# Quantum Theory of Weyl Invariant Scalar-tensor Gravity

Ichiro Oda 111Electronic address<EMAIL_ADDRESS>
Department of Physics, Faculty of Science, University of the Ryukyus,
Nishihara, Okinawa 903-0213, Japan

###### Abstract

We perform a manifestly covariant quantization of a Weyl invariant, i.e., a
locally scale invariant, scalar-tensor gravity in the extended de Donder gauge
condition (or harmonic gauge condition) for general coordinate invariance and
a new scalar gauge for Weyl invariance within the framework of BRST formalism.
It is shown that choral symmetry, which is a Poincar${\rm{\acute{e}}}$-like
${I\kern-0.59998ptO\kern-1.1ptS\kern-0.39993ptp}(8|8)$ supersymmetry in case
of Einstein gravity, is extended to a Poincar${\rm{\acute{e}}}$-like
${I\kern-0.59998ptO\kern-1.1ptS\kern-0.39993ptp}(10|10)$ supersymmetry. We
point out that there is a gravitational conformal symmetry in quantum gravity
and account for how conventional conformal symmetry in a flat Minkowski space-
time is related to the gravitational conformal symmetry. Moreover, we examine
the mechanism of spontaneous symmetry breakdown of the choral symmetry, and
show that the gravitational conformal symmetry is spontaneously broken to the
Poincaré symmetry and the corresponding massless Nambu-Goldstone bosons are
the graviton and the dilaton. We also prove the unitarity of the physical
S-matrix on the basis of the BRST quartet mechanism.

##  1 Introduction

There is no question that symmetry plays the central role in both elementary-
particle physics and quantum gravity. For instance, in the Yang-Mills theory
it has been found that we have the non-abelian gauge symmetry and that this
symmetry gives rise to physically significant effects, such as the asymptotic
freedom and the quark confinement.

It is well known that there are two kinds of symmetries in nature: global
symmetry and gauge symmetry. In order to understand the nature more deeply, it
is necessary to understand the meaning of the both symmetries. The meaning of
the global symmetry is clear in the sense that it operates physical
observables in a direct manner and shows the real symmetry of a physical
system. On the other hand, the meaning of the gauge symmetry is more elusive
than that of the global symmetry since it does not operate on physical
observables directly. To treat the gauge symmetry properly in quantum field
theory, it is essential to fix the gauge symmetry by a suitable gauge
condition and consequently physical observables are defined as BRST invariant
operators. Thus, it is sometimes said that the gauge symmetry is redundancy in
the mathematical description of a physical system rather than the property of
the system itself.

Another important property of the symmetries is that many of the global
symmetry are not exact but only approximate whereas the gauge symmetry is an
exact one. For instance, there is a clear prediction of violation of baryon
and lepton numbers by a quantum anomaly in the standard model. This fact is
also supported by a theory of quantum gravity. In particular, when a black
hole evaporates at the quantum level, the baryon and lepton numbers are not
conserved whereas gauge quantum numbers such as electric and magnetic charges
are precisely conserved since they are measured by the flux integrals at
infinity.

Thus, if a certain global symmetry plays a critical role in physics, it must
be promoted to the gauge symmetry. This statement particularly holds in
constructing theories involving quantum gravity. In our previous work [1], we
have presented a quantum theory of a globally scale invariant gravity with a
real scalar field, which is equivalent to the well-known Brans-Dicke gravity
[2], by constructing its manifestly covariant BRST formalism. Since many of
studies of the Brans-Dicke gravity have been so far limited to a classical
analysis, our theory has provided us with some useful information on quantum
aspects of the Brans-Dicke gravity. Indeed, based on this quantum gravity we
have elucidated a mechanism of how a scale invariance is spontaneously broken
and consequently exactly massless “dilaton” emerges thanks to the Nambu-
Goldstone theorem in quantum gravity [1, 3]. Then, it is natural to generalize
our formulation to a case of a locally scale invariant, or equivalently Weyl
invariant, scalar-tensor gravity and ask if we can get some useful knowledge
about quantum aspects of the theory.

In this article, we perform a manifestly covariant BRST quantization of a Weyl
invariant scalar-tensor gravity with a real scalar field in addition to the
metric tensor field, investigate the remaining global symmetries and their
spontaneous symmetry breakdown, prove the unitarity of the S-matrix, and then
elucidate that there exists a gravitational analog of conformal symmetry in
our theory. Long ago, in a pioneering work by Nakanishi [4, 5], on the basis
of the Einstein-Hilbert action in the de Donder gauge (harmonic gauge) for
general coordinate transformation (GCT), it has been shown that there remains
a huge residual symmetry, which is a Poincaré-like $ISOp(8|8)$ supersymmetry,
called “choral symmetry”, including the BRST symmetry and $GL(4)$ symmetry
etc. In our present formulation, adopting the extended de Donder gauge
condition for the GCT and a new scalar gauge condition for the Weyl
transformation, the choral symmetry is extended to a Poincaré-like
$ISOp(10|10)$ supersymmetry, which includes the scale symmetry and the
gravitational special conformal symmetry. It is of interest that as in a flat
Minkowski space-time, both the scale symmetry and the special conformal
symmetry are spontaneously broken, and the dilation is not only a Nambu-
Goldstone boson for the scale symmetry but also its derivative provides a
Nambu-Goldstone boson for the special conformal transformation.

The paper is organized as follows. In Section 2, we discuss a general
gravitational theory for which there are two local symmetries, those are, the
general coordinate invariance and the Weyl symmetry. It is pointed out that in
such a theory, we must choose a gauge fixing condition for GCT carefully in
such a way that it does not violate the Weyl symmetry, and similarly a gauge
fixing condition for the Weyl transformation should be selected in order not
to break the GCT. In Section 3, beginning with a Weyl invariant scalar-tensor
gravity [6], we fix the GCT and the Weyl transformation by the extended de
Donder gauge and the new scalar gauge conditions, and construct a gauge-fixed,
BRST-invariant quantum Lagrangian. In Section 4, we calculate various equal-
time (anti-)commutation relations (ETCRs) among the fundamental fields, in
particular, the Nakanishi-Lautrup auxiliary field, the Faddeev-Popov (FP)
ghosts. In Section 5, we derive the ETCRs involving the gravitational field.
In Section 6, we prove the unitarity of the physical S-matrix by means of the
BRST quartet mechanism. In Section 7, we show that there is a choral symmetry,
which is an $IOSp(10|10)$ supersymmetry, in our theory. In Section 8, we point
out the existence of a gravitational conformal symmetry even in quantum
gravity, and we investigate its spontaneous symmetry breaking in Section 9.
The final section is devoted to discussion.

Two appendices are put for technical details. In Appendix A, a derivation of
the equation for the $b_{\rho}$ field is given, and in Appendix B we have
accounted for the relationship between the gravitational conformal symmetry
and conventional conformal symmetry.

##  2 Consistency between two BRST symmetries

We wish to perform a manifestly covariant BRST quantization of a gravitational
theory which is invariant under both general coordinate transformation (GCT)
and Weyl transformation, or equivalently local scale transformation. To take a
more general theory into consideration, without specifying the concrete
expression of the gravitational Lagrangian density, we will start with the
following classical Lagrangian density:222We follow the notation and
conventions of MTW textbook [7]. Greek little letters $\mu,\nu,\cdots$ and
Latin ones $i,j,\cdots$ are used for space-time and spatial indices,
respectively; for instance, $\mu=0,1,2,3$ and $i=1,2,3$. The Riemann curvature
tensor and the Ricci tensor are respectively defined by
$R^{\rho}{}_{\sigma\mu\nu}=\partial_{\mu}\Gamma^{\rho}_{\sigma\nu}-\partial_{\nu}\Gamma^{\rho}_{\sigma\mu}+\Gamma^{\rho}_{\lambda\mu}\Gamma^{\lambda}_{\sigma\nu}-\Gamma^{\rho}_{\lambda\nu}\Gamma^{\lambda}_{\sigma\mu}$
and $R_{\mu\nu}=R^{\rho}{}_{\mu\rho\nu}$. The Minkowski metric tensor is
denoted by $\eta_{\mu\nu}$; $\eta_{00}=-\eta_{11}=-\eta_{22}=-\eta_{33}=-1$
and $\eta_{\mu\nu}=0$ for $\mu\neq\nu$.

$\displaystyle{\cal L}_{c}={\cal L}_{c}(g_{\mu\nu},\phi),$ (2.1)

which includes the metric tensor field $g_{\mu\nu}$ and a scalar field $\phi$
as dynamical variables.333It is straightforward to add the other fields such
as gauge fields and spinors. We assume that ${\cal{L}}_{c}$ does not involve
more than first order derivatives of the metric and matter fields.

We have a physical situation in mind that we fix the general coordinate
symmetry and the Weyl symmetry by suitable gauge conditions. It is a familiar
fact that after introducing the gauge conditions, instead of such the two
local gauge symmetries, we are left with two kinds of global symmetries named
as the BRST symmetries. The BRST transformation, which is denoted as
$\delta_{B}$, corresponding to the GCT is defined as

$\displaystyle\delta_{B}g_{\mu\nu}$ $\displaystyle=$
$\displaystyle-(\nabla_{\mu}c_{\nu}+\nabla_{\nu}c_{\mu})$ $\displaystyle=$
$\displaystyle-(c^{\alpha}\partial_{\alpha}g_{\mu\nu}+\partial_{\mu}c^{\alpha}g_{\alpha\nu}+\partial_{\nu}c^{\alpha}g_{\mu\alpha}),$
$\displaystyle\delta_{B}\tilde{g}^{\mu\nu}$ $\displaystyle=$ $\displaystyle
h(\nabla^{\mu}c^{\nu}+\nabla^{\nu}c^{\mu}-g^{\mu\nu}\nabla_{\rho}c^{\rho}),$
$\displaystyle\delta_{B}\phi$ $\displaystyle=$
$\displaystyle-c^{\lambda}\partial_{\lambda}\phi,\quad\delta_{B}c^{\rho}=-c^{\lambda}\partial_{\lambda}c^{\rho},$
$\displaystyle\delta_{B}\bar{c}_{\rho}$ $\displaystyle=$ $\displaystyle
iB_{\rho},\quad\delta_{B}B_{\rho}=0,$ (2.2)

where $c^{\rho}$ and $\bar{c}_{\rho}$ are respectively the Faddeev-Popov (FP)
ghost and anti-ghost, $B_{\rho}$ is the Nakanishi-Lautrup (NL) field, and we
have defined $\tilde{g}^{\mu\nu}\equiv\sqrt{-g}g^{\mu\nu}\equiv hg^{\mu\nu}$.
For later convenience, in place of the NL field $B_{\rho}$ we will introduce a
new NL field defined as

$\displaystyle
b_{\rho}=B_{\rho}-ic^{\lambda}\partial_{\lambda}\bar{c}_{\rho},$ (2.3)

and its BRST transformation reads

$\displaystyle\delta_{B}b_{\rho}=-c^{\lambda}\partial_{\lambda}b_{\rho}.$
(2.4)

The other BRST transformation, which is denoted as $\bar{\delta}_{B}$,
corresponding to the Weyl transformation is defined as

$\displaystyle\bar{\delta}_{B}g_{\mu\nu}$ $\displaystyle=$ $\displaystyle
2cg_{\mu\nu},\quad\bar{\delta}_{B}\tilde{g}^{\mu\nu}=2c\tilde{g}_{\mu\nu},$
$\displaystyle\bar{\delta}_{B}\phi$ $\displaystyle=$
$\displaystyle-c\phi,\quad\bar{\delta}_{B}\bar{c}=iB,\quad\bar{\delta}_{B}c=\bar{\delta}_{B}B=0,$
(2.5)

where $c$ and $\bar{c}$ are respectively the FP ghost and FP anti-ghost, $B$
is the NL field. Note that the two BRST transformations are nilpotent, i.e.,

$\displaystyle\delta_{B}^{2}=\bar{\delta}_{B}^{2}=0.$ (2.6)

To complete the two BRST transformations, we have to fix not only the GCT BRST
transformation $\delta_{B}$ on $c,\bar{c}$ and $B$ but also the Weyl BRST
transformation $\delta_{B}$ on $c^{\rho},\bar{c}_{\rho}$ and $b_{\rho}$. It is
easy to determine the former BRST transformation since the fields $c,\bar{c}$
and $B$ are all scalar fields so their BRST transformations should take the
form:

$\displaystyle\delta_{B}B=-c^{\lambda}\partial_{\lambda}B,\quad\delta_{B}c=-c^{\lambda}\partial_{\lambda}c,\quad\delta_{B}\bar{c}=-c^{\lambda}\partial_{\lambda}\bar{c}.$
(2.7)

On the other hand, there is an ambiguity in fixing the latter BRST
transformation, but we would like to propose a recipe for achieving this goal.
The recipe is to just assume that the two BRST transformations are anti-
commute with each other, that is,

$\displaystyle\\{\delta_{B},\bar{\delta}_{B}\\}\equiv\delta_{B}\bar{\delta}_{B}+\bar{\delta}_{B}\delta_{B}=0,$
(2.8)

which requires us to take

$\displaystyle\bar{\delta}_{B}b_{\rho}=\bar{\delta}_{B}c^{\rho}=\bar{\delta}_{B}\bar{c}_{\rho}=0.$
(2.9)

Now we would like to explain an important point, which is sometimes missed in
the theoretical physics literature, when two BRST transformations coexist in a
theory. Suppose that we fix the GCT by a gauge condition
$F^{\alpha}(g_{\mu\nu},\phi)=0$ and the Weyl transformation by a gauge
conditon $F(g_{\mu\nu},\phi)=0$. Then, the gauge-fixed and BRST invariant
Lagrangian density is given by

$\displaystyle{\cal L}_{q}={\cal
L}_{c}+\delta_{B}(\bar{c}_{\alpha}F^{\alpha})+\bar{\delta}_{B}(\bar{c}F),$
(2.10)

where the first term is the classical Lagrangian density (2.1). Under this
situation, a natural question arises about the gauge-fixing conditions: Can we
take any gauge-fixing conditions if they fix gauge symmetries anyway? If not,
what gauge conditions are suitable for $F^{\alpha}$ and $F$?

In order to answer the questions, let us take the two BRST transformations
separately and check whether the quantum Lagrangian density (2.10) is really
invariant under the BRST transformations up to surface terms. First, taking
the Weyl BRST transformation leads to

$\displaystyle\bar{\delta}_{B}{\cal{L}}_{q}$ $\displaystyle=$
$\displaystyle\bar{\delta}_{B}\delta_{B}(\bar{c}_{\alpha}F^{\alpha})=-\delta_{B}\bar{\delta}_{B}(\bar{c}_{\alpha}F^{\alpha})$
(2.11) $\displaystyle=$
$\displaystyle-\delta_{B}\left[(\bar{\delta}_{B}\bar{c}_{\alpha})F^{\alpha}-\bar{c}_{\alpha}\bar{\delta}_{B}F^{\alpha}\right],$

where we have used $\bar{\delta}_{B}{\cal{L}}_{c}=0$, and Eqs. (2.6) and
(2.8). This equation clearly shows that the conditions

$\displaystyle\bar{\delta}_{B}\bar{c}_{\alpha}=0,\quad\bar{\delta}_{B}F^{\alpha}=0,$
(2.12)

are sufficient conditions such that the Lagrangian density (2.10) is invariant
under the Weyl BRST transformation.

It is of interest to notice that the former condition in (2.12) leads to two
remaining equations in (2.9). To see this fact, let us take the GCT BRST
transformation of the former equation as follows:

$\displaystyle 0$ $\displaystyle=$
$\displaystyle\delta_{B}\bar{\delta}_{B}\bar{c}_{\alpha}=-\bar{\delta}_{B}\delta_{B}\bar{c}_{\alpha}=-i\bar{\delta}_{B}B_{\alpha}$
(2.13) $\displaystyle=$
$\displaystyle-i\left[\bar{\delta}_{B}b_{\alpha}+i(\bar{\delta}_{B}c^{\lambda})\partial_{\lambda}\bar{c}_{\alpha}\right],$

which implies $\bar{\delta}_{B}b_{\alpha}=\bar{\delta}_{B}c^{\lambda}=0$,
which coincide with the remaining two equations in (2.9).

On the other hand, the latter condition in (2.12) gives rise to important
information on the gauge condition for the GCT: The gauge-fixing condition for
the GCT must be invariant under the Weyl transformation. Thus, for instance,
the conventional de Donder gauge condition (or harmonic gauge condition)

$\displaystyle\partial_{\mu}\tilde{g}^{\mu\nu}=0,$ (2.14)

is not suitable when there is the Weyl invariance.444In two space-time
dimensions the de Donder condition is Weyl invariant so it is allowed to use
it as the gauge-fixing condition for the GCT.

Next, let us operate the GCT BRST transformation on ${\cal L}_{q}$. To do
that, since the Lagrangian density is in general a quantity with density, it
is more convenient to write as ${\cal L}_{q}\equiv\sqrt{-g}{\cal
L}_{q}^{\prime}$ and $F=\sqrt{-g}F^{\prime}$ where $F$ and $F^{\prime}$ are
scalars. Then, taking the GCT BRST variation leads to

$\displaystyle\delta_{B}{\cal L}_{q}$ $\displaystyle=$
$\displaystyle\delta_{B}(\sqrt{-g}{\cal
L}_{q}^{\prime})=\delta_{B}\bar{\delta}_{B}(\sqrt{-g}\bar{c}F^{\prime})=-\bar{\delta}_{B}\delta_{B}(\sqrt{-g}\bar{c}F^{\prime})$
(2.15) $\displaystyle=$
$\displaystyle-\bar{\delta}_{B}\left[-\sqrt{-g}\nabla_{\rho}c^{\rho}\bar{c}F^{\prime}+\sqrt{-g}(-c^{\rho}\partial_{\rho}\bar{c})F^{\prime}-\sqrt{-g}\bar{c}(-c^{\rho}\partial_{\rho}F^{\prime})\right]$
$\displaystyle=$
$\displaystyle\partial_{\rho}\bar{\delta}_{B}(c^{\rho}\bar{c}F),$

which means that ${\cal L}_{q}$ is indeed invariant under the GCT BRST
transformation up to a surface term. In obtaining this result, we have assumed

$\displaystyle\delta_{B}F^{\prime}=-c^{\rho}\partial_{\rho}F^{\prime},$ (2.16)

which is nothing but the requirement that the quantity $F^{\prime}$ should be
a scalar under the GCT. Thus, only a scalar function $F^{\prime}$, or
equivalently a scalar density $F$, makes sense as a gauge-fixing condition for
the Weyl invariance. Of course, this scalar function must break the Weyl
invariance. As suitable gauge-fixing conditions, in this paper we will choose
$F^{\nu}=\partial_{\mu}(\tilde{g}^{\mu\nu}\phi^{2})$ and
$F=\partial_{\mu}(\tilde{g}^{\mu\nu}\phi\partial_{\nu}\phi)$.

##  3 Quantum Weyl invariant scalar-tensor gravity

In this section, as a classical Lagrangian555For simplicity, we henceforth
call a Lagrangian density a Lagrangian., we will take a Weyl invariant scalar-
tensor gravity whose Lagrangian is of form [6]

$\displaystyle{\cal
L}_{c}=\sqrt{-g}\left(\frac{1}{12}\phi^{2}R+\frac{1}{2}g^{\mu\nu}\partial_{\mu}\phi\partial_{\nu}\phi\right),$
(3.1)

where $\phi$ is a real scalar field with a ghost-like kinetic term, and $R$
the scalar curvature. In addition to the invariance under the general
coordinate transformation (GCT), this Lagrangian is also invariant under the
Weyl transformation (or the local scale transformation) defined as

$\displaystyle g_{\mu\nu}\rightarrow
g^{\prime}_{\mu\nu}=\Omega^{2}(x)g_{\mu\nu},\qquad\phi\rightarrow\phi^{\prime}=\Omega^{-1}(x)\phi.$
(3.2)

Recall that in order to prove the invariance, we need to use the following
transformation of the scalar curvature under (3.2):

$\displaystyle R\rightarrow R^{\prime}=\Omega^{-2}(R-6\Omega^{-1}\Box\Omega),$
(3.3)

where $\Box\Omega\equiv
h^{-1}\partial_{\mu}(\tilde{g}^{\mu\nu}\partial_{\nu}\Omega)$.

As explained in the previous section, we have to pay attention to what gauge-
fixing conditions should be chosen for the GCT and the Weyl transformation in
a consistent manner. For instance, taking the de Donder condition as a gauge
condition for GCT is not allowed since it breaks the Weyl symmetry in four
space-time dimensions. There are several interesting choices of suitable gauge
conditions for the GCT, but we shall refer to only two representative
examples: The first gauge condition for the GCT is a Weyl invariant version of
the de Donder gauge:

$\displaystyle\partial_{\mu}\left((-g)^{\frac{1}{4}}g^{\mu\nu}\right)=0.$
(3.4)

This gauge choice is invariant under the Weyl transformation (3.2) and is
physically interesting in the sense that it makes use of only the metric
tensor field. However, some fields such as the Nakanishi-Lautrup field become
not a normal vector field but a vector field with density, which makes several
formulas ugly, so we will not adopt (3.4) as a gauge condition for the GCT.
The second gauge condition, which we will take in this article and call it
“the extended de Donder gauge”, is given by

$\displaystyle\partial_{\mu}(\tilde{g}^{\mu\nu}\phi^{2})=0,$ (3.5)

which is also invariant under the Weyl transformation (3.2).

Next, let us consider a gauge-fixing condition for the Weyl transformation.
From the consistency discussed in Section 2, an appropriate gauge condition
must obey the condition that it is invariant under the GCT, that is, a scalar
quantity. Since there are many of scalars constructed out of the real scalar
field $\phi$ and the Riemannian tensors, we might be left in the dark on this
issue. However, surprisingly enough, if we impose the requirement such that
the FP ghost’s Lagrangian should have a Weyl invariant metric
$\tilde{g}^{\mu\nu}\phi^{2}$ instead of the standard metric
$\tilde{g}^{\mu\nu}$, the suitable gauge condition for the GCT can be uniquely
picked up. Such the gauge condition, we will call “the scalar gauge
condition”, reads

$\displaystyle\partial_{\mu}(\tilde{g}^{\mu\nu}\phi\partial_{\nu}\phi)=0,$
(3.6)

which can be alternatively written as

$\displaystyle\Box\,\phi^{2}=0.$ (3.7)

Incidentally, the unitary gauge $\phi=\rm{constant}$ is often taken to show
that the Weyl invariant scalar-tensor gravity (3.1) is equivalent to the
Einstein-Hilbert term, but this gauge choice is not so interesting since there
remains no conformal symmetry behind.

After taking the extended de Donder gauge condition (3.5) for the GCT and the
scalar gauge condition (3.6) for the Weyl transformation, the gauge-fixed and
BRST invariant quantum Lagrangian is given by

$\displaystyle{\cal L}_{q}$ $\displaystyle=$ $\displaystyle{\cal L}_{c}+{\cal
L}_{GF+FP}+\bar{\cal L}_{GF+FP}$ (3.8) $\displaystyle=$ $\displaystyle{\cal
L}_{c}+i\delta_{B}(\tilde{g}^{\mu\nu}\phi^{2}\partial_{\mu}\bar{c}_{\nu})+i\bar{\delta}_{B}\left[\bar{c}\partial_{\mu}(\tilde{g}^{\mu\nu}\phi\partial_{\nu}\phi)\right]$
$\displaystyle=$
$\displaystyle\sqrt{-g}\left(\frac{1}{12}\phi^{2}R+\frac{1}{2}g^{\mu\nu}\partial_{\mu}\phi\partial_{\nu}\phi\right)-\tilde{g}^{\mu\nu}\phi^{2}(\partial_{\mu}b_{\nu}+i\partial_{\mu}\bar{c}_{\lambda}\partial_{\nu}c^{\lambda})$
$\displaystyle+$
$\displaystyle\tilde{g}^{\mu\nu}\phi\partial_{\mu}B\partial_{\nu}\phi-i\tilde{g}^{\mu\nu}\phi^{2}\partial_{\mu}\bar{c}\partial_{\nu}c,$

where surface terms are dropped. Note that the last term, which is the FP
ghost’s term for the Weyl transformation, certainly involves the Weyl
invariant metric $\tilde{g}^{\mu\nu}\phi^{2}$. Let us rewrite this Lagrangian
concisely as

$\displaystyle{\cal
L}_{q}=\sqrt{-g}\frac{1}{12}\phi^{2}R-\frac{1}{2}\tilde{g}^{\mu\nu}E_{\mu\nu},$
(3.9)

where we have defined

$\displaystyle E_{\mu\nu}$ $\displaystyle\equiv$
$\displaystyle-\frac{1}{2}\partial_{\mu}\phi\partial_{\nu}\phi+\phi^{2}(\partial_{\mu}b_{\nu}+i\partial_{\mu}\bar{c}_{\lambda}\partial_{\nu}c^{\lambda})$
(3.10) $\displaystyle-$
$\displaystyle\phi\partial_{\mu}B\partial_{\nu}\phi+i\phi^{2}\partial_{\mu}\bar{c}\partial_{\nu}c+(\mu\leftrightarrow\nu).$

Moreover, it is sometimes more convenient to introduce the dilaton $\sigma(x)$
by defining

$\displaystyle\phi(x)\equiv e^{\sigma(x)},$ (3.11)

and rewrite (3.9) further into the form

$\displaystyle{\cal
L}_{q}=e^{2\sigma(x)}\left(\sqrt{-g}\frac{1}{12}R-\frac{1}{2}\tilde{g}^{\mu\nu}\hat{E}_{\mu\nu}\right),$
(3.12)

where we have defined

$\displaystyle\hat{E}_{\mu\nu}$ $\displaystyle\equiv$
$\displaystyle-\frac{1}{2}\partial_{\mu}\sigma\partial_{\nu}\sigma+\partial_{\mu}b_{\nu}+i\partial_{\mu}\bar{c}_{\lambda}\partial_{\nu}c^{\lambda}$
(3.13) $\displaystyle-$
$\displaystyle\partial_{\mu}B\partial_{\nu}\sigma+i\partial_{\mu}\bar{c}\partial_{\nu}c+(\mu\leftrightarrow\nu).$

Note that the relation between $E_{\mu\nu}$ and $\hat{E}_{\mu\nu}$ is given by

$\displaystyle
E_{\mu\nu}=\phi^{2}\hat{E}_{\mu\nu}=e^{2\sigma}\hat{E}_{\mu\nu}.$ (3.14)

From the Lagrangian ${\cal L}_{q}$, it is straightforward to derive the field
equations by taking the variation with respect to $g_{\mu\nu}$, $\phi$ (or
$\sigma$), $b_{\nu}$, $B$, $c^{\rho}$, $\bar{c}_{\rho}$, $c$ and $\bar{c}$ in
order:

$\displaystyle\frac{1}{12}\phi^{2}G_{\mu\nu}-\frac{1}{12}(\nabla_{\mu}\nabla_{\nu}-g_{\mu\nu}\Box)\phi^{2}-\frac{1}{2}(E_{\mu\nu}-\frac{1}{2}g_{\mu\nu}E)=0,$
$\displaystyle\frac{1}{6}\phi^{2}R-E-2g^{\mu\nu}\phi\partial_{\mu}B\partial_{\nu}\phi-\phi^{2}\Box
B=0,$
$\displaystyle\partial_{\mu}(\tilde{g}^{\mu\nu}\phi^{2})=0,\qquad\partial_{\mu}(\tilde{g}^{\mu\nu}\phi\partial_{\nu}\phi)=0,$
$\displaystyle
g^{\mu\nu}\partial_{\mu}\partial_{\nu}\bar{c}_{\rho}=g^{\mu\nu}\partial_{\mu}\partial_{\nu}c^{\rho}=g^{\mu\nu}\partial_{\mu}\partial_{\nu}\bar{c}=g^{\mu\nu}\partial_{\mu}\partial_{\nu}c=0.$
(3.15)

where we have defined the Einstein tensor $G_{\mu\nu}\equiv
R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R$ and $E\equiv g^{\mu\nu}E_{\mu\nu}$. The two
gauge-fixing conditions in (3.15) lead to a very simple equation for the
dilaton:

$\displaystyle g^{\mu\nu}\partial_{\mu}\partial_{\nu}\sigma=0.$ (3.16)

It is worthwhile to notice that it is not the scalar field $\phi$ but the
dilaton $\sigma$ that satisfies this type of equation. Furthermore, the trace
part of the Einstein equation, i.e., the first field equation in (3.15) and
the field equation for $\phi$ also give us the equation for $B$:

$\displaystyle g^{\mu\nu}\partial_{\mu}\partial_{\nu}B=0.$ (3.17)

Finally, using the field equations obtained thus far, after some calculations,
we can also derive the equation for $b_{\rho}$:666The detail of the
calculation is presented in Appendix A.

$\displaystyle g^{\mu\nu}\partial_{\mu}\partial_{\nu}b_{\rho}=0.$ (3.18)

In other words, setting
$X^{M}=\\{x^{\mu},b_{\mu},\sigma,B,c^{\mu},\bar{c}_{\mu},c,\bar{c}\\}$,
$X^{M}$ turns out to obey the very simple equation:

$\displaystyle g^{\mu\nu}\partial_{\mu}\partial_{\nu}X^{M}=0.$ (3.19)

This fact, together with the gauge condition
$\partial_{\mu}(\tilde{g}^{\mu\nu}\phi^{2})=0$, produces the two kinds of
conserved currents:

$\displaystyle{\cal P}^{\mu M}$ $\displaystyle\equiv$
$\displaystyle\tilde{g}^{\mu\nu}\phi^{2}\partial_{\nu}X^{M}=\tilde{g}^{\mu\nu}\phi^{2}\bigl{(}1\overset{\leftrightarrow}{\partial}_{\nu}X^{M}\bigr{)}$
$\displaystyle{\cal M}^{\mu MN}$ $\displaystyle\equiv$
$\displaystyle\tilde{g}^{\mu\nu}\phi^{2}\bigl{(}X^{M}\overset{\leftrightarrow}{\partial}_{\nu}Y^{N}\bigr{)},$
(3.20)

where we have defined
$X^{M}\overset{\leftrightarrow}{\partial}_{\mu}Y^{N}\equiv
X^{M}\partial_{\mu}Y^{N}-(\partial_{\mu}X^{M})Y^{N}$.

##  4 Canonical quantization and equal-time commutation relations

In this section, after introducing the canonical commutation relations (CCRs),
we will evaluate various equal-time commutation relations (ETCRs) among
fundamental variables. To simplify various expressions, we will obey the
following abbreviations adopted in the textbook of Nakanishi and Ojima [5]:

$\displaystyle[A,B^{\prime}]$ $\displaystyle=$
$\displaystyle[A(x),B(x^{\prime})]|_{x^{0}=x^{\prime
0}},\qquad\delta^{3}=\delta(\vec{x}-\vec{x}^{\prime}),$
$\displaystyle\tilde{f}$ $\displaystyle=$
$\displaystyle\frac{1}{\tilde{g}^{00}}=\frac{1}{\sqrt{-g}g^{00}}=\frac{1}{hg^{00}},$
(4.1)

where we assume that $\tilde{g}^{00}$ is invertible.

Now let us set up the canonical (anti-)commutation relations:

$\displaystyle[g_{\mu\nu},\pi_{g}^{\rho\lambda\prime}]=i\frac{1}{2}(\delta_{\mu}^{\rho}\delta_{\nu}^{\lambda}+\delta_{\mu}^{\lambda}\delta_{\nu}^{\rho})\delta^{3},\quad[\phi,\pi_{\phi}^{\prime}]=+i\delta^{3},\quad[B,\pi_{B}^{\prime}]=+i\delta^{3},$
$\displaystyle\\{c^{\sigma},\pi_{c\lambda}^{\prime},\\}=\\{\bar{c}_{\lambda},\pi_{\bar{c}}^{\sigma\prime}\\}=+i\delta_{\lambda}^{\sigma}\delta^{3},\quad\\{c,\pi_{c}^{\prime}\\}=\\{\bar{c},\pi_{\bar{c}}^{\prime}\\}=+i\delta^{3},$
(4.2)

where the other (anti-)commutation relations vanish. Here the canonical
variables are $g_{\mu\nu},\phi,B,c^{\rho},\bar{c}_{\rho},c,\bar{c}$ and the
corresponding canonical conjugate momenta are
$\pi_{g}^{\mu\nu},\pi_{\phi},\pi_{B},\pi_{c\rho},\pi_{\bar{c}}^{\rho},\pi_{c},\pi_{\bar{c}}$,
respectively and the $b_{\mu}$ field is regarded as not a canonical variable
but a conjugate momentum of $\tilde{g}^{0\mu}$.

To remove second order derivatives of the metric involved in $R$, we perform
the integration by parts once and rewrite the Lagrangian (3.8) as

$\displaystyle{\cal L}_{q}$ $\displaystyle=$
$\displaystyle-\frac{1}{12}\tilde{g}^{\mu\nu}\phi^{2}(\Gamma^{\sigma}_{\mu\nu}\Gamma^{\alpha}_{\sigma\alpha}-\Gamma^{\sigma}_{\mu\alpha}\Gamma^{\alpha}_{\sigma\nu})-\frac{1}{6}\phi\partial_{\mu}\phi(\tilde{g}^{\alpha\beta}\Gamma^{\mu}_{\alpha\beta}-\tilde{g}^{\mu\nu}\Gamma^{\alpha}_{\nu\alpha})$
(4.3) $\displaystyle+$
$\displaystyle\frac{1}{2}\tilde{g}^{\mu\nu}\partial_{\mu}\phi\partial_{\nu}\phi+\partial_{\mu}(\tilde{g}^{\mu\nu}\phi^{2})b_{\nu}-i\tilde{g}^{\mu\nu}\phi^{2}\partial_{\mu}\bar{c}_{\rho}\partial_{\nu}c^{\rho}+\tilde{g}^{\mu\nu}\partial_{\mu}B\phi\partial_{\nu}\phi$
$\displaystyle-$ $\displaystyle
i\tilde{g}^{\mu\nu}\phi^{2}\partial_{\mu}\bar{c}\partial_{\nu}c+\partial_{\mu}{\cal{V}}^{\mu},$

where a surface term ${\cal{V}}^{\mu}$ is defined as

$\displaystyle{\cal{V}}^{\mu}=\frac{1}{12}\phi^{2}(\tilde{g}^{\alpha\beta}\Gamma^{\mu}_{\alpha\beta}-\tilde{g}^{\mu\nu}\Gamma^{\alpha}_{\nu\alpha})-\tilde{g}^{\mu\nu}\phi^{2}b_{\nu}.$
(4.4)

Using this Lagrangian, the concrete expressions for canonical conjugate
momenta become

$\displaystyle\pi_{g}^{\mu\nu}$ $\displaystyle=$
$\displaystyle\frac{\partial{\cal L}_{q}}{\partial\dot{g}_{\mu\nu}}$
$\displaystyle=$
$\displaystyle-\frac{1}{24}\sqrt{-g}\phi^{2}\Bigl{[}-g^{0\lambda}g^{\mu\nu}g^{\sigma\tau}-g^{0\tau}g^{\mu\lambda}g^{\nu\sigma}-g^{0\sigma}g^{\mu\tau}g^{\nu\lambda}+g^{0\lambda}g^{\mu\tau}g^{\nu\sigma}$
$\displaystyle+$ $\displaystyle
g^{0\tau}g^{\mu\nu}g^{\lambda\sigma}+\frac{1}{2}(g^{0\mu}g^{\nu\lambda}+g^{0\nu}g^{\mu\lambda})g^{\sigma\tau}\Bigr{]}\partial_{\lambda}g_{\sigma\tau}$
$\displaystyle-$
$\displaystyle\frac{1}{6}\sqrt{-g}\Bigl{[}\frac{1}{2}(g^{0\mu}g^{\rho\nu}+g^{0\nu}g^{\rho\mu})-g^{\mu\nu}g^{\rho
0}\Bigr{]}\phi\partial_{\rho}\phi$ $\displaystyle-$
$\displaystyle\frac{1}{2}\sqrt{-g}(g^{0\mu}g^{\nu\rho}+g^{0\nu}g^{\mu\rho}-g^{0\rho}g^{\mu\nu})\phi^{2}b_{\rho},$
$\displaystyle\pi_{\phi}$ $\displaystyle=$ $\displaystyle\frac{\partial{\cal
L}_{q}}{\partial\dot{\phi}}=\tilde{g}^{0\mu}\partial_{\mu}\phi+2\tilde{g}^{0\mu}\phi
b_{\mu}+\frac{1}{6}\phi(-\tilde{g}^{\alpha\beta}\Gamma^{0}_{\alpha\beta}+\tilde{g}^{0\alpha}\Gamma^{\beta}_{\alpha\beta})+\tilde{g}^{0\mu}\partial_{\mu}B\phi,$
$\displaystyle\pi_{B}$ $\displaystyle=$ $\displaystyle\frac{\partial{\cal
L}_{q}}{\partial\dot{B}}=\tilde{g}^{0\mu}\phi\partial_{\mu}\phi,$
$\displaystyle\pi_{c\sigma}$ $\displaystyle=$
$\displaystyle\frac{\partial{\cal
L}_{q}}{\partial\dot{c}^{\sigma}}=-i\tilde{g}^{0\mu}\phi^{2}\partial_{\mu}\bar{c}_{\sigma},$
$\displaystyle\pi_{\bar{c}}^{\sigma}$ $\displaystyle=$
$\displaystyle\frac{\partial{\cal
L}_{q}}{\partial\dot{\bar{c}}_{\sigma}}=i\tilde{g}^{0\mu}\phi^{2}\partial_{\mu}c^{\sigma},$
$\displaystyle\pi_{c}$ $\displaystyle=$ $\displaystyle\frac{\partial{\cal
L}_{q}}{\partial\dot{c}}=-i\tilde{g}^{0\mu}\phi^{2}\partial_{\mu}\bar{c},$
$\displaystyle\pi_{\bar{c}}$ $\displaystyle=$
$\displaystyle\frac{\partial{\cal
L}_{q}}{\partial\dot{\bar{c}}}=i\tilde{g}^{0\mu}\phi^{2}\partial_{\mu}c,$
(4.5)

where we have defined the time derivative such as
$\dot{g}_{\mu\nu}\equiv\frac{\partial g_{\mu\nu}}{\partial
t}\equiv\partial_{0}g_{\mu\nu}$, and differentiation of ghosts is taken from
the right.

From now on, we would like to evaluate various nontrivial equal-time
commutation relations (ETCRs) in order. Let us first work with the ETCR in Eq.
(4.2):

$\displaystyle[\pi_{g}^{\alpha
0},g_{\mu\nu}^{\prime}]=-i\frac{1}{2}(\delta_{\mu}^{\alpha}\delta_{\nu}^{0}+\delta_{\mu}^{0}\delta_{\nu}^{\alpha})\delta^{3}.$
(4.6)

The canonical conjugate momentum $\pi_{g}^{\alpha 0}$ has a structure

$\displaystyle\pi_{g}^{\alpha
0}=A^{\alpha}+B^{\alpha\beta}\partial_{\beta}\phi+C^{\alpha\beta}b_{\beta},$
(4.7)

where $A^{\alpha},B^{\alpha\beta}$ and
$C^{\alpha\beta}\equiv-\frac{1}{2}\tilde{g}^{00}g^{\alpha\beta}\phi^{2}$ have
no $\dot{g}_{\mu\nu}$, and $B^{\alpha\beta}\partial_{\beta}\phi$ does not have
$\dot{\phi}$ since $\pi_{g}^{\alpha 0}$ does not include the dynamics of the
metric and the scalar fields. Then, we find that Eq. (4.6) produces

$\displaystyle[g_{\mu\nu},b_{\rho}^{\prime}]=-i\tilde{f}\phi^{-2}(\delta_{\mu}^{0}g_{\rho\nu}+\delta_{\nu}^{0}g_{\rho\mu})\delta^{3}.$
(4.8)

From this ETCR, we can easily derive ETCRs:

$\displaystyle[g^{\mu\nu},b_{\rho}^{\prime}]=i\tilde{f}\phi^{-2}(g^{\mu
0}\delta_{\rho}^{\nu}+g^{\nu 0}\delta_{\rho}^{\mu})\delta^{3},$
$\displaystyle[\tilde{g}^{\mu\nu},b_{\rho}^{\prime}]=i\tilde{f}\phi^{-2}(\tilde{g}^{\mu
0}\delta_{\rho}^{\nu}+\tilde{g}^{\nu
0}\delta_{\rho}^{\mu}-\tilde{g}^{\mu\nu}\delta_{\rho}^{0})\delta^{3}.$ (4.9)

Here we have used the following fact; since a commutator works as a
derivation, we can have formulae:

$\displaystyle[g^{\mu\nu},\Phi^{\prime}]=-g^{\mu\alpha}g^{\nu\beta}[g_{\alpha\beta},\Phi^{\prime}],$
$\displaystyle[\tilde{g}^{\mu\nu},\Phi^{\prime}]=-\left(\tilde{g}^{\mu\alpha}g^{\nu\beta}-\frac{1}{2}\tilde{g}^{\mu\nu}g^{\alpha\beta}\right)[g_{\alpha\beta},\Phi^{\prime}],$
(4.10)

where $\Phi$ is a generic field. Similarly, the ETCR, $[\pi_{g}^{\alpha
0},\phi^{\prime}]=0$ yields

$\displaystyle[\phi,b_{\rho}^{\prime}]=0.$ (4.11)

The ETCR, $[\pi_{g}^{\alpha 0},B^{\prime}]=0$ yields

$\displaystyle[B,b_{\rho}^{\prime}]=0.$ (4.12)

Moreover, the ETCRs, $[\pi_{B},\phi^{\prime}]=0$ and
$[\pi_{B},B^{\prime}]=-i\delta^{3}$ respectively produce

$\displaystyle[\dot{\phi},\phi^{\prime}]=0,\qquad[\dot{\phi},B^{\prime}]=-i\tilde{f}\phi^{-1}\delta^{3}.$
(4.13)

As for the ETCRs involving FP ghosts, let us first consider the anti-ETCRs,
$\\{\pi_{c\lambda},c^{\sigma\prime}\\}=\\{\pi_{\bar{c}}^{\sigma},\bar{c}_{\lambda}^{\prime}\\}=i\delta_{\lambda}^{\sigma}\delta^{3}$.
These anti-ETCRs lead to the same anti-ETCR:

$\displaystyle\\{\dot{\bar{c}}_{\lambda},c^{\sigma\prime}\\}=-\\{\dot{c}^{\sigma},\bar{c}^{\prime}_{\lambda}\\}=-\tilde{f}\phi^{-2}\delta_{\lambda}^{\sigma}\delta^{3},$
(4.14)

where we have used a useful identity for generic variables $\Phi$ and $\Psi$:

$\displaystyle[\Phi,\dot{\Psi}^{\prime}]=\partial_{0}[\Phi,\Psi^{\prime}]-[\dot{\Phi},\Psi^{\prime}],$
(4.15)

which holds for the anti-commutation relation as well. In a similar way, the
anti-ETCRs,
$\\{\pi_{c},c^{\prime}\\}=\\{\pi_{\bar{c}},\bar{c}^{\prime}\\}=i\delta^{3}$
yield

$\displaystyle\\{\dot{\bar{c}},c^{\prime}\\}=-\\{\dot{c},\bar{c}^{\prime}\\}=-\tilde{f}\phi^{-2}\delta^{3}.$
(4.16)

Moreover, $[\pi_{g}^{\alpha 0},c^{\sigma\prime}]=[\pi_{g}^{\alpha
0},\bar{c}_{\lambda}^{\prime}]=0$ give us the ETCRs:

$\displaystyle[b_{\rho},c^{\sigma\prime}]=[b_{\rho},\bar{c}^{\prime}_{\lambda}]=0,$
(4.17)

and similarly $[\pi_{g}^{\alpha 0},c^{\prime}]=[\pi_{g}^{\alpha
0},\bar{c}^{\prime}]=0$ produce

$\displaystyle[b_{\rho},c^{\prime}]=[b_{\rho},\bar{c}^{\prime}]=0.$ (4.18)

To calculate the ETCRs between $B$ and the FP-ghosts, it is necessary to
utilize the ETCRs,
$[B,\pi_{c\lambda}^{\prime}]=[B,\pi_{\bar{c}}^{\sigma\prime}]=[B,\pi_{c}^{\prime}]=[B,\pi_{\bar{c}}^{\prime}]=0$,
and consequently we have

$\displaystyle[B,\dot{\bar{c}}^{\prime}_{\lambda}]=[B,\dot{c}^{\sigma\prime}]=[B,\dot{\bar{c}}^{\prime}]=[B,\dot{c}^{\prime}]=0.$
(4.19)

Furthermore, taking the Weyl BRST transformation of the third ETCR reads777We
define the BRST transformation for the Weyl transformation as
$\bar{\delta}_{B}\Phi\equiv[i\bar{Q}_{B},\Phi\\}$ where $\Phi$ is a generic
field and $[\hskip 5.69054pt,\hskip 5.69054pt\\}$ denotes the graded bracket.
Of course, in case of the GCT BRST transformation, it is replaced by
$\delta_{B}\Phi\equiv[iQ_{B},\Phi\\}$.

$\displaystyle
0=\\{[i\bar{Q}_{B},B],\dot{\bar{c}}^{\prime}\\}+[B,\\{i\bar{Q}_{B},\dot{\bar{c}}^{\prime}\\}]=[B,i\dot{B}^{\prime}],$
(4.20)

where we have used the Weyl BRST transformation (2.5). As a result we have the
ETCR:

$\displaystyle[B,\dot{B}^{\prime}]=0.$ (4.21)

Next, from
$[\pi_{B},\bar{c}^{\prime}_{\lambda}]=[\pi_{B},c^{\sigma\prime}]=0$, we find

$\displaystyle[\dot{\phi},\bar{c}^{\prime}_{\lambda}]=[\dot{\phi},c^{\sigma\prime}]=0.$
(4.22)

Similarly, from $[\pi_{B},\bar{c}^{\prime}]=[\pi_{B},c^{\prime}]=0$, we have

$\displaystyle[\dot{\phi},\bar{c}^{\prime}]=[\dot{\phi},c^{\prime}]=0.$ (4.23)

Using the field equation for $\bar{c}_{\lambda}$ in (3.15), i.e.,
$g^{\mu\nu}\partial_{\mu}\partial_{\nu}\bar{c}_{\lambda}=0$, the ETCR,
$[\phi,\bar{c}^{\prime}_{\lambda}]=0$, the formula (4.15), and Eq. (4.22), it
is easy to derive the equations:

$\displaystyle[\phi,\ddot{\bar{c}}^{\prime}_{\lambda}]=[\dot{\phi},\dot{\bar{c}}^{\prime}_{\lambda}]=[\ddot{\phi},\bar{c}^{\prime}_{\lambda}]=0.$
(4.24)

Similar equations also hold when $\bar{c}^{\prime}_{\lambda}$ is replaced with
$c^{\sigma\prime}$, $c^{\prime}$ or $\bar{c}^{\prime}$.

Now, using the equations obtained above, we are ready to evaluate the type of
the ETCRs, $[\dot{\Phi},b^{\prime}_{\rho}]$ where $\Phi$ is a generic field.
First, let us focus on $[\dot{\phi},b^{\prime}_{\rho}]$. To do that, we start
with $[\dot{\phi},\bar{c}^{\prime}_{\rho}]=0$ in Eq. (4.22) and take its BRST
variation for the GCT as follows:

$\displaystyle 0$ $\displaystyle=$
$\displaystyle\\{iQ_{B},[\dot{\phi},\bar{c}^{\prime}_{\rho}]\\}$ (4.25)
$\displaystyle=$
$\displaystyle\\{[iQ_{B},\dot{\phi}],\bar{c}^{\prime}_{\rho}\\}+[\dot{\phi},\\{iQ_{B},\bar{c}^{\prime}_{\rho}\\}]$
$\displaystyle=$
$\displaystyle\\{-\partial_{0}(c^{\lambda}\partial_{\lambda}\phi),\bar{c}^{\prime}_{\rho}\\}+[\dot{\phi},i(b^{\prime}_{\rho}+ic^{\lambda\prime}\partial_{\lambda}\bar{c}^{\prime}_{\rho})]$
$\displaystyle=$
$\displaystyle-\\{\dot{c}^{\lambda},\bar{c}^{\prime}_{\rho}\\}\partial_{\lambda}\phi+i[\dot{\phi},b^{\prime}_{\rho}].$

Using Eq. (4.14), we are able to obtain

$\displaystyle[\dot{\phi},b^{\prime}_{\rho}]=-i\tilde{f}\phi^{-2}\partial_{\rho}\phi\delta^{3}.$
(4.26)

It turns out that the ETCRs, $[\pi_{c\lambda},\pi_{g}^{\alpha
0\prime}]=[\pi^{\sigma}_{\bar{c}},\pi_{g}^{\alpha 0\prime}]=0$ give rise to

$\displaystyle[\dot{\bar{c}}_{\lambda},b^{\prime}_{\rho}]=-i\tilde{f}\phi^{-2}\partial_{\rho}\bar{c}_{\lambda}\delta^{3},\qquad[\dot{c}^{\sigma},b^{\prime}_{\rho}]=-i\tilde{f}\phi^{-2}\partial_{\rho}c^{\sigma}\delta^{3}.$
(4.27)

Similarly, the ETCRs, $[\pi_{c},\pi_{g}^{\alpha
0\prime}]=[\pi_{\bar{c}},\pi_{g}^{\alpha 0\prime}]=0$ give us

$\displaystyle[\dot{\bar{c}},b^{\prime}_{\rho}]=-i\tilde{f}\phi^{-2}\partial_{\rho}\bar{c}\delta^{3},\qquad[\dot{c},b^{\prime}_{\rho}]=-i\tilde{f}\phi^{-2}\partial_{\rho}c\delta^{3}.$
(4.28)

In order to evaluate $[\dot{B},b^{\prime}_{\rho}]$, we make use of
$[\pi_{c},b^{\prime}_{\rho}]=0$, which can be easily proved. Taking its BRST
transformation for the Weyl transformation leads to the equation:

$\displaystyle[\\{i\bar{Q}_{B},\pi_{c}\\},b^{\prime}_{\rho}]=0,$ (4.29)

where $[i\bar{Q}_{B},b^{\prime}_{\rho}]=0$ was used. We can show that
$\\{i\bar{Q}_{B},\pi_{c}\\}=\tilde{g}^{0\mu}\phi^{2}\partial_{\mu}B$, so using
(4.9) and (4.11), we can calculate

$\displaystyle[\dot{B},b^{\prime}_{\rho}]=-i\tilde{f}\phi^{-2}\partial_{\rho}B\delta^{3}.$
(4.30)

Finally, the ETCR, $[\dot{g}_{\mu\nu},b_{\rho}^{\prime}]$ (or equivalently,
$[g_{\mu\nu},\dot{b}_{\rho}^{\prime}]$) can be obtained by using the method
developed in our previous article [1]. Only the result is written out as

$\displaystyle[\dot{g}_{\mu\nu},b_{\rho}^{\prime}]$ $\displaystyle=$
$\displaystyle-i\Bigl{\\{}\tilde{f}\phi^{-2}(\partial_{\rho}g_{\mu\nu}+\delta_{\mu}^{0}\dot{g}_{\rho\nu}+\delta_{\nu}^{0}\dot{g}_{\rho\mu})\delta^{3}$
(4.31) $\displaystyle+$
$\displaystyle[(\delta_{\mu}^{k}-2\delta_{\mu}^{0}\tilde{f}\tilde{g}^{0k})g_{\rho\nu}+(\mu\leftrightarrow\nu)]\partial_{k}(\tilde{f}\phi^{-2}\delta^{3})\Bigr{\\}},$

or equivalently,

$\displaystyle[g_{\mu\nu},\dot{b}_{\rho}^{\prime}]$ $\displaystyle=$
$\displaystyle
i\Bigl{\\{}[\tilde{f}\phi^{-2}\partial_{\rho}g_{\mu\nu}-\partial_{0}(\tilde{f}\phi^{-2})(\delta_{\mu}^{0}g_{\rho\nu}+\delta_{\nu}^{0}g_{\rho\mu})]\delta^{3}$
(4.32) $\displaystyle+$
$\displaystyle[(\delta_{\mu}^{k}-2\delta_{\mu}^{0}\tilde{f}\tilde{g}^{0k})g_{\rho\nu}+(\mu\leftrightarrow\nu)]\partial_{k}(\tilde{f}\phi^{-2}\delta^{3})\Bigr{\\}}.$

Following our previous calculation [1], we can prove

$\displaystyle[b_{\mu},b_{\nu}^{\prime}]=0,$
$\displaystyle[b_{\mu},\dot{b}_{\nu}^{\prime}]=i\tilde{f}\phi^{-2}(\partial_{\mu}b_{\nu}+\partial_{\nu}b_{\mu})\delta^{3}.$
(4.33)

##  5 Equal-time commutation relations in gravitational sector

The remaining nontrivial ETCRs are related to the time derivative of the
metric field, i.e., the ETCRs, $[\dot{g}_{\mu\nu},\Phi^{\prime}]$ where $\Phi$
is a generic field. In this section, we will evaluate such the ETCRs.

First of all, let us start with the ETCR,
$[\pi_{\phi},g_{\mu\nu}^{\prime}]=0$. From the expression of $\pi_{\phi}$ in
Eq. (4.5), this ETCR can be described as

$\displaystyle\tilde{g}^{00}[\dot{\phi},g_{\mu\nu}^{\prime}]+\frac{1}{6}\phi(\tilde{g}^{00}g^{\rho\sigma}-\tilde{g}^{0\rho}g^{0\sigma})[\dot{g}_{\rho\sigma},g_{\mu\nu}^{\prime}]+\tilde{g}^{00}\phi[\dot{B},g_{\mu\nu}^{\prime}]$
$\displaystyle=-4i\tilde{f}\phi^{-1}\sqrt{-g}\delta^{0}_{\mu}\delta^{0}_{\nu}\delta^{3}.$
(5.1)

Next, the ETCR, $[\pi_{\phi},\phi^{\prime}]=-i\delta^{3}$ produces the
equation:

$\displaystyle(\tilde{g}^{00}g^{\rho\sigma}-\tilde{g}^{0\rho}g^{0\sigma})[\dot{g}_{\rho\sigma},\phi^{\prime}]=0.$
(5.2)

Moreover, the ETCR, $[\pi_{\phi},B^{\prime}]=0$ reads

$\displaystyle(\tilde{g}^{00}g^{\rho\sigma}-\tilde{g}^{0\rho}g^{0\sigma})[\dot{g}_{\rho\sigma},B^{\prime}]=6i\phi^{-2}\delta^{3}.$
(5.3)

The extended de Donder gauge, $\partial_{\mu}(\tilde{g}^{\mu\nu}\phi^{2})=0$,
can be rewritten as

$\displaystyle{\cal
D}^{\lambda\rho\sigma}\dot{g}_{\rho\sigma}+4\phi^{-1}g^{\lambda\rho}\partial_{\rho}\phi=(2g^{\lambda\rho}g^{\sigma
k}-g^{\rho\sigma}g^{\lambda k})\partial_{k}g_{\rho\sigma},$ (5.4)

where ${\cal D}^{\lambda\rho\sigma}\equiv
g^{0\lambda}g^{\rho\sigma}-2g^{\lambda\rho}g^{0\sigma}$. Since the right-hand
side (RHS) of Eq. (5.4) is independent of $\dot{g}_{\mu\nu}$, it commutes with
$g_{\mu\nu},\phi$ or $B$. Thus, we have three identities:

$\displaystyle{\cal
D}^{\lambda\rho\sigma}[\dot{g}_{\rho\sigma},g_{\mu\nu}^{\prime}]+4\phi^{-1}g^{\lambda
0}[\dot{\phi},g_{\mu\nu}^{\prime}]=0.$ (5.5) $\displaystyle{\cal
D}^{\lambda\rho\sigma}[\dot{g}_{\rho\sigma},\phi^{\prime}]=0.$ (5.6)
$\displaystyle{\cal
D}^{\lambda\rho\sigma}[\dot{g}_{\rho\sigma},B^{\prime}]=4i\tilde{f}\phi^{-2}g^{\lambda
0}\delta^{3}.$ (5.7)

In Eqs. (5.6) and (5.7), we have used Eq. (4.13).

Putting $\lambda=0$ in Eq. (5.6) and using Eq. (5.2), we have

$\displaystyle
g^{\rho\sigma}[\dot{g}_{\rho\sigma},\phi^{\prime}]=g^{0\rho}g^{0\sigma}[\dot{g}_{\rho\sigma},\phi^{\prime}]=0.$
(5.8)

In general, from the argument of symmetry,
$[\dot{g}_{\rho\sigma},\phi^{\prime}]$ must be of form:

$\displaystyle[\dot{g}_{\rho\sigma},\phi^{\prime}]=a_{1}(g_{\rho\sigma}+a_{2}\delta^{0}_{\rho}\delta^{0}_{\sigma})\delta^{3},$
(5.9)

where $a_{1},a_{2}$ are constants. Eq. (5.8) then requires us to take
$a_{1}=a_{2}=0$. Thus, we have

$\displaystyle[\dot{g}_{\rho\sigma},\phi^{\prime}]=0.$ (5.10)

Next, in a similar manner, we can set

$\displaystyle[\dot{g}_{\rho\sigma},B^{\prime}]=b_{1}(g_{\rho\sigma}+b_{2}\delta^{0}_{\rho}\delta^{0}_{\sigma})\delta^{3},$
(5.11)

where $b_{1},b_{2}$ are constants. From Eq. (5.3), $b_{1}$ is determined to be
$2i\tilde{f}\phi^{-2}$, and then Eq. (5.7) requires $b_{2}$ to be vanishing,
so we can obtain

$\displaystyle[\dot{g}_{\rho\sigma},B^{\prime}]=2i\tilde{f}\phi^{-2}g_{\rho\sigma}\delta^{3}.$
(5.12)

Finally, we wish to evaluate $[\dot{g}_{\rho\sigma},g^{\prime}_{\mu\nu}]$, for
which we need some calculations. Before doing so, let us rewrite Eq. (5.1) by
means of Eqs. (5.10) and (5.12) into the form:

$\displaystyle(\tilde{g}^{00}g^{\rho\sigma}-\tilde{g}^{0\rho}g^{0\sigma})[\dot{g}_{\rho\sigma},g_{\mu\nu}^{\prime}]=-12i\phi^{-2}\left(g_{\mu\nu}+\frac{2}{g^{00}}\delta^{0}_{\mu}\delta^{0}_{\nu}\right)\delta^{3}.$
(5.13)

Similarly, Eq. (5.5) reduces to

$\displaystyle(g^{0\lambda}g^{\rho\sigma}-2g^{\lambda\rho}g^{0\sigma})[\dot{g}_{\rho\sigma},g_{\mu\nu}^{\prime}]=0.$
(5.14)

We are now willing to evaluate the ETCR,
$[\dot{g}_{\rho\sigma},g_{\mu\nu}^{\prime}]$. This ETCR has a symmetry under
the simultaneous exchange of $(\mu\nu)\leftrightarrow(\rho\sigma)$ and primed
$\leftrightarrow$ unprimed in addition to the usual symmetry
$\mu\leftrightarrow\nu$ and $\rho\leftrightarrow\sigma$. Then, we can write
down its general expression like

$\displaystyle[\dot{g}_{\rho\sigma},g_{\mu\nu}^{\prime}]$ $\displaystyle=$
$\displaystyle\biggl{\\{}c_{1}g_{\rho\sigma}g_{\mu\nu}+c_{2}(g_{\rho\mu}g_{\sigma\nu}+g_{\rho\nu}g_{\sigma\mu})$
(5.15) $\displaystyle+$ $\displaystyle
h\tilde{f}\Bigl{[}c_{3}(\delta_{\rho}^{0}\delta_{\sigma}^{0}g_{\mu\nu}+\delta_{\mu}^{0}\delta_{\nu}^{0}g_{\rho\sigma})+c_{4}(\delta_{\rho}^{0}\delta_{\mu}^{0}g_{\sigma\nu}+\delta_{\rho}^{0}\delta_{\nu}^{0}g_{\sigma\mu}$
$\displaystyle+$
$\displaystyle\delta_{\sigma}^{0}\delta_{\mu}^{0}g_{\rho\nu}+\delta_{\sigma}^{0}\delta_{\nu}^{0}g_{\rho\mu})\Bigr{]}+(h\tilde{f})^{2}c_{5}\delta_{\rho}^{0}\delta_{\sigma}^{0}\delta_{\mu}^{0}\delta_{\nu}^{0}\biggr{\\}}\delta^{3},$

where $c_{i}(i=1,\cdots,5)$ are some coefficients. Imposing Eq. (5.14) on
(5.15) leads to relations among the coefficients:

$\displaystyle c_{3}=2(c_{1}+c_{2}),\qquad c_{4}=-c_{2},\qquad
c_{5}=4(c_{1}+c_{2}).$ (5.16)

Furthermore, imposing Eq. (5.13), we can determine $c_{2},c_{3},c_{4}$ and
$c_{5}$ via $c_{1}$ as

$\displaystyle c_{3}=-c_{1}-12i\tilde{f}\phi^{-2},\qquad
c_{4}=-c_{2}=\frac{3}{2}c_{1}+6i\tilde{f}\phi^{-2},$ $\displaystyle
c_{5}=-2c_{1}-24i\tilde{f}\phi^{-2}.$ (5.17)

In order to fix the coefficient $c_{1}$, we need to calculate the ETCR,
$[\dot{g}_{kl},g_{mn}^{\prime}]$ explicitly in terms of
$[\pi_{g}^{kl},g_{mn}^{\prime}]=-i\frac{1}{2}(\delta_{m}^{k}\delta_{n}^{l}+\delta_{m}^{l}\delta_{n}^{k})\delta^{3}$
in Eq. (4.2) and the concrete expression of $\pi_{g}^{kl}$ in Eq. (4.5). To do
that, from Eq. (4.5), let us write

$\displaystyle\pi_{g}^{kl}=\hat{A}^{kl}+\hat{B}^{kl\rho}b_{\rho}+\hat{C}^{klmn}\dot{g}_{mn}+\hat{D}^{kl}\dot{\phi}.$
(5.18)

Here $\hat{A}^{kl},\hat{B}^{kl\rho},\hat{C}^{klmn}$ and $\hat{D}^{kl}$ commute
with $g_{mn}$, and $\hat{C}^{klmn}$ and $\hat{D}^{kl}$ are defined as888It
turns out that the concrete expressions of $\hat{A}^{kl}$ and
$\hat{B}^{kl\rho}$ are irrelevant to the calculation of
$[\dot{g}_{kl},g_{mn}^{\prime}]$.

$\displaystyle\hat{C}^{klmn}=\frac{1}{24}h\phi^{2}K^{klmn},\qquad\hat{D}^{kl}=\frac{1}{6}\phi(\tilde{g}^{00}g^{kl}-\tilde{g}^{0k}g^{0l}),$
(5.19)

where the definition of $K^{klmn}$ and its property are given by

$\displaystyle K^{klmn}=\left|\begin{array}[]{rrr}g^{00}&g^{0l}&g^{0n}\\\
g^{k0}&g^{kl}&g^{kn}\\\ g^{m0}&g^{ml}&g^{mn}\\\ \end{array}\right|,$ (5.23)
$\displaystyle
K^{klmn}\frac{1}{2}(g^{00})^{-1}(g_{ij}g_{mn}-g_{im}g_{jn}-g_{in}g_{jm})=\frac{1}{2}(\delta_{i}^{k}\delta_{j}^{l}+\delta_{i}^{l}\delta_{j}^{k}).$
(5.24)

From Eq. (5.18), we can calculate

$\displaystyle[\dot{g}_{kl},g_{mn}^{\prime}]$ $\displaystyle=$
$\displaystyle\hat{C}^{-1}_{klpq}\left([\pi_{g}^{pq},g_{mn}^{\prime}]-\hat{B}^{pq\rho}[b_{\rho},g_{mn}^{\prime}]-\hat{D}^{pq}[\dot{\phi},g_{mn}^{\prime}]\right).$
(5.25) $\displaystyle=$
$\displaystyle-i\frac{1}{2}\hat{C}^{-1}_{klpq}(\delta^{p}_{m}\delta^{q}_{n}+\delta^{q}_{m}\delta^{p}_{n})\delta^{3},$

where we have used Eqs. (4.2), (4.8), and (5.10). Since we can calculate

$\displaystyle\hat{C}^{-1}_{klpq}=12\tilde{f}\phi^{-2}(g_{kl}g_{pq}-g_{kp}g_{lq}-g_{kq}g_{lp}),$
(5.26)

we can eventually arrive at the result:

$\displaystyle[\dot{g}_{kl},g_{mn}^{\prime}]=-12i\tilde{f}\phi^{-2}(g_{kl}g_{mn}-g_{km}g_{ln}-g_{kn}g_{lm})\delta^{3}.$
(5.27)

Meanwhile, from Eq. (5.15) we have the ETCR:

$\displaystyle[\dot{g}_{kl},g_{mn}^{\prime}]=\left[c_{1}g_{kl}g_{mn}+c_{2}(g_{km}g_{ln}+g_{kn}g_{lm})\right]\delta^{3}$
(5.28)

Hence, comparing (5.27) with (5.28), we can obtain

$\displaystyle c_{1}=-12i\tilde{f}\phi^{-2},\qquad
c_{2}=12i\tilde{f}\phi^{-2}.$ (5.29)

Note that these values satisfy the relation in Eq. (5.17),
$-c_{2}=\frac{3}{2}c_{1}+6i\tilde{f}\phi^{-2}$, which gives us a nontrivial
verification of our result. In this way, we have succeeded in getting the
following ETCR:

$\displaystyle[\dot{g}_{\rho\sigma},g_{\mu\nu}^{\prime}]$ $\displaystyle=$
$\displaystyle-12i\tilde{f}\phi^{-2}[g_{\rho\sigma}g_{\mu\nu}-g_{\rho\mu}g_{\sigma\nu}-g_{\rho\nu}g_{\sigma\mu}+h\tilde{f}(\delta_{\rho}^{0}\delta_{\mu}^{0}g_{\sigma\nu}$
(5.30) $\displaystyle+$
$\displaystyle\delta_{\rho}^{0}\delta_{\nu}^{0}g_{\sigma\mu}+\delta_{\sigma}^{0}\delta_{\mu}^{0}g_{\rho\nu}+\delta_{\sigma}^{0}\delta_{\nu}^{0}g_{\rho\mu})]\delta^{3}.$

##  6 Unitarity of physical S-matrix

As in the conventional BRST formalism, the physical state $|\rm{phys}\rangle$
is defined by imposing two subsidiary conditions [8]:

$\displaystyle Q_{B}|\rm{phys}\rangle=\bar{Q}_{B}|\rm{phys}\rangle=0.$ (6.1)

It is then well known that the physical S-matrix is unitary under the
assumption that all the BRST singlet states have positive norm. In this
section, we would like to prove the unitarity of the physical S-matrix. Since
there is a ghost-like scalar field $\phi$ as well as timelike and longitudinal
components of the metric field in our formalism, this is not a trivial
problem.

In analysing the unitarity, it is enough to take account of asymptotic fields
of all the fundamental fields and the free part of the Lagrangian. Let us
first assume the asymptotic fields as

$\displaystyle g_{\mu\nu}$ $\displaystyle=$
$\displaystyle\eta_{\mu\nu}+\varphi_{\mu\nu},\qquad\phi=\phi_{0}+\tilde{\phi},\qquad
b_{\mu}=\beta_{\mu},\qquad B=\beta,$ $\displaystyle c^{\mu}$ $\displaystyle=$
$\displaystyle\gamma^{\mu},\qquad\bar{c}_{\mu}=\bar{\gamma}_{\mu},\qquad
c=\gamma,\qquad\bar{c}=\bar{\gamma},$ (6.2)

where $\eta_{\mu\nu}(=\eta^{\mu\nu})$ is the flat Minkowski metric with the
mostly positive signature and $\phi_{0}$ is a constant. In this section, the
Minkowski metric is used to lower or raise the Lorentz indices. Using these
asymptotic fields, the free part of the Lagrangian reads

$\displaystyle{\cal L}_{q}$ $\displaystyle=$
$\displaystyle\frac{1}{12}\phi_{0}^{2}\left(\frac{1}{4}\varphi_{\mu\nu}\Box\varphi^{\mu\nu}-\frac{1}{4}\varphi\Box\varphi-\frac{1}{2}\varphi^{\mu\nu}\partial_{\mu}\partial_{\rho}\varphi_{\nu}{}^{\rho}+\frac{1}{2}\varphi^{\mu\nu}\partial_{\mu}\partial_{\nu}\varphi\right)$
(6.3) $\displaystyle+$
$\displaystyle\frac{1}{6}\phi_{0}\tilde{\phi}\left(-\Box\varphi+\partial_{\mu}\partial_{\nu}\varphi^{\mu\nu}\right)+\frac{1}{2}\partial_{\mu}\tilde{\phi}\partial^{\mu}\tilde{\phi}-i\phi_{0}^{2}\partial_{\mu}\bar{\gamma}_{\rho}\partial^{\mu}\gamma^{\rho}$
$\displaystyle-$
$\displaystyle\left(2\eta^{\mu\nu}\phi_{0}\tilde{\phi}-\phi_{0}^{2}\varphi^{\mu\nu}+\frac{1}{2}\phi_{0}^{2}\eta^{\mu\nu}\varphi\right)\partial_{\mu}\beta_{\nu}$
$\displaystyle+$
$\displaystyle\phi_{0}\partial_{\mu}\beta\partial^{\mu}\tilde{\phi}-i\phi_{0}^{2}\partial_{\mu}\bar{\gamma}\partial^{\mu}\gamma,$

where $\Box\equiv\eta^{\mu\nu}\partial_{\mu}\partial_{\nu}$ and
$\varphi\equiv\eta^{\mu\nu}\varphi_{\mu\nu}$. Based on this Lagrangian, it is
easy to derive the linearized field equations:

$\displaystyle\frac{1}{12}\phi_{0}\biggl{(}\frac{1}{2}\Box\varphi_{\mu\nu}-\frac{1}{2}\eta_{\mu\nu}\Box\varphi-\partial_{\rho}\partial_{(\mu}\varphi_{\nu)}{}^{\rho}+\frac{1}{2}\partial_{\mu}\partial_{\nu}\varphi+\frac{1}{2}\eta_{\mu\nu}\partial_{\rho}\partial_{\sigma}\varphi^{\rho\sigma}\biggr{)}$
$\displaystyle+\frac{1}{6}\left(-\eta_{\mu\nu}\Box+\partial_{\mu}\partial_{\nu}\right)\tilde{\phi}+\phi_{0}\partial_{(\mu}\beta_{\nu)}-\frac{1}{2}\phi_{0}\eta_{\mu\nu}\partial_{\rho}\beta^{\rho}=0.$
(6.4)
$\displaystyle\frac{1}{6}(\Box\varphi-\partial_{\mu}\partial_{\nu}\varphi^{\mu\nu})+2\partial_{\rho}\beta^{\rho}+\Box\beta=0.$
(6.5)
$\displaystyle\partial_{\mu}\tilde{\phi}-\frac{1}{2}\phi_{0}(\partial^{\nu}\varphi_{\mu\nu}-\frac{1}{2}\partial_{\mu}\varphi)=0.$
(6.6)
$\displaystyle\Box\tilde{\phi}=\Box\gamma^{\mu}=\Box\bar{\gamma}_{\mu}=\Box\gamma=\Box\bar{\gamma}=0.$
(6.7)

Here we have introduced the symmetrization notation
$A_{(\mu}B_{\nu)}\equiv\frac{1}{2}(A_{\mu}B_{\nu}+A_{\nu}B_{\mu})$. Now,
operating $\partial^{\mu}$ on Eq. (6.6) and using Eq. (6.7), we can obtain

$\displaystyle\partial_{\mu}\partial_{\nu}\varphi^{\mu\nu}-\frac{1}{2}\Box\varphi=0.$
(6.8)

Next, taking the trace of Eq. (6.4) with the help of Eqs. (6.7) and (6.8)
leads to

$\displaystyle\Box\varphi+24\partial_{\rho}\beta^{\rho}=0.$ (6.9)

Then, with the help of Eqs. (6.8) ans (6.9), Eq. (6.5) can be rewritten as

$\displaystyle\Box\beta=0.$ (6.10)

Moreover, acting $\partial^{\mu}$ on Eq. (6.4) yields

$\displaystyle\Box\beta_{\mu}=0.$ (6.11)

Finally, using various equations obtained thus far, Eq. (6.4) is reduced to
the form:

$\displaystyle\Box\varphi_{\mu\nu}+24\partial_{(\mu}\beta_{\nu)}=0,$ (6.12)

which means that the field $\varphi_{\mu\nu}$ is a dipole field:

$\displaystyle\Box^{2}\varphi_{\mu\nu}=0.$ (6.13)

On the other hand, the other fields are all simple pole fields:

$\displaystyle\Box\tilde{\phi}=\Box\beta_{\mu}=\Box\beta=\Box\gamma^{\mu}=\Box\bar{\gamma}_{\mu}=\Box\gamma=\Box\bar{\gamma}=0.$
(6.14)

Note that Eq. (6.14) corresponds to Eq. (3.19) in a curved space-time.

Following the standard technique, we can calculate the four-dimensional
(anti-)commutation relations (4D CRs) between asymptotic fields. The point is
that the simple pole fields, for instance, the Nakanishi-Lautrup field
$\beta(x)$ can be expressed in terms of the invariant delta function $D(x)$ as

$\displaystyle\beta_{\mu}(x)=\int
d^{3}zD(x-z)\overleftrightarrow{\partial}_{0}^{z}\beta_{\mu}(z),$ (6.15)

whereas the dipole field $\varphi_{\mu\nu}(x)$ can be done as

$\displaystyle\varphi_{\mu\nu}(x)=\int
d^{3}z\left[D(x-z)\overleftrightarrow{\partial}_{0}^{z}\varphi_{\mu\nu}(z)+E(x-z)\overleftrightarrow{\partial}_{0}^{z}\Box\varphi_{\mu\nu}(z)\right]$
$\displaystyle=\int
d^{3}z\left[D(x-z)\overleftrightarrow{\partial}_{0}^{z}\varphi_{\mu\nu}(z)-24E(x-z)\overleftrightarrow{\partial}_{0}^{z}\partial_{(\mu}\beta_{\nu)}(z)\right],$
(6.16)

where in the last equality we have used Eq. (6.12). Here the invariant delta
function $D(x)$ for massless simple pole fields and its properties are
described as

$\displaystyle D(x)=-\frac{i}{(2\pi)^{3}}\int
d^{4}k\,\epsilon(k_{0})\delta(k^{2})e^{ikx},\qquad\Box D(x)=0,$ $\displaystyle
D(-x)=-D(x),\qquad
D(0,\vec{x})=0,\qquad\partial_{0}D(0,\vec{x})=\delta^{3}(x),$ (6.17)

where $\epsilon(k_{0})\equiv\frac{k_{0}}{|k_{0}|}$. Similarly, the invariant
delta function $E(x)$ for massless dipole fields and its properties are given
by

$\displaystyle E(x)=-\frac{i}{(2\pi)^{3}}\int
d^{4}k\,\epsilon(k_{0})\delta^{\prime}(k^{2})e^{ikx},\qquad\Box E(x)=D(x),$
$\displaystyle E(-x)=-E(x),\qquad
E(0,\vec{x})=\partial_{0}E(0,\vec{x})=\partial_{0}^{2}E(0,\vec{x})=0,$
$\displaystyle\partial_{0}^{3}E(0,\vec{x})=-\delta^{3}(x).$ (6.18)

It is easy to show that the RHS of Eqs. (6.15) and (6.16) is independent of
$z^{0}$. Thus, for instance, when we evaluate the four-dimensional commutation
relation $[\varphi_{\mu\nu}(x),\varphi_{\sigma\tau}(y)]$, we can put
$z^{0}=y^{0}$ and use the three-dimensional commutation relations among
asymptotic fields. The resultant 4D CRs are summarized as

$\displaystyle[\varphi_{\mu\nu}(x),\varphi_{\sigma\tau}(y)]=12i\phi_{0}^{-2}[(\eta_{\mu\nu}\eta_{\sigma\tau}-\eta_{\mu\sigma}\eta_{\nu\tau}-\eta_{\mu\tau}\eta_{\nu\sigma})D(x-y)$
$\displaystyle+(\eta_{\mu\sigma}\partial_{\nu}\partial_{\tau}+\eta_{\nu\sigma}\partial_{\mu}\partial_{\tau}+\eta_{\mu\tau}\partial_{\nu}\partial_{\sigma}+\eta_{\nu\tau}\partial_{\mu}\partial_{\sigma})E(x-y)],$
(6.19)
$\displaystyle[\varphi_{\mu\nu}(x),\beta_{\rho}(y)]=i\phi_{0}^{-2}(\eta_{\mu\rho}\partial_{\nu}+\eta_{\nu\rho}\partial_{\mu})D(x-y).$
(6.20)
$\displaystyle[\varphi_{\mu\nu}(x),\beta(y)]=-2i\phi_{0}^{-1}\eta_{\mu\nu}D(x-y).$
(6.21) $\displaystyle[\tilde{\phi}(x),\beta(y)]=i\phi_{0}^{-1}D(x-y).$ (6.22)
$\displaystyle\\{\gamma^{\sigma}(x),\bar{\gamma}_{\tau}(y)\\}=-\phi_{0}^{-2}\delta_{\tau}^{\sigma}D(x-y).$
(6.23) $\displaystyle\\{\gamma(x),\bar{\gamma}(y)\\}=-\phi_{0}^{-2}D(x-y).$
(6.24)

The other 4D CRs vanish identically.

Now we would like to discuss the issue of the unitarity of the theory in hand.
To do that, it is convenient to perform the Fourier transformation of Eqs.
(6.19)-(6.24). However, for the dipole field we cannot use the three-
dimensional Fourier expansion to define the creation and annihilation
operators. We therefore make use of the four-dimensional Fourier expansion
[5]:999The Fourier transform of a field is denoted by the same field except
for the argument $x$ or $p$, for simplicity.

$\displaystyle\varphi_{\mu\nu}(x)=\frac{1}{(2\pi)^{\frac{3}{2}}}\int
d^{4}p\,\theta(p_{0})[\varphi_{\mu\nu}(p)e^{ipx}+\varphi_{\mu\nu}^{\dagger}(p)e^{-ipx}],$
(6.25)

where $\theta(p_{0})$ is the step function. For any simple pole fields, we
adopt the same Fourier expansion, for instance,

$\displaystyle\beta_{\mu}(x)=\frac{1}{(2\pi)^{\frac{3}{2}}}\int
d^{4}p\,\theta(p_{0})[\beta_{\mu}(p)e^{ipx}+\beta_{\mu}^{\dagger}(p)e^{-ipx}].$
(6.26)

Incidentally, for a generic simple pole field $\Phi$, the three-dimensional
Fourier expansion is defined as

$\displaystyle\Phi(x)=\frac{1}{(2\pi)^{\frac{3}{2}}}\int
d^{3}p\,\frac{1}{\sqrt{2|\vec{p}|}}[\Phi(\vec{p})e^{ipx}+\Phi^{\dagger}(\vec{p})e^{-ipx}],$
(6.27)

where the on-shell relation $p_{0}=|\vec{p}|$ must be satisfied ($\vec{p}$
denotes the three-dimensional momentum) whereas the four-dimensional Fourier
expansion reads

$\displaystyle\Phi(x)=\frac{1}{(2\pi)^{\frac{3}{2}}}\int
d^{4}p\,\theta(p_{0})[\Phi(p)e^{ipx}+\Phi^{\dagger}(p)(p)e^{-ipx}].$ (6.28)

Thus, the annihilation operator $\Phi(p)$ in the four-dimensional Fourier
expansion has connection with the annihilation operator $\Phi(\vec{p})$ in the
three-dimensional Fourier expansion like

$\displaystyle\Phi(p)=\theta(p_{0})\delta(p^{2})\sqrt{2|\vec{p}|}\Phi(\vec{p}).$
(6.29)

Based on these Fourier expansions, we can calculate the Fourier transform of
Eqs. (6.19)-(6.24):

$\displaystyle[\varphi_{\mu\nu}(p),\varphi_{\sigma\tau}^{\dagger}(q)]=12\phi_{0}^{-2}\theta(p_{0})\delta^{4}(p-q)[\delta(p^{2})(\eta_{\mu\nu}\eta_{\sigma\tau}-\eta_{\mu\sigma}\eta_{\nu\tau}-\eta_{\mu\tau}\eta_{\nu\sigma})$
$\displaystyle-3\delta^{\prime}(p^{2})(\eta_{\mu\sigma}p_{\nu}p_{\tau}+\eta_{\nu\sigma}p_{\mu}p_{\tau}+\eta_{\mu\tau}p_{\nu}p_{\sigma}+\eta_{\nu\tau}p_{\mu}p_{\sigma})].$
(6.30)
$\displaystyle[\varphi_{\mu\nu}(p),\beta_{\rho}^{\dagger}(q)]=i\phi_{0}^{-2}(\eta_{\mu\rho}p_{\nu}+\eta_{\nu\rho}p_{\mu})\theta(p_{0})\delta(p^{2})\delta^{4}(p-q).$
(6.31)
$\displaystyle[\varphi_{\mu\nu}(p),\beta^{\dagger}(q)]=-2\phi_{0}^{-1}\eta_{\mu\nu}\theta(p_{0})\delta(p^{2})\delta^{4}(p-q).$
(6.32)
$\displaystyle[\tilde{\phi}(p),\beta^{\dagger}(q)]=\phi_{0}^{-1}\theta(p_{0})\delta(p^{2})\delta^{4}(p-q).$
(6.33)
$\displaystyle\\{\gamma^{\sigma}(p),\bar{\gamma}^{\dagger}_{\tau}(q)\\}=i\phi_{0}^{-2}\delta_{\tau}^{\sigma}\theta(p_{0})\delta(p^{2})\delta^{4}(p-q).$
(6.34)
$\displaystyle\\{\gamma(p),\bar{\gamma}^{\dagger}(q)\\}=i\phi_{0}^{-2}\theta(p_{0})\delta(p^{2})\delta^{4}(p-q).$
(6.35)

Next, let us turn our attention to the linearized field equations. In the
Fourier transformation, Eq. (6.6) takes the form:

$\displaystyle
p^{\nu}\varphi_{\mu\nu}-\frac{1}{2}p_{\mu}\varphi=2\phi_{0}^{-1}p_{\mu}\tilde{\phi}.$
(6.36)

If we fix the degree of freedom associated with $\tilde{\phi}$, which will be
discussed later, this equation gives us four independent relations in ten
components of $\varphi_{\mu\nu}(p)$, thereby reducing the independent
components of $\varphi_{\mu\nu}(p)$ to be six. To deal with six independent
components of $\varphi_{\mu\nu}(p)$, it is convenient to take a specific
Lorentz frame such that $p_{1}=p_{2}=0$ and $p_{3}>0$, and choose the six
components as follows:

$\displaystyle\varphi_{1}(p)=\frac{1}{2}[\varphi_{11}(p)-\varphi_{22}(p)],\qquad\varphi_{2}(p)=\varphi_{12}(p),\qquad\omega_{0}(p)=-\frac{1}{2p_{0}}\varphi_{00}(p),$
$\displaystyle\omega_{I}(p)=-\frac{1}{p_{0}}\varphi_{0I}(p),\qquad\omega_{3}(p)=-\frac{1}{2p_{3}}\varphi_{33}(p),$
(6.37)

where the index $I$ takes the transverse components $I=1,2$.

In this respect, it is worthwhile to consider the GCT BRST transformation for
these components. First, let us write down the GCT BRST transformation for the
Fourier expansion of the asymptotic fields, which reads

$\displaystyle\delta_{B}\varphi_{\mu\nu}(p)=-i[p_{\mu}\gamma_{\nu}(p)+p_{\nu}\gamma_{\mu}(p)],\quad\delta_{B}\gamma^{\mu}(p)=0,\quad\delta_{B}\bar{\gamma}_{\mu}(p)=i\beta_{\mu}(p),$
$\displaystyle\delta_{B}\tilde{\phi}(p)=\delta_{B}\beta_{\mu}(p)=\delta_{B}\beta(p)=\delta_{B}\gamma(p)=\delta_{B}\bar{\gamma}(p)=0.$
(6.38)

Using this BRST transformation, the GCT BRST transformation for the components
in (6.37) takes the form:

$\displaystyle\delta_{B}\varphi_{I}(p)=0,\qquad\delta_{B}\omega_{\mu}(p)=i\gamma_{\mu}(p),$
$\displaystyle\delta_{B}\bar{\gamma}_{\mu}(p)=i\beta_{\mu}(p),\qquad\delta_{B}\gamma_{\mu}(p)=\delta_{B}\beta_{\mu}(p)=0,$
(6.39)

where $p_{1}=p_{2}=0$ was used. This BRST transformation implies that
$\varphi_{I}(p)$ could be the physical observable while a set of fields,
$\\{\omega_{\mu}(p),\beta_{\mu}(p),\gamma_{\mu}(p),\bar{\gamma}_{\mu}(p)\\}$
might belong to the BRST quartet, which are dropped from the physical state by
the Kugo-Ojima subsidiary condition, $Q_{B}|\rm{phys}\rangle=0$ [8]. However,
note that $\beta_{\mu}(p),\gamma_{\mu}(p)$ and $\bar{\gamma}_{\mu}(p)$ are
simple pole fields obeying
$p^{2}\beta_{\mu}(p)=p^{2}\gamma_{\mu}(p)=p^{2}\bar{\gamma}_{\mu}(p)=0$, but
$\varphi_{\mu\nu}(p)$ is a dipole field satisfying
$(p^{2})^{2}\varphi_{\mu\nu}(p)=0$, so that a naive Kugo-Ojima’s quartet
mechanism does not work.

To clarify the BRST quartet mechanism, let us calculate their 4D CRs. From
Eqs. (6.30)-(6.35) and the definition (6.37), it is straightforward to derive
the following 4D CRs:

$\displaystyle[\varphi_{I}(p),\varphi_{J}^{\dagger}(q)]=-12\phi_{0}^{-2}\delta_{IJ}\theta(p_{0})\delta(p^{2})\delta^{4}(p-q).$
(6.40)
$\displaystyle[\varphi_{I}(p),\omega_{\mu}^{\dagger}(q)]=[\varphi_{I}(p),\beta_{\mu}^{\dagger}(q)]=[\beta_{\mu}(p),\beta_{\nu}^{\dagger}(q)]=0.$
(6.41)
$\displaystyle[\omega_{\mu}(p),\beta_{\nu}^{\dagger}(q)]=-i\phi_{0}^{-2}\eta_{\mu\nu}\theta(p_{0})\delta(p^{2})\delta^{4}(p-q).$
(6.42)
$\displaystyle\\{\gamma_{\mu}(p),\bar{\gamma}_{\nu}^{\dagger}(q)\\}=i\phi_{0}^{-2}\eta_{\mu\nu}\theta(p_{0})\delta(p^{2})\delta^{4}(p-q).$
(6.43)

In addition to them, we have a bit complicated expression for
$[\omega_{\mu}(p),\omega_{\nu}^{\dagger}(q)]$ because $\varphi_{\mu\nu}(p)$ is
a dipole field, but luckily enough this expression is not necessary for our
aim [8]. It is known how to take out a simple pole field from a dipole field,
which amounts to using an operator defined by [8]

$\displaystyle{\cal
D}_{p}=\frac{1}{2|\vec{p}|^{2}}p_{0}\frac{\partial}{\partial p_{0}}+c,$ (6.44)

where $c$ is a constant. Using this operator, we can define a simple pole
field $\hat{\varphi}_{\mu\nu}(p)$ from the dipole field $\varphi_{\mu\nu}(p)$,
which obeys $(p^{2})^{2}\varphi_{\mu\nu}(p)=0$, as

$\displaystyle\hat{\varphi}_{\mu\nu}(p)$ $\displaystyle\equiv$
$\displaystyle\varphi_{\mu\nu}(p)-{\cal D}_{p}p^{2}\varphi_{\mu\nu}(p)$ (6.45)
$\displaystyle=$ $\displaystyle\varphi_{\mu\nu}(p)-24i{\cal
D}_{p}p_{(\mu}\beta_{\nu)}(p),$

where in the last equality we have used the Fourier transform of the
linearized field equation (6.12). It is then easy to verify the equation:

$\displaystyle p^{2}\hat{\varphi}_{\mu\nu}(p)=0.$ (6.46)

Then, in (6.37) we replace $\varphi_{\mu\nu}$ of $\omega_{\mu}$ with
$\hat{\varphi}_{\mu\nu}$, and we redefine $\omega_{\mu}$ by
$\hat{\omega}_{\mu}$ as

$\displaystyle\hat{\omega}_{0}(p)=-\frac{1}{2p_{0}}\hat{\varphi}_{00}(p),\quad\hat{\omega}_{I}(p)=-\frac{1}{p_{0}}\hat{\varphi}_{0I}(p),\quad\hat{\omega}_{3}(p)=-\frac{1}{2p_{3}}\hat{\varphi}_{33}(p).$
(6.47)

The key point is that with this redefinition from $\omega_{\mu}$ to
$\hat{\omega}_{\mu}$, the BRST transformation and the 4D CRs remain unchanged
owing to $\delta_{B}\beta_{\mu}=0$ and
$[\beta_{\mu}(p),\beta^{\dagger}_{\nu}(q)]=[\varphi_{I}(p),\beta^{\dagger}_{\mu}(q)]=0$,
those are,

$\displaystyle\delta_{B}\hat{\omega}_{\mu}(p)=i\gamma_{\mu}(p),\quad[\hat{\omega}_{\mu}(p),\beta^{\dagger}_{\nu}(q)]=[\omega_{\mu}(p),\beta^{\dagger}_{\nu}(q)],$
$\displaystyle[\varphi_{I}(p),\hat{\omega}^{\dagger}_{\mu}(q)]=[\varphi_{I}(p),\omega^{\dagger}_{\mu}(q)].$
(6.48)

Now it turns out that all the fields,
$\\{\varphi_{I},\hat{\omega}_{\mu},\beta_{\mu},\gamma_{\mu},\bar{\gamma}_{\mu}\\}$
are simple pole fields.101010Without the redefinition, $\varphi_{I}(p)$ is
already a simple pole field as can be seen in Eq. (6.40). Since all the fields
become simple pole fields, we can obtain the standard creation and
annihilation operators in the three-dimensional Fourier expansion from those
in the four-dimensional one through the relation (6.29). As a result, the
three-dimensional (anti-)commutation relations, which are denoted as
$[\Phi(\vec{p}),\Phi^{\dagger}(\vec{q})\\}$ with
$\Phi(\vec{p})\equiv\\{\varphi_{I}(\vec{p}),\hat{\omega}_{\mu}(\vec{p}),\beta_{\mu}(\vec{p}),\gamma_{\mu}(\vec{p}),\bar{\gamma}_{\mu}(\vec{p})\\}$,
are given by111111The bracket $[A,B\\}$ is the graded commutation relation
denoting either commutator or anti-commutator, according to the Grassmann-even
or odd character of $A$ and $B$, i.e., $[A,B\\}=AB-(-)^{|A||B|}BA$.

$\displaystyle[\Phi(\vec{p}),\Phi^{\dagger}(\vec{q})\\}$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{cc|cc|cc}-12\phi_{0}^{-2}\delta_{IJ}&&&&\\\
\hline\cr&&[\hat{\omega}_{\mu}(\vec{p}),\hat{\omega}_{\nu}^{\dagger}(\vec{q})]&-i\phi_{0}^{-2}\eta_{\mu\nu}&\\\
&&i\phi_{0}^{-2}\eta_{\mu\nu}&0&\\\
\hline\cr&&&&&i\phi_{0}^{-2}\eta_{\mu\nu}\\\
&&&&-i\phi_{0}^{-2}\eta_{\mu\nu}&\\\ \end{array}\right)$ (6.55)
$\displaystyle\times\delta(\vec{p}-\vec{q})_{.}$

The (anti-)commuatation relations (6.55) have in essence the same structure as
those of the Yang-Mills theory [8]. Hence, we find that $\varphi_{I}$ could be
the physical observable while a set of fields
$\\{\hat{\omega}_{\mu},\beta_{\mu},\gamma_{\mu},\bar{\gamma}_{\mu}\\}$ belongs
to the BRST quartet.

Next, let us move on to another BRST transformation, which is the BRST
transformation for the Weyl transformation. The Weyl BRST transformation for
the asymptotic fields is of form:

$\displaystyle\bar{\delta}_{B}\varphi_{\mu\nu}=2c\eta_{\mu\nu},\quad\bar{\delta}_{B}\tilde{\phi}=-\phi_{0}\gamma,\quad\bar{\delta}_{B}\gamma=0,\quad\bar{\delta}_{B}\bar{\gamma}=i\beta,$
$\displaystyle\bar{\delta}_{B}\beta=\bar{\delta}_{B}\beta_{\mu}=\bar{\delta}_{B}\gamma_{\mu}=\bar{\delta}_{B}\bar{\gamma}_{\mu}=0.$
(6.56)

The Weyl BRST transformation of $\varphi_{I}$ is vanishing:

$\displaystyle\bar{\delta}_{B}\varphi_{I}=0,$ (6.57)

which means that together with $\delta_{B}\varphi_{I}=0$, $\varphi_{I}$ is
truely the physical observable. The four-dimensional (anti-)commutation
relations among the fields $\\{\tilde{\phi},\beta,\gamma,\bar{\gamma}\\}$ read

$\displaystyle[\tilde{\phi}(p),\tilde{\phi}^{\dagger}(q)]=0,$
$\displaystyle[\tilde{\phi}(p),\beta^{\dagger}(q)]=\phi_{0}^{-1}\theta(p_{0})\delta(p^{2})\delta^{4}(p-q),$
$\displaystyle\\{\gamma(p),\bar{\gamma}^{\dagger}(q)]=i\phi_{0}^{-2}\theta(p_{0})\delta(p^{2})\delta^{4}(p-q).$
(6.58)

As can be also seen in these 4D CRs, all the fields
$\\{\varphi_{I},\tilde{\phi},\beta,\gamma,\bar{\gamma}\\}$ are massless simple
pole fields. Via the relation (6.29), the three-dimensional (anti-)commutation
relations $[\Phi(\vec{p}),\Phi^{\dagger}(\vec{q})\\}$ with
$\Phi(\vec{p})\equiv\\{\varphi_{I}(\vec{p}),\tilde{\phi}(\vec{p}),\beta(\vec{p}),\gamma(\vec{p}),\bar{\gamma}(\vec{p})\\}$,
are of form

$\displaystyle[\Phi(\vec{p}),\Phi^{\dagger}(\vec{q})\\}=\left(\begin{array}[]{cc|cc|cc}-12\phi_{0}^{-2}\delta_{IJ}&&&&\\\
\hline\cr&&0&\phi_{0}^{-1}&\\\ &&\phi_{0}^{-1}&0&\\\
\hline\cr&&&&&i\phi_{0}^{-2}\\\ &&&&-i\phi_{0}^{-2}&\\\
\end{array}\right)\delta(\vec{p}-\vec{q})_{.}$ (6.64)

Thus, $\varphi_{I}$ is the physical observable while a set of fields,
$\\{\tilde{\phi},\beta,\gamma,\bar{\gamma}\\}$ consists of the BRST quartet
and is the unphysical mode by the Kugo-Ojima’s subsidiary condition [8]. Here
it is worth mentioning that the ghost-like scalar field $\phi$ belongs to the
unphysical mode so together with the result obtained in the analysis of the
GCT BRST cohomology, the physical S-matrix is found to be unitary.

##  7 Choral symmetry

As mentioned in Section 3, a set of fields (including the space-time
coordinates $x^{\mu}$)
$X^{M}\equiv\\{x^{\mu},b_{\mu},\sigma,B,c^{\mu},\bar{c}_{\mu},c,\bar{c}\\}$
obeys a very simple equation:

$\displaystyle g^{\mu\nu}\partial_{\mu}\partial_{\nu}X^{M}=0.$ (7.1)

This equation holds if and only if we adopt the extended de Donder gauge and
the new scalar gauge as gauge-fixing conditions for the GCT and the Weyl
transformation, respectively. The existence of this simple equation suggests
that there could be many of conserved currents defined in Eq. (3.20). In this
section, we shall show explicitly that there exist such currents and we have a
huge global symmetry called choral symmetry, which is the $IOSp(10|10)$ in the
present theory.

Let us start with the Lagrangian (3.12), which can be cast to the form:

$\displaystyle{\cal
L}_{q}=\tilde{g}^{\mu\nu}\phi^{2}\left(\frac{1}{12}R_{\mu\nu}-\frac{1}{2}\hat{E}_{\mu\nu}\right).$
(7.2)

Here note that $\tilde{g}^{\mu\nu}\phi^{2}$ is a Weyl invariant metric and the
Ricci tensor is invariant under only a global scale transformation. We can
further rewrite it into the form:

$\displaystyle{\cal L}_{q}$ $\displaystyle=$
$\displaystyle\tilde{g}^{\mu\nu}\phi^{2}\left(\frac{1}{12}R_{\mu\nu}-\frac{1}{2}\eta_{NM}\partial_{\mu}X^{M}\partial_{\nu}X^{N}\right)$
(7.3) $\displaystyle=$
$\displaystyle\tilde{g}^{\mu\nu}\phi^{2}\left(\frac{1}{12}R_{\mu\nu}-\frac{1}{2}\partial_{\mu}X^{M}\tilde{\eta}_{MN}\partial_{\nu}X^{N}\right),$

where we have introduced an $OSp(10|10)$ metric
$\eta_{NM}=\eta_{MN}^{T}\equiv\tilde{\eta}_{MN}$ defined as [9]

$\displaystyle\eta_{NM}=\tilde{\eta}_{MN}=\left(\begin{array}[]{cc|cc|cc|cc}&\delta_{\mu}^{\nu}&&&&\\\
\delta^{\mu}_{\nu}&&&&&\\\ \hline\cr&&-1&-1&&&\\\ &&-1&0&&&\\\
\hline\cr&&&&&-i\delta_{\mu}^{\nu}&&\\\ &&&&i\delta^{\mu}_{\nu}&&&\\\
\hline\cr&&&&&&&-i\\\ &&&&&&i&\\\ \end{array}\right)_{.}$ (7.12)

Let us note that this $OSp(10|10)$ metric $\eta_{NM}$, which is a c-number
quantity, has the symmetry property such that

$\displaystyle\eta_{MN}=(-)^{|M|\cdot|N|}\eta_{NM}=(-)^{|M|}\eta_{NM}=(-)^{|N|}\eta_{NM},$
(7.13)

where the statistics index $|M|$ is 0 or 1 when $X^{M}$ is Grassmann-even or
Grassmann-odd, respectively. This property comes from the fact that
$\eta_{MN}$ is ‘diagonal’ in the sense that its off-diagonal, Grassmann-even
and Grassmann-odd, and vice versa, matrix elements vanish, i.e., $\eta_{MN}=0$
when $|M|\neq|N|$, thereby being $|M|=|N|=|M|\cdot|N|$ in front of $\eta_{MN}$
[9].

Now that the quantum Lagrangian (7.3) is expressed in a manifestly
$IOSp(10|10)$ invariant form except for the Weyl invariant metric
$\tilde{g}^{\mu\nu}\phi^{2}$, which will be discussed later, there could exist
an $IOSp(10|10)$ as a global symmetry in our theory. Let us show this fact
first. The infinitesimal $OSp$ rotation is defined by

$\displaystyle\delta
X^{M}=\eta^{ML}\varepsilon_{LN}X^{N}\equiv\varepsilon^{M}{}_{N}X^{N},$ (7.14)

where $\eta^{MN}$ is the inverse matrix of $\eta_{MN}$, and the infinitesimal
parameter $\varepsilon_{MN}$ has the following properties:

$\displaystyle\varepsilon_{MN}=(-)^{1+|M|\cdot|N|}\varepsilon_{NM},\qquad\varepsilon_{MN}X^{L}=(-)^{|L|(|M|+|N|)}X^{L}\varepsilon_{MN}.$
(7.15)

Moreover, in order to find the conserved current, we assume that the
infinitesimal parameter $\varepsilon_{MN}$ depends on the space-time
coordinates $x^{\mu}$, i.e., $\varepsilon_{MN}=\varepsilon_{MN}(x^{\mu})$.

Assuming for a while that the metric $\tilde{g}^{\mu\nu}\phi^{2}$ and
$R_{\mu\nu}$ is invariant, the infinitesimal variation of the quantum
Lagrangian (7.3) under the $OSp$ rotation (7.14) is given by

$\displaystyle\delta{\cal
L}_{q}=-\tilde{g}^{\mu\nu}\phi^{2}\left(\partial_{\mu}\varepsilon_{NM}X^{M}\partial_{\nu}X^{N}+\varepsilon_{NM}\partial_{\mu}X^{M}\partial_{\nu}X^{N}\right).$
(7.16)

It is easy to prove that the second term on the RHS vanishes owing to the
first property in Eq. (7.15). Thus, ${\cal L}_{q}$ is invariant under the
infinitesimal $OSp$ rotation. The conserved current is then calculated as

$\displaystyle\delta{\cal L}_{q}$ $\displaystyle=$
$\displaystyle-\tilde{g}^{\mu\nu}\phi^{2}\partial_{\mu}\varepsilon_{NM}X^{M}\partial_{\nu}X^{N}$
(7.17) $\displaystyle=$
$\displaystyle-\frac{1}{2}\tilde{g}^{\mu\nu}\phi^{2}\partial_{\mu}\varepsilon_{NM}\left[X^{M}\partial_{\nu}X^{N}-(-)^{|M|\cdot|N|}X^{N}\partial_{\nu}X^{M}\right]$
$\displaystyle=$
$\displaystyle-\frac{1}{2}\tilde{g}^{\mu\nu}\phi^{2}\partial_{\mu}\varepsilon_{NM}\left(X^{M}\partial_{\nu}X^{N}-\partial_{\nu}X^{M}X^{N}\right)$
$\displaystyle=$
$\displaystyle-\frac{1}{2}\tilde{g}^{\mu\nu}\phi^{2}\partial_{\mu}\varepsilon_{NM}X^{M}\overset{\leftrightarrow}{\partial}_{\nu}X^{N}$
$\displaystyle\equiv$
$\displaystyle-\frac{1}{2}\partial_{\mu}\varepsilon_{NM}{\cal M}^{\mu MN},$

from which the conserved current ${\cal M}^{\mu MN}$ for the $OSp$ rotation
takes the form:

$\displaystyle{\cal M}^{\mu
MN}=\tilde{g}^{\mu\nu}\phi^{2}X^{M}\overset{\leftrightarrow}{\partial}_{\nu}X^{N}.$
(7.18)

In a similar way, we can derive the conserved current for the infinitesimal
translation

$\displaystyle\delta X^{M}=\varepsilon^{M},$ (7.19)

where $\varepsilon^{M}$ is the infinitesimal parameter and assume that it is a
local one for deriving the corresponding conserved current. Indeed, assuming
again that the metric $\tilde{g}^{\mu\nu}\phi^{2}$ and $R_{\mu\nu}$ are
invariant under the translation, we can show that ${\cal L}_{q}$ is invariant
under the infinitesimal translation as follows:

$\displaystyle\delta{\cal L}_{q}$ $\displaystyle=$
$\displaystyle-\tilde{g}^{\mu\nu}\phi^{2}\eta_{NM}\partial_{\mu}\varepsilon^{M}\partial_{\nu}X^{N}$
(7.20) $\displaystyle=$
$\displaystyle-\tilde{g}^{\mu\nu}\phi^{2}\partial_{\mu}\varepsilon_{N}\partial_{\nu}X^{N}$
$\displaystyle\equiv$ $\displaystyle-\partial_{\mu}\varepsilon_{M}{\cal
P}^{\mu M},$

which implies that the conserved current ${\cal P}^{\mu M}$ for the
translation reads

$\displaystyle{\cal P}^{\mu
M}=\tilde{g}^{\mu\nu}\phi^{2}\partial_{\nu}X^{M}=\tilde{g}^{\mu\nu}\phi^{2}\left(1\overset{\leftrightarrow}{\partial}_{\nu}X^{M}\right).$
(7.21)

The above proofs make sense only under the assumption that the metric
$\tilde{g}^{\mu\nu}\phi^{2}$ and $R_{\mu\nu}$ are invariant under the
$IOSp(10|10)$. So the problem reduces to a question: Is this assumption
correct? The answer is obviously not true, but the non-invariant terms can be
compensated by a suitable Weyl transformation. To show this fact, let us
consider only the case of the infinitesimal $OSp$ rotation since we can treat
the case of the translation in a perfectly similar manner. Under the
infinitesimal $OSp$ rotation (7.14), the dilaton $\sigma(x)$, which is defined
as $\phi=e^{\sigma}$, transforms as

$\displaystyle\delta\sigma=\eta^{\sigma
L}\varepsilon_{LN}X^{N}=-\varepsilon_{BN}X^{N},$ (7.22)

where we have used (7.12) and

$\displaystyle\begin{pmatrix}-1&-1\\\
-1&0\end{pmatrix}^{-1}=\begin{pmatrix}0&-1\\\ -1&1\end{pmatrix},$ (7.23)

where recall that the matrix $\eta^{ML}$ is the inverse matrix of $\eta_{ML}$.
As for the scalar field $\phi(x)$, this transformation for the dilaton can be
interpreted as a Weyl transformation:

$\displaystyle\phi\rightarrow\phi^{\prime}=e^{\epsilon(x)}\phi,$ (7.24)

where the infinitesimal parameter is defined as
$\epsilon(x)=-\varepsilon_{BN}X^{N}$. This Weyl transformation induces the
Weyl transformation for the metric tensor field at the same time:

$\displaystyle g_{\mu\nu}\rightarrow
g_{\mu\nu}^{\prime}=e^{-2\epsilon(x)}g_{\mu\nu}.$ (7.25)

Let us recall that the metric $\tilde{g}^{\mu\nu}\phi^{2}$ is the Weyl
invariant metric so that it is invariant under the Weyl transformation (7.24)
and (7.25). This implies that $\tilde{g}^{\mu\nu}\phi^{2}$ is essentially
invariant under the $OSp$ rotation if an appropriate Weyl transformation is
achieved.

How about $R_{\mu\nu}$? Even if $R_{\mu\nu}$ is not invariant under the Weyl
transformation in itself, this object comes from the classical Lagrangian of
the Weyl invariant scalar-tensor gravity in (3.1), so together with the metric
tensor and the scalar field it essentially becomes invariant under the Weyl
transformation (7.24) and (7.25). Thus, in this sense, $R_{\mu\nu}$ is also
invariant under the $OSp$ rotation. In any case, it is worthwhile to stress
that in the present formulation, the choral symmetry $IOSp(10|10)$ is not a
symmetry of only the FP ghosts and the Nakanishi-Latrup fields but closely
related to classical fields $g_{\mu\nu}$ and $\phi$ which lie in the classical
Lagrangian.

An important remark is relevant to the expression of the conserved currents
(7.18) and (7.21). To make the quantum Lagrangian ${\cal L}_{q}$ be invariant
under the choral symmetry $IOSp(10|10)$, it is necessary to perform the Weyl
transformation (7.24) and (7.25). Then, it is natural to ask if because of
this associated Weyl transformation, the expression of the currents would be
modified or not. Here a miracle happens. As shown in Refs. [10, 11], the
current for the Weyl transformation identically vanishes in the Weyl invariant
scalar-tensor gravity. Thus, although we make the Weyl transformation (7.24)
and (7.25), the conserved currents (7.18) and (7.21) are unchanged.

From the conserved currents (7.18) and (7.21), the corresponding conserved
charges become

$\displaystyle M^{MN}$ $\displaystyle\equiv$ $\displaystyle\int d^{3}x\,{\cal
M}^{0MN}=\int
d^{3}x\,\tilde{g}^{0\nu}\phi^{2}X^{M}\overset{\leftrightarrow}{\partial}_{\nu}X^{N},$
$\displaystyle P^{M}$ $\displaystyle\equiv$ $\displaystyle\int d^{3}x\,{\cal
P}^{0M}=\int d^{3}x\,\tilde{g}^{0\nu}\phi^{2}\partial_{\nu}X^{M}.$ (7.26)

It then turns out that using various ETCRs obtained so far, the $IOSp(10|10)$
generators $\\{M^{MN},P^{M}\\}$ generate an $IOSp(10|10)$ algebra:

$\displaystyle[P^{M},P^{N}\\}=0,$
$\displaystyle[M^{MN},P^{R}\\}=i\bigl{[}P^{M}\tilde{\eta}^{NR}-(-)^{|N||R|}P^{N}\tilde{\eta}^{MR}\bigr{]},$
$\displaystyle[M^{MN},M^{RS}\\}=i\bigl{[}M^{MS}\tilde{\eta}^{NR}-(-)^{|N||R|}M^{MR}\tilde{\eta}^{NS}-(-)^{|N||R|}M^{NS}\tilde{\eta}^{MR}$
$\displaystyle+(-)^{|M||R|+|N||S|}M^{NR}\tilde{\eta}^{MS}\bigr{]}.$ (7.27)

As a final remark, it is worthwhile to point out that all the global
symmetries existing in the present theory are expressed in terms of the
generators of the choral symmetry. For instance, the BRST charges for the GCT
and Weyl transformation are respectively expressed as

$\displaystyle Q_{B}\equiv M(b_{\rho},c^{\rho})=\int
d^{3}x\,\tilde{g}^{0\nu}\phi^{2}b_{\rho}\overset{\leftrightarrow}{\partial}_{\nu}c^{\rho},$
$\displaystyle\bar{Q}_{B}\equiv M(B,c)=\int
d^{3}x\,\tilde{g}^{0\nu}\phi^{2}B\overset{\leftrightarrow}{\partial}_{\nu}c.$
(7.28)

##  8 Gravitational conformal symmetry

Even if we have already fixed the Weyl symmetry by the scalar gauge condition
(3.6), we still have its linearized, residual symmetries. In order to look for
the residual symmetries, it is convenient to take the extended de Donder gauge
(3.5) into consideration simultaneously.121212The same strategy has been
adopted in different theories in Refs. [12, 13, 14]. With the help of the
extended de Donder gauge (3.5), the scalar gauge condition (3.6) can be
rewritten as

$\displaystyle
0=\partial_{\mu}(\tilde{g}^{\mu\nu}\phi\partial_{\nu}\phi)=\partial_{\mu}(\tilde{g}^{\mu\nu}\phi^{2}\partial_{\nu}\sigma)=\tilde{g}^{\mu\nu}\phi^{2}\partial_{\mu}\partial_{\nu}\sigma,$
(8.1)

where we have used the relation between the scalar field and dilaton,
$\phi=e^{\sigma}$. Under the Weyl transformation (3.2) with $\Omega(x)\equiv
e^{\Lambda(x)}$, the dilaton $\sigma$ transforms as

$\displaystyle\sigma\rightarrow\sigma^{\prime}=\sigma-\log\Omega=\sigma-\Lambda,$
(8.2)

where we have used the Weyl transformation (3.2) for the scalar field. Since
$\tilde{g}^{\mu\nu}\phi^{2}$ is a Weyl invariant quantity, the Weyl
transformation makes Eq. (8.1) change to

$\displaystyle
0=\tilde{g}^{\mu\nu}\phi^{2}\partial_{\mu}\partial_{\nu}\sigma\rightarrow
0=\tilde{g}^{\mu\nu}\phi^{2}\partial_{\mu}\partial_{\nu}\sigma^{\prime}=\tilde{g}^{\mu\nu}\phi^{2}\partial_{\mu}\partial_{\nu}(\sigma-\Lambda).$
(8.3)

This equation shows that when we use the extended de Donder gauge, the scalar
gauge condition is still invariant under the Weyl transformation as long as

$\displaystyle g^{\mu\nu}\partial_{\mu}\partial_{\nu}\Lambda=0,$ (8.4)

is satisfied, thereby implying the existence of the residual symmetries [12,
13, 14]. Selecting the coefficients appropriately for later convenience, the
solution to Eq. (8.4) is given by

$\displaystyle\Lambda=\lambda-2k_{\mu}x^{\mu},$ (8.5)

where $\lambda,k_{\mu}$ are constants.131313It is shown in Appendix B that the
transformations associated with the parameters $\lambda$ and $k_{\mu}$,
respectively, correpond to dilatation and special conformal transformation in
a flat Minkowski background.

We can also verify the invariance of the quantum Lagrangian under the residual
symmetries more directly. To do that, let us assume that $\Lambda$, or
equivalently, $\lambda$ and $k_{\mu}$, are the infinitesimal parameters. It
then turns out that the quantum Lagrangian (3.12) is invariant under the
residual symmetries:

$\displaystyle\delta g_{\mu\nu}=2(\lambda-2k_{\rho}x^{\rho})g_{\mu\nu},$
$\displaystyle\delta\sigma=-(\lambda-2k_{\rho}x^{\rho}),\qquad\delta
b_{\mu}=2k_{\mu}B,$ (8.6)

where the other fields are unchanged. The generators corresponding to the
transformation parameters $\lambda$ and $k_{\mu}$ are respectively constructed
out of those of the choral symmetry as

$\displaystyle D_{0}$ $\displaystyle\equiv$ $\displaystyle-P(B)=-\int
d^{3}x\,\tilde{g}^{0\nu}\phi^{2}\partial_{\nu}B,$ $\displaystyle K^{\mu}$
$\displaystyle\equiv$ $\displaystyle 2M^{\mu}(x,B)=2\int
d^{3}x\,\tilde{g}^{0\nu}\phi^{2}x^{\mu}\overset{\leftrightarrow}{\partial}_{\nu}B.$
(8.7)

In addition to the generators $D_{0}$ and $K^{\mu}$, one can construct the
translation generator $P_{\mu}$ and $GL(4)$ generator $G^{\mu}{}_{\nu}$ from
those of the choral symmetry $IOSp(10|10)$ as

$\displaystyle P_{\mu}$ $\displaystyle\equiv$ $\displaystyle P_{\mu}(b)=\int
d^{3}x\,\tilde{g}^{0\nu}\phi^{2}\partial_{\nu}b_{\mu},$ $\displaystyle
G^{\mu}{}_{\nu}$ $\displaystyle\equiv$ $\displaystyle
M^{\mu}{}_{\nu}(x,b)-iM^{\mu}{}_{\nu}(c^{\tau},\bar{c}_{\tau})$ (8.8)
$\displaystyle=$ $\displaystyle\int
d^{3}x\,\tilde{g}^{0\lambda}\phi^{2}(x^{\mu}\overset{\leftrightarrow}{\partial}_{\lambda}b_{\nu}-ic^{\mu}\overset{\leftrightarrow}{\partial}_{\lambda}\bar{c}_{\nu}).$

Now we would like to show that in our theory there is a gravitational
conformal algebra which is slightly different from conformal algebra in a flat
Minkowski space-time. For this aim, let us consider a set of generators,
$\\{P_{\mu},G^{\mu}{}_{\nu},K^{\mu},D_{0}\\}$. From these generators, we wish
to construct the generator $D$ for a scale transformation. Recall that in
conformal field theory in the four-dimensional Minkowski space-time, the
dilatation generator obeys the following algebra for an local operator
$O_{i}(x)$ of conformal dimension $\Delta_{i}$ [15, 16]:141414For clarity, we
will call a global scale transformation in a flat Minkowski space-time
“dilatation”. Dilatation is usually interpreted as a subgroup of the general
coordinate transformation in a such way that the space-time coordinates are
transformed as $x^{\mu}\rightarrow\Omega x^{\mu}$ in the flat space-time where
$\Omega$ is a constant scale factor, whereas the global scale transformation
is a rescaling of all lengths by the same $\Omega$ by
$g_{\mu\nu}\rightarrow\Omega^{2}g_{\mu\nu}$. The two viewpoints are completely
equivalent since all the lengths are defined via the line element
$ds^{2}=g_{\mu\nu}dx^{\mu}dx^{\nu}$.

$\displaystyle[iD,O_{i}(x)]=x^{\mu}\partial_{\mu}O_{i}(x)+\Delta_{i}O_{i}(x).$
(8.9)

Since the scalar field $\phi(x)$ has conformal dimension $1$, it must satisfy
the equation:

$\displaystyle[iD,\phi(x)]=x^{\mu}\partial_{\mu}\phi(x)+\phi(x).$ (8.10)

To be consistent with this equation, we shall make a generator for the scale
transformation. From the definitions (8.7) and (8.8), we find

$\displaystyle[iG^{\mu}{}_{\nu},\phi(x)]=x^{\mu}\partial_{\mu}\phi(x),\qquad[iD_{0},\phi(x)]=-\phi(x).$
(8.11)

The following linear combination of $G^{\mu}{}_{\nu}$ and $D_{0}$ does the
job:

$\displaystyle D\equiv G^{\mu}{}_{\mu}-D_{0}.$ (8.12)

As a consistency check, it is valuable to see how this operator $D$ acts on
the metric field whose result reads:

$\displaystyle[iD,g_{\sigma\tau}]=[iG^{\mu}{}_{\mu},g_{\sigma\tau}]-[iD_{0},g_{\sigma\tau}]$
$\displaystyle=(x^{\mu}\partial_{\mu}g_{\sigma\tau}+2g_{\sigma\tau})-2g_{\sigma\tau}=x^{\mu}\partial_{\mu}g_{\sigma\tau},$
(8.13)

which implies that the metric field has conformal dimension $0$ as desired and
this result will be used later when discussing spontaneous symmetry breakdown.

Next, let us calculate an algebra among the generators
$\\{P_{\mu},G^{\mu}{}_{\nu},K^{\mu},D\\}$. After some calculations, we find
that the algebra closes and takes the form:

$\displaystyle[P_{\mu},P_{\nu}]=0,\quad[P_{\mu},G^{\rho}{}_{\sigma}]=iP_{\sigma}\delta^{\rho}_{\mu},\quad[P_{\mu},K^{\nu}]=-2i(G^{\rho}{}_{\rho}-D)\delta^{\nu}_{\mu},\quad$
$\displaystyle[P_{\mu},D]=iP_{\mu},\quad[G^{\mu}{}_{\nu},G^{\rho}{}_{\sigma}]=i(G^{\mu}{}_{\sigma}\delta^{\rho}_{\nu}-G^{\rho}{}_{\nu}\delta^{\mu}_{\sigma}),$
$\displaystyle[G^{\mu}{}_{\nu},K^{\rho}]=iK^{\mu}\delta^{\rho}_{\nu},\quad[G^{\mu}{}_{\nu},D]=[K^{\mu},K^{\nu}]=0,$
$\displaystyle[K^{\mu},D]=-iK^{\mu},\quad[D,D]=0.$ (8.14)

To extract the gravitational conformal algebra in quantum gravity, it is
necessary to introduce the “Lorentz” generator, which can be contructed from
the $GL(4)$ generator as

$\displaystyle
M_{\mu\nu}\equiv-\eta_{\mu\rho}G^{\rho}{}_{\nu}+\eta_{\nu\rho}G^{\rho}{}_{\mu}.$
(8.15)

In terms of the generator $M_{\mu\nu}$, the algebra (8.14) can be cast to the
form:

$\displaystyle[P_{\mu},P_{\nu}]=0,\quad[P_{\mu},M_{\rho\sigma}]=i(P_{\rho}\eta_{\mu\sigma}-P_{\sigma}\eta_{\mu\rho}),$
$\displaystyle[P_{\mu},K^{\nu}]=-2i(G^{\rho}{}_{\rho}-D)\delta^{\nu}_{\mu},\quad[P_{\mu},D]=iP_{\mu},$
$\displaystyle[M_{\mu\nu},M_{\rho\sigma}]=-i(M_{\mu\sigma}\eta_{\nu\rho}-M_{\nu\sigma}\eta_{\mu\rho}+M_{\rho\mu}\eta_{\sigma\nu}-M_{\rho\nu}\eta_{\sigma\mu}),$
$\displaystyle[M_{\mu\nu},K^{\rho}]=i(-K_{\mu}\delta^{\rho}_{\nu}+K_{\nu}\delta^{\rho}_{\mu}),\quad[M_{\mu\nu},D]=[K^{\mu},K^{\nu}]=0,$
$\displaystyle[K^{\mu},D]=-iK^{\mu},\quad[D,D]=0.$ (8.16)

where we have defined $K_{\mu}\equiv\eta_{\mu\nu}K^{\nu}$. It is of interest
that the the algebra (8.16) in quantum gravity, which we call “gravitational
conformal algebra”, formally resembles conformal algebra in the flat Minkowski
space-time except for the expression of $[P_{\mu},K^{\nu}]$.151515In case of
conformal algebra in the flat space-time,
$[P_{\mu},K^{\nu}]=-2i(\delta^{\nu}_{\mu}D+M_{\mu}{}^{\nu})$. This difference
reflects from the difference of the definition of conformal dimension in both
gravity and conformal field theory, for which the metric tensor field
$g_{\mu\nu}$ has $2$ in gravity as seen in Eq. (3.2) while it has $0$ in
conformal field theory as seen in Eq. (8.13).

##  9 Spontaneous breakdown of symmetries

In the theory in hand, there are huge global symmetries, which are
$IOSp(10|10)$ supersymmetry, so it is valuable to investigate which symmetries
are spontaneously broken or survive even in quantum regime. In this section,
we postulate the existence of a unique vacuum $|0\rangle$, which is normalized
to be the unity:

$\displaystyle\langle 0|0\rangle=1.$ (9.1)

Furthermore, we assume that the vacuum is translation invariant:

$\displaystyle P_{\mu}|0\rangle=0,$ (9.2)

and the vacuum expectation values (VEVs) of the metric tensor $g_{\mu\nu}$ and
the scalar field $\phi$ are respectively the Minkowski metric $\eta_{\mu\nu}$
and a non-zero constant $\phi_{0}\neq 0$:

$\displaystyle\langle 0|g_{\mu\nu}|0\rangle=\eta_{\mu\nu},\qquad\langle
0|\phi|0\rangle=\phi_{0}.$ (9.3)

By a straightforward calculation, we can obtain the following VEVs:

$\displaystyle\langle
0|[iP^{\mu}(x),b_{\rho}]|0\rangle=-\delta^{\mu}_{\rho},\quad\langle
0|\\{iP^{\mu}(c^{\tau}),\bar{c}_{\rho}\\}|0\rangle=i\delta^{\mu}_{\rho},\quad$
$\displaystyle\langle
0|\\{iP_{\mu}(\bar{c}_{\tau}),c^{\rho}\\}|0\rangle=-i\delta^{\rho}_{\mu},$
$\displaystyle\langle
0|[iM^{\mu\nu}(x,x),\frac{1}{2}(\partial_{\lambda}b_{\rho}-\partial_{\rho}b_{\lambda})]|0\rangle=-(\delta^{\mu}_{\lambda}\delta^{\nu}_{\rho}-\delta^{\nu}_{\lambda}\delta^{\mu}_{\rho}),$
$\displaystyle\langle
0|\\{iM^{\mu\nu}(x,c^{\tau}),\partial_{\lambda}\bar{c}_{\rho}\\}|0\rangle=i\delta^{\mu}_{\lambda}\delta^{\nu}_{\rho},\quad\langle
0|\\{iM^{\mu}{}_{\nu}(x,\bar{c}_{\tau}),\partial_{\lambda}c^{\rho}\\}|0\rangle=-i\delta^{\mu}_{\lambda}\delta^{\rho}_{\nu},$
$\displaystyle\langle 0|[iP(\sigma),B]|0\rangle=1,\quad\langle
0|\\{iP(c),\bar{c}\\}|0\rangle=i,\quad\langle
0|\\{iP(\bar{c}),c\\}|0\rangle=-i,$ $\displaystyle\langle
0|\\{iM(\sigma,c),\bar{c}\\}|0\rangle=i\sigma_{0},\quad\langle
0|\\{iM(\sigma,\bar{c}),c\\}|0\rangle=-i\sigma_{0},$ (9.4)

where $\langle 0|\sigma(x)|0\rangle\equiv\sigma_{0}$. Eq. (9.4) shows that the
symmetries generated by the conserved charges

$\displaystyle\\{P^{\mu}(x),P^{\mu}(c^{\tau}),P_{\mu}(\bar{c}_{\tau}),M^{\mu\nu}(x,x),M^{\mu\nu}(x,c^{\tau}),M^{\mu}{}_{\nu}(x,\bar{c}_{\tau}),$
$\displaystyle P(\sigma),P(c),P(\bar{c}),M(\sigma,c),M(\sigma,\bar{c})\\}$

are necessarily broken spontaneously, thereby
$b_{\mu},c^{\mu},\bar{c}_{\mu},B,c$ and $\bar{c}$ acquiring massless Nambu-
Goldstone modes. Note that the exact masslessness of the dilaton $\sigma$
cannot be proved in this way.

Next, on the basis of the gravitational conformal symmetry, we will show that
$GL(4)$, special conformal symmetry and scale symmetry are spontaneously
broken down to the Poincaré symmetry. We find that the VEV of a commutator
between the $GL(4)$ generator and the metric field reads

$\displaystyle\langle
0|[iG^{\mu}{}_{\nu},g_{\sigma\tau}]|0\rangle=\delta^{\mu}_{\sigma}\eta_{\nu\tau}+\delta^{\mu}_{\tau}\eta_{\nu\sigma}.$
(9.5)

Thus, the Lorentz generator, which is defined in Eq. (8.15), has the vanishing
VEV:

$\displaystyle\langle 0|[iM_{\mu\nu},g_{\sigma\tau}]|0\rangle=0.$ (9.6)

On the other hand, the symmetric part, which is defined as
$\bar{M}_{\mu\nu}\equiv\eta_{\mu\rho}G^{\rho}{}_{\nu}+\eta_{\nu\rho}G^{\rho}{}_{\mu}$,
has the non-vanishing VEV:

$\displaystyle\langle
0|[i\bar{M}_{\mu\nu},g_{\sigma\tau}]|0\rangle=2(\eta_{\mu\sigma}\eta_{\nu\tau}+\eta_{\mu\tau}\eta_{\nu\sigma}).$
(9.7)

Thus, the $GL(4)$ symmetry is spontaneously broken to the Lorentz symmetry
where the corresponding Nambu-Goldstone boson with ten independent components
is nothing but the massless graviton [17]. Here it is interesting that in a
sector of the scalar field, the $GL(4)$ symmetry and of course the Lorentz
symmetry as well, do not give rise to a symmetry breaking as can be seen in
the commutators:

$\displaystyle\langle 0|[iG^{\mu}{}_{\nu},\phi]|0\rangle=\langle
0|[iM_{\mu\nu},\phi]|0\rangle=\langle 0|[i\bar{M}_{\mu\nu},\phi]|0\rangle=0.$
(9.8)

Now we wish to clarify how the scale symmetry and special conformal symmetry
are spontaneously broken and what the corresponding Nambu-Goldstone bosons
are. As for the scale symmetry, it is not the gravitational field but the
dilaton that gives rise to spontaneous symmetry breaking. Indeed, Eq. (8.13)
gives us

$\displaystyle\langle 0|[iD,g_{\sigma\tau}]|0\rangle=0.$ (9.9)

On the other hand, for the dilaton, from Eq. (8.10) we have

$\displaystyle\langle 0|[iD,\sigma]|0\rangle=1,$ (9.10)

which elucidates the spontaneous symmetry breakdown of the scale symmetry
whose Nambu-Goldstone boson is just the massless dilaton $\sigma(x)$.

Regarding the special conformal symmetry, we find

$\displaystyle\langle
0|[iK^{\mu},\partial_{\nu}\sigma]|0\rangle=2\delta^{\mu}_{\nu}.$ (9.11)

This equation means that the special conformal symmetry is certainly broken
spontaneously and its Nambu-Goldstone boson is the derivative of the dilaton.
This interpretation can be also verified from the gravitational conformal
algebra. In the algebra (8.16), we have a commutator between $P_{\mu}$ and
$K^{\nu}$:

$\displaystyle[P_{\mu},K^{\nu}]=-2i(G^{\rho}{}_{\rho}-D)\delta^{\nu}_{\mu}.$
(9.12)

Let us consider the Jacobi identity:

$\displaystyle[[P_{\mu},K^{\nu}],\sigma]+[[K^{\nu},\sigma],P_{\mu}]+[[\sigma,P_{\mu}],K^{\nu}]=0.$
(9.13)

Using the translational invariance of the vacuum in Eq. (9.2) and the equation

$\displaystyle[P_{\mu},\sigma]=-i\partial_{\mu}\sigma,$ (9.14)

and taking the VEV of the Jacobi identity (9.13), we can obtain the VEV:

$\displaystyle\langle 0|[K^{\nu},\partial_{\mu}\sigma]|0\rangle$
$\displaystyle=$ $\displaystyle-2\delta^{\nu}_{\mu}\langle
0|[G^{\rho}{}_{\rho}-D,\sigma]|0\rangle$ (9.15) $\displaystyle=$
$\displaystyle-2i\delta^{\nu}_{\mu},$

which coincides with Eq. (9.11) as promised. In other words, the $GL(4)$
symmetry is spontaneously broken to the Poincaré symmetry whose Nambu-
Goldstone boson is the graviton, the scale symmetry and the special conformal
symmetry are also spontaneously broken and the corresponding Nambu-Goldstone
bosons are the dilaton and the derivative of the dilaton, respectively. It is
of interest that the Nambu-Goldstone boson associated with the special
conformal symmetry is not an independent field in quantum gravity as in
conformal field theory [18].

##  10 Conclusion

In this article, we have performed a manifestly covariant quantization and
contructed a quantum theory of the Weyl invariant scalar-tensor gravity within
the framework of the BRST formalism. In the past, Nakanishi has made a similar
quantum gravitational theory of Einstein’s general relativity [4, 5], and the
present work provides its natural generalization in the sense that the Weyl
symmetry is treated on the same footing as the general coordinate symmetry.

Since the Weyl invariant scalar-tensor gravity has been known to be equivalent
to general relativity in the unitary gauge where the scalar field is gauge-
fixed to be a constant, it is natural to expect that our present theory shares
several characteristic features with the Nakanishi’s quantum gravity. In
particular, the both theories have a huge global symmetry called “choral
symmetry”, but our choral symmetry $ISOp(10|10)$ is larger than that of the
Nakanishi’s theory, which is an $ISOp(8|8)$, owing to the presence of the Weyl
symmetry in our formulation. Compared with the case of general relativity, one
peculiar feature of our choral symmetry is that the choral symmetry needs the
Weyl symmetry in proving its invariance of the quantum Lagrangian so that it
is closely related to a gravitational sector while in the case of general
relativity the choral symmetry is isolated from classical Lagrangian and comes
from purely the Lagrangian involving the Nakanishi-Lautrup field and the FP
ghosts.

It is worth mentioning that in our quantum gravity there is a gravitational
conformal algebra which is relevant to conventional conformal algebra in a
flat Minkowski space-time. According to the Zumino theorem [19], the theories
which are invariant under the GCT and Weyl transformation have conformal
invariance in the flat Minkowski background at the classical level. The
present study supports a conjecture that the Zumino theorem could be valid
even in quantum gravity.

Last but not least, we should comment on a Weyl anomaly. In this respect, let
us recall that in the manifestly scale invariant regularization method [20] \-
[25], the scale invariance is free of scale anomaly. Though a completely
satisfying formalism is still missing, we believe that in the Weyl invariant
regularization method, the Weyl invariance would be also kept in the operator
level without the Weyl anomaly, and is spontaneously broken in considering
states in the Hilbert space.

There are a lot of works to be done in future. First of all, we should make a
manifestly Weyl invariant regularization methods by introducing an additional
scalar field which plays a role for the renormalization mass scale $\mu$.
Secondly, we should prove a quantum Zumino theorem in case that a classical
Lagrangian is an arbitrary Lagrangian which is invariant under the Weyl
transformation. Thirdly, we should add the Lagrangian of conformal gravity,
that is, ${\cal L}\sim\sqrt{-g}\,C_{\mu\nu\rho\sigma}^{2}$ with conformal
tensor $C_{\mu\nu\rho\sigma}$, and investigate if the similar analysis to the
present work could be done or not. Finally, it has been known that the Weyl
invariant scalar-tensor gravity reduces to the Weyl transverse gravity when
the longitudinal general coordinate transformation is gauge-fixed [26] \-
[29]. The Weyl transverse gravity possesses the Weyl symmetry, to which we
could apply the present formulation and investigate various quantum aspects.
We hope to return these problems in the near future.

## Acknowledgment

This work is supported in part by the JSPS Kakenhi Grant No. 21K03539.

## Appendix

##  Appendix A Derivation of Eq. (3.18)

In this appendix, we present a derivation of Eq. (3.18). First of all, let us
notice that the scalar gauge condition (3.6) is equivalent to the equation:

$\displaystyle\Box\phi^{2}=0.$ (A.1)

Then, the Einstein equation in (3.15) reads

$\displaystyle
G_{\mu\nu}-\phi^{-2}\nabla_{\mu}\nabla_{\nu}\phi^{2}-6\phi^{-2}(E_{\mu\nu}-\frac{1}{2}g_{\mu\nu}E)=0.$
(A.2)

With the help of Eq. (A.1), the trace part of this equation becomes

$\displaystyle R=6\phi^{-2}E.$ (A.3)

Inserting Eq. (A.3) to Eq. (A.2) leads to

$\displaystyle
R_{\mu\nu}=\phi^{-2}(\nabla_{\mu}\nabla_{\nu}\phi^{2}+6E_{\mu\nu}).$ (A.4)

Next, operating a covariant derivative $\nabla^{\mu}$ on Eq. (A.2) and using
the Bianchi identity $\nabla^{\mu}G_{\mu\nu}=0$, we have

$\displaystyle 2\nabla^{\mu}\phi\nabla_{\mu}\nabla_{\nu}\phi^{2}-\phi
R_{\nu}{}^{\mu}\nabla_{\mu}\phi^{2}+12\nabla^{\mu}\phi\,(E_{\mu\nu}-\frac{1}{2}g_{\mu\nu}E)$
$\displaystyle-6\phi\nabla^{\mu}(E_{\mu\nu}-\frac{1}{2}g_{\mu\nu}E)=0,$ (A.5)

where Eq. (A.1) was used. Substituting Eq. (A.4) into (A.5) produces

$\displaystyle\nabla^{\mu}(E_{\mu\nu}-\frac{1}{2}g_{\mu\nu}E)+\phi^{-1}\nabla_{\nu}\phi\,E=0.$
(A.6)

At this point, we make use of an identity which holds for any symmetric tensor
$S^{\mu\nu}=S^{\nu\mu}$ [1]:

$\displaystyle\nabla_{\nu}S^{\nu}{}_{\mu}=h^{-1}\partial_{\nu}(hS^{\nu}{}_{\mu})+\frac{1}{2}S_{\alpha\beta}\partial_{\mu}g^{\alpha\beta}.$
(A.7)

Identifying $S^{\mu\nu}$ with $E^{\mu\nu}$, and using the relation (3.14), we
can obtain

$\displaystyle
g^{\rho\nu}\partial_{\rho}\hat{E}_{\mu\nu}-\frac{1}{2}g^{\alpha\beta}\partial_{\mu}\hat{E}_{\alpha\beta}=0,$
(A.8)

where we used the extended de Donder gauge condition (3.5). Finally, when we
calculate the LHS of Eq. (A.8) by using the definition of $\hat{E}_{\mu\nu}$
in (3.13), we can arrive at the desired equation Eq. (3.18).

##  Appendix B Residual symmetry and conformal symmetry

In this Appendix, we would like to explain that the residual symmetries found
in Eq. (8.5) in a curved space-time reduces to a dilatational invariance and
special conformal invariance in a flat Minkowski space-time.

Before doing so, let us first recall that conformal transformation [15, 16]
can be defined as the general coordinate transformation which can be undone by
the Weyl transformation when the space-time metric is the flat Minkowski one.
With this definition, the conformal transformation is described by the
equation:

$\displaystyle\partial_{\mu}\epsilon_{\nu}+\partial_{\nu}\epsilon_{\mu}=2\Lambda(x)\eta_{\mu\nu},$
(B.1)

where $\Lambda(x)$ is the infinitesimal transformation parameter of the Weyl
transformation, i.e., $\Omega(x)\equiv e^{\Lambda(x)}\approx 1+\Lambda(x)$.

Taking the trace of Eq. (B.1) enables us to determine $\Lambda(x)$ to be

$\displaystyle\Lambda=\frac{1}{4}\partial^{\rho}\epsilon_{\rho}.$ (B.2)

Inserting this $\Lambda$ to Eq. (B.1) yields

$\displaystyle\partial_{\mu}\epsilon_{\nu}+\partial_{\nu}\epsilon_{\mu}=\frac{1}{2}\partial^{\rho}\epsilon_{\rho}\eta_{\mu\nu},$
(B.3)

which is often called the “conformal Killing equation” in the Minkowski space-
time. It is worth stressing that Eq. (B.3) implies the following fact: The
flat Minkowski metric $g_{\mu\nu}=\eta_{\mu\nu}$ is invariant in the space of
the metric functions under a suitable combination of the general coordinate
transformation and the Weyl transformation in such a way that

$\displaystyle\delta(\epsilon_{\mu})=\delta_{GCT}(\epsilon_{\mu})-\delta_{W}(\Lambda=\frac{1}{4}\partial^{\rho}\epsilon_{\rho}),$
(B.4)

when the vector field $\epsilon_{\mu}(x)$ obeys the conformal Killing equation
(B.3). To put it differently, the characteristic feature of the theory under
consideration is that the Lagrangian (3.1) possesses the conformal symmetry
with 15 global parameters which is a subgroup of the general coordinate
transformation and the Weyl transformation.

Multiplying it by $\partial^{\mu}\partial^{\nu}$, we obtain

$\displaystyle\Box\partial^{\mu}\epsilon_{\mu}=0.$ (B.5)

Moreover, multiplying Eq. (B.3) by $\partial^{\mu}\partial_{\lambda}$ and then
symmetrizing the indices $\lambda$ and $\nu$ leads to the equation:

$\displaystyle\partial_{\nu}\partial_{\lambda}\partial^{\mu}\epsilon_{\mu}=0,$
(B.6)

where we have used Eqs. (B.3) and (B.5). It turns out that a general solution
to Eq. (B.6) reads

$\displaystyle\epsilon^{\mu}=a^{\mu}+\omega^{\mu\nu}x_{\nu}+\lambda
x^{\mu}+k^{\mu}x^{2}-2x^{\mu}k_{\rho}x^{\rho},$ (B.7)

where $a^{\mu},\omega^{\mu\nu}=-\omega^{\nu\mu},\lambda$ and $k^{\mu}$ are all
constant parameters and they correspond to the translation, the Lorentz
transformation, the dilatation and the special conformal transformation,
respectively.

At this point, it is useful to verify what expression the infinitesimal
parameter $\Lambda$ generated by the “conformal Killing vector”
$\epsilon^{\mu}$ in Eq. (B.7) takes. Actually, substituting Eq. (B.7) into Eq.
(B.2), we have

$\displaystyle\Lambda=\lambda-2k_{\mu}x^{\mu}.$ (B.8)

This is nothing but zero-mode solutions in Eq. (8.5). This result implies that
finding the residual symmetries (8.5) amounts to solving the conformal Killing
equation in a flat Minkowski space-time.

To summarize, we have explicitly shown that in our quantum gravity the Weyl
symmetry, together with the general coordinate invariance, generates the
conformal symmetry in the flat Minkowski background. This result is a quantum-
mechanical generalization of the well-known Zumino’s theorem [19] which
insists that the theories invariant under both the general coordinate
transformation and the Weyl transformation (or local scale transformation)
possess conformal symmetry in the flat Minkowski background. Even if we used a
Weyl invariant classical Lagrangian (3.1), we think that the result obtained
here holds for any theories which are invariant under the GCT and the Weyl
transformation if we adopt the extended de Donder gauge and the scalar gauge
for these invariances.

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