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# One-Loop Partition Function,
Gauge Accessibility and Spectra in AdS3 Gravity

Joel Acosta†, Alan Garbarz‡, Andrés Goya§, Mauricio Leston§

###### Abstract

We continue the study of the one-loop partition function of AdS3 gravity with
focus on the square-integrability condition on the fluctuating fields. In a
previous work we found that the Brown-Henneaux boundary conditions follow
directly from the $L^{2}$ condition. Here we rederive the partition function
as a ratio of Laplacian determinants by performing a suitable decomposition of
the metric fluctuations. We pay special attention to the asymptotics of the
fields appearing in the partition function. We also show that in the usual
computation using ghost fields for the de Donder gauge, such gauge condition
is accessible precisely for square-integrable ghost fields. Finally, we
compute the spectrum of the relevant Laplacians in thermal AdS3, in particular
noticing that there are no isolated eigenvalues, only essential spectrum. This
last result supports the analytic continuation approach of David, Gaberdiel
and Gopakumar. The purely essential spectra found are consistent with the
independent results of Lee and Delay of the essential spectrum of the TT
rank-2 tensor Lichnerowickz Laplacian on asymptotically hyperbolic spaces.

E-mails<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>
† Departamento de Matemática-FCEN-UBA & IMAS-CONICET, Ciudad Universitaria,
Pabellón 1, 1428, Buenos Aires, Argentina.

‡ Departamento de Física-FCEN-UBA & IFIBA-CONICET, Ciudad Universitaria,
Pabellón 1, 1428, Buenos Aires, Argentina.

§ Instituto de Astronomía y Física del Espacio (IAFE), Pabellón IAFE-CONICET,
Ciudad Universitaria, C.C. 67 Suc. 28, Buenos Aires, Argentina.

##  1 Introduction

It is a major goal in high energy theoretical physics to fully understand the
partition function of AdS3 quantum gravity. It encodes the spectrum of the
putative quantum theory and can be used to gain insight about a possible dual
CFT. In [1], Giombi et al. took an important step by computing from the
gravitational theory the one-loop partition function
$Z^{\text{(1-loop)}}(\tau,\bar{\tau})$ on a solid torus. More precisely, they
departed from the quadratic action around thermal AdS3, chose a gauge,
introduced ghosts fields, and by the heat-kernel method computed the relevant
functional determinants in the solid torus $\mathbb{H}^{3}/\Gamma$. Here
$\Gamma$ implements the appropriate quotient which we will explain in more
detail later. They landed at the following result,

$Z^{\text{(1-loop)}}(\tau,\bar{\tau})=\prod_{m=2}^{\infty}\frac{1}{|1-q^{m}|^{2}},\qquad
q:=e^{2\pi i\tau}\,,$ (1)

where $\tau$ is the modular parameter of the (conformal) boundary torus. Such
a partition function is the product of two characters over an irreducible
representation of the Virasoro algebra. This was a robust indicative of the
existence of an underlying two-dimensional CFT. This fact has a classical
precursor: Brown and Henneaux in [2] showed that, with sensible asymptotic
boundary conditions on Lorentzian metrics in General Relativity with negative
cosmological constant, the algebra of conserved charges is nothing but two
copies of the Virasoro algebra with central charge $c=\frac{3\ell}{2G}$. It is
worth noticing that in [1] the expectation and confirmation of the appearance
of the Virasoro algebra in the partition function did not came from the
imposition of Brown-Henneaux boundary conditions, but from an argument
borrowed by from [3].

In 2007 Maloney and Witten had already argued that if Brown-Henneaux boundary
conditions are imposed, the computation of the gravity partition function as a
trace of $e^{-\beta H-i\theta J}$ should give a Virasoro character. In short,
the reason is that with Brown-Henneaux asymptotic boundary conditions the
physically different gravitational configurations are organized in Virasoro
coadjoint representations [4, 5]. Coming back to the work [1], we see that
although it was reasonable to expect that a Virasoro character would show up
in the Euclidean computation of the partition function, the Brown-Henneaux
boundary conditions were not really imposed at any moment. And moreover, we
can wonder how is it that a Virasoro character makes its appearance by a
purely Euclidean path-integral computation.

Motivated by the question of where are implicitly assumed/imposed the
asymptotic boundary conditions of Brown-Henneaux in the path-integral of [1],
we found in [6] that the square integrability of the metric perturbations
directly imply the (Euclidean version of) Brown-Henneaux boundary conditions.
Even more, it was possible to make a distinction, within the space of
asymptotic symmetry generators, between proper asymptotic vectors (those
square integrable, which are trivial under the coadjoint representation) and
improper asymptotic vectors (those which are not square integrable, implying
that are not trivial under the coadjoint representation, modulo proper ones).
This distinction is essentially the same as that of [2].

However, a few but important points were left to be clarified in our previous
work. First of all, the ghost fields in [1] appear from the start in the path
integral but is not a priori clear if they should belong to the space of
$L^{2}$ vectors or to the space of vectors which generate $L^{2}$ metric
perturbations. This issue is relevant because of the following: if the space
of vector fields is too small, it might not include the proper asymptotic
vectors. On the contrary, if it is too big, it might exceed the domain of the
operator whose determinant appears in the partition function (see [7] for a
discussion on these possibilities and the implications on partition
functions).

Second, and relevant for any gauge theory, is the matter of the accessibility
of the gauge condition: can the de Donder gauge be reached from any $L^{2}$
perturbation by an infinitesimal diffeomorphism? This is crucial since ensures
that the Faddeev-Popov method actually works and it is also needed in order to
implement a “constraint first” quantization procedure. As we will see, this is
closely related to the question about the space to which the ghost fields
belong.

In third place, we have the issue of describing in detail the spectrum of the
relevant operators entering in the 1-loop partition function. This is crucial
for understanding the computation of the functional determinant of such
operators. In order to be as clear as possible, let us first introduce these
operators. The result (1) of [1] comes from the expression of the 1-loop
partition function as a quotient of functional determinants,

$Z^{\text{(1-loop)}}(\tau,\bar{\tau})=\frac{\text{det}\left(-{\Delta}^{(1)}\right)}{\sqrt{\text{det}\left(-{\Delta}^{(2)}\right)\text{det}\left(-{\Delta}^{(0)}\right)}}\,,$
(2)

where $\Delta^{(i)}$ are Laplacian-like operators with different masses acting
on tensor fields of rank $i$. The operator $\Delta^{(2)}$ acts on traceless-
symmetric tensors. In [1] the determinants were computed using a combination
of the heat-kernel method for $\mathbb{H}^{3}$ and the method of images to
find the heat kernel in $\mathbb{H}^{3}/\Gamma$. It is hard to say if this
procedure manages to capture the full spectrum of the operators in
$\mathbb{H}^{3}/\Gamma$ when it starts from the essential spectrum in
$\mathbb{H}^{3}$. It is known in the mathematical literature that the spectrum
of Lichnerowicz Laplacians on $\mathbb{H}^{3}$ is a purely essential spectrum
[8, 9]. In order to simplify the previous expression, it is possible to work
with one less operator, as showed for example in [10],

$Z^{\text{(1-loop)}}(\tau,\bar{\tau})=\frac{\text{det}^{1/2}_{T}\left(-\nabla^{2}+2\right)}{\text{det}^{1/2}_{TT}\left(-\nabla^{2}-2\right)}\,.$
(3)

Here the determinant in the numerator is computed over transverse vector
fields, while the one in the denominator is computed over transverse-traceless
symmetric rank-2 tensors. In [10] the heat kernels in $\mathbb{H}^{3}/\Gamma$
are obtained from a clever analytic continuation from the heat kernel on a
quotient of $S^{3}$. Again, it is hard to say whether this procedure captures
the full spectrum of the operators, or may be it misses some possible isolated
points in the spectrum.

Although there are available computations related to the spectrum of the
operators in (3) for the BTZ [11], and as we mentioned the heat kernel of
thermal AdS3 was computed in [10] and [1], to the best of our knowledge there
is no known result of the full spectrum of the relevant operators in Thermal
AdS3. In the mathematics community it has been studied in great depth the
essential spectrum of Laplacian-like operators in asymptotically hyperbolic
manifolds in arbitrary dimensions. In particular, in [8] and [9] it was shown
that the essential spectrum of the Lichnerowicz Laplacian of a symmetric-
traceless rank-2 tensor is the ray $[(n-1)(n-9)/4,\infty)$. Also they showed
that for $\mathbb{H}^{n}$ this is the whole spectrum. For $n=3$ this
dimensions implies that $-\nabla^{2}-2$ has essential spectrum $[1,\infty)$,
for any asymptotically hyperbolic 3-manifold. What is not clear if there is a
non-empty discrete spectrum, in other words isolated eigenvalues. A similar
situation holds for vector fields [12, 13].

Having this in mind, in this paper we address the following questions:

  1. 1.

What are the sensible boundary conditions on the ghost fields?

  2. 2.

Is de Donder gauge accessible with proper asymptotic diffs?

  3. 3.

What is the complete spectrum of the relevant operators in thermal AdS3?

These questions guide the organization of this work. In Section 2 we show that
the ghost fields must belong to $L^{2}$ by scrutinizing the differential
operators acting on a convenient decomposition of the metric fluctuations
inside the path integral. In addition, we show that indeed de Donder gauge is
accessible by gauge transformations generated by $L^{2}$ vector fields. This
is another way of seeing that ghost fields must belong to $L^{2}$. Finally, in
Section 3 we compute the full spectrum of the relevant operator in the path
integral in thermal AdS3, following and improving [11].

##  2 The Partition Function

###  2.1 Identifications and Coordinates of Hyperbolic Spaces

The hyperbolic space $\mathbb{H}^{3}$ can be described by the half-space model
as $(X,Y,Z)\in\mathbb{R}^{3}$ with $Z>0$ and with metric

$ds^{2}=\frac{dX^{2}+dY^{2}+dZ^{2}}{Z^{2}}\,.$ (4)

We are going to consider the quotient $\mathbb{H}^{3}/\Gamma$ where $\Gamma$
is the group generated by an isometry $\gamma$ which combines a rotation in
the $X,Y$ plane by a fixed amount $\theta$ and a global dilation by a factor
$e^{\beta}$

$\begin{pmatrix}X\\\ Y\\\ Z\end{pmatrix}\sim\gamma\cdot\begin{pmatrix}X\\\
Y\\\ Z\end{pmatrix}=e^{\beta}\begin{pmatrix}\cos(\theta)X-\sin(\theta)Y\\\
\sin(\theta)X+\cos(\theta)Y\\\ Z\end{pmatrix}\,.$ (5)

This identifies the semispheres of radius $1$ with the semisphere of radius
$e^{\beta}$, $e^{2\beta}$ and so on. We are going to consider the fundamental
region inside the semispheres of radius $1$ and $e^{\beta}$ (see Figure (1)).
Usually, the parameters $\theta$ and $\beta$ are combined into a new one
$\tau=\frac{1}{2\pi}(\theta+i\beta$). The quotient $\mathbb{H}^{3}/\Gamma$ can
be thought of as a solid torus where $\tau$ is the modular parameter of its
conformal boundary.

Performing the following change in the coordinates of $\mathbb{H}^{3}$

$\displaystyle\lambda$ $\displaystyle=\sqrt{\dfrac{X^{2}+Y^{2}}{Z^{2}}}\,,$
(6) $\displaystyle\psi$
$\displaystyle=\frac{1}{2}\log\left(X^{2}+Y^{2}+Z^{2}\right)\,,$
$\displaystyle\varphi$ $\displaystyle=\arctan\left(Y/X\right)\,,$

we arrive to the familar (Euclidean) AdS3 metric

$ds^{2}=\dfrac{1}{\left(\rho^{2}+1\right)}d\rho^{2}+\left(\rho^{2}+1\right)d\psi^{2}+\rho^{2}d\varphi^{2}\,,$
(7)

here $\varphi$ is the $2\pi$-periodic angular coordinate. The identification
(5) acts on the coordinates as

$(\psi\,,\,\varphi)\sim(\psi\,,\,\varphi+2\pi)\sim(\psi+\beta\,,\,\varphi+\theta)\,,$
(8)

On the other hand, performing the following change in the coordinates

$\begin{array}[]{rcccl}\tanh^{2}(\zeta)&=&\dfrac{r^{2}-r_{+}^{2}}{r^{2}+r_{-}^{2}}&=&\dfrac{X^{2}+Y^{2}}{X^{2}+Y^{2}+Z^{2}}\,,\\\
x_{-}&=&r_{+}\tau+r_{-}\phi&=&\arctan\left(Y/X\right)\,,\\\ x_{+}&=&r_{+}\phi-
r_{-}\tau&=&\dfrac{1}{2}\log\left(X^{2}+Y^{2}+Z^{2}\right)\,,\end{array}$ (9)

we obtain the metric

$\displaystyle ds^{2}$
$\displaystyle=d\zeta^{2}+\sinh^{2}(\zeta)dx_{-}^{2}+\cosh^{2}(\zeta)dx_{+}^{2}$
(10)
$\displaystyle=\dfrac{r^{2}}{\left(r^{2}-r_{+}^{2}\right)\left(r^{2}+r^{2}_{-}\right)}dr^{2}+\dfrac{\left(r^{2}-r_{+}^{2}\right)\left(r^{2}+r_{-}^{2}\right)}{r^{2}}d\tau^{2}+r^{2}\left(d\phi-\dfrac{r_{+}r_{-}}{r^{2}}d\tau\right)^{2}\,,$

which corresponds to the Euclidean version of the so-called BTZ black hole
[14, 15]. The $(x_{+},x_{-})$ coordinates are now identified as

$(x_{+}\,,\,x_{-})\sim(x_{+}\,,\,x_{-}+2\pi)\sim(x_{+}+\beta\,,\,x_{-}+\theta)\,.$
(11)

The parameters $\theta$ and $\beta$ can be expressed in terms of $r_{+}$ and
$r_{-}$ by writing $(\tau,\phi)$ in terms of $x_{\pm}$ and imposing the
periodicity condition $\phi\sim\phi+2\pi$. We obtain $\beta=2\pi r_{+}$,
$\theta=2\pi r_{-}$ and the periodicity of the coordinates become

$\displaystyle(x_{+}\,,\,x_{-})$
$\displaystyle\sim(x_{+}\,,\,x_{-}+2\pi)\sim(x_{+}+2\pi r_{+}\,,\,x_{-}+2\pi
r_{-})\,,$ (12) $\displaystyle(\tau\,,\,\phi)$
$\displaystyle\sim(\tau\,,\,\phi+2\pi)\sim\left(\tau+\frac{2\pi
r_{+}}{r_{+}^{2}+r_{-}^{2}}\,,\,\phi+\frac{2\pi
r_{-}}{r_{+}^{2}+r_{-}^{2}}\right)\,.$

To visualize the coordinates $(r,x_{-},x_{+})$ in the half-space model it is
useful to invert the transformations (9)

$\displaystyle X$
$\displaystyle=\left(\dfrac{r^{2}-r_{+}^{2}}{r^{2}+r_{-}^{2}}\right)^{1/2}\cos(x_{-})e^{x_{+}}\,,$
(13) $\displaystyle Y$
$\displaystyle=\left(\dfrac{r^{2}-r_{+}^{2}}{r^{2}+r_{-}^{2}}\right)^{1/2}\sin(x_{-})e^{x_{+}}\,,$
$\displaystyle Z$
$\displaystyle=\left(\dfrac{r_{+}^{2}+r_{-}^{2}}{r^{2}+r_{-}^{2}}\right)^{1/2}e^{x_{+}}\,.$

For a fixed $r$ these are the equations of cones through the origin, which
means $r$ can be thought of as a zenithal angle. The coordinates $x_{-}$ and
$x_{+}$ are coordinates on those cones, acting as angular and radial
coordinates respectively.

Figure 1: The fundamental region of $\mathbb{H}^{3}$/$\Gamma$ lies between the
two domes. The surface of the inner dome is identified with the surface of the
exterior one. The red line indicates the contractible cycle and the blue line
indicates the non-contractible one.

After the identification, the intersections between the $r$-constant cones and
the fundamental region are tori. Any such torus serves as a boundary dividing
two regions inside the fundamental region, one inner and one reaching the
conformal boundary. Because of this, an $r-$constant torus can be used to
evaluate boundary terms by computing them at a fixed value of $r$ and then
taking limit $r\to\infty$ (see [16] for a broader discussion). One might be
tempted do this procedure by taking horizontal planes at a fixed value of $Z$
and then take the limit $Z\to 0$. However, due to the identification, the
surfaces of constant $Z$ are not the boundaries of any region inside the
fundamental region. Put differently, constant $Z$ surfaces are not invariant
under the quotient.

###  2.2 Decomposition of Metric Perturbations

Here we describe the decomposition that we use for the rank-2 tensors in order
to evaluate the action. It is customary to split the perturbations in a
transverse-traceless component, a longitudinal component and a tracefull
component. Following [17] we further impose the trace component to be
transverse, this cannot be done in general but it is possible when the
background spacetime is maximally symmetric. This extra condition guarantees
the orthogonality between the three components and allows us to rewrite the
action in a very convenient form.

We are going to consider tensors that are square integrable with respect to
the inner product

$\langle T\mid T^{\prime}\rangle=\int d^{3}x\sqrt{g}\,\bar{T}\cdot
T^{\prime}\,,$ (14)

where $\bar{T}\cdot T^{\prime}$ refers to the full contraction with respect of
the metric $g$. Thus, for example, for rank-2 tensors the inner product is

$\langle T\mid T^{\prime}\rangle=\int
d^{3}x\sqrt{g}\,\bar{T}^{\mu\nu}T^{\prime}_{\mu\nu}=\int
d^{3}x\sqrt{g}\,g^{\mu\alpha}g^{\nu\beta}\bar{T}_{\mu\nu}T^{\prime}_{\alpha\beta}\,.$
(15)

A square-integrable symmetric tensor $T_{\mu\nu}$ can be decomposed into a
transverse and a longitudinal part as follows

$T_{\mu\nu}=T^{t}_{\mu\nu}+\left(\nabla_{\mu}V_{\nu}+\nabla_{\nu}V_{\mu}\right)=T^{t}_{\mu\nu}+\mathcal{L}_{\mu\nu}V\,,\quad\text{with}\quad\nabla^{\mu}T^{t}_{\mu\nu}=0\,,$
(16)

where the vector $V$ is also square integrable with respect to (14). The
divergence of the original tensor is a square-integrable vector and it is
carried by the longitudinal part

$\nabla^{\nu}T_{\mu\nu}=B_{\mu}=\nabla^{\nu}\left(\nabla_{\mu}V_{\nu}+\nabla_{\nu}V_{\mu}\right)\,.$
(17)

By integrating by parts it can be shown that the operator that takes $V_{\mu}$
into $B_{\mu}$ is negative-semi-definite

$\int
dx^{3}\sqrt{g}V^{\nu}\nabla^{\mu}(\nabla_{\nu}V_{\mu}+\nabla_{\mu}V_{\nu})=-\dfrac{1}{2}\int
dx^{3}\sqrt{g}(\nabla_{\nu}V_{\mu}+\nabla_{\mu}V_{\nu})^{2}\leq 0\,,$ (18)

where the boundary term vanishes if the vectors are square integrable. Thus,
it can be inverted and we can always find the vector $V_{\mu}$ of the
decomposition (16).

Now, we want to decompose the transverse part $T^{t}_{\mu\nu}$ into a
traceless and a tracefull part while keeping the transversality property. This
can be done by defining

$T^{t}_{\mu\nu}=T^{TT}_{\mu\nu}+\dfrac{1}{2}\left(g_{\mu\nu}\nabla^{2}-\nabla_{\mu}\nabla_{\nu}\right)\chi-
g_{\mu\nu}\chi=T^{TT}_{\mu\nu}+\Theta_{\mu\nu}\chi\,,$ (19)

where $\Theta_{\mu\nu}\chi$ is given by

$\Theta_{\mu\nu}\chi=\dfrac{1}{2}\left(g_{\mu\nu}\nabla^{2}-\nabla_{\mu}\nabla_{\nu}\right)\chi\,,$
(20)

and $\chi$ is a square-integrable function. Taking trace in the equation (19)
we find that the relevant operator is

$g^{\mu\nu}T^{t}_{\mu\nu}=T^{t}=(\nabla^{2}-3)\chi\,,$ (21)

where the trace $T^{t}$ is a square-integrable function. By integrating by
parts again, we can see that this operator is also negative semi-definite.
Therefore, any square-integrable symmetric tensor can be decomposed as

$T_{\mu\nu}=T^{TT}_{\mu\nu}+\Theta_{\mu\nu}\chi+\mathcal{L}_{\mu\nu}V=T_{TT}+T_{\Theta}+T_{L}\,,$
(22)

where we denote the transverse-traceless part as $T_{TT}$, the transverse-
tracefull part by $T_{\Theta}$, and the longitudinal part by $T_{L}$. To find
the vector field $V$ and the scalar field $\chi$ one has to invert the
operators appearing in (17) and (21) respectively. The orthogonality between
the components in (22) follows from the respective transverse and traceless
properties when implemented in the inner product (14).

###  2.3 Computation of the Partition Function

In this section we proceed to compute the one-loop partition function around
AdS3

$Z=\int Dh\,e^{-S^{(2)}\left[h\right]}\,,$ (23)

where $S^{(2)}$ is the Einstein-Hilbert action expanded to quadratic order in
the perturbation $h$ around the Euclidean AdS3 background, i.e., $\bar{g}=g+h$
with $g$ the background metric. From now on, all contractions, raising and
lowering of indices, and inner products such as (14) are taken with respect to
the background metric $g$. The quadratic action, fixing $\Lambda=-1$ and
taking into account the Gibbons-Hawking term, is [18]

$\displaystyle S^{(2)}=S$ $\displaystyle=-\int
d^{3}x\sqrt{g}\left(\dfrac{1}{4}h_{\mu\nu}(\nabla^{2}+2)h^{\mu\nu}-\dfrac{1}{8}h\nabla^{2}h+\dfrac{1}{2}G(h)^{2}\right)\,.$
(24)

The last term in the above expression invites to define the de Donder gauge
functional

$G_{\nu}(h)=\nabla^{\mu}h_{\mu\nu}-\dfrac{1}{2}\nabla_{\nu}h\,.$ (25)

As expected, the quadratic action (24) is invariant under linearized
diffeomorphisms, i.e., $h_{\mu\nu}\to h_{\mu\nu}+\mathcal{L}_{\xi}g_{\mu\nu}$,
for $h$ and $\xi$ in $L^{2}$. As a result of this and the orthogonality
properties of the decomposition (22), (24) factorizes into two independent
pieces and no longer depends on the longitudinal component, that is

$S[h_{\mu\nu}]=S[h_{TT}+h_{\Theta}+h_{L}]=S[h_{TT}]+S[h_{\Theta}]\,.$ (26)

The explicit form of the actions can be easily written as quadratic operators
acting of each component as follows

$\displaystyle S[h_{TT}]$ $\displaystyle=\int
d^{3}x\sqrt{g}\left(\dfrac{1}{4}(h_{TT})_{\mu\nu}\left(-\nabla^{2}-2\right)(h_{TT})^{\mu\nu}\right)\,,$
(27) $\displaystyle S[h_{\Theta}]$ $\displaystyle=-\int
d^{3}x\sqrt{g}\left(\dfrac{1}{8}\chi\left(-\nabla^{2}+3\right)\left(-\nabla^{2}+2\right)^{2}\chi\right)\,.$
(28)

We see that the action for the trace mode is negative definite. This is a
well-known problem for the Euclidean action [19] and the standard procedure to
deal with it is to rotate the contour of integration to be parallel to the
imaginary axis. The partition function can be easily computed using the
decomposition (22) and the property (26)

$\displaystyle Z$ $\displaystyle=\int Dh\,e^{-S\left[h\right]}=\int
Dh_{TT}Dh_{V}Dh_{\Theta}\,e^{-S\left[h_{TT}+h_{V}+h_{\Theta}\right]}$ (29)
$\displaystyle=\int
Dh_{TT}DVD\chi\,J_{V}J_{\chi}e^{-S\left[h_{TT}+h_{\Theta}(\chi)+\mathcal{L}_{V}g\right]}$
$\displaystyle=J_{V}J_{\chi}\left(\int DV\right)\left(\int
Dh_{TT}\,e^{-S[h_{TT}]}\right)\left(\int
D\chi\,e^{-S[h_{\Theta}(\chi))]}\right)\,,$

where $J_{V}$ and $J_{\chi}$ are the Jacobians associated to the change in the
integration variables $h_{V}\to V$ and $h_{\Theta}\to\chi$ respectively. . The
integral over the vectors diverges and it is nothing but the volume of the
space of square-integrable vectors which can be identified with proper
diffeomorphims [6]. The last two integrals are Gaussian and they can be
written in terms of the determinants of the operators appearing in (27). The
explicit computation of the Jacobians $J_{V}$ and $J_{\chi}$ is carried out in
the appendix (A). Collecting everything together we obtain

$Z=\dfrac{\text{det}^{1/2}_{V_{T}}\left(-\nabla^{2}+2\right)}{\text{det}^{1/2}_{h_{TT}}\left(-\nabla^{2}-2\right)}\,,$
(30)

where the operator in the numerator acts on square-integrable transverse
vectors and the operator in the denominator on square-integrable transverse-
traceless rank-2 tensors. This is precisely the result of [20] using the de
Donder gauge-fixing condition.

####  2.3.1 Accessibility of the Gauge Choice

Even though we have not explicitly chosen a gauge in the previous computation,
let us discuss briefly what would have happened if we had chosen one. The
usual gauge-fixing procedure consists in picking a gauge condition $G(h)=0$
and, using the Faddeev-Popov procedure, write an identity operator as follows

$1=J_{\eta}\int D\eta\,\delta(G(h+\mathcal{L}_{\eta}g))\,,$ (31)

were $J_{\eta}$ is the Jacobian associated to the change of variables from the
metric $h$ to the ghost vector field $\eta$. The next step is to insert (31)
in the path integral with the objective of factorize the gauge redundancy. In
order to do this, the previous expression has to be valid for all square-
integrable perturbations $h$. That is, the gauge has to be accessible, which
means that for every square-integrable $h$ there must exist a perturbation
$h_{0}$ which fulfills the gauge condition and a vector $\eta$ such that

$h_{\mu\nu}=h_{0}+\mathcal{L}_{\eta}g\,.$ (32)

Using the decomposition (22) for $h$, the de Donder gauge condition (25) turns
out to be

$G(h^{TT}_{\mu\nu})+G(\Theta_{\mu\nu}\chi)+G(\mathcal{L}_{\mu\nu}V)=G(h_{0})+G(\mathcal{L}_{\mu\nu}\eta)\,.$
(33)

Given that $h^{TT}$ and $h_{0}$ fulfill the gauge condition, the above
expression is simply given by

$-\dfrac{1}{2}\nabla^{\nu}\left(\nabla^{2}-3\right)\chi-P^{\nu}(V)=-P^{\nu}(\eta)\,,$
(34)

where $P^{\nu}$ is defined by

$P^{\nu}(\eta)=\left(-\nabla^{2}+2\right)\eta^{\nu}\,.$ (35)

In order to guarantee the accessibility of the gauge choice we need to be able
to solve (35) for the auxiliary ghost vector field $\eta$. Notice that $P$ is
definite positive when acting on square-integrable vectors and therefore it
can, at least formally, be inverted111In Section 3.1.2 we prove that $0$ does
not belong to the spectrum of $P$ and therefore $P$ is invertible.. Thus,
because the two terms in the LHS of (35) are square integrable, it suffices to
demand the ghost fields to belong to the space of $L^{2}$ vector fields and
not to the space of vectors which generate $L^{2}$ metric perturbations.

##  3 Spectrum of Laplacian Operators

In this section we are going to analyze the $\mathbb{H}^{3}/\Gamma$ spectrum
of the operators appearing in (30) acting on rank-2 symmetrical transverse-
traceless tensors and transverse vector fields respectively. The spectra of
these operators are fully understood in $\mathbb{H}^{3}$, for example, it is a
well-known fact that the full spectrum of the Lichnerowicz Laplacian
$\Delta_{LL}$ acting on rank-2 symmetrical traceless tensors in
$\mathbb{H}^{3}$ is $[-3,\infty)$ with no eigenvalues below it [9, 21]. The
Lichnerowicz Laplacian in this case is related with the ordinary Laplacian by
[22]

$\left(\Delta_{LL}+4\right)T_{\mu\nu}=\left(-\nabla^{2}-2\right)T_{\mu\nu}\,,$
(36)

which means the spectrum of the operator $(-\nabla^{2}-2)$ is $[1,+\infty)$.
If the space is asymptotically hyperbolic (as it is $\mathbb{H}^{3}/\Gamma$)
with dimension $n=3$, then $[-3,\infty)$ is the essential spectrum of the
Lichnerowicz Laplacian and there might exist some eigenvalues below $-3$. A
similar scenario holds for the case of the vector fields [12, 13].

In order to study the spectra of the operators we are going to use an
auxiliary first-order differential equation and prove that their spectra are
closely related. We use coordinates $(z,x_{+},x_{-})$ which are related to the
coordinates described in (9) by

$z=\tanh^{2}(\zeta)=\dfrac{r^{2}-r_{+}^{2}}{r^{2}+r_{-}^{2}}\,.$ (37)

In these coordinates the metric is

$\displaystyle ds^{2}$
$\displaystyle=d\zeta^{2}+\sinh^{2}(\zeta)dx^{2}_{-}+\cosh^{2}(\zeta)dx^{2}_{+}$
(38)
$\displaystyle=\dfrac{1}{4z(1-z)^{2}}dz^{2}+\dfrac{z}{1-z}dx^{2}_{-}+\dfrac{1}{1-z}dx^{2}_{+}\,,$

where the $x_{+}$ and $x_{-}$ have the periodicity (12). In these coordinates
it turns out that the components of the solutions of the differential
equations can be written in terms of the boundary components
$\\{\xi_{+}\,,\xi_{-}\\}$ for vectors, and $\\{h_{++}\,,h_{--}\,,h_{+-}\\}$
for the rank-2 tensors. The resulting differential equations can be solved
with hypergeometric functions. As usual, we are going to impose a delta-like
normalization condition upon the solutions to analyze the spectra of the
operators222Notice that it is possible to obtain $L^{2}$ functions by
multiplying by a bump-like function with an appropriate localized support in
the interior of the space and as close to the asymptotic region as one wants
[23].

###  3.1 Rank-1 Tensor Field

####  3.1.1 Vector Laplacian and its Equivalent First-Order Operator

Our goal is to determine the spectra of the operators appearing in (30)
subject to tranversality conditions. We will start with the rank-1 case in
which we have

$\left(-\nabla^{2}+2\right)\xi_{\mu}=\alpha^{2}\xi_{\mu}\,,$ (39)

with the constraint

$\nabla^{\mu}\xi_{\mu}=0\,.$ (40)

Let us notice that the above system of differential equations can be formally
factorized in terms of first-order differential operators as follows

$m^{2}\mathcal{P}_{\mu}^{\nu}(-m)\mathcal{P}_{\nu}^{\rho}(m)\xi_{\rho}=(\nabla^{2}+2+m^{2})\xi_{\mu}\,,$
(41)

where $\mathcal{P}(\pm m)$ is defined by333Since we will find later on that
$m$ belongs to the spectrum of
$\epsilon\indices{{}_{\mu}^{\nu}{}^{\rho}}\nabla_{\nu}$ if
$m\in\mathbb{R}-\left\\{0\right\\}$, and the spectrum is a closed set of the
complex plane, $m=0$ necessarily belongs to the spectrum of
$\epsilon\indices{{}_{\mu}^{\nu}{}^{\rho}}\nabla_{\nu}$. So we do not need to
consider the case $m=0$ independently.,444We denote the Levi-Civita symbol by
$\varepsilon_{\mu\nu\rho}$ and the Levi-Civita pseudo tensor by
$\epsilon_{\mu\nu\rho}=\sqrt{g}\varepsilon_{\mu\nu\rho}$.

$\mathcal{P}_{\mu}^{\nu}(\pm
m)\xi_{\nu}=\left(\delta_{\mu}^{\nu}\pm\dfrac{1}{m}\epsilon\indices{{}_{\mu}^{\rho}{}^{\nu}}\nabla_{\rho}\right)\xi_{\nu}\,,$
(42)

and $m$ is given by

$m^{2}=\alpha^{2}-4\,.$ (43)

The factorization from above suggests that, instead of solving the second-
order problem of (39), we might solve the first-order one given by555It is
common to find in the literature that this is a shortcut first shown in [24].
However, as there are unbounded operators involved, it seems to be more
complicated than just finding the eigenvalues of $\mathcal{P}$. For example,
it is obvious that an eigenfunction of $\mathcal{P}(m)$ with zero eigenvalue
must be an eigenfunction of (39) by the identity (42). However, how do we know
that the spectrum of the second order differential operator comes completely
from that of the first order differential operator? It will turn out to be the
case, but as far as we know, it has not been analyzed at all in the present
context.

$\epsilon\indices{{}_{\mu}^{\nu}{}^{\rho}}\nabla_{\nu}\xi_{\rho}=-m\xi_{\mu}\,,$
(44)

or, in terms of the operator $\mathcal{P}(\pm m)$,

$\mathcal{P}_{\mu}^{\nu}(\pm m)\xi_{\nu}=0\,.$ (45)

As we said before, we will prove that if one obtains the spectra of (45) then
one automatically obtains the spectra of (39).

As a first step, we will show that the eigenfunctions of (44) are
eigenfunctions of (39) too. Notice that taking derivative in both sides of
(44) we obtain

$\displaystyle-m\nabla_{\mu}\xi^{\mu}=\epsilon\indices{{}^{\mu}{}^{\nu}{}^{\rho}}\nabla_{\mu}\nabla_{\nu}\xi_{\rho}$
$\displaystyle=\dfrac{1}{2}\epsilon\indices{{}^{\mu}{}^{\nu}{}^{\rho}}\left[\nabla_{\mu},\nabla_{\nu}\right]\xi_{\rho}=-\dfrac{1}{2}\epsilon\indices{{}^{\mu}{}^{\nu}{}^{\rho}}R\indices{{}^{\lambda}_{\rho}{}_{\mu}{}_{\nu}}\xi_{\lambda}$
(46)
$\displaystyle=-\dfrac{1}{2}\epsilon\indices{{}^{\mu}{}^{\nu}{}^{\rho}}\left(g_{\lambda\nu}g_{\rho\mu}-g_{\lambda\mu}g_{\rho\nu}\right)\xi^{\lambda}=0\,,$

therefore, solutions of (44) are automatically transverse.

To recover the second-order differential equation (39) we just need to replace
in (44) a solution $\xi$ of the equation itself. What we obtain is

$\displaystyle m^{2}\xi_{\mu}$
$\displaystyle=\epsilon\indices{{}_{\mu}^{\nu}{}^{\rho}}\epsilon\indices{{}_{\rho}^{\alpha}{}^{\beta}}\nabla_{\nu}\nabla_{\alpha}\xi_{\beta}=\left(g_{\mu\alpha}g_{\nu\beta}-g_{\mu\beta}g_{\nu\alpha}\right)\nabla^{\nu}\nabla^{\alpha}\xi^{\beta}=\nabla_{\beta}\nabla_{\mu}\xi^{\beta}-\nabla^{2}\xi_{\mu}$
(47)
$\displaystyle=\nabla_{\mu}\nabla_{\beta}\xi^{\beta}+\left[\nabla_{\beta},\nabla_{\mu}\right]\xi^{\beta}-\nabla^{2}\xi_{\mu}=\left(-\nabla^{2}-2\right)\xi_{\mu}\,,$

where we have used the transverse condition (40) in the last step. Therefore,
the eigenfunctions of (44) also satisfy

$\left(-\nabla^{2}+2\right)\xi_{\mu}=\left(4+m^{2}\right)\xi_{\mu}\,,$ (48)

which is the second-order equation (39) with $\alpha^{2}=4+m^{2}$.

The above computations give us a hint that we might find the spectra of (39)
by finding the spectra of (44) instead. However, knowing the eigenfunctions of
$\mathcal{P}$ are eigenfunctions of the second-order operator in (39) is not
enough. We can ask ourselves if the converse holds (and it does by a simple
argument). But the difficult question is whether the essential spectrum (the
part of the spectrum that are not isolated eigenvalues) of $\mathcal{P}$ is
enough to reconstruct the essential spectrum of the operator in (39). In the
following subsection we will show that this is indeed the case by relating
both spectra.

####  3.1.2 Proof of the Relation between the Spectra of $\mathcal{P}(\pm m)$
and $-\nabla^{2}-2$

Some clarifications related to the computations performed in the last section
are in place. First, we had always implicitly consider the self-adjoint
extensions of the differential operators. Second, we are going to use
repeatedly that $\mathcal{P}(m)\mathcal{P}(-m)=\mathcal{P}(-m)\mathcal{P}(m)$,
which holds since we can think that the dense domain where the LHS and RHS are
originally defined is that of compactly supported vector fields, which is a
core for the Laplacian [25]. Therefore, taking closure on both sides gives
self-adjoint operators, which must coincide since self-adjoint extensions are
unique.

Starting with the isolated eigenvalues, we can say that if $m^{2}$ is an
isolated eigenvalue of $-\nabla^{2}-2$, then both square roots $\pm|m|$ are
eigenvalues of $\epsilon\indices{{}_{\mu}^{\nu}{}^{\rho}}\nabla_{\nu}$. This
is because if $\mathcal{P}(m)\mathcal{P}(-m)\phi_{n}=0$ holds with finite
multiplicity for some vectors $\phi_{n}$, it follows that if $\phi$ is one of
these eigenvectors associated to $m^{2}$, then $\mathcal{P}(\pm m)\phi$ is an
eigenvector of $\mathcal{P}(\mp m)$, implying that both $\pm m$ are
eigenvalues of $\epsilon\indices{{}_{\mu}^{\nu}{}^{\rho}}\nabla_{\nu}$.
Conversely, if $m$ is an eigenvalue of
$\epsilon\indices{{}_{\mu}^{\nu}{}^{\rho}}\nabla_{\nu}$, by (41) $m^{2}$ is an
eigenvalue of $-\nabla^{2}-2$. This means that if we characterize in detail
the spectrum of the first order differential operator
$\epsilon\indices{{}_{\mu}^{\nu}{}^{\rho}}\nabla_{\nu}$, we have a full
characterization of the spectrum of the original operator $-\nabla^{2}-2$.
However, so far we have analyzed the discrete spectrum, it remains to do the
same for the essential spectrum to complete the reasoning.

Let us now show that if $m^{2}\in\mathbb{R}$ is in the essential spectrum of
$-\nabla^{2}-2$, then one of its square roots $\pm|m|$ is in the essential
spectrum of $\epsilon\indices{{}_{\mu}^{\nu}{}^{\rho}}\nabla_{\nu}$.
Conversely, for any $m$ in the essential spectrum of
$\epsilon\indices{{}_{\mu}^{\nu}{}^{\rho}}\nabla_{\nu}$, $m^{2}$ belongs to
the essential spectrum of $-\nabla^{2}-2$.

First we need to recall a few facts about spectral theory of self-adjoint
operators, which we denote as $A$ [23], and Dom$(A)$ refers to its domain:

  1. 1.

$\lambda\in$ Spec$(A)$ if and only if
$\exists\left\\{\phi_{n}\right\\}\subset$ Dom$(A),||\phi_{n}||=1,$ such that
$\lim_{n\rightarrow\infty}||(A-\lambda)\phi_{n}||=0$

  2. 2.

If the previous sequence $\left\\{\phi_{n}\right\\}\subset$ Dom$(A)$ satisfies
that $\lim_{n\rightarrow\infty}\langle\psi,\phi_{n}\rangle=0$ for any $\psi\in
L^{2}$, it is called a Weyl sequence and
$\lambda\in\text{Spec}_{\text{ess.}}(A)$, the essential spectrum of $A$.
Conversely, if $\lambda\in\text{Spec}_{\text{ess.}}(A)$ there exists a
corresponding Weyl sequence. This is the Weyl criterion.

  3. 3.

The remaining sequences of item 1., namely those which are not Weyl sequences
(do not have a zero weak limit) correspond to a $\lambda$ in the discrete
spectrum and have finite degeneracy.

What the second statement is telling us is that $\lambda$ is in the essential
spectrum if and only if there is a Weyl sequence associated to it. We then
have to show that if we have a Weyl sequence associated to $m^{2}$ for the
operator $-\nabla^{2}-2$, then we have a Weyl sequence for $\mathcal{P}(m)$ or
$\mathcal{P}(-m)$ and $\lambda=0$ and vice versa.

Let us start and assume that there is such Weyl sequence of vector fields
$\left\\{\phi_{n}\right\\}$, so we have

$||(-\nabla^{2}-2-m^{2})\phi_{n}||\rightarrow 0\,,$

which by (41) is equivalent to

$||\mathcal{P}(m)\mathcal{P}(-m)\phi_{n}||\rightarrow 0\,.$

We are going to show next that one can choose either
$\left\\{\phi_{n}\right\\}$ or
$\left\\{\frac{\mathcal{P}(-m)\phi_{n}}{||\mathcal{P}(-m)\phi_{n}||}\right\\}$
as Weyl sequence for $\lambda=0$ and $\mathcal{P}(-m)$ or $\mathcal{P}(m)$
respectively, depending on whether the limit
$\lim_{n\rightarrow\infty}||\mathcal{P}(-m)\phi_{n}||$ is zero or not. Let us
first assume that $\lim_{n\rightarrow\infty}||\mathcal{P}(-m)\phi_{n}||=0$.
Then the original Weyl sequence $\left\\{\phi_{n}\right\\}$ is by hypothesis a
Weyl sequence for $\mathcal{P}(-m)$ and $\lambda=0$. Namely, $m$ belongs to
the spectrum of $\epsilon\indices{{}_{\mu}^{\nu}{}^{\rho}}\nabla_{\nu}$.

Now, let us consider the case in which
$\lim_{n\rightarrow\infty}||\mathcal{P}(-m)\phi_{n}||=k\neq{0}$. The vectors
$\psi_{n}:=\frac{\mathcal{P}(-m)\phi_{n}}{||\mathcal{P}(-m)\phi_{n}||}$ are
normalized vectors and have a weak zero limit since666If
$\lim_{n\rightarrow\infty}||\mathcal{P}(-m)\phi_{n}||$ does not exist, then
there is a subsequence
$\psi_{n_{k}}:=\frac{\mathcal{P}(-m)\phi_{n_{k}}}{||\mathcal{P}(-m)\phi_{n_{k}}||}$
which satisfies $\epsilon<||\psi_{n}||$ for some $\epsilon>0$ and by the same
arguments it is a Weyl sequence for $\mathcal{P}(m)$ and $\lambda=0$. :

$\lim_{n\rightarrow{\infty}}\langle\psi,\psi_{n}\rangle=\lim_{n\rightarrow{\infty}}\langle\mathcal{P}(-m)\psi,\frac{1}{||\mathcal{P}(-m)\phi_{n}||}\phi_{n}\rangle=1/k\lim_{n\rightarrow{\infty}}\langle\mathcal{P}(-m)\psi,\phi_{n}\rangle=0\,.$

It remains to show that
$\lim_{n\rightarrow{\infty}}||\mathcal{P}(m)\psi_{n}||=0$ but that follows
from the assumption that $\left\\{\phi_{n}\right\\}$ is a Weyl sequence for
$\mathcal{P}(m)\mathcal{P}(-m)$.

The converse also holds . This comes from another way of
characterizing/defining the essential spectrum of a self-adjoint operator:
$\lambda$ is in the essential spectrum of $A$ if and only if
$A-\lambda\mathbb{I}$ has either infinite dimensional kernel or cokernel [26].

Using this characterization of the essential spectrum, if we assume that 0 in
the essential spectrum of $\mathcal{P}(-m)$, then $\mathcal{P}(-m)$ has either
infinite dimensional kernel or cokernel. We have to show that the same holds
for $\mathcal{P}(m)\mathcal{P}(-m)$. Let us suppose that dim
Ker$\mathcal{P}(-m)=\infty$, then $\mathcal{P}(m)\mathcal{P}(-m)$ has infinite
dimensional kernel as well. On the other hand, if dim
Coker$\mathcal{P}(-m)=\infty$, since
Ran$(\mathcal{P}(-m)\mathcal{P}(m))\subseteq$ Ran$(\mathcal{P}(-m))$, we have

$\text{dim Coker}(\mathcal{P}(m)\mathcal{P}(-m))=\text{dim
Coker}(\mathcal{P}(-m)\mathcal{P}(m))\geq\text{dim
Coker}(\mathcal{P}(-m))=\infty\,,$

which implies that dim Coker$(\mathcal{P}(m)\mathcal{P}(-m))=\infty$. We have
called Ran$(\mathcal{P}(-m))$ to the range of $\mathcal{P}(-m)$.

With the discussion above, we are now certain that by studying the spectrum of
the operators $\mathcal{P}(m)$ for any $m$, we have complete control over the
spectrum of $-\nabla^{2}-2$. It is also important to note that it did not
matter the tensorial structure of the operators, so we can apply the same
logic later to analyze the case of the rank-2 tensor.

####  3.1.3 Solutions to the First-Order Differential Operator and its
Essential Spectrum

In the previous section we showed that the spectrum of $-\nabla^{2}-2$ can be
found by solving the spectrum of a simpler first-order differential operator.
Now, we will proceed to solve $\mathcal{P}(m)\xi=0$ and find its
eigenfunctions for different values of $m$.

It will be useful to rewrite the equation (44) for 1-forms using coordinates
(38)

$\displaystyle\epsilon[\xi]_{\mu}$
$\displaystyle=\epsilon\indices{{}_{\mu}^{\nu}{}^{\rho}}\nabla_{\nu}\xi_{\rho}=-m\xi_{\mu}$
(49)
$\displaystyle=-\frac{1}{2z}(\partial_{+}\xi_{+}-\partial_{-}\xi_{-})dz-2z(z-1)(\partial_{+}\xi_{z}-\partial_{z}\xi_{+})dx_{-}-2(z-1)(\partial_{z}\xi_{-}-\partial_{-}\xi_{z})dx_{+}\,.$

The solutions of this equation also have to satisfy the equation (47). Writing
down the components of the Laplacian acting on 1-forms, it can be shown that

$\displaystyle(\nabla^{2}\xi)_{-}-\Delta\xi_{-}+2\epsilon[\xi]_{+}=-2\xi_{-}\,,$
(50)
$\displaystyle(\nabla^{2}\xi)_{+}-\Delta\xi_{+}-2\epsilon[\xi]_{-}=-2\xi_{+}\,,$

where $\Delta$ is the scalar Laplacian acting on the components of the 1-form.
Using the fact that $\xi$ is a solution of (49) the equation (47) can be
written in terms of the following matrix equation

$\left(\nabla^{2}+2\right)\begin{pmatrix}\xi_{-}\\\
\xi_{+}\end{pmatrix}=\Delta\begin{pmatrix}\xi_{-}\\\
\xi_{+}\end{pmatrix}+\begin{pmatrix}0&2m\\\
-2m&0\end{pmatrix}\begin{pmatrix}\xi_{-}\\\
\xi_{+}\end{pmatrix}=-m^{2}\begin{pmatrix}\xi_{-}\\\ \xi_{+}\end{pmatrix}\,.$
(51)

This can be diagonalized using $\xi_{1,2}=1/2(\xi_{+}\mp i\xi_{-})$ obtaining

$\begin{pmatrix}\Delta+(-2im+m^{2})&0\\\ 0&\Delta+(2im+m^{2})\\\
\end{pmatrix}\begin{pmatrix}\xi_{1}\\\ \xi_{2}\\\ \end{pmatrix}=0\,.$ (52)

To solve this for $\xi_{1}$ and $\xi_{2}$ we use the ansatz

$\displaystyle\xi_{1}=e^{i(k_{+}x^{+}+k_{-}x^{-})}z^{\alpha_{1}}(1-z)^{\beta_{1}}F_{1}(z)\,,$
(53)
$\displaystyle\xi_{2}=e^{i(k_{+}x^{+}+k_{-}x^{-})}z^{\alpha_{2}}(1-z)^{\beta_{2}}F_{2}(z)\,,$

where $k_{-}$ has to be an integer and $k_{+}=(n-k_{-}r_{-})/r_{+}$ with $n$
an integer due to the periodicity conditions (12). Explicitly, the
differential equation for the functions $F_{1,2}(z)$ is given by

$\displaystyle\left[z(1-z)\dfrac{\partial^{2}F_{i}}{\partial
z^{2}}+((2\alpha_{i}+1)-z(2\alpha_{i}+2\beta_{i}+1))\dfrac{\partial
F_{i}}{\partial
z}-\left(\frac{k_{+}^{2}}{4}+(\alpha_{i}+\beta_{i})^{2}\right)F_{i}\right]$
(54)
$\displaystyle-\dfrac{1}{z}\left(\dfrac{k_{-}^{2}}{4}-\alpha_{i}^{2}\right)F_{i}+\dfrac{4\beta_{i}(\beta_{i}-1)+m(m-2i\epsilon_{i})}{4(1-z)}F_{i}=0\,,$

where $i=1,2$ and $(\epsilon_{1},\epsilon_{2})=(1,-1)$. We see that it is
enough to study the case $k_{+}\geq 0$ and $k_{-}\geq 0$. If we choose the
coefficients of the $1/z$ and the $1/(1-z)$ terms to be zero, i.e. if we
choose $\alpha_{i}$ and $\beta_{i}$ to be

$\alpha_{1,2}=\alpha=\dfrac{k_{-}}{2}\,,\qquad\beta_{1}=\left\\{1+i\frac{m}{2}\
;\ -i\frac{m}{2}\right\\}\,,\qquad\beta_{2}=\left\\{1-i\frac{m}{2}\ ;\
i\frac{m}{2}\right\\}\,,$ (55)

the above differential equation becomes the hypergeometric differential
equation

$z(1-z)\dfrac{\partial^{2}F_{i}}{\partial
z}+(c-(a_{i}+b_{i}+1)z)\dfrac{\partial F_{i}}{\partial
z}-a_{i}b_{i}F_{i}=0\,,$ (56)

with

$c=2\alpha+1=k_{-}+1\,,\qquad
a_{i}=\frac{k_{-}}{2}+i\frac{k_{+}}{2}+\beta_{i}\,,\qquad
b_{i}=\frac{k_{-}}{2}-i\frac{k_{+}}{2}+\beta_{i}\,.$ (57)

The solutions are hypergeometric functions ${}_{2}F_{1}(a_{i},b_{i};c_{i};z)$
or $F(a_{i},b_{i},c_{i},z)$ for simplicity. It is important to note that
$k_{-}$ is an integer because of the periodicity of $x_{-}$. This is a
pathological case for the differential equation where the two usual solutions
are not independent and a second independent solution can be constructed by
Frobenius’ method and it looks like (see for example [27])

$F(a,b,k_{-}+1,z)\log(z)-\sum_{i=0}^{k_{-}}a_{i}(-z)^{-i}+\sum_{i=0}^{\infty}b_{i}z^{i}\,.$
(58)

We are going to neglect these solutions because they make the norm of the
vectors diverge at $z\sim 0$. For the solutions of the form $F(a,b,c,z)$
because of the transformation rules of the hypergeometric function, the two
possibilities of $\beta$ yield to the same result. We are going to choose

$\beta_{1}=-i\dfrac{m}{2}=\beta\,,\qquad\beta_{2}=-i\dfrac{m}{2}+1=\beta+1\,.$
(59)

Then the solutions to the equations (52) are

$\displaystyle\xi_{1}$
$\displaystyle=Ae^{i(k_{+}x^{+}+k_{-}x^{-})}z^{\alpha}(1-z)^{\beta}F(a,b,c,z)\,,$
(60) $\displaystyle\xi_{2}$
$\displaystyle=Be^{i(k_{+}x^{+}+k_{-}x^{-})}z^{\alpha}(1-z)^{\beta+1}F(a+1,b+1,c,z)\,,$

with

$\alpha=\frac{k_{-}}{2}\,,\qquad c=k_{-}+1\,,\qquad
a=\frac{k_{-}}{2}+i\frac{k_{+}}{2}+\beta\,,\qquad
b=\frac{k_{-}}{2}-i\frac{k_{+}}{2}+\beta\,.$ (61)

The constants $A$ and $B$ are not independent. This is because $\xi_{+}$ and
$\xi_{-}$ still have to satisfy the first order equation (49). The deduction
of the relation is in the appendix (B.1) and the result is

$\dfrac{B}{A}=\dfrac{2\beta-k}{2\beta+\bar{k}}\,,$ (62)

with $k=k_{-}+ik_{+}$. To obtain the $\xi_{+}$ and $\xi_{-}$ components we
have to invert the relation that we used to diagonalize the matrix (52). Then
the $z$ component of the equation (49) allows us to find the $\xi_{z}$
component in terms of the two others

$\displaystyle\xi_{-}$
$\displaystyle=iAz^{\alpha}(1-z)^{\beta}e^{i(k_{+}x^{+}+k_{-}x^{-})}\left[F(a,b,c,z)-(1-z)\left(\dfrac{2\beta-k}{2\beta+\bar{k}}\right)F(a+1,b+1,c,z)\right]\,,$
(63) $\displaystyle\xi_{+}$
$\displaystyle=Az^{\alpha}(1-z)^{\beta}e^{i(k_{+}x^{+}+k_{-}x^{-})}\left[F(a,b,c,z)+(1-z)\left(\dfrac{2\beta-k}{2\beta+\bar{k}}\right)F(a+1,b+1,c,z)\right]\,,$
$\displaystyle\xi_{z}$ $\displaystyle=\dfrac{1}{4\beta
z}\left(k_{+}\xi_{+}-k_{-}\xi_{-}\right)=-\dfrac{1}{2\beta
z}(k\xi_{2}+\bar{k}\xi_{1})\,.$

For the above solutions the norm does not diverge for $z\simeq 0$ therefore we
only777It might seem, by looking at (64), that if $\alpha=0$ (i.e. $k_{-}=0$)
the contribution of the $\xi_{-}$ component gives a problem near $z=0$ in the
norm, but it is not the case. If $k_{-}=0$, then the constant (62) becomes $1$
and the terms $\mathcal{O}(0)$ of the two hypergeometrics in $\xi_{-}$ cancel
each other. need to check the behavior for $z\simeq 1$. It is easier to
analyze the asymptotic behavior of these solutions using the coordinate
$z=\tanh^{2}(\zeta)$ where the norm of the vectors is

$\displaystyle\langle\xi,\xi\rangle$ $\displaystyle=\int
dzdx_{-}dx_{+}\left(\left|\xi_{z}\right|^{2}z+\dfrac{\left|\xi_{-}\right|^{2}}{(1-z)z}+\dfrac{\left|\xi_{+}\right|^{2}}{(1-z)}\right)\,,$
(64) $\displaystyle=\int d\zeta
dx_{-}dx_{+}\left(\left|\xi_{\zeta}\right|^{2}\sinh(\xi)\cosh(\xi)+\left|\xi_{-}\right|^{2}\dfrac{\cosh(\xi)}{\sinh(\xi)}+\left|\xi_{+}\right|^{2}\dfrac{\sinh(\xi)}{\cosh(\xi)}\right)\,.$

Because of the decomposition (53) the integral in $x_{+}$ and $x_{-}$ are
deltas. The radial part of the integral can be divergent depending on the
asymptotic behavior of the functions $\xi_{1}$ and $\xi_{2}$. Using the
asymptotic behavior computed in (B.3) we obtain

$\displaystyle\xi_{1}\sim c_{1}e^{(-2-im)\zeta}+c_{2}e^{im\zeta}\,,$ (65)
$\displaystyle\xi_{2}\sim c_{3}e^{-im\zeta}+c_{4}e^{(-2+im)\zeta}\,,$
$\displaystyle\xi_{z}\sim e^{-2\zeta}\left(k\xi_{1}+\bar{k}\xi_{2}\right)\,.$

First, it is important to note that both choices of $\beta$ in (55) give the
same asymptotic behavior. It is easy to see that if $\text{Im}(m)\neq 0$ this
norm is infinite. The only case where the form $\xi_{\mu}$ can be delta-like
normalizable is when $m\in\mathbb{R}$. Therefore, the equation (48) turns out
to be

$(-\nabla^{2}+2)\xi_{\mu}=\alpha^{2}\xi_{\mu}=(m^{2}+4)\xi_{\mu}\,.$ (66)

This means that the spectrum of the operator $-\nabla^{2}+2$ is $[4,+\infty)$
with no eigenvalues below. It is consistent with the spectral function
computed it [28] and with the spectrum of the Hodge Laplacian acting on forms
computed in [12, 13].

###  3.2 Rank-2 Tensor Field: First-Order Differential Operator, its Solutions
and the Essential Spectrum

For the rank-2 tensor case, the equation we want to solve is

$\left(-\nabla^{2}-2\right)h_{\mu\nu}=\,\alpha^{2}h_{\mu\nu}\,,$ (67)

subject to the constraints

$\displaystyle\nabla^{\mu}h_{\mu\nu}=$ $\displaystyle\,0\,,$ (68)
$\displaystyle g^{\mu\nu}h_{\mu\nu}=$ $\displaystyle\,0\,,$

where $h_{\mu\nu}$ is symmetric. Analogously as in the vector case, it can be
shown that for symmetric tensors the eigenfunctions and the spectrum of these
equations are related to the eigenfunctions and the spectrum of the following
first order equation (see proof in section (3.1.2))

$\epsilon[h]_{\mu\sigma}=\epsilon\indices{{}_{\mu}^{\nu}{}^{\rho}}\nabla_{\nu}h_{\rho\sigma}=-mh_{\mu\sigma}\,.$
(69)

It can be shown that the solutions of (69) are transverse and traceless. Using
(69) a second time we can show that its solutions satisfy the original
differential equation with $\alpha=m^{2}+1$,

$(-\nabla^{2}-2)h_{\mu\nu}=(1+m^{2})h_{\mu\nu}\,.$ (70)

Looking at the $(z-)$ and $(z+)$ components of the equation (69) we have

$\displaystyle\epsilon[h]_{z-}=-h_{z+}+\dfrac{1}{2z}\left(\dfrac{\partial
h_{--}}{\partial x^{+}}-\dfrac{\partial h_{+-}}{\partial
x^{-}}\right)=-mh_{z-}\,,$ (71)
$\displaystyle\epsilon[h]_{z+}=h_{z-}+\dfrac{1}{2z}\left(\dfrac{\partial
h_{+-}}{\partial x^{+}}-\dfrac{\partial h_{++}}{\partial
x^{-}}\right)=-mh_{z+}\,.$

These equations together with the traceless condition allow us to write all
the components in terms of the boundary components $h_{++},\,h_{+-},h_{--}$.
With help of the equation (69) we can write

$\displaystyle(\nabla^{2}h)_{--}-\Delta h_{--}-4\epsilon[h]_{+-}$
$\displaystyle=$ $\displaystyle-4h_{--}+2h_{++}\,,$ (72)
$\displaystyle(\nabla^{2}h)_{+-}-\Delta
h_{+-}+2\epsilon[h]_{--}-2\epsilon[h]_{++}$ $\displaystyle=$
$\displaystyle-6h_{+-}\,,$ $\displaystyle(\nabla^{2}h)_{++}-\Delta
h_{++}+4\epsilon[h]_{+-}$ $\displaystyle=$ $\displaystyle+2h_{--}-4h_{++}\,,$

where $\Delta$ stands for the scalar Laplacian acting on the tensor
components. Using that $h_{\mu\nu}$ is solution of the first order equation we
can write the following matrix equation

$\left(\nabla^{2}+3\right)\begin{pmatrix}h_{--}\\\ h_{+-}\\\
h_{++}\end{pmatrix}=\Delta\begin{pmatrix}h_{--}\\\ h_{+-}\\\
h_{++}\end{pmatrix}+\begin{pmatrix}-1&-4m&2\\\ 2m&-3&-2m\\\
2&4m&-1\end{pmatrix}\begin{pmatrix}h_{--}\\\ h_{+-}\\\
h_{++}\end{pmatrix}=-m^{2}\begin{pmatrix}h_{--}\\\ h_{+-}\\\
h_{++}\end{pmatrix}\,,$ (73)

which can be diagonalized with

$\begin{pmatrix}h_{0}\\\ h_{-}\\\
h_{+}\end{pmatrix}=\begin{pmatrix}1/2&0&1/2\\\ -1/4&i/2&1/4&\\\
-1/4&-i/2&1/4\\\ \end{pmatrix}\begin{pmatrix}h_{--}\\\ h_{+-}\\\
h_{++}\end{pmatrix}\,,$ (74)

to obtain

$\begin{pmatrix}\Delta+m^{2}+1&0&0\\\ 0&\Delta+(m-2i)^{2}+1&0\\\
0&0&\Delta+(m+2i)^{2}+1\\\ \end{pmatrix}\begin{pmatrix}h_{0}\\\ h_{-}\\\
h_{+}\\\ \end{pmatrix}=0\,.$ (75)

The solutions to these equations are obtained in the same way that in the
vector case. Each component has two equivalent elections for its constant
$\beta_{i}$, choosing them properly we can write the final result,

$\displaystyle h_{+}$
$\displaystyle=A_{+}z^{\alpha}(1-z)^{\beta}F(a,b,c;z)e^{i(k_{+}x^{+}+k_{-}x^{-})}\,,$
(76) $\displaystyle h_{0}$
$\displaystyle=A_{0}z^{\alpha}(1-z)^{\beta+1}F(a+1,b+1,c;z)e^{i(k_{+}x^{+}+k_{-}x^{-})}\,,$
$\displaystyle h_{-}$
$\displaystyle=A_{-}z^{\alpha}(1-z)^{\beta+2}F(a+2,b+2,c;z)e^{i(k_{+}x^{+}+k_{-}x^{-})}\,,$

with

$\begin{array}[]{rcccl}\alpha&=&\dfrac{k_{-}}{2}\,,&&\\\
\beta&=&\frac{im}{2}-\frac{1}{2}\,,&&\\\ c&=&2\alpha+1&=&k_{-}+1\,,\\\
a&=&\alpha+\beta+i\frac{k_{+}}{2}&=&\frac{k_{-}}{2}+i\frac{k_{+}}{2}+\beta\,,\\\
b&=&\alpha+\beta-i\frac{k_{+}}{2}&=&\frac{k_{-}}{2}-i\frac{k_{+}}{2}+\beta\,.\\\
\end{array}$ (77)

The constants $A_{0},\,A_{+},\,A_{-}$ are not free and they have to fulfill
the first order equation (69). The relation between them is computed in the
appendix (B.2) and the result is

$\dfrac{A_{0}}{2A_{+}}=\dfrac{2\beta-k}{2\beta+\bar{k}}\,;\qquad\dfrac{2A_{-}}{A_{0}}=\dfrac{2(\beta+1)-k}{2(\beta+1)+\bar{k}}\,,$
(78)

with $k=k_{-}+ik_{+}$. To write down the components of the tensor we have to
invert (74) and use (76) together with (78) to obtain the boundary components
$h_{++},h_{+-},h_{--}$. Then, inverting (71) we can obtain the components
$h_{z+}$ and $h_{z-}$. Finally, the traceless conditions gives the $h_{zz}$
component.

Asymptotically the norm of the tensors is

$\langle h,h\rangle\sim\int d\zeta
dx_{+}dx_{-}\left(\left|h_{++}e^{-\xi}\right|^{2}+\left|h_{--}e^{-\xi}\right|^{2}+2\left|h_{+-}e^{-\xi}\right|^{2}+2\left|h_{\xi+}\right|^{2}+2\left|h_{\xi-}\right|^{2}+\left|h_{\zeta\zeta}\right|^{2}\right)\,,$
(79)

and the asymptotic behaviors of the solutions are

$\displaystyle h_{+}(\xi)$ $\displaystyle\sim c_{1}e^{im\zeta}e^{-3\zeta}$
$\displaystyle+$ $\displaystyle\quad c_{2}e^{-im\zeta}e^{\zeta}\,,$ (80)
$\displaystyle h_{0}(\xi)$ $\displaystyle\sim c_{3}e^{im\zeta}e^{-\zeta}$
$\displaystyle+$ $\displaystyle\quad c_{4}e^{-im\zeta}e^{-\zeta}\,,$
$\displaystyle h_{-}(\xi)$ $\displaystyle\sim c_{5}e^{im\zeta}e^{\zeta}\quad$
$\displaystyle+$ $\displaystyle\quad c_{6}e^{-im\zeta}e^{-3\zeta}\,.$

It is easy to see that as well as in the vector case, this norm is badly
divergent if $m$ has an imaginary part. The equation (70) says that the
spectrum of the operator $-\nabla^{2}-2$ acting on symmetric transverse
traceless rank-2 tensors is $[1,+\infty)$ with no eigenvalues below. This is
consistent with the the well-known results for the essential spectrum of the
Lichnerowicz Laplacian on $\mathbb{H}^{3}$ and $\mathbb{H}^{3}/\Gamma$
computed in [9, 21]. It is also consistent with the spectral functions
computed in [28].

##  4 Summary

We have shown that vector (ghost) fields must belong to the space of square-
integrable vectors fields rather than to the space of vectors fields which
generate square-integrable metric perturbations. This distinction is important
because it is precisely this condition that splits the space of generators of
asymptotic symmetries into proper and improper asymptotic vectors [6]. We also
have shown that this condition is enough to guarantee the accessibility of the
de Donder gauge, which means that the gauge-fixing procedure in the
computation of the partition function only removes the redundancies generated
by proper diffeomorphisms, leaving intact the metric perturbations with a
Brown-Henneaux fall-off behavior.

Then we switched to study the operators that appear in the final form of the
one-loop partition function. We have proven that the spectrum these operators
can be studied by looking at the spectrum of first-order operators. We
concluded that there are no strictly square-integrable eigenfunctions for
these operators in $\mathbb{H}^{3}/\Gamma$. By imposing the solutions to be
delta-like normalizable, we have found the full essential spectra of the
operators which are, in fact, the full spectra. Our results are consistent
with several spectra computed on $\mathbb{H}^{3}$ and $\mathbb{H}^{3}/\Gamma$
[9, 12, 13, 21].

## Acknowledgments

The authors thank Gaston Giribet and Martín Mereb for enlightening
discussions. AG thanks Michael Anderson, Erwann Delay and Rafe Mazzeo for
correspondence. This work was partially supported by grants PIP and PICT from
CONICET and ANPCyT.

##  Appendix A Jacobians

###  A.1 The Computation of $J_{\chi}$

The Jacobian $J_{\chi}$ can be obtained by imposing the following
normalization for the integral

$1=\int Dh_{\Theta}\ e^{-\langle h_{\Theta}\mid h_{\Theta}\rangle}\,.$ (81)

Using the definition of the transverse-tracefull part $h_{\Theta}$ (20) and
performing a change in the integration variable we obtain

$1=\int Dh_{\Theta}\ e^{-\langle h_{\Theta}\mid h_{\Theta}\rangle}=\int D\chi
J_{\chi}\,e^{-\langle\Theta_{\mu\nu}\chi\mid\Theta_{\mu\nu}\chi\rangle}=J_{\chi}\int
D\chi\,e^{-\frac{1}{2}\langle\chi\mid(-\nabla^{2}+3)(-\nabla^{2}+2)\chi\rangle}\,.$
(82)

In the last equality we rewrite the inner product as a quadratic operator
acting on the field $\chi$. The boundary behaviour of $\chi$ guarantees that
the boundary term vanishes. The integral is Gaussian and we can write it as a
determinant. The result is

$J_{\chi}=\text{det}_{\chi}^{1/2}(-\nabla^{2}+3)\times\text{det}_{\chi}^{1/2}(-\nabla^{2}+2)\,.$
(83)

###  A.2 The Computation of $J_{V}$

The factor $J_{V}$ is computed in a similar way to $J_{\chi}$. We impose the
normalization

$1=\int Dh_{L}\,e^{-\langle h_{L}\mid h_{L}\rangle}=\int
DvJ_{V}e^{-\langle\mathcal{L}_{V}\mid\mathcal{L}_{V}\rangle}=J_{V}\int
Dv\,e^{-2\langle V\mid(-\nabla^{\mu}\mathcal{L}_{\mu\nu})[V]\rangle}\,.$ (84)

Now, we can decompose $V=V_{T}+V_{\varphi}$ where the component $V_{T}$ is
transverse and $V_{\varphi}={\rm grad}(\varphi)$. These two components are
orthogonal to each other. Up to a vanishing boundary term, we can rewrite the
exponent in the last term of (84) as the sum of two quadratic operators acting
solely on $V_{T}$ and $V_{\varphi}$

$\langle V\mid-\nabla^{\mu}\mathcal{L}_{\mu\nu}[V]\rangle=\langle
V_{T}\mid\left(-\nabla^{2}+2\right)V_{T}\rangle+2\langle\varphi\mid\nabla^{2}(\nabla^{2}-2)\varphi\rangle\,.$
(85)

Performing a change of variables from $V_{\varphi}$ to $\varphi$ with
$J_{\varphi}$ the corresponding Jacobian, we obtain

$1=J_{V}\left(\int D{V_{T}}e^{-2\langle
V_{T}\mid\left(-\nabla^{2}+2\right)V_{T}\rangle}\right)\left(D\varphi
J_{\varphi}e^{-4\langle\varphi\mid\nabla^{2}(\nabla^{2}-2)\varphi\rangle}\right)\,,$
(86)

where both integrals are Gaussian. The Jacobian $J_{\varphi}$ can be easily
computed with the same procedure as before giving

$J_{\varphi}=\text{det}^{1/2}_{\varphi}(-\nabla^{2})\,.$ (87)

Collecting everything together we finally obtain

$J_{V}=\dfrac{\text{det}^{1/2}_{V_{T}}(-\nabla^{2}+2)\times\text{det}^{1/2}_{\varphi}(-\nabla^{2})\times\text{det}^{1/2}_{\varphi}(-\nabla^{2}+2)}{\text{det}^{1/2}_{\varphi}(-\nabla^{2})}=\text{det}^{1/2}_{V_{T}}(-\nabla^{2}+2)\times\text{det}^{1/2}_{\varphi}(-\nabla^{2}+2)\,.$
(88)

##  Appendix B The Polarization Constants

###  B.1 Rank-1 Tensor

The solutions given in (60) must satisfy the first-order equations (49), in
particular the $\xi_{z}$ component fulfills the equation

$k_{+}\partial_{-}\xi_{z}-k_{-}\partial_{+}\xi_{z}=0\,.$ (89)

The derivatives of $\xi_{z}$ are given in terms of $\xi_{\pm}$ and they can be
found in the $\pm$ components of the first order equation (49). Writing them
in terms of $\xi_{1}$ and $\xi_{2}$ leads to the equation

$\left(-2\beta\bar{k}-2\beta
ik_{+}(1-z)-2\bar{k}z(1-z)\dfrac{d}{dz}\right)\xi_{1}=\left(-2\beta k+2\beta
ik_{+}(1-z)+2kz(1-z)\dfrac{d}{dz}\right)\xi_{2}\,.$ (90)

where $k=k_{-}+ik_{+}$ and $\bar{k}$ its complex conjugate. The equation above
relates the constants $A$ and $B$. To solve it we can use the following
properties of the hypergeometric functions,

$\displaystyle z(1-z)\dfrac{d}{dz}F(a,b,c,z)$
$\displaystyle=(1-z)b\left(F(a,b+1,c,z)-F(a,b,c,z)\right)\,,$ (91)
$\displaystyle z(1-z)\dfrac{d}{dz}F(a+1,b+1,c,z)$
$\displaystyle=(c-a-1)F(a,b+1,c,z)$
$\displaystyle\qquad+(a+1-c+(b+1)z)F(a+1,b+1,c,z)\,,$ (92) $\displaystyle
a(1-z)F(a+1,b+1,c,z)$
$\displaystyle=(c-b-1)F(a,b,c,z)-(c-a-b-1)F(a,b+1,c,z)\,.$ (93)

The LHS of the equation (90) can be written with the help of (91) as

$LHS=-2Abz^{\alpha}(1-z)^{\beta+1}\left[ik_{+}F(a,b,c,z)+\bar{k}F(a,b+1,c,z)\right]\,.$
(94)

For the RHS of (90) we use (B.1) first and then (93) to obtain

$\displaystyle RHS$
$\displaystyle=2Bz^{\alpha}(1-z)^{\beta+1}\left[ik_{+}a(1-z)F(a+1,b+1,c,z)+k(c-a-1)F(a,b+1,c,z)\right]$
(95)
$\displaystyle=2B(c-b-1)z^{\alpha}(1-z)^{\beta+1}\left[ik_{+}F(a,b,c,z)+\bar{k}F(a,b+1,c,z)\right]\,,$

which means $A$ and $B$ satisfy

$\dfrac{B}{A}=-\dfrac{c-b-1}{b}=\frac{2\beta-k}{2\beta+\bar{k}}\,.$ (96)

###  B.2 Rank-2 Symmetric Tensor

The boundary components of the first order equation (69) in this case are

$\begin{split}-mh_{--}&=-h_{+-}+2z(1-z)\left(\partial_{z}h_{+-}-\partial_{+}h_{z-}\right)\,,\\\
-mh_{++}&=h_{+-}+2(1-z)\left(\partial_{-}h_{z+}-\partial_{z}h_{z+}\right)\,,\\\
-mh_{+-}&=h_{--}+2z(1-z)\left(\partial_{z}h_{++}-\partial_{+}h_{z+}\right)\\\
&=-h_{++}+2(1-z)\left(\partial_{-}h_{z-}-\partial_{z}h_{--}\right)\,,\end{split}$
(97)

where the transversality condition for $h_{\mu\nu}$ was used to write down the
last equation. The $h_{z+}$ and $h_{z-}$ component must fulfill equations
similar to (89)

$\displaystyle
0=2z(1-z)\left(k_{-}\partial_{+}h_{z-}-k_{+}\partial_{-}h_{z-}\right)\,,$ (98)
$\displaystyle
0=2z(1-z)\left(k_{+}\partial_{-}h_{z+}-k_{-}\partial_{+}h_{z+}\right)\,.$

Using the relations (97) we can write this in terms of the following matrix
equation

$0=\left[\begin{pmatrix}k_{-}m&-k_{-}+k_{+}mz&-zk_{+}\\\
k_{-}&k_{+}z+k_{-}m&k_{+}mz\end{pmatrix}+2z(1-z)\begin{pmatrix}-k_{+}\partial_{z}&k_{-}\partial_{z}&0\\\
0&-k_{+}\partial_{z}&k_{-}\partial_{z}\end{pmatrix}\right]\begin{pmatrix}h_{--}\\\
h_{+-}\\\ h_{++}\end{pmatrix}\,.$ (99)

In order to write this in terms of the solutions of the hypergeometric
equation $h_{0},\,h_{-}\,h_{+}$ listed in (76) we need to invert the equation
(74), giving

$\begin{pmatrix}h_{--}\\\ h_{+-}\\\
h_{++}\end{pmatrix}=\begin{pmatrix}1&-1&-1\\\ 0&-i&i\\\ 1&1&1\\\
\end{pmatrix}\begin{pmatrix}h_{0}\\\ h_{-}\\\ h_{+}\end{pmatrix}\,.$ (100)

After using this, the easiest way to write down equations that only contain
two of the functions $h_{0},\,h_{-}\,h_{+}$ is to multiply the equation (99)
by the matrix

$c=\begin{pmatrix}1&-i\\\ 1&i\end{pmatrix}\,,$ (101)

whose result is

$0=\begin{pmatrix}a_{11}&a_{12}&0\\\
a_{21}&0&a_{23}\end{pmatrix}\begin{pmatrix}h_{0}\\\ h_{-}\\\
h_{+}\end{pmatrix}\,.$ (102)

Then, the second component of this equation involves $h_{+}$ and $h_{0}$ only,
and it can be written as

$2\left(-2\beta\bar{k}-2\beta
ik_{+}(1-z)-2\bar{k}z(1-z)\dfrac{d}{dz}\right)h_{+}=\left(-2\beta k+2\beta
ik_{+}(1-z)+2kz(1-z)\dfrac{d}{dz}\right)h_{0}\,,$ (103)

which is the same equation as in the vector case (90). Therefore, by (96)
$A_{0}$ and $A_{+}$ satisfy

$\dfrac{A_{0}}{2A_{+}}=\dfrac{2\beta-k}{2\beta+\bar{k}}=\dfrac{im-1-(k_{-}+ik_{+})}{im-1+(k_{-}-ik_{+})}\,.$
(104)

The first component of the matrix equation (102) can also be put in the form
of (90) and then

$\dfrac{2A_{-}}{A_{0}}=\dfrac{2(\beta+1)-k}{2(\beta+1)+\bar{k}}=\dfrac{im+1-(k_{-}+ik_{+})}{im+1+(k_{-}-ik_{+})}\,.$
(105)

###  B.3 Asymptotic Behaviour

We want to analyze the behavior of functions of the form

$R(z)=z^{\alpha}(1-z)^{\beta}F(a,b,c,z)\,,$ (106)

near $z\sim 1$, where the constants $\alpha,\,\beta,\,a,\,b$ and $c$ satisfy
the relations (55) and (57). First, let us suppose that $a+b-c=2\beta-1$ is
not an integer or zero888The same procedure can be done with help of the
functions $\textbf{F}(a,b,c,z)=F(a,b,c,z)/\Gamma(c)$ which are well-defined
solutions of the hypergeometric differential equation for all values of $a$,
$b$ and $c$. A more general connection formula can be written in terms of
$\textbf{F}(a,b,c,z)$. Such formula reduces to (107) when $a+b-c$ is not an
integer or zero [27]. , then we can use the connection formula

$\displaystyle F(a,b,c,z)=$
$\displaystyle\dfrac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}F(a,b,1+a+b-c,1-z)$
(107)
$\displaystyle+\dfrac{\Gamma(c)\Gamma(a+b-c)}{\Gamma(a)\Gamma(b)}(1-z)^{c-a-b}F(c-a,c-b,1+c-a-b,1-z)\,.$

Performing a change of variables $z=\tanh^{2}(\zeta)$ and using the connection
formula we obtain the following expression

$\displaystyle R(\zeta)=\tanh^{2}(\zeta)$
$\displaystyle\left[\dfrac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}\dfrac{1}{\cosh^{2\beta}(\zeta)}F(a,b,1+a+b-c,1/\cosh^{2}(\zeta))\right.$
(108)
$\displaystyle\left.+\dfrac{\Gamma(c)\Gamma(a+b-c)}{\Gamma(a)\Gamma(b)}\dfrac{1}{\cosh^{2(1-\beta)}(\zeta)}F(c-a,c-b,1+c-a-b,1/\cosh^{2}(\zeta))\right]\,,$

for which it is easier to evaluate its behavior when $\zeta\sim\infty$

$\displaystyle
R(\zeta)\sim\dfrac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}e^{-2\beta\zeta}\left(1+\mathcal{O}(e^{-2\zeta})\right)+\dfrac{\Gamma(c)\Gamma(a+b-c)}{\Gamma(a)\Gamma(b)}e^{2(\beta-1)\zeta}\left(1+\mathcal{O}(e^{-2\zeta})\right)\,.$
(109)

It is clear that if $\beta$ has a real part, one of the exponential is
divergent and it may cause problems in the norm of the tensors.

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$\displaystyle=\widetilde{\cal P}_{3}((0,2\kappa_{3}\delta
N),\mathop{}\mathopen{}\mathrm{d}t){-}\widetilde{\cal
P}_{5}((0,\kappa_{5}D_{N}(t{-})),\mathop{}\mathopen{}\mathrm{d}t),$

with $C_{N}(0){=}1$ and $D_{N}(0){=}0$, and where, for $i{\in}\\{0,3,4,5\\}$,
$\widetilde{\cal
P}_{i}(\mathop{}\mathopen{}\mathrm{d}s,\mathop{}\mathopen{}\mathrm{d}t)$ is
the Poisson process ${\cal P}_{i}$ shifted at $t_{1}^{3}$, i.e. ${\cal
P}_{i}(\mathop{}\mathopen{}\mathrm{d}s,t_{1}^{3}{+}\mathop{}\mathopen{}\mathrm{d}t)$.
A simple coupling can be constructed so that the relations
$D_{N}(t){\leq}Z_{4}^{N}(t_{1}^{3}{+}t)$ and
$Z_{3}^{N}(t){\leq}C_{N}(t_{1}^{3}{+}t){+}1$ hold for all
$0{\leq}t{\leq}\tau_{Z}^{N}$.

Standard arguments of stochastic calculus, give that, for the convergence in
distribution, the relation

$\lim_{N\to+\infty}\left(\frac{D_{N}(t/\sqrt{N})}{\sqrt{N}}\right)=(2\kappa_{3}\delta
t)$

holds and if $\tau_{D}^{N}$ is the hitting time of
$\lceil\kappa_{3}\delta\sqrt{N}\rceil$, then

$\limsup_{N\to+\infty}\sqrt{N}\mathbb{E}(\tau_{D}^{N})<{+}\infty,$

and, since $\left(C_{N}(\tau_{D}^{N}){-}1\right)^{+}$ is bounded by the number
of new arrivals for $(C_{N}(t))$ on the time interval $[0,\tau_{D}^{N})$

$\limsup_{N\to+\infty}\sqrt{N}\mathbb{E}\left(\left(C_{N}(\tau_{D}^{N}){-}1\right)^{+}\right)<{+}\infty.$

For $T{>}0$, starting from $\tau_{D}^{N}$, the process $(D_{N}(t))$ stays
above $\kappa_{3}\delta\sqrt{N}$ with high probability on a time interval
$[\tau_{D}^{N},\tau_{D}^{N}{+}T/\sqrt{N}]$. With the same argument as before,
we assume that this relation holds almost surely.

If $\tau_{C}^{N}$ is the hitting time of $0$ by $(C_{N}(\tau_{D}^{N}{+}t))$,
then $\tau_{Z}^{N}{\leq}\tau_{D}^{N}{+}\tau_{C}^{N}$. The integration of the
SDE for $(C_{N}(t))$ gives the relation

$\mathbb{E}\left(C_{N}(\tau_{D}^{N})\right){+}\kappa_{0}\mathbb{E}(\tau_{C}^{N}{\wedge}t){-}\kappa_{4}\kappa_{3}\delta\sqrt{N}\mathbb{E}\left(\int_{0}^{\tau_{C}^{N}{\wedge}t}(1{+}C_{N}(\tau_{D}^{N}{+}s))\mathop{}\mathopen{}\mathrm{d}s\right)\\\
=\mathbb{E}\left(C_{N}(\tau_{D}^{N}{+}\tau_{C}^{N}{\wedge}t)\right),$

hence, since $C_{N}(t){\geq}1$ on the time interval
$[\tau_{D}^{N},\tau_{D}^{N}{+}\tau_{C}^{N})$, we obtain

$\left(\kappa_{4}\kappa_{3}\delta\sqrt{N}{-}\kappa_{0}\right)\mathbb{E}\left(\tau_{C}^{N}{\wedge}t\right)\leq\mathbb{E}\left(C_{N}(\tau_{D}^{N})\right)\leq
1{+}\mathbb{E}\left(\left(C_{N}(\tau_{D}^{N}){-}1\right)^{+}\right),$

hence

$\limsup_{N\to+\infty}\sqrt{N}\mathbb{E}(\tau_{C}^{N})<{+}\infty.$

By gathering these results, and since, for $t{\geq}0$,

$Z_{4}^{N}(t_{1}^{3}{+}t/\sqrt{N})-Z_{4}^{N}(t_{1}^{3}){\leq}\widetilde{\cal
P}_{3}((0,2\kappa_{3}\delta N),[0,t/\sqrt{N}]),$

we have finally obtained that

$\limsup_{N\to+\infty}\sqrt{N}\mathbb{E}(\tau_{Z}^{N})<{+}\infty,\quad\text{and}\quad\limsup_{N\to+\infty}\frac{\mathbb{E}(Z_{4}^{N}(t_{1}^{3}+\tau_{Z}^{N})-Z_{4}^{N}(t_{1}^{3}))}{\sqrt{N}}<{+}\infty.$

This shows that, if $f$ is a bounded Borelian function on
$\mathbb{R}_{+}^{2}$, the sequences of random variables

$\left(\int_{t_{1}^{3}}^{t_{1}^{3}{+}\tau_{Z}^{N}}\mathbbm{1}_{\left\\{Z_{3}^{N}(s){\geq}2\right\\}}f\left(\frac{Z_{2}^{N}(s)}{N},\frac{Z_{4}^{N}(s)}{\sqrt{N}}\right)\mathop{}\mathopen{}\mathrm{d}s\right)\text{
and
}\left(\left|\frac{Z_{2}^{N}(t_{1}^{3}{+}\tau_{Z}^{N}){-}Z_{2}^{N}(t_{1}^{3})}{N}\right|\right)$

converge in distribution to $0$. We can now apply the convergence result,
Theorem 12, for the processes $(Z_{2}^{N}(t),Z_{4}^{N}(t))$ starting from time
$t_{1}^{3}{+}\tau_{Z}^{N}$, as if it was starting from time $t_{1}^{3}$ with
$Z_{3}^{N}(t_{1}^{3}){=}1$, since their first coordinate is of the same order
of magnitude in $N$, and that the limit result for the occupation measure does
not depend on the initial value of $Z_{4}^{N}(t_{1}^{3}{+}\tau_{Z}^{N})$ as
long as it is of the order of $N$ at most.

We conclude the proof of the proposition by induction on the successive jumps
$(t_{i}^{3})$ of ${\cal
P}_{0}((0,\kappa_{0}),\mathop{}\mathopen{}\mathrm{d}t)$ on the time interval
$[0,T]$. ∎

Note that the induction holds because the averaging result of Theorem 12 do
not need for $Z_{4}^{N}(t)$ to start from a state of the order of $\sqrt{N}$,
but only $O(N)$. This particular property is vital here, since we do not have
the control of the process $(Z_{4}^{N}(t)/\sqrt{N})$ over some time interval,
but only of its time integrals.

We have seen in the proof that the process $(Z_{3}^{N}(t))$ is identically
$1$, _as long as we consider time integrals_. The following Corollary is
easily deduced from Relations (37) and (53).

######  Corollary 21.

Let $F{\in}{\cal C}(\mathbb{R}_{+}^{2})$ with support on
$[0,T]{\times}\mathbb{R}_{+}$ such that there exist $c_{1}$, $c_{2}{>}0$ such
that for any $s,x\in\mathbb{R}_{+}$, $F(s,x)\leq c_{1}{+}c_{2}x$ holds for
$(s,x){\in}[0,T]{\times}\mathbb{R}_{+}$, then under the assumptions of Theorem
20, the sequence

$\left(\int_{0}^{+\infty}\mathbbm{1}_{\left\\{Z_{3}^{N}(s)=1\right\\}}F\left(s,\frac{Z_{4}^{N}(s)}{\sqrt{N}}\right)\mathop{}\mathopen{}\mathrm{d}s\right)$

converges in distribution to $\left\langle\mu_{Z}^{\infty},F\right\rangle$.

###  7.2. Time Evolution on the Timescale $(\sqrt{N}t)$

We are now going to express $(U_{N}(t))$ in terms of $(Z_{N}(t))$. Recall that
the process $(Z_{2}^{N}(t),Z_{4}^{N}(t))$ analyzed in the last section is just
the process $(U_{2}^{N}(t),U_{4}^{N}(t))$ with the time intervals during which
$(U_{3}^{N}(t))$ is $0$ removed. See Proposition 18.

We denote by $(Z_{N}(t))$ the solution of the SDE (52) starting from
$(y_{N},1,v_{N})$. We define

(54)
$\left(H_{N}(t)\right){=}\left(t{+}\int_{(0,t]\times\mathbb{R}_{+}{\times}\mathbb{R}_{+}^{\mathbb{N}}}a\mathbbm{1}_{\left\\{Z_{3}^{N}(s-){=}1\right\\}}{\cal
N}\left(\left(0,\kappa_{4}Z_{4}^{N}(s{-})\right],\mathop{}\mathopen{}\mathrm{d}s,\mathop{}\mathopen{}\mathrm{d}a,\mathop{}\mathopen{}\mathrm{d}b\right)\right),$

and the (potential) hitting time of $0$ for $(Z_{3}^{N}(t))$ is defined as
$\tau_{N}$, i.e.

$\tau_{N}=\inf\left\\{t:\int_{(0,t]\times\mathbb{R}_{+}{\times}\mathbb{R}_{+}^{\mathbb{N}}}\mathbbm{1}_{\left\\{Z_{3}^{N}(s-){=}1\right\\}}{\cal
N}((0,\kappa_{4}Z_{4}^{N}(s{-})],\mathop{}\mathopen{}\mathrm{d}s,\mathop{}\mathopen{}\mathrm{d}a,\mathop{}\mathopen{}\mathrm{d}b){\neq}0\right\\}.$

Strictly speaking, this is an incorrect presentation for $\tau_{N}$ since
$(Z_{3}^{N}(t))$ never visits $0$. This is in fact meant for $(U_{3}^{N}(t))$,
as long as the two processes coincide. Additionally,
$(a_{N},b_{N}){\in}\mathbb{R}_{+}{\times}\mathbb{R}_{+}^{\mathbb{N}}$ is the
mark associated to $\tau_{N}$, i.e.

${\cal
N}((0,\kappa_{4}Z_{4}^{N}(\tau_{N}{-})],\\{\tau_{N}\\},\mathop{}\mathopen{}\mathrm{d}a,\mathop{}\mathopen{}\mathrm{d}b)=\delta_{(a_{N},b_{N})},$

$a_{N}$ and $b_{N}{=}(b_{N,i})$ are independent and independent of ${\cal
F}_{\tau_{n}}$, with respective distributions, an exponential law with
parameter $\kappa_{0}$ and the law of an i.i.d. sequence of exponential random
variables with parameter $\kappa_{5}$. We define $t_{N}{=}\tau_{N}{+}a_{N}$.

We now construct a process
$(\widetilde{U}_{N}(t)){=}(\widetilde{U}_{2}^{N}(t),\widetilde{U}_{3}^{N}(t),\widetilde{U}_{4}^{N}(t))$
with initial state $(y_{N},1,v_{N})$ and

(55)
$\begin{cases}t{<}\tau_{N},&{\scriptstyle(\widetilde{U}_{2}^{N},\widetilde{U}_{3}^{N},\widetilde{U}_{4}^{N})(t)=(Z_{2}^{N},Z_{3}^{N},Z_{4}^{N})(t),}\\\
\ \\\
\tau_{N}{\leq}t{<}t_{N},&\begin{cases}{\scriptstyle(\widetilde{U}_{2}^{N},\widetilde{U}_{3}^{N})(t){=}(Z_{2}^{N}(\tau_{N}{-}),0),}\\\
{\scriptstyle\widetilde{U}_{4}^{N}(t){=}\sum_{i=1}^{Z_{4}^{N}(\tau_{N}{-})}\mathbbm{1}_{\left\\{b_{N,i}{>}t{-}\tau_{N}\right\\}}{=}Z_{4}^{N}(\tau_{N}{-}){-}S\left(Z_{4}^{N}(\tau_{N}{-}),t{-}\tau_{N},b_{N}\right)},\end{cases}\\\
\ \\\
t{=}t_{N},&{\scriptstyle(\widetilde{U}_{2}^{N},\widetilde{U}_{3}^{N},\widetilde{U}_{4}^{N})(t_{N}){=}\left(\widetilde{U}_{2}^{N}(t_{N}{-}),1,Z_{4}^{N}(\tau_{N}{-}){-}S\left(Z_{4}^{N}(\tau_{N}{-}),a_{N},b_{N}\right)\right)}.\end{cases}$

We have constructed the process $(\widetilde{U}_{N}(t))$ between two visits of
the third coordinate to $1$. We can construct by induction the process on the
whole real half-line.

######  Proposition 22.

For $N{\geq}1$, the processes $(U_{N}(t))$ and $(\widetilde{U}_{N}(t))$
defined respectively by Relations (51) and (55) have the same distribution.
Furthermore the relations

$\left(\int_{0}^{H_{N}(t)}\mathbbm{1}_{\left\\{\widetilde{U}_{3}^{N}(s)\geq
1\right\\}}\mathop{}\mathopen{}\mathrm{d}s\right)=(t)\text{ and
}\left(\widetilde{U}_{2}^{N}\left(H_{N}(t)\right)\right)=\left(Z_{2}^{N}\left(t\right)\right)$

hold.

We have in particular the identity
$(H_{N}(t)){\stackrel{{\scriptstyle\text{\rm dist.}}}{{=}}}(\ell_{1}^{N}(t))$.
See Definition 17.

###### Proof.

The proof of the identity in distribution is analogous to the proof of
Proposition 18. It relies again on strong Markov properties of Poisson
processes and the representation of the process $(A_{N}(t))$ used in this
proof.

The first relation comes directly from the construction of
$(\widetilde{U}_{N}(t))$. The last relation is a consequence of the fact that
$(\widetilde{U}_{2}^{N}(t))$ does not change on the times intervals where
$(\widetilde{U}^{N}_{3}(t))$ is null. ∎

######  Proposition 23.

If $(Z_{N}(t))$ is the solution of the SDEs (52) with the initial condition
$(y_{N},1,v_{N})$ and $(y_{N},v_{N})$ satisfying Relation (41) then, for the
convergence in distribution,

$\lim_{N\to+\infty}\left(\frac{H_{N}(t)}{\sqrt{N}}\right)=\left(t_{\infty}\left(1{-}e^{-\kappa_{3}t/2}\right)\right),$

with

(56) $t_{\infty}\stackrel{{\scriptstyle\text{\rm
def.}}}{{=}}\sqrt{y}\frac{\sqrt{2\kappa_{4}}}{\kappa_{5}\sqrt{\kappa_{3}}}\frac{\Gamma({\kappa_{0}}/{(2\kappa_{5})})}{\Gamma({\kappa_{0}}/{(2\kappa_{5})}{+}{1}/{2})}.$

###### Proof.

For $t{\geq}0$,

$\frac{H_{N}(t)}{\sqrt{N}}=\frac{t}{\sqrt{N}}{+}M_{N}(t){+}\frac{\kappa_{4}}{\kappa_{0}}\int_{0}^{t}\mathbbm{1}_{\left\\{Z_{3}^{N}(s-){=}1\right\\}}\frac{Z_{4}^{N}(s)}{\sqrt{N}}\mathop{}\mathopen{}\mathrm{d}s,$

where $(M_{N}(t))$ is a martingale whose previsible increasing process is

$\left(\left\langle
M_{N}\right\rangle(t)\right)=\left(2\frac{\kappa_{4}}{\kappa_{0}}\int_{0}^{t}\mathbbm{1}_{\left\\{Z_{3}^{N}(s-){=}1\right\\}}\frac{Z_{4}^{N}(s)}{N}\mathop{}\mathopen{}\mathrm{d}s\right).$

Theorem 20 and Doob’s Inequality give that $(M_{N}(t))$ is converging in
distribution to $0$ and, with Corollary 21, that $(H_{N}(t)/\sqrt{N})$
converges to

$\left(\frac{\kappa_{4}}{\kappa_{0}}\int_{0}^{t}\int_{0}^{+\infty}\sqrt{x}\Gamma\left(\frac{\kappa_{0}}{2\kappa_{5}}{+}\frac{1}{2},\frac{\kappa_{4}}{2\kappa_{3}z_{2}(s)}\right)(\mathop{}\mathopen{}\mathrm{d}x)\mathop{}\mathopen{}\mathrm{d}s\right)\\\
=\left(\frac{\kappa_{4}}{\kappa_{0}}\frac{\Gamma({\kappa_{0}}/{(2\kappa_{5})}{+}{1})}{\Gamma({(\kappa_{0})}/{(2\kappa_{5})}{+}{1}/{2})}\sqrt{2}\sqrt{\frac{\kappa_{3}}{\kappa_{4}}}\int_{0}^{t}\sqrt{z_{2}(s)}\mathop{}\mathopen{}\mathrm{d}s\right),$

by Relation (28), with $(z_{2}(t)){=}(y\exp({-}\kappa_{3}t))$. The proposition
is proved. ∎

######  Theorem 24.

If $(U_{N}(t))$ is the solution of the SDEs (51) with the initial condition
$(y_{N},1,v_{N})$ and $(y_{N},v_{N})$ satisfying Relation (41) then, for the
convergence in distribution,

$\lim_{N\to+\infty}\left(\frac{U_{2}^{N}\left(\sqrt{N}t\right)}{N},t{<}t_{\infty}\right)=(u_{2}(t))\stackrel{{\scriptstyle\text{\rm
def.}}}{{=}}\left(y\left(1{-}\frac{t}{t_{\infty}}\right)^{2},t{<}t_{\infty}\right)$

with $t_{\infty}$ is defined in Proposition 23.

The quadratic decay can be seen in the simulations of Figure 3 (A).

###### Proof.

For $0{\leq}t{<}t_{\infty}$, define

$s_{N}(t)=\inf\\{u{>}0:H_{N}(u)>\sqrt{N}t\\},$

Proposition 23 gives the convergence in distribution

(57)
$\lim_{N\to+\infty}\left(s_{N}(t)\right)=\left({-}\frac{2}{\kappa_{3}}\ln\left(1{-}\frac{t}{t_{\infty}}\right)\right).$

We have the identities

$\left(\frac{\widetilde{U}_{2}^{N}(\sqrt{N}t)}{N}\right)=\left(\frac{\widetilde{U}_{2}^{N}(H_{N}(s_{N}(t)))}{N}\right)=\left(\frac{Z_{2}^{N}((s_{N}(t)))}{N}\right),$

the first one is due to the fact that $\widetilde{U}_{2}^{N}$ does not change
just after a jump of $(H_{N}(t))$ and the second holds by Proposition 23.

We conclude the proof of the theorem with the convergence in distribution of
Theorem 20 and Theorem 3.3 of [KurtzEthier]. ∎

######  Definition 25.

The occupation measure $\mu_{U}^{N}$ of

$\left(\frac{U_{4}^{N}(\sqrt{N}t)}{\sqrt{N}}\right)\text{ is
}\left\langle\mu_{U}^{N},F\right\rangle\stackrel{{\scriptstyle\text{\rm
def.}}}{{=}}\int_{0}^{+\infty}F\left(s,\frac{U_{4}^{N}(s\sqrt{N})}{\sqrt{N}}\right)\mathop{}\mathopen{}\mathrm{d}s,$

for $F{\in}{\cal C}_{c}(\mathbb{R}_{+}^{2})$.

######  Theorem 26.

If $(U_{N}(t))$ is the solution of the SDEs (51) with the initial condition
$(y_{N},1,v_{N})$ and $(y_{N},v_{N})$ satisfying Relation (41), then the
sequence $(\mu_{U}^{N})$ converges in distribution to $\mu_{U}^{\infty}$
defined by, for any $F{\in}{\cal
C}_{c}([0,t_{\infty}){\times}\mathbb{R}_{+})$,

$\int
F(s,x)\mu_{U}^{\infty}(\mathop{}\mathopen{}\mathrm{d}s,\mathop{}\mathopen{}\mathrm{d}x)=\int
F(s,\sqrt{x})\Gamma_{0}\left(\frac{\kappa_{0}}{2\kappa_{5}},\frac{\kappa_{4}}{2\kappa_{3}y(1-t/t_{\infty})^{2}}\right)(\mathop{}\mathopen{}\mathrm{d}x)\mathop{}\mathopen{}\mathrm{d}s,$

where $t_{\infty}$ is defined by Relation (56) and $\Gamma_{0}$ is the
distribution of Definition 9.

By using Proposition 11 of Section 5.2, note that, for
$0{\leq}t{<}t_{\infty}$,

$\Gamma_{0}\left(\frac{\kappa_{0}}{2\kappa_{5}},\frac{\kappa_{4}}{2\kappa_{3}y(1-t/t_{\infty})^{2}}\right)$

is the invariant distribution of the infinitesimal generator ${\cal
A}_{u_{2}(t)}$ of Proposition 15, where $(u_{2}(t))$ is defined by Theorem 24.

###### Proof.

Let $f{\in}{\cal C}_{c}([0,t_{\infty}){\times}\mathbb{R}_{+})$, we have

$\left\langle\mu_{U}^{N},F\right\rangle=\frac{1}{\sqrt{N}}\int_{0}^{+\infty}f\left(\frac{s}{\sqrt{N}},\frac{U_{4}^{N}(s)}{\sqrt{N}}\right)\mathop{}\mathopen{}\mathrm{d}s.$

For $F{\in}{\cal C}_{c}(\mathbb{R}_{+})$ and $S{\geq}0$, by using the
definition of $(\widetilde{U}_{4}^{N}(s))$, the relation

(58)
$\frac{1}{\sqrt{N}}\int_{0}^{H_{N}(S)}F\left(\frac{\widetilde{U}_{4}^{N}(s)}{\sqrt{N}}\right)\mathop{}\mathopen{}\mathrm{d}s=\frac{1}{\sqrt{N}}\int_{0}^{S}F\left(\frac{Z_{4}^{N}(s)}{\sqrt{N}}\right)\mathop{}\mathopen{}\mathrm{d}s\\\
+\frac{1}{\sqrt{N}}\int_{0}^{S}\mathbbm{1}_{\left\\{Z_{3}^{N}(s{-})=1\right\\}}\int_{(a,b)\in\mathbb{R}_{+}\times\mathbb{R}_{+}^{\mathbb{N}}}\left(\int_{0}^{a}F\left(\frac{Z_{4}^{N}(s{-}){-}S(Z_{4}^{N}(s{-}),u,b)}{\sqrt{N}}\right)\mathop{}\mathopen{}\mathrm{d}u\right)\\\
\times{\cal
N}((0,\kappa_{4}Z_{4}^{N}(s{-})],\mathop{}\mathopen{}\mathrm{d}s,\mathop{}\mathopen{}\mathrm{d}a,\mathop{}\mathopen{}\mathrm{d}b),$

holds on the event $\\{H_{N}(S){<}\sqrt{N}t_{\infty}\\}$ whose probability is
converging to $1$ as $N$ gets large.

Tightness.
For $\varepsilon{>}0$ and $0{\leq}t_{0}{<}t_{\infty}$, with Proposition 23, we
get that there exists $S{>}0$ such that, for $N$ sufficiently large

$\mathbb{P}\left(\frac{H_{N}(S)}{\sqrt{N}}\not\in[t_{0},t_{\infty})\right)\leq\varepsilon,$

hence

$\mathbb{E}\left(\left\langle\mu_{U}^{N},[0,t_{0}]{\times}[K,{+}\infty]\right\rangle\right)=\frac{1}{\sqrt{N}}\int_{0}^{\sqrt{N}t_{0}}\mathbb{P}\left(U_{4}^{N}(s)\geq
K{\sqrt{N}}\right)\mathop{}\mathopen{}\mathrm{d}s\\\ \leq\varepsilon
t_{0}{+}\frac{1}{\sqrt{N}}\mathbb{E}\left(\int_{0}^{H_{N}(S)}\mathbbm{1}_{\left\\{{U_{4}^{N}(s)}\geq
K{\sqrt{N}}\right\\}}\mathop{}\mathopen{}\mathrm{d}s\right).$

With Relation (58), we obtain that

$\frac{1}{\sqrt{N}}\mathbb{E}\left(\int_{0}^{H_{N}(S)}\mathbbm{1}_{\left\\{{U_{4}^{N}(s)}\geq
K{\sqrt{N}}\right\\}}\mathop{}\mathopen{}\mathrm{d}s\right)\leq\frac{1}{\sqrt{N}}\int_{0}^{S}\mathbb{P}\left(\frac{Z_{4}^{N}(s)}{\sqrt{N}}{\geq}K\right)\mathop{}\mathopen{}\mathrm{d}s\\\
\qquad+\kappa_{4}\int_{0}^{S}\mathbb{E}\left(\mathbbm{1}_{\left\\{Z_{3}^{N}(s{-})=1\right\\}}\frac{Z_{4}^{N}(s)}{\sqrt{N}}\mathbbm{1}_{\left\\{{Z_{4}^{N}(s)}{\geq}K{\sqrt{N}}\right\\}}\right)\mathop{}\mathopen{}\mathrm{d}s\\\
\leq\frac{S}{\sqrt{N}}+\frac{1}{K}\int_{0}^{S}\mathbb{E}\left(\mathbbm{1}_{\left\\{Z_{3}^{N}(s{-})=1\right\\}}\left(\frac{Z_{4}^{N}(s)}{\sqrt{N}}\right)^{2}\right)\mathop{}\mathopen{}\mathrm{d}s$

holds.

Relations (37) and (53) show that

$\limsup_{N\to+\infty}\mathbb{E}\left(\int_{0}^{S}\mathbbm{1}_{\left\\{Z_{3}^{N}(s{-})=1\right\\}}\left(\frac{Z_{4}^{N}(s)}{\sqrt{N}}\right)^{2}\mathop{}\mathopen{}\mathrm{d}s\right)<+\infty.$

Therefore, one can choose $K$ sufficiently large so that the quantity

$\mathbb{E}\left(\left\langle\mu_{U}^{N},[0,t_{0}]{\times}[K,{+}\infty]\right\rangle\right)$

is arbitrarily small for $N$ sufficiently large. Lemma 1.3 of [Kurtz1992]
gives the tightness of the sequence $(\mu_{U}^{N})$ of random measures on
$[0,t_{\infty}]{\times}\mathbb{R}_{+}$.

Identification of the Limit.

With the tightness property and since any limiting point can be represented as
in Relation (14), it is enough to identify the limit of

$\frac{1}{\sqrt{N}}\int_{0}^{\sqrt{N}t}F\left(\frac{\widetilde{U}_{4}^{N}(s)}{\sqrt{N}}\right)\mathop{}\mathopen{}\mathrm{d}s,$

for any $t{\in}[0,t_{\infty})$. Since, by Proposition 23, the process
$(H_{N}(t)/\sqrt{N})$ converges in distribution to a deterministic function,
one has to obtain the limit of the sequence

$\left(\frac{1}{\sqrt{N}}\int_{0}^{H_{N}(S)}F\left(\frac{\widetilde{U}_{4}^{N}(s)}{\sqrt{N}}\right)\mathop{}\mathopen{}\mathrm{d}s\right).$

The first term of the right-hand side of Relation (58) converges clearly in
distribution to $0$. The second term can be written as $J_{N}(S){+}M_{N}(S)$,
where, for $t{\geq}0$,

$J_{N}(t){=}\kappa_{4}\int_{0}^{t}\mathbbm{1}_{\left\\{Z_{3}^{N}(s=1\right\\}}\\\
\times\mathbb{E}\left.\left(\int_{0}^{E_{\kappa_{0}}}F\left(\frac{Z_{4}^{N}(s)-S(Z_{4}^{N}(s),u,(E_{\kappa_{5},i}))}{\sqrt{N}}\right)\mathop{}\mathopen{}\mathrm{d}u\right|{\cal
F}_{s}\right)\frac{Z_{4}^{N}(s)}{\sqrt{N}}\mathop{}\mathopen{}\mathrm{d}s.$

It is easily checked that $(M_{N}(t))$ is a martingale whose previsible
increasing process is given by, for $t{\geq}0$,

$\left\langle
M\right\rangle_{N}(t)=\frac{\kappa_{4}}{\sqrt{N}}\int_{0}^{t}\mathbbm{1}_{\left\\{Z_{3}^{N}(s)=1\right\\}}\\\
{\times}\left(\mathbb{E}\left.\left(\int_{0}^{E_{\kappa_{0}}}F\left(\frac{Z_{4}^{N}(s)-S(Z_{4}^{N}(s),u,(E_{\kappa_{5},i}))}{\sqrt{N}}\right)^{2}\mathop{}\mathopen{}\mathrm{d}u\right|{\cal
F}_{s}\right)\right)\frac{Z_{4}^{N}(s)}{\sqrt{N}}\mathop{}\mathopen{}\mathrm{d}s.$

Using Corollary (21) one can show that there exists some finite constant
$C_{0}$ such that

$\mathbb{E}\left(\left\langle
M\right\rangle_{N}(t)\right)\leq\frac{C_{0}}{\sqrt{N}},$

the martingale $(M_{N}(t))$ is converging in distribution to $0$.

We are now investigating the asymptotic behavior of $(J_{N}(t))$.

By using again Corollary 21, for $T{>}0$ and $\varepsilon{>}0$, there exist
constants $0{<}d_{0}{\leq}D_{0}$ such that

$\limsup_{N\to+\infty}\mathbb{E}\left(\int_{0}^{T}\frac{Z_{4}^{N}(s)}{\sqrt{N}}\mathbbm{1}_{\left\\{Z_{4}^{N}(s){\not\in}[d_{0}\sqrt{N},D_{0}\sqrt{N}]\right\\}}\mathop{}\mathopen{}\mathrm{d}s\right)\leq\varepsilon,$

this is due to the fact that the limit of the occupation measure of
$(Z_{4}^{N}(s)/\sqrt{N})$ is expressed with a distribution $\Gamma_{0}$ of
Definition 9 and, in particular, without a mass at $0$.

Let $z{\geq}1$, then

(59)
$\mathbb{E}\left(\int_{0}^{E_{\kappa_{0}}}F\left(\frac{z-S(z,u,(E_{\kappa_{5},i}))}{\sqrt{N}}\right)\mathop{}\mathopen{}\mathrm{d}u\right)\\\
=\int_{0}^{+\infty}\mathbb{E}\left(F\left(\frac{z}{\sqrt{N}}\frac{1}{z}\sum_{i=1}^{z}\mathbbm{1}_{\left\\{E_{\kappa_{5},i}{>}u\right\\}}\right)\right)\mathbb{P}(E_{\kappa_{0}}{\geq}u)\mathop{}\mathopen{}\mathrm{d}u,$

for $\eta{>}0$ and $u{\geq}0$, the relation

$\mathbb{P}\left(\left|\frac{1}{z}\sum_{i=1}^{z}\mathbbm{1}_{\left\\{E_{\kappa_{5},i}{>}u\right\\}}{-}e^{-\kappa_{5}u}\right|\geq\eta\right)\leq\frac{1}{z\eta^{2}}$

holds. With the uniform continuity of $F$, we therefore obtain the relation

$\lim_{N\to+\infty}\sup_{z{\in}[d_{0}\sqrt{N},D_{0}\sqrt{N}]}\mathbb{E}\left(\left|F\left(\frac{z}{\sqrt{N}}\frac{1}{z}\sum_{i=1}^{z}\mathbbm{1}_{\left\\{E_{\kappa_{5},i}{>}u\right\\}}\right){-}F\left(\frac{z}{\sqrt{N}}e^{-\kappa_{5}u}\right)\right|\right)=0.$

We have therefore that, for the convergence in distribution, the sequence
$(J_{N}(t))$ has the same asymptotic behavior as

$\left(\kappa_{4}\int_{0}^{t}\mathbbm{1}_{\left\\{Z_{3}^{N}(s{-})=1\right\\}}\int_{0}^{+\infty}F\left(\frac{Z_{4}^{N}(s)}{\sqrt{N}}e^{-\kappa_{5}u}\right)e^{-\kappa_{0}u}\mathop{}\mathopen{}\mathrm{d}u\frac{Z_{4}^{N}(s)}{\sqrt{N}}\mathop{}\mathopen{}\mathrm{d}s\right).$

We can now use Corollary (21) and standard calculus to complete the proof of
the theorem. ∎

We now establish the fact that the process $(X_{2}^{N}(t))$ has indeed the
same asymptotic behavior as $(U_{2}^{N}(t))$.

######  Proposition 27.

Let $(X_{N}(t))$ and $(U_{N}(t))$ be the solutions of the SDEs (40) and (51)
with initial point $(0,y_{N},1,v_{N})$ and $(y_{N},v_{N})$ satisfies Relation
(41), then the two sequences

$\left(\frac{X_{2}^{N}(\sqrt{N}t)}{N}\right)\text{ and
}\left(\frac{U_{2}^{N}(\sqrt{N}t)}{N}\right)$

have the same limit for the convergence in distribution.

###### Proof.

The proof follows the arguments used for the convergence in distribution of
$(U_{2}^{N}(\sqrt{N}t)/N)$. Some adjustments are nevertheless necessary but
the main ideas are essentially the same. We sketch the main lines of the
proof.

If $(X_{N}(t))$ is a solution of SDEs (40), on the time interval
$[0,\sqrt{N}T]$ the contribution of the Poisson process ${\cal P}_{1}$ to the
coordinate $(X_{2}^{N}(t))$ is of the order of $\sqrt{N}$ which is negligible
since the order of magnitude considered for $(X_{2}^{N}(t))$ is $N$.
Therefore, we can take this Poisson process out of the set of SDEs for
$(X_{N}(t))$.

For $T{>}0$, it is not difficult to show that, with high probability, the
values of the process $(X_{2}(\sqrt{N}t)/N)$ are in $(\delta,3/2)$ on the time
interval $[0,T]$. When there is a new arrival for the chemical species
$S_{1}$, it is transformed into chemical species $S_{3}$ at rate at least
$\kappa_{2}\delta N$. Hence, with high probability, on the time interval
$[0,\sqrt{N}T]$, the values of the process $(X_{1}^{N}(t))$ are in the set
$\\{0,1\\}$.

A central argument for the convergence of $(U_{2}^{N}(\sqrt{N}t)/N)$ is
Theorem 20. We define by $(\widetilde{Z}_{N}(t))$ the analogue of
$(Z_{N}(t))$, i.e. the time-changed process $(X_{N}(t))$ with all time
intervals where $X_{3}^{N}$ is null are removed. The process
$(\widetilde{Z}_{N}(t))$ satisfies the analogue of the SDEs (52) where the
last SDE is replaced by

$\mathop{}\mathopen{}\mathrm{d}\widetilde{Z}_{4}^{N}(t)={\cal
P}_{3}\left((0,\kappa_{3}\widetilde{Z}_{2}^{N}\widetilde{Z}_{3}^{N}(t{-})),\mathop{}\mathopen{}\mathrm{d}t\right){-}{\cal
P}_{5}\left((0,\kappa_{5}\widetilde{Z}_{4}^{N}(t{-})),\mathop{}\mathopen{}\mathrm{d}t\right),\\\
\qquad{-}\mathbbm{1}_{\left\\{\widetilde{Z}_{3}^{N}(t{-}){=}1\right\\}}\displaystyle\int_{a,b}S\left(\widetilde{Z}_{4}^{N}(t{-}),a{+}\frac{c}{\kappa_{2}\widetilde{Z}_{2}^{N}(t{-})},b\right)\\\
\times\widetilde{\cal
N}\left(\left(\rule{0.0pt}{11.38109pt}0,\kappa_{4}Z_{4}^{N}(t{-})\right],\mathop{}\mathopen{}\mathrm{d}t,\mathop{}\mathopen{}\mathrm{d}a,\mathop{}\mathopen{}\mathrm{d}b,\mathop{}\mathopen{}\mathrm{d}c\right),$

where $\widetilde{\cal N}$ be a Poisson marked point process on
$\mathbb{R}_{+}^{2}{\times}\mathbb{R}_{+}{\times}\mathbb{R}_{+}^{\mathbb{N}}{\times}\mathbb{R}_{+}{\times}$
with intensity measure

$\mathop{}\mathopen{}\mathrm{d}s\otimes\mathop{}\mathopen{}\mathrm{d}t\otimes\kappa_{0}e^{{-}\kappa_{0}a}\mathop{}\mathopen{}\mathrm{d}a\otimes
Q(\mathop{}\mathopen{}\mathrm{d}b)\otimes
e^{{-}c}\mathop{}\mathopen{}\mathrm{d}c,$

where, as before, $Q$ is the distribution of $(E_{i})$ on
$\mathbb{R}_{+}^{\mathbb{N}}$. The additional variable $c$ of the Poisson
process $\widetilde{\cal N}$ in this SDE is due to the fact that when
$\widetilde{Z}_{1}^{N}(t){=}1$ and $\widetilde{Z}_{3}^{N}(t){=}0$ ,
$\widetilde{Z}_{1}$ leaves the state $1$ at rate
$\kappa_{2}\widetilde{Z}_{2}^{N}(t{-})$. Because of the assumption on
$(\widetilde{Z}_{2}^{N}(t{-}/N)$ in $(\delta,3/2)$, a glance at the proof of
Theorem 12 shows that, even with this extra term, the limit result of this
theorem still holds with the same limits. The proof is then concluded as in
the proof of Proposition 23. ∎