# An Inner Product for 4D Quantum Gravity and the Chern-Simons-Kodama State

Stephon Alexander<EMAIL_ADDRESS>Brown Theoretical Physics
Center Department of Physics, Brown University, Providence, RI 02912, USA
Laurent Freidel<EMAIL_ADDRESS>Perimeter Institute for
Theoretical Physics, 31 Caroline Street North, Waterloo, Ontario, Canada N2L
2Y5  Gabriel Herczeg111Corresponding author<EMAIL_ADDRESS>Brown
Theoretical Physics Center Department of Physics, Brown University,
Providence, RI 02912, USA

###### Abstract

We demonstrate that reality conditions for the Ashtekar connection imply a
non-trivial measure for the inner product of gravitational states in the
polarization where the Ashtekar connection is diagonal, and we express the
measure as the determinant of a certain first-order differential operator.
This result opens the possibility to perform a non-perturbative analysis of
the quantum gravity scalar product. In this polarization, the Chern-Simons-
Kodama state, which solves the constraints of quantum gravity for a certain
factor ordering, and which has de Sitter space as a semiclassical limit, is
perturbatively non-normalizable with respect to the naïve inner product. Our
work reopens the question of whether this state might be normalizable when the
correct non-perturbative inner product and choice of integration contour are
taken into account. As a first step, we perform a semi-classical treatment of
the measure by evaluating it on the round three-sphere, viewed as a closed
spatial slice of de Sitter. The result is a simple, albeit divergent, infinite
product that might serve as a regulator for a more complete treatment of the
problem. Additionally, our results suggest deep connections between the
problem of computing the norm of the CSK state in quantum gravity and
computing the Chern-Simons partition function for a complex group.

###### Contents

  1. I Introduction
  2. II Review of the CSK State
  3. III Inner product from reality conditions
  4. IV Evaluation of the effective action
  5. V Discussion
  6. VI Acknowledgments
  7. A Evaluation of variations
  8. B Spin connection
  9. C Calculation of the pseudo-determinant of $D_{\Gamma}$ on $\mathbb{S}^{3}$

##  I Introduction

Only a handful of attempts to unite quantum mechanics with gravity rely solely
on consistency with the principles of general relativity and quantum
mechanics. Moreover, astrophysical observations can inform which theory of
quantum gravity is correct, where black holes and the early universe have
provided some guidance. For example, cosmic inflation is a phase of the early
universe that is dominated by a constant part of the energy-momentum tensor,
which is indistinguishable from a cosmological constant. Furthermore, our
current universe is dominated by a cosmological constant, $\Omega_{\Lambda}$.
Both situations suggest the semiclassical state of our past and present
universe consists of a de Sitter geometry.

The paradigm of cosmic inflation relies on quantum fluctuations of the metric
in de Sitter space to seed the observed anisotropies in the CMB and the
subsequent cosmic large-scale structure. While the observed amplitudes of
these super-horizon quantum fluctuations are perturbative and infrared, they
emerge from a trans-Planckian epoch at the beginning of inflation where
quantum effects in gravity are important and non-perturbative BM . The
assumption of quantum field theory in a fixed semiclassical spacetime is no
longer valid in this setting, and one needs a well-controlled quantum
gravitational description of nature to understand the nature of these trans-
Planckian degrees of freedom.

It has been known for some time that the Ashtekar variables provide a gauge-
theoretic framework that recovers the full phase space of general relativity.
In these variables, Kodama found an exact solution to the Wheeler-de Witt
equation and the other constraints of general relativity Kodama90 .
Remarkably, this Chern-Simons-Kodama (CSK) state yields a semiclassical
solution that describes gravitons propagating on an inflating (de Sitter)
background freidel2004linearization ; Smolin02 . It was shown in
Alexander:2003wb  that this framework can be extended to inflation driven by a
scalar field. Moreover, it has been established that the CSK state reproduces
both the Vilenkin and Hartle-Hawking wavefunctions of the universe under
Fourier transform depending on the choice of integration contour for the
Ashtekar variable magueijo2020equivalence ; alexander2021generalized ;
magueijo2021real ; paternoga2000triad . These cosmological solutions were
extended to include fermions, leading to a resolution of the big-bang
singularityHoward . These results promise a valuable connection between a non-
perturbative statement of quantum gravity and the real universe contained as a
semiclassical approximation. However, some serious problems linger with the
CSK state: it is not invariant under large gauge transformations or CPT
transformations—the latter of which precludes energy positivity—and perhaps
worst of all, it is not normalizable with respect to the standard inner
product Witten03 ; marugan1995exponential ; freidel2004linearization .

While the full Lorentzian CSK state requires a non-perturbative inner product,
previous studies have demonstrated that the linearized CSK state is non-
normalizable. The mechanism for the instability is simple: The perturbations
split into positive and negative helicity states and while the positive
helicity states have positive energy, the negative helicity ones correspond to
negative energy states freidel2004linearization . One can also speculate that
the non-normalizibility of the CSK state is related to the issue of the
quantum infrared-driven instability of de Sitter spacePolyakov ; Tsamis:1996qm
; Anderson ; Akh . Subsequent studies argued that the negative frequency
tensor perturbations about the CSK state correspond to anti-gravitonsJB ; BJ .
Half of the full set of these modes corresponds to negative energy states. The
authors propose a physical condition that sub-horizon momentum modes that
propagate perpendicular to their polarization should only exit, leading to the
projecting out of negative energy modes222This physical condition translates
into no graviton-mediated particle production for sub-horizon modes. However,
this condition of particle production will exist for super-horizon modes. .
This condition is consistent with the CMB fluctuations considered in standard
cosmological perturbation theory, and defines a positive energy, linearized
physical inner product for gravitons. Nonetheless, if these negative energy
states do persist beyond quadratic order, it is imperative to identify a non-
perturbative inner product, if one exists.

The results we present in this article do not fully resolve these issues.
However, by identifying an alternative, non-perturbative inner product that
follows naturally from reality conditions on the quantum Ashtekar connection,
we pave the way for a non-perturbative analysis of these questions. In
particular, the possibility to choose integration contours different from the
naive ones may offer a possible resolution to the issue of non-normalizability
of the CSK state. In section II, we provide a review of the CSK state from a
covariant formulation. In section III, we define the new inner product and
prove that it implies the reality conditions of the connection operator. In
section IV, we compute the determinant of the operator that determines the
effective action. We conclude with a discussion of open issues and future
directions to be pursued.

##  II Review of the CSK State

Here we give a brief review of the CSK state, and show that it is annihilated
by the quantum Hamiltonian constraint for a particular choice of factor
ordering. Much of what appears in this section can also be found in Howard
and similar discussions appear in many other places, but we repeat it here for
the purposes of making the article as self-contained as possible.333Note that
the definition of the Planck length $\ell_{\mathrm{Pl}}$ used in this section
and throughout the article differs from the one used in Howard  by a constant
numerical factor, a choice that we make for the sake of notational
convenience. Our starting point in deriving the CSK state is the four-
dimensional Holst Action, which is an extension of the Einstein-Cartan action
of General Relativity, with a topological term:

$S=\int_{{\cal M}_{4}}[*(e^{I}\wedge e^{J})\wedge R_{IJ}+{\rm i}\hskip
1.42262pte^{I}\wedge e^{J}\wedge R_{IJ}]\,,$ (1)

where $*$ is the Hodge dual, the first term is the Einstein-Cartan action, and
the second is the Holst term, which is, after integration by part,
proportional to the first Bianchi identity. Gravitational dynamics on a four-
dimensional manifold ${\cal M}_{4}$ is described by the frame field
$e_{\mu}^{I}(x)$, mapping a vector $v^{\mu}$ in the tangent space of ${\cal
M}_{4}$ at the point $x$ into Minkowski spacetime $M_{4}$ (with metric
$\eta_{IJ}={\rm diag}(-1,1,1,1)_{IJ}$). In this formulation, the Lorentz
connection $\omega_{\mu I}{}^{J}$ is initially taken to be independent of the
frame field. Its equation of motion imposes that it becomes torsionless.

Here we are interested in the Hamiltonian formulation in Ashtekar variables.
In the Hamiltonian formulation one chooses a codimension one slice $\Sigma$ of
spacetime and expresses the gravitational phase space variables as pull-backs
on the hypersurface. One also imposes the normal gauge $e^{0}_{a}=0$, where
$a,b,c$ denote indices tangent to $\Sigma$. This forces the coframe $e^{0}$ to
be the hypersurface normal. The frame field variable is then encoded into
their value on $\Sigma$. It is convenient to repack this information into the
densitized triad defined by
$E^{a}_{i}=\frac{1}{2}\epsilon_{ijk}\epsilon^{abc}e^{j}_{b}e^{k}_{c}$. This
quantity is conjugate to the pull-back of the self-dual connection on $\Sigma$

$A_{a}^{i}(x)\equiv-\tfrac{1}{2}\epsilon^{ij}_{\phantom{ij}k}\omega_{aj}^{\phantom{aj}k}-{\rm
i}\hskip 1.42262pt\omega_{a0}^{\phantom{a0}i}.$ (2)

Since the Lorentz connection (and, in particular, the spin connection
$\Gamma_{\alpha}^{i}\equiv-\tfrac{1}{2}\epsilon^{ij}_{\phantom{ij}k}\omega_{\alpha
j}^{\phantom{\alpha j}k}$) is real in real gravity, $A$ is complex-valued and
obeys a non-trivial reality condition. After imposition of the torsion
constraints, the real part of $A$ is determined by the hypersurface triad,
while the imaginary part is proportional to the extrinsic curvature of
$\Sigma$ thiemann2008modern . In other words we have the reality condition

$\bar{A}_{a}^{i}+A_{a}^{i}=2\Gamma_{a}^{i}[E]\,,$ (3)

where $\bar{A}_{\alpha}^{i}$ denotes the complex conjugate of
${A}_{\alpha}^{i}$, and $\Gamma[E]$ denotes the triad spin connection which
solves the equation

${\rm d}e^{i}+\epsilon^{i}{}_{jk}\Gamma^{j}[E]\wedge e^{k}=0.$

The Poisson bracket of the elementary variables $A$ and $E$ is

$\\{A_{a}^{i}({\bf x}),\,E_{j}^{b}({\bf y})\\}={\rm i}\hskip
1.42262ptG\delta_{a}^{b}\delta_{j}^{i}\delta({\bf x}-{\bf y})\,.$ (4)

Introducing the “magnetic” field and the gauge field strength

$\displaystyle B^{ai}$ $\displaystyle\equiv$
$\displaystyle\tfrac{1}{2}\epsilon^{abc}F_{bc}^{i}\,,$ (5) $\displaystyle
F_{ab}^{k}$ $\displaystyle=$
$\displaystyle\partial_{a}A_{b}^{k}-\partial_{b}A_{a}^{k}+\epsilon_{ij}\phantom{}\phantom{}^{k}A_{a}^{i}A_{b}^{j}\,,$
(6)

one can show that the Hamiltonian scalar constraint following from Eq. (1) is

$\displaystyle{\cal H}$ $\displaystyle\equiv$
$\displaystyle\epsilon_{ijk}E^{i}\cdot
E^{j}\times\left(B^{k}+\frac{\Lambda}{3}E^{k}\right)$ (7) $\displaystyle=$
$\displaystyle\epsilon_{ijk}E^{ai}E^{bj}\left(F_{ab}^{k}+\frac{\Lambda}{3}\epsilon_{abc}E^{ck}\right)=0,$

where $\Lambda$ is a cosmological constant (of any sign) and $\times$ is the
vector spatial product defined as $({\bf a}\times{\bf
b})_{a}=\epsilon_{abc}a^{b}b^{c}$. In exterior algebra notation, the gauge
field is $A\equiv A_{\alpha}{\rm d}x^{\alpha}\equiv A_{\alpha}^{i}\tau_{i}{\rm
d}x^{\alpha}$ ($\tau_{i}$ being an $su(2)$ generator), the covariant
derivative is $D={\rm d}+A\wedge$, and Eq. (6) can be compactly recast as
$F={\rm d}A+A\wedge A$. Under a local gauge transformation, the Ashtekar
connection transforms as

$A\rightarrow A^{\prime}=gAg^{-1}-g^{-1}{\rm d}g\,.$ (8)

The full invariance group of the theory is the semidirect product of the
diffeomorphism and gauge groups. Invariance under small gauge transformations
is guaranteed by the Gauss constraint:

${\cal G}_{i}\equiv
D_{a}E^{a}_{i}=\partial_{a}E^{a}_{i}+\epsilon_{ijk}A_{a}^{j}E^{ak}=0,$ (9)

while spatial diffeomorphism invariance is imposed by the vector constraint

${\cal V}_{a}\equiv(E_{i}\times B^{i})_{a}=F_{ab}^{i}E_{i}^{b}=0\,.$ (10)

The total Hamiltonian is a linear combination of the constraints: up to
constants, we have

$H=G^{-1}\int_{\Sigma}{\rm d}^{3}x(N{\cal H}+\lambda^{j}{\cal
G}_{j}+N^{a}{\cal V}_{a}),$ (11)

where $\Sigma$ is the spatial submanifold and $N$, $\lambda^{j}$, and
$N^{\alpha}$ are Lagrange multipliers.

Now let us construct the CSK state by solving the Wheeler-DeWitt equation. We
have the Hamiltonian (constraint)

$\displaystyle{\cal
H}_{WdW}=\epsilon_{ijk}E^{ai}E^{bj}(F^{k}_{ab}+\frac{\Lambda}{3}\epsilon_{abc}E^{ck})=0,$
(12)

which acts on some wave function $\psi[A]$, and we want to find the form of
$\psi[A]$ that satisfies the above relation. Applying the standard canonical
quantization procedure, we get

$\displaystyle\hat{E}^{ai}\rightarrow\ell^{2}_{\mathrm{Pl}}\frac{\delta}{\delta
A_{ai}},$ (13)

where $\ell^{2}_{\mathrm{Pl}}=\hbar G$. The annihilation of the quantum state
becomes

$\displaystyle\widehat{\cal{H}}_{WdW}\,\psi[A]=\ell^{4}_{\mathrm{Pl}}\epsilon_{ijk}\frac{\delta}{\delta
A_{ai}}\frac{\delta}{\delta
A_{bj}}\Big{(}F^{k}_{ab}+\frac{\ell^{2}_{\mathrm{Pl}}\Lambda}{3}\epsilon_{abc}\frac{\delta}{\delta
A_{ck}}\Big{)}\psi[A]=0.$ (14)

Of course, the Wheeler-DeWitt equation will be satisfied if the self-duality
condition

$\Big{(}F^{k}_{ab}+\frac{\ell^{2}_{\mathrm{Pl}}\Lambda}{3}\epsilon_{abc}\frac{\delta}{\delta
A_{ck}}\Big{)}\psi[A]=0$ (15)

is satisfied, and the latter equation has a unique solution:

$\Psi_{CSK}[A]=\mathcal{N}e^{-\frac{3}{\ell^{2}_{\mathrm{Pl}}\Lambda}Y_{CS}[A]},$
(16)

where $Y_{CS}[A]$ is the Chern-Simons functional

$Y_{CS}[A]=\int\mathrm{tr}\left(\tfrac{1}{2}A\wedge dA+\tfrac{1}{3}A\wedge
A\wedge A\right).$ (17)

##  III Inner product from reality conditions

To summarize, we work, at the quantum level, in the polarisation where the
Ashtekar connection is diagonal. In this polarisation, the states are
$\Psi(A)$ and the operators are given by

$\hat{E}^{a}_{i}\Psi=\ell^{2}_{\mathrm{Pl}}\frac{\delta\Psi}{\delta
A_{a}^{i}},\qquad\hat{A}_{a}^{i}\Psi(A)={A}_{a}^{i}\Psi(A).$ (18)

We now want to implement the reality conditions at the quantum level

$\displaystyle(\hat{E}^{a}_{i})^{\dagger}$ $\displaystyle=$
$\displaystyle\hat{E}^{a}_{i},$ (19)
$\displaystyle\hat{A}_{a}^{i}+(\hat{A}_{a}^{i})^{\dagger}$ $\displaystyle=$
$\displaystyle 2\hat{\Gamma}_{a}^{i}.$ (20)

Here $\hat{\Gamma}_{a}^{i}:=\Gamma_{a}^{i}(\hat{E})$ is the operator
corresponding to the 3d spin connection $\Gamma_{a}^{i}(E)$ which is the spin
connection preserving the frame $e$:

$\mathrm{d}e^{i}+[\Gamma,e]^{i}=0.$ (21)

The Lie algebra commutators are SL$(2,\mathbb{C})$ commutators given by
$[A,B]^{i}:=\epsilon^{i}_{\,jk}A^{j}B^{k}.$

The main claim of this article is that the scalar product solving the reality
condition is given by

$\langle\Psi|\Psi\rangle=\int_{{\cal{C}}\times\bar{\cal{C}}}\\!\\!\\!DAD\bar{A}\,\,\exp\left({-S(\Re
A)}\right)\left|\Psi(A)\right|^{2},$ (22)

where we denote by $\Re A=\frac{1}{2}(A+\bar{A})$, the real part of the self-
dual connection $A$. We assume that the integration contours $\cal C$ and
$\bar{\cal C}$ for the holomorphic and antiholomorphic components allow
integration by parts: i.e., they are such that

$\int_{\cal C}DA\frac{\delta\Psi(A)}{\delta A_{a}^{i}}=0=\int_{\bar{\cal
C}}D\bar{A}\frac{\delta\bar{\Psi}(\bar{A})}{\delta\bar{A}_{a}^{i}}\,.$ (23)

It is important to note that these are the only conditions necessary for our
proof of the reality condition, leaving us some freedom in the choice of
contours that could be used to fix the normalizability issue. Finally, $S(\Re
A)$ is an effective action defined by the following path integral

$\exp\left(-S(\Re A)\right):=\int
DE\,\exp\left(-\frac{1}{\ell_{\mathrm{Pl}}^{2}}\int e_{i}\wedge\mathrm{d}_{\Re
A}e^{i}\right),$ (24)

where $\mathrm{d}_{A}e^{i}=\mathrm{d}e^{i}+[A,e]^{i}$. The main argument is
that the first reality condition (19) forces the measure to depend only on
$\Re A$, while the second reality condition (20) uniquely determines the form
of the effective action $S(\Re A)$.

The detailed proof of the statement goes as follows: first one uses that

$\displaystyle\frac{\delta}{\delta
A_{a}^{i}}\left(\int_{\Sigma}e_{i}\wedge\mathrm{d}_{\Re A}e^{i}\right)$
$\displaystyle=$ $\displaystyle
E_{i}^{a}=\frac{\delta}{\delta\bar{A}_{a}^{i}}\left(\int_{\Sigma}e_{i}\wedge\mathrm{d}_{\Re
A}e^{i}\right),$ (25) $\displaystyle\frac{\delta}{\delta
E_{i}^{a}}\left(\int_{\Sigma}e_{i}\wedge\mathrm{d}_{\Re A}e^{i}\right)$
$\displaystyle=$ $\displaystyle A_{a}^{i}+\bar{A}_{a}^{i}-2\Gamma_{a}^{i}(E).$
(26)

A proof of the second equality is given in appendix A. From this and the
definition (18) of the operators $(\hat{A}_{a}^{i},\hat{E}^{a}_{i})$, we get
that

$\left(\frac{\delta}{\delta\bar{A}_{a}^{i}}-\frac{\delta}{\delta
A_{a}^{i}}\right)e^{-S(\Re A)}=0,$ (27)

and

$\displaystyle\left[A_{a}^{i}+\bar{A}_{a}^{i}-2\hat{\Gamma}_{a}^{i}(-\hat{E})\right]e^{-S(\Re
A)}$ $\displaystyle=$ $\displaystyle\int
DE((A_{a}^{i}+\bar{A}_{a}^{i}-2\hat{\Gamma}_{a}^{i}(E))\exp\left(-\frac{1}{\ell_{\mathrm{Pl}^{2}}}\int_{\Sigma}e_{i}\wedge\mathrm{d}_{\Re
A}e^{i}\right)$ (28) $\displaystyle=$ $\displaystyle\int
DE\frac{\delta}{\delta{E}^{a}_{i}}\exp\left(-\frac{1}{\ell_{\mathrm{Pl}^{2}}}\int_{\Sigma}e_{i}\wedge\mathrm{d}_{\Re
A}e^{i}\right)=0.$ (29)

From which it follows simply that

$\displaystyle\langle\Psi|\hat{E}^{a}_{i}-(\hat{E}^{a}_{i})^{\dagger}|\Psi\rangle$
$\displaystyle=$
$\displaystyle\ell^{2}_{\mathrm{Pl}}\int_{{\cal{C}}\times\bar{\cal{C}}}DAD\bar{A}\,e^{-S(\Re
A)}\left(\Psi(\bar{A})\frac{\delta\Psi(A)}{\delta
A_{a}^{i}}-\frac{\delta\Psi(\bar{A})}{\delta\bar{A}_{a}^{i}}\Psi(A)\right)$
(30) $\displaystyle=$
$\displaystyle\ell^{2}_{\mathrm{Pl}}\int_{{\cal{C}}\times\bar{\cal{C}}}DAD\bar{A}\,\left(\frac{\delta
e^{-S(\Re A)}}{\delta\bar{A}_{a}^{i}}-\frac{\delta e^{-S(\Re A)}}{\delta
A_{a}^{i}}\right)|\Psi(A)|^{2}=0,$ (31)

and that

$\displaystyle\langle\Psi|\hat{A}_{a}^{i}+(\hat{A}_{a}^{i})^{\dagger}|\Psi\rangle$
$\displaystyle=$
$\displaystyle\int_{{\cal{C}}\times\bar{\cal{C}}}DAD\bar{A}\,(A_{a}^{i}+\bar{A}_{a}^{i})e^{-S(\Re
A)}|\Psi(A)|^{2}$ (32) $\displaystyle=$
$\displaystyle\int_{{\cal{C}}\times\bar{\cal{C}}}DAD\bar{A}\,\left(2{\Gamma}_{a}^{i}(-\hat{E})e^{-S(\Re
A)}\right)|\Psi(A)|^{2}.$ (33)

In the last line, the operator $\hat{E}_{i}^{a}$ can be represented as either
$\ell^{2}_{\mathrm{Pl}}\frac{\delta}{\delta A_{a}^{i}}$ or
$\ell^{2}_{\mathrm{Pl}}\frac{\delta}{\delta\bar{A}_{a}^{i}}$ since we have the
Hermiticity condition (27). Integrating by parts, we obtain that this is equal
to

$\int_{{\cal{C}}\times\bar{\cal{C}}}DAD\bar{A}\,e^{-S(\Re
A)}\Psi^{*}(\bar{A})\left(2{\Gamma}_{a}^{i}(\hat{E})\Psi(A)\right).$ (34)

This concludes the proof. We see that the inner product involves an effective
action $S(\Gamma)$, which is evaluated as the log of a determinant in the next
section. It also involves a choice of closed contours
$(\cal{C},\bar{\cal{C}})$ which affects the definition of the product.

##  IV Evaluation of the effective action

In this section, we seek to understand the properties of the effective action
$S(\Gamma)$, where $\Gamma:=\Re A$ denotes the real part of the self-dual
connection. In order to do so, it is convenient to restrict our analysis to
connections $\Gamma$ which are non-degenerate. These satisfy the non-
degeneracy condition

$\nu_{abc}B^{a}_{i}(\Gamma)B^{b}_{j}(\Gamma)B^{c}_{k}(\Gamma)\epsilon^{ijk}\neq
0,$ (35)

where $\nu_{abc}=\nu\epsilon_{abc}$, with $\nu>0$, is a volume form and
$\nu_{abc}B^{c}_{i}(\Gamma):=R_{ab}^{j}(\Gamma)\delta_{ij}$ is the magnetic
field associated with the curvature of $\Gamma$.

In appendix B we show that when the connection is non-degenerate, there exists
a frame field $e_{\Gamma}$ such that

$F(\Gamma)=\pm\frac{1}{2}[e_{\Gamma},e_{\Gamma}],$ (36)

where $\pm$ is the sign of (35). The frame field $e_{\Gamma}$ is uniquely
defined by this condition.

Since the action $\int(e_{i}\wedge\mathrm{d}_{\Gamma}e^{i})$ is purely
quadratic, the effective action simply computes the determinant of
$\mathrm{d}_{\Gamma}$. More precisely, the operator
$\mathrm{d}_{\Gamma}:\Omega_{1}(\mathfrak{su}(2))\to\Omega_{2}(\mathfrak{su}(2))$
maps Lie algebra valued one-forms onto Lie algebra valued two forms. We can
compose this operation with the Hodge duality operator
$*_{\Gamma}:\Omega_{2}(\mathfrak{su}(2))\to\Omega_{1}(\mathfrak{su}(2))$ which
is constructed from the metric
$(q_{\Gamma})_{ab}=(e_{\Gamma})_{a}^{i}(e_{\Gamma})_{b}^{j}\delta_{ij}$. This
gives an endomorphism
$D_{\Gamma}:\Omega_{1}(\mathfrak{su}(2))\to\Omega_{1}(\mathfrak{su}(2))$ where
$D_{\Gamma}=*_{\Gamma}\mathrm{d}_{\Gamma}$, which is explicitely given by

$D_{\Gamma}:e_{a}^{i}\to(D_{\Gamma}e)_{a}^{i}=\nu_{a}{}^{bc}\nabla^{\Gamma}_{b}e_{c}^{i}\,.$
(37)

Here
$\nu_{\Gamma}=\nu_{abc}\,\mathrm{d}x^{a}\wedge\mathrm{d}x^{b}\wedge\mathrm{d}x^{c}$
is the measure associated with $e_{\Gamma}$, while spacetime indices are
raised with the metric $q_{\Gamma}$. Given this definition, we then get that

$e^{S(\Gamma)}=\frac{\mathrm{det}(D_{\Gamma})}{\mathrm{det}(e_{\Gamma})}.$
(38)

The goal is now to evaluate this determinant in a homogeneously curved
background where $(\Gamma_{R})_{a}^{i}:=R\delta_{a}^{i}$ is constant. The
simplest background that we can consider is the flat slicing of de Sitter. In
this case, the determinant can be computed by decomposing the fields in
Fourier modes. However, in a flat background $D_{\Gamma}$ is completely
singular, so we consider a closed slicing of de Sitter with spacetime metric

$ds^{2}=-dt^{2}+a(t)^{2}d\Omega_{3}^{2}$ (39)

where
$a(t)^{2}=\frac{3}{\Lambda}\cosh^{2}\left(\sqrt{\frac{\Lambda}{3}}t\right)$,
and $d\Omega_{3}^{2}$ is the round metric on the three-sphere
$\mathbb{S}^{3}$. A convenient orthonormal frame for the unit three-sphere is
given by the left-invariant one-forms of SU(2)

$\displaystyle e_{0}^{1}$ $\displaystyle=$ $\displaystyle\frac{1}{2}(\cos\psi
d\theta+\sin\psi\sin\theta d\phi)$ $\displaystyle e_{0}^{2}$ $\displaystyle=$
$\displaystyle\frac{1}{2}(\sin\psi d\theta-\cos\psi\sin\theta d\phi)$
$\displaystyle e_{0}^{3}$ $\displaystyle=$
$\displaystyle\frac{1}{2}(d\psi+\cos\theta d\phi).$

This frame satisfies $de_{0}^{i}=\epsilon^{i}_{\,jk}e_{0}^{j}\wedge
e_{0}^{k}.$ We can define a physical frame $e^{i}=a(t)e_{0}^{i}$, which
satisfies $de^{i}=\frac{1}{a}\epsilon^{i}_{\,jk}e^{j}\wedge e^{k}$, from which
we can read off the spin connection
$\omega^{i}_{\,j}=\frac{1}{a}\epsilon^{i}_{\,jk}e^{k}$. Using
$\Gamma^{i}=-\frac{1}{2}\epsilon^{ijk}\omega_{jk}$ leads to

$\Gamma^{i}=-\frac{1}{a}e^{i}.$

In this coordinate system, the extrinsic curvature is just the time derivative
of the spatial metric, so we have $K_{ij}=2H\delta_{ij}$, with
$H=\frac{\dot{a}}{a}$, which leads to

$K^{i}=2He^{i}.$

So the Ashtekar connection for de Sitter in closed slicing is

$A^{i}=\left(-\frac{1}{a}+2iH\right)e^{i},$

or

$A^{i}=\left(-1+2i\dot{a}\right)e_{0}^{i}.$

It can be verified that the associated spin connection satisfies the non-
degenaracy condition (35), so its existence is an important check on the self-
consistency of our approach.

Now, we would like to compute the determinant of the operator $D_{\Gamma}$
defined by

$D_{\Gamma}X^{i}{}_{a}=\epsilon_{a}{}^{bc}\nabla_{b}X^{i}{}_{c}.$ (40)

Since any Lie algebra-valued one-form can be identified with a covariant, rank
two tensor via ${X_{ab}=e_{ia}X^{i}{}_{b}}$, we can equivalently view
$D_{\Gamma}$ as an endomorphism on rank two tensors:

$D_{\Gamma}X_{ab}=\epsilon_{a}{}^{cd}\nabla_{c}X_{db}.$ (41)

Note that this operator has a kernel since $D_{\Gamma}g_{ab}=0$. We therefore
expect that its determinant is zero, and we verify this explicitly below.
However, we can still calculate a non-zero pseudo-determinant for this
operator by considering the product of its non-zero eigenvalues, and this is
what we would expect the path integral to compute. The details of this
procedure can also be found in appendix C, where we diagonalize $D_{\Gamma}$
by expanding it in vector and tensor harmonics on $\mathbb{S}^{3}$. The three-
sphere spherical Harmonics $Y_{k\ell m}$ depend on 3 integers $k,\ell$, and
$m$, with $k\geq\ell\geq 0$ and $\ell\geq m\geq-\ell$. In coordinates where
the three-sphere metric is
$\mathrm{d}s^{2}=\mathrm{d}\chi^{2}+\sin^{2}\chi(\mathrm{d}\theta^{2}+\sin^{2}\theta\mathrm{d}\varphi^{2})$,
we have that $(\ell,m)$ are the usual two-sphere spherical harmonics that
controls the dependence on $(\theta,\varphi)$, while $k$ is the quantum number
that determines the value of the Laplacian and the dependence on the angle
$\chi$. On the unit sphere we have $\nabla_{a}\nabla^{a}Y_{k\ell
m}=-k(k+2)Y_{k\ell m}$.

To diagonalize $D_{\Gamma}$ we have to expand the 2-tensors $X_{ab}$ in terms
of 2-tensor Harmonics. Since there are 9 such harmonics, we have nine types of
expansion coefficients $\tilde{X}_{(A)}^{{k\ell m}}$, $A=0,\ldots,8$ labelled
by the spherical harmonics numbers $(k,\ell,m)$. One finds that like the
Laplacian, the operator $\tilde{D}_{\Gamma}$—i.e., the coefficients of the
operator $D_{\Gamma}$ written in the basis of tensor harmonics on the round
three-sphere—is simply a $9\times 9$ matrix which depends on $k$, but not
$\ell$ or $m$. The main result of appendix C is the following system of linear
equations:

$\displaystyle\tilde{D}_{\Gamma}\tilde{X}^{k\ell m}_{(0)}$ $\displaystyle=$
$\displaystyle-\frac{\sqrt{k(k+2)}}{\sqrt{3}a}\tilde{X}^{k\ell m}_{(6)},$
$\displaystyle\tilde{D}_{\Gamma}\tilde{X}^{k\ell m}_{(1)}$ $\displaystyle=$
$\displaystyle\frac{k+1}{2a}\tilde{X}^{k\ell
m}_{(2)}+\frac{2-(k+1)^{2}}{a\sqrt{2(k-1)(k+3)}}\tilde{X}^{k\ell m}_{(7)},$
$\displaystyle\tilde{D}_{\Gamma}\tilde{X}^{k\ell m}_{(2)}$ $\displaystyle=$
$\displaystyle\frac{k+1}{2a}\tilde{X}^{k\ell
m}_{(1)}+\frac{2-(k+1)^{2}}{a\sqrt{2(k-1)(k+3)}}\tilde{X}^{k\ell m}_{(8)},$
$\displaystyle\tilde{D}_{\Gamma}\tilde{X}^{k\ell m}_{(3)}$ $\displaystyle=$
$\displaystyle\frac{\sqrt{(k-1)(k+3)}}{\sqrt{6}a}\tilde{X}^{k\ell m}_{(6)},$
$\displaystyle\tilde{D}_{\Gamma}\tilde{X}^{k\ell m}_{(4)}$ $\displaystyle=$
$\displaystyle\frac{k+1}{a}\tilde{X}^{{k\ell m}}_{(5)},$
$\displaystyle\tilde{D}_{\Gamma}\tilde{X}^{k\ell m}_{(5)}$ $\displaystyle=$
$\displaystyle\frac{3k(k+2)}{2a(k+1)}\tilde{X}^{k\ell m}_{(4)},$
$\displaystyle\tilde{D}_{\Gamma}\tilde{X}^{k\ell m}_{(6)}$ $\displaystyle=$
$\displaystyle-\frac{2\sqrt{k(k+2)}}{\sqrt{3}a}\tilde{X}^{k\ell
m}_{(0)}-\frac{\sqrt{2(k-1)(k+3)}}{\sqrt{3}a}\tilde{X}^{k\ell m}_{(3)},$
$\displaystyle\tilde{D}_{\Gamma}\tilde{X}^{k\ell m}_{(7)}$ $\displaystyle=$
$\displaystyle-\frac{\sqrt{(k-1)(k+3)}}{\sqrt{2}a}\tilde{X}^{k\ell
m}_{(1)}+\frac{1+k(k+2)}{a(k+1)}\tilde{X}^{k\ell m}_{(8)},$
$\displaystyle\tilde{D}_{\Gamma}\tilde{X}^{k\ell m}_{(8)}$ $\displaystyle=$
$\displaystyle-\frac{\sqrt{(k-1)(k+3)}}{\sqrt{2}a}\tilde{X}^{k\ell
m}_{(2)}+\frac{1+k(k+2)}{a(k+1)}\tilde{X}^{k\ell m}_{(7)},$ (42)

which defines the matrix $\tilde{D}_{\Gamma}$. As promised,
$\tilde{D}_{\Gamma}$ is singular—one of its eigenvalues is zero, while the
remaining eight are non-zero. In particular, the eigenvalues are
$\\{0,\pm\lambda_{1},\pm\lambda_{2},\pm\lambda_{\pm}\\}$, where

$\displaystyle\lambda_{1}$ $\displaystyle=$
$\displaystyle{\frac{1}{a}\sqrt{{k}^{2}+2\,k-1}},$ $\displaystyle\lambda_{2}$
$\displaystyle=$ $\displaystyle{\frac{1}{2\,a}\sqrt{6\,{k}^{2}+12\,k}},$
$\displaystyle\lambda_{\pm}$ $\displaystyle=$
$\displaystyle{\frac{1}{4\,a}\sqrt{18\,{k}^{2}+36\,k+2\pm
6\,\sqrt{9\,{k}^{4}+36\,{k}^{3}+38\,{k}^{2}+4\,k-7}}}\,.$ (43)

The product $\tilde{\Delta}_{k}$ of the eight non-zero eigenvalues has a
simple form:

$\tilde{\Delta}_{k}=\frac{3}{2a^{8}}\left[(k+1)^{2}-1\right]\left[(k+1)^{2}-2\right].$
(44)

The naïve pseudo-determinant of $D_{\Gamma}$ is then a product over $k,\ell,m$
of this quantity:

$\displaystyle\prod_{k\ell
m}\tilde{\Delta}_{k}=\prod_{k=0}^{\infty}\tilde{\Delta}_{k}^{(k+1)^{2}},$ (45)

where we used that the dimension of the $k$ representation is
$\sum_{\ell=0}^{k}(2\ell+1)=(k+1)^{2}$. However, this quantity also vanishes,
since $\tilde{\Delta}_{0}=0$. We therefore truncate the product to positive
values of $k$ to define the pseudo-determinant of $D_{\Gamma}$:

$\psi\\!\det(D_{\Gamma})=\prod_{k=1}^{\infty}\tilde{\Delta}_{k}^{(k+1)^{2}}.$
(46)

Finally, we can re-index the product by defining $n=k+1$ to obtain

$\psi\\!\det(D_{\Gamma})=\prod_{n=2}^{\infty}\left[\frac{3}{2a^{8}}(n^{2}-1)(n^{2}-2)\right]^{n^{2}}.$
(47)

This expression needs to be regulated, as we now discuss.

##  V Discussion

We have shown that a non-trivial inner product measure for the states of four-
dimensional quantum gravity follows from a simple reality condition on the
quantum Ashetkar connection, and a closure assumption about its integration
contour, leaving room for the possibility that the CSK state might be
normalizable when the correct inner product associated with the right choice
of the contour is taken into account. While we have taken some first steps
toward evaluating the inner product on a homogeneously curved slice of de
Sitter space, which is an appropriate setting for studying the CSK state which
has de Sitter as a semi-classical limit, the work we have presented here is
far from conclusive. Indeed, even the simplified problem of studying the inner
product on a homogeneous background proved to be a non-trivial task. While we
have attempted to keep the presentation of the main results as clean as
possible, the careful reader will appreciate that the calculation of the
expansion coefficients in appendix C was a laborious undertaking, and even
then the resulting infinite product is divergent.

Nevertheless, it may be possible to find a finite determinant without
specifying a background by performing a zeta-function regularization for the
determinant of $D_{\Gamma}$. This prescription was first used by Hawking to
regulate the path integral of Euclidean gravity on de Sitter space
Hawking:1976ja . Given that the determinant is in the form of a divergent
product of eigenvalues $\det(D_{\Gamma})=\prod_{n}\lambda_{n}$, one can form a
zeta function from the eigenvalues $\lambda_{n}(D_{\Gamma})$ of the operator
$D_{\Gamma}$:

$\zeta(s,D_{\Gamma})=\sum_{n}\lambda^{-s}_{n}(D_{\Gamma}).$ (48)

One would expect such an expansion to converge for $\Re(s)>2$, and to define
via analytic continuation a meromorphic function of $s$ on the whole complex
plane that is analytic at $s=0$. We will pursue this procedure in an upcoming
work.

Another fascinating possibility is provided by Witten’s fundamental study of
Chern-Simons theory for complex groups Witten:2010cx . From his work, we know
that the proper definition of the Chern-Simons partition function requires a
lucid understanding of what integration contour is used in defining the
theory. From our analysis, it is clear that there is a connection between
computing the norm of the CSK state in quantum gravity and computing the
Chern-Simons partition function for a complex group. We plan to study this
connection in a forthcoming work. The issue of contour integration for the CSK
state was already touched on in magueijo2020equivalence  in a mini-superspace
context.

As we have noted, and as is often the case with operator determinants derived
from path integrals in quantum field theories, the product (47) diverges. In
order to go further, one needs a prescription for either regulating the
operator determinant (such as a zeta-function regularization), or interpreting
it as the divergent part of some physically meaningful quantity. In the
present case, the latter possibility seems promising, since we have only
evaluated $\psi det(D_{\Gamma})$ on a slice of de Sitter space, while the
full, non-perturbative inner product measure requires us to path-integrate
over all possible configurations. It could well be the case that the pseudo-
determinant we have computed for the round three-sphere might serve as a
regulator for a perturbative treatment of the inner product measure expanded
about this homogeneously curved background in the same manner that the path
integral for a quantum field theory without sources serves as a regulator for
the generating functional including sources. However, even a perturbative
treatment of the inner product measure about the round three-sphere presents
significant challenges, since one either needs to develop a perturbation
theory for the vector and tensor harmonics, or else come up with another
approach altogether for computing the measure perturbatively.

Yet another promising direction that could allow us to make progress with
computing the inner product measure is rooted in the Belinskii Khalatnikov
Lifshitz (BKL) conjecture, which states that as one approaches a space-like
singularity, spatial derivatives become negligible belinskii1970oscillatory .
In the context of the Kodama state, this means that only the cubic, non-
abelian part of the action will be non-vanishing. This was already confirmed
in previous work where the Kodama state was evaluated in Bianchi-type
spacetimes and it was shown that only the non-Abelian term in the action
functional is presentalexander2021generalized . Conversely, it was also shown
that when the spacetime is homogeneous and isotropic, the non-Abelian term in
the action vanishes. This is reminiscent of non-abelian projection proposed by
t’ Hooft, to exploit features of confinement of gauge theories tHooft:1981bkw
. We speculate that in the BKL limit, as the spacetime enters a space-like
singularity, this mix-master behavior, proposed to avoid singularities, could
be described by a Chern-Simons matrix theory, which could perhaps be obtained
from our proposed inner product.

Many of the ingredients in the approach we have taken in this article also
appear in a series of papers by Haggard, Han, Kamiński and Riello on
$SL(2,\mathbb{C})$ Chern-Simons theory and four-dimensional quantum gravity
with a cosmological constant haggard2015sl-1 ; haggard2015sl-2 ;
haggard2016four , and it would be interesting to develop those connections
further. Finally, any observer in a universe with a positive cosmological
constant is limited by their cosmological horizon. It would be interesting to
consider how the inner product we have introduced here might affect the
analysis of local sub-systems as in donnelly2016local  and entanglement
entropy donnelly2008entanglement  for the Kodama state.

##  VI Acknowledgments

We are grateful to Jim Gates, Ibou Bah, João Magueijo, Jim Simons, Dan Freed,
Edward Frenkel, Steve Carlip, and Lee Smolin for discussions and helpful
feedback.

Research at Perimeter Institute is partly supported by the Government of
Canada through the Department of Innovation, Science and Economic Development
Canada and the Province of Ontario through the Ministry of Colleges and
Universities.

##  Appendix A Evaluation of variations

In this section we give the proof of the variational derivative (25,26)

Using that $E^{i}=\frac{1}{2}[e,e]^{i}$, we have.

$\displaystyle\delta\left(\int_{\Sigma}e_{i}\wedge\mathrm{d}_{a}e^{i}\right)$
$\displaystyle=$ $\displaystyle 2\int_{\Sigma}\delta
e_{i}\wedge\mathrm{d}_{a}e^{i}+\int_{\Sigma}e_{i}\wedge[\delta a,e]^{i}$ (49)
$\displaystyle=$ $\displaystyle 2\int_{\Sigma}\delta
e_{i}\wedge[a-\Gamma,e]^{i}+\int_{\Sigma}\delta a_{i}\wedge[e,e]^{i}$ (50)
$\displaystyle=$ $\displaystyle 2\int_{\Sigma}[\delta
e,e]_{i}\wedge(a-\Gamma)^{i}+2\int_{\Sigma}\delta a_{i}\wedge E^{i}$ (51)
$\displaystyle=$ $\displaystyle\int_{\Sigma}\delta
E_{i}\wedge(2a-2\Gamma)^{i}+2\int_{\Sigma}\delta a_{i}\wedge E^{i}.$ (52)

we used that

$\delta E^{i}=\delta\left(\frac{1}{2}[e,e]^{i}\right)=[\delta e,e]^{i}.$ (53)

##  Appendix B Spin connection

In this section we show that given a connection $\Gamma$ we can construct a
frame field $e_{\Gamma}$ solution of $\mathrm{d}_{\Gamma}e_{\Gamma}=0$
Herfray:2016std . Given a 3d real connection $\Gamma_{a}^{i}$ we define its
curvature by
$F(\Gamma)=\mathrm{d}\Gamma+\epsilon^{i}{}_{jk}\Gamma^{j}\wedge\Gamma^{k}$.
Given a volume form
$\bm{\nu}=\nu_{abc}\mathrm{d}x^{a}\wedge\mathrm{d}x^{b}\wedge\mathrm{d}x^{c}=\nu\mathrm{d}^{3}x$
we define the tensor $B^{a}_{j}$ to be such that

$F_{ab}^{i}=\nu_{abc}B^{c}_{j}\delta^{ij}.$ (54)

From this we define

${\det}_{\nu}(B):=\frac{1}{6}\nu_{abc}B^{a}_{i}B^{b}_{j}B^{c}_{k}\epsilon^{ijk}.$
(55)

under rescaling of the volume form $\bm{\nu}\to\lambda\bm{\nu}$ we have

$B^{a}_{i}\to\lambda^{-1}B^{a}_{i},\qquad{\det}_{\nu}(B)\to\lambda^{-2}{\det}_{\nu}(B).$
(56)

this means that $s_{\Gamma}=\mathrm{sign}(\det(B))$ is independent of the
choice of background volume form and we define a volume form $\nu_{\Gamma}$ to
be such that

$\nu_{\Gamma}:=\sqrt{|{\det}_{\nu}(B)|}\nu.$ (57)

$\nu_{\Gamma}$ is independent of the background volume form.

We can now define a frame field $e_{\Gamma}$ such that

$F(\Gamma):=\frac{s_{\Gamma}}{2}[e_{\Gamma},e_{\Gamma}].$ (58)

In coordinates this means that

$\hat{e}^{a}_{i}=\frac{\nu}{\nu_{\Gamma}}B^{a}_{i},\qquad\det(e)=s_{\Gamma}\nu_{\Gamma}.$
(59)

where $\hat{e}_{i}$ is the frame field inverse to the form $e^{i}$. This also
means that

$[F_{[ab}(\Gamma),F_{c]d}(\Gamma)]^{k}:=S_{\Gamma}(\bm{\nu}_{\Gamma})_{abc}({e}_{\Gamma})_{d}^{k}$
(60)
$\frac{1}{3}\frac{\nu}{\nu_{\Gamma}}[B^{c}(\Gamma),F_{cd}(\Gamma)]^{k}:=(\hat{e}_{\Gamma})_{d}^{k}$
(61)

##  Appendix C Calculation of the pseudo-determinant of $D_{\Gamma}$ on
$\mathbb{S}^{3}$

In order to diagonalize $D_{\Gamma}$, it’s useful to introduce the vector and
tensor harmonicson $\mathbb{S}^{3}$, which, following the notations and
conventions in lindblom2017scalar , can be defined in terms of first
derivatives of the scalar harmonics $Y^{k\ell m}$ as follows:

$\displaystyle Y^{k\ell m}_{(0)\,a}$ $\displaystyle=$
$\displaystyle\frac{a}{\sqrt{k(k+2)}}\nabla_{a}Y^{k\ell m},$ (62)
$\displaystyle Y^{k\ell m}_{(1)\,a}$ $\displaystyle=$
$\displaystyle\frac{a^{2}}{\sqrt{\ell(\ell+1)}}\epsilon_{a}{}^{bc}\nabla_{b}Y^{k\ell
m}\nabla_{c}\cos\chi,$ (63) $\displaystyle Y^{k\ell m}_{(2)\,a}$
$\displaystyle=$
$\displaystyle\frac{a}{k+1}\epsilon_{a}{}^{bc}\nabla_{b}Y^{k\ell m}_{(1)\,c},$
(64)

where $\epsilon_{abc}$ is the tensor volume element associated with $g_{ab}$,
and the definition of the angle $\chi$ appearing in the definition of
$Y^{k\ell m}_{(1)\,a}$ can be found in lindblom2017scalar , but is not
important for our purposes. The vector harmonics $Y^{k\ell m}_{(A)\,a}$ are
eigenfunctions of the Laplacian on $\mathbb{S}^{3}$ with the following
eigenvalues:

$\displaystyle\nabla^{b}\nabla_{b}Y^{k\ell m}_{(0)\,a}$ $\displaystyle=$
$\displaystyle\frac{2-k(k+2)}{a^{2}}Y^{k\ell m}_{(0)\,a},$ (65)
$\displaystyle\nabla^{b}\nabla_{b}Y^{k\ell m}_{(1)\,a}$ $\displaystyle=$
$\displaystyle\frac{1-k(k+2)}{a^{2}}Y^{k\ell m}_{(1)\,a},$ (66)
$\displaystyle\nabla^{b}\nabla_{b}Y^{k\ell m}_{(2)\,a}$ $\displaystyle=$
$\displaystyle\frac{1-k(k+2)}{a^{2}}Y^{k\ell m}_{(2)\,a}.$ (67)

For future convenience, we also record the divergence of each of the vector
harmonics:

$\displaystyle\nabla^{a}Y^{k\ell m}_{(0)\,a}$ $\displaystyle=$
$\displaystyle-\frac{\sqrt{k(k+2)}}{a}Y^{k\ell m},$ (68)
$\displaystyle\nabla^{a}Y^{k\ell m}_{(1)\,a}$ $\displaystyle=$ $\displaystyle
0,$ (69) $\displaystyle\nabla^{a}Y^{k\ell m}_{(2)\,a}$ $\displaystyle=$
$\displaystyle 0.$ (70)

Now we can express any one-form $V_{a}$ as an expansion in vector harmonics:

$\displaystyle V_{a}=\sum_{Ak\ell m}\tilde{V}^{k\ell m}_{(A)}\,Y^{k\ell
m}_{(A)\,a},\qquad$ (71)

where the range of allowed values for $k$ depends on $A$, the range of allowed
values for $\ell$ depends on $k$, and the range of allowed values for $m$
depends on $\ell$, but the details are not important at this point. Using the
orthonormality relation

$\displaystyle\delta_{(A)(B)}\delta^{\,kk^{\prime}}\delta^{\,\ell\ell^{\prime}}\delta^{\,mm^{\prime}}=\frac{1}{a^{3}}\int\\!\\!g^{ab}\,Y^{\,k\ell
m}_{(A)\,a}\,Y^{\,*k^{\prime}\ell^{\prime}m^{\prime}}_{(B)\,b}\\!\sqrt{g}\,d^{\,3}x,\\!\\!\\!\\!\\!$

the expansion coefficients can be written as

$\displaystyle\tilde{V}^{k\ell m}_{(A)}=\frac{1}{a^{3}}\int V^{a}\,Y^{*k\ell
m}_{(A)\,a}\sqrt{g}\,d^{\,3}x.$ (72)

For future reference, we compute $\epsilon^{abc}\nabla_{a}Y^{k\ell m}_{(A)b}$
for $A=0,1,2$. For $A=0$, we have

$\epsilon^{abc}\nabla_{a}Y^{k\ell
m}_{(0)b}=\epsilon^{abc}\nabla_{a}\Big{(}\frac{a}{\sqrt{k(k+2)}}\nabla_{b}Y^{k\ell
m}\Big{)}=0.$ (73)

For $A=1$, we have

$\epsilon^{abc}\nabla_{a}Y^{k\ell m}_{(1)b}=\frac{k+1}{a}g^{bc}\,Y^{k\ell
m}_{(2)b},$ (74)

where we used equation (64). For $A=2$ we have to do a calculation:

$\displaystyle\epsilon^{abc}\nabla_{a}Y^{k\ell m}_{(2)b}$ $\displaystyle=$
$\displaystyle\frac{a}{k+1}\epsilon^{bca}\epsilon_{b}{}^{de}\nabla_{a}\nabla_{d}Y^{k\ell
m}_{(1)e}$ (75) $\displaystyle=$
$\displaystyle\frac{a}{k+1}\left(\nabla^{e}\nabla^{c}Y^{k\ell
m}_{(1)e}-g^{ce}\nabla^{a}\nabla_{a}Y^{k\ell m}_{(1)e}\right)$
$\displaystyle=$ $\displaystyle\frac{a}{k+1}\left[R^{ec}{}_{e}{}^{f}Y^{k\ell
m}_{(1)f}+\nabla^{c}\nabla^{e}Y^{k\ell
m}_{(1)e}-g^{ce}\left(\frac{1-k(k+2)}{a^{2}}\right)Y^{k\ell m}_{(1)e}\right],$

where in going from the second line to the third line we used equation (66).
Now, the first term in the last line above is just the Ricci tensor, which for
the three-sphere is proportional to the metric:
$R^{ec}{}_{e}{}^{f}=R^{cf}=\frac{2}{a^{2}}g^{cf}$. Meanwhile, the second term
in the last line vanishes due to equation (69). So we are left with

$\epsilon^{abc}\nabla_{a}Y^{k\ell
m}_{(2)b}=\frac{1+k(k+2)}{a(k+1)}g^{ca}Y^{k\ell
m}_{(1)a}=\frac{k+1}{a}g^{ca}Y^{k\ell m}_{(1)a}.$ (76)

To summarize, we have

$\displaystyle\epsilon^{abc}\nabla_{a}Y^{k\ell m}_{(0)b}$ $\displaystyle=$
$\displaystyle 0,$ (77) $\displaystyle\epsilon^{abc}\nabla_{a}Y^{k\ell
m}_{(1)b}$ $\displaystyle=$ $\displaystyle\frac{k+1}{a}g^{ca}Y^{k\ell
m}_{(2)a},$ (78) $\displaystyle\epsilon^{abc}\nabla_{a}Y^{k\ell m}_{(2)b}$
$\displaystyle=$ $\displaystyle\frac{k+1}{a}g^{ca}Y^{k\ell m}_{(1)a}.$ (79)

The vector harmonics $Y^{k\ell m}_{(A)a}$ can also be used to expand two-forms
$\alpha_{ab}=-\alpha_{ba}$:

$\displaystyle\alpha_{ab}=\sum_{Ak\ell m}\\!\\!\tilde{\alpha}^{k\ell
m}_{(A)}\epsilon_{ab}{}^{c}Y^{k\ell m}_{(A)\,c}.\qquad$ (80)

In order to expand general rank-two tensors, we need to introduce the
(symmetric) tensor harmonics:

$\displaystyle Y^{k\ell m}_{(0)\,ab}=\frac{1}{\sqrt{3}}Y^{k\ell m}g_{ab},$
(81) $\displaystyle Y^{k\ell
m}_{(1)\,ab}=\frac{a}{\sqrt{2(k-1)(k+3)}}\left(\nabla_{a}Y^{k\ell
m}_{(1)\,b}+\nabla_{b}Y^{k\ell m}_{(1)\,a}\right),$ (82) $\displaystyle
Y^{k\ell m}_{(2)\,ab}=\frac{a}{\sqrt{2(k-1)(k+3)}}\left(\nabla_{a}Y^{k\ell
m}_{(2)\,b}+\nabla_{b}Y^{k\ell m}_{(2)\,a}\right),$ (83) $\displaystyle
Y^{k\ell
m}_{(3)\,ab}=\frac{\sqrt{3}a}{\sqrt{2(k-1)(k+3)}}\Biggl{(}\nabla_{a}Y^{k\ell
m}_{(0)\,b}+\frac{\sqrt{k(k+2)}}{\sqrt{3}a}Y^{k\ell m}_{(0)\,ab}\Biggr{)},$
(84) $\displaystyle Y^{k\ell
m}_{(4)ab}=a\sqrt{\frac{(\ell-1)(\ell+2)}{2k(k+2)}}\biggl{\\{}\tfrac{1}{2}E^{k\ell}\left(\nabla_{a}F^{\ell
m}_{b}+\nabla_{b}F^{\ell m}_{a}\right)$
$\displaystyle\qquad+\csc^{2}\\!\chi\left[\tfrac{1}{2}(\ell-1)\cos\chi\,E^{k\ell}+C^{k\ell}\right]\left(F^{\ell
m}_{a}\nabla_{b}\cos\chi+F^{\ell
m}_{b}\nabla_{a}\cos\chi\right)\biggr{\\}},\quad$ (85) $\displaystyle Y^{k\ell
m}_{(5)\,ab}=\frac{a}{2(k+1)}\left(\epsilon_{a}{}^{ce}\nabla_{c}Y^{k\ell
m}_{(4)\,eb}+\epsilon_{b}{}^{ce}\nabla_{c}Y^{k\ell m}_{(4)\,ea}\right),\qquad$
(86)

where the quantities $E^{k\ell},C^{k\ell}$, and $F_{a}^{\ell m}$ appearing in
the definition of $Y^{k\ell m}_{(4)\,ab}$ are defined in lindblom2017scalar
(along with the angle $\chi$) but are not important for our purposes.
The symmetric tensor harmonics $Y^{k\ell m}_{(A)\,ab}$ are eigenfunctions of
the Laplacian with the following eigenvalues:

$\displaystyle\nabla^{n}\nabla_{n}Y^{k\ell m}_{(0)\,ab}$ $\displaystyle=$
$\displaystyle-\frac{k(k+2)}{a^{2}}Y^{k\ell m}_{(0)\,ab},$ (87)
$\displaystyle\nabla^{n}\nabla_{n}Y^{k\ell m}_{(1)\,ab}$ $\displaystyle=$
$\displaystyle\frac{5-k(k+2)}{a^{2}}Y^{k\ell m}_{(1)\,ab},$ (88)
$\displaystyle\nabla^{n}\nabla_{n}Y^{k\ell m}_{(2)\,ab}$ $\displaystyle=$
$\displaystyle\frac{5-k(k+2)}{a^{2}}Y^{k\ell m}_{(2)\,ab},$ (89)
$\displaystyle\nabla^{n}\nabla_{n}Y^{k\ell m}_{(3)\,ab}$ $\displaystyle=$
$\displaystyle\frac{6-k(k+2)}{a^{2}}Y^{k\ell m}_{(3)\,ab},$ (90)
$\displaystyle\nabla^{n}\nabla_{n}Y^{k\ell m}_{(4)\,ab}$ $\displaystyle=$
$\displaystyle\frac{2-k(k+2)}{a^{2}}Y^{k\ell m}_{(4)\,ab},$ (91)
$\displaystyle\nabla^{n}\nabla_{n}Y^{k\ell m}_{(5)\,ab}$ $\displaystyle=$
$\displaystyle\frac{2-k(k+2)}{a^{2}}Y^{k\ell m}_{(5)\,ab}.$ (92)

The divergences of the symmetric tensor harmonics are given by

$\displaystyle\nabla^{a}Y^{k\ell m}_{(0)\,ab}$ $\displaystyle=$
$\displaystyle\frac{\sqrt{k(k+2)}}{\sqrt{3}a}Y^{k\ell m}_{(0)\,b},$ (93)
$\displaystyle\nabla^{a}Y^{k\ell m}_{(1)\,ab}$ $\displaystyle=$
$\displaystyle-\frac{\sqrt{(k-1)(k+3)}}{\sqrt{2}a}Y^{k\ell m}_{(1)\,b},$ (94)
$\displaystyle\nabla^{a}Y^{k\ell m}_{(2)\,ab}$ $\displaystyle=$
$\displaystyle-\frac{\sqrt{(k-1)(k+3)}}{\sqrt{2}a}Y^{k\ell m}_{(2)\,b},$ (95)
$\displaystyle\nabla^{a}Y^{k\ell m}_{(3)\,ab}$ $\displaystyle=$
$\displaystyle-\frac{\sqrt{2(k-1)(k+3)}}{\sqrt{3}a}Y^{k\ell m}_{(0)\,b},$ (96)
$\displaystyle\nabla^{a}Y^{k\ell m}_{(4)\,ab}$ $\displaystyle=$ $\displaystyle
0,$ (97) $\displaystyle\nabla^{a}Y^{k\ell m}_{(5)\,ab}$ $\displaystyle=$
$\displaystyle 0,$ (98)

and their traces are

$\displaystyle g^{ab}Y^{k\ell m}_{(A)\,ab}$ $\displaystyle=$
$\displaystyle\left\\{\begin{array}[]{c l}\sqrt{3}Y^{k\ell
m},&\qquad{\scriptstyle A}=0,\\\ 0,&\qquad 1\leq{\scriptstyle A}\leq 5.\\\
\end{array}\right.$ (101)

Now one can expand an arbitrary symmetric tensor $s_{ab}$ on $\mathbb{S}^{3}$
as:

$\displaystyle s_{ab}=\sum_{Ak\ell m}\\!\\!\tilde{s}^{k\ell m}_{(A)}\,Y^{k\ell
m}_{(A)\,ab},$ (102)

where, as in the vector harmonic expansion, the range of allowed values for
$k$ depends on $A$, the range of allowed values for $\ell$ depends on $k$, and
the range of allowed values for $m$ depends on $\ell$, but again, the details
are not important yet. By inserting the orthonormality relations

$\displaystyle\delta_{(A)(B)}\delta^{\,kk^{\prime}}\delta^{\,\ell\ell^{\prime}}\delta^{\,mm^{\prime}}=\frac{1}{a^{3}}\int
g^{ac}g^{bd}\,Y^{\,k\ell
m}_{(A)\,ab}\,Y^{\,*k^{\prime}\ell^{\prime}m^{\prime}}_{(B)\,cd}\\!\sqrt{g}\,d^{\,3}x,\qquad$
(103)

into (102), the expansion coefficients $\tilde{s}^{k\ell m}_{(A)}$ can be
written as

$\displaystyle\tilde{s}^{k\ell m}_{(A)}=\frac{1}{a^{3}}\int
g^{ac}g^{bd}s_{ab}\,Y^{*k\ell m}_{(A)\,cd}\,\sqrt{g}\,d^{\,3}x.$ (104)

Now we are in a position to compute the expansion coefficients of
$D_{\Gamma}X^{i}_{a}$. First, we identify the Lie-algebra valued one-forms
$X^{i}_{a}$ and $D_{\Gamma}X^{i}_{a}$ with rank-two tensors
$X_{ab}=e_{ib}X^{i}_{a}$, $D_{\Gamma}X_{ab}=e_{ib}D_{\Gamma}X^{i}_{a}$. Now we
can calculate:

$\displaystyle\tilde{D}_{\Gamma}\tilde{X}^{k\ell m}_{(0)}$ $\displaystyle=$
$\displaystyle\frac{1}{a^{3}}\int
g^{ac}g^{bd}\epsilon_{a}{}^{\alpha\beta}\nabla_{\alpha}X_{\beta b}Y^{*\,{k\ell
m}}_{(0)\,cd}\sqrt{g}\,d^{3}x$ (105) $\displaystyle=$
$\displaystyle-\frac{1}{a^{3}}\int
g^{ac}g^{bd}\epsilon_{a}{}^{\alpha\beta}X_{\beta
b}\nabla_{\alpha}\\!\left(\tfrac{1}{\sqrt{3}}Y^{*k\ell
m}g_{cd}\right)\\!\sqrt{g}\,d^{3}x$ (106) $\displaystyle=$
$\displaystyle-\frac{1}{a^{3}}\int\frac{\sqrt{k(k+2)}}{\sqrt{3}a}\epsilon^{\alpha\beta
b}X_{\beta b}Y^{*k\ell m}_{(0)\,\alpha}\\!\sqrt{g}\,d^{3}x$ (107)
$\displaystyle=$
$\displaystyle-\frac{\sqrt{k(k+2)}}{\sqrt{3}a}\tilde{\alpha}^{k\ell m}_{(0)}.$
(108) $\displaystyle\\!\\!\\!\\!\\!\tilde{D}_{\Gamma}\tilde{X}^{k\ell
m}_{(1)}$ $\displaystyle=$
$\displaystyle-\frac{1}{a^{3}}\int\epsilon^{c\alpha\beta}X_{\beta}{}^{d}\nabla_{\alpha}Y^{*k\ell
m}_{(1)\,cd}\sqrt{g}\,d^{3}x$ $\displaystyle=$
$\displaystyle-\frac{1}{a^{3}}\int\epsilon^{a\alpha\beta}X_{\beta}{}^{b}\nabla_{\alpha}\left(\frac{a}{\sqrt{2(k-1)(k+3)}}\left(\nabla_{a}Y^{*k\ell
m}_{(1)\,b}+\nabla_{b}Y^{*k\ell m}_{(1)\,a}\right)\right)\\!\sqrt{g}\,d^{3}x$
$\displaystyle=$
$\displaystyle-\frac{1}{a^{3}}\int\frac{a}{\sqrt{2(k-1)(k+3)}}\epsilon^{a\alpha\beta}X_{\beta}{}^{b}\left(-\tfrac{1}{2}R_{\alpha
a}{}^{\lambda}{}_{b}Y^{*\,k\ell
m}_{(1)\,\lambda}+\nabla_{b}\nabla_{\alpha}Y^{*\,k\ell m}_{(1)\,a}-R_{\alpha
b}{}^{\lambda}{}_{a}Y^{*\,k\ell m}_{(1)\,\lambda}\right)\\!\sqrt{g}\,d^{3}x.$

Using
$R_{ab}{}^{c}{}_{d}=\tfrac{1}{a^{2}}\left(\delta^{c}{}_{a}g_{bd}-\delta^{c}{}_{b}g_{ad}\right)$,
the terms with curvature tensors above combine to give
$\frac{2}{a\sqrt{2(k-1)(k+3)}}\tilde{\alpha}^{k\ell m}_{(1)}$, so we have

$\displaystyle\tilde{D}_{\Gamma}\tilde{X}^{k\ell m}_{(1)}=Q_{(1)}^{k\ell
m}+\frac{2}{a\sqrt{2(k-1)(k+3)}}\,\tilde{\alpha}^{k\ell m}_{(1)},$

where

$\displaystyle Q_{(1)}^{k\ell m}$ $\displaystyle=$
$\displaystyle-\frac{1}{a^{3}}\int\frac{a}{\sqrt{2(k-1)(k+3)}}\epsilon^{a\alpha\beta}X_{\beta}{}^{b}\nabla_{b}\nabla_{\alpha}Y^{*\,k\ell
m}_{(1)\,a}\\!\sqrt{g}\,d^{3}x.$ $\displaystyle=$
$\displaystyle\frac{1}{a^{3}}\int\frac{a}{\sqrt{2(k-1)(k+3)}}X_{\beta}{}^{b}\nabla_{b}\left(\frac{k+1}{a}g^{\beta\lambda}Y^{*\,k\ell
m}_{(2)\,\lambda}\right)\\!\sqrt{g}\,d^{3}x$ $\displaystyle=$
$\displaystyle\frac{1}{a^{3}}\int\frac{k+1}{\sqrt{2(k-1)(k+3)}}(s^{ab}+\alpha^{ab})\nabla_{b}Y^{*\,k\ell
m}_{(2)\,a}\\!\sqrt{g}\,d^{3}x$

where in going from the first line to the second line we used (78), and in the
last line we have decomposed $X^{ab}=s^{ab}+\alpha^{ab}$, where $s^{ab}$ is
the symmetric part (including the trace term) and $\alpha^{ab}$ is the
antisymmtric part of $X^{ab}$. So now we have

$Q_{(1)}^{k\ell m}=R_{(1)}^{k\ell m}+S_{(1)}^{k\ell m}$ (112)

where

$\displaystyle R_{(1)}^{k\ell m}$ $\displaystyle=$
$\displaystyle\frac{1}{a^{3}}\int\frac{k+1}{\sqrt{2(k-1)(k+3)}}s^{ab}\tfrac{1}{2}\left(\nabla_{a}Y^{*\,k\ell
m}_{(2)\,b}+\nabla_{b}Y^{*\,k\ell m}_{(2)\,a}\right)\\!\sqrt{g}\,d^{3}x$ (113)
$\displaystyle=$
$\displaystyle\frac{1}{a^{3}}\int\frac{k+1}{2\sqrt{2(k-1)(k+3)}}s^{ab}\frac{\sqrt{2(k-1)(k+3)}}{a}Y^{*\,k\ell
m}_{(2)\,ab}\\!\sqrt{g}\,d^{3}x$ $\displaystyle=$
$\displaystyle\frac{1}{a^{3}}\int\frac{k+1}{2a}\sigma^{ab}Y^{*\,k\ell
m}_{(2)\,ab}\\!\sqrt{g}\,d^{3}x$ $\displaystyle=$
$\displaystyle\frac{k+1}{2a}\tilde{\sigma}^{k\ell m}_{(2)},$

with $\sigma_{ab}$ being the trace-free symmetric part of $X_{ab}$, and

$\displaystyle S_{(1)}^{k\ell m}$ $\displaystyle=$
$\displaystyle\frac{1}{a^{3}}\int\frac{k+1}{\sqrt{2(k-1)(k+3)}}\nabla_{a}\alpha^{ab}Y^{*\,k\ell
m}_{(2)\,b}\\!\sqrt{g}\,d^{3}x$ (114) $\displaystyle=$
$\displaystyle\frac{1}{a^{3}}\int\frac{k+1}{\sqrt{2(k-1)(k+3)}}\nabla_{a}\left(\sum_{A^{\prime}k^{\prime}\ell^{\prime}m^{\prime}}\\!\\!\\!\\!\epsilon^{abc}\tilde{\alpha}^{k^{\prime}\ell^{\prime}m^{\prime}}_{(A^{\prime})}Y^{k^{\prime}\ell^{\prime}m^{\prime}}_{(A^{\prime})\,c}\right)Y^{*\,k\ell
m}_{(2)\,b}\\!\sqrt{g}\,d^{3}x$ $\displaystyle=$
$\displaystyle\frac{1}{a^{3}}\int\frac{k+1}{\sqrt{2(k-1)(k+3)}}\sum_{k^{\prime}\ell^{\prime}m^{\prime}}\\!\\!\\!\\!\tilde{\alpha}^{k^{\prime}\ell^{\prime}m^{\prime}}_{(1)}\left(-\frac{k+1}{a}g^{ab}Y^{k^{\prime}\ell^{\prime}m^{\prime}}_{(2)\,a}\right)Y^{*\,k\ell
m}_{(2)\,b}\\!\sqrt{g}\,d^{3}x$ $\displaystyle=$
$\displaystyle-\sum_{k^{\prime}\ell^{\prime}m^{\prime}}\frac{(k+1)^{2}}{a\sqrt{2(k-1)(k+3)}}\tilde{\alpha}^{k^{\prime}\ell^{\prime}m^{\prime}}_{(1)}\frac{1}{a^{3}}\int
g^{ab}Y^{k^{\prime}\ell^{\prime}m^{\prime}}_{(2)\,a}Y^{*\,k\ell
m}_{(2)\,b}\\!\sqrt{g}\,d^{3}x$ $\displaystyle=$
$\displaystyle-\frac{(k+1)^{2}}{a\sqrt{2(k-1)(k+3)}}\tilde{\alpha}^{k\ell
m}_{(1)},$

where in going from the second line to the third line we used equations (77),
(78), (79), and the orthonormality condition (C), and in going from the fourth
line to the last line we used the orthonormality condition (C) again. So now
we have

$\displaystyle Q_{(1)}^{k\ell m}$ $\displaystyle=$ $\displaystyle
R_{(1)}^{k\ell m}+S_{(1)}^{k\ell m}$ (115) $\displaystyle=$
$\displaystyle\frac{k+1}{2a}\tilde{\sigma}^{k\ell
m}_{(2)}-\frac{(k+1)^{2}}{a\sqrt{2(k-1)(k+3)}}\tilde{\alpha}^{k\ell m}_{(1)},$
(116)

so that

$\displaystyle\tilde{D}_{\Gamma}\tilde{X}^{k\ell
m}_{(1)}=\frac{k+1}{2a}\tilde{\sigma}^{k\ell
m}_{(2)}+\frac{2-(k+1)^{2}}{a\sqrt{2(k-1)(k+3)}}\tilde{\alpha}^{k\ell
m}_{(1)}.$ (117)

The calculation for $\tilde{D}_{\Gamma}\tilde{X}^{k\ell m}_{(2)}$ proceeds in
the same way as $\tilde{D}_{\Gamma}\tilde{X}^{k\ell m}_{(1)}$, so we will skip
the details and record the result:

$\tilde{D}_{\Gamma}\tilde{X}^{k\ell
m}_{(2)}=\frac{2-(k+1)^{2}}{a\sqrt{2(k-1)(k+3)}}\tilde{\alpha}^{k\ell
m}_{(2)}+\frac{k+1}{2a}\tilde{\sigma}^{k\ell m}_{(1)}.$ (118)
$\displaystyle\\!\\!\\!\\!\\!\\!\tilde{D}_{\Gamma}\tilde{X}^{k\ell m}_{(3)}$
$\displaystyle=$
$\displaystyle-\frac{1}{a^{3}}\int\\!\epsilon^{c\alpha\beta}X_{\beta}{}^{d}\nabla_{\alpha}Y^{*\,k\ell
m}_{(3)\,cd}\sqrt{g}\,d^{3}x$ (119) $\displaystyle=$
$\displaystyle-\frac{1}{a^{3}}\int\\!\frac{\sqrt{3}a}{\sqrt{2(k-1)(k+3)}}\epsilon^{c\alpha\beta}X_{\beta}{}^{d}\nabla_{\alpha}\left(\nabla_{c}Y^{*\,k\ell
m}_{(0)\,d}+\frac{\sqrt{k(k+2)}}{\sqrt{3}a}Y^{*\,k\ell
m}_{(0)\,cd}\right)\sqrt{g}\,d^{3}x$ $\displaystyle=$
$\displaystyle-\frac{1}{a^{3}}\int\\!\frac{\sqrt{3}a}{\sqrt{2(k-1)(k+3)}}\epsilon^{c\alpha\beta}X_{\beta}{}^{d}\left(-\tfrac{1}{2}R_{\alpha
c}{}^{\lambda}{}_{d}Y^{*\,{k\ell
m}}_{(0)\,\lambda}+\frac{k(k+2)}{3a^{2}}g_{cd}Y^{*\,{k\ell
m}}_{(0)\,\alpha}\right)\sqrt{g}\,d^{3}x$ $\displaystyle=$
$\displaystyle\ldots$ $\displaystyle=$
$\displaystyle-\frac{1}{a^{3}}\int\frac{-3+k(k+2)}{\sqrt{3}a\sqrt{2(k-1)(k+3)}}\epsilon^{abc}X_{ab}Y^{*\,{k\ell
m}}_{(0)\,c}\sqrt{g}\,d^{3}x$ $\displaystyle=$
$\displaystyle\frac{\sqrt{(k-1)(k+3)}}{\sqrt{6}a}\tilde{\alpha}^{k\ell
m}_{(0)},$

where in the omitted steps above we again used
$R_{ab}{}^{c}{}_{d}=\frac{1}{a^{2}}(\delta^{c}{}_{a}g_{bd}-\delta^{c}{}_{b}g_{ad})$
and did some algebra.

$\displaystyle\\!\\!\\!\\!\\!\\!\tilde{D}_{\Gamma}\tilde{X}^{k\ell m}_{(4)}$
$\displaystyle=$
$\displaystyle-\frac{1}{a^{3}}\int\epsilon^{c\alpha\beta}X_{\beta}{}^{d}\nabla_{\alpha}Y^{*\,{k\ell
m}}_{(4)\,cd}\sqrt{g}\,d^{3}x$ (120) $\displaystyle=$
$\displaystyle\frac{1}{a^{3}}\int\left(\sigma^{ab}\frac{k+1}{a}Y^{*\,{k\ell
m}}_{(5)\,ab}+\epsilon_{\beta}{}^{c\alpha}\nabla_{\alpha}\alpha^{\beta
d}Y^{*\,{k\ell m}}_{(4)\,cd}\right)\sqrt{g}\,d^{3}x$ $\displaystyle=$
$\displaystyle\frac{k+1}{a}\tilde{\sigma}^{{k\ell m}}_{(5)}+P_{(4)}^{k\ell
m},$

where

$\displaystyle P_{(4)}^{k\ell m}$ $\displaystyle=$
$\displaystyle\frac{1}{a^{3}}\int\epsilon_{\beta}{}^{c\alpha}\nabla_{\alpha}\left(\sum_{Ak^{\prime}\ell^{\prime}m^{\prime}}\\!\\!\\!\\!\epsilon^{\beta
d\lambda}\tilde{\alpha}^{k^{\prime}\ell^{\prime}m^{\prime}}_{(A)}Y^{k^{\prime}\ell^{\prime}m^{\prime}}_{(A)\,\lambda}\right)Y^{*\,{k\ell
m}}_{(4)\,cd}\sqrt{g}\,d^{3}x$ (121) $\displaystyle=$
$\displaystyle\frac{1}{a^{3}}\int\left(g^{cd}g^{\alpha\lambda}-g^{c\lambda}g^{\alpha
d}\right)\\!\\!\\!\sum_{Ak^{\prime}\ell^{\prime}m^{\prime}}\\!\\!\\!\\!\tilde{\alpha}^{k^{\prime}\ell^{\prime}m^{\prime}}_{(A)}\nabla_{\alpha}Y^{k^{\prime}\ell^{\prime}m^{\prime}}_{(A)\,\lambda}Y^{*\,{k\ell
m}}_{(4)\,cd}\sqrt{g}\,d^{3}x$ $\displaystyle=$
$\displaystyle-\frac{1}{a^{3}}\int\sum_{Ak^{\prime}\ell^{\prime}m^{\prime}}\\!\\!\\!\\!\tilde{\alpha}^{k^{\prime}\ell^{\prime}m^{\prime}}_{(A)}\nabla^{d}Y^{k^{\prime}\ell^{\prime}m^{\prime}\,c}_{(A)}Y^{*\,{k\ell
m}}_{(4)\,cd}\sqrt{g}d^{3}x$ $\displaystyle\propto$
$\displaystyle-\frac{1}{a^{3}}\int\sum_{A=0}^{2}\sum_{k^{\prime}\ell^{\prime}m^{\prime}}\\!\\!\\!\\!\tilde{\alpha}^{k^{\prime}\ell^{\prime}m^{\prime}}_{(A)}Y^{k^{\prime}\ell^{\prime}m^{\prime}\,cd}_{(Q(A))}Y^{*\,{k\ell
m}}_{(4)\,cd}\sqrt{g}\,d^{3}x=0,$

where in going from the second line to the third line we used the fact that
$Y^{*\,{k\ell m}}_{(4)\,ab}$ is trace free and in the last line $Q(0)=3$,
$Q(1)=1$ and $Q(2)=2$. Finally, we see that $P^{{k\ell m}}_{(4)}$ vanishes by
virtue of the orthonormality condition (103). So we have

$\tilde{D}_{\Gamma}\tilde{X}^{k\ell
m}_{(4)}=\frac{k+1}{a}\tilde{\sigma}^{{k\ell m}}_{(5)}.$ (122)
$\displaystyle\\!\\!\\!\\!\\!\\!\tilde{D}_{\Gamma}\tilde{X}^{k\ell m}_{(5)}$
$\displaystyle=$
$\displaystyle-\frac{1}{a^{3}}\int\epsilon^{c\alpha\beta}X_{\beta}{}^{d}\nabla_{\alpha}Y^{*\,{k\ell
m}}_{(5)\,cd}\\!\sqrt{g}\,d^{3}x$ (123) $\displaystyle=$
$\displaystyle-\frac{1}{a^{3}}\int\epsilon^{c\alpha\beta}X_{\beta}{}^{d}\frac{a}{2(k+1)}\nabla_{\alpha}\\!\left(\epsilon_{c}{}^{\lambda\mu}\nabla_{\lambda}Y^{*\,{k\ell
m}}_{(4)\,\mu d}+\epsilon_{d}{}^{\lambda\mu}\nabla_{\lambda}Y^{*\,{k\ell
m}}_{(4)\,\mu c}\right)\\!\sqrt{g}\,d^{3}x$ $\displaystyle=$ $\displaystyle
P^{{k\ell m}}_{(5)}+Q^{{k\ell m}}_{(5)},$

where

$\displaystyle P^{{k\ell m}}_{(5)}$ $\displaystyle=$
$\displaystyle-\frac{1}{a^{3}}\int\frac{a}{2(k+1)}\left(g^{\alpha\lambda}g^{\beta\mu}-g^{\alpha\mu}g^{\beta\lambda}\right)X_{\beta}{}^{d}\nabla_{\alpha}\nabla_{\lambda}Y^{*\,{k\ell
m}}_{(4)\,\mu d}\\!\sqrt{g}\,d^{3}x,$ (124) $\displaystyle Q^{{k\ell
m}}_{(5)}$ $\displaystyle=$
$\displaystyle-\frac{1}{a^{3}}\int\frac{a}{2(k+1)}\epsilon^{c\alpha\beta}\epsilon^{d\lambda\mu}X_{\beta
d}\left(-R_{\alpha\lambda}{}^{\rho}{}_{\mu}Y^{*\,{k\ell m}}_{(4)\,\rho
c}-R_{\alpha\lambda}{}^{\rho}{}_{c}Y^{*\,{k\ell
m}}_{(4)\,\mu\rho}+\nabla_{\lambda}\nabla_{\alpha}Y^{*\,{k\ell m}}_{(4)\,\mu
c}\right)\\!\sqrt{g}\,d^{3}x.$

Let’s focus on $P^{{k\ell m}}_{(5)}$ first:

$\displaystyle P^{{k\ell m}}_{(5)}$ $\displaystyle=$
$\displaystyle-\frac{1}{a^{3}}\int\frac{a}{2(k+1)}\bigg{[}X^{\mu
d}\frac{2-k(k+2)}{a(k+1)}Y^{*\,{k\ell m}}_{(4)\,\mu d}$ (126)
$\displaystyle-X^{\lambda
d}g^{\alpha\mu}\left(-R_{\alpha\lambda}{}^{\rho}{}_{\mu}Y^{*\,{k\ell
m}}_{(4)\,\rho d}-R_{\alpha\lambda}{}^{\rho}{}_{d}Y^{*\,{k\ell
m}}_{(4)\,\mu\rho}+\nabla_{\lambda}\nabla_{\alpha}Y^{*\,{k\ell m}}_{(4)\,\mu
d}\right)\bigg{]}$ $\displaystyle=$
$\displaystyle\frac{k(k+2)-2}{a(k+1)}\tilde{\sigma}^{{k\ell m}}_{(4)}+R^{k\ell
m}_{(5)},$

where in the first line we used (91), and where we have defined

$\displaystyle R^{k\ell m}_{(5)}$ $\displaystyle=$
$\displaystyle\frac{1}{a^{3}}\int\frac{a}{2(k+1)}X^{\lambda
d}g^{\alpha\mu}\Big{[}-\tfrac{1}{a^{2}}\left(\delta^{\rho}{}_{\alpha}g_{\lambda\mu}-\delta^{\rho}{}_{\lambda}g_{\alpha\mu}\right)Y^{*\,{k\ell
m}}_{(4)\,\rho d}$ (127)
$\displaystyle-\tfrac{1}{a^{2}}\left(\delta^{\rho}{}_{\alpha}g_{\lambda
d}-\delta^{\rho}{}_{\lambda}g_{\alpha d}\right)Y^{*\,{k\ell
m}}_{(4)\,\mu\rho}+\nabla_{\lambda}\nabla_{\alpha}Y^{*\,{k\ell m}}_{(4)\,\mu
d}\Big{]}\\!\sqrt{g}\,d^{3}x$ $\displaystyle=$
$\displaystyle\frac{1}{a^{3}}\int\frac{3}{2a(k+1)}X^{\lambda d}Y^{*\,{k\ell
m}}_{(4)\,\lambda d}\sqrt{g}\,d^{3}x$ $\displaystyle=$
$\displaystyle\frac{3}{2a(k+1)}\tilde{\sigma}^{{k\ell m}}_{(4)},$

where in going from the first equality to the second equality we used the
trace-free and divergence-free properties of $Y^{*\,{k\ell m}}_{(4)\,ab}$ and
did some algebra. So we have

$\displaystyle P^{{k\ell m}}_{(5)}$ $\displaystyle=$
$\displaystyle\frac{k(k+2)-2}{a(k+1)}\tilde{\sigma}^{{k\ell m}}_{(4)}+R^{k\ell
m}_{(5)},$ (128) $\displaystyle=$
$\displaystyle\frac{k(k+2)-2}{a(k+1)}\tilde{\sigma}^{{k\ell
m}}_{(4)}+\frac{3}{2a(k+1)}\tilde{\sigma}^{{k\ell m}}_{(4)}$ $\displaystyle=$
$\displaystyle\frac{2k(k+2)-1}{2a(k+1)}\tilde{\sigma}^{{k\ell m}}_{(4)}.$

Now let’s turn to $Q^{k\ell m}_{(5)}$:

$Q^{k\ell m}_{(5)}=S^{k\ell m}_{(5)}+T^{k\ell m}_{(5)},$ (129)

where

$\displaystyle\\!\\!\\!\\!S^{k\ell m}_{(5)}$ $\displaystyle=$
$\displaystyle\frac{1}{a^{3}}\int\epsilon^{c\alpha\beta}\epsilon^{d\lambda\mu}X_{\beta
d}\frac{a}{2(k+1)}\bigg{[}\tfrac{1}{a^{2}}\left(\delta^{\rho}{}_{\alpha}g_{\lambda\mu}-\delta^{\rho}{}_{\lambda}g_{\alpha\mu}\right)Y^{*\,{k\ell
m}}_{(4)\,\rho c}+\tfrac{1}{a^{2}}\left(\delta^{\rho}{}_{\alpha}g_{\lambda
c}-\delta^{\rho}{}_{\lambda}g_{\alpha c}\right)Y^{*\,{k\ell
m}}_{(4)\,\mu\rho}\bigg{]}\sqrt{g}\,d^{3}x$ (130) $\displaystyle=$
$\displaystyle\frac{1}{a^{3}}\int\frac{1}{2a(k+1)}X_{\beta
d}\left(\epsilon^{c\beta}{}_{\mu}\epsilon^{d\rho\mu}Y^{*\,{k\ell
m}}_{(4)\,\rho c}-\epsilon^{c\rho\beta}\epsilon_{c}{}^{d\mu}Y^{*\,{k\ell
m}}_{(4)\,\mu\rho}\right)\sqrt{g}\,d^{3}x$ $\displaystyle=$
$\displaystyle\frac{1}{a^{3}}\int\frac{1}{2a(k+1)}X_{\beta
d}\left[\left(g^{cd}g^{\beta\rho}-g^{c\rho}g^{\beta d}\right)Y^{*\,{k\ell
m}}_{(4)\,\rho c}-\left(g^{\rho d}g^{\beta\mu}-g^{\rho\mu}g^{\beta
d}\right)Y^{*\,{k\ell m}}_{(4)\,\rho c}\right]$ $\displaystyle=$
$\displaystyle\frac{1}{a^{3}}\int\frac{1}{2a(k+1)}(X^{\rho c}Y^{*\,{k\ell
m}}_{(4)\,\rho c}-X^{\mu\rho}Y^{*\,{k\ell
m}}_{(4)\,\mu\rho})\sqrt{g}\,d^{3}x=0,$

where in going from the first line to the second line we eliminated traces of
the epsilon tensors, and in going from the third line to the fourth line we
used the trace-free property of $Y^{*\,{k\ell m}}_{(4)\,ab}$. Now let’s focus
on $T^{{k\ell m}}_{(5)}$:

$\displaystyle T^{{k\ell m}}_{(5)}$ $\displaystyle=$
$\displaystyle-\frac{1}{a^{3}}\int\frac{a}{2(k+1)}\bigg{[}\delta^{\alpha}{}_{d}\left(\delta^{\beta}{}_{\lambda}\delta^{c}{}_{\mu}-\delta^{\beta}{}_{\mu}\delta^{c}{}_{\lambda}\right)+\delta^{\alpha}{}_{\lambda}\left(\delta^{\beta}{}_{\mu}\delta^{c}{}_{d}-\delta^{\beta}{}_{d}\delta^{c}{}_{\mu}\right)$
(131)
$\displaystyle+\delta^{\alpha}{}_{\mu}\left(\delta^{\beta}{}_{d}\delta^{c}{}_{\lambda}-\delta^{\beta}{}_{\lambda}\delta^{c}{}_{d}\right)\bigg{]}X_{\alpha}{}^{d}\nabla^{\lambda}\nabla_{\beta}\left(Y^{*\,{k\ell
m}}_{(4)}\right)_{c}{}^{\mu}\sqrt{g}\,d^{3}x$ $\displaystyle=$
$\displaystyle\frac{1}{a^{3}}\int\frac{a}{2(k+1)}X_{\mu}{}^{d}\left[\nabla^{c}\nabla_{d}\left(Y^{*\,{k\ell
m}}_{(4)}\right)_{c}{}^{\mu}-\frac{2-k(k+2)}{a}\left(Y^{*\,{k\ell
m}}_{(4)}\right)_{d}{}^{\mu}\right]\sqrt{g}\,d^{3}x$ $\displaystyle=$
$\displaystyle\frac{k(k+2)-2}{2a(k+1)}\tilde{\sigma}^{{k\ell
m}}_{(4)}+U^{k\ell m}_{(5)},$

where

$\displaystyle U^{k\ell m}_{(5)}$ $\displaystyle=$
$\displaystyle\frac{1}{a^{3}}\int\frac{a}{2(k+1)}X_{\mu}{}^{d}\left(R_{cd}{}^{c}{}_{\lambda}Y^{*\,{k\ell
m}\,\lambda\mu}_{(4)}+R_{cd}{}^{\mu}{}_{\lambda}Y^{*\,{k\ell
m}\,c\lambda}_{(4)}\right)\sqrt{g}\,d^{3}x$ (132) $\displaystyle=$
$\displaystyle\frac{1}{a^{3}}\int\frac{a}{2(k+1)}X_{\mu}{}^{d}\left[\tfrac{2}{a^{2}}g_{d\lambda}Y^{*\,{k\ell
m}\,\lambda\mu}_{(4)}+\tfrac{1}{a^{2}}\left(\delta^{\mu}{}_{c}g_{d\lambda}-\delta^{\mu}{}_{d}g_{c\lambda}\right)Y^{*\,{k\ell
m}\,c\lambda}_{(4)}\right]\sqrt{g}\,d^{3}x$ $\displaystyle=$
$\displaystyle\frac{1}{a^{3}}\int\frac{a}{2(k+1)}X_{\mu}{}^{d}\tfrac{3}{a^{2}}g_{d\lambda}Y^{*\,{k\ell
m}\,\lambda\mu}_{(4)}\sqrt{g}\,d^{3}x$ $\displaystyle=$
$\displaystyle\frac{3}{2a(k+1)}\tilde{\sigma}^{k\ell m}_{(4)}.$

So

$Q^{k\ell m}_{(5)}=T^{k\ell
m}_{(5)}=\frac{k(k+2)+1}{2a(k+1)}\tilde{\sigma}^{{k\ell m}}_{(4)},$ (133)

and finally

$\displaystyle\tilde{D}_{\Gamma}\tilde{X}^{k\ell m}_{(5)}$ $\displaystyle=$
$\displaystyle P^{k\ell m}_{(5)}+Q^{k\ell m}_{(5)}$ (134) $\displaystyle=$
$\displaystyle\frac{2k(k+2)-1}{2a(k+1)}\tilde{\sigma}^{{k\ell
m}}_{(4)}+\frac{k(k+2)+1}{2a(k+1)}\tilde{\sigma}^{{k\ell m}}_{(4)}$
$\displaystyle=$ $\displaystyle\frac{3k(k+2)}{2a(k+1)}\tilde{\sigma}^{k\ell
m}_{(4)}.$

We have now computed all of the expansion coefficients corresponding to the
symmetric part of $D_{\Gamma}X_{ab}$. Next we need to compute the coefficients
corresponding to the antisymmetric part. We will denote these coefficients by

$\displaystyle\tilde{D}_{\Gamma}\tilde{X}^{k\ell m}_{(A+6)}$ $\displaystyle=$
$\displaystyle\frac{1}{a^{3}}\int D_{\Gamma}X_{ab}\epsilon^{abc}Y^{*\,{k\ell
m}}_{(A)\,c}\sqrt{g}\,d^{3}x$ (135) $\displaystyle=$
$\displaystyle\frac{1}{a^{3}}\int\epsilon_{a}{}^{de}\nabla_{d}X_{eb}\epsilon^{abc}Y^{*\,{k\ell
m}}_{(A)\,c}\sqrt{g}\,d^{3}x$ $\displaystyle=$
$\displaystyle-\frac{1}{a^{3}}\int\left(g^{db}g^{ec}-g^{dc}g^{be}\right)X_{eb}\nabla_{d}Y^{*\,{k\ell
m}}_{(A)\,c}\sqrt{g}\,d^{3}x$ $\displaystyle=$
$\displaystyle-\frac{1}{a^{3}}\int\left(X^{ab}\nabla_{b}Y^{*\,{k\ell
m}}_{(A)\,a}-g_{cd}X^{cd}\nabla^{a}Y^{*\,{k\ell
m}}_{(A)\,a}\right)\sqrt{g}\,d^{3}x$ $\displaystyle=$
$\displaystyle-\frac{1}{a^{3}}\int X^{ab}\nabla_{b}Y^{*\,{k\ell
m}}_{(A)\,a}\sqrt{g}\,d^{3}x-\delta^{(0)}{}_{\\!\\!(A)}\frac{\sqrt{3k(k+2)}}{a}\tilde{X}^{k\ell
m}_{(0)}$ $\displaystyle=$
$\displaystyle\frac{1}{a^{3}}\int\nabla_{b}X^{ab}Y^{*\,{k\ell
m}}_{(A)\,a}\sqrt{g}\,d^{3}x-\delta^{(0)}{}_{\\!\\!(A)}\frac{\sqrt{3k(k+2)}}{a}\tilde{X}^{k\ell
m}_{(0)},$

where in going from the fourth line to the fifth line we used equations (68),
(69), and (70). Now we can compute

$\tilde{D}_{\Gamma}\tilde{X}^{k\ell
m}_{(6)}=-\frac{\sqrt{3k(k+2)}}{a}\tilde{X}^{k\ell m}_{(0)}+P^{k\ell
m}_{(6)}+Q^{k\ell m}_{(6)},$ (136)

where

$\displaystyle P^{k\ell m}_{(6)}$ $\displaystyle=$
$\displaystyle\frac{1}{a^{3}}\int\nabla_{a}\sigma^{ab}Y^{*{k\ell
m}}_{(0)\,b}\sqrt{g}\,d^{3}x,$ $\displaystyle Q^{k\ell m}_{(6)}$
$\displaystyle=$
$\displaystyle-\frac{1}{a^{3}}\int\nabla_{a}\alpha^{ab}Y^{*{k\ell
m}}_{(0)\,b}\sqrt{g}\,d^{3}x.$ (137)

Let’s focus on $P^{k\ell m}_{(6)}$ first:

$\displaystyle P^{k\ell m}_{(6)}$ $\displaystyle=$
$\displaystyle\frac{1}{a^{3}}\int\left(\sum_{Ak^{\prime}\ell^{\prime}m^{\prime}}\tilde{s}^{k^{\prime}\ell^{\prime}m^{\prime}}_{(A)}\nabla_{a}Y^{k\ell^{\prime}m^{\prime}\,ab}_{(A)}\right)Y^{*{k\ell
m}}_{(0)\,b}\sqrt{g}\,d^{3}x$ (138) $\displaystyle=$
$\displaystyle\frac{1}{a^{3}}\int\left[\sum_{k^{\prime}\ell^{\prime}m^{\prime}}\left(\frac{\sqrt{k(k+2)}}{\sqrt{3}a}\tilde{X}^{k^{\prime}\ell^{\prime}m^{\prime}}_{(0)}-\frac{\sqrt{2(k-1)(k+3)}}{\sqrt{3}a}\tilde{\sigma}^{k^{\prime}\ell^{\prime}m^{\prime}}_{(3)}\right)Y^{k^{\prime}\ell^{\prime}m^{\prime}\,b}_{(0)}+\perp\,\right]Y^{*{k\ell
m}}_{(0)\,b}\sqrt{g}\,d^{3}x$ $\displaystyle=$
$\displaystyle\frac{\sqrt{k(k+2)}}{\sqrt{3}a}\tilde{X}^{k\ell
m}_{(0)}-\frac{\sqrt{2(k-1)(k+3)}}{\sqrt{3}a}\tilde{\sigma}^{k\ell m}_{(3)}.$

where $``\\!\perp\\!"$ denotes terms that vanish when integrated against
$Y^{*{k\ell m}}_{(0)\,b}$, and we used equations (93)-(98) for the divergences
of the symmetric tensor harmonics and the orthonormality relation (C). For the
other term we have

$\displaystyle Q^{k\ell m}_{(6)}$ $\displaystyle=$
$\displaystyle-\frac{1}{a^{3}}\int\left(\sum_{Ak^{\prime}\ell^{\prime}m^{\prime}}\tilde{\alpha}^{k^{\prime}\ell^{\prime}m^{\prime}}_{(A)}\epsilon^{abc}\nabla_{a}Y^{k^{\prime}\ell^{\prime}m^{\prime}}_{(A)\,c}\right)Y^{*{k\ell
m}}_{(0)\,b}\sqrt{g}\,d^{3}x$ (139) $\displaystyle=$
$\displaystyle-\frac{1}{a^{3}}\int\sum_{A=1}^{2}\sum_{k^{\prime}\ell^{\prime}m^{\prime}}\tilde{\alpha}^{k^{\prime}\ell^{\prime}m^{\prime}}_{(A)}\left(-\frac{k+1}{a}Y^{k^{\prime}\ell^{\prime}m^{\prime}\,b}_{(Q(A))}\right)Y^{*{k\ell
m}}_{(0)\,b}\sqrt{g}\,d^{3}x=0,$

where we used equations (77), (78), and (79), and we defined $Q(1)=2$,
$Q(2)=1$. Finally, we see that $Q^{k\ell m}_{(6)}=0$, by virtue of the
orthogonality relation (C). So in the end we have

$\tilde{D}_{\Gamma}\tilde{X}^{k\ell
m}_{(6)}=-\frac{2\sqrt{k(k+2)}}{\sqrt{3}a}\tilde{X}^{k\ell
m}_{(0)}-\frac{\sqrt{2(k-1)(k+3)}}{\sqrt{3}a}\tilde{\sigma}^{k\ell m}_{(3)}.$
(140)

Using the same strategy, we can compute

$\displaystyle\tilde{D}_{\Gamma}\tilde{X}^{k\ell m}_{(7)}$ $\displaystyle=$
$\displaystyle-\frac{\sqrt{(k-1)(k+3)}}{\sqrt{2}a}\tilde{\sigma}^{k\ell
m}_{(1)}+\frac{1+k(k+2)}{a(k+1)}\tilde{\alpha}^{k\ell m}_{(2)},$ (141)
$\displaystyle\tilde{D}_{\Gamma}\tilde{X}^{k\ell m}_{(8)}$ $\displaystyle=$
$\displaystyle-\frac{\sqrt{(k-1)(k+3)}}{\sqrt{2}a}\tilde{\sigma}^{k\ell
m}_{(2)}+\frac{1+k(k+2)}{a(k+1)}\tilde{\alpha}^{k\ell m}_{(1)}.$ (142)

Finally, defining $\tilde{X}^{{k\ell m}}_{(A)}=\tilde{\sigma}^{{k\ell
m}}_{(A)}$, $\tilde{X}^{{k\ell m}}_{(A+6)}=\tilde{\alpha}^{{k\ell m}}_{(A)}$,
we have

$\displaystyle\tilde{D}_{\Gamma}\tilde{X}^{k\ell m}_{(0)}$ $\displaystyle=$
$\displaystyle-\frac{\sqrt{k(k+2)}}{\sqrt{3}a}\tilde{X}^{k\ell m}_{(6)},$
$\displaystyle\tilde{D}_{\Gamma}\tilde{X}^{k\ell m}_{(1)}$ $\displaystyle=$
$\displaystyle\frac{k+1}{2a}\tilde{X}^{k\ell
m}_{(2)}+\frac{2-(k+1)^{2}}{a\sqrt{2(k-1)(k+3)}}\tilde{X}^{k\ell m}_{(7)},$
$\displaystyle\tilde{D}_{\Gamma}\tilde{X}^{k\ell m}_{(2)}$ $\displaystyle=$
$\displaystyle\frac{k+1}{2a}\tilde{X}^{k\ell
m}_{(1)}+\frac{2-(k+1)^{2}}{a\sqrt{2(k-1)(k+3)}}\tilde{X}^{k\ell m}_{(8)},$
$\displaystyle\tilde{D}_{\Gamma}\tilde{X}^{k\ell m}_{(3)}$ $\displaystyle=$
$\displaystyle\frac{\sqrt{(k-1)(k+3)}}{\sqrt{6}a}\tilde{X}^{k\ell m}_{(6)},$
$\displaystyle\tilde{D}_{\Gamma}\tilde{X}^{k\ell m}_{(4)}$ $\displaystyle=$
$\displaystyle\frac{k+1}{a}\tilde{X}^{{k\ell m}}_{(5)},$
$\displaystyle\tilde{D}_{\Gamma}\tilde{X}^{k\ell m}_{(5)}$ $\displaystyle=$
$\displaystyle\frac{3k(k+2)}{2a(k+1)}\tilde{X}^{k\ell m}_{(4)},$
$\displaystyle\tilde{D}_{\Gamma}\tilde{X}^{k\ell m}_{(6)}$ $\displaystyle=$
$\displaystyle-\frac{2\sqrt{k(k+2)}}{\sqrt{3}a}\tilde{X}^{k\ell
m}_{(0)}-\frac{\sqrt{2(k-1)(k+3)}}{\sqrt{3}a}\tilde{X}^{k\ell m}_{(3)},$
$\displaystyle\tilde{D}_{\Gamma}\tilde{X}^{k\ell m}_{(7)}$ $\displaystyle=$
$\displaystyle-\frac{\sqrt{(k-1)(k+3)}}{\sqrt{2}a}\tilde{X}^{k\ell
m}_{(1)}+\frac{1+k(k+2)}{a(k+1)}\tilde{X}^{k\ell m}_{(8)},$
$\displaystyle\tilde{D}_{\Gamma}\tilde{X}^{k\ell m}_{(8)}$ $\displaystyle=$
$\displaystyle-\frac{\sqrt{(k-1)(k+3)}}{\sqrt{2}a}\tilde{X}^{k\ell
m}_{(2)}+\frac{1+k(k+2)}{a(k+1)}\tilde{X}^{k\ell m}_{(7)}.$ (143)

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