YITP-21-87

BRST Quantization of General Relativity in Unimodular Gauge and Unimodular
Gravity

Taichiro Kugo,1,***e-mail address<EMAIL_ADDRESS>Ryuichi
Nakayama,2,†††e-mail address<EMAIL_ADDRESS>and
Nobuyoshi Ohta 3,4,‡‡‡e-mail address<EMAIL_ADDRESS>
1 Yukawa Institute for Theoretical Physics, Kyoto University, Kyoto 606-8502,
Japan

2 Division of Physics, Graduate School of Science, Hokkaido University,
Sapporo 060-0810, Japan

3 Department of Physics, National Central University, Zhongli, Taoyuan 320317,
Taiwan

and

4 Research Institute for Science and Technology, Kindai University, Higashi-
Osaka, Osaka 577-8502, Japan

Abstract

Unimodular gravity (UG) is an important theory which may explain the smallness
of the cosmological constant. To get insight into the covariant quantization
of UG, we discuss the BRST quantization of General Relativity (GR) with a
cosmological constant in the unimodular gauge. We develop a novel way to gauge
fix the transverse diffeomorphism (TDiff) and then further to fulfill the
unimodular gauge. This process requires the introduction of an additional pair
of BRST doublets which decouple from the physical sector together with the
other three pairs of BRST doublets for the TDiff. We show that the physical
spectrum is the same as GR in the usual covariant gauge fixing. We then study
a theory derived by making “Fourier transform” of GR in the unimodular gauge
with respect to the cosmological constant as a candidate of “quantum UG.” We
clarify the difference from GR and point out problems in this theory.

##  1 Introduction

General Relativity (GR) well describes gravitational low-energy phenomena. The
action is given by

$\displaystyle S_{\rm GR}=Z_{N}\int d^{4}x\sqrt{-g}\left(R-2\Lambda\right),$
(1.1)

where $Z_{N}=\frac{1}{16\pi G_{N}}$ with the Newton constant $G_{N}$. Here we
have also included the cosmological constant, and we refer to this theory
simply as GR. Making variation of this theory, we get the Einstein equation
with the cosmological constant:

$\displaystyle R_{\mu\nu}-\frac{1}{2}(R-2\Lambda)g_{\mu\nu}=0.$ (1.2)

Classical GR has several equivalent formulations. An important example is the
Unimodular Gravity (UG) [1, 2, 3, 4, 5, 6], which is defined as a theory of
gravity with the constraint

$\displaystyle\sqrt{-g}=\omega,$ (1.3)

where $\omega$ is a fixed volume form. This constraint can be imposed by using
a Lagrange multiplier field $\lambda$:

$\displaystyle S_{\rm UG}=Z_{N}\int
d^{4}x\left[\sqrt{-g}(R-2\Lambda)+\lambda(\sqrt{-g}-\omega)\right].$ (1.4)

This theory is considered to give a solution to the cosmological constant
problem. Making the variation of the action with respect to the metric, we
obtain

$\displaystyle R_{\mu\nu}-\frac{1}{2}(R-2\Lambda+\lambda)g_{\mu\nu}=0,$ (1.5)

together with the unimodular condition (1.3). Taking the trace of this
equation, we get

$\displaystyle\lambda=2\Lambda-\frac{R}{2}.$ (1.6)

Plugging this into the field equation (1.5), we find

$\displaystyle R_{\mu\nu}-\frac{1}{4}Rg_{\mu\nu}=0,$ (1.7)

which is the traceless part of the Einstein equation. Thus the “cosmological
constant” drops out of our field equation. In this theory, we cannot get the
trace part of the usual Einstein equation, but this can be recovered using the
Bianchi identity. By taking the covariant derivative of (1.7), we get

$\displaystyle\partial_{\mu}R=0,$ (1.8)

which, upon integration yields

$\displaystyle R=4\Lambda_{0}.$ (1.9)

Then eq. (1.7) can be rewritten as

$\displaystyle R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu}+\Lambda_{0}\,g_{\mu\nu}=0,$
(1.10)

recovering the Einstein equation with the cosmological constant $\Lambda_{0}$.
Thus the cosmological constant appears as an integration constant. Note that
$\Lambda_{0}$ has nothing to do with the constant term in the action. In the
presence of matter, this could be repeated together with the conservation of
energy-momentum tensor.

The constant term $\Lambda$ in (1.4) can be absorbed into the definition of
the Lagrange multiplier field $\lambda$:

$\displaystyle S_{\rm UG}=Z_{N}\int
d^{4}x\left[\sqrt{-g}R+\lambda(\sqrt{-g}-\omega)\right].$ (1.11)

Using the unimodular condition, our action can also be written as

$\displaystyle S_{\rm UG}=Z_{N}\int d^{4}x\left[\omega
R+\lambda(\sqrt{-g}-\omega)\right].$ (1.12)

This is the standard formulation of UG, and this shows that the classical
gravity is equivalent to UG modulo the above-mentioned global property.

At the classical level, GR and UG are almost equivalent in the sense that
(1.3) can be seen as a gauge fixing condition in GR. However when this
condition is present, the identity for the global quantity,

$\displaystyle\int d^{4}x\sqrt{-g}=\int d^{4}x\,\omega,$ (1.13)

follows, which is a gauge (GC transformation) invariant relation and cannot be
regarded merely as a gauge condition. Thus classical UG may be considered to
be equivalent to the classical GR with the fixed spacetime volume. Conversely
GR may be considered to be obtained by summing over the spacetime volume.

The fact that the “cosmological constant” in the action does not have any
physical meaning is an important advantage of UG. If we consider the history
of the universe, our universe underwent phase transitions before it settles
down to the spontaneous broken phase of various symmetries including the
electroweak symmetry, and this must have produced huge vacuum energy. In
addition, the quantum fluctuation of the gravity and matter also produces huge
vacuum energy. Both must be canceled out by fine tuning in GR, but it is not
necessary in UG.

The price we have to pay is that the action (1.11) or (1.12) is no longer
invariant under the full diffeomorphism, but only under the transverse (or,
volume-preserving) diffeomorphism (TDiff):

$\displaystyle\delta
g_{\mu\nu}=\nabla_{\mu}\epsilon_{\nu}^{T}+\nabla_{\nu}\epsilon_{\mu}^{T},\qquad\nabla^{\mu}\epsilon_{\mu}^{T}=0.$
(1.14)

This gives a source of confusion as to the question if this theory is
equivalent to GR or not at the quantum level. At first sight, since this
theory could be regarded as just a partially gauge fixed theory of GR, one
would expect that it is equivalent to GR, but there has been lots of debate on
the equivalence [7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22].
In fact upon further consideration, this may not be so trivial. Since the
theory has invariance only under TDiff, we expect that this invariance removes
degrees of freedom one less than the full diffeomorphism. This reduction of
degrees of freedom could be done by the Faddeev-Popov (FP) ghosts. In GR, we
have the full diffeomorphism with 4 coordinate parameters, and this introduces
4 sets of FP ghosts and antighosts, leaving $10-8=2$ degrees of freedom,
properly. In UG, however, the gauge fixing can be done only for the transverse
modes, with 3 sets of ghosts. The additional condition is provided by the
unimodularity condition (1.3), but this condition does not seem to require
Faddeev-Popov ghosts. This leaves the question how the remaining degrees of
freedom are removed.

On the other hand, we can use the reparametrization invariance in GR to
partially gauge fix to satisfy the unimodular gauge condition (1.3). It is
then an interesting problem how the above problem on degrees of freedom is
resolved in this gauge, since the remaining reparametrization invariance is
only TDiff. Once understood properly, this is expected to cast light on how to
quantize UG.

In this respect, Baulieu [21] has recently proposed an interesting way of
gauge fixing which seems to realize this unimodular gauge in a manifestly
covariant manner. He introduces a new set of “BRST quartet” fields consisting
of an additional scalar field together with ghosts and a scalar. The reasoning
is, however, not very clear and it is difficult to understand the nature and
the role of the added scalar field as well as the origin of the additional
BRST symmetry.

Here we propose a novel and general way of decomposing a $d$-vector condition
into a scalar plus $d-1$ (i.e., transverse-vector) conditions in a manifestly
covariant and local manner, which naturally leads to the introduction of a
scalar field and the emergence of a new gauge symmetry. This reproduces
essentially the same unimodular gauge-fixed Lagrangian as Baulieu’s one.
Naively one might expect that the newly introduced four fields would be a
“BRST quartet” and decouple by themselves. We find that this is not the case,
and the BRST multiplets rearrange themselves in a very subtle way. We show
that after all they completely decouple and this covariant theory gives the
correct number of degrees of freedom in GR.

Our key observation is that when we gauge fix TDiff, we have to deliberately
impose the gauge condition such that we impose it only on the transverse
modes. In doing this, we introduce an additional longitudinal mode and then
there appears additional invariance, which removes the remaining degrees of
freedom. Since this is the quantization of GR, the cosmological constant is
present in the field equation even though we fix the theory to unimodular
gauge with $\sqrt{-g}=\omega$. We expect that our new way of describing GR in
the unimodular gauge is useful for understanding the formulation of quantum
theory of UG. Our discussion is valid in the presence of matter fields.

We then discuss what implications this formulation of GR may have to the
problem of quantization of UG. The unimodularity condition (1.3) is imposed as
a constraint in UG whereas it is just a gauge choice in GR. Similarly the
spacetime volume is fixed in UG whereas the cosmological constant is fixed in
GR. Considering the dual role of these [23], it would be natural to consider
that the quantum theory of UG may be obtained by the Fourier transform of GR
in the unimodular gauge with respect to the cosmological constant.
Unfortunately we find that such a formulation has problem with unitarity, but
we hope that our discussions clarify the problems in the covariant
quantization of UG.

##  2 BRST quantization of GR in the unimodular gauge

###  2.1 Covariant unimodular gauge fixing in GR

Now let us consider the unimodular gauge fixing in GR. Using the
reparametrization invariance, we can partially gauge fix the theory to the
unimodular gauge with the condition (1.3).

In order to quantize the theory, we have to gauge fix the theory. In the usual
GR, we introduce the usual gauge fixing condition called de Donder gauge:

$\displaystyle
f^{\mu}\equiv\partial_{\rho}\tilde{g}^{\rho\mu}=0.\qquad(\tilde{g}^{\mu\nu}\equiv\sqrt{-g}g^{\mu\nu}).$
(2.1)

This contains $d=4$ conditions and completely fixes all the diffeomorphism
transformations. In our theory where we will take the unimodular gauge
condition (1.3) as a partial gauge fixing, however, Eq. (2.1) imposes too many
conditions, and we have to remove one condition from this. If we gave up
manifest covariance, we could take only $d-1=3$ conditions for the spacial
components $f^{i}=0$ $(i=1,2,3)$ of the de Donder gauge (2.1). However, we
want to keep manifest covariance here.

Now comes our first crucial step. Instead of imposing the $d$-vector condition
(2.1), we introduce an additional field $S$ and impose the condition

$\displaystyle\tilde{g}_{\mu\nu}f^{\nu}=\partial_{\mu}S.$ (2.2)

The meaning of this gauge fixing is as follows: The right-hand side (rhs) of
this equation is a derivative of the scalar field $S$, and so the condition
(2.2) requires the transverse component of the left-hand side (lhs) should
vanish. The longitudinal component simply defines the scalar field $S$, which
is left arbitrary. It is important that we consider full diffeomorphism
transformation here, but this condition (2.2) imposes the gauge condition only
on the transverse modes in $\tilde{g}_{\mu\nu}f^{\nu}$.

Our second crucial observation is the following. Our unimodularity condition
means that the remaining longitudinal mode should be removed from the spectrum
of the theory. We then notice that the new field $S$ does not appear at all in
the original action of our GR theory, and hence the action is trivially
invariant under an arbitrary shift of the scalar field $S$:

$\displaystyle\delta S(x)=d(x).$ (2.3)

Note that this is a (hidden) gauge-invariance. (Such a gauge-invariance under
the arbitrary change of the fields not appearing in the action, is trivial but
was recognized useful by Izawa [24] in the context of BRST gauge-fixing.) We
can lift this new gauge invariance as well as the original GC transformation
invariance to those under the BRST transformation defined by

$\displaystyle\delta_{\rm B}g_{\mu\nu}$ $\displaystyle=$ $\displaystyle
g_{\mu\rho}\partial_{\nu}c^{\rho}+g_{\nu\rho}\partial_{\mu}c^{\rho}+c^{\rho}\partial_{\rho}g_{\mu\nu},$
$\displaystyle\delta_{\rm B}c^{\mu}$ $\displaystyle=$ $\displaystyle
c^{\rho}\partial_{\rho}c^{\mu},$ $\displaystyle\delta_{\rm B}\bar{c}_{\mu}$
$\displaystyle=$ $\displaystyle ib_{\mu},\qquad\delta_{\rm B}b_{\mu}=0,$ (2.4)
$\displaystyle\delta_{\rm B}S$ $\displaystyle=$ $\displaystyle
d,\qquad\delta_{\rm B}d=0,$ $\displaystyle\delta_{\rm B}\bar{d}$
$\displaystyle=$ $\displaystyle ib,\qquad\delta_{\rm B}b=0,$

where $c^{\mu}$ and $\bar{c}_{\mu}$ are the anticommuting ghosts corresponding
to the diffeomorphism transformations, $b_{\mu}$ and $b$ are the Nakanishi-
Lautrup (NL) fields [25], and $d$ and $\bar{d}$ are anticommuting ghosts
corresponding to the invariance (2.3).

Given the invariance and BRST transformation, we have a systematic method to
give the gauge fixing and ghost terms [26, 27]. Thus our gauge-fixing and
ghost terms are given by

$\displaystyle{\cal L}_{{\rm GF}+{\rm FP}}$ $\displaystyle=$
$\displaystyle-i\delta_{\rm
B}\Bigl{[}\bar{c}_{\mu}\Bigl{(}\partial_{\nu}\tilde{g}^{\mu\nu}-\tilde{g}^{\mu\nu}\partial_{\nu}S+\frac{\alpha}{2}\eta^{\mu\nu}b_{\nu}\Bigr{)}+\bar{d}\left(\sqrt{-g}-\omega+\frac{\beta}{2}b\right)\Bigr{]}$
(2.5) $\displaystyle=$ $\displaystyle
b_{\mu}\Bigl{(}\partial_{\nu}\tilde{g}^{\mu\nu}-\tilde{g}^{\mu\nu}\partial_{\nu}S\Bigr{)}+\frac{\alpha}{2}\eta^{\mu\nu}b_{\mu}b_{\nu}+b\left(\sqrt{-g}-\omega\right)+\frac{\beta}{2}b^{2}$
$\displaystyle+i\bar{c}_{\mu}\Bigl{(}\partial_{\nu}\delta_{\rm
B}(\tilde{g}^{\mu\nu})-\delta_{\rm
B}(\tilde{g}^{\mu\nu})\partial_{\nu}S-\tilde{g}^{\mu\nu}\partial_{\nu}d\Bigr{)}+i\bar{d}\left(\sqrt{-g}\nabla_{\mu}c^{\mu}\right),$

where BRST transform of $\tilde{g}^{\mu\nu}=\sqrt{-g}g^{\mu\nu}$ is

$\delta_{\rm
B}(\tilde{g}^{\mu\nu})=-\partial_{\rho}c^{\mu}\cdot\tilde{g}^{\rho\nu}-\partial_{\rho}c^{\nu}\cdot\tilde{g}^{\mu\rho}+\partial_{\rho}(c^{\rho}\tilde{g}^{\mu\nu}),$
(2.6)

and $\alpha$ and $\beta$ are gauge parameters, and $\omega$ is an arbitrary
gauge function fixing the ‘unimodularity’. We see that the first term in (2.5)
only fixes the transverse coordinate transformations, and $d$ and $\bar{d}$
restricts the reparametrization ghost $c^{\mu},\bar{c}_{\mu}$ to transverse
components. The additional set of fields $(S,d,\bar{d},b)$ apparently looks to
form a BRST quartet but actually they do not. We are now going to show that
they rearrange with other ghosts into sets of BRST quartets and all of them
completely decouple.

###  2.2 Propagators

Let us check in more detail if we get nonsingular fully gauge fixed action
with the correct degrees of freedom on the flat background with $\omega=1$.
For the GR theory with $\Lambda=0$, setting the gauge parameters
$\alpha=\beta=0$ for simplicity, our total action is given as

$\displaystyle{\cal L}_{\rm GR}$ $\displaystyle=$
$\displaystyle\sqrt{-g}R+b_{\mu}\Bigl{(}\partial_{\nu}\tilde{g}^{\mu\nu}-\tilde{g}^{\mu\nu}\partial_{\nu}S\Bigr{)}+b\left(\sqrt{-g}-1\right)$
(2.7) $\displaystyle+i\bar{c}_{\mu}\Bigl{(}\partial_{\nu}\delta_{\rm
B}(\tilde{g}^{\mu\nu})-\delta_{\rm
B}(\tilde{g}^{\mu\nu})\partial_{\nu}S-\tilde{g}^{\mu\nu}\partial_{\nu}d\Bigr{)}+i\bar{d}\left(\sqrt{-g}\nabla_{\mu}c^{\mu}\right).$

The fluctuation is defined by

$\displaystyle\tilde{g}^{\mu\nu}=\eta^{\mu\nu}+\tilde{h}^{\mu\nu},$ (2.8)

with $\eta_{\mu\nu}=\mbox{diag}(-1,+1,+1,+1)$ and then to the linear order we
have

$\displaystyle
g_{\mu\nu}=\eta_{\mu\nu}-\tilde{h}_{\mu\nu}+\frac{1}{2}\eta_{\mu\nu}\tilde{h}+\cdots,\qquad\sqrt{-g}=\sqrt{-\tilde{g}}=1+\frac{1}{2}\tilde{h}+\cdots,$
(2.9)

where $\tilde{h}\equiv\tilde{h}_{\mu}^{\mu}$ is the trace of the fluctuation.

The quadratic terms of ${\cal L}_{\rm GR}$ are given by

$\displaystyle{\cal L}_{\rm GR}\Bigr{|}_{\rm quadr}$ $\displaystyle=$
$\displaystyle{\cal L}_{\rm boson}+{\cal L}_{\rm ghost},$ $\displaystyle{\cal
L}_{\rm boson}$ $\displaystyle=$
$\displaystyle\frac{1}{4}\tilde{h}_{\mu\nu}\square\tilde{h}^{\mu\nu}+\frac{1}{2}(\partial_{\nu}\tilde{h}^{\mu\nu})^{2}-\frac{1}{8}\tilde{h}\square\tilde{h}+b_{\mu}(\partial_{\nu}\tilde{h}^{\mu\nu}-\partial^{\mu}S)+\frac{1}{2}b\tilde{h},$
$\displaystyle{\cal L}_{\rm ghost}$ $\displaystyle=$ $\displaystyle
i\bar{c}_{\mu}[\partial_{\nu}(-\partial^{\nu}c^{\mu}-\partial^{\mu}c^{\nu}+\partial_{\rho}c^{\rho}\eta^{\mu\nu})-\partial^{\mu}d]+i\bar{d}\partial_{\mu}c^{\mu}$
(2.10) $\displaystyle=$ $\displaystyle i\bar{c}_{\mu}[-\square
c^{\mu}-\partial^{\mu}d]+i\bar{d}\partial_{\mu}c^{\mu},$

which are written in the matrix form

$\displaystyle\frac{1}{2}\bigl{(}\tilde{h}_{\mu\nu},S,b_{\mu},b\bigr{)}\Gamma^{(2)}_{\rm
boson}\begin{pmatrix}\tilde{h}_{\rho\sigma}\\\ S\\\ b_{\rho}\\\
b\end{pmatrix}+(\bar{c}_{\mu},\bar{d})\Gamma^{(2)}_{\rm
ghost}\left(\begin{array}[]{c}c^{\nu}\\\ d\end{array}\right),$ (2.13)

in terms of the 2-point vertex matrices $\Gamma^{(2)}_{\rm boson}$ and
$\Gamma^{(2)}_{\rm ghost}$, the inverses of which give the propagators. If the
gauge is fixed properly by the present gauge, then the inverses exist and the
propagators are determined unambiguously.

Let us begin with the simpler ghost sector. The 2-point vertex
$\Gamma^{(2)}_{\rm ghost}$ in momentum space reads

$\displaystyle\Gamma^{(2)}_{\rm
ghost}=-i\times\bordermatrix{&c^{\nu}&d\cr\bar{c}_{\mu}&-p^{2}\delta^{\mu}_{\nu}&ip^{\mu}\cr\bar{d}&-ip_{\nu}&0\cr},$
(2.14)

the inverse of which surely exists and is given by

${\Gamma^{(2)}}^{-1}_{\rm
ghost}=i\frac{1}{-p^{2}}\times\bordermatrix{&\bar{c}_{\nu}&\bar{d}\cr
c^{\mu}&\delta^{\mu}_{\nu}-p^{\mu}p_{\nu}/p^{2}&-ip^{\mu}\cr
d&ip_{\nu}&-p^{2}\cr}.$ (2.15)

That is, we find the ghost propagators $i\times{\Gamma^{(2)}}^{-1}_{\rm
ghost}$:

$\displaystyle{\rm F.T.}\langle 0|\,{\rm
T}\,c^{\mu}\,\bar{c}_{\nu}\,|0\rangle$ $\displaystyle=$
$\displaystyle\frac{\delta^{\mu}_{\nu}-\frac{p^{\mu}p_{\nu}}{p^{2}}}{p^{2}},$
(2.16) $\displaystyle{\rm F.T.}\langle 0|\,{\rm
T}\,c^{\mu}\,\bar{d}\,|0\rangle$ $\displaystyle=$
$\displaystyle\frac{-ip^{\mu}}{p^{2}},\qquad{\rm F.T.}\langle 0|\,{\rm
T}\,d\,\bar{c}_{\nu}\,|0\rangle=\frac{ip_{\nu}}{p^{2}},$ (2.17)
$\displaystyle{\rm F.T.}\langle 0|\,{\rm T}\,d\,\bar{d}\,|0\rangle$
$\displaystyle=$ $\displaystyle-1\,,$ (2.18)

where the F.T. means

${\rm F.T.}\langle 0|\,{\rm T}\,\phi_{1}\,\phi_{2}\,|0\rangle\equiv\int
d^{4}xe^{-ip(x-y)}\langle 0|{\rm T}\phi_{1}(x)\,\phi_{2}(y)|0\rangle\,.$
(2.19)

Next is the boson sector: We find the 2-point vertex $\Gamma^{(2)}_{\rm
boson}$ in momentum space

$\displaystyle\Gamma^{(2)}_{\rm boson}=$
$\displaystyle\bordermatrix{&\tilde{h}_{\rho\sigma}&S&b_{\rho}&b\cr\tilde{h}_{\mu\nu}&\begin{matrix}-p^{2}\,\Bigl{[}\frac{1}{2}I^{\mu\nu,\rho\sigma}-\frac{1}{12}d^{\mu\nu}d^{\rho\sigma}\phantom{\frac{3}{4}e^{\mu\nu}e^{\rho\sigma}\frac{3}{4}e^{\mu\nu}e^{\rho\sigma}}\\\
-\frac{1}{4}\left(d^{\mu\nu}e^{\rho\sigma}+e^{\mu\nu}d^{\rho\sigma}\right)-\frac{3}{4}e^{\mu\nu}e^{\rho\sigma}\Bigr{]}\end{matrix}&0&-i\eta^{\rho(\mu}p^{\nu)}&\frac{1}{2}\eta^{\mu\nu}\cr
S&0&0&ip^{\rho}&0\cr b_{\mu}&i\eta^{\mu(\rho}p^{\sigma)}&-ip^{\mu}&0&0\cr
b&\frac{1}{2}\eta^{\rho\sigma}&0&0&0\cr},~{}~{}~{}$ (2.20)

by using the projection operators

$\displaystyle d_{\mu\nu}=\eta_{\mu\nu}-\frac{p_{\mu}p_{\nu}}{p^{2}},\qquad
e_{\mu\nu}=\frac{p_{\mu}p_{\nu}}{p^{2}},$ (2.21) $\displaystyle
I_{\mu\nu,\rho\sigma}=\frac{1}{2}\left(d_{\mu\rho}d_{\nu\sigma}+d_{\mu\sigma}d_{\nu\rho}-\frac{2}{3}d_{\mu\nu}d_{\rho\sigma}\right).$
(2.22)

Noting their projection properties

$\displaystyle p^{\mu}d_{\mu\nu}=0,\quad d_{\mu\nu}\eta^{\mu\nu}=3,\quad
e_{\mu\nu}\eta^{\mu\nu}=1,$ (2.23) $\displaystyle
d_{\mu\alpha}d^{\alpha\nu}=d_{\mu}{}^{\nu},\quad
e_{\mu\alpha}e^{\alpha\nu}=e_{\mu}{}^{\nu},\quad
d_{\mu\alpha}e^{\alpha\nu}=0,$ (2.24) $\displaystyle
I_{\mu\nu,\alpha\beta}d^{\alpha\beta}=0,\quad
I_{\mu\nu,\alpha\beta}e^{\alpha\beta}=0,\quad
I_{\mu\nu,\alpha\beta}I^{\alpha\beta,\rho\sigma}=I_{\mu\nu}{}^{\rho\sigma},$
(2.25)

we can straightforwardly compute the inverse of this matrix,
${\Gamma^{(2)}}^{-1}_{\rm boson}$, with a little effort:

$\displaystyle{\Gamma^{(2)}}^{-1}_{\rm boson}=\frac{1}{-p^{2}}\times$
$\displaystyle\bordermatrix{&\tilde{h}_{\rho\sigma}&S&b_{\rho}&b\cr\tilde{h}_{\mu\nu}&\begin{matrix}\Bigl{[}2I^{\mu\nu,\rho\sigma}-\frac{1}{3}d^{\mu\nu}d^{\rho\sigma}\phantom{\frac{3}{4}e^{\mu\nu}e^{\rho\sigma}\frac{3}{4}e^{\mu\nu}e^{\rho\sigma}}\\\
+\left(d^{\mu\nu}e^{\rho\sigma}+e^{\mu\nu}d^{\rho\sigma}\right)-3e^{\mu\nu}e^{\rho\sigma}\Bigr{]}\end{matrix}&d^{\mu\nu}-3e^{\mu\nu}&2ip^{(\mu}d^{\nu)\rho}&-p^{2}(d^{\mu\nu}-e^{\mu\nu})\cr
S&d^{\rho\sigma}-3e^{\rho\sigma}&-3&-ip^{\rho}&p^{2}\cr
b_{\mu}&-2ip^{(\rho}d^{\sigma)\mu}&ip^{\mu}&0&0\cr
b&-p^{2}(d^{\rho\sigma}-e^{\rho\sigma})&p^{2}&0&0\cr}.$ (2.26)

Thus, we explicitly confirmed that the inverse also exists in the boson sector
and hence that our gauge fixing is complete. The propagators given by
$i\times{\Gamma^{(2)}}^{-1}_{\rm boson}$ read e.g.,

$\displaystyle{\rm F.T.}\langle 0|\,{\rm
T}\,\tilde{h}^{\mu\nu}\,\tilde{h}^{\rho\sigma}\,|0\rangle$ $\displaystyle=$
$\displaystyle
i\,\frac{2I^{\mu\nu,\rho\sigma}-\frac{1}{3}d^{\mu\nu}d^{\rho\sigma}+\left(d^{\mu\nu}e^{\rho\sigma}+e^{\mu\nu}d^{\rho\sigma}\right)-3e^{\mu\nu}e^{\rho\sigma}}{-p^{2}},$
$\displaystyle{\rm F.T.}\langle 0|\,{\rm T}\,\tilde{h}^{\mu\nu}\,S\,|0\rangle$
$\displaystyle=$ $\displaystyle i\,\frac{d^{\mu\nu}-3e^{\mu\nu}}{-p^{2}},$
(2.27)

etc. Note that the graviton propagator has not only a dipole but also a
tripole part.

###  2.3 Mode counting

For completeness, let us also confirm in this gauge that the physical modes
are only two transverse modes with helicity $j=\pm 2$ and all the other modes
properly fall into the unphysical BRST quartets.

To do so, consider the free field equations of motion following from the
quadratic terms of $S_{\rm GR}=\int d^{4}x{\cal L}_{\rm GR}$, which are also
the equations of motion of the asymptotic fields:

$\displaystyle\frac{\delta S_{\rm GR}}{\delta\tilde{h}^{\mu\nu}}:$
$\displaystyle\frac{1}{2}\square\tilde{h}_{\mu\nu}-\partial_{\rho}\partial_{(\nu}\tilde{h}_{\mu)}{}^{\rho}-\frac{1}{4}\eta_{\mu\nu}\square\tilde{h}-\partial_{(\mu}b_{\nu)}+\frac{1}{2}\eta_{\mu\nu}b=0,$
(2.28) $\displaystyle\frac{\delta S_{\rm GR}}{\delta S}:$
$\displaystyle\partial^{\mu}b_{\mu}=0,$ (2.29) $\displaystyle\frac{\delta
S_{\rm GR}}{\delta b_{\mu}}:$
$\displaystyle\partial_{\nu}\tilde{h}^{\mu\nu}-\partial^{\mu}S=0,$ (2.30)
$\displaystyle\frac{\delta S_{\rm GR}}{\delta b}:$ $\displaystyle\tilde{h}=0,$
(2.31) $\displaystyle\frac{\delta S_{\rm
GR}}{\delta\bar{c}_{\mu}},\frac{\delta S_{\rm GR}}{\delta\bar{d}}:$
$\displaystyle\square
c^{\mu}+\partial^{\mu}d=0,\qquad\partial_{\mu}c^{\mu}=0,\quad\rightarrow\quad\square
d=0,$ (2.32) $\displaystyle\frac{\delta S_{\rm GR}}{\delta
c^{\mu}},\frac{\delta S_{\rm GR}}{\delta d}:$
$\displaystyle\square\bar{c}_{\mu}+\partial_{\mu}\bar{d}=0,\qquad\partial^{\mu}\bar{c}_{\mu}=0,\quad\rightarrow\quad\square\bar{d}=0,$
(2.33)

where the bracket ( ) attached to the indices means the weight 1
symmetrization; e.g.,
$A_{(\mu}B_{\nu)}=(1/2)\big{(}A_{\mu}B_{\nu}+A_{\nu}B_{\mu}\big{)}$. Taking
the $\eta^{\mu\nu}$-trace of (2.28) and using (2.30) and (2.31), we find

$\displaystyle-\square S+2b=0\,.$ (2.34)

The divergence of the gravity equation (2.28), combined with (2.29) – (2.31),
yields

$\displaystyle-\frac{1}{2}(\square\partial_{\mu}S+\square
b_{\mu}-\partial_{\mu}b)=0\quad\rightarrow\quad\square
b_{\mu}+\partial_{\mu}b=0\quad\rightarrow\quad\square b=0\,,$ (2.35)

where in the first step, we have used (2.34), and in the next step, we took
the divergence and used (2.29). These equations (2.34) and (2.35) imply that
$S$ and $b_{\mu}$ fields satisfy dipole equations $\square^{2}S=0$ and
$\square^{2}b_{\mu}=0$, and their dipole parts are supplied by the simple pole
$b$ field. The gravity field equation (2.28) is rewritten by using Eqs. (2.30)
and (2.31) into

$\frac{1}{2}\square\tilde{h}_{\mu\nu}-\partial_{\mu}\partial_{\nu}S-\partial_{(\mu}b_{\nu)}+\frac{1}{2}\eta_{\mu\nu}b=0.$
(2.36)

Applying $\square$ to this and using Eqs. (2.34) and (2.35), we find

$\frac{1}{2}\square^{2}\tilde{h}_{\mu\nu}-\partial_{\mu}\partial_{\nu}b=0.$
(2.37)

So, we see that $\tilde{h}_{\mu\nu}$ field is now a tripole field, and the
tripole part is supplied by the simple pole $b$ field and the dipole parts are
supplied by simple pole parts of $S$ and $b_{\mu}$.

The conventional method to count the number of independent modes for such a
system containing multipole fields, is to redefine such multipole fields into
simple pole fields by subtracting the mixed simple pole fields. Since it is
quite cumbersome to do so, however, we here adopt the 4-dimensional Fourier
expansion [25] defined as

$\phi(x)=\frac{1}{\sqrt{(2\pi)^{3}}}\int
d^{4}p\,\theta(p^{0})\left[\phi(p)e^{ipx}+\phi^{\dagger}(p)e^{-ipx}\right].$
(2.38)

Namely, we treat the multipole fields as they stand; the 4-dimensional
operators $\phi(p)$ and $\phi(p)^{\dagger}$ annihilate and create the
multipole particles as they stand. The BRST singlet physical modes must of
course be simple pole fields. Multipole fields are necessarily unphysical and
so will fall into the BRST quartet. We will see that the members of a BRST
quartet have a common multipole structure to decouple among them. We note
that, when $\phi(x)$ is a simple pole field, $\phi(p)$ is given in terms of
the usual annihilation operator $\phi({\boldsymbol{p}})$ by 3-dimensional
Fourier transform as

$\phi(p)=\theta(p^{0})\delta(p^{2})\sqrt{2|{\boldsymbol{p}}|}\phi({\boldsymbol{p}}).$
(2.39)

Now the counting the number of modes by 4-dimensional Fourier modes $\phi(p)$
and $\phi^{\dagger}(p)$ is very easy and the same as the number of field’s
components. We have $10(\tilde{h}_{\mu\nu})+1(S)+4(b_{\mu})+1(b)=16$ component
fields in the boson sector, and $4(c^{\mu})+1(d)=5$ and
$4(\bar{c}_{\mu})+1(\bar{d})=5$ components in the ghost and antighost sectors,
respectively. But they are subject to the constraint equations of motion which
reduce the number of independent modes.

First $\tilde{h}_{\mu\nu}$ field is subject to 4 constraints by Eq. (2.30) and
1 traceless constraint (2.31). Thus, $\tilde{h}_{\mu\nu}$ contains $10-4-1=5$
independent modes. The $b_{\mu}$ field obeys the transverse equation (2.29),
so leaving $4-1=3$ independent modes. $S$ and $b$ each gives an independent
mode. Similarly, in the ghost sector, $c^{\mu}$ and $\bar{c}_{\mu}$ fields
each have 3 independent modes by Eqs. (2.32) and (2.33) and $d$ and $\bar{d}$
each gives an independent mode.

Now recall the BRST transformation at the linearized level (which is also the
BRST transformation of the asymptotic fields under the perturbative
assumption).

$\displaystyle[iQ_{\rm B},\tilde{h}_{\mu\nu}(p)]$ $\displaystyle=$
$\displaystyle-
ip_{\mu}c_{\nu}(p)-ip_{\nu}c_{\mu}(p)\quad(\because\partial_{\rho}c^{\rho}=0),$
(2.40) $\displaystyle{}\\{iQ_{\rm B},\bar{c}_{\mu}(p)\\}$ $\displaystyle=$
$\displaystyle ib_{\mu}(p),$ (2.41) $\displaystyle{}[iQ_{\rm B},S(p)]$
$\displaystyle=$ $\displaystyle d(p),\qquad\\{iQ_{\rm
B},\bar{d}(p)\\}=ib(p)\,.$ (2.42)

As for the 5 independent modes among 10 gravity fields $\tilde{h}_{\mu\nu}$,
we first define the following 6 linear combinations of $\tilde{h}_{\mu\nu}$,
choosing the $x^{3}$ axis along ${\boldsymbol{p}}$:

$\displaystyle\tilde{h}_{{\rm T}1}(p)$ $\displaystyle=$
$\displaystyle\frac{1}{2}(\tilde{h}_{11}(p)-\tilde{h}_{22}(p)),$
$\displaystyle\tilde{h}_{{\rm T}2}(p)$ $\displaystyle=$
$\displaystyle\tilde{h}_{12}(p),$ $\displaystyle\chi_{0}(p)$ $\displaystyle=$
$\displaystyle\frac{1}{2p_{0}}\left(\tilde{h}_{00}(p)-\frac{1}{2}(\tilde{h}_{11}(p)+\tilde{h}_{22}(p))\right),$
$\displaystyle\chi_{i}(p)$ $\displaystyle=$
$\displaystyle\frac{1}{p_{0}}\tilde{h}_{i0}(p)\ \ \hbox{for}\ \ i=1,2,$
$\displaystyle\chi_{3}(p)$ $\displaystyle=$
$\displaystyle\frac{1}{2p_{3}}\left(\tilde{h}_{33}(p)+\frac{1}{2}(\tilde{h}_{11}(p)+\tilde{h}_{22}(p))\right).$
(2.43)

These fields have very simple properties under the BRST transformation (2.40):
The transverse states are BRST singlets,

$[iQ_{\rm B},\tilde{h}_{{\rm T}i}(p)]=0,\qquad(i=1,2),$ (2.44)

and $\chi_{\mu}(p)$’s satisfy

$\displaystyle[iQ_{\rm B},\chi_{\mu}(p)]=-ic_{\mu}(p),\qquad(\mu=0,1,2,3).$
(2.45)

They also satisfy the transversality condition $p^{\mu}\chi_{\mu}(p)=0$ so
that they have only 3 components. Together with $\tilde{h}_{{\rm T}i}$, they
give the 5 independent components from $\tilde{h}_{\mu\nu}$.

Other than these 5 modes $\big{(}\tilde{h}_{{\rm T}i}(p),\chi_{\mu}(p)\big{)}$
from $\tilde{h}_{\mu\nu}$, we only have 5 bosons $S(p),b_{\mu}(p),b(p)$ and 4
ghosts $(c^{\mu},d)$ plus 4 antighosts $(\bar{c}_{\mu},\bar{d})$. We now see
that aside from two physical transverse modes $\tilde{h}_{{\rm T}i}(p)$, all
the other modes fall into eight BRST doublets:

$\displaystyle\begin{array}[]{rcccl}\hbox{vector BRST
doublets}&:&\bigl{(}\chi^{\mu},c^{\mu}\bigr{)}&;&\bigl{(}\bar{c}_{\mu},b_{\mu}\bigr{)}\,,\\\
\hbox{scalar BRST
doublets}&:&\bigl{(}S,d\bigr{)}&;&\bigl{(}\bar{d},b\bigr{)}\,.\end{array}$
(2.48)

Recall that these vectors $\chi^{\mu},c^{\mu},\bar{c}_{\mu},b_{\mu}$ are all
transversal and we count each of them as $d-1=3$ modes, so we have
$(d-1)\times 2+1\times 2=8$ BRST doublets in total. They indeed have the
following BRST transformation laws:

$\left[iQ_{\rm B},\begin{pmatrix}\chi^{\mu}\\\
S\end{pmatrix}\right]=\begin{pmatrix}-ic^{\mu}\\\
d\end{pmatrix},\qquad\left[iQ_{\rm B},\begin{pmatrix}\bar{c}_{\mu}\\\
\bar{d}\end{pmatrix}\right]=i\begin{pmatrix}b_{\mu}\\\ b\end{pmatrix}.$ (2.49)

These eight BRS doublets fall into four BRST quartets which decouple from the
physical subspace, and we are left with the two physical transverse modes with
helicity $j=\pm 2$.

It is, however, worth noting that the metric structure of these eight BRST
doublets is slightly unfamiliar one. In particular, it is not correct to say
that the following pair of the BRST doublets gives a BRST quartet:

$\bigl{[}(S(p),d(p));(\bar{d}(p),b(p))\bigr{]}.$ (2.50)

This is because $\langle 0|d(p)\bar{d}^{\dagger}(q)|0\rangle$ ${}=-i\langle
0|S(p)b^{\dagger}(q)|0\rangle=0$, as is seen shortly in Eq. (2.55), which
implies that the BRST doublet $(S(p),d(p))$ does not have nonvanishing inner-
product with the other BRST doublet $(\bar{d}(p),b(p))$ and so they do not
constitute a BRST quartet in the proper sense of the terminology.

The 4-dimensional commutation relations for these asymptotic fields can be
derived from the expressions for the propagator $i\times{\Gamma^{(2)}}^{-1}$
in Eqs. (2.26) and (2.15) by the standard procedure and are explicitly given
in Eq. (A.5) in the Appendix. From Eq. (A.5), we can see the following
commutation relations for the present independent modes:

$\displaystyle[\tilde{h}_{{\rm T}i}(p),\tilde{h}^{\dagger}_{{\rm T}j}(q)]$
$\displaystyle=$
$\displaystyle\delta_{ij}\theta(p^{0})\delta(p^{2})\delta^{4}(p-q),$ (2.51)
$\displaystyle{}[\tilde{h}_{{\rm T}i}(p),\phi^{\dagger}(q)]$ $\displaystyle=$
$\displaystyle 0,\quad\hbox{for}\ \ \phi=\chi_{\mu},S,b_{\mu},b,$ (2.52)
$\displaystyle{}[\chi_{\mu}(p),b^{\dagger}_{\nu}(q)]$ $\displaystyle=$
$\displaystyle\\{c_{\mu}(p),\bar{c}^{\dagger}_{\nu}(q)\\}$ (2.53)
$\displaystyle=$ $\displaystyle
i\eta_{\mu\nu}\theta(p^{0})\delta(p^{2})\delta^{4}(p-q)-ip_{\mu}p_{\nu}\theta(p^{0})(-\delta^{\prime}(p^{2}))\delta^{4}(p-q),$
$\displaystyle{}[\chi_{\mu}(p),b^{\dagger}(q)]$ $\displaystyle=$
$\displaystyle\\{c_{\mu}(p),\bar{d}^{\dagger}(q)\\}=p_{\mu}\theta(p^{0})\delta(p^{2})\delta^{4}(p-q),$
(2.54) $\displaystyle{}[S(p),b^{\dagger}(q)]$ $\displaystyle=$ $\displaystyle
i\\{d(p),\bar{d}^{\dagger}(q)\\}=0,$ (2.55)
$\displaystyle{}[S(p),b_{\mu}^{\dagger}(q)]$ $\displaystyle=$ $\displaystyle
i\\{d(p),\bar{c}_{\mu}^{\dagger}(q)\\}=-ip_{\mu}\theta(p^{0})\delta(p^{2})\delta^{4}(p-q).$
(2.56)

The first Eq. (2.51) means that the transverse modes $\tilde{h}_{{\rm T}i}$
really are simple pole fields and have positive norms, and the second (2.52)
show that they are orthogonal to all the other modes. Equation (2.55) shows
the vanishing inner-product $\langle 0|d(p)\bar{d}^{\dagger}(p)|0\rangle=0$
mentioned above. Equations (2.54) and (2.56) show the cross inner-product
relations between the scalar and vector BRST doublets; Eq. (2.54) shows that
the partner BRST doublet of the scalar BRST doublet $(\bar{d},b)$ is the
longitudinal component $(\chi_{\rm L},c_{\rm L})$ of the vector BRST doublet
$(\chi_{\mu},c_{\mu})$ (Here, the longitudinal means the mode whose
polarization vector is proportional to $p^{\mu}$):

$\langle 0|c_{\rm L}(p)\bar{d}^{\dagger}(q)|0\rangle=\langle 0|\chi_{\rm
L}(p)\,Q_{\rm B}\,\bar{d}^{\dagger}(q)|0\rangle=\langle 0|\chi_{\rm
L}(p)b^{\dagger}(q)|0\rangle\neq 0.$ (2.57)

In the same way, Eq. (2.56) means that the BRST doublet $(S(p),d(p))$ has
nonvanishing innerproduct with the $(\bar{c}_{\rm L}(p),b_{\rm L}(p))$. Thus
we can identify the following four BRST quartets (pairs of BRST doublets) in
the present case, in the proper sense of terminology of BRST quartet:

$\displaystyle\hbox{transverse quartets}\ (i=1,2)$ $\displaystyle:$
$\displaystyle(\chi_{i}(p),c_{i}(p))\leftrightarrow(\bar{c}_{i}(p),b_{i}(p)),$
longitudinal $\chi$ quartet $\displaystyle:$ $\displaystyle(\chi_{\rm
L}(p),c_{\rm L}(p))\leftrightarrow(\bar{d}(p),b(p)),$ longitudinal $\bar{c}$
quartet $\displaystyle:$ $\displaystyle(S(p),d(p))\leftrightarrow(\bar{c}_{\rm
L}(p),b_{\rm L}(p)),$ (2.58)

with $\leftrightarrow$ indicating the mutual partner doublets with which
nonvanishing inner-products exist like Eq. (2.57). These four sets of fields
are the real BRST quartets and completely decouple from the physical sector,
leaving only 2 transverse modes in GR.

###  2.4 Cosmological constant in GR in the unimodular gauge

Before closing this section, we should comment on the crucial difference on
the cosmological constant between GR in the unimodular gauge and UG.

Our total action $(\ref{GRaction})+(\ref{GR:gf})$ in the present unimodular
gauge GR yields the field equation for the metric

$\displaystyle
R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}(R-2\Lambda+b)-(\partial_{(\mu}b_{\nu)}+b_{(\mu}\partial_{\nu)}S)+\frac{1}{2}g_{\mu\nu}g^{\rho\sigma}(\partial_{\rho}b_{\sigma}+b_{\rho}\partial_{\sigma}S)$
$\displaystyle\hskip 20.00003pt{}+\mbox{(terms involving the FP ghost fields
$c^{\mu},\bar{c}_{\mu},d,\bar{d}$)}=0.$ (2.59)

Let us consider the classical vacuum solution of this gravity equation as the
background field configuration of the quantum theory. If we look for the
solution with vanishing vector NL field $b_{\mu}=0$, then this Eq. (2.60)
reduces to the form

$\displaystyle R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}(R-2\Lambda+b)=0\,,$ (2.60)

which is identical to the field equation (1.5) in UG if we identify $b$ here
with the multiplier field $\lambda$ there. One might then be tempted to
conclude that the cosmological constant in the action has no physical meaning
in GR in the unimodular gauge, just like UG.

However, in GR, there is a crucial physical state condition [28]

$Q_{\rm B}|{\rm phys}\rangle=0,$ (2.61)

and the NL field $b$ is a BRST exact field, BRST transform of the antighost
field $\bar{d}$; $\\{Q_{\rm B},\bar{d}\\}=b$. Since the vacuum must be a
physical state, the NL field $b$ must always have vanishing vacuum expectation
value:

$\langle 0|b(x)|0\rangle=\langle 0|\\{Q_{\rm B},\bar{d}(x)\\}|0\rangle=0.$
(2.62)

This is also true for $b_{\mu}$. At the tree level, the vacuum expectation
value (VEV) of the field operator product equals the product of the VEV of
each field. So all the terms containing FP ghost fields vanish. Therefore, the
above field equation (2.59) correctly reproduces the classical Einstein
equation with the cosmological constant (1.2). This is to be expected since
physical contents should be independent of the gauge fixing, and in GR in
covariant gauge like de Donder gauge, the cosmological constant is really a
physical quantity which determines the scalar curvature of the vacuum.

Then, what is the difference between UG and GR in the unimodular gauge? It is
the field properties of the multiplier field $\lambda$ in UG and the NL field
$b$ in GR that make the difference. For the multiplier field $\lambda$ in UG,
unlike $b$, there is no constraint that its VEV should vanish or any other
matrix element. So, we have the gravity field equation (1.5) and $\lambda$ is
a field that should be eliminated by using the field equation. This requires
the manipulation described in the introduction, and we can get only the
traceless part of the field equation (1.7). We see that $\lambda$ gets VEV to
cancel the cosmological constant.

##  3 Implications for Unimodular Gravity

We would like to discuss if and how the above formulation of GR in the
unimodular gauge may cast any light on the problem of covariant quantization
of UG with the action (1.4). One notable fact is that, as we have discussed in
the preceding section, the condition (1.3) is imposed as a gauge fixing in GR
by one of the member $b$ of the BRST quartet but here it is imposed as a
constraint by the Lagrange multiplier field $\lambda$ independent of the BRST
transformation. This restricts the invariance of the theory to TDiff.

Since UG has fixed spacetime volume, it appears natural to consider that the
usual GR is recovered by summing over the spacetime volume in UG. This reminds
us of the suggestion made long time ago by Hawking that the partition function
for the theory with $\Lambda$ are related by “Laplace transform” with that
with spacetime volume $V$ and vice versa [23]. In the Minkowski space, the
transformation should be Fourier transformation. So let us consider what
happens if we make Fourier transformation of the partition function of GR in
the unimodular gauge.

The partition function for GR in the unimodular gauge is given by

$\displaystyle Z_{\rm GR}[\Lambda,\omega]$ $\displaystyle=$
$\displaystyle\\!\\!\int[Dg_{\mu\nu}][DS][Db_{\mu}][Dc^{\mu}][D\bar{c}_{\mu}][Dd][D\bar{d}][Db]$
(3.1) $\displaystyle\exp\left[i\frac{1}{16\pi G}\int
d^{4}x\,\left\\{\sqrt{-g}(R-2\Lambda)+{\cal L}_{{\rm GF}+{\rm
FP}}\right\\}\right].$

If we multiply this by

$\displaystyle\exp\left[\frac{i\Lambda}{8\pi G}\int d^{4}x\omega\right],$
(3.2)

and integrate over $\Lambda$ first, we get $\delta\left(\int
d^{4}x\sqrt{-g}-\int d^{4}x\,\omega\right)$. This is the global relation
(1.13), which is a consequence in UG, but this does not immediately lead to
the local constraint $\sqrt{-g}-\omega=0$. This is close to the UG, and it may
be instructive to study the theory with this condition changed into a local
one:

$\displaystyle Z_{\rm
UG}\equiv\int[D\lambda]\\!\\!\int[Dg_{\mu\nu}][DS][Db_{\mu}][Dc^{\mu}][D\bar{c}_{\mu}][Dd][D\bar{d}][Db]$
$\displaystyle\hskip 56.9055pt\exp\left[i\frac{1}{16\pi G}\int
d^{4}x\,\left\\{\sqrt{-g}R+{\cal L}_{{\rm GF}+{\rm
FP}}+\lambda(\sqrt{-g}-\omega)\right\\}\right].$ (3.3)

The fact that we obtain this by summing over cosmological constant would be
consistent with the fact that the cosmological constant is not determined in
UG.

Recall that we have the term of the same form $b(\sqrt{-g}-\omega)$ as this
constraint in the gauge fixing terms (2.5), and then this term may be absorbed
into $\lambda$. The integral over $b$ may be performed and yields a trivial
constant

$\int Db\exp\left[\frac{i}{16\pi G}\int d^{4}x\frac{\beta}{2}b^{2}\right]={\rm
const.}\prod_{x}\frac{1}{\sqrt{\beta}}.$ (3.4)

Absorbing this constant into the definition of the partition function, we find
the transformed partition function

$\displaystyle Z_{\rm
UG}=\\!\\!\int[Dg_{\mu\nu}][DS][Db_{\mu}][Dc^{\mu}][D\bar{c}_{\mu}][D\lambda][Dd][D\bar{d}]$
$\displaystyle\hskip 56.9055pt\exp\left[i\frac{1}{16\pi G}\int
d^{4}x\,\left\\{\sqrt{-g}R+{\cal L}_{{\rm GF}+{\rm
FP}}^{\prime}+\lambda(\sqrt{-g}-\omega)\right\\}\right],$ (3.5)

where

$\displaystyle{\cal L}_{{\rm GF}+{\rm FP}}^{\prime}$ $\displaystyle=$
$\displaystyle
b_{\mu}\Bigl{(}\partial_{\nu}\tilde{g}^{\mu\nu}-\tilde{g}^{\mu\nu}\partial_{\nu}S\Bigr{)}+\frac{\alpha}{2}b_{\mu}^{2}$
(3.6) $\displaystyle+i\bar{c}_{\mu}\Bigl{(}\partial_{\nu}\delta_{\rm
B}(\tilde{g}^{\mu\nu})-\delta_{\rm
B}(\tilde{g}^{\mu\nu})\partial_{\nu}S-\tilde{g}^{\mu\nu}\partial_{\nu}d\Bigr{)}+i\bar{d}\left(\sqrt{-g}\nabla_{\mu}c^{\mu}\right).$

This looks like the same as GR in the unimodular gauge. Since it has the same
structure as GR in the unimodular gauge, one would naively expect that the
analysis in the preceding section goes through and the remaining degrees of
freedom are the same as GR. In fact, the Feynman rules for both theories are
the same, and any scattering amplitudes calculated in both theories agree.

Considering all these circumstances, it appears that this theory may be a
possible candidate for the quantum theory of UG. Then what is the difference
between GR in the unimodular gauge and this theory? The difference lies in the
fundamental difference in the nature of the field $b$ and $\lambda$, as we
have discussed in sect. 2.4. In GR, the field $b$ giving the unimodular
constraint is the NL field, a member of BRST quartet, and the theory has the
BRST invariance. As a consequence, it does not have VEV. In contrast, in the
current theory, $\lambda$ is BRST singlet and can have VEV, and may cancel the
cosmological constant. However, since the action in (3.5) is no longer
invariant under the BRST transformation because $\lambda\neq b$, we do not
have the same physical state condition as in GR. If we defined another new
BRST transformation under which our $\lambda$ field were the BRST transform of
the antighost $\bar{d}$, $\\{Q_{\rm B},\bar{d}\\}=\lambda$, and then we would
have the physical state condition but the VEV of $\lambda$ would have to
vanish, reproducing GR in the unimodular gauge.

As stated above, we have the same Feynman rules and get the same amplitudes in
both theories. For the transverse gravitons, these give amplitudes without any
problem. The problem manifests itself when we calculate the amplitudes for the
longitudinal modes. In the BRST invariant theory, the physical state condition
forbids them to appear by themselves but must come together with the FP
ghosts, and their contributions vanish. Without the physical state condition,
nothing tells us such amplitudes should be absent.

Another point to be noticed is that in the present theory, the invariance
under TDiff

$\displaystyle\delta_{B}g_{\mu\nu}=\nabla_{\mu}c_{\nu}^{\rm
T}+\nabla_{\nu}c_{\mu}^{\rm T},\qquad\nabla^{\mu}c_{\mu}^{\rm T}=0,$ (3.7)

is realized only on shell. We can see that this constraint is imposed by the
field equations for the ghosts $d$ and $\bar{d}$. If one started with the UG
with the action (1.11), we would have the invariance only under (3.7) off
shell, but then we could not see the necessity of introduction of the
additional fields $(S,d,\bar{d})$, which are necessary for the decoupling of
the modes other than those in GR. It is plausible that the formulation of
quantum UG must be realized by fixing only TDiff. This should be related to
the question how to specify the physical states in this theory.

To summarize, as far as the Feynman rules and other properties are concerned,
both theories describe the same theory except for the treatment of the
cosmological constant, but here the criterion for choosing our physical
asymptotic states is lacking in the above theory. We may get the same
scattering amplitudes among the same set of particles in both theories, but it
is not specified in this theory which scattering amplitudes are physical.
These problems of identifying physical asymptotic states and gauge fixing the
invariance under TDiff off shell in the theory are under study. We hope to
report on this in the near future.

## Acknowledgment

We thank Roberto Percacci for valuable discussions. T.K. is supported in part
by the JSPS KAKENHI Grant No. JP18K03659. N.O. is supported in part by the
Grant-in-Aid for Scientific Research Fund of the JSPS (C) No. 16K05331, No.
20K03980, and Taiwan MOST No. 110-2811-M-008-510.

##  Appendix A 4-dimensional commutation relations from propagators

The propagators, or the inverse of the 2-point vertices
$i\times{\Gamma^{(2)}}^{-1}$ of the boson fields and FP ghost fields in the
present system are computed in Eqs. (2.26) and (2.15), respectively. We can
rewrite those expression for ${\rm F.T.}\langle 0|\,{\rm
T}\,\phi_{i}(x)\phi_{j}(0)\,|0\rangle=i\times{\Gamma^{(2)}}_{ij}^{-1}$ in
$x$-space in terms of the invariant functions

$\displaystyle D_{F}(x)$ $\displaystyle=$
$\displaystyle\Delta_{F}(x;m^{2})\bigr{|}_{m^{2}=0}$ $\displaystyle E_{F}(x)$
$\displaystyle=$ $\displaystyle-\frac{\partial}{\partial
m^{2}}\Delta_{F}(x;m^{2})\bigr{|}_{m^{2}=0}$ $\displaystyle F_{F}(x)$
$\displaystyle=$ $\displaystyle\frac{1}{2}\left(\frac{\partial}{\partial
m^{2}}\right)^{2}\Delta_{F}(x;m^{2})\bigr{|}_{m^{2}=0}$ (A.1)
$\displaystyle\Delta_{F}(x;m^{2})$ $\displaystyle=$
$\displaystyle\int\frac{d^{4}p}{i(2\pi)^{4}}\frac{e^{ipx}}{m^{2}+p^{2}-i\epsilon}$
$\displaystyle\Delta(x;m^{2})$ $\displaystyle=$
$\displaystyle\int\frac{d^{4}p}{i(2\pi)^{3}}\epsilon(p^{0})\delta(m^{2}+p^{2})e^{ipx}$
(A.2)

with suffix $F$ meaning Feynman’s causal functions. Then, for any free theory,
the commutation relations $[\phi_{i}(x),\phi_{j}(0)]$ can be obtained simply
by replacing the invariant functions by the same invariant functions without
the suffix $F$ multiplied by $i$; that is, replacing $\Delta_{F}(x;m^{2})$ by
$i\Delta(x;m^{2})$, $D_{F}(x)$ by $iD(x)$ etc. (This rule actually holds even
for interacting Heisenberg fields if we use the spectral function
representation.)

Applying this rule to the $x$-space representation of the propagators obtained
in Eqs. (2.26) and (2.15), we find the following 4-dimensional commutation
relations for our asymptotic fields:

$\displaystyle[\tilde{h}_{\mu\nu}(x),\tilde{h}_{\rho\sigma}(y)]$
$\displaystyle=$
$\displaystyle(\eta_{\mu\rho}\eta_{\nu\sigma}+\eta_{\mu\sigma}\eta_{\nu\rho}-\eta_{\mu\nu}\eta_{\rho\sigma})iD(x-y)$
$\displaystyle{}+\\{{\cal A}_{\mu\nu,\rho\sigma}-2{\cal
B}_{\mu\nu,\rho\sigma}\\}iE(x-y)-4\partial_{\mu}\partial_{\nu}\partial_{\rho}\partial_{\sigma}iF(x-y)$
$\displaystyle{}[\tilde{h}_{\mu\nu}(x),S(y)]$ $\displaystyle=$
$\displaystyle\eta_{\mu\nu}iD(x-y)+4\partial_{\mu}\partial_{\nu}iE(x-y),$
$\displaystyle{}[\tilde{h}_{\mu\nu}(x),b_{\rho}(y)]$ $\displaystyle=$
$\displaystyle\left(\eta_{\mu\rho}\partial_{\nu}+\eta_{\nu\rho}\partial_{\mu}\right)iD(x-y)+2\partial_{\mu}\partial_{\nu}\partial_{\rho}iE(x-y),$
$\displaystyle{}[\tilde{h}_{\mu\nu}(x),b(y)]$ $\displaystyle=$
$\displaystyle-2\partial_{\mu}\partial_{\nu}iD(x-y),$
$\displaystyle{}[S(x),S(y)]$ $\displaystyle=$ $\displaystyle-3iD(x-y),$
$\displaystyle{}[S(x),b_{\rho}(y)]$ $\displaystyle=$
$\displaystyle-\partial_{\rho}iD(x-y),$
$\displaystyle{}[b_{\mu}(x),b_{\nu}(y)]$ $\displaystyle=$
$\displaystyle[b_{\mu}(x),b(y)]=[b(x),b(y)]=0$
$\displaystyle\\{c_{\mu}(x),\bar{c}_{\nu}(y)\\}$ $\displaystyle=$
$\displaystyle-\eta_{\mu\nu}D(x-y)-\partial_{\mu}\partial_{\nu}E(x-y),$
$\displaystyle\\{c_{\mu}(x),\bar{d}(y)\\}$ $\displaystyle=$
$\displaystyle\partial_{\mu}D(x-y),$
$\displaystyle\\{d(x),\bar{c}_{\mu}(y)\\}$ $\displaystyle=$
$\displaystyle-\partial_{\mu}D(x-y),$ $\displaystyle\\{d(x),\bar{d}(y)\\}$
$\displaystyle=$ $\displaystyle 0,$ (A.3)

where

$\displaystyle{\cal A}_{\mu\nu,\rho\sigma}$ $\displaystyle=$
$\displaystyle\eta_{\mu\rho}\partial_{\nu}\partial_{\sigma}+\eta_{\mu\sigma}\partial_{\nu}\partial_{\rho}+\eta_{\nu\rho}\partial_{\mu}\partial_{\sigma}+\eta_{\nu\sigma}\partial_{\mu}\partial_{\rho},$
$\displaystyle{\cal B}_{\mu\nu,\rho\sigma}$ $\displaystyle=$
$\displaystyle\eta_{\mu\nu}\partial_{\rho}\partial_{\sigma}+\eta_{\rho\sigma}\partial_{\mu}\partial_{\nu}.$
(A.4)

The Fourier transforms of these give the commutation relations of the
‘creation-annihilation’ operators $\phi(p)$ and $\phi^{\dagger}(p)$ defined by
4-dimensional Fourier expansion in Eq. (2.38). We find

$\displaystyle[\tilde{h}_{\mu\nu}(p),\tilde{h}^{\dagger}_{\rho\sigma}(q)]$
$\displaystyle=$
$\displaystyle(\eta_{\mu\rho}\eta_{\nu\sigma}+\eta_{\mu\sigma}\eta_{\nu\rho}-\eta_{\mu\nu}\eta_{\rho\sigma})\theta(p^{0})\delta(p^{2})\delta^{4}(p-q)$
$\displaystyle{}+\\{{\cal A}_{\mu\nu,\rho\sigma}\bigr{|}_{p}-2{\cal
B}_{\mu\nu,\rho\sigma}\bigr{|}_{p}\\}\theta(p^{0})(-\delta^{\prime}(p^{2}))\delta^{4}(p-q)$
$\displaystyle{}-2p_{\mu}p_{\nu}p_{\rho}p_{\sigma}\theta(p^{0})\delta^{\prime\prime}(p^{2})\delta^{4}(p-q),$
$\displaystyle{}[\tilde{h}_{\mu\nu}(p),S^{\dagger}(q)]$ $\displaystyle=$
$\displaystyle\eta_{\mu\nu}\theta(p^{0})\delta(p^{2})\delta^{4}(p-q)-4p_{\mu}p_{\nu}\theta(p^{0})(-\delta^{\prime}(p^{2}))\delta^{4}(p-q),$
$\displaystyle{}[\tilde{h}_{\mu\nu}(p),b^{\dagger}_{\rho}(q)]$
$\displaystyle=$
$\displaystyle\left(\eta_{\mu\rho}ip_{\nu}+\eta_{\nu\rho}ip_{\mu}\right)\theta(p^{0})\delta(p^{2})\delta^{4}(p-q)-2ip_{\mu}p_{\nu}p_{\rho}\theta(p^{0})(-\delta^{\prime}(p^{2}))\delta^{4}(p-q),$
$\displaystyle{}[\tilde{h}_{\mu\nu}(p),b^{\dagger}(q)]$ $\displaystyle=$
$\displaystyle 2p_{\mu}p_{\nu}\theta(p^{0})\delta(p^{2})\delta^{4}(p-q),$
$\displaystyle{}[S(p),S^{\dagger}(q)]$ $\displaystyle=$
$\displaystyle-3\theta(p^{0})\delta(p^{2})\delta^{4}(p-q),$
$\displaystyle{}[S(p),b^{\dagger}_{\rho}(q)]$ $\displaystyle=$ $\displaystyle-
ip_{\rho}\theta(p^{0})\delta(p^{2})\delta^{4}(p-q),$
$\displaystyle{}[b_{\mu}(p),b^{\dagger}_{\nu}(q)]$ $\displaystyle=$
$\displaystyle[b_{\mu}(p),b^{\dagger}(q)]=[b(p),b^{\dagger}(q)]=0$
$\displaystyle\\{c_{\mu}(p),\bar{c}^{\dagger}_{\nu}(q)\\}$ $\displaystyle=$
$\displaystyle
i\eta_{\mu\nu}\theta(p^{0})\delta(p^{2})\delta^{4}(p-q)-ip_{\mu}p_{\nu}\theta(p^{0})(-\delta^{\prime}(p^{2}))\delta^{4}(p-q),$
$\displaystyle\\{c_{\mu}(p),\bar{d}^{\dagger}(q)\\}$ $\displaystyle=$
$\displaystyle p_{\mu}\theta(p^{0})\delta(p^{2})\delta^{4}(p-q),$
$\displaystyle\\{d(p),\bar{c}^{\dagger}_{\mu}(q)\\}$ $\displaystyle=$
$\displaystyle-p_{\mu}\theta(p^{0})\delta(p^{2})\delta^{4}(p-q),$
$\displaystyle\\{d(p),\bar{d}^{\dagger}(q)\\}$ $\displaystyle=$ $\displaystyle
0,$ (A.5)

where $X\bigr{|}_{p}$ indicates that all the derivative factors
$\partial_{\mu}$ contained in $X$ should be replaced by $ip_{\mu}$.

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