$. This result was mentioned in the main text in
Section 3.1. By taking the flat space limit, this also implies that the
eigenvalues $\lambda^{0}_{n,k}(\ell)$ of $\Delta_{0}$ are strictly decreasing
functions of the boundary length $\ell$.

The operator $\Delta_{\pm}$, with Dirichlet boundary conditions, are symmetric
operators with respect to the scalar product

$\langle\psi_{1},\psi_{2}\rangle=\int_{\mathscr{D}}{\rm
d}^{2}x\sqrt{g_{\pm}}\,\psi_{1}\psi_{2}\,,$ (A.1)

where $g_{\pm}=e^{2\sigma_{\pm}}({\rm d}r^{2}+r^{2}{\rm d}\theta^{2})$. It is
a positive operator because the Dirichlet boundary condition allows to perform
an integration by part to prove that

$\langle\psi,\Delta_{\pm}\psi\rangle=\int_{\mathscr{D}}{\rm
d}^{2}x\sqrt{g_{\pm}}\,g_{\pm}^{\mu\nu}\partial_{\mu}\psi\partial_{\nu}\psi\geq
0\,.$ (A.2)

Thus, all the eigenvalues of $\Delta_{\pm}$ are positive. Eq. (A.2) actually
shows that an eigenvector with zero eigenvalue must be a constant, which is
impossible because the Dirichlet boundary condition would enforce this
constant to be zero.

Let us denote by $(\psi^{\pm}_{k})_{k\geq 0}$ an orthonormal basis of
eigenvectors, with eigenvalues
$0<\lambda^{\pm}_{0}\leq\lambda^{\pm}_{1}\leq\cdots$ (this is just a
rearrangement of the eigenvalues noted $\lambda_{n,k}^{\pm}$ above). The
eigenvalues and the eigenfunctions are functions of the parameter $r_{0}$. It
is convenient to rescale the radial coordinate and use $\rho=r_{0}r$ instead
of $r$, so that the metrics

$g_{\pm}=\frac{4L^{2}}{(1\pm\rho^{2})^{2}}\bigl{(}{\rm d}\rho^{2}+\rho^{2}{\rm
d}\theta^{2}\bigr{)}$ (A.3)

do not depend explicitly on $r_{0}$ anymore. All the $r_{0}$ dependence is
then in the range of the radial coordinate, $0\leq\rho\leq r_{0}$. We have

$\lambda^{\pm}_{k}(r_{0})=\langle\psi^{\pm}_{k},\Delta_{\pm}\psi^{\pm}_{k}\rangle=\int_{0}^{r_{0}}{\rm
d}\rho\,\rho\int_{0}^{2\pi}{\rm
d}\theta\,\Bigl{[}\bigl{(}\partial_{\rho}\psi^{\pm}_{k}\bigr{)}^{2}+\frac{1}{\rho^{2}}\bigl{(}\partial_{\theta}\psi^{\pm}_{k}\bigr{)}^{2}\Bigr{]}\,.$
(A.4)

The variation of $\lambda^{\pm}_{k}$ with respect to $r_{0}$ is thus

$\frac{{\rm d}\lambda^{\pm}_{k}}{{\rm d}r_{0}}=r_{0}\int_{0}^{2\pi}{\rm
d}\theta\Bigl{[}\bigl{(}\partial_{\rho}\psi^{\pm}_{k}\bigr{)}^{2}+\frac{1}{\rho^{2}}\bigl{(}\partial_{\theta}\psi^{\pm}_{k}\bigr{)}^{2}\Bigr{]}\Big{|}_{r=r_{0}}\\\
+\int_{0}^{r_{0}}{\rm d}\rho\,\rho\int_{0}^{2\pi}{\rm
d}\theta\,\frac{\partial}{\partial
r_{0}}\Bigl{[}\bigl{(}\partial_{\rho}\psi^{\pm}_{k}\bigr{)}^{2}+\frac{1}{\rho^{2}}\bigl{(}\partial_{\theta}\psi^{\pm}_{k}\bigr{)}^{2}\Bigr{]}\,.$
(A.5)

The second term can be massaged as follows:

$\begin{split}\int_{0}^{r_{0}}{\rm d}\rho\,\rho\int_{0}^{2\pi}{\rm
d}\theta\,\frac{\partial}{\partial
r_{0}}&\Bigl{[}\bigl{(}\partial_{\rho}\psi^{\pm}_{k}\bigr{)}^{2}+\frac{1}{\rho^{2}}\bigl{(}\partial_{\theta}\psi^{\pm}_{k}\bigr{)}^{2}\Bigr{]}=\int_{\mathscr{D}}{\rm
d}^{2}x\sqrt{g_{\pm}}\frac{\partial}{\partial
r_{0}}\bigl{(}g_{\pm}^{\mu\nu}\partial_{\mu}\psi^{\pm}_{k}\partial_{\nu}\psi^{\pm}_{k}\bigr{)}\\\
&=2\int_{\mathscr{D}}{\rm
d}^{2}x\sqrt{g_{\pm}}\,g_{\pm}^{\mu\nu}\partial_{\mu}\psi^{\pm}_{k}\partial_{\nu}\frac{\partial\psi^{\pm}_{k}}{\partial
r_{0}}=2\int_{\mathscr{D}}{\rm
d}^{2}x\sqrt{g_{\pm}}\,\psi^{\pm}_{k}\Delta_{\pm}\frac{\partial\psi^{\pm}_{k}}{\partial
r_{0}}\\\ &=2\int_{\mathscr{D}}{\rm
d}^{2}x\sqrt{g_{\pm}}\,\psi^{\pm}_{k}\frac{\partial}{\partial
r_{0}}\Delta_{\pm}\psi^{\pm}_{k}=2\int_{\mathscr{D}}{\rm
d}^{2}x\sqrt{g_{\pm}}\,\psi^{\pm}_{k}\frac{\partial}{\partial
r_{0}}\bigl{(}\lambda^{\pm}_{k}\psi^{\pm}_{k}\bigr{)}\\\ &=2\frac{{\rm
d}\lambda^{\pm}_{k}}{{\rm d}r_{0}}+2\lambda^{\pm}_{k}\int_{\mathscr{D}}{\rm
d}^{2}x\sqrt{g_{\pm}}\,\psi^{\pm}_{k}\frac{\partial\psi^{\pm}_{k}}{\partial
r_{0}}\,\cdotp\end{split}$ (A.6)

Note that, to derive the third equality, we have used an integration by part
and the Dirichlet boundary condition on $\psi^{\pm}_{k}$; and to derive the
last equality we have used the normalisation condition

$\langle\psi^{\pm}_{k},\psi^{\pm}_{k}\rangle=\int_{\mathscr{D}}{\rm
d}^{2}x\sqrt{g_{\pm}}\,(\psi^{\pm}_{k})^{2}=1\,.$ (A.7)

If we take the derivative of this condition with respect to $r_{0}$, we get a
boundary term, that vanishes because $\psi^{\pm}_{k}$ is zero on the boundary,
plus precisely the second term in the last line of (A.6). Thus, this term
vanishes and we obtain

$\int_{0}^{r_{0}}{\rm d}\rho\,\rho\int_{0}^{2\pi}{\rm
d}\theta\,\frac{\partial}{\partial
r_{0}}\Bigl{[}\bigl{(}\partial_{\rho}\psi^{\pm}_{k}\bigr{)}^{2}+\frac{1}{\rho^{2}}\bigl{(}\partial_{\theta}\psi^{\pm}_{k}\bigr{)}^{2}\Bigr{]}=2\frac{{\rm
d}\lambda^{\pm}_{k}}{{\rm d}r_{0}}\,\cdotp$ (A.8)

Using this result in Eq. (A.5), we get

$\frac{{\rm d}\lambda^{\pm}_{k}}{{\rm d}r_{0}}=-r_{0}\int_{0}^{2\pi}\\!{\rm
d}\theta\,\bigl{(}\partial_{\rho}\psi^{\pm}_{k}\bigr{)}^{2}_{|\rho=r_{0}}<0\,.$
(A.9)

We have used the fact that $\partial_{\theta}\psi_{k}^{\pm}=0$ and
$(\partial_{\rho}\psi_{k}^{\pm})^{2}>0$ along the boundary, which are direct
consequences of the explicit form of the eigenfunctions, see Eqs. (3.5) and
(3.6).

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