DPUR/TH/73

January, 2022

Quantum Scale Invariant Gravity with de Donder Gauge

Ichiro Oda111 E-mail address<EMAIL_ADDRESS>
Department of Physics, Faculty of Science, University of the Ryukyus,

Nishihara, Okinawa 903-0213, Japan

###### Abstract

We perform the manifestly covariant quantization of a scale invariant gravity
with a scalar field, which is equivalent to the well-known Brans-Dicke gravity
via a field redefinition of the scalar field, in the de Donder gauge condition
(or harmonic gauge condition) for general coordinate invariance. First,
without specifying the expression of a gravitational theory, we write down
various equal-time (anti-)commutation relations (ETCRs), in particular, those
involving the Nakanishi-Lautrup field, the FP ghost, and the FP antighost only
on the basis of the de Donder gauge condition. It is shown that choral
symmetry, which is a Poincar${\rm{\acute{e}}}$-like $IOSp(8|8)$ supersymmetry,
can be derived from such a general action with the de Donder gauge. Next,
taking the scale invariant gravity with a scalar field as a classical theory,
we derive the ETCRs for the gravitational sector involving the metric tensor
and scalar fields. Moreover, we account for how scale symmetry is
spontaneously broken in quantum gravity, thereby showing that the dilaton is a
massless Nambu-Goldstone particle.

##  1 Introduction

A residual symmetry which is left behind after taking a certain gauge-fixing
condition for the gauge invariance, has thus far played an important role in
quantum field theory. For instance, in string theory, conformal symmetry on
the world sheet can be thought as the fundamental symmetry in perturbative
regime where strings are weakly interacting [1]. The conformal symmetry is a
typical residual symmetry, which is left in a theory after taking the
conformal gauge for the world-sheet diffeomorphism (or general coordinate
transformation (GCT)) and the Weyl symmetry (or a local scale transformation)
[2].

Recently, we have elucidated various aspects of such residual symmetries
existing in some gravitational theories. In particular, in the most recent
study, we have shown that using the simplest scalar-tensor gravity [3] the
restricted Weyl symmetry (RWS) and general coordinate invariance generate
conformal symmetry in four dimensions in a flat Minkowski background [4]-[6].

Also about Einstein gravity, in a pioneering work by Nakanishi [7, 8], on the
basis of the Einstein-Hilbert action in the de Donder gauge (harmonic gauge)
for general coordinate invariance, it has been shown that there remains a huge
residual symmetry, which is a Poincar$\rm{\acute{e}}$-like $ISOp(8|8)$
supersymmetry, called “choral symmetry”, including the BRST symmetry and
$GL(4)$ symmetry etc. It is of interest that in this formulation the graviton
can be identified with a Nambu-Goldstone particle associated with spontaneous
symmetry breakdown of $GL(4)$ symmetry to the Lorentz symmetry $SO(1,3)$,
thereby proving the exact masslessness of the graviton in a nonperturbative
manner [9].

However, in this formulation, the Einstein equation was critically used in
obtaining the equal-time commutation relation
$[b_{\mu},\dot{b}_{\nu}^{\prime}]=i\tilde{f}(\partial_{\mu}b_{\nu}+\partial_{\nu}b_{\mu})\delta^{3}$,
which is needed in proving the closure of the $ISOp(8|8)$ algebra among the
generators, so the formulation depends on the Einstein-Hilbert term in a
classical action. One of our motivations is to relax this situation and show
that the choral symmetry does not depend on the expression of the classical
gravity but completely comes from the de Donder gauge condition for GCT in the
BRST formalism [10]. For this purpose, without the knowledge of the classical
Lagrangian we derive various equal-time (anti-)commutation relations (ETCRs)
for the the Nakanishi-Lautrup field, the Faddeev-Popov (FP) ghost, and the FP
antighost only on the basis of the de Donder gauge condition.

Another motivation behind the study at hand is to construct a quantum theory
of the well-known Brans-Dicke gravity [11] by constructing its manifestly
covariant BRST formalism since many of studies of the Brans-Dicke gravity have
been limited to a classical analysis. As a concrete advantage of our quantum
theory where the gravity as well as a scalar field are quantized, we will show
that scale invariance is spontaneously broken and consequently “dilaton” is
exactly massless thanks to the Nambu-Goldstone theorem even in quantum
gravity.

We close this section with an overview of this article. In Section 2, we
discuss a quantum gravity for which the de Donder gauge is adopted as a gauge
condition for the general coordinate invariance. In Section 3, we calculate
the ETCRs for the Nakanishi-Lautrup auxiliary field, the FP ghost and the FP
antighost based on the quantum gravity made in Section 2. In Section 4, we
calculate the ETCRs involving among the Nakanishi-Lautrup field and its time
derivative without using the information on a classical gravitational action,
and then comment on choral symmetry. In Section 5, a scale invariant scalar-
tensor gravity is briefly reviewed. In Section 6, by selecting the scale
invariant scalar-tensor gravity as a classical theory we calculate the ETCRs
for the gravitational sector, and in Section 7 we discuss spontaneous symmetry
breakdown of the scale invariance and show that the dilaton is exactly
massless owing to the Nambu-Goldstone theorem in quantum gravity. The final
section is devoted to the conclusion. Two appendices are put for technical
details. In Appendix A, a derivation of the tensorlike ETCR is given, and in
Appendix B we have derived the scale current via the Noether theorem.

##  2 Quantum gravity with de Donder gauge

We wish to consider a manifestly covariant canonical formalism of general
gravitational theories where general coordinate invariance is fixed by the de
Donder gauge condition. To take a more general theory into consideration,
without specifying the concrete expression of the gravitational Lagrangian, we
will start with the following classical Lagrangian:222We follow the notation
and conventions of MTW textbook [12]. 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 $i=1,2,3$. Furthermore, the
Riemann curvature tensor and the Ricci tensor are respectively defined by
$R^{\rho}\,_{\sigma\mu\nu}=\partial_{\mu}\Gamma^{\rho}_{\sigma\nu}+\Gamma^{\rho}_{\lambda\mu}\Gamma^{\lambda}_{\sigma\nu}-(\mu\leftrightarrow\nu)$
and $R_{\mu\nu}=R^{\rho}\,_{\mu\rho\nu}$.

$\displaystyle{\cal{L}}_{c}={\cal{L}}_{c}(g_{\mu\nu},\phi),$ (1)

which includes the metric tensor field $g_{\mu\nu}$ and a scalar field $\phi$
as dynamical fields, and is invariant under the general coordinate
transformation (GCT). We assume that ${\cal{L}}_{c}$ does not involve more
than first order derivatives of the metric and matter fields.

Let us fix the general coordinate symmetry by the de Donder gauge condition
(or harmonic gauge condition):

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

where we have defined $\tilde{g}^{\mu\nu}\equiv\sqrt{-g}g^{\mu\nu}\equiv
hg^{\mu\nu}$. Then, the BRST transformation is of form:

$\displaystyle\delta_{B}\bar{c}_{\rho}$ $\displaystyle=$ $\displaystyle
iB_{\rho},\quad\delta_{B}c^{\rho}=-c^{\lambda}\partial_{\lambda}c^{\rho},\quad\delta_{B}\phi=-c^{\lambda}\partial_{\lambda}\phi,$
$\displaystyle\delta_{B}g_{\mu\nu}$ $\displaystyle=$
$\displaystyle-(\nabla_{\mu}c_{\nu}+\nabla_{\nu}c_{\mu})$ $\displaystyle=$
$\displaystyle-(c^{\alpha}\partial_{\alpha}g_{\mu\nu}+\partial_{\mu}c^{\alpha}g_{\alpha\nu}+\partial_{\nu}c^{\alpha}g_{\mu\alpha}),$
$\displaystyle\delta_{B}\tilde{g}^{\mu\nu}$ $\displaystyle=$ $\displaystyle
h(\nabla^{\mu}c^{\nu}+\nabla^{\nu}c^{\mu}-g^{\mu\nu}\nabla_{\rho}c^{\rho}).$
(3)

Using this BRST transformation, the Lagrangian for the gauge-fixing condition
and FP ghosts can be constructed in a standard manner

$\displaystyle{\cal{L}}_{GF+FP}$ $\displaystyle=$
$\displaystyle\delta_{B}(i\tilde{g}^{\mu\nu}\partial_{\mu}\bar{c}_{\nu})$ (4)
$\displaystyle=$
$\displaystyle-\tilde{g}^{\mu\nu}\partial_{\mu}B_{\nu}-i\partial_{\mu}\bar{c}_{\nu}\Bigl{[}\tilde{g}^{\mu\rho}\partial_{\rho}c^{\nu}+\tilde{g}^{\nu\rho}\partial_{\rho}c^{\mu}-\partial_{\rho}(\tilde{g}^{\mu\nu}c^{\rho})\Bigr{]}.$

To simplify this expression, let us introduce a new auxiliary field $b_{\rho}$
defined as

$\displaystyle
b_{\rho}=B_{\rho}-ic^{\lambda}\partial_{\lambda}\bar{c}_{\rho},$ (5)

and its BRST transformation reads

$\displaystyle\delta_{B}b_{\rho}=-c^{\lambda}\partial_{\lambda}b_{\rho}.$ (6)

Then, the Lagrangian (4) can be cast to the form:

$\displaystyle{\cal{L}}_{GF+FP}=-\tilde{g}^{\mu\nu}\partial_{\mu}b_{\nu}-i\tilde{g}^{\mu\nu}\partial_{\mu}\bar{c}_{\rho}\partial_{\nu}c^{\rho}+i\partial_{\rho}(\tilde{g}^{\mu\nu}\partial_{\mu}\bar{c}_{\nu}\cdot
c^{\rho}).$ (7)

As a result, up to a total derivative, the gauge-fixed and BRST-invariant
quantum Lagrangian is given by

$\displaystyle{\cal{L}}_{q}$ $\displaystyle=$
$\displaystyle{\cal{L}}_{c}-\tilde{g}^{\mu\nu}\partial_{\mu}b_{\nu}-i\tilde{g}^{\mu\nu}\partial_{\mu}\bar{c}_{\rho}\partial_{\nu}c^{\rho}$
(8) $\displaystyle\equiv$
$\displaystyle{\cal{L}}_{c}+{\cal{L}}_{GF}+{\cal{L}}_{FP},$

where we have defined

$\displaystyle{\cal{L}}_{GF}\equiv-\tilde{g}^{\mu\nu}\partial_{\mu}b_{\nu},\qquad{\cal{L}}_{FP}\equiv-i\tilde{g}^{\mu\nu}\partial_{\mu}\bar{c}_{\rho}\partial_{\nu}c^{\rho}.$
(9)

By performing the integration by parts once, let us rewrite the Lagrangian (8)
as

$\displaystyle{\cal{L}}_{q}$ $\displaystyle=$
$\displaystyle{\cal{L}}_{c}+\partial_{\mu}\tilde{g}^{\mu\nu}b_{\nu}-i\tilde{g}^{\mu\nu}\partial_{\mu}\bar{c}_{\rho}\partial_{\nu}c^{\rho}+\partial_{\mu}{\cal{V}}^{\mu}$
(10) $\displaystyle\equiv$
$\displaystyle{\cal{L}}_{c}+\bar{\cal{L}}_{GF}+{\cal{L}}_{FP}+\partial_{\mu}{\cal{V}}^{\mu},$

where a surface term ${\cal{V}}^{\mu}$ and $\bar{\cal{L}}_{GF}$ are defined as

$\displaystyle{\cal{V}}^{\mu}\equiv-\tilde{g}^{\mu\nu}b_{\nu},\quad\bar{\cal{L}}_{GF}\equiv\partial_{\mu}\tilde{g}^{\mu\nu}b_{\nu}.$
(11)

From this Lagrangian, we can obtain field equations by taking the variation
with respect to $g_{\mu\nu}$, $\phi$, $b_{\nu}$, $\bar{c}_{\rho}$ and
$c^{\rho}$ in order:

$\displaystyle\frac{1}{\sqrt{-g}}\frac{\delta{\cal{L}}_{c}}{\delta
g^{\mu\nu}}-\frac{1}{2}(E_{\mu\nu}-\frac{1}{2}g_{\mu\nu}E)=0,\quad\frac{\delta{\cal{L}}_{c}}{\delta\phi}=0,$
$\displaystyle\partial_{\mu}\tilde{g}^{\mu\nu}=0,\quad
g^{\mu\nu}\partial_{\mu}\partial_{\nu}c^{\rho}=0,\quad
g^{\mu\nu}\partial_{\mu}\partial_{\nu}\bar{c}_{\rho}=0,$ (12)

where we have defined

$\displaystyle E_{\mu\nu}$ $\displaystyle=$
$\displaystyle\partial_{\mu}b_{\nu}+i\partial_{\mu}\bar{c}_{\rho}\partial_{\nu}c^{\rho}+(\mu\leftrightarrow\nu),$
$\displaystyle E$ $\displaystyle=$ $\displaystyle g^{\mu\nu}E_{\mu\nu}.$ (13)

Next, in order to find the field equation for the $b_{\rho}$ field, let us
take a covariant derivative of the first Einstein equation in Eq. (12). The
result reads

$\displaystyle\nabla_{\mu}\left(E^{\mu\nu}-\frac{1}{2}g^{\mu\nu}E\right)=0,$
(14)

where we have used the equation

$\displaystyle\nabla^{\mu}\frac{\delta{\cal{L}}_{c}}{\delta g^{\mu\nu}}=0.$
(15)

This equation can be shown as follows: Using the GCT invariance of the
classical action, we have

$\displaystyle 0$ $\displaystyle=$
$\displaystyle\delta_{\varepsilon}S_{c}\equiv\int
d^{4}x\delta_{\varepsilon}{\cal{L}}_{c}=\int
d^{4}x\Bigl{(}\frac{\delta{\cal{L}}_{c}}{\delta
g^{\mu\nu}}\delta_{\varepsilon}g^{\mu\nu}+\frac{\delta{\cal{L}}_{c}}{\delta\phi}\delta_{\varepsilon}\phi\Bigr{)}$
(16) $\displaystyle=$ $\displaystyle\int
d^{4}x\frac{\delta{\cal{L}}_{c}}{\delta
g^{\mu\nu}}\delta_{\varepsilon}g^{\mu\nu}=\int
d^{4}x\frac{\delta{\cal{L}}_{c}}{\delta
g^{\mu\nu}}(\nabla^{\mu}\varepsilon^{\nu}+\nabla^{\nu}\varepsilon^{\mu})$
$\displaystyle=$ $\displaystyle-2\int
d^{4}x\Bigl{(}\nabla^{\mu}\frac{\delta{\cal{L}}_{c}}{\delta
g^{\mu\nu}}\Bigr{)}\varepsilon^{\nu},$

where we have used the field equation for $\phi$.

Generally for a symmetric tensor $S^{\mu\nu}$, we have a formula:

$\displaystyle\nabla_{\nu}S^{\nu}{}_{\mu}=h^{-1}\partial_{\nu}(hS^{\nu}{}_{\mu})+\frac{1}{2}S_{\alpha\beta}\partial_{\mu}g^{\alpha\beta}.$
(17)

Using this, we can show an equality:

$\displaystyle\nabla_{\nu}E^{\nu}{}_{\mu}$ $\displaystyle=$ $\displaystyle
h^{-1}\partial_{\alpha}\tilde{g}^{\alpha\nu}\cdot
E_{\nu\mu}+\partial^{2}b_{\mu}+i\left(\partial^{2}\bar{c}_{\lambda}\cdot\partial_{\mu}c^{\lambda}+\partial_{\mu}\bar{c}_{\lambda}\cdot\partial^{2}c^{\lambda}\right)+\frac{1}{2}\partial_{\mu}E,$
(18)

where $\partial^{2}\equiv g^{\mu\nu}\partial_{\mu}\partial_{\nu}$. Then Eq.
(14), together with the help of the other field equations in Eq. (12), is seen
to lead to the field equation for the $b_{\rho}$ field:

$\displaystyle g^{\mu\nu}\partial_{\mu}\partial_{\nu}b_{\rho}=0.$ (19)

In other words, the $b_{\rho}$ field, the ghost field $c^{\rho}$ and the
antighost field $\bar{c}_{\rho}$ all satisfy the d’Alembert equation.
Furthermore, it is of interest to see that the space-time coordinates
$x^{\lambda}$ obey the d’Alembert equation,
$g^{\mu\nu}\partial_{\mu}\partial_{\nu}x^{\lambda}=0$ as well.333Using the de
Donder condition, these d’Alembert equations can be rewritten as
$\partial_{\mu}(\tilde{g}^{\mu\nu}\partial_{\nu}\Phi)=0$ where
$\Phi\equiv\\{x^{\lambda},b_{\rho},c^{\sigma},\bar{c}_{\tau}\\}$.

##  3 Equal-time commutation relations

In this section, after introducing the canonical commutation relations (CCRs),
we will evaluate the equal-time commutation relations (ETCRs) among
fundamental variables, in particular, the Nakanishi-Lautrup field $b_{\mu}$,
the FP ghost $c^{\mu}$ and the FP antighost $\bar{c}_{\mu}$ in detail. To
simplify various expressions, we will obey the following abbreviations adopted
in the textbook of Nakanishi and Ojima [8]:

$\displaystyle[A,B^{\prime}]$ $\displaystyle=$
$\displaystyle[A(x),B(x^{\prime})]|_{x^{0}=x^{\prime
0}},\qquad\delta^{3}=\delta(\vec{x}-\vec{x}^{\prime}),$
$\displaystyle\tilde{f}$ $\displaystyle=$
$\displaystyle\frac{1}{\tilde{g}^{00}}=\frac{1}{\sqrt{-g}g^{00}}=\frac{1}{hg^{00}},$
(20)

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

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

$\displaystyle[g_{\mu\nu},\pi_{g}^{\rho\lambda\prime}]$ $\displaystyle=$
$\displaystyle
i\frac{1}{2}(\delta_{\mu}^{\rho}\delta_{\nu}^{\lambda}+\delta_{\mu}^{\lambda}\delta_{\nu}^{\rho})\delta^{3},\qquad[\phi,\pi_{\phi}^{\prime}]=+i\delta^{3},$
$\displaystyle\\{c^{\sigma},\pi_{c\lambda}^{\prime},\\}$ $\displaystyle=$
$\displaystyle\\{\bar{c}_{\lambda},\pi_{\bar{c}}^{\sigma\prime}\\}=+i\delta_{\lambda}^{\sigma}\delta^{3},$
(21)

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

Based on the Lagrangian (10), the expressions for canonical conjugate momenta
read

$\displaystyle\pi_{g}^{\mu\nu}$ $\displaystyle=$
$\displaystyle\frac{\partial{\cal{L}}}{\partial\dot{g}_{\mu\nu}},$
$\displaystyle\pi_{\phi}$ $\displaystyle=$
$\displaystyle\frac{\partial{\cal{L}}}{\partial\dot{\phi}},$
$\displaystyle\pi_{c\sigma}$ $\displaystyle=$
$\displaystyle\frac{\partial{\cal{L}}}{\partial\dot{c}^{\sigma}}=-i\tilde{g}^{\mu
0}\partial_{\mu}\bar{c}_{\sigma},$ $\displaystyle\pi_{\bar{c}}^{\sigma}$
$\displaystyle=$
$\displaystyle\frac{\partial{\cal{L}}}{\partial\dot{\bar{c}}_{\sigma}}=i\tilde{g}^{\mu
0}\partial_{\mu}c^{\sigma},$ (22)

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.
(21):

$\displaystyle[\pi_{g}^{\alpha
0},g_{\mu\nu}^{\prime}]=-i\frac{1}{2}(\delta_{\mu}^{\alpha}\delta_{\nu}^{0}+\delta_{\mu}^{0}\delta_{\nu}^{\alpha})\delta^{3}.$
(23)

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

$\displaystyle\pi_{g}^{\alpha
0}=A^{\alpha}+B^{\alpha\beta}\partial_{\beta}\phi+C^{\alpha\beta}b_{\beta},$
(24)

where $A^{\alpha},B^{\alpha\beta}$ and
$C^{\alpha\beta}\equiv-\frac{1}{2}\tilde{g}^{00}g^{\alpha\beta}$ 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. (23) produces

$\displaystyle[g_{\mu\nu},b_{\rho}^{\prime}]=-i\tilde{f}(\delta_{\mu}^{0}g_{\rho\nu}+\delta_{\nu}^{0}g_{\rho\mu})\delta^{3}.$
(25)

From this ETCR, we can easily derive ETCRs:

$\displaystyle[g^{\mu\nu},b_{\rho}^{\prime}]=i\tilde{f}(g^{\mu
0}\delta_{\rho}^{\nu}+g^{\nu 0}\delta_{\rho}^{\mu})\delta^{3},$
$\displaystyle[\tilde{g}^{\mu\nu},b_{\rho}^{\prime}]=i\tilde{f}(\tilde{g}^{\mu
0}\delta_{\rho}^{\nu}+\tilde{g}^{\nu
0}\delta_{\rho}^{\mu}-\tilde{g}^{\mu\nu}\delta_{\rho}^{0})\delta^{3}.$ (26)

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

$\displaystyle[g^{\mu\nu},\Phi^{\prime}]=-g^{\mu\alpha}g^{\nu\beta}[g_{\alpha\beta},\Phi^{\prime}],$
$\displaystyle[\tilde{g}^{\mu\nu},\Phi^{\prime}]=-\left(\tilde{g}^{\mu\alpha}g^{\nu\beta}-\frac{1}{2}\tilde{g}^{\mu\nu}g^{\alpha\beta}\right)[g_{\alpha\beta},\Phi^{\prime}],$
(27)

where $\Phi$ is a generic field.

As for the ETCRs involving 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 ETCRs lead to the same ETCR:

$\displaystyle\\{\dot{\bar{c}}_{\lambda},c^{\sigma\prime}\\}=-\tilde{f}\delta_{\lambda}^{\sigma}\delta^{3},$
(28)

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

$\displaystyle[\Phi,\dot{\Psi}^{\prime}]=\partial_{0}[\Phi,\Psi^{\prime}]-[\dot{\Phi},\Psi^{\prime}],$
(29)

which also holds for the anti-commutation relation. Next, it is easy to see
that the ETCRs, $\\{\pi_{g}^{\alpha 0},c^{\sigma\prime}\\}=\\{\pi_{g}^{\alpha
0},\bar{c}_{\lambda}^{\prime}\\}=0$ produce

$\displaystyle[b_{\rho},c^{\sigma\prime}]=[b_{\rho},\bar{c}_{\lambda}^{\prime}]=0.$
(30)

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

$\displaystyle[\dot{\bar{c}}_{\lambda},b_{\rho}^{\prime}]=-i\tilde{f}\partial_{\rho}\bar{c}_{\lambda}\delta^{3},\qquad[\dot{c}^{\sigma},b_{\rho}^{\prime}]=-i\tilde{f}\partial_{\rho}c^{\sigma}\delta^{3}.$
(31)

In this article, we make use of the following ETCR:

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

This ETCR can be in general shown to hold when the system has the
translational invariance and the general coordinate transformation is fixed by
the de Donder gauge as follows: The translational invariance requires the
validity of the following equation for a generic field $\Phi(x)$:

$\displaystyle[\Phi(x),P_{\rho}]=i\partial_{\rho}\Phi(x),$ (33)

where $P_{\rho}$ is the generator of the translation which is now given
by444This translation generator belongs to the generators of choral symmetry
as will be seen in Eq. (67).

$\displaystyle P_{\rho}=\int
d^{3}x\,\tilde{g}^{0\lambda}\partial_{\lambda}b_{\rho}.$ (34)

Now let us consider the specific case $\Phi(x)=g_{\mu\nu}(x)$:

$\displaystyle[g_{\mu\nu}(x),P_{\rho}]=[g_{\mu\nu}(x),\int
d^{3}x^{\prime}\,\tilde{g}^{0\lambda\prime}\partial_{\lambda}b_{\rho}^{\prime}]=i\partial_{\rho}g_{\mu\nu}(x).$
(35)

Taking $x^{0}=x^{\prime 0}$ and using
$[g_{\mu\nu},\tilde{g}^{0\lambda\prime}]=0$, we have

$\displaystyle\int
d^{3}x^{\prime}\,\tilde{g}^{0\lambda}(x^{\prime})[g_{\mu\nu},\partial_{\lambda}b_{\rho}^{\prime}]=i\partial_{\rho}g_{\mu\nu}(x).$
(36)

Using the de Donder gauge condition (2) and Eq. (25), this equation can be
rewritten as

$\displaystyle\int
d^{3}x^{\prime}\,\tilde{g}^{00}(x^{\prime})[g_{\mu\nu},\dot{b}_{\rho}^{\prime}]=i\left[\partial_{\rho}g_{\mu\nu}-\frac{1}{\tilde{f}}\partial_{0}\tilde{f}(\delta_{\mu}^{0}g_{\rho\nu}+\delta_{\nu}^{0}g_{\rho\mu})\right],$
(37)

which is easily solved for $[g_{\mu\nu},\dot{b}_{\rho}^{\prime}]$ to be

$\displaystyle[g_{\mu\nu},\dot{b}_{\rho}^{\prime}]=i\left[\tilde{f}\partial_{\rho}g_{\mu\nu}-\partial_{0}\tilde{f}(\delta_{\mu}^{0}g_{\rho\nu}+\delta_{\nu}^{0}g_{\rho\mu})\right]\delta^{3}+F_{(\mu\nu)\rho}\,^{k}\partial_{k}(\tilde{f}\delta^{3}),$
(38)

where $F_{(\mu\nu)\rho}\,^{k}$ is an arbitrary function which is symmetric
under the exchange of $\mu\leftrightarrow\nu$.

Next, to fix the function $F_{(\mu\nu)\rho}\,^{k}$, let us take account of the
consistency with the de Donder gauge condition (2):

$\displaystyle[\partial_{\mu}\tilde{g}^{\mu\nu},b_{\rho}^{\prime}]=0.$ (39)

After some calculations, Eq. (39) leads to an equation for
$F_{(\mu\nu)\rho}\,^{k}$:

$\displaystyle\left(\tilde{g}^{0\alpha}g^{\nu\beta}-\frac{1}{2}\tilde{g}^{0\nu}g^{\alpha\beta}\right)F_{(\alpha\beta)\rho}\,^{k}=-i(\tilde{g}^{0k}\delta_{\rho}^{\nu}+\tilde{g}^{0\nu}\delta_{\rho}^{k}-\tilde{g}^{k\nu}\delta_{\rho}^{0}).$
(40)

This equation has the unique solution given by

$\displaystyle
F_{(\mu\nu)\rho}\,^{k}=i[(\delta_{\mu}^{k}-2\delta_{\mu}^{0}\tilde{f}\tilde{g}^{0k})g_{\rho\nu}+(\mu\leftrightarrow\nu)].$
(41)

We can therefore obtain

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

Finally, using Eqs. (25) and (29), we can arrive at the desired equation (32).
It is of interest that Eq. (32) can be derived from only the translational
invariance and the de Donder gauge condition without reference to the
classical Lagrangian ${\cal{L}}_{c}$ which knows information of the dynamics
of the gravitational field $g_{\mu\nu}$ and the scalar field $\phi$.

Then, using Eq. (32) together with Eqs. (25) and (26), we can easily show that

$\displaystyle[\Gamma^{\rho}_{\mu\nu},b_{\lambda}^{\prime}]=i\tilde{f}(\delta_{\lambda}^{\rho}\Gamma_{\mu\nu}^{0}-\delta_{\mu}^{0}\Gamma_{\lambda\nu}^{\rho}-\delta_{\nu}^{0}\Gamma_{\mu\lambda}^{\rho})\delta^{3}$
$\displaystyle+i\delta_{\lambda}^{\rho}(2\delta^{0}_{\mu}\delta^{0}_{\nu}\tilde{f}\tilde{g}^{0k}-\delta^{0}_{\mu}\delta^{k}_{\nu}-\delta^{0}_{\nu}\delta^{k}_{\mu})\partial_{k}(\tilde{f}\delta^{3}).$
(43)

Incidentally, this ETCR is also needed in deriving the ETCR
$[b_{\mu},\dot{b}_{\nu}^{\prime}]$ in the next section and the tensorlike ETCR

$\displaystyle[R_{\mu\nu},b_{\rho}^{\prime}]=-i\tilde{f}(\delta_{\mu}^{0}R_{\rho\nu}+\delta_{\nu}^{0}R_{\rho\mu})\delta^{3},$
(44)

which will be derived in the Appendix A.

Finally, for later conveniece we derive the ETCRs, $[\phi,b_{\rho}^{\prime}]$,
$[\dot{\phi},b_{\rho}^{\prime}]$ and $[\dot{\phi},\dot{b}_{\rho}^{\prime}]$.
To do so, we begin by considering $[\phi,\bar{c}_{\rho}^{\prime}]=0$ and take
its BRST transformation as follows:555We define the BRST transformation as
$\delta_{B}\Phi\equiv[iQ_{B},\Phi\\}$ where $\Phi$ is a generic field and
$[\hskip 5.69054pt,\hskip 5.69054pt\\}$ denotes the graded bracket.

$\displaystyle 0$ $\displaystyle=$
$\displaystyle\\{iQ_{B},[\phi,\bar{c}_{\rho}^{\prime}]\\}=\\{[iQ_{B},\phi],\bar{c}_{\rho}^{\prime}\\}+[\phi,\\{iQ_{B},\bar{c}_{\rho}^{\prime}\\}]$
(45) $\displaystyle=$
$\displaystyle\\{-c^{\lambda}\partial_{\lambda}\phi,\bar{c}_{\rho}^{\prime}\\}+[\phi,iB_{\rho}^{\prime}]=[\phi,i(b_{\rho}^{\prime}+ic^{\lambda\prime}\partial_{\lambda}\bar{c}_{\rho}^{\prime})]$
$\displaystyle=$ $\displaystyle i[\phi,b_{\rho}^{\prime}],$

where together with $[\dot{\phi},\bar{c}_{\rho}^{\prime}]=0$, Eqs. (3) and (5)
have been used. Thus, we have shown

$\displaystyle[\phi,b_{\rho}^{\prime}]=0.$ (46)

In a perfectly similar way, we can calculate $[\dot{\phi},b_{\rho}^{\prime}]$
by starting with $[\dot{\phi},\bar{c}_{\rho}^{\prime}]=0$, which holds in the
de Donder gauge, as follows:

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

from which, using Eq. (28) we can obtain

$\displaystyle[\dot{\phi},b_{\rho}^{\prime}]=-i\tilde{f}\partial_{\rho}\phi\delta^{3}.$
(48)

The calculation of $[\dot{\phi},\dot{b}_{\rho}^{\prime}]$ proceeds as follows:
First, we utilize the formula (cf. Eq. (29))

$\displaystyle[\dot{\phi},\dot{b}_{\rho}^{\prime}]=\partial_{0}([\phi,\dot{b}_{\rho}^{\prime}])-[\phi,\ddot{b}_{\rho}^{\prime}].$
(49)

Since the $b$ field obeys the d’Alembert equation as in Eq. (19),
$\ddot{b}_{\rho}^{\prime}$ can be described in terms of $\dot{b}_{\rho}$ and
$b_{\rho}$ like

$\displaystyle\ddot{b}_{\rho}=-\tilde{f}(2\tilde{g}^{0k}\partial_{i}\dot{b}_{\rho}+\tilde{g}^{kl}\partial_{k}\partial_{l}b_{\rho}).$
(50)

Then, we are ready to evaluate

$\displaystyle[\phi,\ddot{b}_{\rho}^{\prime}]$ $\displaystyle=$
$\displaystyle-2\tilde{f}^{\prime}\tilde{g}^{0k\prime}\partial_{k}^{\prime}[\phi,\dot{b}_{\rho}^{\prime}]=-2i\tilde{f}^{\prime}\tilde{g}^{0k\prime}\partial_{k}^{\prime}(\tilde{f}\partial_{\rho}\phi\delta^{3})$
(51) $\displaystyle=$ $\displaystyle
2i\tilde{f}\partial_{\rho}\phi\left[\tilde{f}^{-1}\partial_{0}\tilde{f}\delta^{3}+\tilde{g}^{0k}\partial_{k}(\tilde{f}\delta^{3})\right].$

Thus, we reach a result

$\displaystyle[\dot{\phi},\dot{b}_{\rho}^{\prime}]=i\tilde{f}(\partial_{\rho}\dot{\phi}-\tilde{f}^{-1}\partial_{0}\tilde{f}\partial_{\rho}\phi)\delta^{3}-2i\tilde{f}\partial_{\rho}\phi\tilde{g}^{0k}\partial_{k}(\tilde{f}\delta^{3}).$
(52)

Note that Eqs. (46), (48) and (52) hold as well when we replace $\phi$ by
$\varphi$.

##  4 Derivation of ETCRs involving $b_{\mu}$ field and choral symmetry

We are now in a position to address the novel part of our formulation and
discuss a huge residual symmetry called “choral symmetry” which emerges in
adopting the de Donder gauge for the general coordinate invariance.

In order to derive the commutation relations among the $b_{\mu}$ field in
terms of the BRST transformation, let us start with the latter equation in Eq.
(30) and take the BRST transformation:

$\displaystyle 0$ $\displaystyle=$
$\displaystyle\\{iQ_{B},[b_{\mu},\bar{c}_{\nu}^{\prime}]\\}$ (53)
$\displaystyle=$
$\displaystyle\\{[iQ_{B},b_{\mu}],\bar{c}_{\nu}^{\prime}\\}+[b_{\mu},\\{iQ_{B},\bar{c}_{\nu}^{\prime}\\}]$
$\displaystyle=$
$\displaystyle-\\{c^{\rho}\partial_{\rho}b_{\mu},\bar{c}_{\nu}^{\prime}\\}+i[b_{\mu},b_{\nu}^{\prime}]-[b_{\mu},c^{\rho\prime}\partial_{\rho}\bar{c}_{\nu}^{\prime}],$

where Eqs. (3), (5) and (6) have been used. Using Eq. (31), the first and
third terms precisely cancel so we can obtain

$\displaystyle[b_{\mu},b_{\nu}^{\prime}]=0.$ (54)

Next, let us turn our attention to the derivation of the ETCR:

$\displaystyle[b_{\mu},\dot{b}_{\nu}^{\prime}]=i\tilde{f}(\partial_{\mu}b_{\nu}+\partial_{\nu}b_{\mu})\delta^{3}.$
(55)

To do that, let us start with the ETCR

$\displaystyle[\pi_{c\mu},b_{\nu}^{\prime}]=0,$ (56)

which can be easily shown. Taking its BRST transformation leads to an
equation:

$\displaystyle[\\{iQ_{B},\pi_{c\mu}\\},b_{\nu}^{\prime}]-\\{\pi_{c\mu},[iQ_{B},b_{\nu}^{\prime}]\\}=0.$
(57)

The first term on the LHS can be calculated to be

$\displaystyle[\\{iQ_{B},\pi_{c\mu}\\},b_{\nu}^{\prime}]=-i[(\tilde{g}^{\rho\sigma}\nabla_{\sigma}c^{0}+\tilde{g}^{0\sigma}\nabla_{\sigma}c^{\rho}-\tilde{g}^{\rho
0}\nabla_{\lambda}c^{\lambda})\partial_{\rho}\bar{c}_{\mu},b_{\nu}^{\prime}]$
$\displaystyle+i\partial_{\nu}b_{\mu}\cdot\delta^{3}-\tilde{g}^{00}[b_{\mu},\dot{b}_{\nu}^{\prime}]+i[\tilde{g}^{\rho
0}\partial_{\rho}(c^{\lambda}\partial_{\lambda}\bar{c}_{\mu}),b_{\nu}^{\prime}],$
(58)

where Eqs. (3), (5), (22), (26) and (54) have been used. The second term on
the LHS in Eq. (57) can be also calculated to be

$\displaystyle\\{\pi_{c\mu},[iQ_{B},b_{\nu}^{\prime}]\\}=-i\partial_{\mu}b_{\nu}\cdot\delta^{3}+c^{\lambda\prime}[\pi_{c\mu},\partial_{\lambda}b_{\nu}^{\prime}].$
(59)

Then, Eq. (57) together with Eqs. (58) and (59) gives us an equation:

$\displaystyle[b_{\mu},\dot{b}_{\nu}^{\prime}]$ $\displaystyle=$
$\displaystyle
i\tilde{f}(\partial_{\mu}b_{\nu}+\partial_{\nu}b_{\mu})\delta^{3}-i\tilde{f}[(\tilde{g}^{\rho\sigma}\nabla_{\sigma}c^{0}+\tilde{g}^{0\sigma}\nabla_{\sigma}c^{\rho}-\tilde{g}^{\rho
0}\nabla_{\lambda}c^{\lambda})\partial_{\rho}\bar{c}_{\mu},b_{\nu}^{\prime}]$
(60) $\displaystyle+$ $\displaystyle i\tilde{f}[\tilde{g}^{\rho
0}\partial_{\rho}(c^{\lambda}\partial_{\lambda}\bar{c}_{\mu}),b_{\nu}^{\prime}]-\tilde{f}c^{\lambda\prime}[\pi_{c\mu},\partial_{\lambda}b_{\nu}^{\prime}]$
$\displaystyle\equiv$ $\displaystyle
i\tilde{f}(\partial_{\mu}b_{\nu}+\partial_{\nu}b_{\mu})\delta^{3}+K_{\mu\nu}.$

After some calculations, we can prove $K_{\mu\nu}=0$, which implies that Eq.
(55) is certainly valid. Note that in proving $K_{\mu\nu}=0$, it is necessary
to make use of Eq. (43) and the field equation for the antighost
$\bar{c}_{\mu}$, i.e.,
$\tilde{g}^{\mu\nu}\partial_{\mu}\partial_{\nu}\bar{c}_{\rho}=0$. In
particular, we must use the following ETCRs:

$\displaystyle[\ddot{\bar{c}}_{\mu},b_{\nu}^{\prime}]=-2i\tilde{f}[\partial_{\nu}\dot{\bar{c}}_{\mu}\delta^{3}-\tilde{g}^{0k}\partial_{\nu}\bar{c}_{\mu}\partial_{k}(\tilde{f}\delta^{3})],$
$\displaystyle[\nabla_{\sigma}c^{\rho},b_{\nu}^{\prime}]=i\tilde{f}(-\delta_{\sigma}^{0}\nabla_{\nu}c^{\rho}+\delta_{\nu}^{\rho}\Gamma_{\sigma\lambda}^{0}c^{\lambda}-\Gamma_{\sigma\nu}^{\rho}c^{0})\delta^{3}$
$\displaystyle+i\delta_{\nu}^{\rho}(2\delta_{\sigma}^{0}c^{0}\tilde{f}\tilde{g}^{0k}-\delta_{\sigma}^{0}c^{k}-\delta_{\sigma}^{k}c^{0})\partial_{k}(\tilde{f}\delta^{3}).$
(61)

We end this section with the argument of choral symmetry, which is a huge
residual symmetry $IOSp(8|8)$ involving the BRST symmetry, the rigid
translation and $GL(4)$ symmetry etc. Via the Noether theorem, the $IOSp(8|8)$
generators can be constructed out of ${\cal{L}}_{GF}+{\cal{L}}_{FP}$ in Eq.
(9) as [8]

$\displaystyle M^{MN}$ $\displaystyle\equiv$ $\displaystyle\int
d^{3}x\,\tilde{g}^{0\nu}(X^{M}\vbox{\halign{#\cr$\leftrightarrow$\crcr\kern-1.0pt\nointerlineskip\cr$\hfil\displaystyle{\partial}\hfil$\crcr}}_{\nu}X^{N})\equiv\int
d^{3}x\,\tilde{g}^{0\nu}(X^{M}\partial_{\nu}X^{N}-\partial_{\nu}X^{M}\cdot
X^{N}),$ (64) $\displaystyle P^{M}$ $\displaystyle\equiv$ $\displaystyle\int
d^{3}x\,\tilde{g}^{0\nu}(1\vbox{\halign{#\cr$\leftrightarrow$\crcr\kern-1.0pt\nointerlineskip\cr$\hfil\displaystyle{\partial}\hfil$\crcr}}_{\nu}X^{N})\equiv\int
d^{3}x\,\tilde{g}^{0\nu}\partial_{\nu}X^{M},$ (67)

where $X^{M}\equiv\\{x^{\mu},b_{\mu},c^{\mu},\bar{c}_{\mu}\\}$. The
$IOSp(8|8)$ algebra takes the graded form:

$\displaystyle[iM^{MN},M^{RS}\\}=-M^{MS}\tilde{\eta}^{NR}+(-)^{|R||S|}M^{MR}\tilde{\eta}^{NS}-(-)^{|M||N|}(M\leftrightarrow
N),$
$\displaystyle[iM^{MN},P^{R}\\}=-P^{M}\tilde{\eta}^{NR}+(-)^{|M||N|}P^{N}\tilde{\eta}^{MR},\qquad[P^{M},P^{N}\\}=0,$
(68)

where $\tilde{\eta}^{MN}$ is a $16\times 16$ $IOSp(8|8)$ metric [13].

Since the generators $M^{MN}$ and $P^{M}$ could have one time derivative,
calculating the algebra requires us to use the ETCRs including two time
derivatives such as

$\displaystyle[\dot{b}_{\mu},\dot{b}_{\nu}^{\prime}]=i\tilde{f}\left[\partial_{\mu}\dot{b}_{\nu}+\partial_{\nu}\dot{b}_{\mu}-2\tilde{f}\tilde{g}^{0\rho}\partial_{\rho}\partial_{\mu}b_{\nu}-\frac{\partial_{0}\tilde{f}}{\tilde{f}}(\partial_{\mu}b_{\nu}+\partial_{\nu}b_{\mu})\right]\delta^{3},$
$\displaystyle-2i\tilde{f}\tilde{g}^{0k}(\partial_{\mu}b_{\nu}+\partial_{\nu}b_{\mu})\partial_{k}(\tilde{f}\delta^{3})$
$\displaystyle[\dot{c}^{\mu},\dot{b}_{\nu}^{\prime}]=i\tilde{f}\left[\left(\partial_{\nu}\dot{c}^{\mu}-\frac{\partial_{0}\tilde{f}}{\tilde{f}}\partial_{\nu}c^{\mu}\right)\delta^{3}-2\tilde{g}^{0k}\partial_{\nu}c^{\mu}\partial_{k}(\tilde{f}\delta^{3})\right],$
$\displaystyle[\dot{\bar{c}}_{\mu},\dot{b}_{\nu}^{\prime}]=i\tilde{f}\left[\left(\partial_{\nu}\dot{\bar{c}}_{\mu}-\frac{\partial_{0}\tilde{f}}{\tilde{f}}\partial_{\nu}\bar{c}_{\mu}\right)\delta^{3}-2\tilde{g}^{0k}\partial_{\nu}\bar{c}_{\mu}\partial_{k}(\tilde{f}\delta^{3})\right].$
(69)

These ETCRs can be all derived from the field equations and the ETCRs obtained
so far, for instance, Eqs. (12), (19), (29), (30) and (31) without specifying
the expression of a classical gravitational Lagrangian. This situation should
be contrasted with the previous formulation [8] where Eq. (55) has been
derived by using the Einstein-Hilbert Lagrangian. Thus, it is said that the
choral symmetry uniquely characterizes the expression of
${\cal{L}}_{GF}+{\cal{L}}_{FP}$ [8]. By contrast, we can mention that the
origin of the choral symmetry purely lies in the de Donder gauge and the
corresponding FP ghost Lagrangian irrespective of a specific choice of a
classical gravitational theory as long as there exists the general coordinate
invariance in the classical theory.

##  5 A scale invariant scalar-tensor gravity

In this section, as a classical Lagrangian, we will take the simplest scalar-
tensor gravity [3] whose Lagrangian is given by

$\displaystyle{\cal{L}}_{c}=\sqrt{-g}\left(\frac{1}{2}\xi\phi^{2}R-\frac{1}{2}g^{\mu\nu}\partial_{\mu}\phi\partial_{\nu}\phi\right),$
(70)

where $\xi$ is a constant called the non-minimal coupling constant, $\phi$ a
real scalar field with a normal kinetic term (i.e., not a ghost), and $R$ the
scalar curvature. In addition to the general coordinate transformation (GCT)
and a global scale transformation with $\Omega=\textrm{constant}$, this
Lagrangian is also invariant under the restricted Weyl transformation
[14]-[16], [4]-[6]:

$\displaystyle g_{\mu\nu}\rightarrow
g^{\prime}_{\mu\nu}=\Omega^{2}(x)g_{\mu\nu},\qquad\phi\rightarrow\phi^{\prime}=\Omega^{-1}(x)\phi,$
(71)

where the gauge transformation parameter $\Omega(x)$ obeys a constraint
$\Box\Omega=0$. In order to prove the invariance, we need to use the following
transformation of the scalar curvature under (71):

$\displaystyle R\rightarrow R^{\prime}=\Omega^{-2}(R-6\Omega^{-1}\Box\Omega).$
(72)

For the sake of simplicity, in what follows we will put

$\displaystyle\varphi\equiv\frac{1}{2}\xi\phi^{2}.$ (73)

It is worth recalling that we can rewrite (70) as the Lagrangian of the well-
known Brans-Dicke theory [11]:

$\displaystyle{\cal{L}}_{c}=\sqrt{-g}\left(\varphi
R-\omega\frac{1}{\varphi}g^{\mu\nu}\partial_{\mu}\varphi\partial_{\nu}\varphi\right),$
(74)

where $\omega\equiv\frac{1}{4\xi}$ is known as the Brans-Dicke parameter.
Thus, our classical Lagrangian (70) is at least classically equivalent to that
of Brans-Dicke theory.

Taking the de Donder condition as a gauge-fixing condition for GCT, the gauge-
fixed and BRST-invariant quantum Lagrangian is given by

$\displaystyle{\cal{L}}_{q}=\sqrt{-g}\left(\varphi
R-\frac{1}{2}g^{\mu\nu}\partial_{\mu}\phi\partial_{\nu}\phi\right)-\tilde{g}^{\mu\nu}\partial_{\mu}b_{\nu}-i\tilde{g}^{\mu\nu}\partial_{\mu}\bar{c}_{\rho}\partial_{\nu}c^{\rho}.$
(75)

From this Lagrangian, we can obtain field equations by taking the variation
with respect to $g_{\mu\nu}$, $\phi$, $b_{\nu}$, $\bar{c}_{\rho}$ and
$c^{\rho}$ in order:

$\displaystyle\varphi
G_{\mu\nu}-(\nabla_{\mu}\nabla_{\nu}-g_{\mu\nu}\Box)\varphi-\frac{1}{2}T_{\mu\nu}-\frac{1}{2}(E_{\mu\nu}-\frac{1}{2}g_{\mu\nu}E)=0,$
$\displaystyle\xi\phi R+\Box\phi=0,\qquad\partial_{\mu}\tilde{g}^{\mu\nu}=0,$
$\displaystyle g^{\mu\nu}\partial_{\mu}\partial_{\nu}c^{\rho}=0,\qquad
g^{\mu\nu}\partial_{\mu}\partial_{\nu}\bar{c}_{\rho}=0,$ (76)

where we have defined

$\displaystyle
T_{\mu\nu}=\partial_{\mu}\phi\partial_{\nu}\phi-\frac{1}{2}g_{\mu\nu}(\partial_{\rho}\phi)^{2}.$
(77)

Of course, even in this case the $b_{\rho}$ field satisfies the d’Alembert
equation (19).

##  6 Equal-time commutation relations in gravitational sector

Since we have introduced the classical Lagrangian (70) in a theory at hand, we
are now ready to evaluate the equal-time commutation relations (ETCRs)
involving the metric tensor and the scalar fields in the gravitational sector.

For later convenience, here let us take account of the de Donder gauge
condition (2), from which we have identities:

$\displaystyle g^{\mu\nu}\Gamma^{\lambda}_{\mu\nu}=0,\qquad
g^{\lambda\mu}\partial_{\lambda}g_{\mu\nu}=\Gamma^{\lambda}_{\lambda\nu}.$
(78)

Moreover, since the equation $g^{\mu\nu}\Gamma^{\lambda}_{\mu\nu}=0$ reads

$\displaystyle(2g^{\lambda\mu}g^{\nu\rho}-g^{\mu\nu}g^{\lambda\rho})\partial_{\rho}g_{\mu\nu}=0,$
(79)

it is possible to express the time derivative of the metric field in terms of
its spacial one as

$\displaystyle{\cal{D}}^{\lambda\mu\nu}\dot{g}_{\mu\nu}=(2g^{\lambda\mu}g^{\nu
k}-g^{\mu\nu}g^{\lambda k})\partial_{k}g_{\mu\nu},$ (80)

where the operator ${\cal{D}}^{\lambda\mu\nu}$ is defined by

$\displaystyle{\cal{D}}^{\lambda\mu\nu}=g^{0\lambda}g^{\mu\nu}-2g^{\lambda\mu}g^{0\nu}.$
(81)

To remove second order derivatives of the metric involved in $R$, we perform
the integration by parts once and rewrite the Lagrangian (75) as666See
Appendix B.

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

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

$\displaystyle{\cal{V}}^{\mu}=\varphi(\tilde{g}^{\alpha\beta}\Gamma^{\mu}_{\alpha\beta}-\tilde{g}^{\mu\nu}\Gamma^{\alpha}_{\nu\alpha})-\tilde{g}^{\mu\nu}b_{\nu}.$
(83)

From the Lagrangian (82), the concrete expressions for canonical conjugate
momenta for the metric tensor and the scalar fields read

$\displaystyle\pi_{g}^{\mu\nu}$ $\displaystyle=$
$\displaystyle\frac{\partial{\cal{L}}}{\partial\dot{g}_{\mu\nu}}$
$\displaystyle=$
$\displaystyle-\frac{1}{2}\sqrt{-g}\,\varphi\Bigl{[}-g^{0\lambda}g^{\mu\nu}g^{\sigma\tau}-g^{0\tau}g^{\mu\lambda}g^{\nu\sigma}-g^{0\sigma}g^{\mu\tau}g^{\nu\lambda}+g^{0\lambda}g^{\mu\tau}g^{\nu\sigma}$
$\displaystyle+$ $\displaystyle
g^{0\tau}g^{\mu\nu}g^{\lambda\sigma}+\frac{1}{2}(g^{0\mu}g^{\nu\lambda}+g^{0\nu}g^{\mu\lambda})g^{\sigma\tau}\Biggr{]}\partial_{\lambda}g_{\sigma\tau}$
$\displaystyle-$
$\displaystyle\sqrt{-g}\Biggl{[}\frac{1}{2}(g^{0\mu}g^{\rho\nu}+g^{0\nu}g^{\rho\mu})-g^{\mu\nu}g^{\rho
0}\Bigr{]}\partial_{\rho}\varphi$ $\displaystyle-$
$\displaystyle\frac{1}{2}\sqrt{-g}(g^{0\mu}g^{\nu\rho}+g^{0\nu}g^{\mu\rho}-g^{0\rho}g^{\mu\nu})b_{\rho},$
$\displaystyle\pi_{\phi}$ $\displaystyle=$
$\displaystyle\frac{\partial{\cal{L}}}{\partial\dot{\phi}}=-\tilde{g}^{0\mu}\partial_{\mu}\phi+\xi\phi(-\tilde{g}^{\alpha\beta}\Gamma^{0}_{\alpha\beta}+\tilde{g}^{0\nu}\Gamma^{\alpha}_{\nu\alpha}).$
(84)

From now on, we would like to evaluate several nontrivial equal-time
commutation relations (ETCRs) relevant to the gravitational Lagrangian (82).
For this purpose, let us first write down some equations to be solved in
order. Since $\pi_{\phi}$ in Eq. (84) is rewritten as

$\displaystyle\pi_{\phi}=-\tilde{g}^{00}\dot{\phi}-\tilde{g}^{0k}\partial_{k}\phi+\xi\phi\left[(\tilde{g}^{00}g^{\rho\sigma}-\tilde{g}^{0\rho}g^{0\sigma})\dot{g}_{\rho\sigma}+(\tilde{g}^{0k}g^{\rho\sigma}-\tilde{g}^{0\rho}g^{k\sigma})\partial_{k}g_{\rho\sigma}\right],$
(85)

$[\pi_{\phi},\phi^{\prime}]=-i\delta^{3}$ in Eq. (21) gives rise to an
equation:

$\displaystyle-\tilde{g}^{00}[\dot{\phi},\phi^{\prime}]+\xi\phi(\tilde{g}^{00}g^{\rho\sigma}-\tilde{g}^{0\rho}g^{0\sigma})[\dot{g}_{\rho\sigma},\phi^{\prime}]=-i\delta^{3}.$
(86)

Next, $[\pi_{\phi},g^{\prime}_{\mu\nu}]=0$ produces an equation:

$\displaystyle-\tilde{g}^{00}[\dot{\phi},g_{\mu\nu}^{\prime}]+\xi\phi(\tilde{g}^{00}g^{\rho\sigma}-\tilde{g}^{0\rho}g^{0\sigma})[\dot{g}_{\rho\sigma},g_{\mu\nu}^{\prime}]=0.$
(87)

Moreover,
$[{\cal{D}}^{\lambda\rho\sigma}\dot{g}_{\rho\sigma},g_{\mu\nu}^{\prime}]=0$,
which stems from the ${\cal{D}}$-equation (80), reads

$\displaystyle(g^{0\lambda}g^{\rho\sigma}-2g^{\lambda\rho}g^{0\sigma})[\dot{g}_{\rho\sigma},g_{\mu\nu}^{\prime}]=0.$
(88)

Similarly,
$[{\cal{D}}^{\lambda\rho\sigma}\dot{g}_{\rho\sigma},\phi^{\prime}]=0$ gives us
an equation:

$\displaystyle(g^{0\lambda}g^{\rho\sigma}-2g^{\lambda\rho}g^{0\sigma})[\dot{g}_{\rho\sigma},\phi^{\prime}]=0.$
(89)

Now we are willing to solve Eqs. (86)-(89). First of all, let us focus on Eq.
(89). From the argument of symmetry, $[\dot{g}_{\rho\sigma},\phi^{\prime}]$
must be of form:

$\displaystyle[\dot{g}_{\rho\sigma},\phi^{\prime}]=a_{1}(g_{\rho\sigma}+a_{2}\delta_{\rho}^{0}\delta_{\sigma}^{0})\delta^{3},$
(90)

where $a_{1},a_{2}$ are certain coefficients to be determined sooner. Indeed,
from Eq. (89), we find that $a_{2}=\frac{2}{g^{00}}$, so the ETCR (90) reads

$\displaystyle[\dot{g}_{\rho\sigma},\phi^{\prime}]=a_{1}\left(g_{\rho\sigma}+\frac{2}{g^{00}}\delta_{\rho}^{0}\delta_{\sigma}^{0}\right)\delta^{3}.$
(91)

Next, let us solve Eq. (88). In this case, we also find that
$[\dot{g}_{\rho\sigma},g_{\mu\nu}^{\prime}]$ 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

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

where $c_{i}(i=1,\cdots,5)$ are some coefficients. Imposing Eq. (88) on (92)
leads to relations among the coefficients:

$\displaystyle c_{3}=2(c_{1}+c_{2}),\qquad c_{4}=-c_{2},\qquad
c_{5}=4(c_{1}+c_{2}).$ (93)

Furthermore, using Eq. (91), Eq. (86) gives $[\dot{\phi},\phi^{\prime}]$,
which is of form:

$\displaystyle[\dot{\phi},\phi^{\prime}]=(i\tilde{f}+3a_{1}\xi\phi)\delta^{3}.$
(94)

Finally, with the help of Eqs. (91), (92) and (93), Eq. (87) leads to a
relation:

$\displaystyle a_{1}=(3c_{1}+2c_{2})\xi\phi.$ (95)

In order to fix the coefficients $a_{1},c_{1}$ and $c_{2}$ completely, we need
to have two independent relations among them. Such the relations can be
provided by calculating $[\dot{g}_{kl},g_{mn}^{\prime}]$ 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. (21) and the concrete expression of $\pi_{g}^{kl}$ in Eq. (84). To do
that, from Eq. (84), let us write

$\displaystyle\pi_{g}^{kl}=\hat{A}^{kl}+\hat{B}^{kl\rho}b_{\rho}+\hat{C}^{klmn}\dot{g}_{mn}+\hat{D}^{kl}\dot{\varphi}.$
(96)

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 as777It
turns out that the concrete expressions of $\hat{A}^{kl}$ and
$\hat{B}^{kl\rho}$ are irrelevant to the calculation of
$[\dot{g}_{kl},g_{mn}^{\prime}]$.

$\displaystyle\hat{C}^{klmn}=\frac{1}{2}h\varphi
K^{klmn},\qquad\hat{D}^{kl}=\tilde{g}^{00}g^{kl}-\tilde{g}^{0k}g^{0l},$ (97)

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

$\displaystyle K^{klmn}=\left|\begin{array}[]{rrr}g^{00}&g^{0l}&g^{0n}\\\
g^{k0}&g^{kl}&g^{kn}\\\ g^{m0}&g^{ml}&g^{mn}\\\ \end{array}\right|,$ (101)
$\displaystyle
K^{klmn}\frac{1}{2}(g^{00})^{-1}(g_{ij}g_{mn}-g_{im}g_{jn}-g_{in}g_{jm})=\frac{1}{2}(\delta_{i}^{k}\delta_{j}^{l}+\delta_{i}^{l}\delta_{j}^{k}).$
(102)

From Eq. (96), we can calculate

$\displaystyle[\dot{g}_{kl},g_{mn}^{\prime}]=\hat{C}^{-1}_{klpq}\left([\pi_{g}^{pq},g_{mn}^{\prime}]-\hat{B}^{pq\rho}[b_{\rho},g_{mn}^{\prime}]-\hat{D}^{pq}[\dot{\varphi},g_{mn}^{\prime}]\right).$
(103)

Then, using Eqs. (21), (25), (91) and (102), we find

$\displaystyle[\dot{g}_{kl},g_{mn}^{\prime}]$ $\displaystyle=$
$\displaystyle\tilde{f}\varphi^{-1}\left[(-i-a_{1}\tilde{g}^{00}\xi\phi)g_{kl}g_{mn}+i(g_{km}g_{ln}+g_{kn}g_{lm})\right]\delta^{3}$
(104) $\displaystyle=$
$\displaystyle\left[c_{1}g_{kl}g_{mn}+c_{2}(g_{km}g_{ln}+g_{kn}g_{lm})\right]\delta^{3},$

where the last equality comes from Eq. (92). In this way, we have suceeded in
getting two independent relations among $a_{1},c_{1}$ and $c_{2}$:

$\displaystyle
c_{1}=\tilde{f}\varphi^{-1}(-i-a_{1}\tilde{g}^{00}\xi\phi),\qquad
c_{2}=i\tilde{f}\varphi^{-1}.$ (105)

Using Eqs. (95) and (105), we can fix completely the coefficients as

$\displaystyle c_{1}=-i\frac{4\xi+1}{6\xi+1}\tilde{f}\varphi^{-1},\qquad
c_{2}=i\tilde{f}\varphi^{-1},\qquad
a_{1}=-i\frac{2}{6\xi+1}\tilde{f}\phi^{-1}.$ (106)

Accordingly, we can obtain the following ETCRs:

$\displaystyle[\dot{g}_{\rho\sigma},\phi^{\prime}]=-\frac{2}{6\xi+1}i\tilde{f}\phi^{-1}\left(g_{\rho\sigma}+\frac{2}{g^{00}}\delta_{\rho}^{0}\delta_{\sigma}^{0}\right)\delta^{3}.$
(107)
$\displaystyle[\dot{\phi},\phi^{\prime}]=\frac{1}{6\xi+1}i\tilde{f}\delta^{3}.$
(108) $\displaystyle[\dot{g}_{\rho\sigma},g_{\mu\nu}^{\prime}]$
$\displaystyle=$ $\displaystyle
i\tilde{f}\varphi^{-1}\biggl{\\{}-\frac{4\xi+1}{6\xi+1}g_{\rho\sigma}g_{\mu\nu}+g_{\rho\mu}g_{\sigma\nu}+g_{\rho\nu}g_{\sigma\mu}$
(109) $\displaystyle+$ $\displaystyle
h\tilde{f}\Bigl{[}\frac{4\xi}{6\xi+1}(\delta_{\rho}^{0}\delta_{\sigma}^{0}g_{\mu\nu}+\delta_{\mu}^{0}\delta_{\nu}^{0}g_{\rho\sigma})-(\delta_{\rho}^{0}\delta_{\mu}^{0}g_{\sigma\nu}+\delta_{\rho}^{0}\delta_{\nu}^{0}g_{\sigma\mu}$
$\displaystyle+$
$\displaystyle\delta_{\sigma}^{0}\delta_{\mu}^{0}g_{\rho\nu}+\delta_{\sigma}^{0}\delta_{\nu}^{0}g_{\rho\mu})\Bigr{]}+(h\tilde{f})^{2}\frac{8\xi}{6\xi+1}\delta_{\rho}^{0}\delta_{\sigma}^{0}\delta_{\mu}^{0}\delta_{\nu}^{0}\biggr{\\}}\delta^{3}.$

It is worthwhile to notice that these ETCRs have two peculiar features, one of
which is the presence of the factor $6\xi+1$ in the denominator, thereby
implying that they do not make sense in a theory with a local scale (or Weyl)
symmetry. In other words, in the case of $6\xi+1=0$ corresponding to the Weyl
invariant scalar-tensor gravity, we need to introduce one more gauge condition
such as $R=0$ or $\phi={\rm{constant}}$ to fix the Weyl symmetry. The other
important feature is the existence of the field $\phi$ (or $\varphi$) in the
denominator, which means that an unbroken phase $\langle\phi(x)\rangle=0$
cannot be dealt with in the present formalism. This fact strongly suggests
that a (global) scale invariance must be broken spontaneously even in quantum
gravity as in classical gravity in order to construct a consistent quantum
theory of the scale invariant scalar-tensor gravity.

##  7 Spontaneous symmetry breakdown of scale invariance

In the previous work, we have shown that the scale invariance is in fact
spontaneously broken in classical gravity where the gravitational sector is
not quantized [4]-[6]. In this section, we wish to investigate whether a
(global) scale invariance is spontaneously broken even in quantum gravity or
not.

Taking $\Omega$ in Eq. (71) to be a constant, we can define a scale
transformation as

$\displaystyle g_{\mu\nu}$ $\displaystyle\rightarrow$ $\displaystyle
g^{\prime}_{\mu\nu}=\Omega^{2}g_{\mu\nu},\qquad\phi\rightarrow\phi^{\prime}=\Omega^{-1}\phi,$
$\displaystyle b_{\rho}$ $\displaystyle\rightarrow$ $\displaystyle
b^{\prime}_{\rho}=\Omega^{-2}b_{\rho},\qquad\bar{c}_{\rho}\rightarrow\bar{c}^{\prime}_{\rho}=\Omega^{-1}\bar{c}_{\rho},\qquad\bar{c}^{\rho}\rightarrow\bar{c}^{\rho\prime}=\Omega^{-1}\bar{c}^{\rho},$
(110)

where we have added the scale transformation for the Nakanishi-Lautrup field
and the FP (anti-)ghosts. Then, it is easy to see that the quantum Langrangian
(75) is invariant under the scale transformation (110). This fact implies that
the de Donder gauge is invariant under the scale transformation. Incidentally,
the de Donder gauge is not invariant under a local scale (or Weyl)
transformation. To make a gauge condition for the general coordinate
invariance be invariant under the Weyl transformation requires us to take a
different gauge condition such as
$\partial_{\mu}((-g)^{\frac{1}{4}}g^{\mu\nu})=0$ or
$\partial_{\mu}(\sqrt{-g}\phi^{2}g^{\mu\nu})=0$ from the de Donder gauge.

Since the scale transformation is a global one, we can construct a conserved
Noether current and charge along the standard procedure. After some
calculations, it turns out that the conserved current for the scale symmetry
is given by888See Appendix B.

$\displaystyle
J^{\mu}=\tilde{g}^{\mu\nu}\left[\frac{6\xi+1}{2}\partial_{\nu}(\phi^{2})+2b_{\nu}+i\partial_{\nu}(\bar{c}_{\rho}c^{\rho})\right].$
(111)

It is straightforward to verify that this current is conserved,
$\partial_{\mu}J^{\mu}=0$, in terms of field equations. Note that in the
conformal coupling $\xi=-\frac{1}{6}$, the first term on the RHS coming from
the classical action is identically vanishing [17, 18] while the second and
third terms coming from the gauge-fixing and FP ghost terms, respectively, do
not so.

The Noether charge $Q=\int d^{3}xJ^{0}$ turns out to generate the
infinitesimal scale transformation correctly by using the ETCRs obtained thus
far:

$\displaystyle\delta g_{\mu\nu}$ $\displaystyle=$ $\displaystyle 2\Lambda
g_{\mu\nu},\qquad\delta\phi=-\Lambda\phi,\qquad\delta b_{\rho}=-2\Lambda
b_{\rho},$ $\displaystyle\delta\bar{c}_{\rho}$ $\displaystyle=$
$\displaystyle-\Lambda\bar{c}_{\rho},\qquad\delta c^{\rho}=-\Lambda c^{\rho},$
(112)

where we have set $\Omega=e^{\Lambda}\approx 1+\Lambda$. It might be curious
about why no derivative of the metric tensor $g_{\mu\nu}$ appears in $Q$ since
it usually generates the transformation of the metric tensor [3]. This problem
can be understood for the first time in the present formulation since we have
successfully quantized the metric field and the scalar field.

To check that the charge $Q$ indeed generates the scale transformation (112),
let us calculate $\delta g_{\mu\nu}$ and $\delta b_{\rho}$ explicitly. As for
$\delta g_{\mu\nu}$,

$\displaystyle\delta g_{\mu\nu}$ $\displaystyle\equiv$ $\displaystyle[i\Lambda
Q,g_{\mu\nu}]$ (113) $\displaystyle=$ $\displaystyle i\Lambda\int
d^{3}x^{\prime}\left[\tilde{g}^{0\rho\prime}\left(\frac{6\xi+1}{2}\partial_{\rho}(\phi^{\prime
2})+2b_{\rho}^{\prime}+i\partial_{\rho}(\bar{c}_{\sigma}^{\prime}c^{\sigma\prime})\right),g_{\mu\nu}\right]$
$\displaystyle=$ $\displaystyle i\Lambda\int
d^{3}x^{\prime}\Biggl{\\{}(6\xi+1)\tilde{g}^{00\prime}\phi^{\prime}[\dot{\phi}^{\prime},g_{\mu\nu}]+2\tilde{g}^{0\rho\prime}[b_{\rho}^{\prime},g_{\mu\nu}]\Biggr{\\}}$
$\displaystyle=$ $\displaystyle 2\Lambda g_{\mu\nu},$

where in the third equality we put $x^{0}=x^{\prime 0}$, and used Eqs. (25)
and (107). Against the expectation that the derivative of $\phi$ would play a
role [3], the $b$ field also does the job in generating the scale
transformation. In a similar manner, as for $\delta b_{\rho}$, we have

$\displaystyle\delta b_{\rho}$ $\displaystyle\equiv$ $\displaystyle[i\Lambda
Q,b_{\rho}]$ (114) $\displaystyle=$ $\displaystyle i\Lambda\int
d^{3}x^{\prime}\left[\tilde{g}^{0\nu\prime}\left(\frac{6\xi+1}{2}\partial_{\nu}(\phi^{\prime
2})+2b_{\nu}^{\prime}+i\partial_{\nu}(\bar{c}_{\sigma}^{\prime}c^{\sigma\prime})\right),b_{\rho}\right]$
$\displaystyle=$ $\displaystyle i\Lambda\int
d^{3}x^{\prime}\Biggl{\\{}[\tilde{g}^{0\nu\prime},b_{\rho}]\left(\frac{6\xi+1}{2}\partial_{\nu}(\phi^{\prime
2})+2b_{\nu}^{\prime}+i\partial_{\nu}(\bar{c}_{\sigma}^{\prime}c^{\sigma\prime})\right)$
$\displaystyle+$
$\displaystyle\tilde{g}^{00\prime}\Biggl{(}(6\xi+1)\phi^{\prime}[\dot{\phi}^{\prime},b_{\rho}]+i\biggl{(}[\dot{\bar{c}}_{\sigma}^{\prime},b_{\rho}]c^{\sigma\prime}+\bar{c}_{\sigma}^{\prime}[\dot{c}^{\sigma\prime},b_{\rho}]\biggr{)}\Biggr{)}\Biggr{\\}}$
$\displaystyle=$ $\displaystyle-2\Lambda b_{\rho},$

where we have used Eqs. (26), (31) and (48). We wish to mention again that in
case of classical gravity, one cannot show that the charge $Q$ generates the
scale transformation owing to the absence of the ETCRs relevant to the
Nakanishi-Lautrup field $b_{\rho}$ and the FP ghosts $\bar{c}_{\sigma}$ and
$c^{\sigma}$.

Now let us move to the issue of spontaneous symmetry breakdown of scale
symmetry in quantum gravity. From Eq. (112) and the definition of
$\delta\Phi\equiv[i\Lambda Q,\Phi]$ for a generic field $\Phi$, we obtain that

$\displaystyle[iQ,g_{\mu\nu}]=2g_{\mu\nu},\qquad[iQ,\phi]=-\phi,\qquad[iQ,b_{\rho}]=-2b_{\rho},$
$\displaystyle[iQ,\bar{c}_{\rho}]=-\bar{c}_{\rho},\qquad[iQ,c^{\rho}]=-c^{\rho}.$
(115)

Assuming that the fields take the following vacuum expectation values:

$\displaystyle\langle 0|g_{\mu\nu}|0\rangle=\eta_{\mu\nu},\qquad\langle
0|\phi|0\rangle=\phi_{0},\qquad\langle 0|b_{\rho}|0\rangle=0,$
$\displaystyle\langle 0|\bar{c}_{\rho}|0\rangle=0,\qquad\langle
0|c^{\rho}|0\rangle=0,$ (116)

with $\phi_{0}$ being a constant, Eq. (115) implies that

$\displaystyle\langle 0|[iQ,g_{\mu\nu}]|0\rangle=2\eta_{\mu\nu},\qquad\langle
0|[iQ,\phi]|0\rangle=-\phi_{0},\qquad\langle 0|[iQ,b_{\rho}]|0\rangle=0,$
$\displaystyle\langle 0|[iQ,\bar{c}_{\rho}]|0\rangle=0,\qquad\langle
0|[iQ,c^{\rho}]|0\rangle=0.$ (117)

The second equation in Eq. (117) shows that the scale invariance is
spontaneously broken at the quantum level as long as $\phi_{0}\neq 0$ holds
[4, 6].

Here three important remarks are in order. First, the first assumption in Eq.
(116), $\langle 0|g_{\mu\nu}|0\rangle=\eta_{\mu\nu}$, comes from our postulate
that the vacuum is invariant under translation [8]:

$\displaystyle P_{\mu}|0\rangle=0,$ (118)

which means that translational invariance is not broken spontaneously.
Moreover, with this assumption the graviton can be identified with a Nambu-
Goldstone (NG) boson corresponding to spontaneous symmetry breakdown of
$GL(4)$ symmetry down to $SO(1,3)$ Lorentz symmetry, thereby proving the exact
masslessness of the graviton [9].

Second, let us pay our attention to the second assumption in Eq. (116),
$\langle 0|\phi|0\rangle=\phi_{0}\neq 0$, which is also physically plausible
by the following argument: As usual, let us consider to move from the Jordan
frame to the Einstein frame by implementing a local scale transformation only
for the metric tensor field as

$\displaystyle g_{\mu\nu}\rightarrow
g_{\ast\mu\nu}=\Omega(x)^{2}g_{\mu\nu}=\frac{1}{M_{Pl}^{2}}\xi\phi^{2}g_{\mu\nu},$
(119)

where $M_{Pl}$ is the reduced Planck mass. Then, in the Einstein frame, up to
a surface term the quantum Lagrangian (75) is reduced to the form:

$\displaystyle{\cal{L}}_{q}=\sqrt{-g_{\ast}}\left(\frac{M_{Pl}^{2}}{2}R_{\ast}-\frac{1}{2}g_{\ast}^{\mu\nu}\partial_{\mu}\sigma\partial_{\nu}\sigma\right)-e^{-\frac{2\zeta}{M_{Pl}}\sigma}\left(\tilde{g}_{\ast}^{\mu\nu}\partial_{\mu}b_{\nu}-i\tilde{g}_{\ast}^{\mu\nu}\partial_{\mu}\bar{c}_{\rho}\partial_{\nu}c^{\rho}\right),$
(120)

where we have defined a scalar field $\sigma(x)$ and a constant $\zeta$ as

$\displaystyle\phi=\xi^{-\frac{1}{2}}M_{Pl}e^{\frac{\zeta}{M_{Pl}}\sigma},\qquad\zeta=\sqrt{\frac{\xi}{6\xi+1}}.$
(121)

In this way, we can arrive at the Lagrangian in the Einstein frame by starting
with that in the Jordan frame. The key point for the change of the frames lies
in Eq. (119) where the scale factor $\Omega(x)$ is proportional to
$\frac{1}{M_{Pl}}\phi(x)$. Namely, the existence of the non-vanishing
“dilaton” $\phi\neq 0$, or more precisely, $\langle 0|\phi|0\rangle\neq 0$,
makes it possible to move from the Jordan frame to the Einstein one. In this
sense, our assumption $\langle 0|\phi|0\rangle=\phi_{0}\neq 0$ makes sense
physically.

As a final remark, as mentioned in the previous section, in order to make a
consistent quantum gravity based on the scale invariant scalar-tensor gravity,
it is necessary to require the condition $\langle 0|\phi|0\rangle\neq 0$. Any
physical theories must be formulated within the framework of quantum field
theories, so it is natural to assume such the condition.

To close this section, let us verify more explicitly that the spontaneous
symmetry breakdown of the scale symmetry occurs in the Einstein frame where
the scale symmetry is replaced with a shift symmetry. For this purpose, let us
rewrite the charge for the scale symmetry in the Jordan frame into that in the
Einstein frame as

$\displaystyle Q=\int
d^{3}x\,\tilde{g}_{\ast}^{0\nu}\left[\frac{M_{Pl}}{\zeta}\partial_{\nu}\sigma+e^{-\frac{2\zeta}{M_{Pl}}\sigma}\left(2b_{\nu}+\partial_{\nu}(\bar{c}_{\rho}c^{\rho})\right)\right],$
(122)

where we have used Eqs. (111), (119) and (121). Since $Q$ has a linear term in
$\sigma(x)$, the charge cannot annihilate the vacuum $|0\rangle$:

$\displaystyle Q|0\rangle\neq 0,$ (123)

which means the spontaneous symmetry breakdown of scale symmetry in the Jordan
frame or shift symmetry in the Einstein frame. Actually, from the Lagrangian
(120) the canonical conjugate momentum for the scalar field $\sigma(x)$ reads

$\displaystyle\pi_{\sigma}\equiv\frac{\partial{\cal{L}}}{\partial\partial_{0}\sigma}=-\sqrt{-g_{\ast}}g_{\ast}^{0\nu}\partial_{\nu}\sigma.$
(124)

Then, $Q$ can be rewritten as

$\displaystyle Q=-\frac{M_{Pl}}{\zeta}\int d^{3}x\,\pi_{\sigma}+\cdots,$ (125)

where $\dots$ denote contributions from the Nakanishi-Lautrup field and the FP
ghosts. Using the equal-time commutation relation
$[\sigma,\pi_{\sigma}^{\prime}]=i\delta^{3}$, we obtain

$\displaystyle[iQ,\sigma(x)]=-\frac{M_{Pl}}{\zeta}.$ (126)

Taking the vacuum expectation value of this equation yields

$\displaystyle\langle 0|[iQ,\sigma(x)]|0\rangle=-\frac{M_{Pl}}{\zeta}\neq 0,$
(127)

which clearly means the spontaneous symmetry breakdown and that the scalar
field $\sigma(x)$ is the massless NG boson for the shift symmetry.

##  8 Conclusions

In this article, we have performed a manifestly covariant quantization of a
scale invariant gravity which is equivalent to the Brans-Dicke gravity [11]
via the field redefinition of a scalar field. Many of studies of the Brans-
Dicke gravity have been thus far confined to the classical analysis, so it is
expected that our quantum formulation of the Brans-Dicke gravity could provide
us with some useful information on quantum aspects of the theory.

Actually, we have presented two new results, one of which is that in
classically scale invariant gravitational theories, we have a quantum scale
symmetry in addition to a huge choral symmetry when we choose the de Donder
gauge for the general coordinate invariance. In this respect, it is worthwhile
to recall that in the manifestly scale invariant regularization method
[19]-[24], the scale invariance is free of scale anomaly.

As the second result, we have shown that the scale symmetry is spontaneously
broken by quantum effects, thus proving that the “dilaton” is exactly massless
because of the Nambu-Goldstone theorem. As pointed out in the previous paper
[25], when the dilaton is exactly massless at the quantum level, it mediates a
long-range force between massive objects as in the Newtonian force, which
imposes a severe phenomenological constraint on parameters in the Brans-Dicke
model [11] since the long-range force stemming from the massless dilaton could
affect the perihelion advance of Mercury, for instance.

Furthermore, we have shown that the choral symmetry, which is a
Poincar${\rm{\acute{e}}}$-like $IOSp(8|8)$ supersymmetry, can be derived from
any gravitational theories, which are invariant under the general coordinate
transformation (GCT), if the GCT is gauge-fixed by the de Donder gauge. To put
it differently, the choral symmetry comes from only the de Donder gauge for
the GCT in the BRST formalism.

We still have a lot of works to be done in future. For instance, we would like
to extend the present formalism to gravitational theories with a local scale
invariance (or Weyl invariance) and investigate the resultant residual
symmetries. In the case of the Weyl invariance, it seems that we might prefer
the Weyl-invariant gauge conditions such as
$\partial_{\mu}((-g)^{\frac{1}{4}}g^{\mu\nu})=0$ and
$\partial_{\mu}(\sqrt{-g}\phi^{2}g^{\mu\nu})=0$ to the de Donder gauge
condition $\partial_{\mu}(\sqrt{-g}g^{\mu\nu})=0$ since we have the restricted
Weyl symmetry in these Weyl invariant gauge conditions when we take the gauge
condition $R=0$ for the Weyl invariance. However, then we will lose the choral
symmetry but instead we would have new residual symmetries.

As another problem, it is of interest to construct a manifestly scale-
invariant regularization scheme in the theory at hand and calculate an
effective potential explicitly [26]-[28]. With this regularization scheme, it
is necessary to introduce an additional scalar field in addition to the
dilaton and the two scalar fields might collaborate with each other for
nullifying the scalar force. In fact, such an approach on the basis of the
dilaton and the axion has been recently proposed [29]. In near future, we
would like to report these problems.

Acknowledgements

We would like to be grateful to T. Kugo for valuable discussions, in
particular, on Sections 2 and 3. This work is partly supported by the JSPS
Kakenhi Grant No. 21K03539.

## Appendix

##  Appendix A Derivation of $[R_{\mu\nu},b_{\rho}^{\prime}]$

In this appendix, we wish to prove the tensorlike ETCR:

$\displaystyle[R_{\mu\nu},b_{\rho}^{\prime}]=-i\tilde{f}(\delta_{\mu}^{0}R_{\rho\nu}+\delta_{\nu}^{0}R_{\rho\mu})\delta^{3}.$
(A.1)

Our strategy for the proof is similar to that of [8], but is in essense
different from it since we have been already able to derive the ETCR (55)
without recourse to the Einstein equation. In what follows, we will prove the
following two equations:

$\displaystyle[G^{0}\,_{\nu},b_{\rho}^{\prime}]=i\tilde{f}(\delta_{\rho}^{0}R^{0}\,_{\nu}-\delta_{\nu}^{0}R^{0}\,_{\rho})\delta^{3},\qquad[R_{kl},b_{\rho}^{\prime}]=0,$
(A.2)

from which it is easy to see that we can reach our goal (A.1).

Let us first prove the former equation in (A.2). From the Einstein equation,
which is the first equation in (76), $G^{0}\,_{\nu}$ is

$\displaystyle
G^{0}\,_{\nu}=\frac{1}{\varphi}\left[(\nabla^{0}\nabla_{\nu}-\delta^{0}_{\nu}\Box)\varphi+\frac{1}{2}T^{0}\,_{\nu}+\frac{1}{2}(E^{0}\,_{\nu}-\frac{1}{2}\delta^{0}_{\nu}E)\right].$
(A.3)

Note that the first term
$(\nabla^{0}\nabla_{\nu}-\delta^{0}_{\nu}\Box)\varphi$ contains no
$\ddot{\varphi}$. Then, $[G^{0}\,_{\nu},b_{\rho}^{\prime}]$ reads

$\displaystyle[G^{0}\,_{\nu},b_{\rho}^{\prime}]$ $\displaystyle=$
$\displaystyle\frac{1}{\varphi}\left\\{[(\nabla^{0}\nabla_{\nu}-\delta^{0}_{\nu}\Box)\varphi,b_{\rho}^{\prime}]+\frac{1}{2}[T^{0}\,_{\nu},b_{\rho}^{\prime}]+\frac{1}{2}[E^{0}\,_{\nu}-\frac{1}{2}\delta^{0}_{\nu}E,b_{\rho}^{\prime}]\right\\}$
(A.4) $\displaystyle\equiv$
$\displaystyle\frac{1}{\varphi}(A_{1}+A_{2}+A_{3}).$

After some calculations, $A_{1}$ is found to be

$\displaystyle
A_{1}\equiv[(\nabla^{0}\nabla_{\nu}-\delta^{0}_{\nu}\Box)\varphi,b_{\rho}^{\prime}]=i\tilde{f}(\delta^{0}_{\rho}\nabla^{0}\nabla_{\nu}\varphi-\delta^{0}_{\nu}\nabla^{0}\nabla_{\rho}\varphi)\delta^{3}.$
(A.5)

In order to evaluate $A_{2}$, it is necessary to calculate
$[T_{\mu\nu},b_{\rho}^{\prime}]$ whose result is given by

$\displaystyle[T_{\mu\nu},b_{\rho}^{\prime}]=-i\tilde{f}(\delta_{\mu}^{0}T_{\rho\nu}+\delta_{\nu}^{0}T_{\rho\mu})\delta^{3}.$
(A.6)

Using this result, $A_{2}$ becomes

$\displaystyle
A_{2}\equiv\frac{1}{2}[T^{0}\,_{\nu},b_{\rho}^{\prime}]=\frac{1}{2}i\tilde{f}(\delta^{0}_{\rho}T^{0}\,_{\nu}-\delta^{0}_{\nu}T^{0}\,_{\rho})\delta^{3}.$
(A.7)

Finally, evaluating $A_{3}$ requires us to calculate
$[E_{\mu\nu},b_{\rho}^{\prime}]$, which is found to be

$\displaystyle[E_{\mu\nu},b_{\rho}^{\prime}]=-i\tilde{f}(\delta_{\mu}^{0}E_{\rho\nu}+\delta_{\nu}^{0}E_{\rho\mu})\delta^{3}.$
(A.8)

Since we can calculate $[E,b_{\rho}^{\prime}]=0$, $A_{3}$ reads

$\displaystyle
A_{3}\equiv\frac{1}{2}[E^{0}\,_{\nu}-\frac{1}{2}\delta^{0}_{\nu}E,b_{\rho}^{\prime}]=\frac{1}{2}i\tilde{f}(\delta^{0}_{\rho}E^{0}\,_{\nu}-\delta^{0}_{\nu}E^{0}\,_{\rho})\delta^{3}.$
(A.9)

Adding $A_{1},A_{2}$ and $A_{3}$ gives rise to the former equation in (A.2)
since from Eq. (76) $R^{0}\,_{\nu}$ is described as

$\displaystyle
R^{0}\,_{\nu}=\frac{1}{\varphi}\left[(\nabla^{0}\nabla_{\nu}+\frac{1}{2}\delta^{0}_{\nu}\Box)\varphi+\frac{1}{2}(T^{0}\,_{\nu}-\frac{1}{2}\delta^{0}_{\nu}T)+\frac{1}{2}E^{0}\,_{\nu}\right].$
(A.10)

Next, let us prove the validity of the latter equation in (A.2). From the
Einstein equation in (76), $R_{kl}$ takes the form:

$\displaystyle
R_{kl}=\frac{1}{\varphi}\left[(\nabla_{k}\nabla_{l}+\frac{1}{2}g_{kl}\Box)\varphi+\frac{1}{2}(T_{kl}-\frac{1}{2}g_{kl}T)+\frac{1}{2}E_{kl}\right].$
(A.11)

With the help of Eqs. (A.6) and (A.8), we find

$\displaystyle[R_{kl},b_{\rho}^{\prime}]=\frac{1}{\varphi}[(\nabla_{k}\nabla_{l}+\frac{1}{2}g_{kl}\Box)\varphi,b_{\rho}^{\prime}].$
(A.12)

Using Eq. (43) and the formula:

$\displaystyle[\ddot{\varphi},b_{\rho}^{\prime}]=-2i\tilde{f}\partial_{\rho}\dot{\varphi}\delta^{3}+2i\tilde{f}\partial_{\rho}\varphi\tilde{g}^{0k}\partial_{k}(\tilde{f}\delta^{3}),$
(A.13)

which can be proved through (48) and (52), we can show that the RHS of Eq.
(A.12) is identically vanishing. Accordingly, we have succeeded in proving our
statement (A.1).

##  Appendix B Conserved current for scale symmetry

In this appendix, we will present a derivation of the conserved current for
the scale symmetry in Eq. (111) and show that the current is indeed conserved
on-shell.

First, let us rewrite the Einstein-Hilbert term in order not to include second
order derivatives of the metric by the standard technique [30]:

$\displaystyle\sqrt{-g}R=\tilde{g}^{\mu\nu}(\partial_{\alpha}\Gamma^{\alpha}_{\mu\nu}-\partial_{\nu}\Gamma^{\alpha}_{\mu\alpha}+\Gamma^{\sigma}_{\mu\nu}\Gamma^{\alpha}_{\sigma\alpha}-\Gamma^{\sigma}_{\mu\alpha}\Gamma^{\alpha}_{\nu\sigma}),$
(B.1)

where the first and second terms can be rewritten as

$\displaystyle\tilde{g}^{\mu\nu}\partial_{\alpha}\Gamma^{\alpha}_{\mu\nu}=\partial_{\alpha}(\tilde{g}^{\mu\nu}\Gamma^{\alpha}_{\mu\nu})-\partial_{\alpha}\tilde{g}^{\mu\nu}\cdot\Gamma^{\alpha}_{\mu\nu}=\partial_{\alpha}(\tilde{g}^{\mu\nu}\Gamma^{\alpha}_{\mu\nu})-(\Gamma^{\rho}_{\alpha\rho}\tilde{g}^{\mu\nu}-2\Gamma^{\mu}_{\alpha\rho}\tilde{g}^{\nu\rho})\Gamma^{\alpha}_{\mu\nu}$
$\displaystyle\tilde{g}^{\mu\nu}\partial_{\nu}\Gamma^{\alpha}_{\mu\alpha}=\partial_{\nu}(\tilde{g}^{\mu\nu}\Gamma^{\alpha}_{\mu\alpha})-\partial_{\nu}\tilde{g}^{\mu\nu}\cdot\Gamma^{\alpha}_{\mu\alpha}=\partial_{\nu}(\tilde{g}^{\mu\nu}\Gamma^{\alpha}_{\mu\alpha})+\Gamma^{\mu}_{\rho\sigma}\tilde{g}^{\rho\sigma}\Gamma^{\alpha}_{\mu\alpha}.$
(B.2)

From Eqs. (B.1) and (B.2), we have

$\displaystyle\sqrt{-g}R=\partial_{\alpha}(\tilde{g}^{\mu\nu}\Gamma^{\alpha}_{\mu\nu}-\tilde{g}^{\mu\alpha}\Gamma^{\beta}_{\mu\beta})+\tilde{g}^{\mu\nu}(\Gamma^{\sigma}_{\mu\alpha}\Gamma^{\alpha}_{\nu\sigma}-\Gamma^{\sigma}_{\mu\nu}\Gamma^{\alpha}_{\sigma\alpha}).$
(B.3)

Using the formula (B.3), up to a surface term the part of the scalar-tensor
gravity in the classical Lagrangian (74) can be cast to the form:

$\displaystyle\sqrt{-g}\varphi
R=-\sqrt{-g}(G^{\alpha}\partial_{\alpha}\varphi+G\varphi),$ (B.4)

where $G^{\alpha}$ and $G$ are defined by

$\displaystyle G^{\alpha}\equiv
g^{\mu\nu}\Gamma^{\alpha}_{\mu\nu}-g^{\mu\alpha}\Gamma^{\beta}_{\mu\beta},\qquad
G\equiv
g^{\mu\nu}(\Gamma^{\sigma}_{\mu\nu}\Gamma^{\alpha}_{\sigma\alpha}-\Gamma^{\sigma}_{\mu\alpha}\Gamma^{\alpha}_{\nu\sigma}).$
(B.5)

Then, the quantum Lagrangian (75) takes the form:

$\displaystyle{\cal{L}}_{q}=-\sqrt{-g}\left(G^{\alpha}\partial_{\alpha}\varphi+G\varphi+\frac{1}{2}g^{\mu\nu}\partial_{\mu}\phi\partial_{\nu}\phi\right)-\tilde{g}^{\mu\nu}\partial_{\mu}b_{\nu}-i\tilde{g}^{\mu\nu}\partial_{\mu}\bar{c}_{\rho}\partial_{\nu}c^{\rho}.$
(B.6)

With the infinitesimal scale invariance (112), the Noether theorem provides us
with a formula for the conserved current:

$\displaystyle\Lambda
J^{\mu}=\sum_{i}\frac{\partial^{R}{\cal{L}}_{q}}{\partial\partial_{\mu}\Phi_{i}}\delta\Phi_{i},$
(B.7)

where we use the right-derivative notation and
$\Phi_{i}=\\{g_{\rho\sigma},\phi,b_{\rho},\bar{c}_{\rho},c^{\rho}\\}$. It is
therefore necessary to evaluate each term on the RHS of the following
equation:

$\displaystyle
J^{\mu}=\frac{\partial^{R}{\cal{L}}_{q}}{\partial\partial_{\mu}g_{\rho\sigma}}2g_{\rho\sigma}-\frac{\partial^{R}{\cal{L}}_{q}}{\partial\partial_{\mu}\phi}\phi-\frac{\partial^{R}{\cal{L}}_{q}}{\partial\partial_{\mu}b_{\rho}}2b_{\rho}-\frac{\partial^{R}{\cal{L}}_{q}}{\partial\partial_{\mu}\bar{c}_{\rho}}\bar{c}_{\rho}-\frac{\partial^{R}{\cal{L}}_{q}}{\partial\partial_{\mu}c^{\rho}}c^{\rho}.$
(B.8)

To calculate the first term on the RHS, we need the formulae:

$\displaystyle\frac{\partial(\sqrt{-g}G^{\alpha})}{\partial\partial_{\mu}g_{\rho\sigma}}=\sqrt{-g}(g^{\alpha(\rho}g^{\sigma)\mu}-g^{\rho\sigma}g^{\mu\alpha}),$
(B.9)
$\displaystyle\frac{\partial(\sqrt{-g}G)}{\partial\partial_{\mu}g_{\rho\sigma}}=\sqrt{-g}\Biggl{[}\frac{1}{2}g^{\rho\sigma}g^{\alpha\beta}\Gamma^{\mu}_{\alpha\beta}+\Gamma^{\alpha}_{\alpha\lambda}\left(g^{\lambda(\rho}g^{\sigma)\mu}-\frac{1}{2}g^{\lambda\mu}g^{\rho\sigma}\right)$
$\displaystyle+$
$\displaystyle\frac{1}{2}\left(g^{\lambda(\rho}\partial_{\lambda}g^{\sigma)\mu}-\frac{1}{2}g^{\lambda\mu}\partial_{\lambda}g^{\rho\sigma}-g^{\rho\alpha}g^{\beta\sigma}\Gamma^{\mu}_{\alpha\beta}\right)\Biggr{]},$

where we have used the symmetrization notation,
$A_{(\rho}B_{\sigma)}\equiv\frac{1}{2}(A_{\rho}B_{\sigma}+A_{\sigma}B_{\rho})$.
After a straightforward calculation, we find that the current is given by

$\displaystyle J^{\mu}$ $\displaystyle=$
$\displaystyle\tilde{g}^{\mu\nu}\left[\frac{6\xi+1}{\xi}\partial_{\nu}\varphi+2b_{\nu}+i\partial_{\nu}(\bar{c}_{\rho}c^{\rho})\right]$
(B.10) $\displaystyle=$
$\displaystyle\tilde{g}^{\mu\nu}\left[\frac{6\xi+1}{2}\partial_{\nu}(\phi^{2})+2b_{\nu}+i\partial_{\nu}(\bar{c}_{\rho}c^{\rho})\right],$

where we have used $\varphi\equiv\frac{1}{2}\xi\phi^{2}$. Note that in case of
the Weyl invariant scalar-tensor gravity where $6\xi+1=0$ is satisfied, the
first term, which is a classical contribution, on the RHS in the above
equation identically vanishes [17, 18].

Finally, let us check that this current is in fact conserved in terms of the
field equations (76). Taking the derivative of the former current in (B.10)
leads to

$\displaystyle\partial_{\mu}J^{\mu}=\sqrt{-g}\left(\frac{6\xi+1}{\xi}\Box\varphi+E\right),$
(B.11)

where we have used the de Donder gauge and the field equations for the FP
ghosts. It is easy to see that this expression vanishes by using the field
equations (76). To do so, let us take the trace of the Einstein equation,
i.e., the first equation, in Eq. (76) whose result reads

$\displaystyle(3\Box-R)\varphi-\frac{1}{2}T^{\mu}\,_{\mu}+\frac{1}{2}E=0.$
(B.12)

From the field equation for the scalar field, i.e., the second equation, in
Eq. (76), we obtain

$\displaystyle\varphi R=-\frac{1}{2}\phi\Box\phi.$ (B.13)

Moreoever, taking the trace of $T_{\mu\nu}$ in Eq. (77) gives rise to

$\displaystyle T^{\mu}\,_{\mu}=-(\partial_{\rho}\phi)^{2}.$ (B.14)

Then, substituting Eqs. (B.13) and (B.14) into (B.12), we have

$\displaystyle
3\Box\varphi+\frac{1}{2}[\phi\Box\phi+(\partial_{\rho}\phi)^{2}]+\frac{1}{2}E=0.$
(B.15)

Since the definition $\varphi\equiv\frac{1}{2}\xi\phi^{2}$ gives us an
equation:

$\displaystyle\Box\varphi=\xi[\phi\Box\phi+(\partial_{\rho}\phi)^{2}],$ (B.16)

inserting this equation to Eq. (B.15) yields the desired equation:

$\displaystyle\frac{6\xi+1}{\xi}\Box\varphi+E=0,$ (B.17)

by which we can prove the conservation law of the scale current,
$\partial_{\mu}J^{\mu}=0$ as seen in Eq. (B.11).

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