the $1/c$ and $1/c^{2}$ expansions, the latter being a consistent truncation
to even powers of the former. This allowed us to formulate an algorithm to
compute the $1/c^{2}$ expansion up to order $c^{-2N}$ from the $1/c$ expansion
to order $c^{-N}$. Using this shuffling algorithm we obtained the Lagrangians
describing the $1/c^{2}$ expansion to order $c^{-4}$. At order $c^{-2}$ this
matches the earlier results of [8, 11], at order $c^{-4}$ the result is new. A
further truncation simplifies the $1/c^{2}$ expansion to the PN expansion – in
table 2 one finds all potentials included in the $1/c^{2}$ expansion up to
order $c^{-4}$ listed at the second right and the corresponding potentials in
the PN expansion to 1PN order on the far right – and in that case our result
simplifies to the standard expressions [7].

$\psi$ | $C_{i}$ | $\gamma_{ij}$
---|---|---
${\overset{\,\scalebox{0.5}{($0$)}}{\psi}}{}$ | ${\overset{\,\scalebox{0.5}{($0$)}}{C}}{}_{i}$ | ${\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}_{ij}$
${\overset{\,\scalebox{0.5}{($1$)}}{\psi}}{}$ | ${\overset{\,\scalebox{0.5}{($1$)}}{C}}{}_{i}$ | ${\overset{\,\scalebox{0.5}{($1$)}}{\gamma}}{}_{ij}$
${\overset{\,\scalebox{0.5}{($2$)}}{\psi}}{}$ | ${\overset{\,\scalebox{0.5}{($2$)}}{C}}{}_{i}$ | ${\overset{\,\scalebox{0.5}{($2$)}}{\gamma}}{}_{ij}$
${\overset{\,\scalebox{0.5}{($3$)}}{\psi}}{}$ | ${\overset{\,\scalebox{0.5}{($3$)}}{C}}{}_{i}$ | ${\overset{\,\scalebox{0.5}{($3$)}}{\gamma}}{}_{ij}$
${\overset{\,\scalebox{0.5}{($4$)}}{\psi}}{}$ | ${\overset{\,\scalebox{0.5}{($4$)}}{C}}{}_{i}$ | ${\overset{\,\scalebox{0.5}{($4$)}}{\gamma}}{}_{ij}$

$1/c$ expansion

pedantic notation

---
$c^{0}$
$c^{-1}$
$c^{-2}$
$c^{-3}$
$c^{-4}$

$\psi$ | $C_{i}$ | $\gamma_{ij}$
---|---|---
$\psi$ | $C_{i}$ | $\gamma_{ij}$
$\chi$ | $B_{i}$ | $\alpha_{ij}$
$\phi$ | $A_{i}$ | $\beta_{ij}$
$\upsilon$ | $Z_{i}$ | $\varpi_{ij}$
$\tau$ | $Y_{i}$ | $\epsilon_{ij}$

$\psi$ | $C_{i}$ | $\gamma_{ij}$
---|---|---
$\psi$ | $C_{i}$ | $\gamma_{ij}$
$\chi$ | $B_{i}$ | $\alpha_{ij}$
$\phi$ | $A_{i}$ | $\beta_{ij}$

$\psi$ | $C_{i}$ | $\gamma_{ij}$
---|---|---
$\psi$ | $0$ | $\gamma_{ij}$
$0$ | $B_{i}$ | 0
$\phi$ | $0$ | $\beta_{ij}$
$0$ | $Z_{i}$ | $0$
$\tau$ | 0 | $\epsilon_{ij}$

$\psi$ | $C_{i}$ | $\gamma_{ij}$
---|---|---
$0$ | $0$ | $\delta_{ij}$
$0$ | $0$ | 0
$-2U$ | $0$ | $0$
$0$ | $4U_{i}$ | $0$
$-2\Psi$ | 0 |

$1/c$ expansion $1/c$ expansion  $1/c^{2}$ expansion PN expansion

alt notation up to NNLO up to NNLO up to 1PNO

Table 2: On the top all potentials/fields in the $1/c$ expansion up to order
$c^{-4}$ are listed in a precise but somewhat pedantic notation. On the left
of the second line the same potentials are listed in an alternative notation.
Second to left are those potentials featuring in the $1/c$ expansion up to
NNLO, while second to right are those potentials present in the truncation to
the $1/c^{2}$ expansion up to NNLO. All the way on the right we list those
potentials present in the PN expansion up to first Post-Newtonian order. On
the far left the order at which these fields enter the $1/c$ expansion of
their relativistic relativistic counterparts is indicated.

We believe that in the form introduced in this paper, the $1/c$ expansion is
now finally ready to be applied. Rather than a push to ever increasing order,
the priority in the near future should be to explore non-trivial solutions of
the equations of motion, an interpretation of the physics they describe and to
study if and how this might improve on results obtained by the PN expansion.
That the $1/c$ expansion is not an empty theory is well established. Various
example solutions are discussed in [8, 11, 18] but these all take the form of
expansions of exact solutions of GR. Although such examples are useful to gain
intuition into some features of the expansion they are not teaching us
anything inherently new. Of greater interest would be solutions to the $1/c$
expansion that provide approximations to solutions of GR in situations where
no exact solution is known. Natural situations to think of would be strong
gravitational systems such as Neutron stars or gravitational dynamics close to
merger, where the PN expansion would be put to the limit and the extra
potentials included in the $1/c$ expansion could play a crucial role. We hope
the work in this paper will have paved some of the way for future work in this
direction.

Let us shortly return to the ADM and KS formulations of general relativity.
Our discussion in section 2 showed how both can be implemented in a unified
way, where they appear as each others dual, in that the first is based on a
preferred frame
$e_{i}=\partial_{i},u=N^{-1}(c^{-1}\partial_{t}-N^{i}\partial_{i})$ while the
second is base on the particular co-frame
$e^{i}=dx^{i},n=-M(cdt+C_{i}dx^{i})$. The frame becomes degenerate in the
$c\rightarrow\infty$ limit, but remains generating in the $c\rightarrow 0$
limit, while the opposite is true for the co-frame. This implies the KS
formalism is the natural 3+1 decomposition to use in the non-relativistic or
galilean limit, while the ADM formalism is well suited for the ultra-
relativistic or carrolian limit. This interpretation suggests a link between
KS/ADM duality and Galilei/Carroll duality [32, 33, 34, 35, 36]. In the ADM,
respectively KS, formalism the Lorentzian metric takes the form (2.9),
respectively (2.13) where the fields $(N,N^{i},h_{ij})$ respectively
$(M,C_{i},h_{ij})$ are functions of both time $t$ and space, $x^{i}$. If one
however assumes these fields to be time independent then both metrics (2.9)
and (2.13) are stationary metrics. These two different forms of a stationary
metric are known as the Zermelo and Randers (-Papapetrou) forms. The two forms
can also be argued to be dual [37] and appear in carrolian and galilean fluid
dynamics, see e.g. [38, 39]. A better understanding of the relation between
KS/ADM duality and Galilei/Carroll as well as Randers/Zermelo duality would be
interesting on its own and might be applicable to the $1/c$ expansion or PN
expansion as well, see e.g. [40].

Finally let us mention that in the last few years the study of nonrelativistic
gravity per se – i.e. independent of a relativistic counterpart – has been
very active, see e.g. [41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53],
motivated by quantum gravity, holography and string theory as well as
condensed matter applications. The $1/c^{2}$ expansion has provided some
concrete examples of nonrelativistic gravity theories and provides a rather
generic technique to obtain them. This has influenced advancement in the wider
field of nonrelativistic gravity as well and we hope the same might be true
for the results and insights provided in this paper, in particular as it is
the first development of the $1/c$ expansion beyond leading order keeping all
odd power coefficients.

## Acknowledgements

We thank I. Aras, D. Demirhan, J. Figueroa-O’Farrill, G. Gibbons, P. Horvathy
and N. Obers for discussions and/or correspondence, and the anonymous referee
for a careful reading of the manuscript and valuable comments and suggestions.
DVdB and ME were partially supported by the Boğaziçi University Research Fund
under grant number 21BP2. ME is currently supported by TÜBİTAK under 2236-Co-
Funded Brain Circulation Scheme2 (CoCirculation2) with project number 121C356.
UZ is supported by TÜBİTAK - 2218 National Postdoctoral Research Fellowship
Program with grant number 118C512. DVdB was also partially supported by the
Bilim Akademisi through a BAGEP award.

##  Appendix A Additional technical arguments

###  A.1 Expanded action and equations of motion

The relation between the expanded action222222Since Lagrangians are ambiguous
up to total derivatives we find it convenient to perform the discussion at the
level of the action, but the discussion passes over to Lagrangians
immediately. and expanded equations of motion is independent of the particular
theory under question, so we will keep it general. Apart from setting some
definitions and notation the main result derived in this appendix that is of
use in the main text is (A.3).

Let $S[\Phi;\epsilon]$ be an action functional of some fields $\Phi$, that can
have some explicit dependence on an expansion parameter $\epsilon$. In the
case of interest to this paper $\Phi=(\psi,C_{i},\gamma_{ij})$ and
$\epsilon=1/c$. We’ll denote the Euler-Lagrange equations associated to this
action as ${\cal E}[\Phi;\epsilon]=\frac{\delta S}{\delta\Phi}$. We assume the
fields themselves to be analytic functions of $\epsilon$:
$\Phi=\sum_{n=0}^{\infty}{\overset{\,\scalebox{0.5}{($n$)}}{\Phi}}{}\epsilon^{n}$.

Let us then consider an expansion order by order in $\epsilon$, of the
equations of motion and the action. We can write

$\displaystyle S[\Phi;\epsilon]$
$\displaystyle=\sum_{n=0}^{\infty}{\overset{\,\scalebox{0.5}{($n$)}}{S}}{}[{\overset{\,\scalebox{0.5}{($0$)}}{\Phi}}{},\ldots,{\overset{\,\scalebox{0.5}{($n$)}}{\Phi}}{}]\epsilon^{n}\,,$
(A.1) $\displaystyle{\cal E}[\Phi;\epsilon]$
$\displaystyle=\sum_{n=0}^{\infty}{\overset{\,\scalebox{0.5}{($n$)}}{{\cal
E}}}{}[{\overset{\,\scalebox{0.5}{($0$)}}{\Phi}}{},\ldots,{\overset{\,\scalebox{0.5}{($n$)}}{\Phi}}{}]\epsilon^{n}\,.$
(A.2)

Note that if we expand the theory up to order $N$ then we would have $N$
equations of motion, but also $N$ actions that can be varied with respect to
$N$ fields. At first there might appear a mismatch, but the key insight [11]
is that there is a large degeneracy between the variations of the expanded
actions. Indeed, as is shown below, one has the following relations:

$\frac{\delta{\overset{\,\scalebox{0.5}{($n+m$)}}{S}}{}}{\delta{\overset{\,\scalebox{0.5}{($m$)}}{\Phi}}{}}=\frac{\delta{\overset{\,\scalebox{0.5}{($n$)}}{S}}{}}{\delta{\overset{\,\scalebox{0.5}{($0$)}}{\Phi}}{}}={\overset{\,\scalebox{0.5}{($n$)}}{{\cal
E}}}{}\,.$ (A.3)

To see this, first observe that:

$\delta\Phi=\sum_{n=0}^{\infty}\delta{\overset{\,\scalebox{0.5}{($n$)}}{\Phi}}{}\epsilon^{n}\,,$
(A.4)

which implies that

$\frac{\delta
S}{\delta{\overset{\,\scalebox{0.5}{($n$)}}{\Phi}}{}}=\frac{\delta
S}{\delta\Phi}\frac{\delta\Phi}{\delta{\overset{\,\scalebox{0.5}{($n$)}}{\Phi}}{}}=\frac{\delta
S}{\delta\Phi}\epsilon^{n}\,.$ (A.5)

This equation has two consequences. First note that in the case $n=0$ it gives
$\frac{\delta S}{\delta\Phi}=\frac{\delta
S}{\delta{\overset{\,\scalebox{0.5}{($0$)}}{\Phi}}{}}$ so we get

$\sum_{n=0}^{\infty}{\overset{\,\scalebox{0.5}{($n$)}}{{\cal
E}}}{}\epsilon^{n}={\cal E}=\frac{\delta S}{\delta\Phi}=\frac{\delta
S}{\delta{\overset{\,\scalebox{0.5}{($0$)}}{\Phi}}{}}=\sum_{n=0}^{\infty}\frac{\delta{\overset{\,\scalebox{0.5}{($n$)}}{S}}{}}{\delta{\overset{\,\scalebox{0.5}{($0$)}}{\Phi}}{}}\epsilon^{n}\,,$
(A.6)

and this implies

$\frac{\delta{\overset{\,\scalebox{0.5}{($n$)}}{S}}{}}{\delta{\overset{\,\scalebox{0.5}{($0$)}}{\Phi}}{}}={\overset{\,\scalebox{0.5}{($n$)}}{{\cal
E}}}{}\,.$ (A.7)

Secondly, it directly follows from (A.5) that

$\frac{\delta
S}{\delta{\overset{\,\scalebox{0.5}{($n+1$)}}{\Phi}}{}}=\frac{\delta
S}{\delta{\overset{\,\scalebox{0.5}{($n$)}}{\Phi}}{}}\epsilon\,.$ (A.8)

Via (A.1) this becomes

$\sum_{k=0}^{\infty}\frac{\delta{\overset{\,\scalebox{0.5}{($k$)}}{S}}{}}{\delta{\overset{\,\scalebox{0.5}{($n$)}}{\Phi}}{}}\epsilon^{k+1}=\sum_{k=0}^{\infty}\frac{\delta{\overset{\,\scalebox{0.5}{($k$)}}{S}}{}}{\delta{\overset{\,\scalebox{0.5}{($n+1$)}}{\Phi}}{}}\epsilon^{k}=\sum_{k=1}^{\infty}\frac{\delta{\overset{\,\scalebox{0.5}{($k$)}}{S}}{}}{\delta{\overset{\,\scalebox{0.5}{($n+1$)}}{\Phi}}{}}\epsilon^{k}=\sum_{k=0}^{\infty}\frac{\delta{\overset{\,\scalebox{0.5}{($k+1$)}}{S}}{}}{\delta{\overset{\,\scalebox{0.5}{($n+1$)}}{\Phi}}{}}\epsilon^{k+1}\,.$
(A.9)

Here we used that
${\overset{\,\scalebox{0.5}{($0$)}}{S}}{}[{\overset{\,\scalebox{0.5}{($0$)}}{\Phi}}{}]$,
i.e. the leading order action is a functional of the leading order fields
only, and that $n\geq 0$. Matching the coefficients then gives

$\frac{\delta{\overset{\,\scalebox{0.5}{($k+1$)}}{S}}{}}{\delta{\overset{\,\scalebox{0.5}{($n+1$)}}{\Phi}}{}}=\frac{\delta{\overset{\,\scalebox{0.5}{($k$)}}{S}}{}}{\delta{\overset{\,\scalebox{0.5}{($n$)}}{\Phi}}{}}\,.$
(A.10)

Applying this relation $m$ times leads to (A.3).

###  A.2 Gauge fixing the traces: details

In this appendix we discuss in detail how the traces
${\overset{\,\scalebox{0.5}{($n$)}}{\gamma}}{}={\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}^{ij}{\overset{\,\scalebox{0.5}{($n$)}}{\gamma}}{}_{ij}$,
$n\geq 1$, can be put to zero by a choice of coordinate gauge and then show
the equivalence (4.9). We conclude with a summary of the result and provide a
detailed description of the variational principle (4.8).

####  A.2.1 Gauge fixing

To show how using the time-redefinition symmetry (3.58) we can always remove
the traces ${\overset{\,\scalebox{0.5}{($n$)}}{\gamma}}{}$, we start from the
time-redefinition symmetry (3.58), which in expanded form gives

$\delta{\overset{\,\scalebox{0.5}{($n$)}}{\gamma}}{}_{ij}=\sum_{m=0}^{n-1}\left(2{\overset{\,\scalebox{0.5}{($n-m-1$)}}{\gamma}}{}\\!\\!\\!\\!\\!\\!{}_{ij}\
\partial_{t}{\overset{\,\scalebox{0.5}{($m$)}}{\Lambda}}{}+\partial_{t}\\!\\!{\overset{\,\scalebox{0.5}{($n-m-1$)}}{\gamma}}{}\\!\\!\\!\\!\\!\\!{}_{ij}\
{\overset{\,\scalebox{0.5}{($m$)}}{\Lambda}}{}\right)\,.$ (A.11)

The first observation is that
${\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}_{ij}$ does not232323Consistency
of (3.58) with our expansion ansatz requires a form
$\Lambda(t,x)=c\,f(t)+\sum_{n=0}^{\infty}{\overset{\,\scalebox{0.5}{($n$)}}{\Lambda}}{}(t,x)c^{-n}$.
Since $f$ is a function only of $t$ it cannot be used to set a field to zero
and so it is irrelevant in the discussion here, so we can put it to zero
without loss of generality. transform, i.e.
$\delta{\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}_{ij}=0$, since the
transformation (3.58) of $\gamma_{ij}$ is suppressed by a factor $c^{-1}$.
Using this we get

$\displaystyle\delta{\overset{\,\scalebox{0.5}{($n$)}}{\gamma}}{}=\delta({\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}^{ij}{\overset{\,\scalebox{0.5}{($n$)}}{\gamma}}{}_{ij})={\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}^{ij}\delta{\overset{\,\scalebox{0.5}{($n$)}}{\gamma}}{}_{ij}=$
$\displaystyle\,6\partial_{t}\\!\\!{\overset{\,\scalebox{0.5}{($n-1$)}}{\Lambda}}{}+{\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}^{ij}\partial_{t}{\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}_{ij}{\overset{\,\scalebox{0.5}{($n-1$)}}{\Lambda}}{}$
$\displaystyle+\sum_{m=0}^{n-2}\left(2{\overset{\,\scalebox{0.5}{($n-m-1$)}}{\gamma}}{}\partial_{t}{\overset{\,\scalebox{0.5}{($m$)}}{\Lambda}}{}+{\overset{\,\scalebox{0.5}{($m$)}}{\Lambda}}{}{\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}^{ij}\partial_{t}{\overset{\,\scalebox{0.5}{($n-m-1$)}}{\gamma}}{}\\!\\!\\!\\!{}_{ij}\right)\,.$

From this it follows that we can iteratively set all
${\overset{\,\scalebox{0.5}{($n$)}}{\gamma}}{}$, $n\geq 1$, to zero. To see
this explicitly let us define

${\overset{\,\scalebox{0.5}{($n$)}}{\Lambda}}{}_{\star}=-\frac{e^{-\frac{1}{6}{\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}}}{6}\int^{t}e^{\frac{1}{6}{\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}}{\overset{\,\scalebox{0.5}{($n+1$)}}{\gamma}}{}dt^{\prime}\,,$
(A.12)

where
${\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}=\det({\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}_{ij})$.
As the first step then, observe that the gauge transformation with parameters
${\overset{\,\scalebox{0.5}{($0$)}}{\Lambda}}{}={\overset{\,\scalebox{0.5}{($0$)}}{\Lambda}}{}_{\star}$
and ${\overset{\,\scalebox{0.5}{($m$)}}{\Lambda}}{}=0$ when $m\neq 0$, puts
${\overset{\,\scalebox{0.5}{($1$)}}{\gamma}}{}$ to zero. In a second step we
can then put (the new) ${\overset{\,\scalebox{0.5}{($2$)}}{\gamma}}{}$ to zero
by a transformation with parameters
${\overset{\,\scalebox{0.5}{($1$)}}{\Lambda}}{}={\overset{\,\scalebox{0.5}{($1$)}}{\Lambda}}{}_{\star}$
and ${\overset{\,\scalebox{0.5}{($m$)}}{\Lambda}}{}=0$ when $m\neq 1$. The key
point is that the transformation of the second step does not alter
${\overset{\,\scalebox{0.5}{($1$)}}{\gamma}}{}$ (which we previously set to
zero), since ${\overset{\,\scalebox{0.5}{($1$)}}{\gamma}}{}$ does only
transform with ${\overset{\,\scalebox{0.5}{($0$)}}{\Lambda}}{}$. Indeed, since
${\overset{\,\scalebox{0.5}{($n$)}}{\gamma}}{}$ only transforms under the
${\overset{\,\scalebox{0.5}{($m$)}}{\Lambda}}{}$ for which $m<n$, we see we
can keep on repeating these steps in such a way that when we put
${\overset{\,\scalebox{0.5}{($n$)}}{\gamma}}{}$ to zero by the transformation
with
${\overset{\,\scalebox{0.5}{($n-1$)}}{\Lambda}}{}={\overset{\,\scalebox{0.5}{($n-1$)}}{\Lambda}}{}_{\star}$
and ${\overset{\,\scalebox{0.5}{($m$)}}{\Lambda}}{}=0$ when $m\neq n-1$, all
the ${\overset{\,\scalebox{0.5}{($m$)}}{\gamma}}{}$ with $m<n$ (that we
previously set to zero) will remain zero.

We thus conclude we can always redefine the time coordinate (while keeping the
metric in the form (3.44)) such that all
${\overset{\,\scalebox{0.5}{($n$)}}{\gamma}}{}$ are zero.

####  A.2.2 Equivalence to gauge fixed action

Here we provide the argument leading to the equivalence (4.9). Since the
${\overset{\,\scalebox{0.5}{($n$)}}{\gamma}}{}$ are independent of
${\overset{\,\scalebox{0.5}{($0$)}}{\psi}}{}$ and
${\overset{\,\scalebox{0.5}{($0$)}}{C}}{}_{i}$ it immediately follows
that242424To ease notation we write simply
${{\overset{\,\scalebox{0.5}{($m$)}}{\gamma}}{}=0}$, rather than
${\overset{\,\scalebox{0.5}{($m$)}}{\gamma}}{}=0,\forall m\geq 1$.

$\displaystyle\left.{\overset{\,\scalebox{0.5}{($n$)}}{{\cal
E}}}{}_{\psi}\right|_{{\overset{\,\scalebox{0.5}{($m$)}}{\gamma}}{}=0}$
$\displaystyle=\left.\frac{\delta{\overset{\,\scalebox{0.5}{($n$)}}{S}}{}}{\delta{\overset{\,\scalebox{0.5}{($0$)}}{\psi}}{}}\right|_{{\overset{\,\scalebox{0.5}{($m$)}}{\gamma}}{}=0}=\frac{\delta\left({\overset{\,\scalebox{0.5}{($n$)}}{S}}{}|_{{\overset{\,\scalebox{0.5}{($m$)}}{\gamma}}{}=0}\right)}{\delta{\overset{\,\scalebox{0.5}{($0$)}}{\psi}}{}}=\frac{\delta{\overset{\,\scalebox{0.5}{($n$)}}{\bar{S}}}{}}{\delta{\overset{\,\scalebox{0.5}{($0$)}}{\psi}}{}}={\overset{\,\scalebox{0.5}{($n$)}}{\bar{\cal
E}}}{}_{\psi}\,,$ (A.13)
$\displaystyle\left.{\overset{\,\scalebox{0.5}{($n$)}}{{\cal
E}}}{}^{i}\right|_{{\overset{\,\scalebox{0.5}{($m$)}}{\gamma}}{}=0}$
$\displaystyle=\left.\frac{\delta{\overset{\,\scalebox{0.5}{($n$)}}{S}}{}}{\delta{\overset{\,\scalebox{0.5}{($0$)}}{C}}{}_{i}}\right|_{{\overset{\,\scalebox{0.5}{($m$)}}{\gamma}}{}=0}=\frac{\delta\left({\overset{\,\scalebox{0.5}{($n$)}}{S}}{}|_{{\overset{\,\scalebox{0.5}{($m$)}}{\gamma}}{}=0}\right)}{\delta{\overset{\,\scalebox{0.5}{($0$)}}{C}}{}_{i}}=\frac{\delta{\overset{\,\scalebox{0.5}{($n$)}}{\bar{S}}}{}}{\delta{\overset{\,\scalebox{0.5}{($0$)}}{C}}{}_{i}}={\overset{\,\scalebox{0.5}{($n$)}}{\bar{\cal
E}}}{}^{i}\,.$ (A.14)

The same is not true for the variation with respect to
${\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}_{ij}$. Since
${\overset{\,\scalebox{0.5}{($m$)}}{\gamma}}{}={\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}^{ij}{\overset{\,\scalebox{0.5}{($n$)}}{\gamma}}{}_{ij}$
it follows that

$\left.\frac{\delta{\overset{\,\scalebox{0.5}{($n$)}}{S}}{}}{\delta{\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}_{ij}}\right|_{{\overset{\,\scalebox{0.5}{($m$)}}{\gamma}}{}=0}\neq\frac{\delta\left({\overset{\,\scalebox{0.5}{($n$)}}{S}}{}|_{{\overset{\,\scalebox{0.5}{($m$)}}{\gamma}}{}=0}\right)}{\delta{\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}_{ij}}\,.$
(A.15)

To understand more precisely how these two objects are related we split
${\overset{\,\scalebox{0.5}{($m$)}}{\gamma}}{}_{ij}$ in its trace and trace-
less part, as defined in (4.7), so that252525In (A.16) again we use some
shorthand notation, where
${\overset{\,\scalebox{0.5}{($n$)}}{S}}{}[{\overset{\,\scalebox{0.5}{($0$)}}{\psi}}{},{\overset{\,\scalebox{0.5}{($0$)}}{C}}{}_{i},{\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}_{ij};{\overset{\,\scalebox{0.5}{($m$)}}{\psi}}{},{\overset{\,\scalebox{0.5}{($m$)}}{C}}{}_{i},{\overset{\,\scalebox{0.5}{($m$)}}{\gamma}}{}_{ij}]$
should be understood as
${\overset{\,\scalebox{0.5}{($n$)}}{S}}{}[{\overset{\,\scalebox{0.5}{($0$)}}{\psi}}{},{\overset{\,\scalebox{0.5}{($0$)}}{C}}{}_{i},{\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}_{ij};{\overset{\,\scalebox{0.5}{($1$)}}{\psi}}{},\ldots,{\overset{\,\scalebox{0.5}{($n$)}}{\psi}}{},{\overset{\,\scalebox{0.5}{($1$)}}{C}}{}_{i},\ldots,{\overset{\,\scalebox{0.5}{($n$)}}{C}}{}_{i},{\overset{\,\scalebox{0.5}{($1$)}}{\gamma}}{}_{ij},\ldots,{\overset{\,\scalebox{0.5}{($n$)}}{\gamma}}{}_{ij}]$
and similarly for ${\overset{\,\scalebox{0.5}{($n$)}}{\tilde{S}}}{}$.

${\overset{\,\scalebox{0.5}{($n$)}}{S}}{}[{\overset{\,\scalebox{0.5}{($0$)}}{\psi}}{},{\overset{\,\scalebox{0.5}{($0$)}}{C}}{}_{i},{\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}_{ij};{\overset{\,\scalebox{0.5}{($m$)}}{\psi}}{},{\overset{\,\scalebox{0.5}{($m$)}}{C}}{}_{i},{\overset{\,\scalebox{0.5}{($m$)}}{\gamma}}{}_{ij}]={\overset{\,\scalebox{0.5}{($n$)}}{\tilde{S}}}{}[{\overset{\,\scalebox{0.5}{($0$)}}{\psi}}{},{\overset{\,\scalebox{0.5}{($0$)}}{C}}{}_{i},{\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}_{ij};{\overset{\,\scalebox{0.5}{($m$)}}{\psi}}{},{\overset{\,\scalebox{0.5}{($m$)}}{C}}{}_{i},{\overset{\,\scalebox{0.5}{($m$)}}{\bar{\gamma}}}{}_{ij},{\overset{\,\scalebox{0.5}{($m$)}}{\bar{\gamma}}}{}]\,.$
(A.16)

One can think of the replacement
$({\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}_{ij};{\overset{\,\scalebox{0.5}{($m$)}}{\gamma}}{}_{ij})$
by
$({\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}_{ij};{\overset{\,\scalebox{0.5}{($m$)}}{\bar{\gamma}}}{}_{ij},{\overset{\,\scalebox{0.5}{($m$)}}{\gamma}}{})$
as a change of variables. There is an important subtlety however, in that the
new variables are redundant and not all independent, indeed the collection of
variables
$({\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}_{ij};{\overset{\,\scalebox{0.5}{($m$)}}{\bar{\gamma}}}{}_{ij},{\overset{\,\scalebox{0.5}{($m$)}}{\gamma}}{})$
satisfy the constraints

${\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}^{ij}{\overset{\,\scalebox{0.5}{($m$)}}{\bar{\gamma}}}{}_{ij}=0\,.$
(A.17)

In practice this has the effect that while we can keep all
${\overset{\,\scalebox{0.5}{($m$)}}{\gamma}}{}$ fixed while varying
${\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}_{ij}$, this is not the case for
the ${\overset{\,\scalebox{0.5}{($m$)}}{\bar{\gamma}}}{}_{kl}$. We can however
require them to vary in as minimal262626Remark that $\eqref{constr}$ only
constrains the trace of the variation of
${\overset{\,\scalebox{0.5}{($m$)}}{\bar{\gamma}}}{}_{ij}$, enforcing it to be
non-zero. A priori one could consider adding a traceless part (in $kl$) to the
right hand side of (A.18). Excluding this, as we do, amounts to taking
${\overset{\,\scalebox{0.5}{($m$)}}{\bar{\gamma}}}{}_{ij}$ as independent as
possible of ${\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}_{kl}$. a way as
possible, while preserving (A.17). This leads to

$\frac{\partial{\overset{\,\scalebox{0.5}{($m$)}}{\bar{\gamma}}}{}_{kl}}{\partial{\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}_{ij}}=\frac{1}{3}{\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}_{kl}{\overset{\,\scalebox{0.5}{($m$)}}{\bar{\gamma}}}{}^{ij}\,.$
(A.18)

In reverse, it is possible to keep
${\overset{\,\scalebox{0.5}{($m$)}}{\gamma}}{}$ and
${\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}_{ij}$ fixed when varying the
${\overset{\,\scalebox{0.5}{($m$)}}{\bar{\gamma}}}{}_{ij}$, if we do so in a
limited fashion. More precisely we only allow the
${\overset{\,\scalebox{0.5}{($m$)}}{\bar{\gamma}}}{}_{ij}$ to vary in such a
way that this preserves (A.17) while
${\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}_{ij}$ is kept fixed. This is
equivalent to

$\frac{\partial{\overset{\,\scalebox{0.5}{($m$)}}{\bar{\gamma}}}{}_{kl}}{\partial{\overset{\,\scalebox{0.5}{($n$)}}{\bar{\gamma}}}{}_{ij}}=\delta_{m,n}\mathbb{P}_{kl}^{ij}\qquad\forall
m,n\geq 1\,.$ (A.19)

where we introduced the following projector on the traceless part:

$\mathbb{P}_{ij}^{kl}=\delta_{(i}^{k}\delta_{j)}^{l}-\frac{1}{3}{\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}_{ij}{\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}^{kl}\,.$
(A.20)

This implies the ${\overset{\,\scalebox{0.5}{($m$)}}{\bar{\gamma}}}{}_{ij}$
are being varied only orthogonal to the direction of
${\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}_{ij}$, i.e.

${\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}^{ij}\frac{\delta}{\delta{\overset{\,\scalebox{0.5}{($m$)}}{\bar{\gamma}_{ij}}}{}}=0\,.$
(A.21)

Taking into account these subtleties, the variational principle for
$\tilde{S}$ can then be related to the variational principle for $S$ via the
standard procedure for changing variables. One finds the following relation
between the two variational principles:

$\displaystyle\frac{\delta{\overset{\,\scalebox{0.5}{($n$)}}{S}}{}}{\delta{\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}_{ij}}$
$\displaystyle=\frac{\delta{\overset{\,\scalebox{0.5}{($n$)}}{\tilde{S}}}{}}{\delta{\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}_{ij}}-\sum_{m=1}^{n}\left({\overset{\,\scalebox{0.5}{($m$)}}{\bar{\gamma}}}{}^{ij}\frac{\delta{\overset{\,\scalebox{0.5}{($n$)}}{\tilde{S}}}{}}{\delta{\overset{\,\scalebox{0.5}{($m$)}}{\gamma}}{}}+\frac{{\overset{\,\scalebox{0.5}{($m$)}}{\gamma}}{}}{3}\left({\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}^{ij}\frac{\delta{\overset{\,\scalebox{0.5}{($n$)}}{\tilde{S}}}{}}{\delta{\overset{\,\scalebox{0.5}{($m$)}}{\gamma}}{}}+\frac{\delta\tilde{S}}{\delta{\overset{\,\scalebox{0.5}{($m$)}}{\bar{\gamma}}}{}_{ij}}\right)\right)\,,$
(A.22)
$\displaystyle\frac{\delta{\overset{\,\scalebox{0.5}{($n$)}}{S}}{}}{\delta{\overset{\,\scalebox{0.5}{($m$)}}{\gamma}}{}_{ij}}$
$\displaystyle={\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}^{ij}\frac{\delta{\overset{\,\scalebox{0.5}{($n$)}}{\tilde{S}}}{}}{\delta{\overset{\,\scalebox{0.5}{($m$)}}{\gamma}}{}}+\frac{\delta{\overset{\,\scalebox{0.5}{($n$)}}{\tilde{S}}}{}}{\delta{\overset{\,\scalebox{0.5}{($m$)}}{\bar{\gamma}}}{}_{ij}}\,.$
(A.23)

Inversely one has

$\displaystyle\frac{\delta{\overset{\,\scalebox{0.5}{($n$)}}{\tilde{S}}}{}}{\delta{\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}_{ij}}$
$\displaystyle=\frac{\delta{\overset{\,\scalebox{0.5}{($n$)}}{S}}{}}{\delta{\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}_{ij}}+\frac{1}{3}\sum_{m=1}^{n}\left({\overset{\,\scalebox{0.5}{($m$)}}{\gamma}}{}^{ij}{\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}_{kl}\frac{\delta{\overset{\,\scalebox{0.5}{($n$)}}{S}}{}}{\delta{\overset{\,\scalebox{0.5}{($m$)}}{\gamma}}{}_{kl}}+{\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}^{mn}{\overset{\,\scalebox{0.5}{($n$)}}{\gamma}}{}_{mn}\mathbb{P}^{ij}_{kl}\frac{\delta{\overset{\,\scalebox{0.5}{($n$)}}{S}}{}}{\delta{\overset{\,\scalebox{0.5}{($m$)}}{\gamma}}{}_{kl}}\right)\,,$
(A.24)
$\displaystyle\frac{\delta{\overset{\,\scalebox{0.5}{($n$)}}{\tilde{S}}}{}}{\delta{\overset{\,\scalebox{0.5}{($m$)}}{\bar{\gamma}}}{}_{ij}}$
$\displaystyle=\mathbb{P}^{ij}_{kl}\frac{\delta{\overset{\,\scalebox{0.5}{($n$)}}{S}}{}}{\delta{\overset{\,\scalebox{0.5}{($m$)}}{\gamma}}{}_{kl}}\,,$
(A.25)
$\displaystyle\frac{\delta{\overset{\,\scalebox{0.5}{($n$)}}{\tilde{S}}}{}}{\delta{\overset{\,\scalebox{0.5}{($m$)}}{\gamma}}{}}$
$\displaystyle=\frac{1}{3}{\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}_{kl}\frac{\delta{\overset{\,\scalebox{0.5}{($n$)}}{S}}{}}{\delta{\overset{\,\scalebox{0.5}{($m$)}}{\gamma}}{}_{kl}}\,.$
(A.26)

After we dealt with these technicalities, it has now become straightforward to
prove (4.9). Using the relations (A.22-A.26) one computes

$\displaystyle\left.{\overset{\,\scalebox{0.5}{($n$)}}{{\cal
E}}}{}^{ij}\right|_{{\overset{\,\scalebox{0.5}{($m$)}}{\gamma}}{}=0}$
$\displaystyle=\left.\frac{\delta{\overset{\,\scalebox{0.5}{($n$)}}{S}}{}}{\delta{\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}_{ij}}\right|_{{\overset{\,\scalebox{0.5}{($m$)}}{\gamma}}{}=0}$
(A.27)
$\displaystyle=\left.\frac{\delta{\overset{\,\scalebox{0.5}{($n$)}}{\tilde{S}}}{}}{\delta{\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}_{ij}}\right|_{{\overset{\,\scalebox{0.5}{($m$)}}{\gamma}}{}=0}-\left.\sum_{p=1}^{n}{\overset{\,\scalebox{0.5}{($p$)}}{\bar{\gamma}}}{}^{ij}\frac{\delta{\overset{\,\scalebox{0.5}{($n$)}}{\tilde{S}}}{}}{\delta{\overset{\,\scalebox{0.5}{($p$)}}{\gamma}}{}}\right|_{{\overset{\,\scalebox{0.5}{($m$)}}{\gamma}}{}=0}$
(A.28)
$\displaystyle=\frac{\delta\bigg{(}{\overset{\,\scalebox{0.5}{($n$)}}{\tilde{S}}}{}\big{|}_{{\overset{\,\scalebox{0.5}{($m$)}}{\gamma}}{}=0}\bigg{)}}{\delta{\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}_{ij}}-\frac{1}{3}\left.\sum_{p=1}^{n}{\overset{\,\scalebox{0.5}{($p$)}}{\bar{\gamma}}}{}^{ij}{\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}_{kl}\frac{\delta{\overset{\,\scalebox{0.5}{($n$)}}{S}}{}}{\delta{\overset{\,\scalebox{0.5}{($p$)}}{\gamma}}{}_{ij}}\right|_{{\overset{\,\scalebox{0.5}{($m$)}}{\gamma}}{}=0}$
(A.29) $\displaystyle={\overset{\,\scalebox{0.5}{($n$)}}{\bar{\cal
E}}}{}^{ij}-\frac{1}{3}\sum_{p=1}^{n}{\overset{\,\scalebox{0.5}{($p$)}}{\bar{\gamma}}}{}^{ij}\left.{\overset{\,\scalebox{0.5}{($n-p$)}}{{\cal
E}}}{}\right|_{{\overset{\,\scalebox{0.5}{($m$)}}{\gamma}}{}=0}\,.$ (A.30)

Summarizing we find

${\overset{\,\scalebox{0.5}{($n$)}}{\bar{\cal
E}}}{}^{ij}=\left.\left({\overset{\,\scalebox{0.5}{($n$)}}{{\cal
E}}}{}^{ij}+\frac{1}{3}\sum_{p=1}^{n}{\overset{\,\scalebox{0.5}{($p$)}}{\bar{\gamma}}}{}^{ij}{\overset{\,\scalebox{0.5}{($n-p$)}}{{\cal
E}}}{}\right)\right|_{{\overset{\,\scalebox{0.5}{($m$)}}{\gamma}}{}=0}\,.$
(A.31)

This relation, together with (A.13, A.14), then immediately implies (4.9).

####  A.2.3 Summary

The upshot is that the Euler-Lagrange equations
$\\{{\overset{\,\scalebox{0.5}{($n$)}}{\bar{{\cal
E}}}}{}_{\psi}=0,{\overset{\,\scalebox{0.5}{($n$)}}{\bar{{\cal
E}}}}{}^{i}=0,{\overset{\,\scalebox{0.5}{($n$)}}{\bar{{\cal E}}}}{}^{ij}=0\\}$
obtained by varying a particular, simpler, gauge-fixed action
${\overset{\,\scalebox{0.5}{($n$)}}{\bar{S}}}{}$ are equivalent to the
expanded equations of motion $\\{{\overset{\,\scalebox{0.5}{($n$)}}{{{\cal
E}}}}{}_{\psi}=0,{\overset{\,\scalebox{0.5}{($n$)}}{{{\cal
E}}}}{}^{i}=0,{\overset{\,\scalebox{0.5}{($n$)}}{{{\cal E}}}}{}^{ij}=0\\}$ in
traceless gauge. We now summarize the precise definition of this alternative
action and the related variational principle.

The gauge fixed actions ${\overset{\,\scalebox{0.5}{($n$)}}{\bar{S}}}{}$ are

$\displaystyle{\overset{\,\scalebox{0.5}{($n$)}}{\bar{S}}}{}[{\overset{\,\scalebox{0.5}{($0$)}}{\psi}}{},{\overset{\,\scalebox{0.5}{($0$)}}{C}}{}_{i},{\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}_{ij};{\overset{\,\scalebox{0.5}{($m$)}}{\psi}}{},{\overset{\,\scalebox{0.5}{($m$)}}{C}}{}_{i},{\overset{\,\scalebox{0.5}{($m$)}}{\bar{\gamma}}}{}_{ij}]$
$\displaystyle=$
$\displaystyle\left.{\overset{\,\scalebox{0.5}{($n$)}}{S}}{}[{\overset{\,\scalebox{0.5}{($0$)}}{\psi}}{},{\overset{\,\scalebox{0.5}{($0$)}}{C}}{}_{i},{\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}_{ij};{\overset{\,\scalebox{0.5}{($m$)}}{\psi}}{},{\overset{\,\scalebox{0.5}{($m$)}}{C}}{}_{i},{\overset{\,\scalebox{0.5}{($m$)}}{\bar{\gamma}}}{}_{ij}]\right|_{{\overset{\,\scalebox{0.5}{($n$)}}{\gamma}}{}=0}$
(A.32) $\displaystyle=$
$\displaystyle{\overset{\,\scalebox{0.5}{($n$)}}{\tilde{S}}}{}[{\overset{\,\scalebox{0.5}{($0$)}}{\psi}}{},{\overset{\,\scalebox{0.5}{($0$)}}{C}}{}_{i},{\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}_{ij};{\overset{\,\scalebox{0.5}{($m$)}}{\psi}}{},{\overset{\,\scalebox{0.5}{($m$)}}{C}}{}_{i},{\overset{\,\scalebox{0.5}{($m$)}}{\bar{\gamma}}}{}_{ij},0]\,.$
(A.33)

In practice the ${\overset{\,\scalebox{0.5}{($n$)}}{\bar{S}}}{}$ can be
obtained by expanding the KS action (3.45, 3.57), replacing all
${\overset{\,\scalebox{0.5}{($m$)}}{\gamma}}{}_{ij}$ by
${\overset{\,\scalebox{0.5}{($m$)}}{\bar{\gamma}}}{}_{ij}$ and using that
${\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}^{ij}{\overset{\,\scalebox{0.5}{($m$)}}{\bar{\gamma}}}{}_{ij}=0$.

The associated Euler-Lagrange equations are then defined as

${\overset{\,\scalebox{0.5}{($n$)}}{\bar{\cal
E}}}{}_{\psi}=\frac{\delta{\overset{\,\scalebox{0.5}{($n$)}}{\bar{S}}}{}}{\delta{\overset{\,\scalebox{0.5}{($0$)}}{\psi}}{}}\,,\qquad{\overset{\,\scalebox{0.5}{($n$)}}{\bar{\cal
E}}}{}^{i}=\frac{\delta{\overset{\,\scalebox{0.5}{($n$)}}{\bar{S}}}{}}{\delta{\overset{\,\scalebox{0.5}{($0$)}}{C}}{}_{i}}\,,\qquad{\overset{\,\scalebox{0.5}{($n$)}}{\bar{\cal
E}}}{}^{ij}=\frac{\delta{\overset{\,\scalebox{0.5}{($n$)}}{\bar{S}}}{}}{\delta{\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}_{ij}}+\frac{1}{3}\sum_{m=1}^{n}\frac{\delta{\overset{\,\scalebox{0.5}{($n$)}}{\bar{S}}}{}}{\delta{\overset{\,\scalebox{0.5}{($m$)}}{\bar{\gamma}}}{}_{kl}}{\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}_{kl}{\overset{\,\scalebox{0.5}{($m$)}}{\bar{\gamma}}}{}^{ij}\,.$
(A.34)

We might need to clarify the presence of the last terms in
${\overset{\,\scalebox{0.5}{($n$)}}{\bar{\cal E}}}{}^{ij}$. From our argument
above, see in particular $\eqref{keystep}$, one see that the variational
principle used is that associated to
${\overset{\,\scalebox{0.5}{($n$)}}{\tilde{S}}}{}$. As pointed out in (A.18),
in that variational principle
${\overset{\,\scalebox{0.5}{($m$)}}{\bar{\gamma}}}{}_{ij}$ and
${\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}_{ij}$ are not fully
independent, i.e. under variation of
${\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}_{ij}$ one has

$\delta{\overset{\,\scalebox{0.5}{($m$)}}{\bar{\gamma}}}{}_{ij}=\frac{1}{3}{\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}_{ij}{\overset{\,\scalebox{0.5}{($m$)}}{\bar{\gamma}}}{}^{kl}\delta{\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}_{kl}\,.$
(A.35)

It is this effect that is responsible for the extra terms in
${\overset{\,\scalebox{0.5}{($n$)}}{\bar{\cal E}}}{}^{ij}$. There is another
point of view, which based on (A.21) implies that variations of
${\overset{\,\scalebox{0.5}{($m$)}}{\bar{\gamma}}}{}_{ij}$ are such that the
extra terms vanish. At first thought these two point of views might seem
contradictory, but they are actually perfectly equivalent if interpreted
correctly.

Let us consider an example of varying
${\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}_{ij}$ from the first point of
view:

$\displaystyle\delta({\overset{\,\scalebox{0.5}{($n$)}}{\bar{\gamma}}}{}_{ij}F^{ij})$
$\displaystyle=\delta{\overset{\,\scalebox{0.5}{($n$)}}{\bar{\gamma}}}{}_{ij}F^{ij}+{\overset{\,\scalebox{0.5}{($n$)}}{\bar{\gamma}}}{}_{ij}\delta
F^{ij}$
$\displaystyle=\frac{1}{3}F{\overset{\,\scalebox{0.5}{($m$)}}{\bar{\gamma}}}{}^{kl}\delta{\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}_{kl}+{\overset{\,\scalebox{0.5}{($n$)}}{\bar{\gamma}}}{}_{ij}\delta(\bar{F}^{ij}+\frac{1}{3}{\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}^{ij}F)$
(A.36)
$\displaystyle={\overset{\,\scalebox{0.5}{($n$)}}{\bar{\gamma}}}{}_{ij}\delta\bar{F}^{ij}\,.$

We see that the extra terms, the
$F^{ij}\delta{\overset{\,\scalebox{0.5}{($n$)}}{\bar{\gamma}}}{}_{ij}$
originating from (A.35), seem to play a key role in removing the trace of
$F_{ij}$ in the final result. But looking at the same calculation differently,
we could observe that since
${\overset{\,\scalebox{0.5}{($n$)}}{\bar{\gamma}}}{}_{ij}$ is traceless we
could have from the beginning written

$\displaystyle\delta({\overset{\,\scalebox{0.5}{($n$)}}{\bar{\gamma}}}{}_{ij}F^{ij})$
$\displaystyle=$
$\displaystyle\delta({\overset{\,\scalebox{0.5}{($n$)}}{\bar{\gamma}}}{}_{ij}\bar{F}^{ij})$
$\displaystyle=$
$\displaystyle\delta{\overset{\,\scalebox{0.5}{($n$)}}{\bar{\gamma}}}{}_{ij}\bar{F}^{ij}+{\overset{\,\scalebox{0.5}{($n$)}}{\bar{\gamma}}}{}_{ij}\delta\bar{F}^{ij}$
$\displaystyle=$
$\displaystyle{\overset{\,\scalebox{0.5}{($n$)}}{\bar{\gamma}}}{}_{ij}\delta\bar{F}^{ij}$

In this way of doing the calculation the extra terms, now
$\delta{\overset{\,\scalebox{0.5}{($n$)}}{\bar{\gamma}}}{}_{ij}\bar{F}^{ij}$,
actually vanish.

Of the two approaches the second might seem the most efficient, namely to
project from the beginning everything contracted with the
${\overset{\,\scalebox{0.5}{($n$)}}{\bar{\gamma}}}{}_{ij}$ on its traceless
part so that one can ignore the extra contribution from
$\delta{\overset{\,\scalebox{0.5}{($n$)}}{\bar{\gamma}}}{}_{ij}$. On the other
hand, the first approach has the advantage of being more straightforward. In
particular when time derivatives are present the first approach seems more
tractable. In that case one can use, by (A.35), that

$\delta{\overset{\,\scalebox{0.5}{($m$)}}{\dot{\bar{\gamma}}}}{}_{ij}=\frac{1}{3}{\overset{\,\scalebox{0.5}{($0$)}}{\dot{\gamma}}}{}_{ij}{\overset{\,\scalebox{0.5}{($m$)}}{\bar{\gamma}}}{}^{kl}\delta{\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}_{kl}-\frac{1}{3}{\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}_{ij}{\overset{\,\scalebox{0.5}{($m$)}}{\dot{\bar{\gamma}}}}{}^{kl}\delta{\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}_{kl}+\frac{1}{3}{\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}_{ij}{\overset{\,\scalebox{0.5}{($m$)}}{\bar{\gamma}}}{}^{kl}\delta{\overset{\,\scalebox{0.5}{($0$)}}{\dot{\gamma}}}{}_{kl}$
(A.38)

This illustrates that a number of further terms appear, but partial
integration will only be needed for the last one.

Taking into account (A.35, A.38) then gives a precise way to compute all the
equations of motion ${\overset{\,\scalebox{0.5}{($n$)}}{\bar{\cal E}}}{}^{ij}$
by varying ${\overset{\,\scalebox{0.5}{($n$)}}{\bar{S}}}{}$ with respect to
$\delta{\overset{\,\scalebox{0.5}{($0$)}}{\gamma}}{}_{ij}$.

##  Appendix B Extra trace terms

In the main text we have chosen to present a number of results in a gauge
where we have put to zero the traces
${\overset{\,\scalebox{0.5}{($n$)}}{\gamma}}{}$ of the subleading tensor
fields ${\overset{\,\scalebox{0.5}{($n$)}}{\gamma}}{}_{ij}$, $n\geq 1$, see
section 4.1.2 and appendix A.2 for details on how and why this is done. For
completeness we provide in this appendix the additional terms that would
appear in these results if they were not gauge-fixed. This could be useful in
case one would like to study certain features in a gauge independent fashion,
or in another gauge.

### Additions to results in section 4.1.4

The linear part of the Lagrangian gets the additional trace terms in the $1/c$
expansion:

${\overset{\,\scalebox{0.5}{($n$)}}{{\cal
L}}}{}_{\mathrm{lin}}=\left.{\overset{\,\scalebox{0.5}{($n$)}}{\bar{\cal
L}}}{}_{\mathrm{lin}}\right|_{{\overset{\,\scalebox{0.5}{($n$)}}{\bar{\gamma}}}{}\\!{}_{ij}\rightarrow{\overset{\,\scalebox{0.5}{($n$)}}{\gamma}}{}\\!{}_{ij}}+\frac{1}{2}{\overset{\,\scalebox{0.5}{($n$)}}{\gamma}}{}{\overset{\,\scalebox{0.5}{($0$)}}{{\cal
L}}}{}\,,$ (B.1)

where ${\overset{\,\scalebox{0.5}{($0$)}}{{\cal L}}}{}$ can be found in
(4.13).

The 2nd order linear differential operators appearing in the linear part in
the $1/c$ expansion get the additional trace terms:

$\displaystyle\mathbb{D}^{2}{\overset{\,\scalebox{0.5}{($n$)}}{\psi}}{}=\,$
$\displaystyle\left.\overline{\mathbb{D}}^{2}{\overset{\,\scalebox{0.5}{($n$)}}{\psi}}{}\right|_{{\overset{\,\scalebox{0.5}{($n$)}}{\bar{\gamma}}}{}\\!{}_{ij}\rightarrow{\overset{\,\scalebox{0.5}{($n$)}}{\gamma}}{}\\!{}_{ij}}+\frac{1}{2}{\overset{\,\scalebox{0.5}{($n$)}}{\gamma}}{}{\overset{\,\scalebox{0.5}{($0$)}}{E}}{}_{\psi}+\frac{1}{2}\partial^{i}\psi
D_{i}{\overset{\,\scalebox{0.5}{($n$)}}{\gamma}}{}\,,$ (B.2)
$\displaystyle\mathbb{D}^{2}{\overset{\,\scalebox{0.5}{($n$)}}{C}}{}^{i}=\,$
$\displaystyle\left.\overline{\mathbb{D}}^{2}{\overset{\,\scalebox{0.5}{($n$)}}{C}}{}^{i}\right|_{{\overset{\,\scalebox{0.5}{($n$)}}{\bar{\gamma}}}{}\\!{}_{kl}\rightarrow{\overset{\,\scalebox{0.5}{($n$)}}{\gamma}}{}\\!{}_{kl}}+\frac{1}{2}{\overset{\,\scalebox{0.5}{($n$)}}{\gamma}}{}{\overset{\,\scalebox{0.5}{($0$)}}{E}}{}^{i}+\frac{1}{2}C^{ji}D_{j}{\overset{\,\scalebox{0.5}{($n$)}}{\gamma}}{}\,,$
(B.3)
$\displaystyle\mathbb{D}^{2}{\overset{\,\scalebox{0.5}{($n$)}}{\gamma}}{}^{ij}=\,$
$\displaystyle\left.\overline{\mathbb{D}}^{2}{\overset{\,\scalebox{0.5}{($n$)}}{{\bar{\gamma}}}}{}^{ij}\right|_{{\overset{\,\scalebox{0.5}{($n$)}}{\bar{\gamma}}}{}\\!{}_{kl}\rightarrow{\overset{\,\scalebox{0.5}{($n$)}}{\gamma}}{}\\!{}_{kl}}-\frac{1}{3}{\overset{\,\scalebox{0.5}{($n$)}}{\gamma}}{}^{ij}\gamma_{kl}{\overset{\,\scalebox{0.5}{($0$)}}{E}}{}^{kl}+\frac{1}{2}{\overset{\,\scalebox{0.5}{($n$)}}{\gamma}}{}{\overset{\,\scalebox{0.5}{($0$)}}{E}}{}^{ij}$
(B.4)
$\displaystyle+\frac{1}{2}(D^{(i}D^{j)}{\overset{\,\scalebox{0.5}{($n$)}}{\gamma}}{}-\gamma^{ij}D_{k}D^{k}{\overset{\,\scalebox{0.5}{($n$)}}{\gamma}}{})\,,$

where

$\displaystyle{\overset{\,\scalebox{0.5}{($0$)}}{E}}{}^{ij}=$
$\displaystyle\,-(R^{ij}+\frac{e^{2\psi}}{2}C_{k}{}^{i}C^{kj}-\frac{1}{2}\partial^{i}\psi\partial^{j}\psi)$
(B.5)
$\displaystyle+\frac{\gamma^{ij}}{2}(R+\frac{e^{2\psi}}{4}C_{kl}C^{kl}-\frac{1}{2}\partial_{k}\psi\partial^{k}\psi)$

are leading order equations of motion, and for this reason can be ignored on-
shell.

### Additions to results in section 4.2

The extra contributions to the linear parts of the Lagrangian are already
given at all order in (B.1). In addition to those one has that
${\overset{\,\scalebox{0.5}{($1$)}}{L}}{}_{1\mathrm{td}}={\overset{\,\scalebox{0.5}{($1$)}}{\bar{L}}}{}_{1\mathrm{td}}$
and

$\displaystyle{\overset{\,\scalebox{0.5}{($2$)}}{L}}{}_{0\mathrm{td}}=$
$\displaystyle\left.{\overset{\,\scalebox{0.5}{($2$)}}{\bar{L}}}{}_{0\mathrm{td}}\right|_{\bar{\alpha}_{kl}\rightarrow\alpha_{kl}}\\!\\!\\!\\!+\frac{\alpha}{2}\left(\frac{\alpha}{4}(R-\frac{1}{2}\partial_{i}\psi\partial^{i}\psi+\frac{e^{2\psi}}{4}C_{ij}C^{ij})\right.$
(B.6)
$\displaystyle\left.-\alpha^{ij}(R_{ij}-\frac{1}{2}\partial_{i}\psi\partial_{j}\psi+\frac{e^{2\psi}}{2}C_{i}{}^{l}C_{jl})-\partial_{i}\psi\partial^{i}\chi+\frac{\chi}{2}e^{2\psi}C_{ij}C^{ij}+\frac{e^{2\psi}}{4}B_{ij}C^{ij}\right)\,,$
$\displaystyle{\overset{\,\scalebox{0.5}{($2$)}}{L}}{}_{1\mathrm{td}}=$
$\displaystyle\left.{\overset{\,\scalebox{0.5}{($2$)}}{\bar{L}}}{}_{1\mathrm{td}}\right|_{\bar{\alpha}_{kl}\rightarrow\alpha_{kl}}\\!\\!\\!\\!+\frac{\alpha}{2}\left(\Gamma^{ij\,kl}D_{i}C_{j}\dot{\gamma}_{kl}+C_{i}(\dot{\psi}\partial^{i}\psi-e^{2\psi}\dot{C}_{j}C^{ij})+2\dot{C}_{i}\partial^{i}\psi\right)$
(B.7) $\displaystyle-\frac{1}{2}C_{i}\dot{\gamma}D^{i}\alpha\,,$
$\displaystyle{\overset{\,\scalebox{0.5}{($2$)}}{L}}{}_{2\mathrm{td}}=$
$\displaystyle\left.{\overset{\,\scalebox{0.5}{($2$)}}{\bar{L}}}{}_{2\mathrm{td}}\right|_{\bar{\alpha}_{kl}\rightarrow\alpha_{kl}}\\!\\!\\!\\!+\frac{\alpha}{2}e^{-2\psi}\left(\frac{1}{4}\Gamma^{kl\,mn}\dot{\gamma}_{kl}\dot{\gamma}_{mn}+\dot{\gamma}\dot{\psi}-\frac{3}{2}\dot{\psi}^{2}\right)\,.$
(B.8)

### Additions to results in section 5.1.3

The linear part of the Lagrangian gets the additional trace terms in the
$1/c^{2}$ expansion:

${\overset{\,\scalebox{0.5}{($2n$)}}{{\cal
L}}}{}_{\mathrm{lin}}^{\mathrm{e}}=\left.{\overset{\,\scalebox{0.5}{($2n$)}}{\bar{\cal
L}}}{}_{\mathrm{lin}}^{\mathrm{e}}\right|_{{\overset{\,\scalebox{0.5}{($n$)}}{\bar{\gamma}}}{}\\!{}_{ij}\rightarrow{\overset{\,\scalebox{0.5}{($2n$)}}{\gamma}}{}\\!{}_{ij}}+\frac{1}{2}{\overset{\,\scalebox{0.5}{($2n$)}}{\gamma}}{}{\overset{\,\scalebox{0.5}{($0$)}}{{\cal
L}^{\mathrm{e}}}}{}\,,$ (B.9)

where ${\overset{\,\scalebox{0.5}{($0$)}}{{\cal L}}}{}^{\mathrm{e}}$ can be
found in (5.12).

The 2nd order linear differential operators appearing in the linear part in
the $1/c^{2}$ expansion get the additional trace terms:

$\displaystyle\mathbb{D}^{2}{\overset{\,\scalebox{0.5}{($2n$)}}{\psi}}{}=\,$
$\displaystyle\left.\overline{\mathbb{D}}^{2}{\overset{\,\scalebox{0.5}{($2n$)}}{\psi}}{}\right|_{{\overset{\,\scalebox{0.5}{($2n$)}}{\bar{\gamma}}}{}\\!{}_{ij}\rightarrow{\overset{\,\scalebox{0.5}{($2n$)}}{\gamma}}{}\\!{}_{ij}}+\frac{1}{2}{\overset{\,\scalebox{0.5}{($2n$)}}{\gamma}}{}{\overset{\,\scalebox{0.5}{($0$)}}{E}}{}_{\psi}^{\mathrm{e}}+\frac{1}{2}\partial^{i}\psi
D_{i}{\overset{\,\scalebox{0.5}{($2n$)}}{\gamma}}{}\,,$ (B.10)
$\displaystyle\mathbb{D}^{2}{\overset{\,\scalebox{0.5}{($2n$)}}{\gamma}}{}^{ij}=\,$
$\displaystyle\left.\overline{\mathbb{D}}^{2}{\overset{\,\scalebox{0.5}{($2n$)}}{{\bar{\gamma}}}}{}^{ij}\right|_{{\overset{\,\scalebox{0.5}{($2n$)}}{\bar{\gamma}}}{}\\!{}_{kl}\rightarrow{\overset{\,\scalebox{0.5}{($2n$)}}{\gamma}}{}\\!{}_{kl}}-\frac{1}{3}{\overset{\,\scalebox{0.5}{($2n$)}}{\gamma}}{}^{ij}\gamma_{kl}{\overset{\,\scalebox{0.5}{($0$)}}{E}}{}^{kl}_{\mathrm{e}}+\frac{1}{2}{\overset{\,\scalebox{0.5}{($2n$)}}{\gamma}}{}{\overset{\,\scalebox{0.5}{($0$)}}{E}}{}^{ij}_{\mathrm{e}}$
(B.11)
$\displaystyle+\frac{1}{2}(D^{(i}D^{j)}{\overset{\,\scalebox{0.5}{($2n$)}}{\gamma}}{}-\gamma^{ij}D_{k}D^{k}{\overset{\,\scalebox{0.5}{($2n$)}}{\gamma}}{})\,,$
$\displaystyle{\cal
D}_{\mathrm{e}}^{2}{\overset{\,\scalebox{0.5}{($2n+2$)}}{B}}{}^{i}$
$\displaystyle=\left.\overline{{\cal
D}}_{\mathrm{e}}^{2}{\overset{\,\scalebox{0.5}{($2n+2$)}}{B}}{}^{i}\right|_{{\overset{\,\scalebox{0.5}{($2n$)}}{\bar{\gamma}}}{}\\!{}_{kl}\rightarrow{\overset{\,\scalebox{0.5}{($2n$)}}{\gamma}}{}\\!{}_{kl}}\,,$
(B.12)

where

${\overset{\,\scalebox{0.5}{($0$)}}{E}}{}^{ij}_{\mathrm{e}}=-(R^{ij}-\frac{1}{2}\partial^{i}\psi\partial^{j}\psi)+\frac{\gamma^{ij}}{2}(R-\frac{1}{2}\partial_{k}\psi\partial^{k}\psi)$
(B.13)

are leading order equations of motion.

### Additions to results in section 5.2

The extra contributions to the linear parts of the even Lagrangian are already
given at all order in (B.9). Outside the traceless gauge, the Lagrangians
(5.28, 5.40) become

$\displaystyle{\overset{\,\scalebox{0.5}{($2$)}}{L}}{}^{\mathrm{e}}=$
$\displaystyle\left.{\overset{\,\scalebox{0.5}{($2$)}}{\bar{L}}}{}^{\mathrm{e}}\right|_{\bar{\beta}_{kl}\rightarrow\beta_{kl}}\\!\\!\\!\\!+\frac{\beta}{2}\left(R-\frac{1}{2}\partial_{i}\psi\partial^{i}\psi\right)\,,$
(B.14) $\displaystyle{\overset{\,\scalebox{0.5}{($4$)}}{L}}{}^{\mathrm{e}}=$
$\displaystyle\left.{\overset{\,\scalebox{0.5}{($4$)}}{\bar{L}}}{}^{\mathrm{e}}\right|_{\begin{subarray}{c}\bar{\beta}_{kl}\rightarrow\beta_{kl}\\\
\bar{\epsilon}_{kl}\rightarrow\epsilon_{kl}\end{subarray}}\\!\\!\\!\\!+\frac{\epsilon}{2}\left(R-\frac{1}{2}\partial_{i}\psi\partial^{i}\psi\right)$
(B.15)
$\displaystyle+\frac{\beta}{2}\left(\frac{\beta}{4}(R-\frac{1}{2}\partial_{i}\psi\partial^{i}\psi)-\beta^{ij}(R_{ij}-\frac{1}{2}\partial_{i}\psi\partial_{j}\psi)-\partial_{i}\psi\partial^{i}\phi+\frac{e^{2\psi}}{4}B_{ij}B^{ij}\right.$
$\displaystyle\left.+\Gamma^{ij\,kl}D_{i}B_{j}\dot{\gamma}_{kl}+B_{i}\dot{\psi}\partial^{i}\psi+2\dot{B}_{i}\partial^{i}\psi+\frac{e^{-2\psi}}{4}\left(\Gamma^{ij\,kl}\dot{\gamma}_{ij}\dot{\gamma}_{kl}+4\dot{\gamma}\dot{\psi}-6\dot{\psi}^{2}\right)\right)$
$\displaystyle-\frac{1}{2}B_{i}\dot{\gamma}D^{i}\beta\,.$

##  Appendix C Equations of motion

Here we present the equations of motion of the (un-expanded) KS action. As in
(3.45) the Lagrangian can be split as

${\cal L}={\cal L}_{0}+c^{-1}{\cal L}_{1}+c^{-2}{\cal L}_{2}\qquad{\cal
L}_{a}=\sqrt{\gamma}L_{a}\,,$ (C.1)

with the $L_{a}$ given in (3.57). Similarly we can split the equations of
motion (in the absence of matter) as

$E_{0}=-c^{-1}E_{1}-c^{-2}E_{2}\qquad
E_{a}=\frac{1}{\sqrt{\gamma}}\frac{\delta{\cal L}_{a}}{\delta\Phi}\,.$ (C.2)

We present the different parts of the equations of motion obtained from
variation by each of the fields $\Phi=(\psi,C_{i},\gamma_{ij})$ below.

### Variation by $\psi$

Variation of the KS Lagrangian with $\psi$ yields

$\displaystyle E_{0}^{\psi}=$
$\displaystyle\,\,D_{i}\partial^{i}\psi+\frac{e^{2\psi}}{2}C_{ij}C^{ij}\,,$
(C.3) $\displaystyle-E_{1}^{\psi}=$
$\displaystyle\,\,D_{i}(C^{i}\dot{\psi})+\dot{C}^{i}\partial_{i}\psi+2D_{i}\dot{C}^{i}+C^{i}\Big{(}\partial_{i}\dot{\psi}+\frac{\dot{\gamma}}{2}\partial_{i}\psi-\dot{\gamma}_{ij}\partial^{j}\psi\Big{)}$
(C.4) $\displaystyle+2e^{2\psi}C_{ij}C^{i}\dot{C}^{j}\,,$ $\displaystyle-
E_{2}^{\psi}=$
$\displaystyle-2\dot{\psi}C^{i}\dot{C}_{i}-2C^{i}\ddot{C}_{i}-2\dot{C}^{i}\dot{C}_{i}-C^{i}C_{i}\ddot{\psi}$
(C.5)
$\displaystyle-\Big{(}C_{i}\frac{\dot{\gamma}}{2}-C^{j}\dot{\gamma}_{ij}\Big{)}(\dot{\psi}C^{i}+2\dot{C}^{i})+e^{2\psi}\Gamma^{ij\,kl}C_{i}C_{j}\dot{C}_{k}\dot{C}_{l}$
$\displaystyle-\frac{1}{2}e^{-2\psi}\Big{(}\dot{\gamma}^{ij}\dot{\gamma}_{ij}+3\dot{\gamma}\dot{\psi}-6\dot{\psi}^{2}+6\ddot{\psi}-2\gamma^{ij}\ddot{\gamma}_{ij}\Big{)}\,.$

### Variation by $C_{i}$

Next, one finds the equation of motion for the gauge field $C_{i}$ to be

$\displaystyle E_{0}^{i}=$ $\displaystyle\,\,D_{j}(e^{2\psi}C^{ij})\,,$ (C.6)
$\displaystyle-E_{1}^{i}=$
$\displaystyle\,\,\Gamma^{ij\,kl}D_{j}\dot{\gamma}_{kl}-\dot{\psi}\partial^{i}\psi+\dot{\gamma}\partial^{i}\psi+2\partial^{i}\dot{\psi}-2\dot{\gamma}^{ij}\partial_{j}\psi+2D_{j}(e^{2\psi}C^{[i}\dot{C}^{j]})$
(C.7)
$\displaystyle+e^{2\psi}\left(\Big{(}2\dot{\psi}C_{j}+2\dot{C}_{j}+\frac{1}{2}\dot{\gamma}C_{j}-C^{k}\dot{\gamma}_{jk}\Big{)}C^{ij}+C_{j}\dot{C}^{ij}+C_{j}\dot{\gamma}_{k}{}^{i}C^{jk}\right)\,,$
$\displaystyle-E_{2}^{i}=$
$\displaystyle-\Gamma^{ij\,kl}C_{j}\ddot{\gamma}_{kl}-\frac{1}{2}(C^{i}\dot{\gamma}^{jk}\dot{\gamma}_{jk}+\dot{\gamma}^{ij}C_{j}\dot{\gamma})+C^{k}\dot{\gamma}^{ij}\dot{\gamma}_{jk}$
(C.8)
$\displaystyle+\dot{\psi}^{2}C^{i}-\dot{\gamma}C^{i}\dot{\psi}-2C^{i}\ddot{\psi}+2C_{j}\dot{\psi}\dot{\gamma}^{ij}$
$\displaystyle-2e^{2\psi}\left(\Big{(}2C_{j}\dot{\psi}+2\dot{C}_{j}-C^{k}\dot{\gamma}_{jk}+\frac{1}{2}C_{j}\dot{\gamma}\Big{)}C^{[i}\dot{C}^{j]}+C_{j}\Big{(}C^{[i}\ddot{C}^{j]}+\dot{\gamma}^{i}{}_{k}C^{[j}\dot{C}^{k]}\Big{)}\right)\,.$

### Variation by $\gamma_{ij}$

It is a bit cumbersome but straightforward to obtain the contributions:

$\displaystyle E_{0}^{ij}=$
$\displaystyle\,\,\frac{\gamma^{ij}}{2}\Big{(}R-\frac{1}{2}\partial_{k}\psi\partial^{k}\psi+\frac{e^{2\psi}}{4}C_{kl}C^{kl}\Big{)}-\Big{(}R^{ij}-\frac{1}{2}\partial^{i}\psi\partial^{j}\psi+\frac{e^{2\psi}}{2}C^{i}{}_{k}C^{jk}\Big{)},$
(C.9) $\displaystyle-E_{1}^{ij}=$
$\displaystyle\,\,\frac{\gamma^{ij}}{2}\Gamma^{kl\,mn}(\dot{\gamma}_{mn}D_{k}C_{l}+2C_{k}D_{l}\dot{\gamma}_{mn})+C^{k}D_{k}\dot{\gamma}^{ij}-C^{(i}D_{k}\dot{\gamma}^{j)k}$
(C.10) $\displaystyle-
C_{k}D^{(i}\dot{\gamma}^{j)}{}_{k}+C^{(i}D^{j)}\dot{\gamma}+\frac{\dot{\gamma}^{ij}}{2}D_{k}C^{k}-\dot{\gamma}^{k(i}D_{k}C^{j)}+\frac{\dot{\gamma}}{2}D^{(i}C^{j)}$
$\displaystyle+\frac{e^{2\psi}}{2}\Big{(}C_{kl}(\frac{\gamma^{ij}}{2}C^{k}\dot{C}^{l}-2\gamma^{ki}C^{[j}\dot{C}^{l]})+C_{[k}\dot{C}_{l]}(\frac{\gamma^{ij}}{2}C^{kl}-2\gamma^{k(i}C^{j)l})\Big{)}$
$\displaystyle+\Gamma^{ij\,kl}D_{k}\dot{C}_{l}-\gamma^{ij}\dot{C}^{k}\partial_{k}\psi+2\dot{C}^{(i}\partial^{j)}\psi-\dot{\psi}\Big{(}\frac{{\gamma}^{ij}}{2}C^{k}\partial_{k}\psi-C^{(i}\partial^{j)}\psi\Big{)}\,,$
$\displaystyle-E_{2}^{ij}=$
$\displaystyle\,\,C_{k}C^{(i}\ddot{\gamma}^{j)k}-\frac{1}{2}C^{k}C_{k}\ddot{\gamma}^{ij}-\frac{1}{2}C^{i}C^{j}\ddot{\gamma}+C_{k}\dot{C}^{(i}\dot{\gamma}^{j)k}-\frac{1}{2}C^{k}\dot{C}_{k}\dot{\gamma}^{ij}$
(C.11)
$\displaystyle+\frac{1}{4}C^{(i}C^{j)}\dot{\gamma}^{kl}\dot{\gamma}_{kl}-\frac{1}{2}\dot{C}^{(i}C^{j)}\dot{\gamma}+\frac{1}{2}C_{k}C_{l}\dot{\gamma}^{kl}\dot{\gamma}^{ij}-\frac{1}{4}C_{k}C^{k}\dot{\gamma}\dot{\gamma}^{ij}-C^{k}C^{(i}\dot{\gamma}^{j)l}\dot{\gamma}_{kl}$
$\displaystyle+\frac{1}{2}C_{k}C^{(i}\dot{\gamma}^{j)k}\dot{\gamma}-\frac{1}{2}C_{k}C_{l}\dot{\gamma}^{k(i}\dot{\gamma}^{j)l}+\frac{1}{2}C_{k}C^{k}\dot{\gamma}^{l(i}\dot{\gamma}^{j)}{}_{l}$
$\displaystyle-\Gamma^{(ij)\,kl}\Big{(}C_{k}\ddot{C}_{l}+\frac{1}{2}\dot{C}^{m}(C_{m}\dot{\gamma}_{kl}-2C_{k}\dot{\gamma}_{lm})+\dot{C}_{k}\dot{C}_{l}\Big{)}+2\dot{\psi}(\frac{\gamma^{ij}}{2}\dot{C}_{k}C^{k}-\dot{C}^{(i}C^{j)})$
$\displaystyle+\frac{1}{4}(\gamma^{ij}C_{k}C^{k}-2C^{i}C^{j})\dot{\psi}^{2}-e^{2\psi}C_{[k}\dot{C}_{l]}(\frac{\gamma^{ij}}{2}C^{k}\dot{C}^{l}-2\gamma^{ik}C^{[j}\dot{C}^{l]})$
$\displaystyle-\frac{e^{-2\psi}}{2}\gamma^{ij}\Big{(}\frac{\dot{\gamma}^{2}}{4}-\frac{3}{4}\dot{\gamma}_{kl}\dot{\gamma}^{kl}-2\dot{\psi}\dot{\gamma}+\frac{5}{2}\dot{\psi}^{2}\Big{)}$
$\displaystyle-\frac{e^{-2\psi}}{2}\Big{(}\dot{\gamma}^{k(i}\dot{\gamma}^{j)}{}_{k}+(2\dot{\psi}-\frac{\dot{\gamma}}{2})\dot{\gamma}^{ij}-2\gamma^{ij}\ddot{\psi}-\Gamma^{ij\,kl}\ddot{\gamma}_{kl}\Big{)}$
$\displaystyle-\frac{\gamma^{ij}}{2}\Big{(}\Gamma^{kl\,mn}(C_{k}\dot{C}_{l}\dot{\gamma}_{mn}+C_{k}C_{l}\ddot{\gamma}_{mn})-\frac{1}{4}C_{k}C^{k}\dot{\gamma}^{2}+C_{k}C_{l}\dot{\gamma}\dot{\gamma}^{kl}\Big{)}$
$\displaystyle-\frac{\gamma^{ij}}{2}\Big{(}\frac{3}{4}C_{k}C^{k}\dot{\gamma}^{lm}\dot{\gamma}_{lm}-\frac{3}{2}C^{k}C^{l}\dot{\gamma}_{km}\dot{\gamma}^{m}{}_{l}\Big{)}\,.$

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