DPUR/TH/79

March, 2024

Conformal Symmetry in Quantum Gravity

Ichiro Oda111 E-mail address<EMAIL_ADDRESS>
Department of Physics, Faculty of Science, University of the Ryukyus,
Nishihara, Okinawa 903-0213, Japan

###### Abstract

We study the problem of how to derive conformal symmetry in the framework of
quantum gravity. We start with a generic gravitational theory which is
invariant under both the general coordinate transformation (GCT) and Weyl
transformation (or equivalently, local scale transformation), and then
construct its BRST formalism by fixing the gauge symmetries by the extended de
Donder gauge and scalar gauge conditions. These gauge-fixing conditions are
invariant under global $GL(4)$ and global scale transformations. The gauge-
fixed and BRST invariant quantum action possesses a huge Poincaré-like
$IOSp(10|10)$ global symmetry, from which we can construct an extended
conformal symmetry in a flat Minkowski background in the sense that the
Lorentz symmetry is replaced with the $GL(4)$ symmetry. Moreover, we construct
the conventional conformal symmetry out of this extended symmetry. With a flat
Minkowski background $\langle g_{\mu\nu}\rangle=\eta_{\mu\nu}$ and a non-zero
scalar field $\langle\phi\rangle\neq 0$, the $GL(4)$ and global scale
symmetries are spontaneously broken to the Lorentz symmetry, thereby proving
that the graviton and the dilaton are respectively the corresponding Nambu-
Goldstone bosons, and therefore they must be exactly massless at
nonperturbative level. One of remarkable aspects in our findings is that in
quantum gravity, a derivation of conformal symmetry does not depend on a
classical action, and its generators are built from only the gauge-fixing and
the FP ghost actions. Finally, we address a generalized Zumino theorem in
quantum gravity.

##  1 Introduction

There is no question that both local and global symmetries play an important
role in elementary-particle physics and quantum gravity. For instance, in
quantum chromodynamics (QCD) it has been found that a local symmetry is the
nonabelian gauge symmetry based on the gauge group $SU(3)$, and that this
local symmetry gives rise to physically significant effects, such as the
asymptotic freedom and the confinement of quarks and gluons. In addition to
the $SU(3)$ gauge symmetry, there is a $U(1)_{V}\times U(1)_{A}$ global
symmetry of the quark action. The $U(1)_{A}$ symmetry is anomalous and the
effect of a $U(1)_{A}$ transformation is to change the value of the theta
angle, which requires us to consider an instanton to make the vacuum energy
density to depend on the theta angle.

On the other hand, in quantum gravity, the meaning of symmetries is more
subtle than that in elementary-particle physics. Although general relativity
has been beautifully established in the Riemannian geometry on the basis of
the general coordinate invariance and the equivalence principle, it seems that
the feature of the non-renormalizability of general relativity would need a
more huge local symmetry such as a local supersymmetry or a Weyl symmetry (or
equivalently, a local scale symmetry). As for a global symmetry, in classical
general relativity, it could be broken by the no-hair theorem of black holes
[1]. Moreover, in string theory, which is a strong candidate of quantum
gravity, we never get any additive conservation laws and at least in known
string vacua, the additive global symmetries turn out to be either gauge
symmetries or explicitly violated [2].

In such a situation, it might appear to be strange to seek for a new global
symmetry in quantum gravity, but even in quantum gravity, we have important
global symmetries such as the BRST symmetry and the conservation law of the
ghost number. The BRST symmetry, which is a residual global symmetry emerging
after the gauge-fixing procedure, plays a role in proving the unitarity of the
theory and deriving the Ward identities among the Green functions [3]. On the
other hand, the conservation of the ghost number is violated in a two-
dimensional quantum theory, and leads to the ghost number anomaly, which is
closely related to the Riemann-Roch theorem on the closed Riemann surfaces
[4].

From this viewpoint, it is valuable to derive global symmetries, which include
the BRST symmetry and the symmetry of the FP ghost number, in quantum gravity.
If such the symmetries are purely built from quantum fields such as the FP
(anti)ghosts and the Nakanishi-Lautrup auxiliary fields, they could escape
from the no-hair theorem of black holes, and might have some important
applications for the study of their anomalies and the confinement of massive
ghost in higher-derivative gravities.

In a flat Minkowski space-time, conformal symmetry occupies a special position
among many global symmetries in that conformal symmetry is ubiquitous in
physics ranging from elementary-particle physics to condensed matter physics
[5, 6]. Even if conformal symmetry plays an important role, it must be
spontaneously broken in reality since our world has a built-in scale in it.
Moreover, it is of interest that conformal symmetry is connected with a
classical gravity formulated in a curved space-time. In fact, the Zumino
theorem advocates that the theories which are invariant under the GCT and Weyl
transformation have conformal invariance in the flat Minkowski background at
the classical level [7]. Then, it is natural to ask ourselves whether the
Zumino theorem can be generalized even in quantum gravity or not. In our past
works [8, 9, 10, 11], we have attacked this question. We have shown that in
Weyl invariant scalar-tensor gravity [9], the Zumino theorem is valid whereas
in conformal gravity [11] it is not so. In these works, since we have
specified a classical theory, it is not clear if the answer is independent of
the choice of the classical action or not. It is therefore desirable to start
with not a specific action but a more general gravitational action, which is
invariant under both the GCT and Weyl transformation, and investigate the
validity of the Zumino theorem in quantum gravity. This is one of our
motivations behind the present study. Actually, we will find that it is not
necessary to fix such a classical action at all in order to understand the
Zumino theorem in quantum gravity.

The outline of this article is as follows: In the next section we construct a
BRST formalism of our theory. In Section 3, we carry out the canonical
quantization and calculate the equal-time (anti)commutation relations (ETCRs).
In Section 4, we prove the existence of a Poincaré-like $IOSp(10|10)$ global
symmetry in a gauge-fixed and BRST invariant quantum Lagrangian and compute
its algebra. In Section 5, we derive conformal symmetry in a flat Minkowski
space-time from the Poincaré-like $IOSp(10|10)$ symmetry. In Section 6, we
investigate spontaneous symmetry breakdown of $GL(4)$ symmetry and global
scale symmetry to Lorentz symmetry. Through this mechanism of the symmetry
breaking, we can precisely prove that the graviton and the dilaton are exactly
massless owing to the Nambu-Goldstone theorem. The final section is devoted to
a conclusion.

Four appendices are put for technical details. In Appendix A, we show that the
extended de Donder gauge condition is invariant under the global $GL(4)$
transformation. In Appendix B, we derive the ETCR,
$[\dot{g}_{\mu\nu},b_{\rho}^{\prime}]$. In Appendix C, we present a proof of
the ETCR,
$[\dot{b}_{\mu},b_{\nu}^{\prime}]=-i\tilde{f}\phi^{-2}(\partial_{\mu}b_{\nu}+\partial_{\nu}b_{\mu})\delta^{3}$
without recourse to the Einstein’s equation. In Appendix D, we give two
different proofs for an algebra
$[P_{\mu},K^{\nu}]=-2i(G^{\rho}\,_{\rho}-D)\delta_{\mu}^{\nu}$.

##  2 BRST formalism

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 a concrete
expression of the gravitational Lagrangian, we will start with the following
classical Lagrangian222We follow the notation and conventions of MTW textbook
[1]. 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 $\mu\neq\nu$.

$\displaystyle{\cal L}_{c}={\cal L}_{c}(g_{\mu\nu},\phi),$ (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 spinor fields. We assume that ${\cal{L}}_{c}$ does not
involve more than first order derivatives of the metric and matter fields.

We are ready to 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

$\displaystyle\delta_{B}g_{\mu\nu}$ $\displaystyle=$
$\displaystyle-(\nabla_{\mu}c_{\nu}+\nabla_{\nu}c_{\mu})$ $\displaystyle=$
$\displaystyle-(c^{\alpha}\partial_{\alpha}g_{\mu\nu}+\partial_{\mu}c^{\alpha}g_{\alpha\nu}+\partial_{\nu}c^{\alpha}g_{\mu\alpha}),$
$\displaystyle\delta_{B}\tilde{g}^{\mu\nu}$ $\displaystyle=$
$\displaystyle\sqrt{-g}(\nabla^{\mu}c^{\nu}+\nabla^{\nu}c^{\mu}-g^{\mu\nu}\nabla_{\rho}c^{\rho}),$
$\displaystyle\delta_{B}\phi$ $\displaystyle=$
$\displaystyle-c^{\lambda}\partial_{\lambda}\phi,\quad\delta_{B}c^{\rho}=-c^{\lambda}\partial_{\lambda}c^{\rho},$
$\displaystyle\delta_{B}\bar{c}_{\rho}$ $\displaystyle=$ $\displaystyle
iB_{\rho},\quad\delta_{B}B_{\rho}=0,$ (2)

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

$\displaystyle
b_{\rho}=B_{\rho}-ic^{\lambda}\partial_{\lambda}\bar{c}_{\rho},$ (3)

and its BRST transformation reads

$\displaystyle\delta_{B}b_{\rho}=-c^{\lambda}\partial_{\lambda}b_{\rho}.$ (4)

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

$\displaystyle\bar{\delta}_{B}g_{\mu\nu}$ $\displaystyle=$ $\displaystyle
2cg_{\mu\nu},\quad\bar{\delta}_{B}\tilde{g}^{\mu\nu}=2c\tilde{g}^{\mu\nu},$
$\displaystyle\bar{\delta}_{B}\phi$ $\displaystyle=$
$\displaystyle-c\phi,\quad\bar{\delta}_{B}\bar{c}=iB,\quad\bar{\delta}_{B}c=\bar{\delta}_{B}B=0,$
(5)

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

$\displaystyle\delta_{B}^{2}=\bar{\delta}_{B}^{2}=0.$ (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 $\bar{\delta}_{B}$ 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 such scalar fields that to match the transformation
law for scalar fields their BRST transformations should take the form:

$\displaystyle\delta_{B}B=-c^{\lambda}\partial_{\lambda}B,\quad\delta_{B}c=-c^{\lambda}\partial_{\lambda}c,\quad\delta_{B}\bar{c}=-c^{\lambda}\partial_{\lambda}\bar{c}.$
(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 [9] is to just assume that the two BRST transformations are anti-
commute with each other, that is,

$\displaystyle\\{\delta_{B},\bar{\delta}_{B}\\}\equiv\delta_{B}\bar{\delta}_{B}+\bar{\delta}_{B}\delta_{B}=0,$
(8)

which requires us to take

$\displaystyle\bar{\delta}_{B}b_{\rho}=\bar{\delta}_{B}c^{\rho}=\bar{\delta}_{B}\bar{c}_{\rho}=0.$
(9)

As the gauge condition for the GCT, we will take “the extended de Donder
gauge” [9]

$\displaystyle\partial_{\mu}(\tilde{g}^{\mu\nu}\phi^{2})=0,$ (10)

which is invariant under the Weyl transformation (5) as required from the
condition (8). As for the Weyl transformation, we have to choose an
appropriate gauge condition, which is invariant under the GCT, that is, a
scalar quantity. As such a gauge condition, we will take the so-called “the
scalar gauge condition” [9]

$\displaystyle\partial_{\mu}(\tilde{g}^{\mu\nu}\phi\partial_{\nu}\phi)=0,$
(11)

which can be alternatively written as

$\displaystyle\Box\,\phi^{2}=0.$ (12)

The key observation here is that both the extended de Donder gauge condition
(10) and the scalar gauge condition (11) are invariant under both a
${\it{global}}$ $GL(4)$ transformation and a ${\it{global}}$ scale
transformation.444The unitary gauge condition $\phi=\rm{constant}$ and the
Lorenz condition $\nabla_{\mu}S^{\mu}=0$ [10], where $S_{\mu}$ is a Weyl gauge
field, are also invariant under these global transformations. For instance, a
proof of the $GL(4)$ invariance of the extended Donder gauge condition (10) is
given in Appendix A.

After taking the extended de Donder gauge condition (10) for the GCT and the
scalar gauge condition (11) for the Weyl transformation, the gauge-fixed and
BRST invariant quantum Lagrangian is given by

$\displaystyle{\cal L}_{q}$ $\displaystyle=$ $\displaystyle{\cal L}_{c}+{\cal
L}_{GF+FP}+\bar{\cal L}_{GF+FP}$ (13) $\displaystyle=$ $\displaystyle{\cal
L}_{c}+i\delta_{B}(\tilde{g}^{\mu\nu}\phi^{2}\partial_{\mu}\bar{c}_{\nu})+i\bar{\delta}_{B}\left[\bar{c}\partial_{\mu}(\tilde{g}^{\mu\nu}\phi\partial_{\nu}\phi)\right]$
$\displaystyle=$ $\displaystyle{\cal
L}_{c}-\tilde{g}^{\mu\nu}\phi^{2}(\partial_{\mu}b_{\nu}+i\partial_{\mu}\bar{c}_{\lambda}\partial_{\nu}c^{\lambda})$
$\displaystyle+$
$\displaystyle\tilde{g}^{\mu\nu}\phi\partial_{\mu}B\partial_{\nu}\phi-i\tilde{g}^{\mu\nu}\phi^{2}\partial_{\mu}\bar{c}\partial_{\nu}c,$

where surface terms are dropped. Let us note that this quantum Lagrangian is
also invariant under the global $GL(4)$ transformation and the global scale
transformation.

Let us rewrite this Lagrangian concisely as

$\displaystyle{\cal L}_{q}={\cal
L}_{c}-\frac{1}{2}\tilde{g}^{\mu\nu}E_{\mu\nu},$ (14)

where we have defined

$\displaystyle
E_{\mu\nu}\equiv\phi^{2}(\partial_{\mu}b_{\nu}+i\partial_{\mu}\bar{c}_{\lambda}\partial_{\nu}c^{\lambda})-\phi\partial_{\mu}B\partial_{\nu}\phi+i\phi^{2}\partial_{\mu}\bar{c}\partial_{\nu}c+(\mu\leftrightarrow\nu).$
(15)

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

$\displaystyle\phi(x)\equiv e^{\sigma(x)},$ (16)

and rewrite (14) further into the form

$\displaystyle{\cal L}_{q}={\cal
L}_{c}-\frac{1}{2}\phi^{2}\tilde{g}^{\mu\nu}\hat{E}_{\mu\nu}={\cal
L}_{c}-\frac{1}{2}e^{2\sigma(x)}\tilde{g}^{\mu\nu}\hat{E}_{\mu\nu},$ (17)

where we have defined

$\displaystyle\hat{E}_{\mu\nu}\equiv\partial_{\mu}b_{\nu}+i\partial_{\mu}\bar{c}_{\lambda}\partial_{\nu}c^{\lambda}-\partial_{\mu}B\partial_{\nu}\sigma+i\partial_{\mu}\bar{c}\partial_{\nu}c+(\mu\leftrightarrow\nu).$
(18)

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

$\displaystyle
E_{\mu\nu}=\phi^{2}\hat{E}_{\mu\nu}=e^{2\sigma}\hat{E}_{\mu\nu}.$ (19)

The field equations obtained from variations of $b_{\nu}$, $B$, $c^{\rho}$,
$\bar{c}_{\rho}$, $c$ and $\bar{c}$ in the Lagrangian ${\cal L}_{q}$, take the
form:

$\displaystyle\partial_{\mu}(\tilde{g}^{\mu\nu}\phi^{2})=0,\qquad\partial_{\mu}(\tilde{g}^{\mu\nu}\phi\partial_{\nu}\phi)=0,$
$\displaystyle\partial_{\mu}(\tilde{g}^{\mu\nu}\phi^{2}\partial_{\nu}\bar{c}_{\rho})=0,\qquad\partial_{\mu}(\tilde{g}^{\mu\nu}\phi^{2}\partial_{\nu}c^{\rho})=0,$
$\displaystyle\partial_{\mu}(\tilde{g}^{\mu\nu}\phi^{2}\partial_{\nu}\bar{c})=0,\qquad\partial_{\mu}(\tilde{g}^{\mu\nu}\phi^{2}\partial_{\nu}c)=0.$
(20)

Then, the two gauge-fixing conditions in (20), or equivalently (10) and (11),
lead to a very simple d’Alembert-like equation for the dilaton:

$\displaystyle g^{\mu\nu}\partial_{\mu}\partial_{\nu}\sigma=0.$ (21)

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
extended de Donder gauge condition, i.e., the first field equation in (20)
gives us the similar simple equation for the (anti)ghosts:

$\displaystyle g^{\mu\nu}\partial_{\mu}\partial_{\nu}\bar{c}_{\rho}=0,\qquad
g^{\mu\nu}\partial_{\mu}\partial_{\nu}c^{\rho}=0,$ $\displaystyle
g^{\mu\nu}\partial_{\mu}\partial_{\nu}\bar{c}=0,\qquad
g^{\mu\nu}\partial_{\mu}\partial_{\nu}c=0.$ (22)

In order to derive the field equation for $B$, it is useful to take the Weyl
BRST transformation for the field equation for the antighost $\bar{c}$ and the
result again provides a simple equation:

$\displaystyle g^{\mu\nu}\partial_{\mu}\partial_{\nu}B=0.$ (23)

Finally, the field equation for $b_{\rho}$ can be obtained by taking the GCT
BRST transformation for the field equation for $\bar{c}_{\rho}$ as
follows:555This field equation has been previously obtained from the
Einstein’s equation [12, 8, 9], but this time we use the GCT BRST
transformation since we have no concrete expression of the Einstein’s equation
owing to a generic gravitational action. Taking the GCT BRST transformation
for the field equation for $\bar{c}_{\rho}$, i. e., , i.e.,
$g^{\mu\nu}\partial_{\mu}\partial_{\nu}\bar{c}_{\rho}=0$ leads to

$\displaystyle(-\partial_{\lambda}g^{\mu\nu}c^{\lambda}+2g^{\mu\lambda}\partial_{\lambda}c^{\nu})\partial_{\mu}\partial_{\nu}\bar{c}_{\rho}+ig^{\mu\nu}\partial_{\mu}\partial_{\nu}B_{\rho}=0.$
(24)

Inserting Eq. (3) to this equation and arranging terms, we have

$\displaystyle
g^{\mu\nu}\partial_{\mu}\partial_{\nu}b_{\rho}=-ic^{\lambda}\partial_{\lambda}(g^{\mu\nu}\partial_{\mu}\partial_{\nu}\bar{c}_{\rho})=0.$
(25)

In other words, setting
$X^{M}=\\{x^{\mu},b_{\mu},\sigma,B,c^{\mu},\bar{c}_{\mu},c,\bar{c}\\}$ 666In
the present formalism, it is important to incorporate the mere space-time
coordinates $x^{\mu}$ into a set of fields $X^{M}$. This might suggest that
$x^{\mu}$ should be promoted to a quantum field in a more complete formalism.,
$X^{M}$ turns out to obey the d’Alembert-like equation:

$\displaystyle g^{\mu\nu}\partial_{\mu}\partial_{\nu}X^{M}=0.$ (26)

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

$\displaystyle{\cal P}^{\mu M}$ $\displaystyle\equiv$
$\displaystyle\tilde{g}^{\mu\nu}\phi^{2}\partial_{\nu}X^{M}=\tilde{g}^{\mu\nu}\phi^{2}\bigl{(}1\overset{\leftrightarrow}{\partial}_{\nu}X^{M}\bigr{)},$
$\displaystyle{\cal M}^{\mu MN}$ $\displaystyle\equiv$
$\displaystyle\tilde{g}^{\mu\nu}\phi^{2}\bigl{(}X^{M}\overset{\leftrightarrow}{\partial}_{\nu}Y^{N}\bigr{)},$
(27)

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

##  3 Canonical quantization and equal-time commutation relations

In this section, after introducing the canonical (anti)commutation relations
(CCRs), we will evaluate various equal-time (anti)commutation relations
(ETCRs) among fundamental variables. To simplify various expressions, we will
obey the following abbreviations adopted in the textbook of Nakanishi and
Ojima [12, 13]

$\displaystyle[A,B^{\prime}]$ $\displaystyle=$
$\displaystyle[A(x),B(x^{\prime})]|_{x^{0}=x^{\prime
0}},\qquad\delta^{3}=\delta(\vec{x}-\vec{x}^{\prime}),$
$\displaystyle\tilde{f}$ $\displaystyle=$
$\displaystyle\frac{1}{\tilde{g}^{00}}=\frac{1}{\sqrt{-g}g^{00}},$ (28)

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

Now, in order to carry out the canonical quantization, we perform the
integration by parts once and rewrite the Lagrangian (13) as

$\displaystyle{\cal L}_{q}^{\prime}$ $\displaystyle=$ $\displaystyle{\cal
L}_{c}+\partial_{\mu}(\tilde{g}^{\mu\nu}\phi^{2})b_{\nu}-i\tilde{g}^{\mu\nu}\phi^{2}\partial_{\mu}\bar{c}_{\lambda}\partial_{\nu}c^{\lambda}+\tilde{g}^{\mu\nu}\phi\partial_{\mu}B\partial_{\nu}\phi$
(29) $\displaystyle-$ $\displaystyle
i\tilde{g}^{\mu\nu}\phi^{2}\partial_{\mu}\bar{c}\partial_{\nu}c+\partial_{\mu}{\cal{V}}^{\mu},$

where a surface term ${\cal{V}}^{\mu}$ is defined as
${\cal{V}}^{\mu}\equiv-\tilde{g}^{\mu\nu}\phi^{2}b_{\nu}$. Next, let us set up
the canonical (anti)commutation relations (CCRs)

$\displaystyle[g_{\mu\nu},\pi_{g}^{\rho\lambda\prime}]=i\frac{1}{2}(\delta_{\mu}^{\rho}\delta_{\nu}^{\lambda}+\delta_{\mu}^{\lambda}\delta_{\nu}^{\rho})\delta^{3},\quad[\phi,\pi_{\phi}^{\prime}]=i\delta^{3},\quad[B,\pi_{B}^{\prime}]=i\delta^{3},$
$\displaystyle\\{c^{\sigma},\pi_{c\lambda}^{\prime}\\}=\\{\bar{c}_{\lambda},\pi_{\bar{c}}^{\sigma\prime}\\}=i\delta_{\lambda}^{\sigma}\delta^{3},\quad\\{c,\pi_{c}^{\prime}\\}=\\{\bar{c},\pi_{\bar{c}}^{\prime}\\}=i\delta^{3},$
(30)

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

From the Lagrangian (29), it is easy to derive canonical conjugate momenta

$\displaystyle\pi_{g}^{\mu\nu}$ $\displaystyle=$
$\displaystyle\frac{\partial{\cal
L}_{q}^{\prime}}{\partial\dot{g}_{\mu\nu}}=\frac{\partial{\cal
L}_{c}}{\partial\dot{g}_{\mu\nu}}-\frac{1}{2}\sqrt{-g}\phi^{2}(g^{0\mu}g^{\nu\rho}+g^{0\nu}g^{\mu\rho}-g^{0\rho}g^{\mu\nu})b_{\rho},$
$\displaystyle\pi_{\phi}$ $\displaystyle=$ $\displaystyle\frac{\partial{\cal
L}_{q}^{\prime}}{\partial\dot{\phi}}=\frac{\partial{\cal
L}_{c}}{\partial\dot{\phi}}+2\tilde{g}^{0\mu}\phi
b_{\mu}+\tilde{g}^{0\mu}\phi\partial_{\mu}B,\quad\pi_{B}=\frac{\partial{\cal
L}_{q}^{\prime}}{\partial\dot{B}}=\tilde{g}^{0\mu}\phi\partial_{\mu}\phi,$
$\displaystyle\pi_{c\sigma}$ $\displaystyle=$
$\displaystyle\frac{\partial{\cal
L}_{q}^{\prime}}{\partial\dot{c}^{\sigma}}=-i\tilde{g}^{0\mu}\phi^{2}\partial_{\mu}\bar{c}_{\sigma},\quad\pi_{\bar{c}}^{\sigma}=\frac{\partial{\cal
L}_{q}^{\prime}}{\partial\dot{\bar{c}}_{\sigma}}=i\tilde{g}^{0\mu}\phi^{2}\partial_{\mu}c^{\sigma},$
$\displaystyle\pi_{c}$ $\displaystyle=$ $\displaystyle\frac{\partial{\cal
L}_{q}^{\prime}}{\partial\dot{c}}=-i\tilde{g}^{0\mu}\phi^{2}\partial_{\mu}\bar{c},\quad\pi_{\bar{c}}=\frac{\partial{\cal
L}_{q}^{\prime}}{\partial\dot{\bar{c}}}=i\tilde{g}^{0\mu}\phi^{2}\partial_{\mu}c,$
(31)

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. At this point, it is worthwhile to point out that although we have
not specified the classical Lagrangian ${\cal L}_{c}$ and cannot write down a
concrete expression of the conjugate momentum $\pi_{g}^{\mu\nu}$, the
canonical conjugate momentum $\pi_{g}^{0\mu}$ generally has the following
expression

$\displaystyle\pi_{g}^{0\mu}=A^{\mu}+B^{\mu\nu}\partial_{\nu}\phi+C^{\mu\nu}b_{\nu},$
(32)

where $A^{\mu},B^{\mu\nu}$ and
$C^{\mu\nu}\equiv-\frac{1}{2}\tilde{g}^{00}g^{\mu\nu}\phi^{2}$ have no
$\dot{g}_{\mu\nu}$, and $B^{\mu\nu}\partial_{\nu}\phi$ does not have
$\dot{\phi}$ since $\pi_{g}^{0\mu}$ does not include the dynamics of the
metric and the scalar field.

Henceforth, we will evaluate various nontrivial equal-time (anti)commutation
relations (ETCRs) by using the canonical (anti)commutation relations (CCRs),
field equations and two kinds of BRST transformations. In particular, use of
the GCT and Weyl BRST transformations makes it possible to derive nontrivial
ETCRs without appealing to the Einstein’s equation.

Let us first calculate ETCRs, which can be directly derived from the CCRs in
Eq. (30) and the canonical conjugate momenta in Eq. (31). The CCR,
$[B,\pi_{B}^{\prime}]=i\delta^{3}$ leads to the ETCR:

$\displaystyle[B,\dot{\phi}^{\prime}]=i\tilde{f}\phi^{-1}\delta^{3}.$ (33)

The other CCRs, $[\Phi,\pi_{B}^{\prime}]=0$ produce

$\displaystyle[\Phi,\dot{\phi}^{\prime}]=0,$ (34)

where $\Phi$ denotes a set of canonical variables except $B$, i.e,
$\Phi=\\{g_{\mu\nu},\phi,c^{\mu},\bar{c}_{\mu},c,\bar{c}\\}$. The antiCCRs,
$\\{c^{\nu},\pi_{c\mu}^{\prime}\\}=\\{\bar{c}_{\mu},\pi_{\bar{c}}^{\nu\prime}\\}=i\delta_{\mu}^{\nu}\delta^{3}$
yield

$\displaystyle\\{c^{\nu},\dot{\bar{c}}_{\mu}^{\prime}\\}=-\\{\dot{c}^{\nu},\bar{c}_{\mu}^{\prime}\\}=-\tilde{f}\phi^{-2}\delta_{\mu}^{\nu}\delta^{3},$
(35)

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

$\displaystyle[\Phi,\dot{\Psi}^{\prime}]=\partial_{0}[\Phi,\Psi^{\prime}]-[\dot{\Phi},\Psi^{\prime}],$
(36)

which holds for the anticommutation relation as well. Similarly, the antiCCRs,
$\\{c,\pi_{c}^{\prime}\\}=\\{\bar{c},\pi_{\bar{c}}^{\prime}\\}=i\delta^{3}$
produce

$\displaystyle\\{c,\dot{\bar{c}}^{\prime}\\}=-\\{\dot{c},\bar{c}^{\prime}\\}=-\tilde{f}\phi^{-2}\delta^{3}.$
(37)

Now we are willing to derive a form of the (anti)ETCRs,
$[\Phi,b_{\rho}^{\prime}]$ with $\Phi$ denoting a generic field, which are
important ETCRs since the Nakanishi-Lautrup field $b_{\mu}$ in essence
generates a translation. Particularlly, the ETCR,
$[g_{\mu\nu},b_{\rho}^{\prime}]$ enables us to derive more complicated ETCRs
later, so we present two different derivations, one of which is based on Eq.
(32) while the other derivation relies on the GCT BRST transformation.

Let us first work with the CCR:

$\displaystyle[\pi_{g}^{\alpha
0},g_{\mu\nu}^{\prime}]=-i\frac{1}{2}(\delta_{\mu}^{\alpha}\delta_{\nu}^{0}+\delta_{\mu}^{0}\delta_{\nu}^{\alpha})\delta^{3}.$
(38)

Then, using Eq. (32), we find that this CCR produces

$\displaystyle[g_{\mu\nu},b_{\rho}^{\prime}]=-i\tilde{f}\phi^{-2}(\delta_{\mu}^{0}g_{\rho\nu}+\delta_{\nu}^{0}g_{\rho\mu})\delta^{3}.$
(39)

Incidentally, this ETCR gives us the similar ETCRs:

$\displaystyle[g^{\mu\nu},b_{\rho}^{\prime}]=i\tilde{f}\phi^{-2}(g^{\mu
0}\delta_{\rho}^{\nu}+g^{\nu 0}\delta_{\rho}^{\mu})\delta^{3},$
$\displaystyle[\tilde{g}^{\mu\nu},b_{\rho}^{\prime}]=i\tilde{f}\phi^{-2}(\tilde{g}^{\mu
0}\delta_{\rho}^{\nu}+\tilde{g}^{\nu
0}\delta_{\rho}^{\mu}-\tilde{g}^{\mu\nu}\delta_{\rho}^{0})\delta^{3}.$ (40)

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

$\displaystyle[g^{\mu\nu},\Phi^{\prime}]=-g^{\mu\alpha}g^{\nu\beta}[g_{\alpha\beta},\Phi^{\prime}],$
$\displaystyle[\tilde{g}^{\mu\nu},\Phi^{\prime}]=-\left(\tilde{g}^{\mu\alpha}g^{\nu\beta}-\frac{1}{2}\tilde{g}^{\mu\nu}g^{\alpha\beta}\right)[g_{\alpha\beta},\Phi^{\prime}],$
(41)

where $\Phi$ is a generic field.

As an alternative derivation of Eq. (39), we make use of the GCT BRST
transformation. For this purpose, let us start with the CCR,
$[g_{\mu\nu},\bar{c}_{\rho}^{\prime}]=0$, and take its GCT BRST
transformation:

$\displaystyle-\\{c^{\alpha}\partial_{\alpha}g_{\mu\nu}+\partial_{\mu}c^{\alpha}g_{\alpha\nu}+\partial_{\nu}c^{\alpha}g_{\alpha\mu},\bar{c}_{\rho}^{\prime}\\}+[g_{\mu\nu},iB_{\rho}^{\prime}]=0.$
(42)

Then, we find that the CCR, $[g_{\mu\nu},\pi_{c\rho}^{\prime}]=0$ produces

$\displaystyle[\dot{g}_{\mu\nu},\bar{c}_{\rho}^{\prime}]=0,$ (43)

where Eqs. (31) and (36) were used. Using this ETCR twice, the relation (3),
and (35) in Eq. (42), we can arrive at our desired ETCR (39). Incidentally,
the CCRs,
$[g_{\mu\nu},\pi_{\bar{c}}^{\rho\prime}]=[g_{\mu\nu},\pi_{c}^{\prime}]=[g_{\mu\nu},\pi_{\bar{c}}^{\prime}]=0$
provide the ETCRs between $g_{\mu\nu}$ and FP ghosts:

$\displaystyle[g_{\mu\nu},\dot{c}^{\rho\prime}]=[g_{\mu\nu},\dot{c}^{\prime}]=[g_{\mu\nu},\dot{\bar{c}}^{\prime}]=0.$
(44)

Moreover, with the help of the CCR, $[\pi_{g}^{\alpha 0},\Phi^{\prime}]=0$, it
is easy to derive the remaining ETCRs with the form of
$[\Phi,b_{\rho}^{\prime}]$ except for $\Phi=b_{\mu}$, which will be treated
later:

$\displaystyle[\phi,b_{\rho}^{\prime}]=[B,b_{\rho}^{\prime}]=[c^{\mu},b_{\rho}^{\prime}]=[\bar{c}_{\mu},b_{\rho}^{\prime}]=[c,b_{\rho}^{\prime}]=[\bar{c},b_{\rho}^{\prime}]=0.$
(45)

Next, let us attempt to derive a form of the (anti)ETCRs,
$[\dot{\Phi},b_{\rho}^{\prime}]$ with $\Phi$ denoting a generic field. Let us
begin by showing the following ETCRs:

$\displaystyle[\phi,\dot{\bar{c}}_{\rho}^{\prime}]=[\phi,\ddot{\bar{c}}_{\rho}^{\prime}]=[\dot{\phi},\dot{\bar{c}}_{\rho}^{\prime}]=0.$
(46)

The first equality can be obtained from either Eq. (34) or the CCR,
$[\phi,\pi_{c\rho}^{\prime}]=0$. A proof of the second equality needs the
field equation for $\bar{c}_{\rho}$ in Eq. (22), from which we have

$\displaystyle\ddot{\bar{c}}_{\rho}=-\tilde{f}(2\tilde{g}^{0i}\partial_{i}\dot{\bar{c}}_{\rho}+\tilde{g}^{ij}\partial_{i}\partial_{j}\bar{c}_{\rho}).$
(47)

Together with the first equality in (46), this equation makes it possible to
show the second equality, i.e., $[\phi,\ddot{\bar{c}}_{\rho}^{\prime}]=0$. And
the last equality in (46) can be obtained from the first and second
equalities. Then, taking the GCT BRST transformation of the first equality in
(46), we have

$\displaystyle 0$ $\displaystyle=$
$\displaystyle\\{-c^{\alpha}\partial_{\alpha}\phi,\dot{\bar{c}}_{\rho}^{\prime}\\}+[\phi,i\partial_{0}(b_{\rho}^{\prime}+ic^{\alpha\prime}\partial_{\alpha}\bar{c}_{\rho}^{\prime})]$
(48) $\displaystyle=$
$\displaystyle-\\{c^{\alpha},\dot{\bar{c}}_{\rho}^{\prime}\\}\partial_{\alpha}\phi+i[\phi,\dot{b}_{\rho}^{\prime}],$

where Eq. (46) was used again. Using Eqs. (35), (36) and (45), we are able to
obtain

$\displaystyle[\dot{\phi},b^{\prime}_{\rho}]=-i\tilde{f}\phi^{-2}\partial_{\rho}\phi\delta^{3}.$
(49)

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

$\displaystyle[\dot{\bar{c}}_{\mu},b^{\prime}_{\rho}]=-i\tilde{f}\phi^{-2}\partial_{\rho}\bar{c}_{\mu}\delta^{3},\qquad[\dot{c}^{\mu},b^{\prime}_{\rho}]=-i\tilde{f}\phi^{-2}\partial_{\rho}c^{\mu}\delta^{3}.$
(50)

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

$\displaystyle[\dot{\bar{c}},b^{\prime}_{\rho}]=-i\tilde{f}\phi^{-2}\partial_{\rho}\bar{c}\delta^{3},\qquad[\dot{c},b^{\prime}_{\rho}]=-i\tilde{f}\phi^{-2}\partial_{\rho}c\delta^{3}.$
(51)

In order to evaluate $[\dot{B},b^{\prime}_{\rho}]$, we start with the first
equality in Eq. (51), and take its BRST transformation for the Weyl
transformation, which immediately leads to the equation:

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

Furthermore, the ETCR, $[\dot{g}_{\mu\nu},b_{\rho}^{\prime}]$ (or
equivalently, $[g_{\mu\nu},\dot{b}_{\rho}^{\prime}]$) is derived in Appendix
B, whose result is written out as

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

Similarly, we can obtain that

$\displaystyle[g_{\mu\nu},\dot{b}_{\rho}^{\prime}]=i\Bigl{\\{}[\tilde{f}\phi^{-2}\partial_{\rho}g_{\mu\nu}-\partial_{0}(\tilde{f}\phi^{-2})(\delta_{\mu}^{0}g_{\rho\nu}+\delta_{\nu}^{0}g_{\rho\mu})]\delta^{3}$
$\displaystyle+[(\delta_{\mu}^{k}-2\delta_{\mu}^{0}\tilde{f}\tilde{g}^{0k})g_{\rho\nu}+(\mu\leftrightarrow\nu)]\partial_{k}(\tilde{f}\phi^{-2}\delta^{3})\Bigr{\\}},$
$\displaystyle[\tilde{g}^{\mu\nu},\dot{b}_{\rho}^{\prime}]=i\Bigl{\\{}[\tilde{f}\phi^{-2}\partial_{\rho}\tilde{g}^{\mu\nu}+\partial_{0}(\tilde{f}\phi^{-2})(\tilde{g}^{0\mu}\delta_{\rho}^{\nu}+\tilde{g}^{0\nu}\delta_{\rho}^{\mu}-\tilde{g}^{\mu\nu}\delta_{\rho}^{0})]\delta^{3}$
$\displaystyle-[(\tilde{g}^{\mu
k}-2\tilde{g}^{0\mu}\tilde{f}\tilde{g}^{0k})\delta_{\rho}^{\nu}-\frac{1}{2}\tilde{g}^{\mu\nu}(\delta_{\rho}^{k}-2\delta_{\rho}^{0}\tilde{f}\tilde{g}^{0k})$
$\displaystyle+(\mu\leftrightarrow\nu)]\partial_{k}(\tilde{f}\phi^{-2}\delta^{3})\Bigr{\\}}.$
(54)

In particular, the last equality will be useful in computing an algebra of an
extended conformal algebra later.

At this stage, we would like to prove very important ETCRs:

$\displaystyle[b_{\mu},b_{\nu}^{\prime}]=0,$
$\displaystyle[\dot{b}_{\mu},b_{\nu}^{\prime}]=-i\tilde{f}\phi^{-2}(\partial_{\mu}b_{\nu}+\partial_{\nu}b_{\mu})\delta^{3}.$
(55)

The first equality can be easily shown by taking the GCT BRST transformation
of the ETCR, $[b_{\mu},\bar{c}_{\nu}^{\prime}]=0$. It is a proof of the second
equality that we have to rely on the Einstein’s equation [12]. In the
formalism under consideration, we have not specified the classical action, so
we cannot utilize the Einstein’s equation to prove this equality. However, we
have found that instead of using the Einstein’s equation, we can prove the
second equality in (55) by appealing to the BRST transformation whose proof is
given in Appendix C.

To close this section, we should comment on the ETCRs involving the canonical
variable $B$ since the $B$ field is essentially the generator of a global
scale transformation. As seen in Eq. (33), $\dot{\phi}$ is the canonical
conjugate momentum of the $B$ field, and the other ETCRs, which include
$\dot{B}$, can be calculated from the CCRs and the Weyl BRST transformation.
For instance, the CCRs,
$[B,\pi_{c\mu}^{\prime}]=[B,\pi_{\bar{c}}^{\mu\prime}]=[B,\pi_{c}^{\prime}]=[B,\pi_{\bar{c}}^{\prime}]=0$
give us the ETCRs:

$\displaystyle[\dot{B},c_{\mu}^{\prime}]=[\dot{B},\bar{c}^{\mu\prime}]=[\dot{B},c^{\prime}]=[\dot{B},\bar{c}^{\prime}]=0,$
(56)

where the formula (36) was used. Moreover, taking the Weyl BRST transformation
of the last equality in (56) produces

$\displaystyle[\dot{B},B^{\prime}]=0.$ (57)

Similarly, taking the Weyl BRST transformation of
$[\dot{\bar{c}},g_{\mu\nu}^{\prime}]=0$, which can be obtained from the CCR,
$[\pi_{c},g_{\mu\nu}^{\prime}]=0$, gives rise to the ETCR:

$\displaystyle[\dot{B},g_{\mu\nu}^{\prime}]=2i\tilde{f}\phi^{-2}g_{\mu\nu}\delta^{3}.$
(58)

It is worthwhile to emphasize that via the CCRs, field equations and BRST
transformations, we can calculate all the ECTRs except for the ones relevant
to the metric and its derivatives such as
$[\dot{g}_{\mu\nu},g_{\rho\sigma}^{\prime}]$ etc.

##  4 Poincaré-like $IOSp(10|10)$ global symmetry

As mentioned in Section 2, 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 d’Alembert-like equation:

$\displaystyle g^{\mu\nu}\partial_{\mu}\partial_{\nu}X^{M}=0.$ (59)

This equation holds if and only if we adopt the extended de Donder gauge and
the 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 as defined in Eq. (27). In this
section, we shall show explicitly that there exist such currents and
consequently we have a Poincaré-like $IOSp(10|10)$ global symmetry.

Let us start with the Lagrangian (17), which can be cast to the form:

$\displaystyle{\cal L}_{q}={\cal
L}_{c}-\frac{1}{2}\tilde{g}^{\mu\nu}\phi^{2}\hat{E}_{\mu\nu}.$ (60)

We can further rewrite it into the form

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

where we have introduced an $IOSp(10|10)$ metric
$\eta_{NM}=\eta_{MN}^{T}\equiv\tilde{\eta}_{MN}$ defined as [14]

$\displaystyle\eta_{NM}=\tilde{\eta}_{MN}=\left(\begin{array}[]{cc|cc|cc|cc}&\delta_{\mu}^{\nu}&&&&\\\
\delta^{\mu}_{\nu}&&&&&\\\ \hline\cr&&0&-1&&&\\\ &&-1&0&&&\\\
\hline\cr&&&&&-i\delta_{\mu}^{\nu}&&\\\ &&&&i\delta^{\mu}_{\nu}&&&\\\
\hline\cr&&&&&&&-i\\\ &&&&&&i&\\\ \end{array}\right)_{.}$ (70)

Let us note that this $IOSp(10|10)$ metric $\eta_{NM}$ has the symmetry
property such that

$\displaystyle\eta_{MN}=(-)^{|M|\cdot|N|}\eta_{NM}=(-)^{|M|}\eta_{NM}=(-)^{|N|}\eta_{NM},$
(71)

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 $\eta_{MN}$
[14].

Now we would like to show that the quantum Lagrangian (61) is invariant under
a Poincaré-like $IOSp(10|10)$ global transformation. First, let us focus on
the infinitesimal $OSp$ rotation, which is defined by

$\displaystyle\delta
X^{M}=\eta^{ML}\varepsilon_{LN}X^{N}\equiv\varepsilon^{M}{}_{N}X^{N},$ (72)

where $\eta^{MN}$ is the inverse matrix of $\eta_{MN}$777Note that $\eta_{MN}$
is a usual c-number quantity and satisfies the relation,
$\eta_{MN}=\eta^{MN}$., and the infinitesimal parameter $\varepsilon_{MN}$ has
the following properties:

$\displaystyle\varepsilon_{MN}=(-)^{1+|M|\cdot|N|}\varepsilon_{NM},\qquad\varepsilon_{MN}X^{L}=(-)^{|L|(|M|+|N|)}X^{L}\varepsilon_{MN}.$
(73)

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., $\varepsilon_{MN}=\varepsilon_{MN}(x^{\mu})$.

Before delving into the invariance of the quantum Lagrangian (61), we should
notice that under the infinitesimal $OSp$ rotation (72), the dilaton
$\sigma(x)$, which is defined as $\phi=e^{\sigma}$, transforms as

$\displaystyle\delta\sigma=\eta^{\sigma
L}\varepsilon_{LN}X^{N}=-\varepsilon_{BN}X^{N},$ (74)

where we have used (70) and $\eta_{MN}=\eta^{MN}$. As for the scalar field
$\phi(x)$, this transformation for the dilaton amounts to a Weyl
transformation

$\displaystyle\phi\rightarrow\phi^{\prime}=e^{\epsilon(x)}\phi,$ (75)

where the infinitesimal parameter is defined as
$\epsilon(x)=-\varepsilon_{BN}X^{N}$.

To make the classical Lagrangian ${\cal L}_{c}$ and the compound metric
$\tilde{g}^{\mu\nu}\phi^{2}$ be invariant under the $OSp$ rotation (72), the
transformation (75) should accompany a Weyl transformation for the metric
tensor field:

$\displaystyle g_{\mu\nu}\rightarrow
g_{\mu\nu}^{\prime}=e^{-2\epsilon(x)}g_{\mu\nu}.$ (76)

To put it differently, it is necessary to perform the Weyl transformation (76)
for the metric when we perform the $OSp$ rotation (72).

With the proviso that the Weyl transformation (76) is made, we can prove an
invariance of the quantum Lagrangian (61) under the $OSp$ rotation (72), which
is given by

$\displaystyle\delta{\cal
L}_{q}=-\tilde{g}^{\mu\nu}\phi^{2}(\partial_{\mu}\varepsilon_{NM}X^{M}\partial_{\nu}X^{N}+\varepsilon_{NM}\partial_{\mu}X^{M}\partial_{\nu}X^{N}),$
(77)

where we have used the fact that both the classical Lagrangian ${\cal L}_{c}$
and the metric $\tilde{g}^{\mu\nu}\phi^{2}$ are invariant under the Weyl
transformation. It is easy to prove that the second term on the RHS vanishes
owing to the first property in Eq. (73). Thus, ${\cal L}_{q}$ is invariant
under the infinitesimal $OSp$ rotation. The conserved current is then
calculated as

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

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

$\displaystyle{\cal M}^{\mu
MN}=\tilde{g}^{\mu\nu}\phi^{2}X^{M}\overset{\leftrightarrow}{\partial}_{\nu}X^{N}.$
(79)

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

$\displaystyle\delta X^{M}=\varepsilon^{M},$ (80)

where $\varepsilon^{M}$ is the infinitesimal parameter and assume that it is a
local one for deriving the corresponding conserved current. Indeed, we can
show that ${\cal L}_{q}$ is invariant under the infinitesimal translation as
follows

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

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

$\displaystyle{\cal P}^{\mu
M}=\tilde{g}^{\mu\nu}\phi^{2}\partial_{\nu}X^{M}=\tilde{g}^{\mu\nu}\phi^{2}\left(1\overset{\leftrightarrow}{\partial}_{\nu}X^{M}\right).$
(82)

In this case as well, we need to perform an appropriate Weyl transformation
for the metric since the dilaton transforms as
$\delta\sigma=\varepsilon^{\sigma}(x)$ in Eq. (80).

An important remark is relevant to the expression of the conserved currents
(79) and (82). To make the quantum Lagrangian ${\cal L}_{q}$ be invariant
under the $IOSp(10|10)$ transformation, it is necessary to perform a Weyl
transformation for the metric. Then, there could be a possibility that the
expression of the currents might be modified because of this associated Weyl
transformation. The good news is that as shown in Refs. [15, 16, 17, 18, 19,
20], the current for the Weyl transformation identically vanishes since the
Weyl transformation does not involve the derivative of the transformation
parameter as in $\delta g_{\mu\nu}=2\Lambda g_{\mu\nu}$ unlike the
conventional gauge transformation as in $\delta
A_{\mu}=\partial_{\mu}\Lambda$. Thus, although we perform the Weyl
transformation, the conserved currents (79) and (82) remain unchanged. From
the conserved currents (79) and (82), the corresponding conserved charges
become

$\displaystyle M^{MN}$ $\displaystyle\equiv$ $\displaystyle\int d^{3}x\,{\cal
M}^{0MN}=\int
d^{3}x\,\tilde{g}^{0\nu}\phi^{2}X^{M}\overset{\leftrightarrow}{\partial}_{\nu}X^{N},$
$\displaystyle P^{M}$ $\displaystyle\equiv$ $\displaystyle\int d^{3}x\,{\cal
P}^{0M}=\int d^{3}x\,\tilde{g}^{0\nu}\phi^{2}\partial_{\nu}X^{M}.$ (83)

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

$\displaystyle[P^{M},P^{N}\\}=0,$
$\displaystyle[M^{MN},P^{R}\\}=i\bigl{[}P^{M}\tilde{\eta}^{NR}-(-)^{|N||R|}P^{N}\tilde{\eta}^{MR}\bigr{]},$
$\displaystyle[M^{MN},M^{RS}\\}=i\bigl{[}M^{MS}\tilde{\eta}^{NR}-(-)^{|N||R|}M^{MR}\tilde{\eta}^{NS}-(-)^{|N||R|}M^{NS}\tilde{\eta}^{MR}$
$\displaystyle+(-)^{|M||R|+|N||S|}M^{NR}\tilde{\eta}^{MS}\bigr{]},$ (84)

where the graded bracket is defined as $[A,B\\}=AB-(-)^{|A||B|}BA$.

Actually, it is a little tedious to derive the $IOSp(10|10)$ algebra (84) from
the ETCRs in a direct manner, so we present a different and more concise
derivation here. The key observation is that as mentioned in the proof of the
$IOSp(10|10)$ symmetry of the quantum Lagrangian ${\cal{L}}_{q}$, in order to
make the classical Lagrangian ${\cal{L}}_{c}$ and the composite metric
$\tilde{g}^{\mu\nu}\phi^{2}$ be invariant under the $IOSp(10|10)$
transformation, we simultaneously need to perform an associated Weyl
transformation for the metric. But, then this Weyl transformation makes no
effect on the conserved currents. This fact implies that as long as the
$IOSp(10|10)$ charges and their algebra are concerned, one does not have to
care about such Weyl invariant quantities and can ignore any contribution from
them.

Under such a situation, one can regard $X^{M}$ in the Lagrangian (61) as only
a canonical variable and introduce its corresponding canonical conjugate
momentum $\pi_{M}$ as

$\displaystyle\pi_{M}\equiv\frac{\partial{\cal{L}}_{q}}{\partial\dot{X}^{M}}=-\tilde{g}^{0\mu}\phi^{2}\eta_{MN}\partial_{\mu}X^{N},$
(85)

where the differentiation is taken from the right. The graded CCR,
$[X^{M},\pi_{N}^{\prime}\\}=i\delta_{N}^{M}\delta^{3}$ leads to

$\displaystyle[X^{M},\dot{X}^{N\prime}\\}=-[\dot{X}^{M},X^{N\prime}\\}=-i\tilde{f}\phi^{-2}\tilde{\eta}^{MN}\delta^{3}.$
(86)

From this CCR, it is easy to obtain the following algebra:

$\displaystyle[M^{MN},X^{R}\\}=i\left(X^{M}\tilde{\eta}^{NR}-(-)^{|N||R|}X^{N}\tilde{\eta}^{MR}\right),$
$\displaystyle[M^{MN},\partial_{\nu}X^{R}\\}=i\left(\partial_{\nu}X^{M}\tilde{\eta}^{NR}-(-)^{|N||R|}\partial_{\nu}X^{N}\tilde{\eta}^{MR}\right),$
$\displaystyle[P^{M},X^{R}\\}=i\tilde{\eta}^{MR},\qquad[P^{M},\partial_{\nu}X^{R}\\}=0.$
(87)

Then, on the basis of this algebra, it is straightforward to show the
$IOSp(10|10)$ algebra (84). For instance, $[M^{MN},M^{RS}\\}$ can be
calculated as follows:

$\displaystyle[M^{MN},M^{RS}\\}=[M^{MN},\int
d^{3}x\,\tilde{g}^{0\nu}\phi^{2}X^{R}\overset{\leftrightarrow}{\partial}_{\nu}X^{S}\\}$
$\displaystyle=\int
d^{3}x\,\tilde{g}^{0\nu}\phi^{2}[M^{MN},X^{R}\partial_{\nu}X^{S}-\partial_{\nu}X^{R}X^{S}\\}$
$\displaystyle=\int
d^{3}x\,\tilde{g}^{0\nu}\phi^{2}\Bigl{(}[M^{MN},X^{R}\\}\partial_{\nu}X^{S}+(-)^{(|M|+|N|)|R|}X^{R}[M^{MN},\partial_{\nu}X^{S}\\}$
$\displaystyle-[M^{MN},\partial_{\nu}X^{R}\\}X^{S}-(-)^{(|M|+|N|)|R|}\partial_{\nu}X^{R}[M^{MN},X^{S}\\}\Bigr{)}$
$\displaystyle=i\int
d^{3}x\,\tilde{g}^{0\nu}\phi^{2}\Bigl{[}(X^{M}\tilde{\eta}^{NR}-(-)^{|N||R|}X^{N}\tilde{\eta}^{MR})\partial_{\nu}X^{S}$
$\displaystyle+(-)^{(|M|+|N|)|R|}X^{R}(\partial_{\nu}X^{M}\tilde{\eta}^{NS}-(-)^{|N||S|}\partial_{\nu}X^{N}\tilde{\eta}^{MS})$
$\displaystyle-(\partial_{\nu}X^{M}\tilde{\eta}^{NR}-(-)^{|N||R|}\partial_{\nu}X^{N}\tilde{\eta}^{MR})X^{S}$
$\displaystyle-(-)^{(|M|+|N|)|R|}\partial_{\nu}X^{R}(X^{M}\tilde{\eta}^{NS}-(-)^{|N||S|}X^{N}\tilde{\eta}^{MS})\Bigr{]}$
$\displaystyle=i\int
d^{3}x\,\tilde{g}^{0\nu}\phi^{2}\Bigl{[}X^{M}\overset{\leftrightarrow}{\partial}_{\nu}X^{S}\tilde{\eta}^{NR}+(-)^{(|M|+|N|)|R|}X^{R}\overset{\leftrightarrow}{\partial}_{\nu}X^{M}\tilde{\eta}^{NS}$
$\displaystyle-(-)^{|N||R|}X^{N}\overset{\leftrightarrow}{\partial}_{\nu}X^{S}\tilde{\eta}^{MR}-(-)^{(|M|+|N|)|R|+|N||S|}X^{R}\overset{\leftrightarrow}{\partial}_{\nu}X^{N}\tilde{\eta}^{MS}\Bigr{]}$
$\displaystyle=i\bigl{[}M^{MS}\tilde{\eta}^{NR}+(-)^{(|M|+|N|)|R|}M^{RM}\tilde{\eta}^{NS}-(-)^{|N||R|}M^{NS}\tilde{\eta}^{MR}$
$\displaystyle-(-)^{(|M|+|N|)|R|+|N||S|}M^{RN}\tilde{\eta}^{MS}\bigr{]}.$
$\displaystyle=i\bigl{[}M^{MS}\tilde{\eta}^{NR}-(-)^{|N||R|}M^{MR}\tilde{\eta}^{NS}-(-)^{|N||R|}M^{NS}\tilde{\eta}^{MR}$
$\displaystyle+(-)^{|M||R|+|N||S|}M^{NR}\tilde{\eta}^{MS}\bigr{]}.$ (88)

At the last equality, we have used the relation

$\displaystyle M^{MN}=-(-)^{|M||N|}M^{NM}.$ (89)

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 Poincaré-like $IOSp(10|10)$ global symmetry. For instance,
the BRST charges for the GCT and Weyl transformation are respectively
described as

$\displaystyle Q_{B}$ $\displaystyle\equiv$ $\displaystyle
M(b_{\rho},c^{\rho})=\int
d^{3}x\,\tilde{g}^{0\nu}\phi^{2}b_{\rho}\overset{\leftrightarrow}{\partial}_{\nu}c^{\rho},$
$\displaystyle\bar{Q}_{B}$ $\displaystyle\equiv$ $\displaystyle M(B,c)=\int
d^{3}x\,\tilde{g}^{0\nu}\phi^{2}B\overset{\leftrightarrow}{\partial}_{\nu}c.$
(90)

##  5 Conformal symmetry

In this section, we wish to explain that the Poincaré-like $IOSp(10|10)$
global symmetry, which was established in the previous section, includes
conformal symmetry as a subgroup when the background is fixed to be a flat
Minkowski space-time.

Before doing that, let us review conformal symmetry briefly. Conformal
transformation can be defined as the general coordinate transformation which
can be undone by a Weyl transformation when the space-time metric is the flat
Minkowski one [5, 6]. With this definition, the conformal transformation is
described by the equation

$\displaystyle\partial_{\mu}\epsilon_{\nu}+\partial_{\nu}\epsilon_{\mu}=2\Lambda(x)\eta_{\mu\nu},$
(91)

where $\epsilon_{\mu}(x)$ and $\Lambda(x)$ are the infinitesimal
transformation parameters of the GCT and the Weyl transformation,
respectively. Taking the trace of Eq. (91) enables us to determine
$\Lambda(x)$ to be

$\displaystyle\Lambda=\frac{1}{4}\partial^{\rho}\epsilon_{\rho}.$ (92)

Inserting this $\Lambda$ to Eq. (91) yields

$\displaystyle\partial_{\mu}\epsilon_{\nu}+\partial_{\nu}\epsilon_{\mu}=\frac{1}{2}\partial^{\rho}\epsilon_{\rho}\eta_{\mu\nu},$
(93)

which is often called the “conformal Killing equation” in the Minkowski space-
time. It is well-known that a general solution to the conformal Killing
equation reads

$\displaystyle\epsilon^{\mu}=a^{\mu}+\omega^{\mu\nu}x_{\nu}+\lambda
x^{\mu}+k^{\mu}x^{2}-2x^{\mu}k_{\rho}x^{\rho},$ (94)

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. Substituting Eq. (94) into Eq. (92) gives us

$\displaystyle\Lambda=\lambda-2k_{\mu}x^{\mu}.$ (95)

We are now ready to clarify the presence of conformal symmetry in the
formalism at hand. To do so, let us note that although we have already fixed
general coordinate invariance and Weyl one by the extended de Donder gauge
condition (10) and the scalar gauge one (11), respectively, we are still left
with its linearized, residual symmetries. In order to look for such residual
symmetries, it is convenient to begin by the extended de Donder gauge (10) and
take its variation under the GCT.888A similar strategy has been adopted in
different theories in Refs. [21, 22, 23, 24]. As seen in Eq. (2), the GCT is
defined as

$\displaystyle\delta g_{\mu\nu}$ $\displaystyle=$
$\displaystyle-(\varepsilon^{\alpha}\partial_{\alpha}g_{\mu\nu}+\partial_{\mu}\varepsilon^{\alpha}g_{\alpha\nu}+\partial_{\nu}\varepsilon^{\alpha}g_{\mu\alpha}),$
$\displaystyle\delta\tilde{g}^{\mu\nu}$ $\displaystyle=$
$\displaystyle-\partial_{\alpha}(\varepsilon^{\alpha}\tilde{g}^{\mu\nu})+\partial_{\alpha}\varepsilon^{\mu}\tilde{g}^{\alpha\nu}+\partial_{\alpha}\varepsilon^{\nu}\tilde{g}^{\mu\alpha},$
$\displaystyle\delta\phi$ $\displaystyle=$
$\displaystyle-\varepsilon^{\alpha}\partial_{\alpha}\phi,$ (96)

where $\varepsilon^{\alpha}$ is an infinitesimal transformation parameter
since replacing $\varepsilon^{\alpha}$ with the ghost $c^{\alpha}$ produces
the GCT BRST in (2). By using Eq. (96) and the extended de Donder gauge
condition (10), the variation of the extended de Donder gauge condition (10)
takes the form

$\displaystyle\delta\left[\partial_{\mu}(\tilde{g}^{\mu\nu}\phi^{2})\right]=\partial_{\mu}\partial_{\alpha}\varepsilon^{\nu}\tilde{g}^{\mu\alpha}\phi^{2}.$
(97)

Thus, a residual symmetry for the GCT exists when the following equation is
obeyed:

$\displaystyle\partial_{\mu}\partial_{\alpha}\varepsilon^{\nu}=0.$ (98)

It is obvious that with a flat Minkowski metric $g_{\mu\nu}=\eta_{\mu\nu}$,
there is a zero-mode solution to Eq. (98) which is linear in $x^{\mu}$

$\displaystyle\varepsilon^{\mu}=A^{\mu}{}_{\nu}x^{\nu}+a^{\mu},$ (99)

where $A^{\mu}{}_{\nu}$ and $a^{\mu}$ are constant quantities and here indices
are raised and lowered with the flat Minkowski metric $\eta_{\mu\nu}$. This
equation implies that we have a global $GL(4)$ symmetry and a translation
symmetry associated with $A^{\mu}{}_{\nu}$ and $a^{\mu}$, respectively, as
residual symmetries. To put it differently, the extended de Donder gauge
condition (10) leaves invariances under the GCT with the transformation
parameter $\varepsilon^{\mu}(x)$ linear in the coordinates $x^{\mu}$, which
precisely correspond to the global $GL(4)$ and translational invariances.

Next, along the similar line of arguments, let us start with the scalar gauge
condition (11) and take its variation under Weyl transformation, which is
defined as

$\displaystyle\bar{\delta}g_{\mu\nu}$ $\displaystyle=$ $\displaystyle 2\Lambda
g_{\mu\nu},\qquad\bar{\delta}\tilde{g}^{\mu\nu}=2\Lambda\tilde{g}^{\mu\nu},$
$\displaystyle\bar{\delta}\phi$ $\displaystyle=$
$\displaystyle-\Lambda\phi,\qquad\bar{\delta}\sigma=-\Lambda,$ (100)

where $\Lambda$ is an infinitesimal transformation parameter. Note that the
replacement of $\Lambda$ with the ghost $c$ in Eq. (100) leads to the Weyl
BRST transformation in Eq. (5) as required. The result reads

$\displaystyle\bar{\delta}\left[\partial_{\mu}(\tilde{g}^{\mu\nu}\phi\partial_{\nu}\phi)\right]=\bar{\delta}\left[\partial_{\mu}(\tilde{g}^{\mu\nu}\phi^{2}\partial_{\nu}\sigma)\right]=-\tilde{g}^{\mu\nu}\phi^{2}\partial_{\mu}\partial_{\nu}\Lambda,$
(101)

where the extended de Donder gauge condition (10) was used. We therefore find
that a residual symmetry for the Weyl transformation exists when the equation

$\displaystyle g^{\mu\nu}\partial_{\mu}\partial_{\nu}\Lambda=0,$ (102)

is satisfied. With the flat Minkowski metric $g_{\mu\nu}=\eta_{\mu\nu}$, this
equation has a linearized zero-mode solution

$\displaystyle\Lambda=\lambda-2k_{\mu}x^{\mu},$ (103)

where $\lambda$ and $k_{\mu}$ are constants. Here the coefficient $-2$ in
front of the last term is chosen for convenience. Comparing (103) with (95),
we find that the transformations associated with the parameters $\lambda$ and
$k_{\mu}$, respectively, correspond to dilatation and special conformal
transformation in a flat Minkowski background.999For 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
$ds^{2}=g_{\mu\nu}dx^{\mu}dx^{\nu}$. In other words, in this case, finding the
residual symmetries is equivalent to solving the conformal Killing equation.

We can also verify the invariance of the quantum Lagrangian under the residual
symmetries more directly. As an illustration, let us take the residual
symmetries (103) stemmed from the Weyl transformation into consideration. It
is easy to check that the quantum Lagrangian (17) is invariant under the
residual symmetries

$\displaystyle\delta g_{\mu\nu}=2(\lambda-2k_{\rho}x^{\rho})g_{\mu\nu},$
$\displaystyle\delta\sigma=-(\lambda-2k_{\rho}x^{\rho}),\qquad\delta
b_{\mu}=2k_{\mu}B,$ (104)

where the other fields are unchanged. Then, generators corresponding to the
transformation parameters $\lambda$ and $k_{\mu}$ are respectively constructed
out of those of the Poincaré-like $IOSp(10|10)$ symmetry as

$\displaystyle D_{0}$ $\displaystyle\equiv$
$\displaystyle-P^{B}\equiv-P(B)=-\int
d^{3}x\,\tilde{g}^{0\nu}\phi^{2}\partial_{\nu}B,$ $\displaystyle K^{\mu}$
$\displaystyle\equiv$ $\displaystyle 2M^{x^{\mu}B}\equiv 2M^{\mu}(x,B)=2\int
d^{3}x\,\tilde{g}^{0\nu}\phi^{2}x^{\mu}\overset{\leftrightarrow}{\partial}_{\nu}B.$
(105)

For instance, we can explicitly show that these generators generate the former
transformation in Eq. (104) as follows

$\displaystyle\delta g_{\mu\nu}(x)$ $\displaystyle=$ $\displaystyle[i(\lambda
D_{0}+k_{\rho}K^{\rho}),g_{\mu\nu}(x)]$ (106) $\displaystyle=$ $\displaystyle
i\int
d^{3}y[\tilde{g}^{0\sigma}\phi^{2}(y)(-\lambda\partial_{\sigma}B(y)+2k_{\rho}y^{\rho}\overset{\leftrightarrow}{\partial}_{\sigma}B(y)),g_{\mu\nu}(x)]$
$\displaystyle=$ $\displaystyle i\int
d^{3}y\,\tilde{g}^{00}\phi^{2}(y)(-\lambda+2k_{\rho}y^{\rho})[\dot{B}(y),g_{\mu\nu}(x)]$
$\displaystyle=$ $\displaystyle 2(\lambda-2k_{\rho}x^{\rho})g_{\mu\nu}(x),$

where the ETCR (58) was used in the last equality.

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

$\displaystyle P_{\mu}$ $\displaystyle\equiv$ $\displaystyle P_{b_{\mu}}\equiv
P_{\mu}(b)=\int d^{3}x\,\tilde{g}^{0\nu}\phi^{2}\partial_{\nu}b_{\mu},$
$\displaystyle G^{\mu}{}_{\nu}$ $\displaystyle\equiv$ $\displaystyle
M^{x^{\mu}}{}_{b_{\nu}}-iM^{c^{\mu}}{}_{\bar{c}_{\nu}}\equiv
M^{\mu}{}_{\nu}(x,b)-iM^{\mu}{}_{\nu}(c^{\tau},\bar{c}_{\lambda})$ (107)
$\displaystyle=$ $\displaystyle\int
d^{3}x\,\tilde{g}^{0\lambda}\phi^{2}(x^{\mu}\overset{\leftrightarrow}{\partial}_{\lambda}b_{\nu}-ic^{\mu}\overset{\leftrightarrow}{\partial}_{\lambda}\bar{c}_{\nu}).$

At this point, let us elaborate a bit on the special conformal generator
$K^{\mu}$. It is well-known that under the special conformal transformation an
arbitrary “primary” operators ${\cal{O}}_{i}(x)$ of conformal dimension
$\Delta_{i}$ transforms as [5, 6]

$\displaystyle[iK_{\mu},{\cal{O}}_{i}(x)]=(2x_{\mu}x^{\nu}\partial_{\nu}-x^{2}\partial_{\mu}){\cal{O}}_{i}(x)+2\Delta_{i}x_{\mu}{\cal{O}}_{i}(x)-2x^{\nu}(S_{\mu\nu})_{i}{}^{j}{\cal{O}}_{j}(x),$
(108)

where $(S_{\mu\nu})_{i}{}^{j}$ is a spin matrix for the Lorentz
transformation. In our formalism, it is easy to show that under the special
conformal transformation the fundamental variables
$\Phi\equiv\\{g_{\mu\nu},\phi,b_{\mu},B,c_{\mu},\bar{c}^{\mu},c,\bar{c}\\}$
transform as

$\displaystyle[iK^{\rho},g_{\mu\nu}]=-4x^{\rho}g_{\mu\nu},\qquad[iK^{\rho},\phi]=2x^{\rho}\phi,\qquad[iK^{\rho},b_{\mu}]=2\delta_{\mu}^{\rho}B,\qquad$
$\displaystyle[iK^{\rho},B]=[iK^{\rho},c_{\mu}]=[iK^{\rho},\bar{c}^{\mu}]=[iK^{\rho},c]=[iK^{\rho},\bar{c}]=0.$
(109)

Comparing Eq. (109) to Eq. (108), we can immediately notice three things:
Firstly, $g_{\mu\nu}$ and $\phi$ are primary fields of conformal dimension
$-2$ and $1$, respectively, and $B,c_{\mu},\bar{c}^{\mu},c$ and $\bar{c}$ are
all primary fields of conformal dimension $0$ whereas $b_{\mu}$ is not a
primary field. Secondly, there is no term involving the terms quadratic in
$x^{\mu}$ in Eq. (109) since we have confined ourselves to linearized
symmetries in $x^{\mu}$. Finally, there is not the term including the spin
matrix $(S_{\mu\nu})_{i}{}^{j}$ in Eq. (109) since we do not consider spinor
fields.

Now we would like to show that in our theory there is an extended conformal
algebra in a flat Minkowski space-time where the Lorentz symmetry is replaced
with the $GL(4)$ symmetry. For this aim, let us consider a set of generators,
$\\{P_{\mu},G^{\mu}{}_{\nu},K^{\mu},D_{0}\\}$, which has 25 independent
generators and consists of a larger symmetry group than conformal symmetry
constructed in terms of 15 independent generators,
$\\{P_{\mu},M^{\mu}{}_{\nu},K^{\mu},D\\}$.

Let us begin by making the generator $D$ for dilatation. Recall that in
conformal field theory in the four-dimensional Minkowski space-time, the
dilatation generator obeys the following algebra for a local primary operator
$O_{i}(x)$ of conformal dimension $\Delta_{i}$ [5, 6]:

$\displaystyle[iD,O_{i}(x)]=x^{\mu}\partial_{\mu}O_{i}(x)+\Delta_{i}O_{i}(x).$
(110)

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

$\displaystyle[iD,\phi(x)]=x^{\mu}\partial_{\mu}\phi(x)+\phi(x).$ (111)

To be consistent with this equation, we shall make a generator $D$ for the
dilatation. From the definitions (105) and (107), we find

$\displaystyle[iG^{\mu}{}_{\mu},\phi(x)]=x^{\mu}\partial_{\mu}\phi(x),\qquad[iD_{0},\phi(x)]=-\phi(x),$
(112)

where Eqs. (33) and (49) were used. The following linear combination of
$G^{\mu}{}_{\nu}$ and $D_{0}$ does the job:

$\displaystyle D\equiv G^{\mu}{}_{\mu}-D_{0}.$ (113)

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

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

where we have used Eqs. (10), (39), (54) and (58). This equation implies that
the metric field is a conformal field of conformal dimension $0$ as expected
in the conventional conformal field theory.

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:

$\displaystyle[P_{\mu},P_{\nu}]=0,\quad[P_{\mu},G^{\rho}{}_{\sigma}]=iP_{\sigma}\delta^{\rho}_{\mu},\quad[P_{\mu},K^{\nu}]=-2i(G^{\rho}{}_{\rho}-D)\delta^{\nu}_{\mu},\quad$
$\displaystyle[P_{\mu},D]=iP_{\mu},\quad[G^{\mu}{}_{\nu},G^{\rho}{}_{\sigma}]=i(G^{\mu}{}_{\sigma}\delta^{\rho}_{\nu}-G^{\rho}{}_{\nu}\delta^{\mu}_{\sigma}),$
$\displaystyle[G^{\mu}{}_{\nu},K^{\rho}]=iK^{\mu}\delta^{\rho}_{\nu},\quad[G^{\mu}{}_{\nu},D]=[K^{\mu},K^{\nu}]=0,$
$\displaystyle[K^{\mu},D]=-iK^{\mu},\quad[D,D]=0.$ (115)

In this way, we have succeeded in showing that we have an extended conformal
algebra in the formalism at hand. As mentioned above, the essential difference
between the extended conformal symmetry and the conventional conformal
symmetry is the presence of the $GL(4)$ symmetry instead of the $SO(1,3)$
Lorentz symmetry, and this fact provides us with an important proof of exact
masslessness of the graviton [25].

We mention two remarks, one of which is that to show this closure we do not
have to use the gravitational ETCRs such as
$[\dot{g}_{\mu\nu},g_{\rho\sigma}^{\prime}]$ which is missing in the formalism
under consideration owing to an unspecified gravitational classical action. As
the second remark, let us note that we have two different ways for deriving
the algebra (115). The direct way is to use almost all the ETCRs obtained in
this article. In addition to it, we need some nontrivial ETCRs which are not
written out explicitly but can be obtained through the ETCRs appeared thus
far. For instance, we sometimes have to make use of the ETCR:

$\displaystyle[\dot{b}_{\mu},\dot{b}_{\nu}^{\prime}]=i\tilde{f}\phi^{-2}(\delta_{\nu}^{0}\ddot{b}_{\mu}+2\delta_{\nu}^{k}\partial_{k}\dot{b}_{\mu})\delta^{3}-2i\tilde{f}\tilde{g}^{0k}\partial_{k}[\tilde{f}\phi^{-2}(\partial_{\mu}b_{\nu}+\partial_{\nu}b_{\mu})\delta^{3}]$
$\displaystyle+i\tilde{f}^{2}\phi^{-2}(2\tilde{g}^{0k}\delta_{\nu}^{l}-\tilde{g}^{kl}\delta_{\nu}^{0})\partial_{k}\partial_{l}b_{\mu}\delta^{3}-i\partial_{0}[\tilde{f}\phi^{-2}(\partial_{\mu}b_{\nu}+\partial_{\nu}b_{\mu})]\delta^{3}.$
(116)

This ETCR is obtained by taking time-derivative of the latter equation in (55)
and then using the field equation (25). The other easier way of the derivation
of (115) is to appeal to the $IOSp(10|10)$ algebra in Eq. (84). To illustrate
the two different ways of derivation, we will explicitly evaluate the algebra
$[P_{\mu},K^{\nu}]=-2i(G^{\rho}{}_{\rho}-D)\delta_{\mu}^{\nu}$ in Appendix D.

To extract conformal algebra from the extended conformal algebra (115), it is
necessary to introduce the “Lorentz generator”, which can be contructed from
the $GL(4)$ generator as

$\displaystyle
M_{\mu\nu}\equiv-\eta_{\mu\rho}G^{\rho}{}_{\nu}+\eta_{\nu\rho}G^{\rho}{}_{\mu}.$
(117)

In terms of the generator $M_{\mu\nu}$, the algebra (115) can be cast to the
form:

$\displaystyle[P_{\mu},P_{\nu}]=0,\quad[P_{\mu},M_{\rho\sigma}]=i(P_{\rho}\eta_{\mu\sigma}-P_{\sigma}\eta_{\mu\rho}),$
$\displaystyle[P_{\mu},K^{\nu}]=-2i(G^{\rho}{}_{\rho}-D)\delta^{\nu}_{\mu},\quad[P_{\mu},D]=iP_{\mu},$
$\displaystyle[M_{\mu\nu},M_{\rho\sigma}]=-i(M_{\mu\sigma}\eta_{\nu\rho}-M_{\nu\sigma}\eta_{\mu\rho}+M_{\rho\mu}\eta_{\sigma\nu}-M_{\rho\nu}\eta_{\sigma\mu}),$
$\displaystyle[M_{\mu\nu},K^{\rho}]=i(-K_{\mu}\delta^{\rho}_{\nu}+K_{\nu}\delta^{\rho}_{\mu}),\quad[M_{\mu\nu},D]=[K^{\mu},K^{\nu}]=0,$
$\displaystyle[K^{\mu},D]=-iK^{\mu},\quad[D,D]=0.$ (118)

where we have defined $K_{\mu}\equiv\eta_{\mu\nu}K^{\nu}$. It is worthwhile to
point out that the the algebra (118) in quantum gravity coincides with the
well-known conformal algebra except for the expression of $[P_{\mu},K^{\nu}]$,
which takes the form,
$[P_{\mu},K^{\nu}]=-2i(\delta^{\nu}_{\mu}D+M_{\mu}{}^{\nu})$ in the conformal
algebra. However, this difference reflects only 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. (100) or Eq. (109) while it has $0$ in conformal field theory as seen in
Eq. (114). Moreover, the absence of the term $M_{\mu}{}^{\nu}$ on the RHS of
$[P_{\mu},K^{\nu}]$ in (118) can be understood from the fact that this term
comes from the quadratic terms in $x^{\mu}$ in Eq. (108), which is now missing
at present owing to our linear approximation as seen in Eq. (109).
Accordingly, in this sense the algebra (118) is essentially equivalent to the
conventional conformal algebra, and thus we have succeeded in obtaining the
conformal symmetry within the framework of quantum gravity.

##  6 Spontaneous breakdown of symmetries

In the theory under consideration, there is a huge global symmetry called the
Poincaré-like $IOSp(10|10)$ symmetry where we have $20$ $P^{M}$ generators and
$200$ $M^{MN}$ generators so totally $220$ symmetry generators. Then, it is
valuable to investigate which symmetries are spontaneously broken or survive
even in quantum regime. In particular, since our world has a built-in scale in
it, a global scale symmetry or dilatation must be spontaneously broken if our
theory makes sense.

In order to clarify the mechanism of symmetry breakdown, let us first
postulate the existence of a unique vacuum $|0\rangle$, which is normalized to
be the unity:

$\displaystyle\langle 0|0\rangle=1.$ (119)

It is also natural to assume that the vacuum is translation invariant

$\displaystyle P_{\mu}|0\rangle\equiv P_{\mu}(b)|0\rangle=0.$ (120)

Furthermore, it is supposed that 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 $\phi_{0}\neq 0$:

$\displaystyle\langle 0|g_{\mu\nu}|0\rangle=\eta_{\mu\nu},\qquad\langle
0|\phi|0\rangle=\phi_{0}.$ (121)

With these assumptions, a straightforward calculation reveals the following
nonvanishing VEVs for $P^{M}$

$\displaystyle\langle 0|[iP^{\mu}(x),b_{\rho}]|0\rangle=-\delta^{\mu}_{\rho},$
$\displaystyle\langle 0|[iP(\sigma),B]|0\rangle=-\langle
0|[iP(B),\sigma]|0\rangle=1,$ $\displaystyle\langle
0|\\{iP^{\mu}(c^{\tau}),\bar{c}_{\rho}\\}|0\rangle=i\delta^{\mu}_{\rho},\quad\langle
0|\\{iP_{\mu}(\bar{c}_{\tau}),c^{\rho}\\}|0\rangle=-i\delta^{\rho}_{\mu},$
$\displaystyle\langle 0|\\{iP(c),\bar{c}\\}|0\rangle=-\langle
0|\\{iP(\bar{c}),c\\}|0\rangle=i,$ (122)

and the ones for $M^{MN}$

$\displaystyle\langle
0|[iM^{\mu\nu}(x,x),\frac{1}{2}(\partial_{\lambda}b_{\rho}-\partial_{\rho}b_{\lambda})]|0\rangle=-(\delta^{\mu}_{\lambda}\delta^{\nu}_{\rho}-\delta^{\nu}_{\lambda}\delta^{\mu}_{\rho}),$
$\displaystyle\langle
0|[iM^{\mu}{}_{\nu}(x,b),g_{\sigma\tau}]|0\rangle=\delta^{\mu}_{\sigma}\eta_{\nu\tau}+\delta^{\mu}_{\tau}\eta_{\nu\sigma},$
$\displaystyle\langle
0|[iM^{\mu}(x,\sigma),\partial_{\lambda}B]|0\rangle=\langle
0|[iM^{\mu}(x,B),\partial_{\lambda}\sigma]|0\rangle=\delta^{\mu}_{\lambda},$
$\displaystyle\langle
0|\\{iM^{\mu\nu}(x,c^{\tau}),\partial_{\lambda}\bar{c}_{\rho}\\}|0\rangle=i\delta^{\mu}_{\lambda}\delta^{\nu}_{\rho},\quad\langle
0|\\{iM^{\mu}{}_{\nu}(x,\bar{c}_{\tau}),\partial_{\lambda}c^{\rho}\\}|0\rangle=-i\delta^{\mu}_{\lambda}\delta^{\rho}_{\nu},$
$\displaystyle\langle
0|\\{iM^{\mu}(x,c),\partial_{\lambda}\bar{c}\\}|0\rangle=i\delta^{\mu}_{\lambda},\quad\langle
0|\\{iM^{\mu}(x,\bar{c}),\partial_{\lambda}c\\}|0\rangle=-i\delta^{\mu}_{\lambda},$
$\displaystyle\langle 0|[iM(\sigma,B),\sigma]|0\rangle=\sigma_{0},$
$\displaystyle\langle
0|\\{iM^{\mu}(\sigma,c^{\tau}),\bar{c}_{\rho}\\}|0\rangle=i\sigma_{0}\delta^{\mu}_{\rho},\quad\langle
0|\\{iM_{\mu}(\sigma,\bar{c}_{\tau}),c^{\rho}\\}|0\rangle=-i\sigma_{0}\delta^{\rho}_{\mu},$
$\displaystyle\langle 0|\\{iM(\sigma,c),\bar{c}\\}|0\rangle=-\langle
0|\\{iM(\sigma,\bar{c}),c\\}|0\rangle=i\sigma_{0},$ (123)

where $\langle 0|\sigma(x)|0\rangle\equiv\sigma_{0}$. These VEVs clearly show
that the symmetries generated by the conserved charges

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

are necessarily broken spontaneously, thereby
$g_{\mu\nu},b_{\mu},\sigma,B,c^{\mu},\bar{c}_{\mu},c$ and $\bar{c}$ acquiring
massless Nambu-Goldstone modes. On the other hand, the remaining generators

$\displaystyle\\{M_{\mu\nu}(b,b),M_{\mu}(b,\sigma),M_{\mu}(b,B),M_{\mu}{}^{\nu}(b,c^{\tau}),M_{\mu\nu}(b,\bar{c}_{\tau}),M_{\mu}(b,c),M_{\mu}(b,\bar{c}),$
$\displaystyle
M^{\mu}(B,c^{\tau}),M_{\mu}(B,\bar{c}_{\tau}),M(B,c),M(B,\bar{c}),M^{\mu\nu}(c^{\tau},c^{\lambda}),M^{\mu}{}_{\nu}(c^{\tau},\bar{c}_{\lambda}),$
$\displaystyle
M^{\mu}(c^{\tau},c),M^{\mu}(c^{\tau},\bar{c}),M_{\mu\nu}(\bar{c}_{\tau},\bar{c}_{\lambda}),M_{\mu}(\bar{c}_{\tau},c),M_{\mu}(\bar{c}_{\tau},\bar{c}),$
$\displaystyle M(c,c),M(c,\bar{c}),M(\bar{c},\bar{c})\\},$

are found to be unbroken.

We are now ready to show that $GL(4)$, special conformal symmetry and
dilatation are spontaneously broken down to the Poincaré symmetry. For this
aim, it is more convenient to utilize the extended conformal algebra (115)
rather than the Poincaré-like $IOSp(10|10)$ algebra (84) and the conformal
algebra (118). It is then easy to see that the VEV of a commutator between the
$GL(4)$ generator and the metric field reads

$\displaystyle\langle
0|[iG^{\mu}{}_{\nu},g_{\sigma\tau}]|0\rangle=\delta^{\mu}_{\sigma}\eta_{\nu\tau}+\delta^{\mu}_{\tau}\eta_{\nu\sigma}.$
(124)

Thus, the Lorentz generator, which is defined in Eq. (117), has the vanishing
VEV:

$\displaystyle\langle 0|[iM_{\mu\nu},g_{\sigma\tau}]|0\rangle=0.$ (125)

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:

$\displaystyle\langle
0|[i\bar{M}_{\mu\nu},g_{\sigma\tau}]|0\rangle=2(\eta_{\mu\sigma}\eta_{\nu\tau}+\eta_{\mu\tau}\eta_{\nu\sigma}).$
(126)

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 [25]. Here it is interesting that in a
sector of the scalar field, the $GL(4)$ symmetry and the Lorentz symmetry as
well, do not give rise to a symmetry breaking as can be checked in the
commutators:

$\displaystyle\langle 0|[iG^{\mu}{}_{\nu},\phi]|0\rangle=\langle
0|[iM_{\mu\nu},\phi]|0\rangle=\langle 0|[i\bar{M}_{\mu\nu},\phi]|0\rangle=0.$
(127)

Next, we wish to clarify how the dilatation and special conformal symmetry are
spontaneously broken and what the corresponding Nambu-Goldstone bosons are. As
for the dilatation, it is not the gravitational field but the dilaton that
triggers a spontaneous symmetry breaking since for the metric Eq. (114)
provides us with

$\displaystyle\langle 0|[iD,g_{\sigma\tau}]|0\rangle=0,$ (128)

while, for the dilaton, from Eq. (111) we have

$\displaystyle\langle 0|[iD,\sigma]|0\rangle=1.$ (129)

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

Regarding the special conformal symmetry, we find

$\displaystyle\langle
0|[iK^{\mu},\partial_{\nu}\sigma]|0\rangle=2\delta^{\mu}_{\nu}.$ (130)

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 extended conformal algebra.
In the algebra (115), we have a commutator between $P_{\mu}$ and $K^{\nu}$:

$\displaystyle[P_{\mu},K^{\nu}]=-2i(G^{\rho}{}_{\rho}-D)\delta^{\nu}_{\mu}.$
(131)

Let us consider the Jacobi identity:

$\displaystyle[[P_{\mu},K^{\nu}],\sigma]+[[K^{\nu},\sigma],P_{\mu}]+[[\sigma,P_{\mu}],K^{\nu}]=0.$
(132)

Using the translational invariance of the vacuum in Eq. (120) and the equation

$\displaystyle[P_{\mu},\sigma]=-i\partial_{\mu}\sigma,$ (133)

and taking the VEV of the Jacobi identity (132), we can obtain the VEV

$\displaystyle\langle 0|[K^{\nu},\partial_{\mu}\sigma]|0\rangle$
$\displaystyle=$ $\displaystyle-2\delta^{\nu}_{\mu}\langle
0|[G^{\rho}{}_{\rho}-D,\sigma]|0\rangle$ (134) $\displaystyle=$
$\displaystyle-2i\delta^{\nu}_{\mu},$

which coincides with Eq. (130) as promised. In other words, the $GL(4)$
symmetry is spontaneously broken to the Poincaré symmetry whose Nambu-
Goldstone boson is the graviton, the dilatation and the special conformal
symmetry are at the same time spontaneously broken and the corresponding
Nambu-Goldstone bosons are the dilaton and the derivative of the dilaton,
respectively. It is not surprising that this pattern of the symmetry breaking
for the dilatation and special conformal symmetry is the same as that of the
conventional conformal field theory [26] since we share the same conformal
algebra.

##  7 Conclusion

In this article, we have performed a manifestly covariant quantization and
contructed a quantum theory of a general gravitational theory, which is
invariant under both general coordinate transformation (GCT) and Weyl
transformation, within the framework of the BRST formalism. In the present
formulation, since we have started with such a general action, it is
impossible to derive a concrete expression of the Einstein’s equation as well
as equal-time commutation relations (ETCRs) relevant to the metric tensor and
its derivatives such as $[\dot{g}_{\mu\nu},g_{\rho\sigma}^{\prime}]$. The key
point is, however, that without the Einstein’s equation and those ETCRs one
can derive the field equation for $b_{\mu}$ and the other nontrivial ETCRs, in
particular, $[\dot{b}_{\mu},b_{\nu}^{\prime}]$, which have been so far derived
from the Einstein’s equation and the metric’s ETCRs [12], by the help of two
kinds of BRST transformations and field equations except for the Einstein’s
equation.

Although we do not fix the classical gravitational action and consider a
general one, it is remarkable that we have a huge global symmetry, called a
Poincaré-like $IOSp(10|10)$ global transformation with $220$ symmetry
generators, which is larger than an $ISOp(8|8)$ with $144$ symmetry generators
in case of general relativity [12], owing to the presence of the Weyl symmetry
in our formulation. In a flat Minkowski background, the Poincaré-like
$IOSp(10|10)$ symmetry naturally includes the extended conformal algebra where
the usual Lorentz symmetry is extended to a larger $GL(4)$ symmetry. Then, it
is obvious that this extended conformal algebra includes conformal algebra as
a subgroup. In other words, in our formulation of quantum gravity, the
conformal symmetry emerges from not the classical action but the gauge-fixing
and FP ghost actions.

According to the Zumino theorem [7], 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 partially
supports a conjecture that the Zumino theorem could be valid even in quantum
gravity. However, there is a loophole in our argument. In this article, we
have assumed that a classical action does not involve more than first order
derivatives of the metric and matter fields. If the classical theory includes
the higher-derivative terms as in quadratic gravity, we have to rewrite the
theory in order not to include more than first order derivatives by
introducing suitabe auxiliary fields when we perform the canonical
quantization. Then, the introduction of the auxiliary fields sometimes gives
rise to a new local symmetry such as the Stückelberg symmetry [27, 28, 29, 30,
31, 11]. In such a situation, our argument is not valid. Thus, we can address
a generalized Zumino theorem in quantum gravity as follows: The Zumino theorem
is certainly valid in quantum gravity when a classical action does not involve
more than first order derivatives.

As future works, we would like to list up two important issues. First, it is
well-known that Weyl symmetry is explicitly broken by conformal anomaly at the
quantum level [32]. However, since a scalar field is now at our disposal, it
is not obvious that the Weyl symmetry is anomalous or not in our theory.
Actually, in a Weyl invariant scalar-tensor gravity, there is no conformal
anomaly [33].

Another important issue is related to the non-renormalizability of general
relativity. To remedy the problem, we are accustomed to adding quadratic terms
in the Riemann curvatures and consider the Einstein-Hilbert term supplemented
with the terms containing $R^{2}$ and $R_{\mu\nu}R^{\mu\nu}$, but then we meet
another serious problem of non-unitarity. We conjecture that the non-unitarity
problem in the higher-derivative gravities might be solved by a mechanism of
ghost confinement [34] and then the Poincaré-like global symmetry would play a
critical role, as in the confinement of quarks and gluons by the BRST symmetry
in QCD [35]. We shall return these two issues in near future.

## Acknowledgment

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

## Appendix

##  Appendix A $GL(4)$ invariance of the extended de Donder gauge condition

In this appendix, we present a proof of the global $GL(4)$ invariance of the
extended de Donder gauge condition (10). To do that, let us first recall the
$GL(4)$ transformation:

$\displaystyle x^{\prime\mu}=A^{\mu}{}_{\nu}x^{\nu},$ (A.1)

where $A^{\mu}{}_{\nu}$ is a constant and invertible $4\times 4$ matrix
belonging to a $GL(4,R)$. From this definition, $A^{\mu}{}_{\nu}$ and its
inverse $A^{-1\mu}{}_{\nu}$ can be expressed as

$\displaystyle A^{\mu}{}_{\nu}=\frac{\partial x^{\prime\mu}}{\partial
x^{\nu}},\qquad A^{-1\mu}{}_{\nu}=\frac{\partial x^{\mu}}{\partial
x^{\prime\nu}}.$ (A.2)

Under the $GL(4)$ transformation, the metric and the scalar field transform as

$\displaystyle
g^{\prime}_{\mu\nu}(x^{\prime})=A^{-1\rho}{}_{\mu}A^{-1\sigma}{}_{\nu}g_{\rho\sigma}(x),\qquad
g^{\prime\mu\nu}(x^{\prime})=A^{\mu}{}_{\rho}A^{\nu}{}_{\sigma}g^{\rho\sigma}(x),$
$\displaystyle\sqrt{-g^{\prime}(x^{\prime})}=(\det
A)^{-1}\sqrt{-g(x)},\qquad\phi^{\prime}(x^{\prime})=\phi(x).$ (A.3)

Using the transformation law and
$\partial^{\prime}_{\mu}=A^{-1\nu}{}_{\mu}\partial_{\nu}$, we can show that

$\displaystyle\partial^{\prime}_{\mu}(\sqrt{-g^{\prime}(x^{\prime})}g^{\prime\mu\nu}(x^{\prime})\phi^{\prime
2}(x^{\prime}))=(\det
A)^{-1}A^{\nu}{}_{\sigma}\partial_{\rho}(\sqrt{-g(x)}g^{\rho\sigma}(x)\phi^{2}(x)).$
(A.4)

Thus, the extended de Donder gauge condition (10) is invariant under the
$GL(4)$ transformation in the sense that when the extended de Donder gauge
condition is imposed in a frame it holds true in a $GL(4)$ transformed frame
as well. In a similar manner, the $GL(4)$ invariance of the unitary gauge
$\phi=\rm{constant}$ and the Lorenz gauge $\nabla_{\mu}S^{\mu}=0$ can be
shown. Precisely speaking, the extended de Donder gauge condition is
transformed as a vector density under the $GL(4)$ transformation as seen in
Eq. (A.4).

Indeed, it is a gauge-fixing action that is exactly invariant under the
$GL(4)$ transformation. To see this fact, let us consider a gauge-fixed action
corresponding to the extended de Donder gauge condition (10):

$\displaystyle S_{GF}=-\int
d^{4}x\sqrt{-g}g^{\mu\nu}\phi^{2}\partial_{\mu}b_{\nu}.$ (A.5)

Under the $GL(4)$ transformation, this action transforms as

$\displaystyle S_{GF}^{\prime}$ $\displaystyle=$ $\displaystyle-\int
d^{4}x^{\prime}\sqrt{-g^{\prime}}g^{\prime\mu\nu}\phi^{\prime
2}\partial_{\mu}^{\prime}b_{\nu}^{\prime}$ (A.6) $\displaystyle=$
$\displaystyle-\int d^{4}x\det A(\det
A)^{-1}A^{\mu}{}_{\rho}A^{\nu}{}_{\sigma}\sqrt{-g}g^{\rho\sigma}\phi^{2}A^{-1\tau}{}_{\mu}\partial_{\tau}(A^{-1\lambda}{}_{\nu}b_{\lambda})$
$\displaystyle=$ $\displaystyle S_{GF},$

where we have used

$\displaystyle b_{\mu}^{\prime}(x^{\prime})=A^{-1\nu}{}_{\mu}b_{\nu}(x),\qquad
d^{4}x^{\prime}=\det A\,d^{4}x.$ (A.7)

Hence, the gauge-fixed action is precisely invariant under the $GL(4)$
transformation.

##  Appendix B Proof of Eqs. (53) and (54)

In this Appendix, we derive Eqs. (53) and (54) by following the method
developed previously in case of the de Donder gauge condition [8].

The translational invariance in general requires the validity of the following
equation for a generic field $\Phi(x)$:

$\displaystyle[\Phi(x),P_{\rho}]=i\partial_{\rho}\Phi(x),$ (B.1)

where $P_{\rho}$ is the generator of the translation which is now given by

$\displaystyle P_{\rho}=\int
d^{3}x\,\tilde{g}^{0\lambda}\phi^{2}\partial_{\lambda}b_{\rho}.$ (B.2)

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

$\displaystyle[g_{\mu\nu}(x),P_{\rho}]=[g_{\mu\nu}(x),\int
d^{3}x^{\prime}\,\tilde{g}^{0\lambda\prime}\phi^{2\prime}\partial_{\lambda}b_{\rho}^{\prime}]=i\partial_{\rho}g_{\mu\nu}(x).$
(B.3)

Taking $x^{0}=x^{\prime 0}$ and using
$[g_{\mu\nu},\tilde{g}^{0\lambda\prime}\phi^{2\prime}]=0$, we have

$\displaystyle\int
d^{3}x^{\prime}\,\tilde{g}^{0\lambda}(x^{\prime})\phi^{2}(x^{\prime})[g_{\mu\nu},\partial_{\lambda}b_{\rho}^{\prime}]=i\partial_{\rho}g_{\mu\nu}(x).$
(B.4)

Using the extended de Donder gauge condition (10) and Eq. (39), this equation
can be rewritten as

$\displaystyle\int
d^{3}x^{\prime}\,\tilde{g}^{00}(x^{\prime})\phi^{2}(x^{\prime})[g_{\mu\nu},\dot{b}_{\rho}^{\prime}]=i\left[\partial_{\rho}g_{\mu\nu}+\partial_{0}(\tilde{g}^{00}\phi^{2})\tilde{f}\phi^{-2}(\delta_{\mu}^{0}g_{\rho\nu}+\delta_{\nu}^{0}g_{\rho\mu})\right],$
(B.5)

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

$\displaystyle[g_{\mu\nu},\dot{b}_{\rho}^{\prime}]$ $\displaystyle=$
$\displaystyle
i\tilde{f}\phi^{-2}\left[\partial_{\rho}g_{\mu\nu}+\partial_{0}(\tilde{g}^{00}\phi^{2})\tilde{f}\phi^{-2}(\delta_{\mu}^{0}g_{\rho\nu}+\delta_{\nu}^{0}g_{\rho\mu})\right]\delta^{3}$
(B.6) $\displaystyle+$ $\displaystyle
F_{(\mu\nu)\rho}{}^{k}\partial_{k}(\tilde{f}\phi^{-2}\delta^{3}),$

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 extended de Donder gauge condition (10):

$\displaystyle[\partial_{\mu}(\tilde{g}^{\mu\nu}\phi^{2}),b_{\rho}^{\prime}]=0.$
(B.7)

An explicit calculation reveals that Eq. (B.7) leads to an equation for
$F_{(\mu\nu)\rho}{}^{k}$:

$\displaystyle\left(\tilde{g}^{0\alpha}g^{\nu\beta}-\frac{1}{2}\tilde{g}^{0\nu}g^{\alpha\beta}\right)F_{(\alpha\beta)\rho}{}^{k}=-i(\tilde{g}^{0k}\delta_{\rho}^{\nu}+\tilde{g}^{0\nu}\delta_{\rho}^{k}-\tilde{g}^{k\nu}\delta_{\rho}^{0}).$
(B.8)

This equation has the unique solution given by

$\displaystyle
F_{(\mu\nu)\rho}{}^{k}=i[(\delta_{\mu}^{k}-2\delta_{\mu}^{0}\tilde{f}\tilde{g}^{0k})g_{\rho\nu}+(\mu\leftrightarrow\nu)].$
(B.9)

We can therefore obtain

$\displaystyle[g_{\mu\nu},\dot{b}_{\rho}^{\prime}]$ $\displaystyle=$
$\displaystyle
i\Bigl{\\{}[\tilde{f}\phi^{-2}\partial_{\rho}g_{\mu\nu}-\partial_{0}(\tilde{f}\phi^{-2})(\delta_{\mu}^{0}g_{\rho\nu}+\delta_{\nu}^{0}g_{\rho\mu})]\delta^{3}$
(B.10) $\displaystyle+$
$\displaystyle[(\delta_{\mu}^{k}-2\delta_{\mu}^{0}\tilde{f}\tilde{g}^{0k})g_{\rho\nu}+(\mu\leftrightarrow\nu)]\partial_{k}(\tilde{f}\phi^{-2}\delta^{3})\Bigr{\\}},$

which is nothing but the former equation in Eq. (54). The latter equation in
Eq. (54) can be easily obtained from Eq. (B.10). Finally, using Eqs. (36) and
(39), we can arrive at another equation (53). It is of interest that Eq. (53)
can be derived from only the translational invariance and the extended 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 $\phi$.

##  Appendix C Proof of
$[\dot{b}_{\mu},b_{\nu}^{\prime}]=-i\tilde{f}\phi^{-2}(\partial_{\mu}b_{\nu}+\partial_{\nu}b_{\mu})\delta^{3}$

In this Appendix, we present a proof of the ETCR,
$[\dot{b}_{\mu},b_{\nu}^{\prime}]=-i\tilde{f}\phi^{-2}(\partial_{\mu}b_{\nu}+\partial_{\nu}b_{\mu})\delta^{3}$
on the basis of the BRST transformation without using the Einstein’s equation.
This calculation exhibits that the BRST transformation offers a very powerful
method for deriving a nontrivial ETCR.

Let us begin by the ETCR in Eq. (50):

$\displaystyle[\dot{\bar{c}}_{\mu},b_{\nu}^{\prime}]=-i\tilde{f}\phi^{-2}\partial_{\nu}\bar{c}_{\mu}\delta^{3}.$
(C.1)

Then, taking its GCT BRST transformation yields

$\displaystyle[i\dot{B}_{\mu},b_{\nu}^{\prime}]-\\{\dot{\bar{c}}_{\mu},-c^{\alpha\prime}\partial_{\alpha}b_{\nu}^{\prime}\\}$
$\displaystyle=$
$\displaystyle-i(\delta_{B}\tilde{f}\phi^{-2}-2\tilde{f}\phi^{-3}\delta_{B}\phi)\partial_{\nu}\bar{c}_{\mu}\delta^{3}$
(C.2) $\displaystyle-$ $\displaystyle
i\tilde{f}\phi^{-2}\partial_{\nu}(iB_{\mu})\delta^{3}.$

Using Eq. (3), the first term on the left-hand side (LHS) and the last term on
the right-hand side (RHS) can be rewritten as

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

Furthermore, we can calculate $\delta_{B}\tilde{f}$ to be

$\displaystyle\delta_{B}\tilde{f}=-\tilde{f}^{2}\delta_{B}\tilde{g}^{00}=\tilde{f}\partial_{\rho}c^{\rho}-\partial_{\rho}\tilde{f}c^{\rho}-2\tilde{f}^{2}\tilde{g}^{0\rho}\partial_{\rho}c^{0}.$
(C.4)

Using these equations and $\delta_{B}\phi=-c^{\rho}\partial_{\rho}\phi$, Eq.
(C.2) can be further cast to the form:

$\displaystyle[\dot{b}_{\mu},b_{\nu}^{\prime}]=-i[\partial_{0}(c^{\alpha}\partial_{\alpha}\bar{c}_{\mu}),b_{\nu}^{\prime}]+i\\{\dot{\bar{c}}_{\mu},c^{\alpha\prime}\partial_{\alpha}b_{\nu}^{\prime}\\}-if\phi^{-2}\partial_{\nu}b_{\mu}\delta^{3}$
$\displaystyle+\tilde{f}\phi^{-2}\partial_{\nu}(c^{\alpha}\partial_{\alpha}\bar{c}_{\mu})\delta^{3}-\tilde{f}\phi^{-2}(\partial_{\rho}c^{\rho}-\tilde{f}^{-1}\partial_{\rho}\tilde{f}c^{\rho}-2\tilde{f}\tilde{g}^{0\rho}\partial_{\rho}c^{0}$
$\displaystyle+2\phi^{-1}c^{\rho}\partial_{\rho}\phi)\partial_{\nu}\bar{c}_{\mu}\delta^{3}.$
(C.5)

The remaining work is to calculate two (anti)commutators on the RHS. It is
straightforward to carry out such calculations by using various ETCRs obtained
thus far. In particular, we make use of the following ETCRs

$\displaystyle[\ddot{\bar{c}}_{\mu},b_{\nu}^{\prime}]=-i\tilde{f}\phi^{-2}\delta_{\nu}^{0}\ddot{\bar{c}}_{\mu}\delta^{3}-2i\tilde{f}[\phi^{-2}\delta_{\nu}^{k}\partial_{k}\dot{\bar{c}}_{\mu}\delta^{3}-\tilde{g}^{0k}\partial_{k}(\tilde{f}\phi^{-2}\partial_{\nu}\bar{c}_{\mu}\delta^{3})]$
$\displaystyle-i\tilde{f}^{2}\phi^{-2}(2\tilde{g}^{0k}\delta_{\nu}^{l}-\tilde{g}^{kl}\delta_{\nu}^{0})\partial_{k}\partial_{l}\bar{c}_{\mu}\delta^{3},$
(C.6)

and

$\displaystyle[\dot{\bar{c}}_{\mu},\dot{b}_{\nu}^{\prime}]=-[\ddot{\bar{c}}_{\mu},b_{\nu}^{\prime}]-i\partial_{0}(\tilde{f}\phi^{-2}\partial_{\nu}\bar{c}_{\mu})\delta^{3}.$
(C.7)

Eq. (C.6) can be obtained from Eqs. (40), (45), (47) and (50).

As a result, the two (anti)commutators on the RHS in Eq. (C.5) are given by

$\displaystyle-i[\partial_{0}(c^{\alpha}\partial_{\alpha}\bar{c}_{\mu}),b_{\nu}^{\prime}]=-\tilde{f}\phi^{-2}(\partial_{\nu}c^{\alpha}\partial_{\alpha}\bar{c}_{\mu}+\dot{c}^{0}\partial_{\nu}\bar{c}_{\mu}+\delta_{\nu}^{0}c^{0}\ddot{\bar{c}}_{\mu})\delta^{3}$
$\displaystyle-2\tilde{f}c^{0}[\phi^{-2}\delta_{\nu}^{k}\partial_{k}\dot{\bar{c}}_{\mu}\delta^{3}-\tilde{g}^{0k}\partial_{k}(\tilde{f}\phi^{-2}\partial_{\nu}\bar{c}_{\mu}\delta^{3})]$
$\displaystyle-\tilde{f}^{2}\phi^{-2}c^{0}(2\tilde{g}^{0k}\delta_{\nu}^{l}-\tilde{g}^{kl}\delta_{\nu}^{0})\partial_{k}\partial_{l}\bar{c}_{\mu}\delta^{3}-c^{k}\partial_{k}(\tilde{f}\phi^{-2}\partial_{\nu}\bar{c}_{\mu}\delta^{3}),$
(C.8)

and

$\displaystyle
i\\{\dot{\bar{c}}_{\mu},c^{\alpha\prime}\partial_{\alpha}b_{\nu}^{\prime}\\}=-i\tilde{f}\phi^{-2}\partial_{\mu}b_{\nu}\delta^{3}+\tilde{f}\phi^{-2}\delta_{\nu}^{0}c^{0}\ddot{\bar{c}}_{\mu}\delta^{3}$
$\displaystyle+2\tilde{f}[\phi^{-2}\delta_{\nu}^{k}c^{0}\partial_{k}\dot{\bar{c}}_{\mu}\delta^{3}-\tilde{g}^{0k}\partial_{k}(\tilde{f}\phi^{-2}c^{0}\partial_{\nu}\bar{c}_{\mu}\delta^{3})]$
$\displaystyle+\tilde{f}^{2}\phi^{-2}(2\tilde{g}^{0k}\delta_{\nu}^{l}-\tilde{g}^{kl}\delta_{\nu}^{0})c^{0}\partial_{k}\partial_{l}\bar{c}_{\mu}\delta^{3}$
$\displaystyle-c^{0}\partial_{0}(\tilde{f}\phi^{-2}\partial_{\nu}\bar{c}_{\mu})\delta^{3}+\tilde{f}\phi^{-2}\partial_{k}(c^{k}\delta^{3})\partial_{\nu}\bar{c}_{\mu}.$
(C.9)

Substituting Eqs. (C.8) and (C.9) into Eq. (C.5), we can reach the desired
equation:

$\displaystyle[\dot{b}_{\mu},b_{\nu}^{\prime}]=-i\tilde{f}\phi^{-2}(\partial_{\mu}b_{\nu}+\partial_{\nu}b_{\mu})\delta^{3}.$
(C.10)

##  Appendix D Two derivations of
$[P_{\mu},K^{\nu}]=-2i(G^{\rho}{}_{\rho}-D)\delta_{\mu}^{\nu}$

For illustrative purposes, in this appendix we present two different
derivations for one of conformal algebra,
$[P_{\mu},K^{\nu}]=-2i(G^{\rho}{}_{\rho}-D)\delta_{\mu}^{\nu}$, one of which
relies on the $IOSp(10|10)$ algebra in Eq. (84). Another derivation is a
direct calculation based on the ETCRs obtained in this article. We find that
the former derivation method is much easier than that of the latter and this
fact elucidates the power of the Poincaré-like $IOSp(10|10)$ algebra.

For the translation generator $P_{\mu}$ in (107) and the special conformal
generator $K^{\mu}$ in (105), the $IOSp(10|10)$ algebra Eq. (84) provides us
with

$\displaystyle[P_{\mu},K^{\nu}]$ $\displaystyle=$
$\displaystyle-[K^{\nu},P_{\mu}]=-2[M^{\nu}(x,B),P_{\mu}(b)]=-2[M^{x^{\nu}B},P_{b_{\mu}}]$
(D.1) $\displaystyle=$
$\displaystyle-2i[P^{x^{\nu}}\tilde{\eta}^{B}{}_{b_{\mu}}-(-)^{|B||b_{\mu}|}P^{B}\tilde{\eta}^{x^{\nu}}{}_{b_{\mu}}]=2iP^{B}\delta_{\mu}^{\nu}$
$\displaystyle=$ $\displaystyle
2iP(B)\delta_{\mu}^{\nu}=-2iD_{0}\delta_{\mu}^{\nu}=-2i(G^{\rho}{}_{\rho}-D)\delta_{\mu}^{\nu},$

where we have used $\tilde{\eta}^{B}{}_{b_{\mu}}=0$ and
$\tilde{\eta}^{x^{\nu}}{}_{b_{\mu}}=\delta_{\mu}^{\nu}$ at the fifth equality.
Note that this derivation is very transparent.

As an alternative derivation of this algebra, one can rely on the ETCRs
obtained thus far. This derivation method is very direct but it is a bit
complicated since we have to use the extended de Donder gauge condition (10),
field equations and the integration by parts etc. repeatedly.

From Eqs. (105) and (107), we have

$\displaystyle[P_{\mu},K^{\nu}]=\int
d^{3}xd^{3}x^{\prime}[\tilde{g}^{0\lambda}\phi^{2}\partial_{\lambda}b_{\mu},2\tilde{g}^{0\rho\prime}\phi^{2\prime}x^{\nu\prime}\overset{\leftrightarrow}{\partial}_{\rho}B^{\prime}]$
$\displaystyle=2\int
d^{3}xd^{3}x^{\prime}\Bigl{(}x^{\nu\prime}[\tilde{g}^{0\lambda}\phi^{2}\partial_{\lambda}b_{\mu},\tilde{g}^{0\rho\prime}\phi^{2\prime}\partial_{\rho}B^{\prime}]-[\tilde{g}^{0\lambda}\phi^{2}\partial_{\lambda}b_{\mu},\tilde{g}^{0\nu\prime}\phi^{2\prime}B^{\prime}]\Bigr{)}$
$\displaystyle\equiv 2(I+J).$ (D.2)

It is straightforward to compute $J$ whose final result takes the form

$\displaystyle J=i\int
d^{3}x\,\delta_{\mu}^{0}\tilde{g}^{\lambda\nu}\phi^{2}\partial_{\lambda}B.$
(D.3)

Here, as well as the extended de Donder gauge condition (10), we have used the
following ETCRs:

$\displaystyle[\dot{b}_{\mu},\tilde{g}^{0\nu\prime}]=-i\Bigl{\\{}[\tilde{f}\phi^{-2}\partial_{\mu}\tilde{g}^{0\nu}+\partial_{0}(\tilde{f}\phi^{-2})\tilde{g}^{00}\delta_{\mu}^{\nu}]\delta^{3}+(\tilde{g}^{0k\prime}\delta_{\mu}^{\nu}-\tilde{g}^{\nu
k\prime}\delta_{\mu}^{0}+\tilde{g}^{0\nu\prime}\delta_{\mu}^{k})\partial_{k}^{\prime}(\tilde{f}\phi^{-2}\delta^{3})\Bigr{\\}},$
$\displaystyle[\dot{b}_{\mu},\phi^{2\prime}]=-2i\tilde{f}\phi^{-1}\partial_{\mu}\phi\delta^{3},\qquad[\dot{b}_{\mu},B^{\prime}]=-i\tilde{f}\phi^{-2}\partial_{\mu}B\delta^{3}.$
(D.4)

Next, let us pay our attention to an evaluation of $I$, which can be divided
into four parts:

$\displaystyle I=\int
d^{3}xd^{3}x^{\prime}x^{\nu\prime}\Bigl{(}[\tilde{g}^{00}\phi^{2}\dot{b}_{\mu},\tilde{g}^{00\prime}\phi^{2\prime}\dot{B}^{\prime}]+[\tilde{g}^{0i}\phi^{2}\partial_{i}b_{\mu},\tilde{g}^{00\prime}\phi^{2\prime}\dot{B}^{\prime}]$
$\displaystyle+[\tilde{g}^{00}\phi^{2}\dot{b}_{\mu},\tilde{g}^{0j\prime}\phi^{2\prime}\partial_{j}B^{\prime}]+[\tilde{g}^{0i}\phi^{2}\partial_{i}b_{\mu},\tilde{g}^{0j\prime}\phi^{2\prime}\partial_{j}B^{\prime}]\Bigr{)}$
$\displaystyle\equiv I_{1}+I_{2}+I_{3}+I_{4}.$ (D.5)

The calculation of $I_{4}$ is easiest since it involves only one ETCR as
follows:

$\displaystyle I_{4}$ $\displaystyle=$ $\displaystyle\int
d^{3}xd^{3}x^{\prime}x^{\nu\prime}\tilde{g}^{0i}\phi^{2}\phi^{2\prime}\partial_{j}B^{\prime}[\partial_{i}b_{\mu},\tilde{g}^{0j\prime}]$
(D.6) $\displaystyle=$ $\displaystyle-i\int
d^{3}x\delta_{\mu}^{i}x^{\nu}\partial_{0}(\tilde{g}^{00}\phi^{2})\partial_{i}B.$

In a similar manner, it is straightforward to carry out the calculations of
$I_{2}$ and $I_{3}$, which are summarized as

$\displaystyle I_{2}$ $\displaystyle=$ $\displaystyle i\int
d^{3}x\tilde{g}^{0i}\phi^{2}\delta_{\mu}^{j}(\delta_{i}^{\nu}\partial_{j}B+x^{\nu}\partial_{i}\partial_{j}B),$
$\displaystyle I_{3}$ $\displaystyle=$ $\displaystyle i\int
d^{3}x\Bigl{[}x^{\nu}\phi^{2}(\tilde{g}^{00}\ddot{B}\delta_{\mu}^{0}+\tilde{g}^{0i}\partial_{i}\partial_{\mu}B)-\delta_{i}^{\nu}\phi^{2}(-\tilde{g}^{0i}\delta_{\mu}^{j}$
(D.7) $\displaystyle+$
$\displaystyle\tilde{g}^{ij}\delta_{\mu}^{0}-\tilde{g}^{0j}\delta_{\mu}^{i})\partial_{j}B\Bigr{]}.$

The evaluation of $I_{1}$ is more difficult than those of the other
$I_{i}(i=2,3,4)$, but is straightforward whose result is given by

$\displaystyle I_{1}=-i\int
d^{3}xx^{\nu}\Bigl{[}\partial_{\mu}(\tilde{g}^{00}\phi^{2}\dot{B})+\delta_{\mu}^{i}\partial_{0}(\tilde{g}^{00}\phi^{2})\partial_{i}B\Bigr{]},$
(D.8)

where in particular we have used the ETCR:

$\displaystyle[\dot{b}_{\mu},\dot{B}^{\prime}]=-i\partial_{0}(\tilde{f}\phi^{-2}\partial_{\mu}B)\delta^{3}-2i\tilde{f}\tilde{g}^{0i}\partial_{i}(\tilde{f}\phi^{-2}\partial_{\mu}B\delta^{3}).$
(D.9)

Then, summing up $I$ and $J$, and multiplying it by 2, we can obtain

$\displaystyle[P_{\mu},K^{\nu}]$ $\displaystyle\equiv$ $\displaystyle 2(I+J)$
(D.10) $\displaystyle=$ $\displaystyle 2i\int
d^{3}x\,\tilde{g}^{0\lambda}\phi^{2}\partial_{\lambda}B\delta_{\mu}^{\nu}$
$\displaystyle=$ $\displaystyle-2iD_{0}\delta_{\mu}^{\nu}$ $\displaystyle=$
$\displaystyle-2i(G^{\rho}{}_{\rho}-D)\delta_{\mu}^{\nu}.$

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