aainstitutetext: Department of Physics and Astronomy
Uppsala University, Box 516
SE-751 20 Uppsala, Swedenbbinstitutetext: Department of Physics
Chalmers University of Technology
SE-412 96 Göteborg, Sweden

# On the squashed seven-sphere operator spectrum

Simon Ekhammar b and Bengt E.W. Nilsson<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract

We derive major parts of the eigenvalue spectrum of the operators on the
squashed seven-sphere that appear in the compactification of eleven-
dimensional supergravity. These spectra determine the mass spectrum of the
fields in $AdS_{4}$ and are important for the corresponding $\mathcal{N}=1$
supermultiplet structure. This work is a continuation of the work in
Nilsson:2018lof  where the complete spectrum of irreducible isometry
representations of the fields in $AdS_{4}$ was derived for this
compactification. Some comments are also made concerning the $G_{2}$ holonomy
and its implications on the structure of the operator equations on the
squashed seven-sphere.

###### Keywords:

Kaluza-Klein, weak $G_{2}$ manifolds, operator spectra on squashed spheres.

##  1 Introduction

The purpose of this paper is to continue the work in Nilsson:2018lof  and
derive the eigenvalue spectrum of some of the operators that determine the
masses and supermultiplet structure of the entire Kaluza-Klein spectrum of the
squashed seven-sphere compactification of eleven-dimensional supergravity. The
list of previously known eigenvalue spectra, containing $\Delta_{0}$,
${\not{D}}_{1/2}$ and $\Delta_{1}$, is in this paper extended by those of
$\Delta_{2}$, $\Delta_{3}$ and $\Delta_{L}$ while ${\not{D}}_{3/2}$ remains to
be done. Note that parts of the spectrum of the Lichnerowicz operator
$\Delta_{L}$ are derived below but they were quoted already in Duff:1986hr
111See Ref. [198], unpublished work by Nilsson and Pope..

With the intention to keep this paper as brief as possible we refer the reader
to the review Duff:1986hr  for a Kaluza-Klein background on the problem and
for some of the necessary details and conventions needed in the derivations
below. For the full structure of irreducible isometry representations on
$AdS_{4}$ stemming from the squashed seven-sphere compactification we refer to
Nilsson:2018lof . The latter paper summarises in a few pages the most relevant
information from Duff:1986hr  in particular several tables that are spread out
in various chapters of Duff:1986hr . In this spirit we present below just the
most crucial formulas that are needed to define the problem and derive the
spectra. In the Appendix we collect a number of useful octonionic identities
and other formulas, as well as some for our purposes crucial Weyl tensor
calculations.

The coset description of the squashed seven-sphere is

$G/H=Sp_{2}\times Sp_{1}^{C}/Sp_{1}^{A}\times Sp_{1}^{B+C},$ (1)

where, in order to define the denominator, $Sp_{2}$ is split into
$Sp_{1}^{A}\times Sp_{1}^{B}$ and $Sp_{1}^{B+C}$ is the diagonal subgroup of
$Sp_{1}^{B}$ and $Sp_{1}^{C}$. The $G$ irreps specifying the mode functions
$Y$ of a general Fourier expansion on $G/H$ are denoted $(p,q;r)$ Duff:1986hr
and the entire Kaluza-Klein irrep spectrum is derived and tabulated in terms
of cross diagrams in Nilsson:2018lof . The present paper will fill in some
gaps in our knowledge of the eigenvalue spectrum of the relevant operators on
the squashed seven-sphere but some are still missing. The remaining issues
that we need to address to complete the eigenvalue spectra will be elaborated
upon in a forthcoming publication Nilsson:2021kn .

The interest in deriving complete spectra in various Kaluza-Klein
compactifications has recently been revived due to the discovery of new
powerful versions of the embedding tensor technique. However, these methods
can be applied directly only when the vacuum is an extremum of some gauged
supergravity theory in $AdS_{4}$ which can be lifted to ten or eleven
dimensions, see for instance Malek:2019eaz ; Cesaro:2021haf  and references
therein. The point we want to emphasise here is that, due to the space
invaders scenario Duff:1983ajq ; Duff:1986hr , the squashed seven-sphere
solution is not of this kind and it therefore seems clear that other methods
are required to obtain the $AdS_{4}$ spectrum in this case.

There are also potential applications of this work in the context of the
swampland program222We are grateful to M.J. Duff for raising some interesting
aspects of this question that eventually led to this work., see for instance
Vafa:2005ui ; ArkaniHamed:2006dz ; Ooguri:2016pdq  and references therein.
This particular connection will be addressed elsewhere.

In the next section we very briefly review the background of the problem and
the method that is used in this paper to derive the eigenvalue spectrum of
operators on coset manifolds. The method is explained in many places, e.g.
Salam:1981xd ; Duff:1986hr , and in Section 3 we first apply it to obtain the
spectra of the spin 0 and 1 operators giving straightforwardly the well known
results cited in Duff:1986hr . Already when applied to the spin 1/2 Dirac
operator complications arise which get further pronounced when we subsequently
turn to more and more complicated operators. A summary of the obtained
eigenvalues is provided in the Conclusions, together with comments on some of
the remaining issues. Some technical aspects needed in the derivations below
are explained in the Appendix.

##  2 Compactification on the seven-sphere

The Fourier expansion technique that we will apply is, following the general
strategy explained in Salam:1981xd  and summarised in Duff:1986hr , based on
the coset master equation for the $Spin(n)$ covariant derivative on a
$n$-dimensional coset space $G/H$:

$\nabla_{a}Y+\tfrac{1}{2}f^{bc}{}_{a}\,\Sigma_{bc}Y=-T_{a}Y.$ (2)

The mode functions $Y$ have two suppressed indices: One corresponding to the
$Spin(n)$ tangent space irrep of the field that is being Fourier
expanded333Note, however, that the whole matrix of the $G$ group element is
involved in this equation. and one that specifies the mode, that is a $G$
irrep. This equation is derived in Duff:1986hr  from group elements of the
isometry group $G$. In the conventions used there the Lie algebra of the group
$G$ has generators $T_{A}$, satisfying $[T_{A},T_{B}]=f_{AB}{}^{C}T_{C}$,
which are divided into $T_{\bar{a}}$ of the subgroup $H$ and $T_{a}$ in the
complement of $H$ in $G$. Thus, if the Lie algebra of $G$ is reductive it
splits as follows:

$[T_{\bar{a}},T_{\bar{b}}]=f_{\bar{a}\bar{b}}{}^{\bar{c}}T_{\bar{c}},\,\,\,[T_{\bar{a}},T_{b}]=f_{\bar{a}b}{}^{c}T_{c},\,\,\,\,\,\,\,[T_{a},T_{b}]=f_{ab}{}^{\bar{c}}T_{\bar{c}}+f_{ab}{}^{c}T_{c}.$
(3)

The indices $a,b,..$ are vector indices in the tangent space of $G/H$ and
$\nabla_{a}$ in (2) is an $Spin(n)$ covariant derivative with a torsion free
spin connection while the $\Sigma_{ab}$ are the generators of $Spin(n)$ in the
representation relevant for the operator equation we are solving. Note that it
is only the structure constants $f_{ab}{}^{c}$ that appear in the coset master
equation (2). For symmetric spaces like the round seven-sphere
$f_{ab}{}^{c}=0$. Furthermore, the second algebraic relation in (3) defines
the imbedding of $H$ in the tangent space group $Spin(n)$.

For symmetric coset spaces the coset master equations thus reads
$\nabla_{a}Y=-T_{a}Y$ and the eigenvalue spectrum for the operators on $G/H$
are rather easily derived. The complications arising in the squashed seven-
sphere case therefore stem entirely from the structure constant dependent term
in (2). As first shown in Bais:1983wc  for the squashed seven-sphere they are
given by the structure constants of the octonions $a_{abc}$. In the
normalisation used in Duff:1986hr  they read

$f_{abc}=-\tfrac{1}{\sqrt{5}}a_{abc}=-\tfrac{2}{3}m\,a_{abc}.$ (4)

Since octonions will play a key role in the rest of this paper we have listed
some useful octonionic identities in the Appendix. Furthermore, in the last
expression for these structure constants we have introduced the scale
parameter $m$ arising from the ansatz
$F_{\mu\nu\rho\sigma}=3m\epsilon_{\mu\nu\rho\sigma}$ in the compactification
of eleven-dimensional supergravity. This is useful since we are dealing with
dimensionful quantities, like the covariant derivatives and Hodge-de Rham
operators. The conventions used in Duff:1986hr  correspond to setting

$m^{2}=\tfrac{9}{20}.$ (5)

We will assume $m>0$ and also use $m=\tfrac{3}{2\sqrt{5}}$ as for $f_{abc}$
above. The squashed seven-sphere discussed in this work has the orientation
that gives rise to $\mathcal{N}=1$ supersymmetry in $AdS_{4}$ after
compactification. This fact will be used below when we discuss the
corresponding supermultiplets and the $SO(3,2)$ irreps entering these
multiplets.

With this preparation we can readily attack the eigenvalue problems
$\Delta_{p}Y_{p}=\kappa_{p}^{2}Y_{p}$ where the Hodge-de Rham444This operator
generalises the Laplace-Beltrami operator to $p$-forms with $p>0$. operator on
$p$-forms is defined as $\Delta_{p}=\text{d}\delta+\delta\text{d}$. Here d is
the exterior derivative and $\delta=(-1)^{p}\star\text{d}\star$ its adjoint.
These operators act on forms according to

$(\text{d}Y)_{a_{1}\dots a_{p}}=p\nabla_{[a_{1}}Y_{a_{2}\dots
a_{p}]},\quad(\delta Y)_{a_{1}\dots a_{p}}=-\nabla^{b}Y_{ba_{1}\dots a_{p}},$
(6)

where brackets here and in the following are weighted antisymmetrisations,
$Y_{[a_{1}\dots a_{p}]}=\frac{1}{p!}(Y_{a_{1}\dots a_{p}}+\text{permutations
with sign})$. The explicit form of $\Delta_{p}$ can for all operators
considered in this paper be expressed using the Riemann tensor, $R_{abcd}$,
and the d’Alembertian $\Box\equiv\nabla^{a}\nabla_{a}$. On the squashed seven-
sphere, which is an Einstein space with $R=42m^{2}$,

$R_{ab}{}^{cd}=W_{ab}{}^{cd}+2m^{2}\delta_{ab}^{cd},$ (7)

where $W_{abcd}$ is the Weyl tensor given in the Appendix.

The main use of the coset master equation (2) will be to replace derivatives
by group-theoretical algebraic data. In particular, for the squashed sphere,
by squaring (2) we obtain an expression for $\Box$ which can then be used to
replace $\Box$ by algebraic data plus terms linear in $\nabla_{a}$. This will
be clear below.

##  3 Eigenvalue spectra on the squashed seven-sphere

In this section we start by briefly reviewing some of the results previously
obtained with the coset space techniques mentioned above. These results are
then generalised to the more complicated operators for which we present a
number of new eigenvalues.

###  3.1 Review of the method applied to forms of rank 0 and 1

The method outlined above becomes rather trivial when applied to 0-forms. In
this case we want to solve the eigenvalue equation

$\Delta_{0}Y=\kappa_{0}^{2}Y,$ (8)

where the positive semi-definite Hodge-de Rham operator on 0-forms is
$\Delta_{0}=-\Box$. Thus, following Duff:1986hr , the spectrum is obtained
directly by squaring the coset master equation (2):

$\Delta_{0}Y=-\Box Y=-T_{a}T_{a}Y=(C_{G}-C_{H})Y=\kappa_{0}^{2}Y,$ (9)

where the two second order Casimir operators for $G=Sp_{2}\times Sp_{1}^{C}$
and $H=Sp_{1}^{A}\times Sp_{1}^{B+C}$ have eigenvalues

$C_{G}=C(p,q)+3C^{C}(r)=\tfrac{1}{2}(p^{2}+2q^{2}+4q+6q+2pq)+\tfrac{3}{4}r(r+2),$
(10)

and

$C_{H}=2C^{A}(s)+\tfrac{6}{5}C^{B+C}(t)=\tfrac{1}{2}\,s(s+2)+\tfrac{3}{10}\,t(t+2).$
(11)

Here $(p,q)$ are Dynkin labels for $Sp_{2}$ irreps and $r,s,t$ those for the
irreps of the various $Sp_{1}$ groups that occur in the squashed seven-sphere
coset description.

The whole problem of deriving the eigenvalue spectrum of harmonics is reduced
to determining first which $H$ irreps are contained in the $Spin(7)$ irrep of
Y (trivial for 0-forms) and then finding all $G$ irreps that when split into
$H$ irreps contain any of the $H$ irreps found in the $Spin(7)$ irrep of Y.
Some partial results on the spectrum of $G$ irreps are obtained in
Nilsson:1983ru ; Duff:1986hr  by breaking up the spectrum on the round sphere
according to $Spin(8)\rightarrow Sp_{2}\times Sp_{1}$ while the entire
spectrum of all operators is derived directly from the squashed sphere coset
in Nilsson:2018lof  using the coset method just described.

For 0-forms $C_{H}=0$ and the spectrum therefore becomes
$\kappa^{2}_{0}=C_{G}$. However, as explained above, we should introduce the
scale parameter $m$. In view of this we can use $m^{2}=\tfrac{9}{20}$ to write
the eigenvalues as follows:555We will in the rest of this paper display
eigenvalues either as $\kappa_{p}^{2}$ or, equivalently, as $\Delta_{p}$.

$\Delta_{0}=\kappa^{2}_{0}=\frac{m^{2}}{9}20\,C_{G}(p,q;r).$ (12)

As an example how this result is used let us consider the spectrum in the
graviton sector666As in Nilsson:2018lof  we will use the notation of
Breitenlohner:1982jf  which includes the parity of the field, as in, e.g.,
$2^{+}$.: In units of $m$ the $SO(3,2)$ irrep defining energy $E_{0}(spin)$ is
Heidenreich:1982rz ; Breitenlohner:1982bm ; Englert:1983rn

$E_{0}(2^{+})=\frac{3}{2}+\frac{1}{2}\sqrt{\frac{M_{2}^{2}}{m^{2}}+9}=\frac{3}{2}+\frac{1}{2}\sqrt{\frac{20C_{G}}{9}+9}=\frac{3}{2}+\frac{1}{6}\sqrt{20C_{G}+81},$
(13)

where we have used the fact that the mass-square operator for spin 2 fields in
$AdS_{4}$ is $M^{2}_{2}=\Delta_{0}$. We will occasionally refer to results
like this as being of square-root form, here a $\sqrt{81}$-form. The reason
for this is that all the fields in a supermultiplet must be of the same
square-root form.

We now turn to the 1-form case and review the result and the derivation in
Duff:1986hr 777For a different derivation see Yamagishi:1983ri .. In this case
the structure constant term in the coset master equation comes in and
complicates the calculation somewhat. As for the 0-form modes we write out the
$\Delta_{1}$ eigenvalue equation explicitly:

$\Delta_{1}:\,\,\,\,\Delta_{1}Y_{a}=-\Box Y_{a}+R_{a}{}^{b}Y_{b}=-\Box
Y_{a}+6m^{2}Y_{a}=\kappa_{1}^{2}Y_{a},$ (14)

We want to use the square of the coset master equation to eliminate the $\Box$
term: Inserting $f_{abc}=-\tfrac{2}{3}ma_{abc}$ into (2) gives
$\nabla_{a}Y_{b}-\frac{m}{3}a_{abc}Y_{c}=-T_{a}Y_{b}$ which when squared
yields

$G/H:\,\,\,\,\,\Box
Y_{a}+\frac{2m}{3}a_{abc}\nabla_{b}Y_{c}+\frac{m}{3}(\nabla_{b}a_{abc})Y_{c}-\frac{m^{2}}{9}a_{abc}a_{bcd}Y_{d}=T_{b}T_{b}Y_{a}.$
(15)

Using the octonionic identities $\nabla_{a}a_{bcd}=mc_{abcd}$ and
$a_{abe}a^{cd}{}_{e}=2\delta_{ab}^{cd}+c_{ab}{}^{cd}$ (see the Appendix)
together with $T_{a}T_{a}=-(C_{G}-C_{H})$, the $G/H$ equation above simplifies
to

$-\Box Y_{a}-\frac{2m}{3}a_{abc}\nabla_{b}Y_{c}+6m^{2}Y_{a}=C_{G}Y_{a}.$ (16)

Here we have also used the fact that the irrep ${\bf 7}$ of $SO(7)$ splits
into $(1,1)\oplus(0,2)$ of $H=Sp_{1}^{A}\times Sp_{1}^{B+C}$ and that
$C_{H}=\tfrac{12}{5}$ for both of these $H$ irreps. So eliminating the $\Box$
term from the $\Delta_{1}$ and $G/H$ equations above gives

$(\kappa_{1}^{2}-C_{G})Y_{a}=\frac{2m}{3}a_{abc}\nabla_{b}Y_{c}.$ (17)

In order to extract the eigenvalues $\kappa_{1}^{2}$ from this equation we
will have to square it. Let us define the operator

$DY_{a}\equiv a_{abc}\nabla_{b}Y_{c}.$ (18)

Taking the square of $D$ then goes as follows:

$D^{2}Y_{a}=a_{abc}\nabla_{b}a_{cde}\nabla_{d}Y_{e}=a_{abc}(\nabla_{b}a_{cde})\nabla_{d}Y_{e}+a_{abc}a_{cde}\nabla_{b}\nabla_{d}Y_{e}.$
(19)

Using identities from the Appendix this equation becomes

$D^{2}Y_{a}=ma_{abc}c_{bcde}\nabla_{d}Y_{e}+(2\delta_{ab}^{de}+c_{ab}{}^{de})\nabla_{b}\nabla_{d}Y_{e}=4mDY_{a}+(6m^{2}-\Box)Y_{a}.$
(20)

Here we have also used the Ricci identity on 1-forms
$[\nabla_{a},\nabla_{b}]Y_{c}=R_{abc}{}^{d}Y_{d}$ and the fact that the
Riemann tensor term in the computation of $D^{2}$ above drops due to its
contraction with $c_{eabc}$. Then since the last term in the equation for
$D^{2}Y_{a}$ above is just $\Delta_{1}$, the equation can be written

$D^{2}Y_{a}-4mDY_{a}-\kappa_{1}^{2}Y_{a}=0.$ (21)

If we now use $DY_{a}=\frac{3}{2m}(\kappa_{1}^{2}-C_{G})Y_{a}$ and
$D^{2}Y_{a}=(\frac{3}{2m}(\kappa_{1}^{2}-C_{G}))^{2}Y_{a}$ we get the final
result for the 1-form eigenvalues

$\Delta_{1}=\frac{m^{2}}{9}(20C_{G}+14\pm
2\sqrt{20C_{G}+49})=\frac{m^{2}}{9}(\sqrt{20C_{G}+49}\pm 1)^{2}-4m^{2}.$ (22)

The last form of the answer will be useful later.

###  3.2 Spin 1/2 by squaring

Before turning our attention to forms of rank two and three, and after that
second rank symmetric tensors, we will check that our methods are able to
produce the known result for the Dirac operator acting on spin 1/2 modes. The
spectrum of $\not{D}_{1/2}\equiv-i\not{\nabla}$ was derived in Nilsson:1983ru
by a different method based on the fibre bundle description of the squashed
seven-sphere. The virtue of the method in Nilsson:1983ru  is that one can
follow the eigenvalues as one turns the squashing parameter from the round
sphere value to the squashed Einstein space value. This nice feature is
unfortunately lacking for the methods used in this paper.

To apply our present techniques we start by squaring the Dirac operator in
order to obtain a situation that is similar to the one for the Hodge-de Rham
operators on $p$-forms. Using
$\Gamma^{a}\nabla_{a}\Gamma^{b}\nabla_{b}\psi=(\Box+\tfrac{1}{2}\Gamma^{ab}[\nabla_{a},\nabla_{b}])\psi$
we find

$-i\not{\nabla}\psi=\lambda\psi\Rightarrow(-\Box+\frac{R}{4})\psi=\lambda^{2}\psi.$
(23)

It is now possible to use the $G_{2}$ structure to split the tangent space
spinor irrep into $G_{2}$ irreps as ${\bf 8}\rightarrow{\bf 7}\oplus{\bf 1}$,
or in terms of indices $A=(a,8)$. Then by defining two spinors $\eta$ and
$\eta_{a}$ we can expand a general Dirac spinor as follows

$\psi=V^{a}\eta_{a}+f\,\eta,$ (24)

where $V^{a}$ is a vector field and $f$ a scalar field on the seven-sphere.
While $\eta$ is the standard Killing spinor with components
$\eta_{B}=\delta_{B}^{8}$, $\eta_{a}$ is defined by
$\eta_{a}\equiv-i\Gamma_{a}\eta$ which implies the its explicit form is
$(\eta_{a})_{B}=\delta_{aB}$. These spinors are linearly independent and
satisfy

$\nabla_{a}\eta=\tfrac{m}{2}\eta_{a},\,\,\,\,\,\nabla_{a}\eta_{b}=-\tfrac{m}{2}\delta_{ab}\eta+\tfrac{m}{2}a_{abc}\eta_{c}.$
(25)

These equations imply

$\Box\eta=-\tfrac{7m^{2}}{4}\eta,\,\,\,\,\Box\eta_{a}=-\tfrac{7m^{2}}{4}\eta_{a},$
(26)

from which we obtain the equations

$\displaystyle\Box(f\eta)$ $\displaystyle=$ $\displaystyle(\Box
f-\tfrac{7m^{2}}{4}f)\eta+m(\nabla^{a}f)\eta_{a},$ (27)
$\displaystyle\,\,\,\,\Box(V^{a}\eta_{a})$ $\displaystyle=$
$\displaystyle(\Box
V^{a}-\tfrac{7m^{2}}{4}V^{a})\eta_{a}+m(\nabla^{a}V^{b})a_{abc}\eta_{c}.$ (28)

Consider now the coset master equation for Dirac spinor modes. Squaring it
gives

$\Box\psi-\tfrac{m}{6}a_{abc}\Gamma_{ab}\nabla_{c}\psi-\tfrac{7m^{2}}{12}\psi+\tfrac{m^{2}}{144}c_{abcd}\Gamma_{abcd}\psi=T_{a}T_{a}\psi.$
(29)

So eliminating the $\Box$ term and inserting $R=42m^{2}$ we find that the
equation we need to solve reads

$-\lambda^{2}\psi-\tfrac{m}{6}a_{abc}\Gamma_{ab}\nabla_{c}\psi+\tfrac{119m^{2}}{12}\psi+\tfrac{m^{2}}{144}c_{abcd}\Gamma_{abcd}\psi=T_{a}T_{a}\psi.$
(30)

In order to get the last term on the LHS in a nice form we introduce the
projection operators for the $G_{2}$ split $\psi=\psi_{1}+\psi_{7}$
corresponding to ${\bf 8}\rightarrow{\bf 1}\oplus{\bf 7}$. They are (see
Appendix for more details)

$P^{s}_{1}=\tfrac{1}{8}(1-\tfrac{1}{24}c_{abcd}\Gamma_{abcd}),\,\,\,\,P^{s}_{7}=1-P_{1}=\tfrac{1}{8}(7+\tfrac{1}{24}c_{abcd}\Gamma_{abcd}).$
(31)

In terms of these projectors the sum of the last two terms on the LHS above
becomes

$\tfrac{119}{12}\psi+\tfrac{1}{144}c_{abcd}\Gamma_{abcd}\psi=\tfrac{35}{4}P^{s}_{1}+\tfrac{121}{12}P^{s}_{7}.$
(32)

Then using the octonionic version of the seven-dimensional gamma matrices
given in the Appendix, some algebra gives

$a_{abc}\Gamma_{ab}\nabla_{c}\psi=(-6\nabla_{a}V^{a}-21mf)\eta+(2a^{abc}\nabla_{a}V_{b}-9mV^{c}+6(\nabla^{c}f))\eta_{c}.$
(33)

Inserting these results into (30) we get an equation that can be hit by the
projectors $P_{1}^{s}$ and $P_{7}^{s}$ leading to the two equations

$\lambda^{2}f=\tfrac{20}{9}m^{2}C_{G}f+\tfrac{35}{4}m^{2}f+m(\nabla_{a}V^{a}+\tfrac{7}{2}mf),$
(34)

and

$\lambda^{2}V_{a}=\tfrac{20}{9}m^{2}(C_{G}-\tfrac{12}{5})V_{a}+(\tfrac{121}{12}+\tfrac{3}{2})m^{2}V_{a}-\tfrac{1}{3}ma_{abc}\nabla_{b}V_{c}-m\nabla_{a}f.$
(35)

To solve these equations we either have $f=0$ or $f\neq 0$. In the former case
the first equation gives $\nabla^{a}V_{a}=0$ which implies that the second
equation for $V_{a}$ can be analysed in exactly the same way as for the
1-forms discussed previously. Not surprisingly the result is also of
$\sqrt{49}$-form and reads

$\lambda^{2}=m^{2}(\tfrac{1}{6}\pm\tfrac{1}{3}\sqrt{20C_{G}+49})^{2}.$ (36)

Turning to the latter case, that is $f\neq 0$, we start by taking the
divergence of the second equation and use the fact that $\Box
f=-\frac{20m^{2}}{9}C_{G}f$ to find

$(\tfrac{20}{9}m^{2}C_{G}+\tfrac{25}{4}m^{2}-\lambda^{2})(\nabla_{a}V^{a})=-\tfrac{20}{9}m^{3}C_{G}\,f.$
(37)

Now the system of two equations for the functions $f$ and $\nabla^{a}V_{a}$
has the solutions

$\lambda^{2}=m^{2}(\tfrac{1}{2}\pm\tfrac{1}{3}\sqrt{20C_{G}+81})^{2}.$ (38)

These two sets of eigenvalues are consistent with the known result from
Nilsson:1983ru . However, in that paper there is no sign ambiguity since there
one obtains $\lambda$ instead of $\lambda^{2}$. This problem is easily
eliminated by applying our present result to the spin $2^{+}$ supermultiplet.

This completes the review. We now turn to 2-forms where the results are new.

###  3.3 2-forms

In terms of the Hodge-de Rham operator on 2-forms, the equation to be solved
is

$\Delta_{2}Y_{ab}=-\Box
Y_{ab}-2R_{acbd}Y^{cd}-2R_{[a}{}^{c}Y_{b]c}=\kappa_{2}^{2}Y_{ab}.$ (39)

Using (7) for the Riemann tensor gives the following equation for the box
operator

$\Delta_{2}:\,\,\,\,\Box
Y_{ab}=-\kappa^{2}_{2}Y_{ab}-2W_{acbd}Y^{cd}+10m^{2}Y_{ab}.$ (40)

As for the previous cases the next step is to express the box operator in
terms of algebraic objects on the coset manifold. The master coset formula (2)
for 2-forms reads

$\nabla_{a}Y_{bc}=-T_{a}Y_{bc}-f_{a}{}^{d}{}_{[b}Y_{c]d},$ (41)

which can be written

$\tilde{\nabla}_{a}Y_{bc}=-T_{a}Y_{bc},$ (42)

by indroducing the ”$G_{2}$-derivative” $\tilde{\nabla}_{a}$ defined on
2-forms by

$\tilde{\nabla}_{a}Y_{bc}\equiv\nabla_{a}Y_{bc}+\frac{2m}{3}a_{a[b}{}^{d}Y_{c]d}.$
(43)

This derivative is ”$G_{2}$” in the sense that it satisfies
$\tilde{\nabla}_{a}a_{bcd}=0$ and $\tilde{\nabla}_{a}c_{bcde}=0$. This step is
not crucial here but we will find more implications later of the presence of
the $G_{2}$ holonomy so we will have reason to return to this derivative then.
(For more details about $G_{2}$ in this context, see for instance House:2004pm
.) To get an algebraic expression for the box operator we can now square the
master coset equation, $\tilde{\nabla}_{a}Y_{bc}=-T_{a}Y_{bc}$, to get
$\tilde{\Box}Y_{ab}=T_{c}T_{c}Y_{ab}$. This equation can also be written as

$G/H:\,\,\,\,\,\tilde{\Box}Y_{ab}=(\Box-\frac{10}{9}m^{2})Y_{ab}+\frac{4}{3}ma_{cd[a}\nabla_{|c}Y_{d|b]}-\frac{2}{9}m^{2}c_{abcd}Y_{cd}=T_{c}T_{c}Y_{ab}.$
(44)

Finally, to get the equation that needs to be solved to find the eigenvalues
we just eliminate the box operator from (40) and (44). Using the projectors
defined in the Appendix gives the following useful form of the resulting
equation

$(\kappa^{2}_{2}-12m^{2}+T_{c}T_{c})Y_{ab}=-W_{abcd}Y^{cd}-\frac{4}{3}m^{2}(P_{14}Y)_{ab}+\frac{4}{3}m(a_{cd[a}\tilde{\nabla}_{|c}Y_{d|b]}),$
(45)

which, again, is not entirely algebraic due to the appearance of the operator
$\tilde{\nabla}_{a}$ in the last term. This also happened in the case of the
1-form where it was easy to handle by a squaring procedure. This trick is a
bit harder to apply in the case of 2-forms (as well as for the other operators
to be discussed later) as will become clear shortly. Compared to the 1-form
case there is also a new feature here namely the presence of the Weyl tensor
in one of the terms which will cause additional complications. The approach
used in this paper to deal with the Weyl tensor terms, in this and the other
cases discussed below, is explained in the Appendix.

Thus there are two new issues when trying to solve the 2-form equation (45):
The Weyl tensor term and the $\tilde{\nabla}$ term which now tends to lead to
symmetrised derivatives when squared. To deal with the former issue we recall
how $T_{a}T_{a}$ can be expressed in terms of the Casimir operators for the
groups involved: $T_{a}T_{a}=-(C_{G}-C_{H})$. This, however, has implications
for the how the spectrum is organised in terms of towers. To see this we use
the tangent space decomposition $SO(7)\rightarrow G_{2}\rightarrow
Sp^{A}_{1}\times Sp^{B+C}_{1}$ which makes it possible to read off the
relevant decompositions directly from the McKay and Patera tables Mckay:1981mp
. In the case of the 2-form the composition reads, see Duff:1986hr  or the
summary in Nilsson:2018lof ,

$\bf{21}\rightarrow\bf{7}\oplus\bf{14}\rightarrow(\bf{(1,1)}\oplus\bf{(0,2)})\oplus((0,2)\oplus(1,3)\oplus(2,0)).$
(46)

This means that the towers will be tabulated according to their $H$ irrep
which will thus obscure the $G_{2}$ structure of the spectrum. As is explained
in the Appendix this problem is automatically eliminated once the Weyl tensor
term is analysed and the result combined with the result from the $C_{H}$
term. In fact, adding these two terms gives
$\tfrac{24}{5}Y_{ab}=\tfrac{32}{3}m^{2}Y_{ab}$ for all the $H$ irreps in the
$G_{2}$ irrep ${\bf 14}$. Recall that the corresponding answer for the irrep
${\bf 7}$ is $\tfrac{12}{5}=\tfrac{16}{3}m^{2}$.

Using this insight from the Appendix, namely that the sum $C_{H}+Weyl$ has a
common value on all $H$ irreps in the decomposition of each $G_{2}$ irrep
arising from the $SO(7)$ representation ${\bf 21}$, it becomes possible to
split the 2-form equation into two by acting on it with the projectors $P_{7}$
and $P_{14}$ and insert the respective values of $C_{H}+Weyl$. We find, using
the definitions $Y^{(7)}_{ab}\equiv(P_{7}Y)_{ab}$ and $Y_{a}\equiv
a_{abc}Y_{bc}^{(7)}$,

${\bf
7}:\,\,\,\,\,\,\kappa_{2}^{2}Y_{a}=C_{G}Y_{a}+(12+\frac{4}{3}-\frac{16}{3})m^{2}Y_{a}-\frac{4m}{3}(a_{abc}\nabla_{b}Y_{c}),$
(47) ${\bf
14}:\,\,\,\,\,\,\kappa_{2}^{2}Y^{(14)}_{ab}=C_{G}Y^{(14)}_{ab}+(12-\frac{4}{3}-\frac{32}{3})m^{2}Y^{(14)}_{ab}+\frac{2m}{3}(P_{14})_{ab}^{cd}(\nabla_{c}Y_{d}).$
(48)

Note that the last term in the second equation contains $Y_{a}$ and thus seems
to mix with the first equation. However, the structure of the derivative terms
as a 2 by 2 matrix shows that this term will have no role in determining the
eigenvalues for the 2-form as will be clear below. To proceed we write these
two equations as

${\bf
7}:\,\,\,\,\,\,\kappa_{2}^{2}Y_{a}=\frac{m^{2}}{9}(20C_{G}+72)Y_{a}-\frac{4}{3}m(a_{abc}\nabla_{b}Y_{c}),$
(49) ${\bf
14}:\,\,\,\,\,\,\kappa_{2}^{2}Y^{(14)}_{ab}=\frac{m^{2}}{9}\,20C_{G}Y^{(14)}_{ab}+\frac{2}{3}m(P_{14}\nabla
Y)_{ab}.$ (50)

Consider first the possibility to take the square of the first equation above,
the one for $Y_{a}$. However, although this will give rise to a calculation
quite similar to the one for the 1-form, there is one important difference.
While the 1-form is transverse (i.e., divergence free) this is not the case
for $Y_{a}$ coming from the 2-form. Therefore one needs to analyse the two
equations for the 2-form in stages:

1) Either $Y_{a}=0$ or $Y_{a}\neq 0$, 2) when $Y_{a}\neq 0$ then either
$\nabla_{a}Y_{a}=0$ or $\nabla_{a}Y_{a}\neq 0$.

So, when $Y_{a}=0$ the equation left to solve is the ${\bf 14}$-equation with
the derivative term set to zero. This equation therefore gives the eigenvalues

$\Delta_{2}=\frac{m^{2}}{9}\,20\,C_{G}.$ (51)

According to the characterisation mention above, this result leads to an
energy $E_{0}$ of the $\sqrt{9}$-form. This fits nicely with a supermultiplet
containing spins $(1^{+},1/2,1/2,0^{+})$ with masses related to the operators
$(\Delta_{2},{\not{D}}_{3/2},{\not{D}}_{3/2},\Delta_{L})$ since we know from
Ref. [198] in Duff:1986hr  that $\Delta_{L}$ also has an eigenvalue leading to
a $\sqrt{9}$-form. One can check that the corresponding values of $E_{0}$ work
out as they should in relation to supersymmetry. This eigenvalue of
$\Delta_{L}$ leading to the $\sqrt{9}$-form will be derived in detail later in
this section.

We now turn to the cases with $Y_{a}\neq 0$, namely $\nabla_{a}Y_{a}=0$ and
$\nabla_{a}Y_{a}\neq 0$. Clearly, when $\nabla_{a}Y_{a}=0$ the calculation
will follow the one for 1-forms very closely. Indeed, the result is also of
$\sqrt{49}$-form and reads

$\Delta_{2}=\frac{m^{2}}{9}(\sqrt{20C_{G}+49}\pm 2)^{2}-m^{2}.$ (52)

Although being of $\sqrt{49}$-form, this 2-form eigenvalue formula differs
from the one for 1-forms given in Duff:1986hr , and also derived above,

$\Delta_{1}=\frac{m^{2}}{9}(20C_{G}+14\pm
2\sqrt{20C_{G}+49})=\frac{m^{2}}{9}(\sqrt{20C_{G}+49}\pm 1)^{2}-4m^{2}.$ (53)

The differences are partly compensated for by the different relations between
the Hodge-de Rham operators and the respective $M^{2}$ operators Duff:1986hr :

$M^{2}(1^{+})=\Delta_{2},$ (54) $M^{2}(1^{-})=\Delta_{1}+12m^{2}\pm
6m\sqrt{\Delta_{1}+4m^{2}}=(\sqrt{\Delta_{1}+4m^{2}}\pm 3m)^{2}-m^{2}.$ (55)

If the $\Delta_{1}$ eigenvalues are inserted into this formula for
$M^{2}(1^{-})$ we see that the last term, that is $-4m^{2}$ in $\Delta_{1}$,
is really required to have a chance for supersymmetry to work in a
supermultiplet containing both $1^{+}$ and $1^{-}$ fields. In fact, the field
content of the spin 3/2 supermultiplets has this property and requires
$E_{0}(1^{+})=E_{0}(1^{-})\pm 1$. To verify this relation we need the energy
formula for spin 1 unitary $SO(3,2)$ irreps

$E_{0}(1^{\pm})=\frac{3}{2}+\frac{1}{2}\sqrt{\frac{M^{2}}{m^{2}}+1}.$ (56)

This expression gives the energy values

$E_{0}(1^{-(1),(2)})=\frac{3}{2}\pm\frac{3}{2}\mp\frac{1}{6}+\frac{1}{6}\sqrt{20C_{G}+49},$
(57)

where the first $\pm$ sign refers to the two towers of spin $1^{-(1),(2)}$
fields in $AdS_{4}$ supergravity theory. Comparing these to

$E_{0}(1^{+})=\frac{3}{2}\pm\frac{1}{3}+\frac{1}{6}\sqrt{20C_{G}+49},$ (58)

one finds (using also the spectrum of the spin 3/2 fields in these
supermultiplets) that it is possible to eliminate the sign ambiguities in the
spectra of $\Delta_{1}$ and $\Delta_{2}$. For more details see Nilsson:2021kn
.

Finally, when $\nabla_{a}Y_{a}\neq 0$ we take the divergence of the ${\bf 7}$
projected part of the equation and use $a_{abc}\nabla_{a}\nabla_{b}Y_{c}=0$ to
obtain a simple equation giving the eigenvalue

$\kappa^{2}_{2}=\frac{m^{2}}{9}(20C_{G}+72)$ (59)

This completes the analysis of the 2-form eigenvalue spectrum. A list of the
obtained eigenvalues can be found in the Conclusions.

###  3.4 3-forms

We now turn to 3-forms leaving the discussion of the Lichnerowicz modes to the
next subsection. The reason for doing the analysis in this order is that we
will obtain some equations in the 3-form case that can be applied also for
Lichnerowicz.

The equation for 3-forms that we wish to solve reads

$\Delta_{3}Y_{abc}=-\Box
Y_{abc}+6R^{d}{}_{[ab}{}^{e}Y_{c]de}+3R_{[a}{}^{d}Y_{bc]d}=\kappa_{3}^{2}Y_{abc}.$
(60)

Inserting the Riemann and Ricci tensors of the squashed seven-sphere this
equation becomes

$\Delta_{3}:\,\,\,\,\,\Box
Y_{abc}=-\kappa_{3}^{2}Y_{abc}+6W^{d}{}_{[ab}{}^{e}Y_{c]de}+12m^{2}Y_{abc}.$
(61)

Squaring the coset master equation for 3-forms gives

$G/H:\,\,\,\,\,-(C_{G}-C_{H})Y_{abc}=\Box
Y_{abc}-2ma_{d[a}{}^{e}\tilde{\nabla}_{|d]}Y_{bc]e}+\frac{m^{2}}{3}(4Y_{abc}+2c_{[ab}{}^{de}Y_{c]de}),$
(62)

where for later convenience we have used the $G_{2}$ derivative

$\tilde{\nabla}_{d}Y_{abc}\equiv\nabla_{d}Y_{abc}-ma_{d[a}{}^{e}Y_{bc]e}.$
(63)

Note that this implies that the divergence

$\tilde{\nabla}^{a}Y_{abc}=\nabla^{a}Y_{abc}-\tfrac{m}{3}(a_{aa}{}^{e}Y_{bce}+2a_{a[b}{}^{e}Y_{c]ae})=\tfrac{2m}{3}a_{[b}{}^{de}Y_{c]de}.$
(64)

Eliminating the box-operator term from the above equations gives

$\kappa_{3}^{2}Y_{abc}=C_{G}Y_{abc}-(C_{H}Y_{abc}-6W^{d}{}_{[ab}{}^{e}Y_{c]de})+\frac{m^{2}}{3}(40Y_{abc}+2c_{[ab}{}^{de}Y_{c]de})-2ma_{d[a}{}^{e}\tilde{\nabla}^{d}Y_{bc]e}.$
(65)

To proceed we decompose the 3-form into $G_{2}$ irreps as ${\bf 35}={\bf
1}\oplus{\bf 7}\oplus{\bf 27}$. By defining $Y\equiv a_{abc}Y_{abc}$ and
$Y_{a}\equiv c_{abcd}Y_{bcd}$ we can split the 3-form as follows

$Y_{abc}=Y^{(1)}_{abc}+Y^{(7)}_{abc}+Y^{(27)}_{abc}=\frac{1}{42}a_{abc}Y-\frac{1}{24}c_{abcd}Y_{d}+(P_{27}Y)_{abc},$
(66)

where we have utilised the 3-form projectors defined in the Appendix. The term
$C_{H}Y_{abc}-6W^{d}{}_{[ab}{}^{e}Y_{c]de}$ can be written (see Appendix)

$C_{H}Y_{abc}-6W^{d}{}_{[ab}{}^{e}Y_{c]de}=-\frac{12}{5}(\frac{1}{24}c_{abcd}Y_{d})+\frac{28}{5}(P_{27}Y)_{abc}.$
(67)

Apart from this there are two more problematic terms, the $c_{abcd}$-term and
the derivative term that will force us to square also this equation.

However, before doing that we will follow the strategy in the 2-form case and
start by splitting the equation into $G_{2}$ pieces. The singlet term ${\bf
1}$ is obtained by contracting all three indices by $a_{abc}$ which gives

$\kappa_{3}^{2}Y=C_{G}Y+\frac{m^{2}}{3}(40Y+2a_{abc}c_{[ab}{}^{de}Y_{c]de})-2ma_{abc}a_{d[a}{}^{e}\tilde{\nabla}^{d}Y_{bc]e}.$
(68)

Cleaning up the structure constant factors this equation becomes

${\bf
1}:\,\,\,\,\,\,\kappa_{3}^{2}Y=(C_{G}+16m^{2})Y+2m\tilde{\nabla}_{a}Y_{a}.$
(69)

Turning to the ${\bf 7}$ part, we find after some algebra

${\bf
7}:\,\,\,\,\,\,\kappa_{3}^{2}Y_{a}=(C_{G}+6m^{2})Y_{a}-\frac{12m}{7}\nabla_{a}Y+\frac{m}{3}a_{abc}\nabla_{b}Y_{c}+2ma_{bcd}\nabla_{b}Y^{(27)}_{cda}.$
(70)

When we now come to the last part, the ${\bf 27}$, it will require some new
steps that will relate it to the metric and the Lichnerowicz equation. This
connection is in fact already indicated by the last term in the ${\bf 7}$
equation just discussed: $a_{bcd}\nabla_{b}Y^{(27)}_{cda}$. This expression
suggests that the 2-index tensor $a_{acd}Y^{(27)}_{bcd}$ will be useful as we
now elaborate upon (see House:2004pm  for a closely related discussion). Let
us define a 2-index tensor from the ${\bf 27}$ part of the 3-form by

$\tilde{Y}_{ab}\equiv a_{acd}Y^{(27)}_{bcd},$ (71)

where we use a tilde to avoid confusing it with the antisymmetric 2-form
discussed previously. Clearly, the symmetric and traceless part of
$\tilde{Y}_{ab}$ is also in the irrep ${\bf 27}$. However, a nice and indeed
very useful fact about this definition of $\tilde{Y}_{ab}$ is that it is
automatically symmetric and traceless. The tracelessness follows immediately
from the identity $a_{abc}c_{abcd}=0$, or, using the 3-form projectors in the
Appendix, from $P_{1}P_{27}=0$, while the vanishing of its antisymmetric part
follows from $a_{abc}\tilde{Y}_{bc}=0$ and $c_{abcd}\tilde{Y}_{cd}=0$888Note
that these two conditions can be combined to the identity since
$\delta_{ab}^{cd}=\tfrac{1}{2}(a_{abe}a^{cde}-c_{ab}{}^{cd})$.. These two
results can be checked as follows:

$a_{abc}\tilde{Y}_{bc}=a_{abc}a_{bde}Y^{(27)}_{cde}=-(2\delta_{ac}^{de}+c_{acde})Y^{(27)}_{cde}=-c_{acde}Y^{(27)}_{cde}=0,$
(72)

where we have used $P_{7}P_{27}=0$ in the last step, and

$c_{abcd}\tilde{Y}_{cd}=c_{abcd}a_{cef}Y^{(27)}_{def}=-6a_{[ab}{}^{[e}\delta_{d]}^{f]}Y^{(27)}_{def}=4a_{[a}{}^{de}Y^{(27)}_{b]de}=0.$
(73)

Here the very last equality is a consequence of the identity

$a_{abc}(P_{27})_{bcd}{}^{efg}=a_{(a}{}^{[ef}\delta_{d)}^{g]}-\frac{1}{7}\delta_{ad}a^{efg}.$
(74)

Clearly, also $a_{abc}(P_{27})_{abc}{}^{efg}=0$ again showing that the trace
of $\tilde{Y}_{ab}$ vanishes.

An alternative, and perhaps more direct, way to see that $\tilde{Y}_{ab}$ is
symmetric and traceless is to just insert the expression for the $P_{27}$
projector which gives

$a_{abc}Y^{(27)}_{bcd}=\frac{1}{2}(a_{abc}Y_{bcd}+a_{dbc}Y_{bca})-\frac{1}{7}\delta_{ad}Y.$
(75)

Finally, for the definition $\tilde{Y}_{ab}\equiv a_{acd}Y^{(27)}_{bcd}$ to be
useful we need to give also the inverse relation999Here one may use the fact
that $Y^{(27)}_{abc}=(P_{27}Y^{(27)})_{abc}$ implies the identity
$Y^{(27)}_{abc}=-\tfrac{3}{2}c_{de[ab}Y_{c]de}^{(27)}$.:

$Y^{(27)}_{abc}=\frac{3}{4}a_{d[ab}\tilde{Y}_{c]d}.$ (76)

The decomposition of the 3-form therefore takes the following simple form in
terms of $Y,Y_{a},\tilde{Y}_{ab}$, respectively in irreps ${\bf 1},{\bf
7},{\bf 27}$,

$Y_{abc}=\frac{1}{42}a_{abc}Y-\frac{1}{24}c_{abcd}Y_{d}+\frac{3}{4}a_{d[ab}\tilde{Y}_{c]d}.$
(77)

With these preliminary results at hand it is now a rather straightforward
matter to project the equation (65) onto its ${\bf 27}$ component. One finds

$\kappa_{3}^{2}\tilde{Y}_{ab}=(C_{G}-\frac{m^{2}}{3}(40-\frac{4}{3}))\tilde{Y}_{ab}-(C_{H}\delta_{(ab)}^{cd}+2W_{acbd})\tilde{Y}_{cd}-\frac{2}{3}ma_{cd(a}\tilde{\nabla}^{c}\tilde{Y}_{b)}{}^{d}-\frac{m}{3}(\tilde{\nabla}_{(a}Y_{b)}-\frac{1}{7}\delta_{ab}\tilde{\nabla}^{c}Y_{c}).$
(78)

Again we can replace the second bracket by its common eigenvalue on all $H$
irreps in the $G_{2}$ irrep ${\bf 27}$ as explained in the Appendix, that is
with $\tfrac{28}{5}=\tfrac{112m^{2}}{9}$.

We now insert this into the ${\bf 27}$ equation and sum up what we have found
so far:

${\bf 1}:\,\,\,\,\kappa_{3}^{2}Y=(C_{G}+16m^{2})Y+2m\tilde{\nabla}_{a}Y_{a},$
(79) ${\bf
7}:\,\,\,\,\kappa_{3}^{2}Y_{a}=(C_{G}+6m^{2})Y_{a}-\frac{12m}{7}\nabla_{a}Y+\frac{m}{3}a_{abc}\nabla_{b}Y_{c}+2ma_{bcd}\nabla_{b}Y^{(27)}_{cda}.$
(80) ${\bf
27}:\,\,\,\,\kappa_{3}^{2}\tilde{Y}_{ab}=(C_{G}+\frac{4m^{2}}{9})\tilde{Y}_{ab}-\frac{2}{3}ma_{cd(a}\tilde{\nabla}^{c}\tilde{Y}_{b)}{}^{d}-\frac{m}{3}(\tilde{\nabla}_{(a}Y_{b)}-\frac{1}{7}\delta_{ab}\tilde{\nabla}^{c}Y_{c}).$
(81)

Note that the term $2ma_{bcd}\nabla_{b}Y^{(27)}_{cda}$ appearing in the ${\bf
7}$ equation can be replaced by terms containing only $Y$ and $Y_{a}$. This
relation is derived from the gauge condition $\nabla^{a}Y_{abc}=0$ as follows:

$\nabla^{a}(Y^{1}_{abc}+Y_{abc}^{7}+Y_{abc}^{27})=0.$ (82)

Contracting this equation with $a_{dbc}$ and using
$\nabla_{a}a_{bcd}=mc_{abcd}$ we get

$\nabla^{a}a_{dbc}(Y^{1}_{abc}+Y_{abc}^{7}+Y_{abc}^{27})-mc_{adbc}(Y^{1}_{abc}+Y_{abc}^{7}+Y_{abc}^{27})=0.$
(83)

However, since $c_{abcd}$ projects onto $Y_{a}$ we can use the relations

$a_{dbc}Y_{abc}^{1}=\frac{1}{42}a_{dbc}a_{abc}Y=\frac{1}{7}\delta_{da}Y,\,\,\,\,a_{dbc}Y_{abc}^{7}=-\frac{1}{24}a_{dbc}c_{abce}Y_{e}=-\frac{1}{6}a_{dab}Y_{b},\,\,\,a_{dbc}Y_{abc}^{27}=\tilde{Y}_{da},$
(84)

to rewrite the above 3-form gauge condition as

$\nabla^{a}\tilde{Y}_{ab}+\frac{1}{7}\nabla_{b}Y-\frac{1}{6}a_{bcd}\nabla_{c}Y_{d}+mY_{b}=0.$
(85)

This equation will be used when we summarise the system of equations to be
solved below.

Finally, we show that the last bracket in the ${\bf 27}$ equation will not
play a role in the analysis of the spectrum. This can be seen by writing the
three equations above in matrix form as

$\kappa_{3}^{2}\begin{pmatrix}Y^{1}\\\ Y^{7}\\\
Y^{27}\end{pmatrix}=\begin{pmatrix}A&B&0\\\ C&D&0\\\
0&E&F\end{pmatrix}\begin{pmatrix}Y^{1}\\\ Y^{7}\\\ Y^{27}\end{pmatrix}$ (86)

The eigenvalues must be the roots of the equation $\det(X-\kappa_{3}^{2}{\bf
1})=0$, where $X$ is the matrix appearing above. Since the element $E$ does
not enter this equation we can proceed and solve the three equations in two
steps, first the two coupled equations for $Y$ and $Y_{a}$

${\bf 1}:\,\,\,\,\kappa_{3}^{2}Y=(C_{G}+16m^{2})Y+2m\nabla_{a}Y_{a},$ (87)
${\bf
7}:\,\,\,\,\kappa_{3}^{2}Y_{a}=(C_{G}+4m^{2})Y_{a}-2m\nabla_{a}Y+\frac{2m}{3}a_{abc}\nabla_{b}Y_{c},$
(88)

and then the single equation involving only $\tilde{Y}_{ab}$

${\bf
27}:\,\,\,\,\kappa_{3}^{2}\tilde{Y}_{ab}=(C_{G}+\frac{4m^{2}}{9})\tilde{Y}_{ab}-\frac{2m}{3}a_{cd(a}\tilde{\nabla}^{c}\tilde{Y}_{b)}{}^{d}.$
(89)

We start by solving the first two coupled equations. There are two distinct
cases: either 1) $Y=0$ or 2) $Y\neq 0$.

Case 1): Setting $Y=0$ in the ${\bf 1}$ equation gives $\nabla_{a}Y_{a}=0$
which implies that the ${\bf 7}$ equation is of exactly of the same type as
the equation solved previously for 1-forms, but with different coefficients.
However, one has to pay attention to the fact that the eigenvalues dealt with
here are for 3-forms, not 1-forms, so solving the present equation for $Y_{a}$
involves some new steps. This fact will become clear directly when squaring
the operator $(DY)_{a}\equiv a_{abc}\nabla_{b}Y_{c}$:

$\displaystyle(DDY)_{a}:$ $\displaystyle=$ $\displaystyle
a_{abc}\nabla_{b}a_{cde}\nabla_{d}Y_{e}=a_{abc}a_{cde}\nabla_{b}\nabla_{d}Y_{e}+ma_{abc}c_{bcde}\nabla_{d}Y_{e}=$
(91)
$\displaystyle(2\delta_{ab}^{de}+c_{abde})\nabla_{b}\nabla_{d}Y_{e}+4ma_{ade}\nabla_{d}Y_{e}.$

Since the $c$-term vanishes this equation simplifies to

$(DDY)_{a}=4m(DY)_{a}-\Box Y_{a}+6m^{2}Y_{a}.$ (92)

At this point the $\Box Y_{a}$ term must be related to the 3-form eigenvalue
$\kappa_{3}^{2}$. Using $Y=0$ and $\nabla^{a}Y_{abc}=0$, we find that

$\displaystyle\Box Y_{a}$ $\displaystyle=$ $\displaystyle\Box
c_{abcd}Y_{bcd}=c_{abcd}\Box Y_{bcd}-2m^{2}Y_{a}=$ (94) $\displaystyle
c_{abcd}(-\kappa_{3}^{2}Y_{bcd}+6W^{e}{}_{[bc}{}^{f}Y_{d]ef}+12m^{2}Y_{bcd})-2mY_{a}=-(\kappa_{3}^{2}-10m^{2})Y_{a},$

where we have also used
$c_{abcd}W^{e}{}_{[bc}{}^{f}Y_{d]ef}=W_{adef}Y_{def}=0$ (where the first
equality follows from $P_{7}^{(2)}W=0$). The $(DDY)_{a}$ equation above then
reads

$(DDY)_{a}=4m(DY)_{a}+(\kappa_{3}^{2}-4m^{2})Y_{a}.$ (95)

Inserting the expression for $(DY)_{a}$ from the eigenvalue equation ${\bf 7}$
above we find

$(\kappa_{3}^{2}-C_{G}-4m^{2})^{2}-\frac{8m^{2}}{3}(\kappa_{3}^{2}-C_{G}-4m^{2})-\frac{4m^{2}}{9}(\kappa_{3}^{2}-4m^{2})=0,$
(96)

which has the following two solutions (inserting $1=\tfrac{20}{9}m^{2}$ in
front of $C_{G}$)

$\Delta_{3}=\frac{m^{2}}{9}(\sqrt{20C_{G}+49}\pm 1)^{2}.$ (97)

These eigenvalues are obtained for $\Delta_{3}=Q^{2}$ so it is gratifying to
find its eigenvalues come out as a square.

Case 2): Now we turn to the second case where $Y\neq 0$. Then either
$\nabla_{a}Y_{a}=0$ or $\nabla_{a}Y_{a}\neq 0$. The former case gives a simple
equation for the eigenvalue directly from the ${\bf 1}$ equation. However,
taking the divergence of the ${\bf 7}$ equation tells us that $\Box Y=0$ so
the ${\bf 1}$ equation reduces to

$\kappa_{3}^{2}=16m^{2},$ (98)

which is the eigenvalue of the singlet constant mode $Y_{abc}=a_{abc}$ as is
easily verified. So let us turn to the latter case with $\nabla_{a}Y_{a}\neq
0$. Taking the divergence of the ${\bf 7}$ equation gives

$\Box Y=\frac{1}{2m}(-\kappa_{3}^{2}+C_{G}+4m^{2})\nabla_{a}Y_{a}.$ (99)

Inserting this equation back into the ${\bf 1}$ equation, recalling that $\Box
Y=-C_{G}Y$, gives

$(\kappa_{3}^{2}-C_{G}-16m^{2})(\kappa_{3}^{2}-C_{G}-4m^{2})=\frac{9}{5}C_{G}.$
(100)

This equation has the following solutions (inserting again
$1=\tfrac{20}{9}m^{2}$ in front of $C_{G}$):

$\Delta_{3}=\tfrac{m^{2}}{9}(\sqrt{20C_{G}+81}\pm 3)^{2}.$ (101)

Note that the single eigenvalue found above, $\kappa_{3}^{2}=16m^{2}$, belongs
to the plus branch.

We now turn to the ${\bf 27}$ part of the 3-form equation

${\bf
27}:\,\,\,\,\kappa_{3}^{2}\tilde{Y}_{ab}=(C_{G}+\frac{4m^{2}}{9})\tilde{Y}_{ab}+\frac{2m}{3}a_{c(a}{}^{d}\tilde{\nabla}^{c}\tilde{Y}_{|d|b)}.$
(102)

The derivative term requires as usual a squaring of the whole equation. The
$G_{2}$ derivative on $\tilde{Y}_{ab}$ is given by

$(D\tilde{Y})_{ab}\equiv
2a_{cd(a}\tilde{\nabla}^{c}\tilde{Y}_{b)}{}^{d}=\tilde{\nabla}^{c}(a_{ca}{}^{d}\tilde{Y}_{db}+a_{cb}{}^{d}\tilde{Y}_{ad}),$
(103)

and computing its square, using the fact that $\tilde{\nabla}_{a}$ is zero on
both $a_{abc}$ and $c_{abcd}$, gives

$(DD\tilde{Y})_{ab}\equiv\tilde{\nabla}^{e}\tilde{\nabla}^{f}(a_{ea}{}^{c}a_{fc}{}^{d}\tilde{Y}_{db}+a_{eb}{}^{c}a_{fc}{}^{d}\tilde{Y}_{ad}+a_{ea}{}^{c}a_{fb}{}^{d}\tilde{Y}_{cd}+a_{eb}{}^{c}a_{fa}{}^{d}\tilde{Y}_{cd}).$
(104)

Clearly, the last two terms in this expression contain a symmetric combination
of the two $G_{2}$ covariant derivatives. Such terms can be a real obstacle to
carrying through the calculation but we will see below that there is a trick
that can be used to eliminate this issue. Before applying this trick we
simplify the other terms to get

$(DD\tilde{Y})_{ab}=-2\tilde{\Box}\tilde{Y}_{ab}+2\tilde{\nabla}^{c}\tilde{\nabla}_{(a}\tilde{Y}_{b)c}-2c_{cde(a}\tilde{\nabla}^{c}\tilde{\nabla}^{d}\tilde{Y}_{b)}{}^{e}+2a_{e(a}{}^{c}a_{|f|b)}{}^{d}\tilde{\nabla}^{e}\tilde{\nabla}^{f}\tilde{Y}_{cd}.$
(105)

We now address the four terms on the RHS of this equation starting with last
one. The trick used to deal with the symmetric combination of derivatives in
this term is to reintroduce the 3-form via
$Y^{(27)}_{abc}=\tfrac{3}{4}a_{d[ab}\tilde{Y}_{c]d}$ temporarily giving

$a_{e(a}{}^{c}a_{|f|b)}{}^{d}\tilde{\nabla}^{e}\tilde{\nabla}^{f}\tilde{Y}_{cd}=a_{e(a}{}^{c}\tilde{\nabla}^{e}\tilde{\nabla}^{f}a_{|f|b)}{}^{d}\tilde{Y}_{cd}=a_{e(a}{}^{c}\tilde{\nabla}^{e}\tilde{\nabla}^{f}(4Y^{(27)}_{|f|b)c}-a_{b)c}{}^{d}\tilde{Y}_{fd}-a_{|cf|}{}^{d}\tilde{Y}_{b)d}).$
(106)

The first term in the last expression is just the $G_{2}$ covariant divergence
of the 3-form which can be seen to satisfy

$\tilde{\nabla}^{a}Y_{abc}^{(27)}=\nabla^{a}Y_{abc}^{(27)}=\nabla^{a}Y_{abc}=0,$
(107)

where in the first equality we have used the fact that $\tilde{Y}_{ab}$ is
symmetric and in the second that
$\nabla^{a}Y_{abc}^{(1)}=\nabla^{a}Y_{abc}^{(7)}=0$. Thus
$\nabla^{a}Y_{abc}=0$ is implies $\tilde{\nabla}^{a}\tilde{Y}_{ab}=0$ which
gives the following much simpler form of the expression above

$a_{e(a}{}^{c}a_{|f|b)}{}^{d}\tilde{\nabla}^{e}\tilde{\nabla}^{f}\tilde{Y}_{cd}=-a_{e(a}{}^{c}\tilde{\nabla}^{e}\tilde{\nabla}^{f}a_{|cf|}{}^{d}\tilde{Y}_{b)d}=-\tilde{\Box}\tilde{Y}_{ab}+\tilde{\nabla}^{c}\tilde{\nabla}_{(a}\tilde{Y}_{b)c}+c_{(a}{}^{cde}\tilde{\nabla}^{c}\tilde{\nabla}^{d}\tilde{Y}_{b)}{}^{e}.$
(108)

Inserting this result into the above expression for $(DD\tilde{Y})_{ab}$ we
get

$(DD\tilde{Y})_{ab}=4(-\tilde{\Box}\tilde{Y}_{ab}+\tilde{\nabla}^{c}\tilde{\nabla}_{(a}\tilde{Y}_{b)c}-c_{cde(a}\tilde{\nabla}^{c}\tilde{\nabla}^{d}\tilde{Y}_{b)}{}^{e}).$
(109)

Thus we have managed to eliminate the symmetric combination of covariant
derivatives that has caused a bit of headache until now. The terms in the last
formula for $(DD\tilde{Y})_{ab}$ can be dealt with rather easily as we now
show. The first term is simply the coset master equation squared, i.e.,

$\tilde{\Box}\tilde{Y}_{ab}=-(C_{G}-C_{H})\tilde{Y}_{ab}.$ (110)

The other two terms both involve the commutator of two covariant derivatives
(since $\tilde{\nabla}^{a}\tilde{Y}_{ab}=0$). So we need to compute

$[\tilde{\nabla}_{c},\tilde{\nabla}_{d}]\tilde{Y}_{ab}=\tilde{\nabla}_{c}(\nabla_{d}\tilde{Y}_{ab}-\frac{m}{3}a_{da}{}^{e}\tilde{Y}_{eb}-\frac{m}{3}a_{db}{}^{e}\tilde{Y}_{ae})-(c\leftrightarrow
d)=$
$\nabla_{c}\nabla_{d}\tilde{Y}_{ab}-\frac{m}{3}a_{cd}{}^{e}\nabla_{e}\tilde{Y}_{ab}-\frac{m}{3}a_{ca}{}^{e}\nabla_{d}\tilde{Y}_{eb}-\frac{m}{3}a_{cb}{}^{e}\nabla_{d}\tilde{Y}_{ae}-\frac{m}{3}a_{da}{}^{e}\tilde{\nabla}_{c}\tilde{Y}_{eb}-\frac{m}{3}a_{db}{}^{e}\tilde{\nabla}_{c}\tilde{Y}_{ae}-(c\leftrightarrow
d)=$
$[\nabla_{c},\nabla_{d}]\tilde{Y}_{ab}-\frac{2m}{3}a_{cd}{}^{e}\nabla_{e}\tilde{Y}_{ab}-\frac{2m^{2}}{9}(a_{[c|a|}{}^{e}a_{d]e}{}^{f}\tilde{Y}_{fb}+a_{[c|a|}{}^{e}a_{d]b}{}^{f}\tilde{Y}_{ef}+a_{[c|b|}{}^{e}a_{d]a}{}^{f}\tilde{Y}_{fe}+a_{[c|b|}{}^{e}a_{d]e}{}^{f}\tilde{Y}_{af}).$
(111)

Thus, all single derivative terms cancel except one. Noting also that the two
non-derivative terms in the middle of the bracket cancel the commutator
becomes

$[\tilde{\nabla}_{c},\tilde{\nabla}_{d}]\tilde{Y}_{ab}=[\nabla_{c},\nabla_{d}]\tilde{Y}_{ab}-\frac{2m}{3}a_{cd}{}^{e}\nabla_{e}\tilde{Y}_{ab}-\frac{4m^{2}}{9}\,a_{[c}{}^{(a|e|}a_{d]e}{}^{|f|}\tilde{Y}_{f}{}^{b)}.$
(112)

Simplifying the non-derivative terms finally gives

$[\tilde{\nabla}_{c},\tilde{\nabla}_{d}]\tilde{Y}_{ab}=[\nabla_{c},\nabla_{d}]\tilde{Y}_{ab}-\frac{2m}{3}a_{cd}{}^{e}\nabla_{e}\tilde{Y}_{ab}+\frac{4m^{2}}{9}(\delta_{(a}^{[c}\tilde{Y}_{b)}{}^{d]}-c_{cd(a}{}^{e}\tilde{Y}_{b)e}).$
(113)

Since we are interested in expressing the right hand side in terms of the $D$
operator defined above using $\tilde{\nabla}_{a}$ we rewrite the last equation
as

$[\tilde{\nabla}_{c},\tilde{\nabla}_{d}]\tilde{Y}_{ab}=[\nabla_{c},\nabla_{d}]\tilde{Y}_{ab}-\frac{2m}{3}a_{cd}{}^{e}\tilde{\nabla}_{e}\tilde{Y}_{ab}-\frac{4m^{2}}{9}(\delta_{(a}^{[c}\tilde{Y}_{b)}{}^{d]}+2c_{cd(a}{}^{e}\tilde{Y}_{b)e}).$
(114)

To get the final expression that will be useful here we replace the commutator
on the right hand side by the Riemann tensor. This gives

$[\tilde{\nabla}_{c},\tilde{\nabla}_{d}]\tilde{Y}_{ab}=2W_{cd(a}{}^{e}\tilde{Y}_{b)e}-\frac{2m}{3}a_{cd}{}^{e}\tilde{\nabla}_{e}\tilde{Y}_{ab}+\frac{32m^{2}}{9}(\delta_{(a}^{[c}\tilde{Y}_{b)}{}^{d]}-\frac{1}{4}c_{cd(a}{}^{e}\tilde{Y}_{b)e}).$
(115)

The first term we need to compute in the $DD\tilde{Y}$ equation contains the
contracted expression $[\tilde{\nabla}^{b},\tilde{\nabla}_{d}]\tilde{Y}_{ab}$.
Setting $b=c$ in the last equation above we get

$[\tilde{\nabla}^{c},\tilde{\nabla}_{a}]\tilde{Y}_{bc}=W^{c}{}_{ab}{}^{e}\tilde{Y}_{ce}-\frac{2m}{3}a^{c}{}_{a}{}^{e}\tilde{\nabla}_{e}\tilde{Y}_{bc}+\frac{56m^{2}}{9}\tilde{Y}_{ab}.$
(116)

The second term in the $DD\tilde{Y}$ we need is the one with a contraction of
the commutator with the $c$ symbol:

$c_{cd(a}{}^{f}[\tilde{\nabla}_{c},\tilde{\nabla}_{d}]\tilde{Y}_{b)f}=-2W_{(a}{}^{e}{}_{b)}{}^{f}\tilde{Y}_{ef}+\frac{8m}{3}a_{(a}{}^{ef}\tilde{\nabla}^{e}\tilde{Y}_{b)}{}^{f}+\frac{112m^{2}}{9}\tilde{Y}_{ab}.$
(117)

Inserting the two results in (116) and (117) into (109) we find

$\frac{1}{4}(DD\tilde{Y})_{ab}=-\tilde{\Box}\tilde{Y}_{ab}+\frac{m}{3}(D\tilde{Y})_{ab}+\frac{112m^{2}}{9}\tilde{Y}_{ab}-2W_{(a}{}^{e}{}_{b)}{}^{f}\tilde{Y}_{ef}.$
(118)

Replacing the $G_{2}$ covariant box with Casimirs gives

$\frac{1}{4}(DD\tilde{Y})_{ab}=C_{G}Y_{ab}+\frac{m}{3}(D\tilde{Y})_{ab}+\frac{112m^{2}}{9}\tilde{Y}_{ab}-(C_{H}\tilde{Y}_{ab}+2W_{(a}{}^{e}{}_{b)}{}^{f}\tilde{Y}_{ef}).$
(119)

Then from the Appendix we know that the last bracket gives
$\tfrac{112m^{2}}{9}$ which implies the amazingly simple equation

$(DD\tilde{Y})_{ab}-\frac{4m}{3}(D\tilde{Y})_{ab}-4C_{G}\tilde{Y}_{ab}=0.$
(120)

In view of the eigenvalue equation (102), we may express the ”solution” to the
last equation as

$D\tilde{Y}=\frac{2m}{3}\pm
2\sqrt{C_{G}+\frac{m^{2}}{9}}=\frac{2m}{3}(1\pm\sqrt{20C_{G}+1}).$ (121)

Then by replacing $(D\tilde{Y})_{ab}$ with the expression coming from the
3-form (102), we find

${\bf 27}:\,\,\,\,\kappa_{3}^{2}=\frac{m^{2}}{9}(20C_{G}+2\pm
2\sqrt{20C_{G}+1}).$ (122)

Since $\kappa_{3}^{2}$ is the eigenvalue of $\Delta_{3}=Q^{2}$ this must be a
square. Indeed, it can be written

${\bf 27}:\,\,\,\,\Delta_{3}=\frac{m^{2}}{9}(\sqrt{20C_{G}+1}\pm 1)^{2}.$
(123)

###  3.5 Lichnerowicz

When we now turn to the transverse and traceless metric modes $h_{ab}$ on the
squashed seven-sphere we can take advantage of the results obtained in the
previous case of the 3-form. To see how to do this let us write out the
Lichnerowicz equation explicitly

$\Delta_{L}:\,\,\,\,\,\Delta_{L}h_{ab}=-\Box
h_{ab}-2W_{acbd}h^{cd}+14m^{2}h_{ab}=\kappa_{L}^{2}h_{ab}.$ (124)

The coset master equation reads in this case

$\nabla_{a}h_{bc}+\frac{2m}{3}a_{ad(b}h_{c)d}=-T_{a}h_{bc},$ (125)

which when squared gives

$G/H:\,\,\,\,\,\,\Box
h_{ab}+\frac{4m}{3}a_{cd(a}\nabla^{c}h_{b)}{}^{d}-\frac{14m^{2}}{9}h_{ab}=T_{c}T_{c}h_{ab}.$
(126)

Eliminating the box operator from the above $\Delta_{L}$ and $G/H$ equations
and using $T_{c}T_{c}=-(C_{H}-C_{H})$ gives

$\kappa_{L}^{2}h_{ab}=C_{H}h_{ab}-(C_{H}h_{ab}+2W_{acbd}h^{cd})+(14+\frac{14}{9})m^{2}h_{ab}+\frac{4m}{3}a_{cd(a}\tilde{\nabla}^{c}h_{b)}{}^{d},$
(127)

where we have used the $G_{2}$ covariant derivative

$\tilde{\nabla}_{a}h_{bc}=\nabla_{a}h_{bc}-\frac{2m}{3}a_{a(b}{}^{d}h_{c)}{}^{d}.$
(128)

As in the previous cases we now use the result for the Weyl tensor term from
the Appendix, i.e., that $C_{H}+2W$ acting on the different $H$ irreps in
${\bf 27}$ gives the universal value $\tfrac{28}{5}=\tfrac{112}{5}m^{2}$. This
gives us the rather simple equation

$\kappa_{L}^{2}h_{ab}=\tfrac{m^{2}}{9}(20C_{G}+28)h_{ab}+\frac{4m}{3}a_{cd(a}\tilde{\nabla}^{c}h_{b)}{}^{d}.$
(129)

There is one crucial difference between the Lichnerowicz modes and the 3-form
modes analysed in the last subsection: Transversality of the metric modes does
not imply that the associated 3-form $Y_{abc}\equiv\frac{4}{3}a_{d[ab}h_{c]d}$
is transverse (recall the result $\tilde{\nabla}^{a}\tilde{Y}_{ab}=0$ derived
in the previous subsection). Thus when squaring $(Dh)_{ab}\equiv
2a_{cd(a}\tilde{\nabla}^{c}h_{b)}{}^{d}$ we cannot simply use (120) since this
equation was derived with the assumption of a transverse $Y^{(27)}_{abc}$. The
change is however relatively small; we only need to add the term in (106)
proportional to $\nabla^{a}Y^{(27)}_{abc}$ to proceed. We find

$(DDh)_{ab}-\frac{4m}{3}(Dh)_{ab}-4C_{G}h_{ab}-8a_{cd(a}\tilde{\nabla}^{c}\tilde{\nabla}^{e}Y^{(27)}_{|ed|b)}=0.$
(130)

To deal with the last term we apply the $D$ operator yet another time. To
simplify the computation we introduce the notation
$Y_{ab}\equiv\tilde{\nabla}^{e}Y_{eab}=\nabla^{e}Y_{eab}$ where
$\nabla^{a}Y_{ab}=0$ and define $(DY)_{ab}\equiv
2a_{cd(a}\tilde{\nabla}^{c}Y_{|d|b)}=2a_{cd(a}\nabla^{c}Y_{|d|b)}$. It is then
immediately found that

$(DDY)_{ab}=2a_{cd(a}\tilde{\nabla}^{c}(DY)_{|d|b)}=-\frac{14m}{3}(DY)_{ab}+2a_{cd(a}\nabla^{c}(DY)_{|d|b)}$
(131)

Expanding out the nontrivial last term on the right side gives

$2a_{cd(a}\nabla^{c}(DY)_{|d|b)}=2a_{cd(a}\nabla^{c}a_{|efd}\nabla^{e}Y_{f|b)}+2a_{cd(a}\nabla^{c}a_{|ef|b)}\nabla^{e}Y_{fd}\,.$
(132)

We start by analysing the first term in (132). When the first covariant
derivative passes $a_{efd}$ we need to compute

$\displaystyle a_{cd}{}^{(a}a_{efd}\nabla_{c}\nabla_{e}Y_{f}{}^{b)}$
$\displaystyle=-2\delta^{c(a}_{ef}\nabla_{c}\nabla_{e}Y_{f}{}^{b)}-c_{cef}{}^{(a}\nabla_{c}\nabla_{e}Y_{f}{}^{b)}$
(133)
$\displaystyle=\nabla_{f}\nabla^{(a}Y_{f}{}^{b)}=6mY^{(ab)}+R^{f(ab)g}Y_{fg}$
$\displaystyle=0$

In the second equality we have used the antisymmetry of $Y_{ab}$ and the fact
that the 2-form projector $P_{7}$ vanishes when acting on the Weyl tensor
(which implies that $c_{ab}{}^{cd}\propto\delta_{ab}^{cd}$). The last equality
is true due to the antisymmetry of $Y_{ab}$ together with the symmetrisation
$(ab)$. When the covariant derivative acts on $a_{efg}$ we find

$ma_{cd}{}^{(a}c_{cefd}\nabla^{e}Y_{f}{}^{b)}=4ma_{ef}{}^{(a}\nabla^{|e|}Y_{f}{}^{b)}=2m(DY)_{ab}.$
(134)

We conclude that

$2a_{cd(a}\nabla^{c}a_{|efd}\nabla^{e}Y_{f|b)}=4m(DY)_{ab}.$ (135)

We then turn to the second term in (132). Here the covariant derivative can
act in two different ways. First, moving the derivative past $a_{efb}$ gives
the expression

$a_{cd}{}^{(a}a_{ef}{}^{b)}\nabla^{c}\nabla^{e}Y_{fd}=\tfrac{1}{2}a_{cd}{}^{(a}a_{ef}{}^{b)}[\nabla^{c},\nabla^{e}]Y_{fd}\,.$
(136)

The action of the antisymmetrized covariant derivatives will give two terms
which are essentially identical and we only show how to deal with the first
one. Using the squashed sphere Riemann tensor it follows that

$\displaystyle a_{cd}{}^{(a}a_{ef}{}^{b)}R^{ce}{}_{fg}Y^{g}{}_{d}$
$\displaystyle=2m^{2}\delta^{ce}_{fg}\,a_{cd}{}^{(a}a_{ef}{}^{b)}Y^{g}{}_{d}+a_{cd}{}^{(a}a_{ef}{}^{b)}W^{ce}{}_{fg}Y^{g}{}_{d}$
(137) $\displaystyle=m^{2}a_{fd}{}^{(a}a_{gf}{}^{b)}Y^{g}{}_{d}$
$\displaystyle=0,$

where we have used that
$a_{efb}W^{ce}{}_{fg}=-\frac{1}{2}a_{efb}W^{cg}{}_{ef}=0$ (see the Appendix).
The other term vanishes in exactly the same way. Finally, we have a term
coming from the covariant derivative hitting $a_{efb}$. After some algebra,
this term becomes

$2ma_{cd}{}^{(a}c_{cef}{}^{b)}\nabla^{e}Y_{fd}=2ma_{cd}{}^{(a}\nabla_{c}Y_{d}{}^{b)}=2m(DY)_{ab},$
(138)

and so $a_{cd}{}^{(a}a_{ef}{}^{b)}\nabla^{c}\nabla^{e}Y_{fd}=2m(DY)_{ab}$.
Putting these results together we find that

$(DDY)_{ab}=(6m-\frac{14m}{3})(DY)_{ab}=\frac{4m}{3}(DY)_{ab}.$ (139)

Having calculated the action of $D$ on $(DY)_{ab}$ we can apply $D$ on (130)
and then subtracting the previous equation to eliminate the $(DY)_{ab}$ terms.
From this procedure we find the third order equation

$(D-\frac{4m}{3})(D^{2}h-\frac{4m}{3}Dh-4C_{G}h)=0.$ (140)

”Solving” for $(Dh)_{ab}$ and plugging the result back into (129) gives the
following three different eigenvalues:

$\Delta_{L}=\frac{m^{2}}{9}(20C_{G}+36),\,\,\,\,\Delta_{L}=\frac{m^{2}}{9}\left(20C_{G}+32\pm
4\sqrt{20C_{G}+1}\right).$ (141)

##  4 Conclusions

Let us summarise what we know so far about the spectrum of operators on the
squashed seven-sphere including the new results for $\Delta_{2}$ and
$\Delta_{3}$ obtained in this paper. The previously known eigenvalues for
$\Delta_{0}$, $\not{D}_{1/2}$ Nilsson:1983ru , $\Delta_{1}$ Yamagishi:1983ri ,
and $\Delta_{L}$ Duff:1986hr  are

$\displaystyle\Delta_{0}$ $\displaystyle=$
$\displaystyle\tfrac{m^{2}}{9}\,20C_{G},$ (142) $\displaystyle{\not{D}}_{1/2}$
$\displaystyle=$ $\displaystyle-\tfrac{m}{2}\pm\tfrac{m}{3}\sqrt{20C_{G}+81},$
(143) $\displaystyle{\not{D}}_{1/2}$ $\displaystyle=$
$\displaystyle\tfrac{m}{6}\pm\tfrac{m}{3}\sqrt{20C_{G}+49},$ (144)
$\displaystyle\Delta_{1}$ $\displaystyle=$
$\displaystyle\tfrac{m^{2}}{9}\,(20C_{G}+14\pm
2\sqrt{20C_{G}+49})=\tfrac{m^{2}}{9}\,(\sqrt{20C_{G}+49}\pm 1)^{2}-4m^{2},$
(145) $\displaystyle\Delta_{L}$ $\displaystyle=$
$\displaystyle\tfrac{m^{2}}{9}\,(20C_{G}+36),$ (146) $\displaystyle\Delta_{L}$
$\displaystyle=$ $\displaystyle\tfrac{m^{2}}{9}\,(20C_{G}+32\pm
4\sqrt{20C_{G}+1})=\tfrac{m^{2}}{9}\,(\sqrt{20C_{G}+1}\pm 2)^{2}+3m^{2},$
(147)

while the new ones obtained in this paper are

$\displaystyle\Delta_{2}$ $\displaystyle=$
$\displaystyle\tfrac{m^{2}}{9}\,(20C_{G}+72),$ (148) $\displaystyle\Delta_{2}$
$\displaystyle=$ $\displaystyle\tfrac{m^{2}}{9}\,(20C_{G}+44\pm
4\sqrt{20C_{G}+49})=\tfrac{m^{2}}{9}\,(\sqrt{20C_{G}+49}\pm 2)^{2}-m^{2},$
(149) $\displaystyle\Delta_{2}$ $\displaystyle=$
$\displaystyle\tfrac{m^{2}}{9}\,20C_{G},$ (150) $\displaystyle\Delta_{3}$
$\displaystyle=$ $\displaystyle\tfrac{m^{2}}{9}\,(\sqrt{20C_{G}+49}\pm
1)^{2},$ (151) $\displaystyle\Delta_{3}$ $\displaystyle=$
$\displaystyle\tfrac{m^{2}}{9}\,(\sqrt{20C_{G}+81}\pm 3)^{2},$ (152)
$\displaystyle\Delta_{3}$ $\displaystyle=$
$\displaystyle\tfrac{m^{2}}{9}\,(\sqrt{20C_{G}+1}\pm 1)^{2}.$ (153)

Here it is appropriate to make some comments on the limitations of the
obtained results. First, we have so far no results for the eigenvalues of the
spin $3/2$ operator ${\not{D}}_{3/2}$, although some can easily be extracted
from supersymmetry. Secondly, for $\Delta_{2}$ and $\Delta_{L}$ we seem to
lack some eigenvalues. This is indicated by the degeneracies in the cross
diagrams for these two operators derived in Nilsson:2018lof , as well as by
supersymmetry. For $\Delta_{3}$ there may already be too many available
eigenvalues but if one is supposed to pick only one sign when removing the
square (as done for ${\not{D}}_{1/2}$) one is instead lacking two eigenvalues.
These problematic features are partly due to the fact that although we have
extensive knowledge of the eigenvalue spectra from the list above, the method
applied here does not provide direct information how to associate these
eigenvalues with the cross diagrams of Nilsson:2018lof . These and other
issues will be elaborated upon in a forthcoming publication Nilsson:2021kn .

The results of this paper clearly demonstrate the important role of weak
$G_{2}$ holonomy when solving the eigenvalue equations. There are, however,
deeper issues in the context of holonomy and string/M theory that might be
interesting to study in relation to the squashed seven-sphere, for instance
the notion of generalised holonomy discussed in Batrachenko:2003ng .

## Acknowledgement

One of us (BEWN) thanks M.J. Duff and C.N. Pope for a number of discussions on
issues related to the theory analysed in this paper and for collaboration at
an early stage of this project. We are also grateful to Joel Karlsson for some
useful comments on the manuscript. The work of S.E. is partially supported by
the Knut and Alice Wallenberg Foundation under the grant: ”Exact Results in
Gauge and String Theories”, Dnr KAW 2015.0083.

##  Appendix A Octonions: Conventions and some identities

The octonions satisfy a non-associative algebra defined by the totally
antisymmetric structure constants $a_{abc}$ ($a,b,c,..=0,1,...,6$):

$a_{abc}=1\,\,\text{for}\,\,abc=456,041,052,063,162,135,243.$ (154)

By splitting $a=(0,i,\hat{i})$ (with
$i=1,2,3,\hat{i}=4,5,6=\hat{1},\hat{2},\hat{3}$) they can be written more
compactly as

$a_{0i\hat{j}}=-\delta_{ij},\quad a_{ij\hat{k}}=-\epsilon_{ijk},\quad
a_{\hat{i}\hat{j}\hat{k}}=\epsilon_{ijk}.$ (155)

We define its dual, denoted $c_{abcd}$, by

$c_{abcd}=\tfrac{1}{6}\epsilon_{abcdefg}\,a_{efg},$ (156)

where the epsilon tensor is totally antisymmetric with $\epsilon_{0123456}=1$.

Sometimes it is convenient to write the gamma matrices $\Gamma_{a}$ in seven
dimensions in terms of the structure constants defined above and a Killing
spinor $\eta$ satisfying

$\nabla_{a}\,\eta=-\frac{i}{2}m\Gamma_{a}\,\eta,\,\,\,\bar{\eta}\,\eta=1.$
(157)

The conventions used here are

$a_{abc}=i\bar{\eta}\Gamma_{abc}\eta,\,\,\,\,c_{abcd}=-\bar{\eta}\Gamma_{abcd}\eta,$
(158)

and

$\\{\Gamma_{a},\Gamma_{b}\\}=2\delta_{ab},\,\,\,-i\Gamma_{abcdefg}=\epsilon_{abcdefg}\bf{1}.$
(159)

These gamma matrices are $8\times 8$ matrices with spinor indices $A,B,...$
taking the eight values $A=(a,8)$ etc. The consistency of these conventions
can then be verified by explicitly writing out the gamma matrices in terms of
the structure constants as follows101010Note that the indices for both
vectors, $a,b,c,...$, and spinors, $A,B,C,...$, are raised and lowered by a
unit matrix which means that equations may appear with indices in the wrong up
or down position.:

$(\Gamma_{a})_{B}{}^{C}:\\\
(\Gamma_{a})_{b}{}^{c}=ia_{abc},\,\,\,\,(\Gamma_{a})_{b}{}^{8}=i\delta_{ab},\,\,\,\,(\Gamma_{a})_{8}{}^{b}=-i\delta_{ab}.$
(160) $(\Gamma_{ab})_{B}{}^{C}:\\\
(\Gamma_{ab})^{cd}=2\delta^{cd}_{ab}-c_{ab}{}^{cd},\,\,\,\,(\Gamma_{ab})_{c}{}^{8}=a_{abc}.$
(161)

Note that all seven gamma matrices are antisymmetric and imaginary. With these
definitions one can check that $\bar{\eta}\,\Gamma_{abc}\,\eta=-ia_{abc}$ as
stipulated above. However, this calculation requires some octonionic structure
constant identities that are seldom given in the literature. For the
convenience of the reader we list the identities used in this paper below:

$a_{abe}\,a^{cde}=2\delta_{ab}^{cd}+c_{ab}{}^{cd},$ (162)
$c_{abc}{}^{g}c^{defg}+a_{abc}a^{def}=6\delta_{abc}^{def}+9\delta_{[a}^{[d}c_{bc]}{}^{ef]},$
(163) $c_{abcf}a^{def}=6a_{[ab}{}^{[d}\delta^{e]}_{c]}.$ (164)

These identities may be verified directly by inserting the actual values of
the constants, or by using the above expressions in terms of the Killing
spinor $\eta$ and the gamma matrices. The latter approach requires the Fierz
formula

$\Gamma_{a}\eta\bar{\eta}\Gamma_{a}={\bf 1}-\eta\bar{\eta}.$ (165)

Various contractions of indices in these identities lead directly to a number
of additional identities that are used frequently in the main text. A quite
extensive list of identities is presented in House:2004pm .

Using the Killing spinor equation we furthermore find the derivative
identities

$\nabla_{a}a_{bcd}=-\frac{m}{2}\bar{\eta}(\Gamma_{a}\Gamma_{bcd}-\Gamma_{bcd}\Gamma_{a})\eta=-m\bar{\eta}(\Gamma_{abcd})\eta=mc_{abcd},$
(166)
$\nabla_{a}c_{bcde}=-\frac{im}{2}\bar{\eta}(\Gamma_{a}\Gamma_{bcde}-\Gamma_{bcde}\Gamma_{a})\eta=-4im\delta_{a[b}\bar{\eta}(\Gamma_{cde]})\eta=-4m\delta_{a[b}a_{cde]}.$
(167)

##  Appendix B Projection operators

###  B.1 2-forms

$(P_{7})_{ab}{}^{cd}=\frac{1}{6}a_{ab}{}^{e}a^{cde}=\tfrac{1}{6}(2\delta_{ab}{}^{cd}+c_{ab}{}^{cd}),$
(168) $(P_{14})_{ab}{}^{cd}=\frac{1}{6}(4\delta_{ab}{}^{cd}-c_{ab}{}^{cd}),$
(169)

which implies the useful relation

$c_{ab}{}^{cd}=2(2P_{7}-P_{14})_{ab}{}^{cd}.$ (170)

###  B.2 3-forms

$(P_{1})_{abc}{}^{def}=\frac{1}{42}a_{abc}a^{def},$ (171)
$(P_{7})_{abc}{}^{def}=\frac{1}{24}c_{abc}{}^{g}c^{defg}=\frac{1}{24}(6\delta_{abc}^{def}+9\delta_{[a}^{[d}c_{bc]}{}^{ef]}-a_{abc}a^{def}),$
(172)
$(P_{27})_{abc}{}^{def}=(1-P_{1}-P_{7})_{abc}{}^{def}=\frac{1}{56}(42\delta_{abc}^{def}-21\delta_{[a}^{[d}c_{bc]}{}^{ef]}+a_{abc}a^{def}).$
(173)

###  B.3 Spin $1/2$

The purpose of the projection operators in this case is to split the $SO(7)$
spinor representation into two $G_{2}$ irreps as follows ${\bf
8}\rightarrow{\bf 1}\oplus{\bf 7}$:

$P^{s}_{1}=\frac{1}{8}({\bf 1}-\frac{1}{24}c_{abcd}\Gamma^{abcd}),$ (174)
$P^{s}_{7}={\bf 1}-P^{s}_{1}=\frac{1}{8}(7\cdot{\bf
1}+\frac{1}{24}c_{abcd}\Gamma^{abcd}).$ (175)

While they trivially sum to the unit matrix, the fact that they both square to
themselves and are orthogonal to each other require some algebra to show. In
fact, all these properties follow directly if we show that
$(P^{s}_{1})^{2}=P_{1}^{s}$. This is done as follows

$(P^{s}_{1})^{2}=\frac{1}{64}({\bf
1}-\frac{1}{12}c_{abcd}\Gamma^{abcd}+\frac{1}{(24)^{2}}c_{abcd}c_{efgh}\Gamma^{abcd}\Gamma^{efgh})=P_{1}^{s}.$
(176)

The key calculation in the last step is to expand $\Gamma^{abcd}\Gamma^{efgh}$
in the gamma basis $\Gamma^{(n)}$, $n=0,....,7$, in seven dimensions:

$\Gamma^{abcd}\Gamma_{efgh}=-16\delta^{[a}_{[e}\Gamma^{bcd]}{}_{fgh]}-72\delta^{[ab}_{[ef}\Gamma^{cd]}{}_{gh]}+96\delta^{[abc}_{[efg}\Gamma^{d]}{}_{h]}+24\delta^{abcd}_{efgh}.$
(177)

However, when contracted with $c_{abcd}c_{efgh}$ the first and third terms
vanish since they are antisymmetric under interchange of the two index sets
$abcd$ and $efgh$. The remaining two terms are easily computed using the two
identities for $c_{abcd}$: $c^{abcd}c_{efcd}=8\delta^{ab}_{ef}+2c^{ab}{}_{ef}$
and $c^{abcd}c_{abcd}=168$. $(P^{s}_{1})^{2}=P_{1}^{s}$ then follows directly.

##  Appendix C The role of the Weyl tensor

The Weyl tensor of the squashed seven-sphere is given in Duff:1986hr . The
explicit expression given there can be written in a more compact form in terms
of ’t Hooft symbols. By setting $\alpha=(0,1,2,3)=(0,i)$ and introducing the
’t Hooft symbols by

$\displaystyle\eta^{k}_{\alpha\beta}:$
$\displaystyle\,\,\,\,\eta^{k}_{ij}=\epsilon_{ijk},\,\,\,\,\,\eta^{k}_{0j}=-\eta^{k}_{j0}=-\delta_{jk},$
(178) $\displaystyle\bar{\eta}^{i}_{\alpha\beta}:$
$\displaystyle\,\,\,\,\bar{\eta}^{k}_{ij}=\epsilon_{ijk},\,\,\,\,\,\bar{\eta}^{k}_{0j}=-\bar{\eta}^{k}_{j0}=\delta_{jk},$
(179)

the Weyl tensor can be explicitly expressed as follows

$W_{\alpha\beta}{}^{\gamma\delta}=\frac{4}{5}\delta_{\alpha\beta}^{\gamma\delta},\,\,\,W_{\hat{i}\hat{j}}{}^{\hat{k}\hat{l}}=\frac{8}{5}\delta_{\hat{i}\hat{j}}^{\hat{k}\hat{l}},\,\,\,\,W_{\alpha\beta}{}^{\hat{k}\hat{l}}=\frac{2}{5}\bar{\eta}_{\alpha\beta}^{m}\epsilon^{mkl},\,\,\,\,W_{\alpha}{}^{\hat{j}}{}_{\gamma}{}^{\hat{l}}=\frac{1}{5}\bar{\eta}_{\alpha\gamma}^{m}\epsilon^{mjl}-\frac{2}{5}\delta_{\alpha\gamma}\delta^{\hat{j}\hat{l}}.$
(180)

The demonstration that the squashed seven-sphere has holonomy $G_{2}$ relies
on the fact that defining $W_{ab}\equiv\tfrac{1}{4}W_{ab}{}^{cd}\Gamma_{cd}$
we find

$W_{0i}=\tfrac{1}{5}(\Gamma_{0i}+\tfrac{1}{2}\epsilon_{ijk}\Gamma_{\hat{j}\hat{k}}),$
(181) $W_{ij}=\tfrac{1}{5}(\Gamma_{ij}+\Gamma_{\hat{i}\hat{j}}),$ (182)
$W_{i\hat{j}}=\tfrac{1}{5}(-\Gamma_{i\hat{j}}-\tfrac{1}{2}\Gamma_{j\hat{i}}+\tfrac{1}{2}\delta_{ij}\Gamma_{k\hat{k}}-\tfrac{1}{2}\epsilon_{ijk}\Gamma_{0\hat{k}}),$
(183)
$W_{0\hat{i}}=-\tfrac{1}{5}(\Gamma_{0\hat{i}}+\tfrac{1}{2}\epsilon_{ijk}\Gamma_{j\hat{k}}),$
(184)
$W_{\hat{i}\hat{j}}=\tfrac{1}{5}(2\Gamma_{\hat{i}\hat{j}}+\Gamma_{ij}+\epsilon_{ijk}\Gamma_{0k}),$
(185)

while the remaining components are not independent but instead given by

$W_{i\hat{i}}=0,\,\,\,\,\,W_{0\hat{i}}=\epsilon_{ijk}W_{j\hat{k}},\,\,\,W_{\hat{i}\hat{j}}=W_{ij}+\epsilon_{ijk}W_{0k}.$
(186)

Thus only 14 components are linearly independent which was shown in
Duff:1986hr  to result in $G_{2}$ holonomy. An immediate consequence, as can
be easily verified using the explicit expressions, is that

$a_{abc}W_{bc}{}^{de}=0\,.$ (187)

To obtain identities of this kind it is convenient to express the octonionic
structure constants in terms of ’t Hooft symbols. The non-zero components are
then (recall that $a=(\alpha,\hat{i})$ and
$\hat{i}=(\hat{1},\hat{2},\hat{3})=(4,5,6)$):

$\displaystyle a_{abc}$
$\displaystyle:\,\,\,\,a_{\alpha\beta\hat{k}}=-\bar{\eta}_{\alpha\beta}^{k},\,\,\,\,\,a_{\hat{i}\hat{j}\hat{k}}=\epsilon_{ijk},$
(188) $\displaystyle c_{abcd}:$
$\displaystyle\,\,\,\,c_{\alpha\beta\gamma\delta}=\epsilon_{\alpha\beta\gamma\delta}=\bar{\eta}_{[\alpha\beta}^{k}\bar{\eta}_{\gamma\delta]}^{k},\,\,\,\,\,c_{\alpha\beta\hat{i}\hat{j}}=-\bar{\eta}^{k}_{\alpha\beta}\epsilon_{ijk},$
(189)

where $\epsilon_{\alpha\beta\gamma\delta}$ is antisymmetric with
$\epsilon_{0123}=1$.

In our present context we will for instance use these results to express the
2-form modes $Y^{(14)}_{ab}$ in a way that makes its projections onto the $H$
irreps clear, i.e., the decomposition

$G_{2}\rightarrow Sp_{1}^{A}\times Sp_{1}^{B+C}:\,\,\,\,\,\,{\bf
14}\rightarrow(1,3)\oplus(2,0)\oplus(0,2).$ (190)

To this end we will parametrise the $Spin(4)\times SU(2)$ subgroup of
$G=Sp_{2}\times Sp_{1}^{C}$ using

$\Gamma^{a}=(\Gamma^{\alpha},\Gamma^{\hat{i}}):\,\,\,\,\,\,\,\,\,\,\,\,\Gamma^{\alpha}=\begin{pmatrix}0&(\sigma^{\alpha})_{A\dot{B}}\\\
(\bar{\sigma}^{\alpha})^{\dot{A}B}&0\\\
\end{pmatrix}\otimes\delta^{\dot{A}}_{\dot{B}},\,\,\,\,\,\Gamma^{\hat{i}}={\bf
1}_{4\times 4}\otimes(\sigma^{\hat{i}})^{\dot{A}}{}_{\dot{B}},$ (191)

where

$\sigma^{\alpha}\equiv(-i{\bf
1},\sigma^{i}),\,\,\,\bar{\sigma}^{\alpha}\equiv(i{\bf 1},\sigma^{i}).$ (192)

Note the dot notation on the indices in the second tensor factor. This
reflects the fact that we are using the diagonal subgroup of $Sp_{1}^{B}$ and
$Sp_{1}^{C}$ where the former is a subgroup of $Sp_{2}$ when split into
$Sp_{1}^{A}\times Sp_{1}^{B}$ according to the first tensor factor.

The ’t Hooft symbols $(\eta_{\alpha\beta}^{m},\bar{\eta}_{\alpha\beta}^{m})$
introduced above can also be defined from the Pauli matrices by

$\sigma^{\alpha\beta}\equiv
i\eta_{\alpha\beta}^{m}\sigma^{m},\,\,\,\,\,\bar{\sigma}^{\alpha\beta}\equiv
i\bar{\eta}_{\alpha\beta}^{m}\sigma^{m}.$ (193)

###  C.1 2-form modes

From the above expressions we immediately obtain the projectors onto irreps
$(s,t)$ of $H=Sp_{1}^{A}\times Sp_{1}^{B+C}$

$(P^{ab}_{{\bf
14}\rightarrow(2,0)}Y_{ab})_{A}{}^{B}=(\sigma^{\alpha\beta}Y_{\alpha\beta})_{A}{}^{B},$
(194) $(P^{ab}_{{\bf
14}\rightarrow(0,2)}Y_{ab})^{\dot{A}}{}_{\dot{B}}=(\bar{\sigma}^{\alpha\beta}Y_{\alpha\beta}+2\sigma^{\hat{i}\hat{j}}Y_{\hat{i}\hat{j}})^{\dot{A}}{}_{\dot{B}},$
(195) $(P^{ab}_{{\bf
14}\rightarrow(1,3)}Y_{ab})_{A(\dot{B}\dot{C}\dot{D})}=(\sigma^{\alpha})_{A(\dot{B}}\sigma^{\hat{j}}_{\dot{C}\dot{D})}Y_{\alpha\hat{j}}.$
(196)

As two examples of how to use these projectors we compute the Weyl tensor
eigenvalue of the $(2,0)$ and $(0,2)$ irreps. For $(2,0)$

$P_{{\bf
14}\rightarrow(2,0)}^{ab}W_{abcd}Y_{cd}=\sigma^{\alpha\beta}W_{\alpha\beta
cd}Y_{cd}=\sigma^{\alpha\beta}(\tfrac{4}{5}\delta_{\alpha\beta}^{\gamma\delta}Y_{\gamma\delta}+\tfrac{2}{5}\bar{\eta}^{m}_{\alpha\beta}\epsilon^{mkl}Y_{\hat{k}\hat{l}})=\tfrac{4}{5}\sigma^{\alpha\beta}Y_{\alpha\beta},$
(197)

showing that the sought for eigenvalue is $\tfrac{4}{5}$. Note that in the
last equality we used the fact that
$\sigma^{\alpha\beta}\bar{\eta}^{m}_{\alpha\beta}=0$. For $(0,2)$ we find

$2\sigma^{\hat{i}\hat{j}}W_{\hat{i}\hat{j}cd}Y^{cd}=2\sigma^{\hat{i}\hat{j}}(\frac{8}{5}Y_{\hat{i}\hat{j}}+\frac{2}{5}\bar{\eta}^{m}_{\alpha\beta}\epsilon^{m\hat{i}\hat{j}}Y^{\alpha\beta})=\frac{8}{5}(2\sigma^{\hat{i}\hat{j}}Y_{\hat{i}\hat{j}}+\bar{\sigma}^{\alpha\beta}Y_{\alpha\beta})$
(198) $\bar{\sigma}^{\alpha\beta}W_{\alpha\beta
ab}Y^{ab}=\bar{\sigma}^{\alpha\beta}(\frac{4}{5}Y_{\alpha\beta}+\frac{2}{5}\bar{\eta}^{m}_{\alpha\beta}\epsilon^{mij}Y_{\hat{i}\hat{j}})=\frac{4}{5}(\bar{\sigma}^{\alpha\beta}Y_{\alpha\beta}+2\sigma^{\hat{i}\hat{j}}Y_{\hat{i}\hat{j}})$
(199)

where we have used (193). Then adding together (198) and (199) implies
$P_{\mathbf{14}\rightarrow(0,2)}WY=\frac{12}{5}P_{\mathbf{14}\rightarrow(0,2)}Y$.
Repeating this for the last case one finds the values tabulated below together
with the corresponding values of the $H$ Casimir operator $C_{H}$.

H-irrep | H Casimir | Weyl eigenvalue | Sum
---|---|---|---
$(2,0)$ | $4$ |  $\frac{4}{5}$ | $\frac{24}{5}$
$(0,2)$ | $\frac{12}{5}$ |  $\frac{12}{5}$ | $\frac{24}{5}$
$(1,3)$ | $6$ | $-\frac{6}{5}$ | $\frac{24}{5}$
Table 1: Weyl tensor eigenvalues for the $H$ irreps of the $G_{2}$ ${\bf
14}$-part of the 2-form.

Note that the two eigenvalues do not individually respect the $G_{2}$ holonomy
but that their sums do. Similarly, we know from before that the Weyl tensor
does not enter the equations for the $G_{2}$ ${\bf 7}$-part of the 2-form but
that the $H$ irreps both give $\tfrac{12}{5}$ which thus automatically
respects $G_{2}$.

###  C.2 3-form and metric modes: the 27 irrep

Here we need two more projectors, namely111111In the first formula below the
use of the symmetrisation brackets is not optimal: The two pairs of dotted and
undotted indices are symmetrised independently of each other.

$(P^{ab}_{{\bf
27}\rightarrow(2,2)}Y_{ab})_{(A(\dot{B}C)\dot{D})}=(\bar{\sigma}^{\alpha})_{(A(\dot{B}}(\bar{\sigma}^{\beta})_{C)\dot{D})},$
(200) $(P^{ab}_{{\bf
27}\rightarrow(0,4)}Y_{ab})_{(\dot{A}\dot{B}\dot{C}\dot{D})}=(\bar{\sigma}^{\hat{i}})_{(\dot{A}\dot{B}}(\bar{\sigma}^{\hat{j}})_{\dot{C}\dot{D})}.$
(201)

As an example how to get the Weyl eigenvalues we consider the second
projector. The calculation works as follows:

$(P^{ab}_{{\bf
27}\rightarrow(0,4)}W_{acbd}h^{cd})=\sigma^{\hat{i}}\sigma^{\hat{j}}W_{\hat{i}c\hat{j}d}h^{cd}=\sigma^{\hat{i}}\sigma^{\hat{j}}(\tfrac{8}{5}\delta^{\hat{i}\hat{k}}_{\hat{j}\hat{l}}h^{\hat{k}\hat{l}}+\tfrac{1}{5}\bar{\eta}^{m}_{\alpha\beta}\epsilon_{m\hat{i}\hat{j}}h^{\alpha\beta}-\tfrac{2}{5}\delta_{\hat{i}\hat{j}}h_{\alpha\alpha}).$
(202)

The term containing the ’t Hooft symbol vanishes since
$\bar{\eta}^{m}_{\alpha\beta}$ is antisymmetric in the two lower indices. The
remaining terms simplify immediately to

$(P^{ab}_{{\bf
27}\rightarrow(0,4)}W_{acbd}h^{cd})=-\tfrac{4}{5}(\sigma^{\hat{i}}\sigma^{\hat{j}})_{(0,4)}h_{\hat{i}\hat{j}},$
(203)

where we also used that tracing over the hatted indices on
$\sigma^{\hat{i}}\sigma^{\hat{j}}$ gives zero since this expression is in the
irrep ${\bf 5}$ of $SU(2)$ given by the indices
$(\dot{A}\dot{B}\dot{C}\dot{D})$.

H-irrep | H Casimir | 2 $\times$ Weyl eigenvalue | Sum
---|---|---|---
$(0,0)$ | $0$ |  $\frac{28}{5}$ | $\frac{28}{5}$
$(2,2)$ | $\frac{32}{5}$ | $-\frac{4}{5}$ | $\frac{28}{5}$
$(0,4)$ | $\frac{36}{5}$ | $-\frac{8}{5}$ | $\frac{28}{5}$
$(1,1)$ | $\frac{12}{5}$ |  $\frac{16}{5}$ | $\frac{28}{5}$
$(1,3)$ | $6$ | $-\frac{2}{5}$ | $\frac{28}{5}$
Table 2: Weyl tensor eigenvalues for the $H$ irreps of the $G_{2}$ ${\bf
27}$-part of the 3-form.

Again we find the striking result that all $H$ irrep pieces a $G_{2}$ irrep
produce the same value for $C_{H}+2\times Weyl$, this time for the ${\bf 27}$.

In the main text we found the expression
$C_{H}Y_{abc}-6W^{d}{}_{[ab}{}^{e}Y_{c]de}$. Using the explicit form of the
projection operators we can verify that
$W^{d}{}_{[ab}{}^{e}Y_{c]de}=W^{d}{}_{[ab}{}^{e}Y^{(27)}_{c]de}$. Applying
$(\tilde{P}^{27})_{ab}{}^{fgh}\equiv a_{acd}(P^{27})_{cdb}{}^{fgh}$ gives

$(\tilde{P}^{27})_{ab}{}^{fgh}W^{d}{}_{fg}{}^{e}Y^{(27)}_{gde}=-\frac{1}{3}W_{(a}{}^{c}{}_{b)}{}^{d}\tilde{Y}_{cd}.$
(204)

Using the eigenvalues listed in Table 2 it follows that

$(\tilde{P}^{27})_{ab}{}^{fgh}(C_{H}Y_{abc}-6W^{d}{}_{[ab}{}^{e}Y_{c]de})=(C_{H}\tilde{Y}_{ab}+2W_{(a}{}^{c}{}_{b)}{}^{e}\tilde{Y}_{ce})=\frac{28}{5}\tilde{Y}_{ab}.$
(205)

Going back to the 3-form notation and using that $C_{H}$ has eigenvalue
$\frac{12}{5}$ on $\mathbf{7}$ we find

$C_{H}Y_{abc}-6W^{d}{}_{[ab}{}^{e}Y_{c]de}=-\frac{1}{10}c_{abcd}Y_{d}+\frac{28}{5}Y_{abc}.$
(206)

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