# Supersymmetric domain walls in maximal 6D gauged supergravity II

Parinya Karndumri  Patharadanai Nuchino<EMAIL_ADDRESS>and
<EMAIL_ADDRESS>String Theory and Supergravity Group, Department of
Physics, Faculty of Science, Chulalongkorn University, 254 Phayathai Road,
Pathumwan, Bangkok 10330, Thailand

###### Abstract

We continue the study of supersymmetric domain wall solutions in six-
dimensional maximal gauged supergravity. We first give a classification of
viable gauge groups with the embedding tensor in $\mathbf{5}^{+7}$,
$\bar{\mathbf{5}}^{+3}$, $\mathbf{10}^{-1}$, $\mathbf{24}^{-5}$, and
$\overline{\mathbf{45}}^{+3}$ representations of the off-shell symmetry
$GL(5)\subset SO(5,5)$. We determine an explicit form of the embedding tensor
for gauge groups arising from each representation together with some examples
of possible combinations among them. All of the resulting gauge groups are of
a non-semisimple type with abelian factors and translational groups of
different dimensions. We find $\frac{1}{2}$\- and $\frac{1}{4}$-supersymmetric
domain walls with $SO(2)$ symmetry in $SO(2)\ltimes\mathbb{R}^{8}$ and
$SO(2)\ltimes\mathbb{R}^{6}$ gauge groups from the embedding tensor in
$\mathbf{24}^{-5}$ representation and in $CSO(2,0,2)\ltimes\mathbb{R}^{4}$,
$CSO(2,0,2)\ltimes\mathbb{R}^{2}$, and $CSO(2,0,1)\ltimes\mathbb{R}^{4}$ gauge
groups with the embedding tensor in $\overline{\mathbf{45}}^{+3}$
representations. These gauge groups are parametrized by a traceless matrix and
electrically and magnetically embedded in $SO(5,5)$ global symmetry,
respectively.

##  I Introduction

Domain wall solutions in gauged supergravities play an important role in high-
energy physics. In the AdS/CFT correspondence maldacena ; Gubser_AdS_CFT ;
Witten_AdS_CFT  and a generalization to non-conformal field theories called
the DW/QFT correspondence DW_QFT1 ; DW_QFT2 ; DW_QFT3 ; DW_QFT4 , these
solutions give holographic descriptions of RG flows in the dual conformal and
non-conformal field theories at strong coupling, see for example GPPZ ; FGPP ;
KW_5Dflow ; N2_IIB_flow ; AdS_RG_flow ; N1_SYM_fixed_point . Domain walls are
also useful in studying cosmology DW_cosmology1 ; DW_cosmology2 ;
DW_cosmology3 ; DW_cosmology4 , see also DW_inflation  for a recent result.
Most of the solutions are supersymmetric and have been found within the
framework of gauged supergravities in various dimensions, see Eric_DW_10D ;
9D_DW_Cowdall ; 9D_DW ; 8D_DW1 ; 8D_DW2 ; 7D_DW ; 6D_DW_I ; Pope_DW_massive ;
DW_algebraic_curve ; Eric_DW_maximal_SUGRA ; Pope_symmetric_poten ; DW_Hull ;
DW_Cvetic ; 4D_DW ; 3DN4_DW ; 3DN10_DW ; 2D_DW1 ; 2D_DW2 ; Eric_SUSY_DW  for
an incomplete list.
In this paper, we are interested in supersymmetric domain walls from the
maximal $N=(2,2)$ gauged supergravity in six dimensions constructed in
6D_Max_Gauging , see 6D_SO(5)  for an earlier result. The result of
6D_Max_Gauging  describes the most general gaugings of $N=(2,2)$ supergravity
in six dimensions using the embedding tensor formalism. The embedding tensor
lives in representation $\mathbf{144}_{c}$ of the global symmetry $SO(5,5)$
and determines a viable gauge group $G_{0}\subset SO(5,5)$. In general, there
are a large number of possible gauge groups. In this work, we will consider
only gauge groups classified under $GL(5)\subset SO(5,5)$ which is an off-
shell symmetry of the $N=(2,2)$ supergravity Lagrangian in a particular
symplectic frame.
We will study various possible gaugings and explicitly construct the
corresponding embedding tensors for the resulting gauge groups. We will also
look for supersymmetric domain wall solutions. According to the DW/QFT
correspondence, these solutions are dual to maximally supersymmetric Yang-
Mills theory (SYM) in five dimensions which recently plays an important role
in studying dynamics of (conformal) field theories in higher and lower
dimensions via a number of dualities, see for example Douglas_5D_MSYM ;
5D_MSYM_1 ; 5D_MSYM_2 ; 5D_MSYM_3 ; 5D_MSYM_4 ; 5D_MSYM_5 . The five-
dimensional SYM could be used to define $N=(2,0)$ superconformal field theory
(SCFT) in six dimensions compactified on $S^{1}$. Since the latter is well-
known to describe dynamics of strongly coupled theory on M5-branes, we expect
that supersymmetric domain walls of the maximal gauged supergravity in six
dimensions could be useful in studying various aspects of the maximal SYM in
five dimensions as well as six-dimensional SCFT and physics of M5-branes at
strong coupling. In addition, five-dimensional maximal SYM and
compactifications on $S^{1}$ and $S^{2}$ can lead to some insights to
S-duality of twisted gauge theories in four dimensions and monopoles in
Aharony-Bergman-Jafferis-Maldacena (ABJM) theory. Therefore, the resulting
domain walls could also be useful in this context as well. However, it should
be pointed out that all gaugings classified here currently have no known
higher dimensional origins.
Since $N=4$ superconformal symmetry in five dimensions does not exist Nahm_res
, see also a recent classification of maximally supersymmetric $AdS$ vacua
given in Max_SUSY_AdS , there is no AdS6/CFT5 duality with $32$ supercharges.
Accordingly, supersymmetric domain walls in $N=(2,2)$ gauged supergravity are
expected to play a prominent role in holographic study in this case. A domain
wall solution with $SO(5)$ symmetry in $SO(5)$ gauging has been found in
6D_SO(5)  with the holographic interpretation given in Bobev_spherical_brane1
and Bobev_spherical_brane2 . Moreover, a large number of domain wall solutions
has been given recently in 6D_DW_I  with various gauge groups obtained from
the embedding tensor in $\mathbf{15}^{-1}$ and $\overline{\mathbf{40}}^{-1}$
representations of $GL(5)$. We will extend this investigation by considering
the embedding tensor in other representations of $GL(5)$. These are given by
$\mathbf{5}^{+7}$, $\bar{\mathbf{5}}^{+3}$, $\mathbf{10}^{-1}$,
$\mathbf{24}^{-5}$, and $\overline{\mathbf{45}}^{+3}$ representations. It
turns out that most of the resulting gauge groups are of non-semisimple type
without any compact subgroup. Due to the complexity of working with the full
$25$-dimensional scalar manifold $SO(5,5)/SO(5)\times SO(5)$, we mainly follow
the method introduced in New_Extrema  to reduce the number of scalar fields to
make the analysis more traceable. However, this approach requires the
existence of a non-trivial symmetry $H_{0}\subset G_{0}$. In many of the
gaugings classified here, the residual $H_{0}$ is absent, so there are too
many scalars to take into account. Accordingly, we will give domain wall
solutions only for gauge groups with at least $SO(2)$ subgroup. As in 6D_DW_I
, there exist both $\frac{1}{2}$\- and $\frac{1}{4}$-supersymmetric domain
wall solutions in accord with the general classification of supersymmetric
domain walls given in Eric_SUSY_DW .
The paper is organized as follows. In section II, we briefly review the
construction of six-dimensional maximal gauged supergravity in the embedding
tensor formalism. Gaugings in $\mathbf{5}^{+7}$, $\bar{\mathbf{5}}^{+3}$,
$\mathbf{10}^{-1}$, $\mathbf{24}^{-5}$, and $\overline{\mathbf{45}}^{+3}$
representations of $GL(5)$ are classified in section III. In section IV, we
explicitly construct a number of supersymmetric domain wall solutions.
Conclusions and discussions are given in section V, and useful formulae are
given in the appendix.

##  II $N=(2,2)$ gauged supergravity in six dimensions

We first give a review of six-dimensional $N=(2,2)$ gauged supergravity in the
embedding tensor formalism constructed in 6D_Max_Gauging . We will mainly
collect relevant formulae for constructing the embedding tensor in order to
classify various gauge groups and find supersymmetric domain wall solutions.
For more details, the reader is referred to the original construction in
6D_Max_Gauging .

There is only one supermultiplet in $N=(2,2)$ supersymmetry in six dimensions,
the graviton supermultiplet, with the following field content

$\left(e^{\hat{\mu}}_{\mu},B_{\mu\nu
m},A^{A}_{\mu},{V_{A}}^{\alpha\dot{\alpha}},\psi_{+\mu\alpha},\psi_{-\mu\dot{\alpha}},\chi_{+a\dot{\alpha}},\chi_{-\dot{a}\alpha}\right).$
(1)

We note here all the conventions used throughout the paper. These mostly
follow those used in 6D_Max_Gauging . Curved and flat space-time indices are
respectively denoted by $\mu,\nu,\ldots=0,1,\ldots,5$ and
$\hat{\mu},\hat{\nu},\ldots=0,1,\ldots,5$. Lower and upper
$m,n,\ldots=1,\ldots,5$ indices label fundamental and anti-fundamental
representations of $GL(5)\subset SO(5,5)$, respectively. Indices
$A,B,\ldots=1,\ldots,16$ describe Majorana-Weyl spinors of the $SO(5,5)$
duality symmetry. We also note that the electric two-form potentials
$B_{\mu\nu m}$, appearing in the ungauged Lagrangian, transform as
$\mathbf{5}$ under $GL(5)$ while the vector fields $A^{A}_{\mu}$ transform as
$\mathbf{16}_{c}$ under $SO(5,5)$. Together with the magnetic duals
${B_{\mu\nu}}^{m}$ transforming in $\overline{\mathbf{5}}$ representation of
$GL(5)$, the electric two-forms $B_{\mu\nu m}$ transform in a vector
representation $\mathbf{10}$ of the full global symmetry group $SO(5,5)$
denoted by $B_{\mu\nu M}=(B_{\mu\nu m},{B_{\mu\nu}}^{m})$. Therefore, only the
subgroup $GL(5)\subset SO(5,5)$ is a manifest off-shell symmetry of the
theory. Indices $M,N,\ldots=1,\ldots,10$ denote fundamental or vector
representation of $SO(5,5)$. Finally, there are $25$ scalar fields
parametrizing the coset space $SO(5,5)/SO(5)\times SO(5)$.

Fermionic fields, transforming under the local $SO(5)\times SO(5)$ symmetry,
are symplectic Majorana-Weyl (SMW) spinors. Indices $\alpha,\ldots=1,\ldots,4$
and $\dot{\alpha},\ldots=\dot{1},\ldots,\dot{4}$ are respectively two sets of
$SO(5)$ spinor indices in $SO(5)\times SO(5)$. Similarly, vector indices of
the two $SO(5)$ factors are denoted by $a,\ldots=1,\ldots,5$ and
$\dot{a},\ldots=\dot{1},\ldots,\dot{5}$. We use $\pm$ to indicate space-time
chiralities of the spinors. Under the local $SO(5)\times SO(5)$ symmetry, the
two sets of gravitini $\psi_{+\mu\alpha}$ and $\psi_{-\mu\dot{\alpha}}$
transform as $(\mathbf{4},\mathbf{1})$ and $(\mathbf{1},\mathbf{4})$ while the
spin-$\frac{1}{2}$ fields $\chi_{+a\dot{\alpha}}$ and $\chi_{-\dot{a}\alpha}$
transform as $(\mathbf{5},\mathbf{4})$ and $(\mathbf{4},\mathbf{5})$,
respectively.

In chiral spinor representation, the coset manifold $SO(5,5)/SO(5)\times
SO(5)$ is described by a coset representative ${V_{A}}^{\alpha\dot{\beta}}$
transforming under the global $SO(5,5)$ and local $SO(5)\times SO(5)$
symmetries by left and right multiplications, respectively. The inverse
elements ${(V^{-1})_{\alpha\dot{\beta}}}^{A}$ will be denoted by
${V^{A}}_{\alpha\dot{\beta}}$ satisfying the relations

${V_{A}}^{\alpha\dot{\beta}}{V^{B}}_{\alpha\dot{\beta}}=\delta^{B}_{A}\qquad\textrm{and}\qquad{V_{A}}^{\alpha\dot{\beta}}{V^{A}}_{\gamma\dot{\delta}}=\delta^{\alpha}_{\gamma}\delta^{\dot{\beta}}_{\dot{\delta}}\,.$
(2)

On the other hand, in vector representation, the coset representative is given
by a $10\times 10$ matrix
${\mathcal{V}_{M}}^{\underline{A}}=({\mathcal{V}_{M}}^{a},{\mathcal{V}_{M}}^{\dot{a}})$
with $\underline{A}=(a,\dot{a})$. This is related to the coset representative
in chiral spinor representation by the following relations

$\displaystyle{\mathcal{V}_{M}}^{a}$ $\displaystyle=$
$\displaystyle\frac{1}{16}V^{A\alpha\dot{\alpha}}(\Gamma_{M})_{AB}{(\gamma^{a})_{\alpha\dot{\alpha}}}^{\beta\dot{\beta}}{V^{B}}_{\beta\dot{\beta}},$
(3) $\displaystyle{\mathcal{V}_{M}}^{\dot{a}}$ $\displaystyle=$
$\displaystyle-\frac{1}{16}V^{A\alpha\dot{\alpha}}(\Gamma_{M})_{AB}{(\gamma^{\dot{a}})_{\alpha\dot{\alpha}}}^{\beta\dot{\beta}}{V^{B}}_{\beta\dot{\beta}}\,.$
(4)

In these equations, $(\Gamma_{M})_{AB}$ and
${(\Gamma_{\underline{A}})_{\alpha\dot{\alpha}}}^{\beta\dot{\beta}}=({(\gamma_{a})_{\alpha\dot{\alpha}}}^{\beta\dot{\beta}},{(\gamma_{\dot{a}})_{\alpha\dot{\alpha}}}^{\beta\dot{\beta}})$
are respectively $SO(5,5)$ gamma matrices in non-diagonal $\eta_{MN}$ and
diagonal $\eta_{\underline{A}\underline{B}}$ bases.
The inverse will be denoted by $\mathcal{V}^{M\underline{A}}$ satisfying the
following relations

$\mathcal{V}^{Ma}{\mathcal{V}_{M}}^{b}=\delta^{ab},\qquad\mathcal{V}^{M\dot{a}}{\mathcal{V}_{M}}^{\dot{b}}=\delta^{\dot{a}\dot{b}},\qquad\mathcal{V}^{Ma}{\mathcal{V}_{M}}^{\dot{a}}=0$
(5)

and

${\mathcal{V}_{M}}^{a}\mathcal{V}^{Na}-{\mathcal{V}_{M}}^{\dot{a}}\mathcal{V}^{N\dot{a}}=\delta^{N}_{M}\,.$
(6)

In these equations, we have explicitly raised the $SO(5)\times SO(5)$ vector
index $\underline{A}=(a,\dot{a})$ resulting in a minus sign in equation (6).

The most general gaugings of six-dimensional $N=(2,2)$ supergravity are
described by the embedding tensor ${\Theta_{A}}^{MN}$ leading to the following
covariant derivative

$D_{\mu}=\partial_{\mu}-gA^{A}_{\mu}\ {\Theta_{A}}^{MN}\boldsymbol{t}_{MN}$
(7)

where $g$ is a gauge coupling constant. The embedding tensor identifies
generators $X_{A}={\Theta_{A}}^{MN}\boldsymbol{t}_{MN}$ of the gauge group
$G_{0}\subset SO(5,5)$ with particular linear combinations of the $SO(5,5)$
generators $\boldsymbol{t}_{MN}$. Supersymmetry requires the embedding tensor
to transform as $\mathbf{144}_{c}$ representation of $SO(5,5)$. Accordingly,
${\Theta_{A}}^{MN}$ can be parametrized in terms of a vector-spinor
$\theta^{AM}$ of $SO(5,5)$ as

${\Theta_{A}}^{MN}\ =\ -\theta^{B[M}(\Gamma^{N]})_{BA}\ \equiv\
\left(\Gamma^{[M}\theta^{N]}\right)_{A}$ (8)

with $\theta^{AM}$ subject to the constraint

$(\Gamma_{M})_{AB}\,\theta^{BM}\ =\ 0\,.$ (9)

The $SO(5,5)$ generators in vector and spinor representations can be chosen as

${(\boldsymbol{t}_{MN})_{P}}^{Q}=4\eta_{P[M}\delta^{Q}_{N]}\qquad\text{and}\qquad{(\boldsymbol{t}_{MN})_{A}}^{B}\
=\ {(\Gamma_{MN})_{A}}^{B}$ (10)

with $\eta_{MN}$ being the off-diagonal $SO(5,5)$ invariant tensor given by

$\eta_{MN}\ =\ \eta^{MN}\ =\ \begin{pmatrix}0&\mathds{1}_{5}\\\
\mathds{1}_{5}&0\end{pmatrix}$ (11)

and

${(\Gamma_{MN})_{A}}^{B}=\frac{1}{2}\left[{(\Gamma_{M})_{A}}^{C}{(\Gamma_{N})_{C}}^{B}-{(\Gamma_{N})_{A}}^{C}{(\Gamma_{M})_{C}}^{B}\right].$
(12)

We also note that the notation $\mathds{1}_{n}$ denotes an $n\times n$
identity matrix. ${(\Gamma_{M})_{A}}^{B}$ are chirally projected $SO(5,5)$
gamma matrices.
The corresponding gauge generators in these representations then take the
forms

${(X_{A})_{M}}^{N}=2\left(\Gamma_{M}\theta^{N}\right)_{A}+2\left(\Gamma^{N}\theta_{M}\right)_{A}\qquad\text{and}\qquad{(X_{A})_{B}}^{C}=\left(\Gamma^{M}\theta^{N}\right)_{A}{(\Gamma_{MN})_{B}}^{C}\,.$
(13)

Furthermore, consistency requires the gauge generators to form a closed
subalgebra of $SO(5,5)$ implying the quadratic constraint

$\left[X_{A},X_{B}\right]\ =\ -{(X_{A})_{B}}^{C}\,X_{C}\,.$ (14)

In terms of $\theta^{AM}$, the quadratic constraint reduces to the following
two conditions

$\displaystyle\theta^{AM}\theta^{BN}\eta_{MN}$ $\displaystyle=$ $\displaystyle
0,$ (15) $\displaystyle\theta^{AM}\theta^{B[N}(\Gamma^{P]})_{AB}$
$\displaystyle=$ $\displaystyle 0.$ (16)

It follows that any $\theta^{AM}\in\mathbf{144}_{c}$ satisfying this quadratic
constraint defines a consistent gauging.
In this work, we are only interested in the classification of gauge groups
under $GL(5)\subset SO(5,5)$ and domain wall solutions which only involve the
metric and scalar fields. Therefore, we will, from now on, set all vector and
tensor fields to zero with the bosonic Lagrangian of the maximal $N=(2,2)$
gauged supergravity given by

$e^{-1}\mathcal{L}=\frac{1}{4}R-\frac{1}{16}P_{\mu}^{a\dot{a}}P^{\mu}_{a\dot{a}}-\mathbf{V}\,.$
(17)

We also need supersymmetry transformations of fermionic fields which, for
vanishing fermions and vector/tensor fields, are given by

$\displaystyle\delta\psi_{+\mu\alpha}$ $\displaystyle=$ $\displaystyle
D_{\mu}\epsilon_{+\alpha}+\frac{g}{4}\hat{\gamma}_{\mu}{T_{\alpha}}^{\dot{\beta}}\epsilon_{-\dot{\beta}},$
(18) $\displaystyle\delta\psi_{-\mu\dot{\alpha}}$ $\displaystyle=$
$\displaystyle
D_{\mu}\epsilon_{-\dot{\alpha}}-\frac{g}{4}\hat{\gamma}_{\mu}{T^{\beta}}_{\dot{\alpha}}\epsilon_{+\beta},$
(19) $\displaystyle\delta\chi_{+a\dot{\alpha}}$ $\displaystyle=$
$\displaystyle\frac{1}{4}P^{\mu}_{a\dot{a}}\hat{\gamma}_{\mu}{(\gamma^{\dot{a}})_{\dot{\alpha}}}^{\dot{\beta}}\epsilon_{-\dot{\beta}}+2g{(T_{a})^{\beta}}_{\dot{\alpha}}\epsilon_{+\beta}-\frac{g}{2}{T^{\alpha}}_{\dot{\alpha}}{(\gamma_{a})_{\alpha}}^{\beta}\epsilon_{+\beta},$
(20) $\displaystyle\delta\chi_{-\dot{a}\alpha}$ $\displaystyle=$
$\displaystyle\frac{1}{4}P^{\mu}_{a\dot{a}}\hat{\gamma}_{\mu}{(\gamma^{a})_{\alpha}}^{\beta}\epsilon_{+\beta}+2g{(T_{\dot{a}})_{\alpha}}^{\dot{\beta}}\epsilon_{-\dot{\beta}}+\frac{g}{2}{T_{\alpha}}^{\dot{\alpha}}{(\gamma_{\dot{a}})_{\dot{\alpha}}}^{\dot{\beta}}\epsilon_{-\dot{\beta}}.$
(21)

The covariant derivatives of supersymmetry parameters, $\epsilon_{+\alpha}$
and $\epsilon_{-\dot{\alpha}}$, are defined by

$\displaystyle D_{\mu}\epsilon_{+\alpha}$ $\displaystyle=$
$\displaystyle\partial_{\mu}\epsilon_{+\alpha}+\frac{1}{4}{\omega_{\mu}}^{\nu\rho}\hat{\gamma}_{\nu\rho}\epsilon_{+\alpha}+\frac{1}{4}Q_{\mu}^{ab}{(\gamma_{ab})_{\alpha}}^{\beta}\epsilon_{+\beta},$
(22) $\displaystyle D_{\mu}\epsilon_{-\dot{\alpha}}$ $\displaystyle=$
$\displaystyle\partial_{\mu}\epsilon_{-\dot{\alpha}}+\frac{1}{4}{\omega_{\mu}}^{\nu\rho}\hat{\gamma}_{\nu\rho}\epsilon_{-\dot{\alpha}}+\frac{1}{4}Q_{\mu}^{\dot{a}\dot{b}}{(\gamma_{\dot{a}\dot{b}})_{\dot{\alpha}}}^{\dot{\beta}}\epsilon_{-\dot{\beta}}$
(23)

with $\hat{\gamma}_{\mu}=e_{\mu}^{\hat{\mu}}\hat{\gamma}_{\hat{\mu}}$.
$\hat{\gamma}_{\hat{\mu}}$ are space-time gamma matrices, and for simplicity,
we will suppress space-time spinor indices.

The scalar vielbein $P_{\mu}^{a\dot{a}}$ and $SO(5)\times SO(5)$ composite
connections, $Q_{\mu}^{ab}$ and $Q_{\mu}^{\dot{a}\dot{b}}$, are given by

$\displaystyle P_{\mu}^{a\dot{a}}$ $\displaystyle=$
$\displaystyle\frac{1}{4}{(\gamma^{a})}^{\alpha\beta}{(\gamma^{\dot{a}})}^{\dot{\alpha}\dot{\beta}}{V^{A}}_{\alpha\dot{\alpha}}\partial_{\mu}V_{A\beta\dot{\beta}},$
(24) $\displaystyle Q_{\mu}^{ab}$ $\displaystyle=$
$\displaystyle\frac{1}{8}{(\gamma^{ab})}^{\alpha\beta}\Omega^{\dot{\alpha}\dot{\beta}}{V^{A}}_{\alpha\dot{\alpha}}\partial_{\mu}V_{A\beta\dot{\beta}},$
(25) $\displaystyle Q_{\mu}^{\dot{a}\dot{b}}$ $\displaystyle=$
$\displaystyle\frac{1}{8}\Omega^{\alpha\beta}{(\gamma^{\dot{a}\dot{b}})}^{\dot{\alpha}\dot{\beta}}{V^{A}}_{\alpha\dot{\alpha}}\partial_{\mu}V_{A\beta\dot{\beta}}\,.$
(26)

In these equations, $\Omega^{\alpha\beta}$ and
$\Omega^{\dot{\alpha}\dot{\beta}}$ are the $USp(4)$ symplectic forms that
satisfy the following relations

$\Omega_{\beta\alpha}\ =\ -\Omega_{\alpha\beta},\qquad\Omega^{\alpha\beta}\ =\
(\Omega_{\alpha\beta})^{*},\qquad\Omega_{\alpha\beta}\Omega^{\beta\gamma}\ =\
-\delta^{\gamma}_{\alpha}$ (27)

and similarly for $\Omega_{\dot{\alpha}\dot{\beta}}$. We will use the explicit
form of $\Omega_{\alpha\beta}$ and $\Omega_{\dot{\alpha}\dot{\beta}}$ given by

$\Omega_{\alpha\beta}\ =\ \Omega_{\dot{\alpha}\dot{\beta}}\ =\
\mathds{1}_{2}\otimes i\sigma_{2}\,.$ (28)

We also note that the definitions (24), (25), and (26) can be derived from the
following relation

${V^{A}}_{\alpha\dot{\alpha}}\partial_{\mu}V_{A\beta\dot{\beta}}=\frac{1}{4}P_{\mu}^{a\dot{a}}(\gamma_{a})_{\alpha\beta}(\gamma_{\dot{a}})_{\dot{\alpha}\dot{\beta}}+\frac{1}{4}Q_{\mu}^{ab}(\gamma_{ab})_{\alpha\beta}\Omega_{\dot{\alpha}\dot{\beta}}+\frac{1}{4}Q_{\mu}^{\dot{a}\dot{b}}\Omega_{\alpha\beta}(\gamma_{\dot{a}\dot{b}})_{\dot{\alpha}\dot{\beta}}\,.$
(29)

The scalar potential is given by

$\displaystyle\mathbf{V}$ $\displaystyle=$
$\displaystyle\frac{g^{2}}{2}\theta^{AM}\theta^{BN}{\mathcal{V}_{M}}^{a}{\mathcal{V}_{N}}^{b}\left[{V_{A}}^{\alpha\dot{\alpha}}{(\gamma_{a})_{\alpha}}^{\beta}{(\gamma_{b})_{\beta}}^{\gamma}V_{B\gamma\dot{\alpha}}\right]$
(30) $\displaystyle=$
$\displaystyle-\frac{g^{2}}{2}\left[T^{\alpha\dot{\alpha}}T_{\alpha\dot{\alpha}}-2(T^{a})^{\alpha\dot{\alpha}}(T_{a})_{\alpha\dot{\alpha}}\right]$

with the T-tensors defined by

$(T^{a})^{\alpha\dot{\alpha}}={\mathcal{V}_{M}}^{a}\theta^{AM}{V_{A}}^{\alpha\dot{\alpha}},\qquad(T^{\dot{a}})^{\alpha\dot{\alpha}}=-{\mathcal{V}_{M}}^{\dot{a}}\theta^{AM}{V_{A}}^{\alpha\dot{\alpha}}$
(31)

and

$T^{\alpha\dot{\alpha}}\equiv(T^{a})^{\beta\dot{\alpha}}{(\gamma_{a})_{\beta}}^{\alpha}=-(T^{\dot{a}})^{\alpha\dot{\beta}}{(\gamma_{\dot{a}})_{\dot{\beta}}}^{\dot{\alpha}}\,.$
(32)

##  III Gaugings of $N=(2,2)$ supergravity under $GL(5)$

In this section, we consider gaugings under $GL(5)\subset SO(5,5)$. The
embedding tensor $\theta^{AM}$ in $\mathbf{144}_{c}$ representation of
$SO(5,5)$ decomposes under $GL(5)$ as

$\mathbf{144}_{c}\ \rightarrow\
\overline{\mathbf{5}}^{+3}\,\oplus\,\mathbf{5}^{+7}\,\oplus\,\mathbf{10}^{-1}\,\oplus\,\mathbf{15}^{-1}\,\oplus\,\mathbf{24}^{-5}\,\oplus\,\overline{\mathbf{40}}^{-1}\,\oplus\,\overline{\mathbf{45}}^{+3}\,.$
(33)

The $SO(5,5)$ spinor representation decomposes as

$\mathbf{16}_{s}\ \rightarrow\
\overline{\mathbf{5}}^{+3}\,\oplus\,\mathbf{10}^{-1}\,\oplus\,\mathbf{1}^{-5}\,.$
(34)

Accordingly, the gauge generators can be written in terms of $X^{m}$,
$X_{mn}$, and $X_{\ast}$ denoting respectively $\overline{\mathbf{5}}^{+3}$,
$\mathbf{10}^{-1}$, and $\mathbf{1}^{-5}$ as

$X_{A}\ =\
\mathbb{T}_{Am}X^{m}+\mathbb{T}_{A}^{mn}X_{mn}+\mathbb{T}_{A\ast}X_{\ast}\,.$
(35)

The decomposition matrices $\mathbb{T}_{Am}$, $\mathbb{T}_{A}^{mn}$, and
$\mathbb{T}_{A\ast}$ are given in the appendix.
By the decomposition of $SO(5,5)$ vector representation

$\mathbf{10}\ \rightarrow\
\mathbf{5}^{+2}\,\oplus\,\overline{\mathbf{5}}^{-2},$ (36)

we can write the embedding tensor as

$\theta^{AM}=(\theta^{Am},\theta^{A}_{m}).$ (37)

The two components $\theta^{Am}$ and $\theta^{A}_{m}$ contain the following
irreducible $GL(5)$ representations

$\displaystyle\theta^{Am}$ $\displaystyle:$
$\displaystyle\qquad\overline{\mathbf{5}}^{+3}\,\oplus\,\mathbf{10}^{-1}\,\oplus\,\mathbf{24}^{-5}\,\oplus\,\overline{\mathbf{40}}^{-1},$
(38) $\displaystyle\theta^{A}_{m}$ $\displaystyle:$
$\displaystyle\qquad\overline{\mathbf{5}}^{+3}\,\oplus\,\mathbf{5}^{+7}\,\oplus\,\mathbf{10}^{-1}\,\oplus\,\mathbf{15}^{-1}\,\oplus\,\overline{\mathbf{45}}^{+3}\,.$
(39)

As pointed out in 6D_Max_Gauging , gaugings triggered by $\theta^{Am}$ are
called electric gaugings in the sense that only electric two-forms participate
in the gauged theory while gaugings triggered by $\theta^{A}_{m}$ are called
magnetic gaugings involving magnetic two-forms together with additional three-
form tensor fields. The decompositions in (38) and (39) imply that gaugings in
$\mathbf{24}^{-5}\,\oplus\,\overline{\mathbf{40}}^{-1}$ and
$\mathbf{5}^{+7}\,\oplus\,\mathbf{15}^{-1}\,\oplus\,\overline{\mathbf{45}}^{+3}$
representations are respectively purely electric and purely magnetic while
those in $\overline{\mathbf{5}}^{+3}\,\oplus\,\mathbf{10}^{-1}$
representations correspond to dyonic gaugings. Many possible dyonic gaugings
can also arise from combinations of various electric and magnetic components.
Finally, we note that the quadratic constraint (15) is automatically satisfied
for purely electric or purely magnetic gaugings that involve only
$\theta^{Am}$ or $\theta^{A}_{m}$ components.
In accord with (38) and (39), we can parametrize the embedding tensor as, see
6D_DW_I  for more detail,

$\displaystyle\theta^{Am}$ $\displaystyle=$
$\displaystyle\mathbb{T}^{An}{S_{n}}^{m}+\mathbb{T}^{A}_{np}\left(U^{np,m}+\frac{1}{3\sqrt{2}}\varepsilon^{mnpqr}Z_{qr}\right)-\frac{2\sqrt{2}}{5}\mathbb{T}^{A}_{\ast}J^{m},$
(40) $\displaystyle\theta^{A}_{m}$ $\displaystyle=$
$\displaystyle\mathbb{T}^{An}(Y_{nm}+Z_{nm})+\mathbb{T}^{A}_{np}(W^{np}_{m}+J^{[n}\delta^{p]}_{m})+\mathbb{T}^{A}_{\ast}K_{m}\,.$
(41)

Matrices $\mathbb{T}^{Am}$, $\mathbb{T}^{A}_{mn}$, and $\mathbb{T}^{A}_{\ast}$
are inverses of the decomposition matrices $\mathbb{T}_{Am}$,
$\mathbb{T}_{A}^{mn}$, and $\mathbb{T}_{A\ast}$ given by complex conjugations,
$\mathbb{T}^{A}=(\mathbb{T}_{A})^{-1}=(\mathbb{T}_{A})^{*}$.
We now look at various components in more detail. The first representation
$\mathbf{24}^{-5}$ is described by a traceless $5\times 5$ matrix
${S_{n}}^{m}$ with ${S_{m}}^{m}=0$. The tensor $U^{np,m}=U^{[np],m}$
satisfying $U^{[np,m]}=0$ corresponds to $\overline{\mathbf{40}}^{-1}$
representation. The symmetric tensor $Y_{mn}=Y_{(mn)}$ and antisymmetric one
$Z_{mn}=Z_{[mn]}$ respectively denote $\mathbf{15}^{-1}$ and
$\mathbf{10}^{-1}$ representations. The $\overline{\mathbf{45}}^{+3}$
representation is written as the tensor $W^{np}_{m}=W^{[np]}_{m}$ satisfying
$W^{nm}_{m}=0$. Finally, the last two $\overline{\mathbf{5}}^{+3}$ and
$\mathbf{5}^{+7}$ representations are parametrized by two $GL(5)$ vectors
$J^{m}$ and $K_{m}$, respectively. The general structure of gaugings under
$GL(5)\subset SO(5,5)$ is shown in Table 1 taken from 6D_Max_Gauging . The
left column represents the sixteen vector fields in $GL(5)$ representations
while the top row corresponds to the decomposition of $SO(5,5)$ generators
under $GL(5)$. The table shows the couplings between $SO(5,5)$ generators and
vector fields by various components of the embedding tensor.

The gaugings from the components $\mathbf{15}^{-1}$ and
$\overline{\mathbf{40}}^{-1}$ have been extensively studied in 6D_DW_I . In
the present work, we will focus on the remaining representations and examples
of possible combinations shown in Table 1. In the following, we will determine
an explicit form of the aforementioned $GL(5)$ tensors by imposing the
quadratic constraints (15) and (16) on the embedding tensor under
consideration.

${\Theta_{A}}^{MN}$ | $\overline{\mathbf{10}}^{-4}$ | $\mathbf{1}^{0}$ | $\mathbf{24}^{0}$ | $\mathbf{10}^{+4}$
---|---|---|---|---
$\mathbf{5}^{-3}$ | $\mathbf{5}^{+7}$ | $\overline{\mathbf{5}}^{+3}$ | $(\overline{\mathbf{5}}+\overline{\mathbf{45}})^{+3}$ | $(\mathbf{10}+\overline{\mathbf{40}})^{-1}$
$\overline{\mathbf{10}}^{+1}$ | $(\overline{\mathbf{5}}+\overline{\mathbf{45}})^{+3}$ | $\mathbf{10}^{-1}$ | $(\mathbf{10}+\mathbf{15}+\overline{\mathbf{40}})^{-1}$ | $\mathbf{24}^{-5}$
$\mathbf{1}^{+5}$ | $\mathbf{10}^{-1}$ |  | $\mathbf{24}^{-5}$ |
Table 1: Gauge couplings between the sixteen vector fields and $SO(5,5)$
generators from various $GL(5)$ components of the embedding tensor.

###  III.1 Gaugings in $\mathbf{24}^{-5}$ representation

We begin with gauge groups arising from the embedding tensor in
$\mathbf{24}^{-5}$ representation. Gaugings in this representation are purely
electric and triggered by

$\theta^{Am}\ =\ \mathbb{T}^{An}{S_{n}}^{m}\,.$ (42)

With $\theta^{A}_{m}=0$, the embedding tensor
$\theta^{AM}=(\,\mathbb{T}^{An}{S_{n}}^{m}\,,\,0\,)$ automatically satisfies
the quadratic constraint. Therefore, every traceless $5\times 5$ matrix
${S_{m}}^{n}$ defines a viable gauge group generated by the following gauge
generators

$X^{m}\ =\ 0,\qquad X_{mn}\ =\
{S_{m}}^{p}\boldsymbol{s}_{pn}-{S_{n}}^{p}\boldsymbol{s}_{pm},\qquad X_{\ast}\
=-\sqrt{2}{S_{m}}^{n}{\boldsymbol{\tau}^{m}}_{n}\,.$ (43)

${\boldsymbol{\tau}^{m}}_{n}$ are $SL(5)$ generators defined as

${\boldsymbol{\tau}^{m}}_{n}\ =\
{\boldsymbol{t}^{m}}_{n}-\frac{1}{5}\,\boldsymbol{d}\,\delta^{m}_{n}$ (44)

with ${\boldsymbol{\tau}^{m}}_{m}=0$ and

$\boldsymbol{d}={\boldsymbol{t}^{m}}_{m}={\boldsymbol{t}^{1}}_{1}+{\boldsymbol{t}^{2}}_{2}+{\boldsymbol{t}^{3}}_{3}+{\boldsymbol{t}^{4}}_{4}+{\boldsymbol{t}^{5}}_{5}$
(45)

being the $SO(1,1)\sim\mathbb{R}^{+}$ generator in
$GL(5)\sim\mathbb{R}^{+}\times SL(5)$. We also use
$\boldsymbol{s}_{mn}=\boldsymbol{t}_{mn}$ to denote the generators
corresponding to shift symmetries on the scalar fields, see more detail in the
appendix.

Commutation relations between the gauge generators read

$[X_{mn},X_{pq}]\ =\ 0\qquad\text{ and }\qquad[X_{mn},X_{\ast}]\ =\
{(X_{\ast})_{mn}}^{pq}X_{pq}$ (46)

where the generators with two antisymmetric pairs of $GL(5)$ vector indices
are defined as

${(X_{A})_{mn}}^{pq}=2{(X_{A})_{[m}}^{[p}\delta^{q]}_{n]}\,.$ (47)

From equation (46), we readily see that the generators $X_{mn}$ form a
translational group and transform non-trivially under a one-dimensional group
generated by $X_{\ast}$, a particular linear combination of $SL(5)$
generators.

In vector or fundamental representation of $SL(5)$, $X_{\ast}$ generator takes
the form

${(X_{\ast})_{m}}^{n}=-2\sqrt{2}{S_{m}}^{n}\,.$ (48)

If ${S_{m}}^{n}$ is antisymmetric, $X_{\ast}$ will also be antisymmetric and
generates a compact $SO(2)$ group. On the other hand, for symmetric
${S_{m}}^{n}$, $X_{\ast}$ generates a non-compact $SO(1,1)$ group. The
resulting gauge groups then take the form of

$SO(2)\ltimes\mathbb{R}^{n}\qquad\textrm{and}\qquad
SO(1,1)\ltimes\mathbb{R}^{n}$ (49)

for $3\leq n\leq 10$. The values of $n$ depend on the choices of
${S_{m}}^{n}$. It has been pointed out in 6D_Max_Gauging  that this type of
gaugings is related to Scherk-Schwarz reductions from seven-dimensional gauged
supergravity, and gaugings in $\mathbf{24}^{-1}$ representation correspond to
choosing a generator from the seven-dimensional symmetry group $SL(5)$, see
also maximal_SUGRA_deWit  for a general discussion on Scherk-Schwarz
reductions and gauged supergravities.

We end this case by giving an explicit example with $X_{\ast}$ antisymmetric.
By choosing

${S_{m}}^{n}=\small\begin{pmatrix}0&\kappa&0&0&0\\\ -\kappa&0&0&0&0\\\
0&0&0&0&0\\\ 0&0&0&0&-\lambda\\\ 0&0&0&\lambda&0\end{pmatrix},$ (50)

we find the following non-vanishing gauge generators

$\displaystyle X_{ix}$ $\displaystyle=$
$\displaystyle{S_{i}}^{j}\boldsymbol{s}_{jx}-{S_{x}}^{y}\boldsymbol{s}_{yi},\qquad
X_{i3}\ =\ {S_{i}}^{j}\boldsymbol{s}_{j3},\qquad X_{x3}\ =\
{S_{x}}^{y}\boldsymbol{s}_{y3},$ $\displaystyle X_{\ast}$ $\displaystyle=$
$\displaystyle-\sqrt{2}(\kappa{\boldsymbol{\tau}^{1}}_{2}-\kappa{\boldsymbol{\tau}^{2}}_{1}-\lambda{\boldsymbol{\tau}^{4}}_{5}+\lambda{\boldsymbol{\tau}^{5}}_{4})$
(51)

in which $i,j=1,2$ and $x,y=4,5$. It can be straightforwardly verified that
these gauge generators satisfy the commutation relations given in (46). The
resulting gauge group is of the form

$G_{0}=SO(2)\ltimes\mathbb{R}^{8}_{\boldsymbol{s}}$ (52)

in which $SO(2)$ is generated by $X_{\ast}$, and the shift translational group
$\mathbb{R}^{8}_{\boldsymbol{s}}$ is generated by $X_{ix}$, $X_{i3}$, and
$X_{x3}$.

###  III.2 Gaugings in $\overline{\mathbf{45}}^{+3}$ representation

We now consider the embedding tensor in $\overline{\mathbf{45}}^{+3}$
representation. Gaugings in this case are purely magnetic and related to
reductions of eleven-dimensional supergravity on twisted tori as pointed out
in 6D_Max_Gauging . The linear constraint (9) and the quadratic constraint
(15) are both satisfied if we parametrize the embedding tensor as

$\theta^{AM}=(\,0\,,\,\mathbb{T}^{A}_{np}W^{np}_{m}\,).$ (53)

In terms of $W^{np}_{m}$, the constraint (16) reduces to

$W^{mn}_{r}\,W^{pq}_{[s}\,\varepsilon_{t]mnpq}\ =\ 0\,.$ (54)

As noted in 6D_Max_Gauging , this constraint is the duality of a similar
condition in gaugings from $\overline{\mathbf{40}}^{-1}$ representation
($U^{mn,r}U^{pq,s}\varepsilon_{mnpqt}\ =\ 0$), considered in 6D_DW_I . To
solve this condition, we then follow the same procedure as in
$\overline{\mathbf{40}}^{-1}$ representation. We first write $W^{np}_{m}$ in
the form of

$W^{np}_{m}\ =\ 2v^{[n}{u_{m}}^{p]}$ (55)

and impose the condition ${u_{m}}^{m}=0$ in order to satisfy the traceless
condition $W^{nm}_{m}=0$. This form of $W^{np}_{m}$ is sufficient to solve the
condition (54).

To give an explicit example, we will choose a basis in which
$v^{m}=\delta^{m}_{5}$ and ${u_{5}}^{m}={u_{m}}^{5}=0$. As a result,
consistent gaugings satisfying the linear and quadratic constraints in
$\overline{\mathbf{45}}^{+3}$ representation are now parametrized by a
traceless $4\times 4$ matrix ${u_{i}}^{j}$ with $i,j=1,...,4$ and
${u_{i}}^{i}=0$. In this case, the gauge generators turn out to be
$X_{\ast}=X_{i5}=0$ and the remaining non-vanishing generators given by

$X_{ij}\ =\ \sqrt{2}\varepsilon_{ijkl}{u_{m}}^{k}\boldsymbol{h}^{lm},\qquad
X^{i}\ =-2{u_{k}}^{i}{\boldsymbol{\tau}^{k}}_{5},\qquad X^{5}\
=2{u_{k}}^{j}{\boldsymbol{\tau}^{k}}_{j}\,.$ (56)

We recall that $\boldsymbol{h}^{mn}=\boldsymbol{t}^{mn}$ are $SO(5,5)$
generators associated with the hidden symmetries that do not constitute
symmetries of the action. Commutation relations between the gauge generators
read

$\displaystyle\left[X_{ij},X_{kl}\right]$ $\displaystyle=$
$\displaystyle\left[X_{ij},X^{k}\right]\ =\ \left[X^{i},X^{j}\right]\ =\ 0,$
$\displaystyle\left[X^{5},X_{ij}\right]$ $\displaystyle=$
$\displaystyle{(X^{5})_{ij}}^{kl}X_{kl},\qquad\quad\left[X^{5},X^{i}\right]={(X^{5})_{j}}^{i}X^{j}.$
(57)

As seen from these relations, $X_{ij}$ generate a six-dimensional
translational group $\mathbb{R}^{6}_{\boldsymbol{h}}$ associated with the
hidden symmetries while $X^{i}$ generate another translational group
$\mathbb{R}^{4}$ commuting with $\mathbb{R}^{6}_{\boldsymbol{h}}$. The $X^{5}$
generator takes the same form as $X_{\ast}$ in $\mathbf{24}^{-1}$
representation. We can similarly use the unbroken $SL(4)$ symmetry to fix
${u_{i}}^{j}$ in the form of

$\small{u_{i}}^{j}=\begin{pmatrix}0&\kappa&0&0\\\ \pm\kappa&0&0&0\\\
0&0&0&-\lambda\\\ 0&0&\mp\lambda&0\end{pmatrix}\,.$ (58)

With this explicit form of ${u_{i}}^{j}$, non-vanishing gauge generators are
now given by

$\displaystyle X_{\bar{i}\bar{x}}$ $\displaystyle=$
$\displaystyle-\sqrt{2}\varepsilon_{\bar{i}\bar{j}}\varepsilon_{\bar{x}\bar{y}}({u_{\bar{k}}}^{\bar{j}}\boldsymbol{h}^{\bar{y}\bar{k}}-{u_{\bar{z}}}^{\bar{y}}\boldsymbol{h}^{\bar{j}\bar{z}}),\qquad
X^{i}\ =-2{u_{k}}^{i}{\boldsymbol{\tau}^{k}}_{5},$ $\displaystyle X^{5}$
$\displaystyle=$ $\displaystyle
2(\kappa{\boldsymbol{\tau}^{1}}_{2}\pm\kappa{\boldsymbol{\tau}^{2}}_{1}-\lambda{\boldsymbol{\tau}^{3}}_{4}\mp\lambda{\boldsymbol{\tau}^{4}}_{3})$
(59)

where $i=(\bar{i},\bar{x})$ with $\bar{i},\bar{j},...=1,2$ and
$\bar{x},\bar{y},...=3,4$. Therefore, for arbitrary values of $\kappa$ and
$\lambda$, the corresponding gauge groups are given by

$G_{0}=SO(1,1)\ltimes\left(\mathbb{R}^{4}\times\mathbb{R}^{4}_{\boldsymbol{h}}\right)\sim
CSO(1,1,2)\ltimes\mathbb{R}^{4}_{\boldsymbol{h}}$ (60)

or

$G_{0}=SO(2)\ltimes\left(\mathbb{R}^{4}\times\mathbb{R}^{4}_{\boldsymbol{h}}\right)\sim
CSO(2,0,2)\ltimes\mathbb{R}^{4}_{\boldsymbol{h}}$ (61)

depending on the choices of the signs for $\kappa$ and $\lambda$, making
${u_{i}}^{j}$ in (58) respectively symmetric or antisymmetric.

###  III.3 Gaugings in $\mathbf{5}^{+7}$ representation

In this section, we look at the smallest representation giving rise to purely
magnetic gaugings. With $\theta^{Am}=0$, we parametrize the embedding tensor,
in this case, by a $GL(5)$ vector $K_{m}$ as

$\theta^{A}_{m}\ =\ \mathbb{T}^{A}_{\ast}K_{m}\,.$ (62)

Gaugings triggered by $\theta^{AM}=(\,0\,,\,\mathbb{T}^{A}_{\ast}K_{m}\,)$
automatically satisfy the linear and quadratic constraints. As pointed out in
6D_Max_Gauging , these gaugings might correspond to reductions from eleven
dimensions with non-trivial four-form fluxes.

Only gauge generators $X^{m}$ are non-vanishing in this representation. They
are explicitly given in terms of the generators of the hidden symmetry by

$X^{m}\ =\ 5\sqrt{2}K_{n}\boldsymbol{h}^{nm}\,.$ (63)

These generators commute with each other

$\left[X^{m},X^{n}\right]\ =\ 0\,.$ (64)

However, due to the antisymmetric property of $\boldsymbol{h}^{nm}$, the gauge
generators satisfy the condition $K_{m}X^{m}=0$. Therefore, only four
generators are linearly independent. The resulting gauge group in
$\mathbf{5}^{+7}$ representation is then given by a four-dimensional
translational group $\mathbb{R}^{4}_{\boldsymbol{h}}$ associated with the
hidden symmetries.

To explicitly parametrize this $\mathbb{R}^{4}_{\boldsymbol{h}}$ gauge group,
we will use the $SL(5)$ symmetry to fix $K_{m}=\kappa\,\delta^{5}_{m}$ with
$\kappa\in\mathbb{R}$. In this case, the gauge group is generated by

$X^{i}\ =\ 5\sqrt{2}\kappa\,\boldsymbol{h}^{5i}$ (65)

for $i=1,2,3,4$.

###  III.4 Gaugings in $\overline{\mathbf{5}}^{+3}$ representation

We now consider dyonic gaugings involving both electric and magnetic
components of the embedding tensor. We begin with gaugings in
$\overline{\mathbf{5}}^{+3}$ representation. The embedding tensor in this
representation, satisfying the linear constraint (9), is given by
$\theta^{AM}=(\theta^{Am},\theta^{A}_{m})$ with

$\theta^{Am}\ =-\frac{2\sqrt{2}}{5}\mathbb{T}^{A}_{\ast}J^{m}\qquad\text{ and
}\qquad\theta^{A}_{m}\ =\ \mathbb{T}^{A}_{np}J^{[n}\delta^{p]}_{m}\,.$ (66)

This form of the embedding tensor automatically satisfies the quadratic
constraints given in (15) and (16). Therefore, any vectors $J^{m}$ define
consistent gaugings in $\overline{\mathbf{5}}^{+3}$ representation.

In this case, we find non-vanishing gauge generators given by

$X^{m}=3{J^{n}\boldsymbol{\tau}^{m}}_{n}+\frac{8}{5}J^{m}\boldsymbol{d}\qquad\text{
and }\qquad
X_{mn}=-\frac{1}{\sqrt{2}}\varepsilon_{mnpqr}J^{p}\boldsymbol{h}^{qr}$ (67)

with the following commutation relations

$\left[X_{mn},X_{pq}\right]=0,\qquad\left[X^{m},X_{np}\right]={(X^{m})_{np}}^{qr}X_{qr},\qquad\left[X^{m},X^{n}\right]=3{(X^{[m})_{p}}^{n]}X^{p}\,.$
(68)

To write down the explicit form of gauge generators, we can use $SL(5)$
symmetry to fix $J^{m}=\kappa\,\delta^{m}_{5}$ as in the previous case. With
this choice, non-vanishing gauge generators are given by

$\displaystyle X_{ij}$ $\displaystyle=$
$\displaystyle-\frac{\kappa}{\sqrt{2}}\varepsilon_{ijkl}\boldsymbol{h}^{kl},\qquad
X^{i}\ =\ 3\kappa\,{\boldsymbol{\tau}^{i}}_{5},$ $\displaystyle X^{5}$
$\displaystyle=$
$\displaystyle\kappa\left({\boldsymbol{\tau}^{1}}_{1}+{\boldsymbol{\tau}^{2}}_{2}+{\boldsymbol{\tau}^{3}}_{3}+{\boldsymbol{\tau}^{4}}_{4}+4\,{\boldsymbol{\tau}^{5}}_{5}\right)$
(69)

with $i,j,...=1,...,4$. In vector representation of $SL(5)$, the generator
$X^{5}$ takes the form

${(X^{5})_{m}}^{n}=2\kappa\,\text{diag}(1,1,1,1,4).$ (70)

Commutation relations between these gauge generators become

$\displaystyle\left[X_{ij},X_{kl}\right]$ $\displaystyle=$
$\displaystyle\left[X^{i},X^{j}\right]\ =\ \left[X^{i},X_{jk}\right]\ =\ 0,$
$\displaystyle\left[X^{5},X^{i}\right]$ $\displaystyle=$
$\displaystyle-3{(X^{5})_{j}}^{i}X^{j},\quad\qquad\left[X^{5},X_{ij}\right]\
=\ {(X^{5})_{ij}}^{kl}X_{kl}\,.$ (71)

These imply the gauge group of the form

$G_{0}=SO(1,1)\ltimes\left(\mathbb{R}^{4}\times\mathbb{R}^{6}_{\boldsymbol{h}}\right)\sim
CSO(1,1,2)\ltimes\mathbb{R}^{6}_{\boldsymbol{h}}$ (72)

in which the non-compact factor $SO(1,1)$ is generated by $X^{5}$.

###  III.5 Gaugings in $\mathbf{10}^{-1}$ representation

In addition to gaugings from $\overline{\mathbf{5}}^{+3}$ representation,
gaugings in $\mathbf{10}^{-1}$ representation also require both $\theta^{Am}$
and $\theta^{A}_{m}$ to be non-vanishing in order to satisfy the linear
constraint. Thus, gaugings in this representation are dyonic and triggered by

$\theta^{Am}\ =\
\frac{1}{3\sqrt{2}}\,\mathbb{T}^{A}_{np}\,\varepsilon^{mnpqr}Z_{qr}\qquad\text{
and }\qquad\theta^{A}_{m}\ =\ \mathbb{T}^{An}Z_{nm}$ (73)

with $Z_{mn}=Z_{[mn]}$. With this embedding tensor, the quadratic constraints
in (15) and (16) reduce to

$Z_{mn}\,Z_{pq}\,\varepsilon^{mnpqr}\ =\ 0\,.$ (74)

As pointed in 6D_Max_Gauging , this condition can be solved by $Z_{mn}$ of the
form

$Z_{mn}=u_{[m}v_{n]}$ (75)

with $u_{m}$ and $v_{n}$ being arbitrary $GL(5)$ vectors.

In this case, the corresponding gauge generators are given by

$\displaystyle X_{\ast}$ $\displaystyle=$
$\displaystyle-\sqrt{2}u_{[m}v_{n]}\boldsymbol{h}^{mn},$ (76) $\displaystyle
X^{m}$ $\displaystyle=$
$\displaystyle-\frac{\sqrt{2}}{3}\varepsilon^{mnpqr}u_{[n}v_{p]}\boldsymbol{s}_{qr},$
(77) $\displaystyle X_{mn}$ $\displaystyle=$
$\displaystyle\frac{1}{3}\left(u_{[m}v_{p]}{\boldsymbol{\tau}^{p}}_{n}-u_{[n}v_{p]}{\boldsymbol{\tau}^{p}}_{m}\right)+\frac{4}{5}u_{[m}v_{n]}\boldsymbol{d}$
(78)

with the commutation relations

$\displaystyle\left[X_{\ast},X^{m}\right]$ $\displaystyle=$
$\displaystyle\left[X^{m},X^{n}\right]\ =\ 0,$
$\displaystyle\left[X_{mn},X_{\ast}\right]$ $\displaystyle=$
$\displaystyle{(X_{mp})_{n}}^{p}X_{\ast},\qquad\left[X^{m},X_{np}\right]\ =\
{2(X_{np})_{q}}^{m}X^{q},$ $\displaystyle\left[X_{mn},X_{pq}\right]$
$\displaystyle=$
$\displaystyle\frac{1}{4}\left[{(X_{mn})_{pq}}^{rs}-{(X_{pq})_{mn}}^{rs}\right]X_{rs}\,.$
(79)

To determine the explicit form of possible gauge groups, we repeat the same
procedure as in the previous cases by considering a particular parametrization
of the two vectors $u_{m}$ and $v_{m}$ in the form of

$u_{m}=(0,0,0,\kappa_{1},\kappa_{2})\quad\text{ and }\quad
v_{m}=(0,0,0,\lambda_{1},\lambda_{2}).$ (80)

The gauge generators in this case become

$\displaystyle X_{\ast}$ $\displaystyle=$
$\displaystyle\sqrt{2}(\kappa_{2}\lambda_{1}-\kappa_{1}\lambda_{2})\boldsymbol{h}^{45},$
(81) $\displaystyle X^{i}$ $\displaystyle=$
$\displaystyle\frac{\sqrt{2}}{3}(\kappa_{2}\lambda_{1}-\kappa_{1}\lambda_{2})\varepsilon^{ijk}\boldsymbol{s}_{jk},$
(82) $\displaystyle X_{ix}$ $\displaystyle=$
$\displaystyle\frac{1}{6}(u_{y}v_{x}-u_{x}v_{y}){\boldsymbol{\tau}^{y}}_{i},$
(83) $\displaystyle X_{45}$ $\displaystyle=$
$\displaystyle\frac{1}{6}(\kappa_{1}\lambda_{2}-\kappa_{2}\lambda_{1})\left({\boldsymbol{\tau}^{4}}_{4}+{\boldsymbol{\tau}^{5}}_{5}+\frac{12}{5}\boldsymbol{d}\right)$
(84)

with $m=(i,x)$ for $i,j,...=1,2,3$ and $x,y,...=4,5$. The commutation
relations read

$\displaystyle\left[X_{\ast},X^{i}\right]$ $\displaystyle=$
$\displaystyle\left[X_{\ast},X_{ix}\right]\ =\ \left[X^{i},X^{j}\right]\ =\
\left[X^{i},X_{jx}\right]\ =\ \left[X_{ix},X_{jy}\right]\ =\ 0,\qquad$
$\displaystyle\left[X_{45},X_{\ast}\right]$ $\displaystyle=$
$\displaystyle{(X_{45})_{i}}^{i}X_{\ast},\qquad\left[X_{45},X^{i}\right]\ =\
-2{(X_{45})_{j}}^{i}X^{j},$ $\displaystyle\left[X_{45},X_{ix}\right]$
$\displaystyle=$ $\displaystyle\frac{2}{5}{(X_{45})_{ix}}^{jy}X_{jy}$ (85)

giving rise to the gauge group of the form

$G_{0}\ =\
SO(1,1)\ltimes(\mathbb{R}^{6}\times\mathbb{R}^{3}_{\boldsymbol{s}}\times\mathbb{R}_{\boldsymbol{h}})\sim
CSO(1,1,3)\ltimes(\mathbb{R}^{3}_{\boldsymbol{s}}\times\mathbb{R}_{\boldsymbol{h}}).$
(86)

The non-compact subgroup $SO(1,1)$ is generated by the gauge generator
$X_{45}$ whose explicit form in $GL(5)$ vector representation is given by

${(X_{45})_{m}}^{n}=\frac{1}{3}(\kappa_{1}\lambda_{2}-\kappa_{2}\lambda_{1})\,\text{diag}(2,2,2,3,3).$
(87)

The three commuting translational groups $\mathbb{R}^{6}$,
$\mathbb{R}^{3}_{\boldsymbol{s}}$, and $\mathbb{R}_{\boldsymbol{h}}$ are
respectively generated by $X_{ix}$, $X^{i}$, and $X_{\ast}$.

###  III.6 Gaugings in $(\overline{\mathbf{5}}+\overline{\mathbf{45}})^{+3}$
representation

We now consider dyonic gaugings arising from combining two components of the
embedding tensor in $\overline{\mathbf{5}}^{+3}$ and
$\overline{\mathbf{45}}^{+3}$ representations. These gaugings are also dyonic
since the embedding tensor in $\overline{\mathbf{5}}^{+3}$ representation
contains both electric and magnetic parts. The linear constraint (9) requires
these components to take the form

$\theta^{Am}\ =-\frac{2\sqrt{2}}{5}\mathbb{T}^{A}_{\ast}J^{m}\qquad\text{ and
}\qquad\theta^{A}_{m}\ =\
\mathbb{T}^{A}_{np}(J^{[n}\delta^{p]}_{m}+W^{np}_{m}).$ (88)

This is just a trivial combination between the embedding tensor from each
representation given in (53) and (66).
With this form of the embedding tensor, the quadratic constraints (15) and
(16) become

$\displaystyle W^{mn}_{p}J^{p}$ $\displaystyle=$ $\displaystyle 0,$ (89)
$\displaystyle
W^{qr}_{m}W^{st}_{[n}\varepsilon_{p]qrst}+\frac{3}{2}W^{qr}_{m}J^{s}\varepsilon_{npqrs}$
$\displaystyle=$ $\displaystyle 0\,.$ (90)

To solve these conditions, we first fix the explicit form of the $GL(5)$
vector $J^{m}=\kappa\delta^{m}_{5}$ with $\kappa\in\mathbb{R}$, leading to a
split of a $GL(5)$ index $m=(i,5)$ with $i,j,...=1,2,3,4$ as in section III.4.
With $J^{m}$ of this form, the condition (89) implies $W^{mn}_{5}=0$, so only
$W^{ij}_{k}$ and $W^{i5}_{j}$ are non-vanishing. Note also that the condition
$W^{mn}_{n}=0$ imposes a traceless condition $W^{ij}_{j}=0$.

With only $W^{ij}_{k}$, $W^{i5}_{j}$, and $J^{5}=\kappa$ non-vanishing, the
condition (90) splits into

$\displaystyle W^{kl}_{i}W^{mn}_{j}\varepsilon_{klmn}$ $\displaystyle=$
$\displaystyle 0,$ (91) $\displaystyle
2(W^{lm}_{i}W^{n5}_{[j}+W^{l5}_{i}W^{mn}_{[j})\varepsilon_{k]lmn}+\frac{3}{2}\kappa
W^{lm}_{i}\varepsilon_{jklm}$ $\displaystyle=$ $\displaystyle 0\,.$ (92)

The first condition takes the same form as the quadratic constraint (54) in
$\overline{\mathbf{45}}^{+3}$ representation and can be similarly solved by
choosing

$W^{ij}_{k}\ =\ 2v^{[i}{w_{k}}^{j]}$ (93)

with a $GL(4)$ vector $v^{i}$ and a traceless $4\times 4$ matrix
${w_{k}}^{j}$.

At this point, we can further use $SL(4)\subset GL(5)$ symmetry to rotate
$v^{i}$ such that $v^{i}=\lambda\delta^{i}_{4}$. We will also split the index
$i=(x,4)$ with $x=1,2,3$. Moreover, the traceless condition $W^{ij}_{j}=0$
requires that ${w_{4}}^{i}=0$. For simplicity, we will also set
${w_{i}}^{4}=0$ since these components do not appear in the resulting
embedding tensor. The remaining components ${w_{x}}^{y}$ can be described by a
traceless $3\times 3$ matrix satisfying ${w_{x}}^{x}=0$. Among various
components of $W^{ij}_{k}=(W^{xy}_{z},\ W^{ij}_{4},\ W^{4y}_{x})$, only the
last components are non-vanishing and given by

$W^{4y}_{x}=\lambda{w_{x}}^{y}\,.$ (94)

With all these, the condition (92) splits into

$\displaystyle\lambda W^{x5}_{4}{w_{z}}^{t}\varepsilon_{xyt}$ $\displaystyle=$
$\displaystyle 0,$ (95) $\displaystyle
2\left[{w_{x}}^{t}W^{5a}_{[y}+W^{5a}_{x}{w_{[y}}^{t}\right]\varepsilon_{z]ta}+\frac{3}{2}\kappa{w_{x}}^{t}\varepsilon_{yzt}$
$\displaystyle=$ $\displaystyle 0.$ (96)

The first condition can be solved by setting $W^{x5}_{4}=0$. The remaining
component $W^{x5}_{y}$ can be written in terms of a $3\times 3$ matrix
${u_{y}}^{x}$ as $W^{5x}_{y}=\tau{u_{y}}^{x}$ with $\tau\in\mathbb{R}$. The
condition (96) then gives rise to

$2\tau\left[({w_{x}}^{t}{u_{y}}^{a}+{w_{y}}^{t}{u_{x}}^{a})\varepsilon_{zta}-({w_{x}}^{t}{u_{z}}^{a}+{w_{z}}^{t}{u_{x}}^{a})\varepsilon_{yta}\right]+3\kappa{w_{x}}^{t}\varepsilon_{yzt}=0\,.$
(97)

With the Schouten identity, ${w_{[x}}^{t}\varepsilon_{yzt]}=0$, this condition
can be solved if and only if $\tau=-\frac{\kappa}{2}$ and
${u_{x}}^{y}=\delta^{y}_{x}$. In conclusion, the quadratic constraint
determines the embedding tensor (88) in terms of $J^{m}$ and $W^{mn}_{p}$
given by

$\displaystyle J^{m}$ $\displaystyle=$
$\displaystyle\kappa\delta^{m}_{5},\qquad W^{ij}_{5}\ =\ W^{ij}_{4}\ =\
W^{xy}_{z}\ =\ W^{x5}_{4}\ =\ 0,$ $\displaystyle W^{4y}_{x}$ $\displaystyle=$
$\displaystyle\lambda{w_{x}}^{y},\ \quad W^{5x}_{y}\ =\
-\frac{\kappa}{2}\delta^{x}_{y},\qquad W^{45}_{x}\ =\ v_{x}\,.$ (98)

We note that $W^{45}_{x}$ written in terms of a three-dimensional vector
$v_{x}$ in the last relation do not appear in the linear and quadratic
constraints. Therefore, this vector is unconstrained.

With all these, non-vanishing gauge generators are given by

$\displaystyle X_{xy}$ $\displaystyle=$
$\displaystyle\sqrt{2}\varepsilon_{xyz}(\lambda{w_{t}}^{z}\boldsymbol{h}^{t5}-v_{t}\boldsymbol{h}^{tz}-2\kappa\boldsymbol{h}^{z4}),$
(99) $\displaystyle X_{x5}$ $\displaystyle=$
$\displaystyle-\frac{\lambda}{\sqrt{2}}{w_{x}}^{y}\varepsilon_{yzt}\boldsymbol{h}^{zt},$
(100) $\displaystyle X^{x}$ $\displaystyle=$ $\displaystyle
4\kappa{\boldsymbol{\tau}^{x}}_{5}-2\lambda{w_{y}}^{x}{\boldsymbol{\tau}^{y}}_{4},$
(101) $\displaystyle X^{4}$ $\displaystyle=$ $\displaystyle
2\lambda{w_{x}}^{y}{\boldsymbol{\tau}^{x}}_{y}+2v_{x}{\boldsymbol{\tau}^{x}}_{5},$
(102) $\displaystyle X^{5}$ $\displaystyle=$ $\displaystyle
4\kappa({\boldsymbol{\tau}^{4}}_{4}+{\boldsymbol{\tau}^{5}}_{5}+\frac{2}{5}\boldsymbol{d})-2v_{x}{\boldsymbol{\tau}^{x}}_{4}$
(103)

with the following commutation relations

$\displaystyle\left[X_{xy},X_{zt}\right]$ $\displaystyle=$
$\displaystyle\left[X_{xy},X_{z5}\right]\ =\ \left[X_{x5},X_{y5}\right]\ =\
\left[X^{x},X^{y}\right]\ =\ \left[X^{x},X_{y5}\right]\ =\ 0,\qquad\quad$
$\displaystyle\left[X^{x},X_{yz}\right]$ $\displaystyle=$ $\displaystyle
16\kappa\delta^{x}_{[y}X_{z]5},\qquad\qquad\quad\left[X^{4},X^{x}\right]\ =\
{(X^{4})_{y}}^{x}X^{y},$ $\displaystyle\left[X^{4},X_{x5}\right]$
$\displaystyle=$ $\displaystyle-{(X^{4})_{x}}^{y}X_{y5},\qquad\quad\ \
\,\left[X^{4},X_{xy}\right]\ =-{(X^{4})_{xy}}^{zt}X_{zt}+8v_{[x}X_{y]5},$
$\displaystyle\left[X^{5},X^{x}\right]$ $\displaystyle=$
$\displaystyle-8\kappa X^{x},\qquad\qquad\qquad\left[X^{5},X_{x5}\right]\ =\
0,$ $\displaystyle\left[X^{5},X_{xy}\right]$ $\displaystyle=$ $\displaystyle
8\kappa X_{xy},\ \ \,\qquad\qquad\qquad\left[X^{4},X^{5}\right]\ =\
4v_{x}X^{x}.$ (104)

We find that the only possible compact subgroup is $SO(2)$ generated by the
gauge generator $X^{4}$ with ${w_{x}}^{y}$ antisymmetric for any value of
$v_{x}$. If the matrix ${w_{x}}^{y}$ is symmetric, the generator $X^{4}$ gives
a non-compact $SO(1,1)$ group. For simplicity, we will set $v_{x}=0$ and
restrict ourselves to the compact $SO(2)$ case since we are mainly interested
in domain wall solutions preserving some symmetry. For definiteness, we choose
the matrix ${w_{x}}^{y}$ to be

${w_{x}}^{y}=\small\begin{pmatrix}0&\sigma&0\\\ -\sigma&0&0\\\
0&0&0\end{pmatrix}$ (105)

with $\sigma\in\mathbb{R}$. Together with $v_{x}=0$, the above gauge
generators reduce to

$\displaystyle X_{12}$
$\displaystyle=-2\sqrt{2}\kappa\boldsymbol{h}^{34},\quad\qquad\qquad
X_{\bar{x}3}$
$\displaystyle=-\sqrt{2}\varepsilon_{\bar{x}\bar{y}}(\lambda{w_{\bar{z}}}^{\bar{y}}\boldsymbol{h}^{\bar{z}5}-2\kappa\boldsymbol{h}^{\bar{y}4}),$
$\displaystyle X_{\bar{x}5}$
$\displaystyle=-\sqrt{2}\lambda{w_{\bar{x}}}^{\bar{y}}\varepsilon_{\bar{y}\bar{z}}\boldsymbol{h}^{\bar{z}3},\quad\qquad
X^{\bar{x}}$ $\displaystyle=\
4\kappa{\boldsymbol{\tau}^{\bar{x}}}_{5}-2\lambda{w_{\bar{y}}}^{\bar{x}}{\boldsymbol{\tau}^{\bar{y}}}_{4},$
$\displaystyle X^{3}$ $\displaystyle=\
4\kappa{\boldsymbol{\tau}^{3}}_{5},\quad\qquad\qquad\qquad\ X^{4}$
$\displaystyle=\
2\lambda\sigma({\boldsymbol{\tau}^{1}}_{2}-{\boldsymbol{\tau}^{2}}_{1}),$
$\displaystyle X^{5}$ $\displaystyle=\
4\kappa({\boldsymbol{\tau}^{4}}_{4}+{\boldsymbol{\tau}^{5}}_{5}+\frac{2}{5}\boldsymbol{d})\qquad\quad$
(106)

with $x=(\bar{x},3)$, $\bar{x}=1,2$.

In vector representation, we can explicitly see that $X^{4}$ and $X^{5}$ are
antisymmetric and symmetric

${(X^{4})_{m}}^{n}=\small\begin{pmatrix}0&4\lambda\sigma&0&0&0\\\
-4\lambda\sigma&0&0&0&0\\\ 0&0&0&0&0\\\ 0&0&0&0&0\\\
0&0&0&0&0\end{pmatrix},\qquad{(X^{5})_{m}}^{n}=\small\begin{pmatrix}0&0&0&0&0\\\
0&0&0&0&0\\\ 0&0&0&0&0\\\ 0&0&0&8\kappa&0\\\ 0&0&0&0&8\kappa\end{pmatrix}$
(107)

generating $SO(2)$ and $SO(1,1)$ subgroups, respectively. The gauge generators
satisfy the following commutation relations

$\displaystyle\left[X_{xy},X_{zt}\right]$ $\displaystyle=$
$\displaystyle\left[X_{xy},X_{\bar{z}5}\right]\ =\
\left[X_{\bar{x}5},X_{\bar{y}5}\right]\ =\ \left[X^{x},X^{y}\right]\ =\
\left[X^{x},X_{\bar{y}5}\right]\ =\ 0,$
$\displaystyle\left[X^{\bar{x}},X_{12}\right]$ $\displaystyle=$ $\displaystyle
16\kappa\delta^{\bar{x}}_{[1}X_{2]5},\qquad\qquad\quad\left[X^{\bar{x}},X_{\bar{y}3}\right]\
=\ 0,$ $\displaystyle\left[X^{3},X_{\bar{x}3}\right]$ $\displaystyle=$
$\displaystyle-8\kappa X_{\bar{x}5},\qquad\qquad\quad\ \
\left[X^{3},X_{12}\right]\ =\ 0,$
$\displaystyle\left[X^{4},X^{\bar{x}}\right]$ $\displaystyle=$
$\displaystyle{(X^{4})_{\bar{y}}}^{\bar{x}}X^{\bar{y}},\qquad\qquad\quad\,\left[X^{4},X^{3}\right]\
=\ 0,$ $\displaystyle\left[X^{4},X_{\bar{x}3}\right]$ $\displaystyle=$
$\displaystyle-{(X^{4})_{\bar{x}3}}^{\bar{y}3}X_{\bar{y}3},\qquad\quad\,\left[X^{4},X_{12}\right]\
=\ 0,$ $\displaystyle\left[X^{4},X_{\bar{x}5}\right]$ $\displaystyle=$
$\displaystyle-{(X^{4})_{\bar{x}}}^{\bar{y}}X_{\bar{y}5},$
$\displaystyle\left[X^{5},X^{x}\right]$ $\displaystyle=$
$\displaystyle-8\kappa X^{x},\qquad\qquad\quad\ \
\,\left[X^{5},X_{\bar{x}5}\right]\ =\ 0,$
$\displaystyle\left[X^{5},X_{xy}\right]$ $\displaystyle=$ $\displaystyle
8\kappa X_{xy},\ \ \qquad\qquad\qquad\left[X^{4},X^{5}\right]\ =\ 0\,.$ (108)

From these relations, we see that $(X^{4},X_{\bar{x}5})$ and
$(X^{3},X^{5},X_{12})$ form two commuting non-semisimple groups $ISO(2)$ and
$ISO(1,1)$, respectively. The remaining generators
$(X^{\bar{x}},X_{3\bar{x}})$ generate a four-dimensional translational group
transforming non-trivially under $ISO(2)\times ISO(1,1)$. The gauge group is
then given by

$G_{0}=\left(ISO(2)\times ISO(1,1)\right)\ltimes\mathbb{R}^{4}\,.$ (109)

For a simpler case of $\lambda=0$, the non-vanishing components $W_{y}^{5x}$
still generate a non-trivial subgroup. In this case, we find
$X^{4}=X_{\bar{x}5}=0$ giving rise to the following gauge group

$SO(1,1)\ltimes\left(\mathbb{R}^{3}\times\mathbb{R}^{3}_{\boldsymbol{h}}\right).$
(110)

The three factors $SO(1,1)$, $\mathbb{R}^{3}$, and
$\mathbb{R}^{3}_{\boldsymbol{h}}$ are respectively generated by $X^{5}$,
$X_{xy}$, and $X^{x}$. We also note that another possibility of setting
$\kappa=0$ is not possible since this choice leads to vanishing $J^{m}$.

###  III.7 Gaugings in $(\mathbf{10}+\mathbf{15})^{-1}$ representation

For gaugings with the embedding tensor in $(\mathbf{10}+\mathbf{15})^{-1}$
representation, there are both electric and magnetic components $\theta^{Am}$
and $\theta^{A}_{m}$ given by

$\theta^{Am}\ =\
\frac{1}{3\sqrt{2}}\,\mathbb{T}^{A}_{np}\,\varepsilon^{mnpqr}Z_{qr}\qquad\text{
and }\qquad\theta^{A}_{m}\ =\ \mathbb{T}^{An}(Y_{nm}+Z_{nm}).$ (111)

We recall that $Y_{mn}=Y_{(mn)}$ and $Z_{mn}=Z_{[mn]}$ are symmetric and
antisymmetric tensors corresponding to $\mathbf{15}^{-1}$ and
$\mathbf{10}^{-1}$ representations, respectively. In terms of $Y_{mn}$ and
$Z_{mn}$, the quadratic constraints (15) and (16) reduce to

$\varepsilon^{npqrs}Y_{mq}Z_{rs}=0\qquad\text{and}\qquad\varepsilon^{mnpqr}Z_{np}Z_{qr}=0\,.$
(112)

We can use the $SL(5)\subset GL(5)$ symmetry to bring $Y_{mn}$ to a diagonal
form

$Y_{mn}\ =\
\text{diag}(\underbrace{1,..,1}_{p},\underbrace{-1,..,-1}_{q},\underbrace{0,..,0}_{r})$
(113)

where $p+q+r=5$. We will split the $GL(5)$ index $m=(i,x)$ and $Y_{mn}$ into
$Y_{ij}=\text{diag}(1,..,1,-1,..,-1)$ with $Y_{xy}=0$. Solving the first
condition in (112) using (113), we find only two possible solutions with both
$Y_{mn}$ and $Z_{mn}$ non-vanishing for ranks of $Y_{mn}$ equal to $1$ and
$2$.

####  III.7.1 Rank$Y=2$

With $Y_{ij}=\text{diag}(1,\pm 1)$ for $i,j,...=1,2$ and $x,y,...=3,4,5$, only
$Z_{12}=\kappa$ with $\kappa\in\mathbb{R}$ is allowed to be non-vanishing.
This automatically solves the other condition in (112) and gives rise to the
following gauge generators

$\displaystyle X_{\ast}$ $\displaystyle=$
$\displaystyle-\sqrt{2}\kappa\boldsymbol{h}^{12},$ (114) $\displaystyle X^{x}$
$\displaystyle=$
$\displaystyle-\frac{\sqrt{2}}{3}\kappa\varepsilon^{xyz}\boldsymbol{s}_{yz},$
(115) $\displaystyle X_{ix}$ $\displaystyle=$ $\displaystyle
2Y_{j[i}{\boldsymbol{\tau}^{j}}_{x]}-\frac{2}{3}\kappa\varepsilon_{j[i}{\boldsymbol{\tau}^{j}}_{x]},$
(116) $\displaystyle X_{12}$ $\displaystyle=$ $\displaystyle
2Y_{i[1}{\boldsymbol{\tau}^{i}}_{2]}+\frac{\kappa}{3}({\boldsymbol{\tau}^{1}}_{1}+{\boldsymbol{\tau}^{2}}_{2})+\frac{4}{5}\kappa\boldsymbol{d}\,.$
(117)

Commutation relations between these generators are given by

$\displaystyle\left[X_{\ast},X^{x}\right]$ $\displaystyle=$
$\displaystyle\left[X_{\ast},X_{ix}\right]\ =\ \left[X^{x},X^{y}\right]\ =\
\left[X^{x},X_{iy}\right]\ =\ \left[X_{ix},X_{jy}\right]\ =\ 0,$
$\displaystyle\left[X_{12},X_{\ast}\right]$ $\displaystyle=$
$\displaystyle\frac{1}{2}{(X_{12})_{m}}^{m}X_{\ast},\qquad\left[X_{12},X^{x}\right]\
=\ -2{(X_{12})_{y}}^{x}X^{y},$ $\displaystyle\left[X_{12},X_{ix}\right]$
$\displaystyle=$
$\displaystyle-{(X_{12})_{i}}^{k}X_{jx}-2{(X_{12})_{x}}^{y}X_{iy}\,.$ (118)

The generators $X_{\ast}$, $X^{x}$, and $X_{ix}$ generate three translational
groups $\mathbb{R}_{\boldsymbol{h}}$, $\mathbb{R}^{3}_{\boldsymbol{s}}$, and
$\mathbb{R}^{6}$ commuting with each other. For $Y_{ij}=\textrm{diag}(1,-1)$,
the generator $X_{12}$ can only generate a non-compact $SO(1,1)$ group giving
rise to the following gauge group

$G_{0}=SO(1,1)\ltimes(\mathbb{R}^{6}\times\mathbb{R}^{3}_{\boldsymbol{s}}\times\mathbb{R}_{\boldsymbol{h}})\sim
CSO(1,1,3)\ltimes(\mathbb{R}^{3}_{\boldsymbol{s}}\times\mathbb{R}_{\boldsymbol{h}}).$
(119)

For $Y_{ij}=\delta_{ij}$, $X_{12}$ can become compact $SO(2)$, non-compact
$SO(1,1)$, or nilpotent generators depending on the values of $\kappa$. For
particular values of $\kappa=\pm\sqrt{\frac{3}{5}}$, $X_{12}$ is nilpotent and
will be denoted by $\mathcal{T}$ resulting in
$\mathcal{T}\ltimes(\mathbb{R}^{6}\times\mathbb{R}^{3}_{\boldsymbol{s}}\times\mathbb{R}_{\boldsymbol{h}})$
gauge group. For $-\sqrt{\frac{3}{5}}<\kappa<\sqrt{\frac{3}{5}}$ and
$\kappa<-\sqrt{\frac{3}{5}}$ or $\kappa>\sqrt{\frac{3}{5}}$, $X_{12}$ is
respectively compact and non-compact. Accordingly, the corresponding gauge
group is given by

$G_{0}=SO(2)\ltimes(\mathbb{R}^{6}\times\mathbb{R}^{3}_{\boldsymbol{s}}\times\mathbb{R}_{\boldsymbol{h}})\sim
CSO(2,0,3)\ltimes(\mathbb{R}^{3}_{\boldsymbol{s}}\times\mathbb{R}_{\boldsymbol{h}})$
(120)

or the gauge group given in (119).

####  III.7.2 Rank$Y=1$

In this case, we will choose only $Y_{11}$ non-vanishing with
$Y_{11}=\kappa=\pm 1$. All components $Z_{xy}$ with $x,y=2,3,4,5$ need to be
zero in order to solve the quadratic constraint (112). Therefore, only the
remaining four components $Z_{1x}=z_{x}$ give rise to the following gauge
generators

$\displaystyle X_{\ast}$ $\displaystyle=$
$\displaystyle-\sqrt{2}z_{x}\boldsymbol{h}^{1x},\qquad X^{x}\ =\
\frac{\sqrt{2}}{3}\varepsilon^{xyzt}z_{y}\boldsymbol{s}_{zt},\qquad X_{xy}\
=-\frac{2}{3}z_{[x}{\boldsymbol{\tau}^{1}}_{y]},$ $\displaystyle X_{1x}$
$\displaystyle=$
$\displaystyle\kappa{\boldsymbol{\tau}^{1}}_{x}+\frac{1}{3}(z_{y}{\boldsymbol{\tau}^{y}}_{x}+z_{x}{\boldsymbol{\tau}^{1}}_{1})+\frac{4}{5}z_{x}\boldsymbol{d}$
(121)

with the commutation relations

$\displaystyle\left[X_{\ast},X^{x}\right]$ $\displaystyle=$
$\displaystyle\left[X_{\ast},X_{xy}\right]\ =\ \left[X^{x},X^{y}\right]\ =\
\left[X^{x},X_{yz}\right]\ =\ \left[X_{xy},X_{zt}\right]\ =\ 0,$
$\displaystyle\left[X_{1x},X_{\ast}\right]$ $\displaystyle=$
$\displaystyle\frac{1}{2}{(X_{1x})_{m}}^{m}X_{\ast},\qquad\left[X_{1x},X^{y}\right]\
=\ -2{(X_{1x})_{z}}^{y}X^{z},$ $\displaystyle\left[X_{1x},X_{yz}\right]$
$\displaystyle=$ $\displaystyle\frac{1}{5}{(X_{1x})_{yz}}^{ta}X_{ta},\
\quad\left[X_{1x},X_{1y}\right]\ =\ -\frac{4}{3}z_{[x}X_{y]1}-2\kappa
X_{xy}\,.$ (122)

To proceed further, we will choose $z_{2}=\zeta$ and $z_{3}=z_{4}=z_{5}=0$.
The above gauge generators simplify to

$\displaystyle X_{\ast}$ $\displaystyle=$
$\displaystyle-\sqrt{2}\zeta\boldsymbol{h}^{12},\qquad X^{\bar{x}}\
=-\frac{\sqrt{2}}{3}\zeta\varepsilon^{\bar{x}\bar{y}\bar{z}}\boldsymbol{s}_{\bar{y}\bar{z}},$
$\displaystyle X_{2\bar{x}}$ $\displaystyle=$
$\displaystyle-\frac{1}{3}\zeta{\boldsymbol{\tau}^{1}}_{\bar{x}},\ \qquad
X_{1\bar{x}}\ =\
\kappa{\boldsymbol{\tau}^{1}}_{\bar{x}}+\frac{1}{3}\zeta{\boldsymbol{\tau}^{2}}_{\bar{x}},$
$\displaystyle X_{12}$ $\displaystyle=$
$\displaystyle\kappa{\boldsymbol{\tau}^{1}}_{2}+\frac{1}{3}\zeta({\boldsymbol{\tau}^{2}}_{2}+{\boldsymbol{\tau}^{1}}_{1})+\frac{4}{5}\zeta\boldsymbol{d}$
(123)

with $\bar{x},\bar{y},...=3,4,5$. The corresponding commutation relations read

$\displaystyle\left[X_{\ast},X^{\bar{x}}\right]$ $\displaystyle=$
$\displaystyle\left[X_{\ast},X_{\bar{i}\bar{x}}\right]\ =\
\left[X^{\bar{x}},X^{\bar{y}}\right]\ =\
\left[X^{\bar{x}},X_{\bar{i}\bar{y}}\right]\ =\
\left[X_{\bar{i}\bar{x}},X_{\bar{j}\bar{y}}\right]\ =\ 0,$
$\displaystyle\left[X_{12},X_{\ast}\right]$ $\displaystyle=$
$\displaystyle\frac{1}{2}{(X_{12})_{m}}^{m}X_{\ast},\qquad\left[X_{12},X^{\bar{x}}\right]\
=\ -2{(X_{12})_{\bar{y}}}^{\bar{x}}X^{\bar{y}},$
$\displaystyle\left[X_{12},X_{2\bar{x}}\right]$ $\displaystyle=$
$\displaystyle\frac{1}{5}{(X_{12})_{2\bar{x}}}^{2\bar{y}}X_{2\bar{y}},\qquad\left[X_{12},X_{1\bar{x}}\right]\
=\ \frac{2}{3}\zeta X_{1\bar{x}}-2\kappa X_{2\bar{x}}$ (124)

with $\bar{i},\bar{j},...=1,2$.

We also note that the generator $X_{12}$ does not commute with other
generators. This generator generates an $SO(1,1)$ group while the remaining
generators form three commuting translational groups. Accordingly, the
resulting gauge group is given by

$G_{0}=SO(1,1)\ltimes(\mathbb{R}^{6}\times\mathbb{R}^{3}_{\boldsymbol{s}}\times\mathbb{R}_{\boldsymbol{h}})$
(125)

in which the translational groups $\mathbb{R}^{6}$,
$\mathbb{R}^{3}_{\boldsymbol{s}}$, and $\mathbb{R}_{\boldsymbol{h}}$ are
respectively generated by $X_{\bar{i}\bar{x}}$, $X^{\bar{x}}$, and $X_{\ast}$.

###  III.8 Gaugings in $(\mathbf{10}+\overline{\mathbf{40}})^{-1}$
representation

We now move to the next case with the embedding tensor in $\mathbf{10}^{-1}$
and $\overline{\mathbf{40}}^{-1}$ representations. These two components are
labelled respectively by $Z_{mn}=Z_{[mn]}$ and $U^{mn,p}=U^{[mn],p}$ with
$U^{[mn,p]}=0$. The embedding tensor solving the linear constraint (9) takes
the form

$\theta^{Am}\ =\
\mathbb{T}^{A}_{np}\left(U^{np,m}+\frac{1}{3\sqrt{2}}\varepsilon^{mnpqr}Z_{qr}\right)\qquad\text{
and }\qquad\theta^{A}_{m}\ =\ \mathbb{T}^{An}Z_{nm}\,.$ (126)

The quadratic constraint (15) imposes the following conditions

$\displaystyle U^{mn,p}Z_{pq}$ $\displaystyle=$ $\displaystyle 0,$ (127)
$\displaystyle\varepsilon^{mnpqr}Z_{np}Z_{qr}$ $\displaystyle=$ $\displaystyle
0$ (128)

while the other condition in (16) gives

$3\varepsilon_{mqrst}U^{qr,n}U^{st,p}+2\sqrt{2}Z_{mq}U^{qn,p}=0\,.$ (129)

The condition (128) is similar to that given in the case of $\mathbf{10}^{-1}$
representation. Therefore, we will use the same ansatz for $Z_{mn}$ as given
in (75). Moreover, as in section III.5, we will further fix each $GL(5)$
vector in the ansatz $Z_{mn}=u_{[m}v_{n]}$ in order to see the corresponding
gauge group explicitly. However, instead of using
$u_{m}=(\kappa_{1},\kappa_{2},0,0,0)$ and
$v_{m}=(\lambda_{1},\lambda_{2},0,0,0)$, we equivalently set $Z_{12}=\kappa$
with other components vanishing. The condition (128) is then automatically
satisfied while the condition (127) can be solved if and only if

$U^{12,m}=U^{ix,j}=U^{xy,i}=0\,.$ (130)

We have split the $GL(5)$ index as $m=(i,x)$ with $i,j,...=1,2$ and
$x,y,...=3,4,5$.

At this point, only $U^{ix,y}$ and $U^{xy,z}$ remain in terms of which the
last condition in (129) leads to

$\displaystyle\varepsilon_{ij}\varepsilon_{xta}U^{it,y}U^{ja,z}$
$\displaystyle=$ $\displaystyle 0,$ (131) $\displaystyle
3\varepsilon_{ij}\varepsilon_{zta}(U^{jz,x}U^{ta,y}+U^{jz,y}U^{ta,x})+\sqrt{2}Z_{ij}U^{jx,y}$
$\displaystyle=$ $\displaystyle 0\,.$ (132)

These conditions can be readily solved by setting $U^{ix,y}=0$. This is very
similar to gaugings from $(\mathbf{15}+\overline{\mathbf{40}})^{-1}$
representation with rank$Y=2$ studied in 6D_DW_I . Therefore, in order to
solve the quadratic constraints for gaugings in
$(\mathbf{10}+\overline{\mathbf{40}})^{-1}$ representation, we are left with
only the following non-vanishing components

$Z_{ij}=\kappa\varepsilon_{ij}\qquad\text{and}\qquad
U^{xy,z}=\frac{1}{2\sqrt{2}}\varepsilon^{xyt}{u_{t}}^{z}\,.$ (133)

We have parametrized the components $U^{xy,z}$ in terms of a traceless
$3\times 3$ matrix ${u_{x}}^{y}$ with ${u_{x}}^{x}=0$.

The corresponding gauge generators are given by

$\displaystyle X_{\ast}$ $\displaystyle=$
$\displaystyle-\sqrt{2}\kappa\boldsymbol{h}^{12},$ (134) $\displaystyle X^{x}$
$\displaystyle=$
$\displaystyle-\frac{1}{\sqrt{2}}\varepsilon^{xyz}{u_{z}}^{t}\boldsymbol{s}_{yt}-\frac{2\sqrt{2}}{3}\kappa\varepsilon^{xyz}\boldsymbol{s}_{yz},$
(135) $\displaystyle X_{ix}$ $\displaystyle=$
$\displaystyle-\frac{1}{2}\varepsilon_{ij}{u_{x}}^{y}{\boldsymbol{\tau}^{j}}_{y}+\frac{1}{3}\kappa\varepsilon_{ij}{\boldsymbol{\tau}^{j}}_{x},$
(136) $\displaystyle X_{12}$ $\displaystyle=$
$\displaystyle\frac{1}{2}{u_{x}}^{y}{\boldsymbol{\tau}^{x}}_{y}+\frac{\kappa}{3}({\boldsymbol{\tau}^{2}}_{2}+{\boldsymbol{\tau}^{1}}_{1})+\frac{4}{5}\kappa\boldsymbol{d}$
(137)

with the following commutation relations

$\displaystyle\left[X_{\ast},X^{x}\right]$ $\displaystyle=$
$\displaystyle\left[X_{\ast},X_{ix}\right]\ =\ \left[X^{x},X^{y}\right]\ =\
\left[X^{x},X_{iy}\right]\ =\ \left[X_{ix},X_{jy}\right]\ =\ 0,$
$\displaystyle\left[X_{12},X_{\ast}\right]$ $\displaystyle=$
$\displaystyle\frac{1}{2}{(X_{12})_{m}}^{m}X_{\ast},\quad\qquad\left[X_{12},X^{x}\right]={(X_{12})_{y}}^{x}X^{y}-4\kappa
X^{x},$ $\displaystyle\left[X_{12},X_{ix}\right]$ $\displaystyle=$
$\displaystyle-2{(X_{12})_{ix}}^{jy}X_{jy}+4\kappa X_{ix}\,.$ (138)

The gauge generator $X_{12}$ given in (137) generates either $SO(1,1)$ or
$SO(2)$ group for symmetric and antisymmetric ${u_{x}}^{y}$, respectively. The
corresponding gauge group is given by

$G_{0}=SO(1,1)\ltimes(\mathbb{R}^{6}\times\mathbb{R}^{3}_{\boldsymbol{s}}\times\mathbb{R}_{\boldsymbol{h}})\sim
CSO(1,1,3)\ltimes(\mathbb{R}^{3}_{\boldsymbol{s}}\times\mathbb{R}_{\boldsymbol{h}})$
(139)

or

$G_{0}=SO(2)\ltimes(\mathbb{R}^{6}\times\mathbb{R}^{3}_{\boldsymbol{s}}\times\mathbb{R}_{\boldsymbol{h}})\sim
CSO(2,0,3)\ltimes(\mathbb{R}^{3}_{\boldsymbol{s}}\times\mathbb{R}_{\boldsymbol{h}})$
(140)

in which $X_{ix}$, $X^{x}$, and $X_{\ast}$ respectively generate the
translational groups $\mathbb{R}^{6}$, $\mathbb{R}^{3}_{\boldsymbol{s}}$, and
$\mathbb{R}_{\boldsymbol{h}}$.

###  III.9 Gaugings in $(\mathbf{10}+\mathbf{15}+\overline{\mathbf{40}})^{-1}$
representation

As a final example, we consider gaugings from
$(\mathbf{10}+\mathbf{15}+\overline{\mathbf{40}})^{-1}$ representation in
which the embedding tensor takes the form of

$\theta^{Am}\ =\
\mathbb{T}^{A}_{np}\left(U^{np,m}+\frac{1}{3\sqrt{2}}\varepsilon^{mnpqr}Z_{qr}\right)\qquad\text{
and }\qquad\theta^{A}_{m}\ =\ \mathbb{T}^{An}(Z_{nm}+Y_{mn}).$ (141)

The quadratic constraints in (15) and (16) give

$\displaystyle
U^{mn,q}(3Z_{pq}+Y_{pq})+\frac{2\sqrt{2}}{3}\varepsilon^{mnqrs}Y_{pq}Z_{rs}$
$\displaystyle=$ $\displaystyle 0,$ (142) $\displaystyle
U^{mn,q}(Z_{pq}+3Y_{pq})-\frac{\sqrt{2}}{3}\delta^{[m}_{p}\varepsilon^{n]qrst}Z_{qr}Z_{st}$
$\displaystyle=$ $\displaystyle 0$ (143)

together with

$\displaystyle
U^{nq,m}Y_{pq}+\varepsilon^{mnqrs}Y_{pq}Z_{rs}-\frac{1}{2\sqrt{2}}U^{qr,m}U^{st,n}\varepsilon_{pqrst}$
$\displaystyle+\frac{1}{3}\left[(2U^{mq,n}-U^{nq,m})Z_{pq}+4\delta^{(m}_{p}U^{n)qr}Z_{qr}\right]$
$\displaystyle+\frac{1}{36\sqrt{2}}(\delta^{m}_{p}\varepsilon^{nqrst}-5\delta^{n}_{p}\varepsilon^{mqrst})Z_{qr}Z_{st}$
$\displaystyle=$ $\displaystyle 0\,.$ (144)

To solve these conditions, we repeat the same procedure as in the previous
case by setting all $Z_{mn}$ components to zero except $Z_{12}=\kappa$
together with $Y_{mn}=\text{diag}(1,-1,0,0,0)$. With this choice together with
the index splitting $m=(i,x)$ for $i,j,...=1,2$ and $x,y,...=3,4,5$, we can
solve all the above conditions with all $U^{mn,p}$ components vanishing except

$U^{xy,z}=\frac{1}{2\sqrt{2}}\varepsilon^{xyt}{u_{t}}^{z}\,.$ (145)

This is the same ansatz (133) as used in the previous case.

These lead to the gauge generators

$\displaystyle X_{\ast}$ $\displaystyle=$
$\displaystyle-\sqrt{2}\kappa\boldsymbol{h}^{12},$ (146) $\displaystyle X^{x}$
$\displaystyle=$
$\displaystyle-\frac{1}{\sqrt{2}}\varepsilon^{xyz}{u_{z}}^{t}\boldsymbol{s}_{yt}-\frac{2\sqrt{2}}{3}\kappa\varepsilon^{xyz}\boldsymbol{s}_{yz},$
(147) $\displaystyle X_{ix}$ $\displaystyle=$ $\displaystyle
Y_{ji}{\boldsymbol{\tau}^{j}}_{x}-\frac{1}{2}\varepsilon_{ij}{u_{x}}^{y}{\boldsymbol{\tau}^{j}}_{y}+\frac{1}{3}\kappa\varepsilon_{ij}{\boldsymbol{\tau}^{j}}_{x},$
(148) $\displaystyle X_{12}$ $\displaystyle=$ $\displaystyle
2Y_{i[1}{\boldsymbol{\tau}^{i}}_{2]}+\frac{1}{2}{u_{x}}^{y}{\boldsymbol{\tau}^{x}}_{y}+\frac{\kappa}{3}({\boldsymbol{\tau}^{2}}_{2}+{\boldsymbol{\tau}^{1}}_{1})+\frac{4}{5}\kappa\boldsymbol{d}$
(149)

with the same commutation relations as given in (138). The corresponding gauge
group is again given by (139) or (140) depending on the values of $\kappa$ and
the form of ${u_{x}}^{y}$ as in the previous case. It could be useful to look
for more complicated solutions leading to new gauge groups.

##  IV Supersymmetric domain wall solutions

In this section, we will find supersymmetric domain wall solutions from six-
dimensional gauged supergravities constructed from the embedding tensors given
in the previous section. We begin with a general procedure of finding
supersymmetric domain walls. The analysis has already been performed in
6D_DW_I , so we will only review relevant formulae and refer to 6D_DW_I  for
more detail. The metric ansatz for domain wall solutions take the general form
of

$ds_{6}^{2}=e^{2A(r)}\eta_{\bar{\mu}\bar{\nu}}dx^{\bar{\mu}}dx^{\bar{\nu}}+dr^{2}$
(150)

with $\bar{\mu},\bar{\nu}=0,1,\ldots,4$.
We then consider an explicit parametrization of the scalar fields in
$SO(5,5)/SO(5)\times SO(5)$ coset space. These scalars correspond to $25$ non-
compact generators of $SO(5,5)$ that decompose under $GL(5)$ as

$25\ \rightarrow\
\underbrace{\mathbf{1}+\mathbf{14}}_{\hat{\boldsymbol{t}}^{+}_{a\dot{b}}}\,+\underbrace{\mathbf{10}}_{\boldsymbol{s}_{mn}}\,.$
(151)

The generators $\hat{\boldsymbol{t}}^{+}_{a\dot{b}}$ are non-compact
generators of $GL(5)$ defined as

$\hat{\boldsymbol{t}}^{+}_{a\dot{b}}=\frac{1}{2}\left({\mathbb{M}_{a}}^{M}{\mathbb{M}_{\dot{b}}}^{N}+{\mathbb{M}_{\dot{b}}}^{M}{\mathbb{M}_{a}}^{N}\right)\boldsymbol{t}_{MN}$
(152)

where
${\mathbb{M}_{\underline{A}}}^{M}=({\mathbb{M}_{a}}^{M},{\mathbb{M}_{\dot{a}}}^{M})$
is a transformation matrix whose explicit form is given by

$\mathbb{M}\ =\
\frac{1}{\sqrt{2}}\begin{pmatrix}\mathds{1}_{5}&\mathds{1}_{5}\\\
\mathds{1}_{5}&-\mathds{1}_{5}\end{pmatrix}.$ (153)

This matrix is used to relate the $SO(5,5)$ metric in non-diagonal
($\eta_{MN}$) and diagonal ($\eta_{\underline{A}\underline{B}}$) forms as

$\eta_{\underline{A}\underline{B}}={\mathbb{M}_{\underline{A}}}^{M}{\mathbb{M}_{\underline{B}}}^{N}\eta_{MN}$
(154)

with
$\eta_{\underline{A}\underline{B}}=\textrm{diag}(1,1,1,1,1,-1,-1,-1,-1,-1)$.
In accord with (151), we will denote all $25$ scalar fields as
$\Phi^{I}=\\{\varphi,\phi_{1},\ldots,\phi_{14},\varsigma_{1},\ldots,\varsigma_{10}\\}$
with $I=1,\ldots,25$. The scalar $\varphi$ is the dilaton corresponding to the
$SO(1,1)$ generator

$\boldsymbol{d}=\hat{\boldsymbol{t}}^{+}_{1\dot{1}}+\hat{\boldsymbol{t}}^{+}_{2\dot{2}}+\hat{\boldsymbol{t}}^{+}_{3\dot{3}}+\hat{\boldsymbol{t}}^{+}_{4\dot{4}}+\hat{\boldsymbol{t}}^{+}_{5\dot{5}}\,.$
(155)

This $SO(1,1)\sim\mathbb{R}^{+}$ symmetry is identified with the $SO(1,1)$
factor of $GL(5)\sim SL(5)\times SO(1,1)$. The other fourteen linear
combinations of generators $\hat{\boldsymbol{t}}^{+}_{a\dot{b}}$ correspond to
the fourteen scalar fields $\\{\phi_{1},...,\phi_{14}\\}$ in the $SL(5)/SO(5)$
coset. The remaining ten scalars $\\{\varsigma_{1},...,\varsigma_{10}\\}$
correspond to the shift generators $\boldsymbol{s}_{mn}$ and will be called
shift scalars. By this decomposition of the scalar fields, we can rewrite the
kinetic terms of the scalar fields in (17) and obtain the following bosonic
Lagrangian

$e^{-1}\mathcal{L}=\frac{1}{4}R-G_{IJ}\partial_{\mu}\Phi^{I}\partial^{\mu}\Phi^{J}-\mathbf{V}$
(156)

with $G_{IJ}$ being a symmetric scalar-dependent matrix.
In order to find supersymmetric domain wall solutions, we consider first-order
Bogomol’nyi-Prasad-Sommerfield (BPS) equations derived from the supersymmetry
transformations of fermionic fields. We begin with the variations of the
gravitini from (18) and (19) which give

$\displaystyle\delta\psi_{+\bar{\mu}\alpha}$ $\displaystyle:$
$\displaystyle\quad
A^{\prime}\hat{\gamma}_{r}\epsilon_{+\alpha}+\frac{g}{2}\Omega_{\alpha\beta}T^{\beta\dot{\alpha}}\epsilon_{-\dot{\alpha}}=0,$
(157) $\displaystyle\delta\psi_{-\bar{\mu}\dot{\alpha}}$ $\displaystyle:$
$\displaystyle\quad
A^{\prime}\hat{\gamma}_{r}\epsilon_{-\dot{\alpha}}-\frac{g}{2}\Omega_{\dot{\alpha}\dot{\beta}}T^{\alpha\dot{\beta}}\epsilon_{+\alpha}=0\,.$
(158)

Throughout the paper, we use the notation ′ to denote an $r$-derivative.
Multiply equation (157) by $A^{\prime}\hat{\gamma}_{r}$ and use equation (158)
or vice-versa, we find the following consistency conditions

$\displaystyle{A^{\prime}}^{2}{\delta_{\alpha}}^{\beta}$ $\displaystyle=$
$\displaystyle-\frac{g^{2}}{4}\Omega_{\alpha\gamma}T^{\gamma\dot{\alpha}}\Omega_{\dot{\alpha}\dot{\beta}}T^{\beta\dot{\beta}}=\mathcal{W}^{2}{\delta_{\alpha}}^{\beta},$
(159) $\displaystyle{A^{\prime}}^{2}{\delta_{\dot{\alpha}}}^{\dot{\beta}}$
$\displaystyle=$
$\displaystyle-\frac{g^{2}}{4}\Omega_{\dot{\alpha}\dot{\gamma}}T^{\alpha\dot{\gamma}}\Omega_{\alpha\beta}T^{\beta\dot{\beta}}=\mathcal{W}^{2}{\delta_{\dot{\alpha}}}^{\dot{\beta}}$
(160)

in which we have introduced the “superpotential” $\mathcal{W}$. We then obtain
the BPS equations for the warped factor

$A^{\prime}=\pm\mathcal{W}\,.$ (161)

With this result, equations (157) and (158) lead to the following projectors
on the Killing spinors

$\hat{\gamma}_{r}\epsilon_{+\alpha}=-\frac{g}{2}\Omega_{\alpha\beta}\frac{T^{\beta\dot{\beta}}}{A^{\prime}}\epsilon_{-\dot{\beta}},\qquad\hat{\gamma}_{r}\epsilon_{-\dot{\alpha}}=\frac{g}{2}\Omega_{\dot{\alpha}\dot{\beta}}\frac{T^{\alpha\dot{\beta}}}{A^{\prime}}\epsilon_{+\alpha}.$
(162)

It should be noted that these projectors are not independent. The conditions
$\delta\psi_{+r\alpha}=0$ and $\delta\psi_{-r\dot{\alpha}}=0$ will determine
the Killing spinors as functions of the radial coordinate $r$ as usual. We
will give the corresponding expressions in explicit solutions obtained in
subsequent analyses.
Using the $\hat{\gamma}_{r}$ projectors in $\delta\chi_{+a\dot{\alpha}}=0$ and
$\delta\chi_{-\dot{a}\alpha}=0$ equations, we eventually obtain the BPS
equations for scalar fields of the form

${\Phi^{I}}^{\prime}=\mp 2G^{IJ}\frac{\partial\mathcal{W}}{\partial\Phi^{J}}$
(163)

in which $G^{IJ}$ is the inverse of the scalar metric $G_{IJ}$. In addition,
the scalar potential can also be written in terms of $\mathcal{W}$ as

$\mathbf{V}=2G^{IJ}\frac{\partial\mathcal{W}}{\partial\Phi^{I}}\frac{\partial\mathcal{W}}{\partial\Phi^{J}}-5\mathcal{W}^{2}\,.$
(164)

We are now in a position to find explicit domain wall solutions. As mentioned
before, the complexity of working with all $25$ scalars prohibits any
traceable analysis. We will consider only solutions preserving a non-trivial
residual symmetry. The solutions invariant under this symmetry are
characterized by a smaller set of singlet scalars. Among the gauge groups
classified in the previous section, only five gauge groups,
$SO(2)\ltimes\mathbb{R}^{8}$, $SO(2)\ltimes\mathbb{R}^{6}$,
$CSO(2,0,2)\ltimes\mathbb{R}^{4}$, $CSO(2,0,2)\ltimes\mathbb{R}^{2}$, and
$CSO(2,0,1)\ltimes\mathbb{R}^{4}$, contain a compact $SO(2)$ subgroup that can
be used to reduce the number of scalar fields and result in consistent sets of
BPS equations. Accordingly, in the following, we will consider only domain
wall solutions for these gauge groups. We should also remark here that for
$\left(ISO(2)\times ISO(1,1)\right)\ltimes\mathbb{R}^{4}$ and
$CSO(2,0,3)\ltimes\mathbb{R}^{4}$ gauge groups given in (109), (120), and
(140), we are not able to find consistent sets of BPS equations. The
supersymmetry conditions from
$(\delta\psi_{+\mu\alpha},\delta\psi_{-\mu\dot{\alpha}})$ and
$(\delta\chi_{+a\dot{\alpha}},\delta\chi_{-\dot{a}\alpha})$ are not compatible
with each other. Therefore, we argue that no supersymmetric domain walls with
$SO(2)$ symmetry exist in these gauge groups.

###  IV.1 Supersymmetric domain walls from $SO(2)\ltimes\mathbb{R}^{8}$ and
$SO(2)\ltimes\mathbb{R}^{6}$ gauge groups

We now consider supersymmetric domain walls from $SO(2)\ltimes\mathbb{R}^{8}$
and $SO(2)\ltimes\mathbb{R}^{6}$ gauge groups with the embedding tensor given
in (42) and the traceless matrix ${S_{m}}^{n}$ in (50). We first consider
$SO(2)\ltimes\mathbb{R}^{8}$ gauge group. There are five $SO(2)$ singlet
scalars corresponding to the following non-compact generators commuting with
$X_{\ast}$

$\displaystyle\mathbf{Y}_{1}$ $\displaystyle=$
$\displaystyle\hat{\boldsymbol{t}}^{+}_{1\dot{1}}+\hat{\boldsymbol{t}}^{+}_{2\dot{2}}+\hat{\boldsymbol{t}}^{+}_{3\dot{3}}+\hat{\boldsymbol{t}}^{+}_{4\dot{4}}+\hat{\boldsymbol{t}}^{+}_{5\dot{5}},$
(165) $\displaystyle\mathbf{Y}_{2}$ $\displaystyle=$
$\displaystyle\hat{\boldsymbol{t}}^{+}_{1\dot{1}}+\hat{\boldsymbol{t}}^{+}_{2\dot{2}}-2\,\hat{\boldsymbol{t}}^{+}_{3\dot{3}},$
(166) $\displaystyle\mathbf{Y}_{3}$ $\displaystyle=$
$\displaystyle\hat{\boldsymbol{t}}^{+}_{4\dot{4}}+\hat{\boldsymbol{t}}^{+}_{5\dot{5}}-2\,\hat{\boldsymbol{t}}^{+}_{3\dot{3}},$
(167) $\displaystyle\mathbf{Y}_{4}$ $\displaystyle=$
$\displaystyle\boldsymbol{s}_{12},$ (168) $\displaystyle\mathbf{Y}_{5}$
$\displaystyle=$ $\displaystyle\boldsymbol{s}_{45}\,.$ (169)

It should be noted that the first non-compact generator corresponds to the
$SO(1,1)$ factor defined in (155). This scalar is actually a singlet under the
full compact subgroup $SO(5)\subset GL(5)$. The scalar field corresponding to
this generator will be called the dilaton $\varphi$. Additionally, there are
two $SL(5)/SO(5)$ scalars and two shift scalars invariant under the $SO(2)$
symmetry. These are associated with $\mathbf{Y}_{2,3}$ and $\mathbf{Y}_{4,5}$,
respectively.

The coset representative can be written as

$V=e^{\varphi\mathbf{Y}_{1}+\phi_{1}\mathbf{Y}_{2}+\phi_{2}\mathbf{Y}_{3}+\varsigma_{1}\mathbf{Y}_{4}+\varsigma_{2}\mathbf{Y}_{5}}\,.$
(170)

Using this coset representative, we find that the scalar potential vanishes
identically, and only the dilaton appears in the T-tensors which take the form
of

$(T^{a})^{\alpha\dot{\beta}}=\frac{1}{2\sqrt{2}}e^{5\varphi}{S_{b}}^{a}(\gamma^{b})^{\alpha\beta}\delta^{\dot{\beta}}_{\beta},\qquad(T^{\dot{a}})^{\alpha\dot{\beta}}=\frac{1}{2\sqrt{2}}e^{5\varphi}{S_{\dot{b}}}^{\dot{a}}(\gamma^{\dot{b}})^{\dot{\alpha}\dot{\beta}}\delta_{\dot{\alpha}}^{\alpha}\,.$
(171)

In these equations, we have written
${S_{b}}^{a}={S_{\dot{b}}}^{\dot{a}}={S_{m}}^{n}$. With ${S_{m}}^{n}$ given in
(50), the T-tensors become

$T^{\alpha\dot{\beta}}=\frac{1}{\sqrt{2}}e^{5\varphi}\left(\lambda\,(\gamma^{45})^{\alpha\beta}-\kappa\,(\gamma^{12})^{\alpha\beta}\right)\delta^{\dot{\beta}}_{\beta}\qquad\qquad$
(172)

or explicitly

$\small
T^{\alpha\dot{\beta}}=\frac{1}{\sqrt{2}}e^{5\varphi}\begin{pmatrix}\lambda+\kappa&0&0&0\\\
0&\lambda-\kappa&0&0\\\ 0&0&\lambda+\kappa&0\\\
0&0&0&\lambda-\kappa\end{pmatrix}.$ (173)

There are two possible superpotentials

$\mathcal{W}_{1}=\frac{g}{2\sqrt{2}}e^{5\varphi}(\kappa+\lambda)\qquad\text{and}\qquad\mathcal{W}_{2}=\frac{g}{2\sqrt{2}}e^{5\varphi}(\kappa-\lambda)\,.$
(174)

Both of them give a valid superpotential in terms of which the scalar
potential can be written as (164) using the matrix $G^{IJ}$ of the form

$G^{IJ}=\small\begin{pmatrix}\frac{1}{10}&0&0&\frac{2\varsigma_{1}}{5}&\frac{2\varsigma_{2}}{5}\\\
0&\frac{3}{20}&-\frac{1}{10}&\frac{3\varsigma_{1}}{5}&-\frac{2\varsigma_{2}}{5}\\\
0&-\frac{1}{10}&\frac{3}{20}&-\frac{2\varsigma_{1}}{5}&\frac{3\varsigma_{2}}{5}\\\
\frac{2\varsigma_{1}}{5}&\frac{3\varsigma_{1}}{5}&-\frac{2\varsigma_{1}}{5}&1+4\varsigma_{1}^{2}&0\\\
\frac{2\varsigma_{2}}{5}&-\frac{2\varsigma_{2}}{5}&\frac{3\varsigma_{2}}{5}&0&1+4\varsigma_{2}^{2}\\\
\end{pmatrix}$ (175)

for $\Phi^{I}=\\{\varphi,\phi_{1},\phi_{2},\varsigma_{1},\varsigma_{2}\\}$
with $I=1,2,3,4,5$. It should be noted that for $\lambda=0$, the two
superpotentials are equal leading to half-supersymmetric domain walls. For
$\lambda\neq 0$, one choice of the superpotentials gives rise to
$\frac{1}{4}$-BPS domain walls since only the supersymmetry along the
directions of this choice is unbroken.

We now consider the cases of $\lambda\neq\pm\kappa$ and $\lambda\neq 0$
corresponding to $SO(2)\ltimes\mathbb{R}^{8}$ gauge group. In this case,
choosing one of the two superpotentials corresponds to imposing an additional
projector on the Killing spinors of the form

$\gamma^{3}\epsilon_{\pm}=\epsilon_{\pm}\qquad\text{ or
}\qquad\gamma^{3}\epsilon_{\pm}=-\epsilon_{\pm}$ (176)

for $\mathcal{W}=\mathcal{W}_{1}$ or $\mathcal{W}=\mathcal{W}_{2}$,
respectively. Together with the $\hat{\gamma}_{r}$ projector in (162), the
resulting solutions will preserve only eight supercharges or $\frac{1}{4}$ of
the original supersymmetry. With all these, we obtain the following BPS
equations

$\displaystyle A^{\prime}$ $\displaystyle=$
$\displaystyle-\varphi^{\prime}=\frac{g}{2\sqrt{2}}e^{5\varphi}(\kappa\pm\lambda),$
(177) $\displaystyle\phi^{\prime}_{1}$ $\displaystyle=$
$\displaystyle\phi^{\prime}_{2}=0,$ (178)
$\displaystyle\varsigma^{\prime}_{1}$ $\displaystyle=$ $\displaystyle
4\varphi^{\prime}\varsigma_{1}\
=-g\sqrt{2}e^{5\varphi}(\kappa\pm\lambda)\varsigma_{1},$ (179)
$\displaystyle\varsigma^{\prime}_{2}$ $\displaystyle=$ $\displaystyle
4\varphi^{\prime}\varsigma_{2}\
=-g\sqrt{2}e^{5\varphi}(\kappa\pm\lambda)\varsigma_{2}\,.$ (180)

It should be noted that although the superpotential does not depend on
$\varsigma_{1}$ and $\varsigma_{2}$, there are non-trivial BPS equations for
these scalars due to the cross terms between these scalars and the dilaton in
$G^{IJ}$.
The plus/minus signs in the BPS equations are correlated with the plus/minus
sign of the $\gamma^{3}$ projector (176). The solution is given by

$\displaystyle A$ $\displaystyle=$
$\displaystyle-\varphi=\frac{1}{5}\ln\left[\frac{5}{2\sqrt{2}}gr(\kappa\pm\lambda)-C\right],$
(181) $\displaystyle\varsigma_{1,2}$ $\displaystyle=$ $\displaystyle
e^{4\varphi}+\varsigma^{(0)}_{1,2}$ (182)

for constants $C$ and $\varsigma^{(0)}_{1,2}$. The remaining two scalars
$\phi_{1}$ and $\phi_{2}$ are constant. It turns out that in this case, all
the composite connections $Q_{\mu}^{ab}$ and $Q_{\mu}^{\dot{a}\dot{b}}$
vanish. The BPS equations from $\delta\psi_{+r\alpha}$ and
$\delta\psi_{-r\dot{\alpha}}$ then give the following Killing spinors

$\epsilon_{\pm}=e^{\frac{A(r)}{2}}\epsilon_{\pm}^{0}$ (183)

with the constant SMW spinors $\epsilon_{\pm}^{0}$ satisfying the projectors
(162) and (176). In this solution, an integration constant for $A$ has been
neglected by rescaling the coordinates $x^{\bar{\mu}}$. Note that the
integration constant $C$ and $\varsigma^{(0)}_{1,2}$ can also be removed by
shifting the radial coordinate $r$ and scalars $\varsigma_{1,2}$, but we keep
it here for generality. We also note that from the BPS equations, we can
consistently set $\phi_{1}=\phi_{2}=\varsigma_{1}=\varsigma_{2}=0$. Indeed, it
can be verified that redefining the shift scalars as
$\varsigma_{1,2}\rightarrow\tilde{\varsigma}_{1,2}=e^{-4\varphi}\varsigma_{1,2}$
results in a set of BPS equations with $\tilde{\varsigma}^{\prime}_{1,2}=0$.

For $\lambda=\pm\kappa$ or $\lambda=0$, the translational group
$\mathbb{R}^{8}_{\boldsymbol{s}}$ reduces to
$\mathbb{R}^{6}_{\boldsymbol{s}}$. In the former case, $X_{14}=\pm X_{25}$ and
$X_{15}=\mp X_{24}$ while in the latter $X_{34}$ and $X_{35}$ vanish. In these
two cases, there are more $SO(2)$ singlet scalars. For $\lambda=\pm\kappa$,
there are four additional singlet scalars corresponding to non-compact
generators

$\displaystyle\mathbf{Y}_{6}$ $\displaystyle=$
$\displaystyle\hat{\boldsymbol{t}}^{+}_{1\dot{4}}\pm\hat{\boldsymbol{t}}^{+}_{2\dot{5}},\qquad\,\mathbf{Y}_{7}\
=\ \hat{\boldsymbol{t}}^{+}_{1\dot{5}}\mp\hat{\boldsymbol{t}}^{+}_{2\dot{4}},$
$\displaystyle\mathbf{Y}_{8}$ $\displaystyle=$
$\displaystyle\boldsymbol{s}_{14}\pm\boldsymbol{s}_{15},\qquad\mathbf{Y}_{9}\
=\ \boldsymbol{s}_{15}\mp\boldsymbol{s}_{24}\,.$ (184)

The upper/lower signs in these generators are related to the upper/lower sign
in $\lambda=\pm\kappa$. For $\lambda=0$, we find six additional singlets given
by

$\displaystyle\widetilde{\mathbf{Y}}_{6}$ $\displaystyle=$
$\displaystyle\hat{\boldsymbol{t}}^{+}_{4\dot{4}}-\hat{\boldsymbol{t}}^{+}_{5\dot{5}},\qquad\widetilde{\mathbf{Y}}_{7}\
=\ \hat{\boldsymbol{t}}^{+}_{3\dot{4}},\qquad\widetilde{\mathbf{Y}}_{8}\ =\
\hat{\boldsymbol{t}}^{+}_{3\dot{5}},$
$\displaystyle\widetilde{\mathbf{Y}}_{9}$ $\displaystyle=$
$\displaystyle\hat{\boldsymbol{t}}^{+}_{4\dot{5}},\qquad\qquad\widetilde{\mathbf{Y}}_{10}\
=\ \boldsymbol{s}_{34},\quad\ \,\widetilde{\mathbf{Y}}_{11}\ =\
\boldsymbol{s}_{35}\,.$ (185)

It turns out that even with all these extra $SO(2)$ singlet scalars, the
scalar potential still vanishes identically. Moreover, the T-tensors depend
only on the dilaton and take the same form as given in (171). The resulting
BPS equations for scalars from $SL(5)/SO(5)$ associated with
$\mathbf{Y}_{6,7}$ and $\widetilde{\mathbf{Y}}_{6,7,8,9}$ give constant
scalars. Similar to the previous case, the shift scalars associated with
$\mathbf{Y}_{8,9}$ and $\widetilde{\mathbf{Y}}_{10,11}$ can be redefined such
that the corresponding BPS equations give constant scalars. All these constant
scalars can in turn be consistently set to zero. Therefore, the solutions for
these two special cases are given by the above solution with the substitution
$\lambda=\pm\kappa$ and $\lambda=0$. However, as mentioned before, solutions
with $\lambda=0$ is $\frac{1}{2}$-supersymmetric since in this case the two
superpotentials $\mathcal{W}_{1}$ and $\mathcal{W}_{2}$ are equal. On the
other hand, solutions with $\lambda=\pm\kappa$ preserve $\frac{1}{4}$ of the
original supersymmetry as in the more general case of $\lambda\neq\pm\kappa$.
Each choice of $\lambda=\pm\kappa$ makes one of the superpotentials vanish
rendering the supersymmetry along the directions of this superpotential
broken.

###  IV.2 Supersymmetric domain walls from $CSO(2,0,2)\ltimes\mathbb{R}^{4}$,
$CSO(2,0,2)\ltimes\mathbb{R}^{2}$, and $CSO(2,0,1)\ltimes\mathbb{R}^{4}$ gauge
groups

We now move to another class of solutions in
$CSO(2,0,2)\ltimes\mathbb{R}^{4}$, $CSO(2,0,2)\ltimes\mathbb{R}^{2}$, and
$CSO(2,0,1)\ltimes\mathbb{R}^{4}$ gauge groups obtained from the embedding
tensor in $\overline{\mathbf{45}}^{+3}$ with the traceless matrix
${u_{i}}^{j}$ given in (58). Since in this case, the three possibilities with
$\lambda\neq\pm\kappa$, $\lambda=\pm\kappa$, and $\lambda=0$ are different in
many aspects, we will separately consider these three cases.

####  IV.2.1 $SO(2)$ symmetric domain walls from
$CSO(2,0,2)\ltimes\mathbb{R}^{4}$ gauge group

With $\lambda\neq\pm\kappa$, the gauge group is given by
$CSO(2,0,2)\ltimes\mathbb{R}^{4}$ as in (61). For convenience, we note the
explicit form of the matrix ${u_{i}}^{j}$ of the gauge group (61)

${u_{i}}^{j}=\small\begin{pmatrix}0&\kappa&0&0\\\ -\kappa&0&0&0\\\
0&0&0&-\lambda\\\ 0&0&\lambda&0\end{pmatrix}$ (186)

where $\kappa$ and $\lambda$ are non-vanishing and $\lambda\neq\pm\kappa$.
There are five $SO(2)$ singlet scalars containing the dilaton, two scalars
from $SL(5)/SO(5)$, and two shift scalars corresponding to the following non-
compact generators

$\displaystyle\overline{\mathbf{Y}}_{1}$ $\displaystyle=$
$\displaystyle\hat{\boldsymbol{t}}^{+}_{1\dot{1}}+\hat{\boldsymbol{t}}^{+}_{2\dot{2}}+\hat{\boldsymbol{t}}^{+}_{3\dot{3}}+\hat{\boldsymbol{t}}^{+}_{4\dot{4}}+\hat{\boldsymbol{t}}^{+}_{5\dot{5}},$
(187) $\displaystyle\overline{\mathbf{Y}}_{2}$ $\displaystyle=$
$\displaystyle\hat{\boldsymbol{t}}^{+}_{1\dot{1}}+\hat{\boldsymbol{t}}^{+}_{2\dot{2}}+\hat{\boldsymbol{t}}^{+}_{3\dot{3}}+\hat{\boldsymbol{t}}^{+}_{4\dot{4}}-4\hat{\boldsymbol{t}}^{+}_{5\dot{5}},$
(188) $\displaystyle\overline{\mathbf{Y}}_{3}$ $\displaystyle=$
$\displaystyle\hat{\boldsymbol{t}}^{+}_{1\dot{1}}+\hat{\boldsymbol{t}}^{+}_{2\dot{2}}-\hat{\boldsymbol{t}}^{+}_{3\dot{3}}-\hat{\boldsymbol{t}}^{+}_{4\dot{4}},$
(189) $\displaystyle\overline{\mathbf{Y}}_{4}$ $\displaystyle=$
$\displaystyle\boldsymbol{s}_{12},$ (190)
$\displaystyle\overline{\mathbf{Y}}_{5}$ $\displaystyle=$
$\displaystyle\boldsymbol{s}_{34}\,.$ (191)

The first non-compact generator corresponds to the dilaton $\varphi$ as in the
previous section. The scalar field corresponding to
$\overline{\mathbf{Y}}_{2}$ is another dilaton coming from further
decomposition of $SL(5)\rightarrow SO(1,1)\times SL(4)$ and will be denoted by
$\phi_{0}$. With the split of $GL(5)$ index $m=(i,5)$ for $i=1,...,4$, the
$SL(5)/SO(5)$ coset also decomposes into $SO(1,1)\times SL(4)/SO(4)$ with the
non-compact generator $\overline{\mathbf{Y}}_{3}$ corresponding to an $SO(2)$
singlet from the $SL(4)/SO(4)$ coset.

With the coset representative given by

$V=e^{\varphi\overline{\mathbf{Y}}_{1}+\phi_{0}\overline{\mathbf{Y}}_{2}+\phi\overline{\mathbf{Y}}_{3}+\varsigma_{1}\overline{\mathbf{Y}}_{4}+\varsigma_{2}\overline{\mathbf{Y}}_{5}},$
(192)

the scalar potential vanishes identically as in the previous section. The
T-tensor is more complicated due to the dependence on the shift scalars with
the following form

$\displaystyle
T^{\alpha\dot{\beta}}=e^{-3\varphi-8\phi_{0}}\left[2(\lambda\varsigma_{1}-\kappa\varsigma_{2})\,\Omega^{\alpha\beta}+2(\kappa\varsigma_{1}-\lambda\varsigma_{2})\,(\gamma^{5})^{\alpha\beta}\phantom{\frac{1}{2}}\right.$
$\displaystyle\left.\phantom{\frac{1}{2}}+(\lambda+4\kappa\varsigma_{1}\varsigma_{2})\,(\gamma^{12})^{\alpha\beta}+(\kappa+4\lambda\varsigma_{1}\varsigma_{2})\,(\gamma^{34})^{\alpha\beta}\right]\delta^{\dot{\beta}}_{\beta}\,.$
(193)

This gives rise to

$\displaystyle\Omega_{\alpha\gamma}T^{\gamma\dot{\alpha}}\Omega_{\dot{\alpha}\dot{\beta}}T^{\beta\dot{\beta}}$
$\displaystyle=$
$\displaystyle-e^{-6\varphi-16\phi_{0}}(1+4\varsigma_{1}^{2})(1+4\varsigma_{2}^{2})\left[(\kappa^{2}+\lambda^{2}){\delta_{\alpha}}^{\beta}+2\kappa\lambda{(\gamma^{5})_{\alpha}}^{\beta}\right],$
$\displaystyle\Omega_{\dot{\alpha}\dot{\gamma}}T^{\alpha\dot{\gamma}}\Omega_{\alpha\beta}T^{\beta\dot{\beta}}$
$\displaystyle=$
$\displaystyle-e^{-6\varphi-16\phi_{0}}(1+4\varsigma_{1}^{2})(1+4\varsigma_{2}^{2})\left[(\kappa^{2}+\lambda^{2}){\delta_{\dot{\alpha}}}^{\dot{\beta}}+2\kappa\lambda{(\gamma^{\dot{5}})_{\dot{\alpha}}}^{\dot{\beta}}\right].$

In this case, the two shift scalars $\varsigma_{1}$ and $\varsigma_{2}$ cannot
be removed by a redefinition of scalar fields, and the composite connections
$Q_{\mu}^{ab}$ and $Q_{\mu}^{\dot{a}\dot{b}}$ are non-vanishing. Therefore, as
in 6D_DW_I , we need to modify the ansatz for the Killing spinors to

$\epsilon_{+}=e^{\frac{A(r)}{2}+B(r)\gamma_{12}}\epsilon_{+}^{0}\qquad\text{and}\qquad\epsilon_{-}=e^{\frac{A(r)}{2}-B(r)\gamma_{\dot{1}\dot{2}}}\epsilon_{-}^{0}$
(196)

in which $B(r)$ is an $r$-dependent function. We now impose the $\gamma^{5}$
projector

$\gamma^{5}\epsilon_{\pm}=\epsilon_{\pm}\qquad\text{ or
}\qquad\gamma^{5}\epsilon_{\pm}=-\epsilon_{\pm}$ (197)

which implies $\gamma_{12}\epsilon_{+}=\mp\gamma_{34}\epsilon_{+}$ and
$\gamma_{\dot{1}\dot{2}}\epsilon_{-}=\mp\gamma_{\dot{3}\dot{4}}\epsilon_{-}$.
The latter two conditions are sufficient to solve the conditions
$\delta\psi_{+r\alpha}=0$ and $\delta\psi_{-r\dot{\alpha}}=0$ with
$\epsilon_{\pm}^{0}$ being constant SMW spinors making $\epsilon_{\pm}$
satisfy the projectors (162) and (197).
With the projector (197), the two conditions in (LABEL:45PreSPot1) and
(LABEL:45PreSPot2) give

$\displaystyle\Omega_{\alpha\gamma}T^{\gamma\dot{\alpha}}\Omega_{\dot{\alpha}\dot{\beta}}T^{\beta\dot{\beta}}\epsilon_{+\beta}$
$\displaystyle=$
$\displaystyle-e^{-6\varphi-16\phi_{0}}(1+4\varsigma_{1}^{2})(1+4\varsigma_{2}^{2})(\kappa\pm\lambda)^{2}\epsilon_{+\alpha},$
(198)
$\displaystyle\Omega_{\dot{\alpha}\dot{\gamma}}T^{\alpha\dot{\gamma}}\Omega_{\alpha\beta}T^{\beta\dot{\beta}}\epsilon_{-\dot{\beta}}$
$\displaystyle=$
$\displaystyle-e^{-6\varphi-16\phi_{0}}(1+4\varsigma_{1}^{2})(1+4\varsigma_{2}^{2})(\kappa\pm\lambda)^{2}\epsilon_{-\dot{\alpha}}$
(199)

in which the plus/minus signs correspond to the sign chosen in the
$\gamma^{5}$ projector. Therefore, the superpotential is then given by

$\mathcal{W}=\frac{g}{2}e^{-3\varphi-8\phi_{0}}(\kappa\pm\lambda)\sqrt{1+4\varsigma_{1}^{2}}\sqrt{1+4\varsigma_{2}^{2}}\,.$
(200)

With all these, we find the following BPS equations

$\displaystyle A^{\prime}$ $\displaystyle=$
$\displaystyle\frac{g}{2}e^{-3\varphi-8\phi_{0}}(\kappa\pm\lambda)\sqrt{1+4\varsigma_{1}^{2}}\sqrt{1+4\varsigma_{2}^{2}},$
(201) $\displaystyle\varphi^{\prime}$ $\displaystyle=$
$\displaystyle\frac{ge^{-3\varphi-8\phi_{0}}(\kappa\pm\lambda)\left[3-4(\varsigma_{1}^{2}+\varsigma_{2}^{2})-80\varsigma_{1}^{2}\varsigma_{2}^{2}\right]}{10\sqrt{1+4\varsigma_{1}^{2}}\sqrt{1+4\varsigma_{2}^{2}}},$
(202) $\displaystyle\phi_{0}^{\prime}$ $\displaystyle=$
$\displaystyle\frac{ge^{-3\varphi-8\phi_{0}}(\kappa\pm\lambda)\left[1+4(\varsigma_{1}^{2}+\varsigma_{2}^{2})\right]}{5\sqrt{1+4\varsigma_{1}^{2}}\sqrt{1+4\varsigma_{2}^{2}}},$
(203) $\displaystyle\phi^{\prime}$ $\displaystyle=$
$\displaystyle\frac{2ge^{-3\varphi-8\phi_{0}}(\kappa\pm\lambda)\left(\varsigma_{2}^{2}-\varsigma_{1}^{2}\right)}{\sqrt{1+4\varsigma_{1}^{2}}\sqrt{1+4\varsigma_{2}^{2}}},$
(204) $\displaystyle\varsigma^{\prime}_{1}$ $\displaystyle=$
$\displaystyle-2ge^{-3\varphi-8\phi_{0}}(\kappa\pm\lambda)\varsigma_{1}\sqrt{1+4\varsigma_{1}^{2}}\sqrt{1+4\varsigma_{2}^{2}},$
(205) $\displaystyle\varsigma^{\prime}_{2}$ $\displaystyle=$
$\displaystyle-2ge^{-3\varphi-8\phi_{0}}(\kappa\pm\lambda)\varsigma_{2}\sqrt{1+4\varsigma_{1}^{2}}\sqrt{1+4\varsigma_{2}^{2}},$
(206) $\displaystyle B^{\prime}$ $\displaystyle=$
$\displaystyle\frac{2ge^{-3\varphi-8\phi_{0}}(\kappa\pm\lambda)(\varsigma_{1}-\varsigma_{2}-4\varsigma^{2}_{1}\varsigma_{2}+4\varsigma_{1}\varsigma^{2}_{2})}{\sqrt{1+4\varsigma_{1}^{2}}\sqrt{1+4\varsigma_{2}^{2}}}\,.$
(207)

The domain wall solution to these equations can be written in terms of the
shift scalar $\varsigma_{1}$ by

$\displaystyle A$ $\displaystyle=$
$\displaystyle-\frac{1}{4}\ln\varsigma_{1},\qquad\qquad\varsigma_{2}\ =\
C_{1}\varsigma_{1},$ (208) $\displaystyle\phi$ $\displaystyle=$ $\displaystyle
C_{2}+\frac{1}{8}\ln\left(1+4\varsigma_{1}^{2}\right)-\frac{1}{8}\ln\left(1+4C^{2}_{1}\varsigma_{1}^{2}\right),$
(209) $\displaystyle\phi_{0}$ $\displaystyle=$ $\displaystyle
C_{3}-\frac{1}{10}\ln\varsigma_{1}+\frac{1}{40}\ln\left(1+4\varsigma_{1}^{2}\right)+\frac{1}{40}\ln\left(1+4C^{2}_{1}\varsigma_{1}^{2}\right),$
(210) $\displaystyle\varphi$ $\displaystyle=$ $\displaystyle
C_{4}-\frac{3}{20}\ln\varsigma_{1}+\frac{1}{10}\ln\left(1+4\varsigma_{1}^{2}\right)+\frac{1}{10}\ln\left(1+4C^{2}_{1}\varsigma_{1}^{2}\right),$
(211) $\displaystyle B$ $\displaystyle=$ $\displaystyle
C_{5}-\frac{1}{2}\tan^{-1}(2\varsigma_{1})+\frac{1}{2}\tan^{-1}(2C_{1}\varsigma_{1})$
(212)

in which $C_{i}$, $i=1,...,5$, are integration constants. Using the new radial
coordinate $\rho$ defined by $\frac{d\rho}{dr}=e^{-3\varphi-8\phi_{0}}$, we
find the solution for $\varsigma_{1}$ given by

$\varsigma_{1}=\frac{\text{sech}[2g(\kappa\pm\lambda)\rho-C]}{2\sqrt{C_{1}^{2}\tanh^{2}[2g(\kappa\pm\lambda)\rho-C]-1}}.$
(213)

It is also interesting to consider a domain wall solution with vanishing shift
scalars. From the above solution, we find a constant $B$ function giving rise
to the same Killing spinors (183) up to a constant phase. For
$\varsigma_{1}=\varsigma_{2}=0$, the BPS equations (201) to (206) reduce to

$\displaystyle A^{\prime}$ $\displaystyle=$
$\displaystyle\frac{g}{2}(\kappa\pm\lambda)e^{-3\varphi-8\phi_{0}},$ (214)
$\displaystyle\varphi^{\prime}$ $\displaystyle=$
$\displaystyle\frac{3}{10}g(\kappa\pm\lambda)e^{-3\varphi-8\phi_{0}},$ (215)
$\displaystyle\phi^{\prime}_{0}$ $\displaystyle=$
$\displaystyle\frac{g}{5}(\kappa\pm\lambda)e^{-3\varphi-8\phi_{0}}$ (216)

together with $\phi^{\prime}=\varsigma^{\prime}_{1}=\varsigma^{\prime}_{2}=0$.
Setting $\phi=\varsigma_{1}=\varsigma_{2}=0$, we readily find the solution

$\displaystyle A$ $\displaystyle=$
$\displaystyle\frac{5}{2}\phi_{0},\qquad\varphi\ =\
\frac{1}{3}\ln\left[\frac{5}{2}[g(\kappa\pm\lambda)r+C]\right]-\frac{8}{3}\phi_{0},$
$\displaystyle\phi_{0}$ $\displaystyle=$ $\displaystyle
C_{0}+\frac{2}{25}\ln\left[g(\kappa\pm\lambda)r+C\right].$ (217)

Since the solutions are subject to two projectors (162) and (197), all the
domain walls, with and without the shift scalars, are
$\frac{1}{4}$-supersymmetric.

####  IV.2.2 $SO(2)$ symmetric domain walls from
$CSO(2,0,2)\ltimes\mathbb{R}^{2}$ gauge group

For $\lambda=\pm\kappa$, we find that the gauge generators $X_{13}=\pm X_{24}$
and $X_{14}=\mp X_{23}$. This reduces the translational group
$\mathbb{R}^{4}_{\boldsymbol{h}}$ to $\mathbb{R}^{2}_{\boldsymbol{h}}$
resulting in the gauge group of the form

$G_{0}=SO(2)\ltimes\left(\mathbb{R}^{4}\times\mathbb{R}^{2}_{\boldsymbol{h}}\right)\sim
CSO(2,0,2)\ltimes\mathbb{R}^{2}_{\boldsymbol{h}}.$ (218)

Apart from the five singlet scalars corresponding to non-compact generators in
(187) to (191), there are additional four scalars invariant under the $SO(2)$
subgroup generated by $X^{5}$ with $\lambda=\pm\kappa$. These extra scalars
correspond to the following non-compact generators

$\displaystyle\overline{\mathbf{Y}}_{6}$ $\displaystyle=$
$\displaystyle\hat{\boldsymbol{t}}^{+}_{1\dot{3}}\mp\hat{\boldsymbol{t}}^{+}_{2\dot{4}},\qquad\overline{\mathbf{Y}}_{7}\
=\ \hat{\boldsymbol{t}}^{+}_{1\dot{4}}\pm\hat{\boldsymbol{t}}^{+}_{2\dot{3}},$
$\displaystyle\overline{\mathbf{Y}}_{8}$ $\displaystyle=$
$\displaystyle\boldsymbol{s}_{13}\mp\boldsymbol{s}_{24},\,\qquad\overline{\mathbf{Y}}_{9}\
=\ \boldsymbol{s}_{14}\pm\boldsymbol{s}_{23}\,.$ (219)

Using the coset representative

$V=e^{\varphi\overline{\mathbf{Y}}_{1}+\phi_{0}\overline{\mathbf{Y}}_{2}+\phi_{2}\overline{\mathbf{Y}}_{6}+\phi_{3}\overline{\mathbf{Y}}_{7}+\phi\overline{\mathbf{Y}}_{3}+\varsigma_{1}\overline{\mathbf{Y}}_{4}+\varsigma_{2}\overline{\mathbf{Y}}_{5}+\varsigma_{3}\overline{\mathbf{Y}}_{8}+\varsigma_{4}\overline{\mathbf{Y}}_{9}},$
(220)

we again find that the scalar potential vanishes identically. Using the
$\gamma^{5}$ projector given in (197), we find that the superpotential is
given by

$\mathcal{W}=ge^{-3\varphi-8\phi_{0}}\kappa\sqrt{4\varsigma^{2}_{1}+4\varsigma^{2}_{2}+16\varsigma^{2}_{1}\varsigma^{2}_{2}+32\varsigma_{1}\varsigma_{2}(\varsigma^{2}_{3}+\varsigma^{2}_{4})+(1+4\varsigma^{2}_{3}+4\varsigma^{2}_{4})^{2}}\,.$
(221)

Consistency of the BPS equations requires either
$\varsigma_{3}=\varsigma_{4}=0$ or $\varsigma_{2}=\varsigma_{1}$. The former
gives the same solution as in the previous case with ($\kappa\pm\lambda$)
replaced by $2\kappa$, so we will not consider this choice any further.

For the other choice of $\varsigma_{2}=\varsigma_{1}$, the superpotential
(221) becomes

$\mathcal{W}=ge^{-3\varphi-8\phi_{0}}\kappa\left(1+4\varsigma^{2}_{1}+4\varsigma^{2}_{3}+4\varsigma^{2}_{4}\right).$
(222)

In this case, the composite connections $Q_{\mu}^{ab}$ and
$Q_{\mu}^{\dot{a}\dot{b}}$ vanish. We can then use the ansatz for the Killing
spinors as given in (183) and end up with the following BPS equations

$\displaystyle A^{\prime}$ $\displaystyle=$ $\displaystyle
ge^{-3\varphi-8\phi_{0}}\kappa\left(1+4\varsigma^{2}_{1}+4\varsigma^{2}_{3}+4\varsigma^{2}_{4}\right),$
(223) $\displaystyle\varphi^{\prime}$ $\displaystyle=$
$\displaystyle\frac{g}{5}e^{-3\varphi-8\phi_{0}}\kappa(3-20\varsigma^{2}_{1}-20\varsigma^{2}_{3}-20\varsigma^{2}_{4}),$
(224) $\displaystyle\phi_{0}^{\prime}$ $\displaystyle=$
$\displaystyle\frac{2g}{5}e^{-3\varphi-8\phi_{0}}\kappa,$ (225)
$\displaystyle\varsigma^{\prime}_{1}$ $\displaystyle=$
$\displaystyle-4ge^{-3\varphi-8\phi_{0}}\kappa\varsigma_{1}\left(1+4\varsigma^{2}_{1}+4\varsigma^{2}_{3}+4\varsigma^{2}_{4}\right),$
(226) $\displaystyle\varsigma^{\prime}_{3}$ $\displaystyle=$
$\displaystyle-4ge^{-3\varphi-8\phi_{0}}\kappa\varsigma_{3}\left(1+4\varsigma^{2}_{1}+4\varsigma^{2}_{3}+4\varsigma^{2}_{4}\right),$
(227) $\displaystyle\varsigma^{\prime}_{4}$ $\displaystyle=$
$\displaystyle-4ge^{-3\varphi-8\phi_{0}}\kappa\varsigma_{4}\left(1+4\varsigma^{2}_{1}+4\varsigma^{2}_{3}+4\varsigma^{2}_{4}\right)$
(228)

together with $\phi^{\prime}=\phi^{\prime}_{2}=\phi^{\prime}_{3}=0$. Solving
these equations gives a domain wall solution

$\displaystyle A$ $\displaystyle=$
$\displaystyle-\frac{1}{4}\ln\varsigma_{1},\qquad\varsigma_{3}\ =\
C_{1}\varsigma_{1},\qquad\varsigma_{4}\ =\ C_{2}\varsigma_{1},$
$\displaystyle\varphi$ $\displaystyle=$ $\displaystyle
C_{3}-\frac{3}{20}\ln{\varsigma_{1}}+\frac{1}{5}\ln\left[1+4\varsigma^{2}_{1}(1+C^{2}_{1}+C^{2}_{2})\right],$
$\displaystyle\phi_{0}$ $\displaystyle=$ $\displaystyle
C_{4}-\frac{1}{10}\ln{\varsigma_{1}}+\frac{1}{20}\ln\left[1+4\varsigma^{2}_{1}(1+C^{2}_{1}+C^{2}_{2})\right],$
$\displaystyle\varsigma_{1}$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{e^{8g\kappa\rho-
C_{5}}-4(1+C^{2}_{1}+C^{2}_{2})}}$ (229)

in which $\rho$ is a new radial coordinate defined by the relation
$\frac{d\rho}{dr}=e^{-3\varphi-8\phi_{0}}$. In this solution, we also
consistently set the constant scalars $\phi$, $\phi_{2}$, and $\phi_{3}$ to
zero.

####  IV.2.3 $SO(2)$ symmetric domain walls from
$CSO(2,0,1)\ltimes\mathbb{R}^{4}$ gauge group

For $\lambda=0$, we find $X^{3}=X^{4}=0$ reducing $\mathbb{R}^{4}$ to
$\mathbb{R}^{2}$ in the gauge group (61). The resulting gauge group then takes
the form of

$G_{0}=SO(2)\ltimes\left(\mathbb{R}^{2}\times\mathbb{R}^{4}_{\boldsymbol{h}}\right)\sim
CSO(2,0,1)\ltimes\mathbb{R}^{4}_{\boldsymbol{h}}\,.$ (230)

In this case, the $\gamma^{5}$ projector (197) is not needed, so unlike the
previous two cases, the solutions will be $\frac{1}{2}$-supersymmetric. There
are additional scalars invariant under the $SO(2)$ subgroup generated by
$X^{5}$ with $\lambda=0$. These are given by two scalars parametrizing the
$SL(4)/SO(4)$ coset, two nilpotent, and two shift scalars corresponding to the
following non-compact generators

$\displaystyle\widehat{\mathbf{Y}}_{6}$ $\displaystyle=$
$\displaystyle\hat{\boldsymbol{t}}^{+}_{3\dot{3}}-\hat{\boldsymbol{t}}^{+}_{4\dot{4}},\qquad\widehat{\mathbf{Y}}_{7}\
=\ \hat{\boldsymbol{t}}^{+}_{3\dot{4}},\qquad\widehat{\mathbf{Y}}_{8}\ =\
\hat{\boldsymbol{t}}^{+}_{3\dot{5}},$ $\displaystyle\widehat{\mathbf{Y}}_{9}$
$\displaystyle=$
$\displaystyle\hat{\boldsymbol{t}}^{+}_{4\dot{5}},\qquad\qquad\widehat{\mathbf{Y}}_{10}\
=\ \boldsymbol{s}_{35},\ \,\quad\widehat{\mathbf{Y}}_{11}\ =\
\boldsymbol{s}_{45}\,.$ (231)

Following 6D_DW_I , we will call the four scalars associated with
$\hat{\boldsymbol{t}}^{+}_{i\dot{5}}$, for $i=1,2,3,4$, nilpotent scalars. It
turns out that the two nilpotent scalars associated with
$\hat{\boldsymbol{t}}^{+}_{3\dot{5}}$ and
$\hat{\boldsymbol{t}}^{+}_{4\dot{5}}$ need to vanish in order to find a
consistent set of BPS equations in this case.

As in all the previous cases, it turns out that the scalar potential vanishes.
The superpotential is still given by (200) for $\lambda=0$. This gives rise to
the same set of BPS equations (201) to (206) with all additional scalar fields
constant. The ansatz for the Killing spinors, in this case, takes the form of

$\epsilon_{+}=e^{\frac{A(r)}{2}+B_{1}(r)\gamma_{12}+B_{2}(r)\gamma_{34}}\epsilon_{+}^{0}\qquad\text{and}\qquad\epsilon_{-}=e^{\frac{A(r)}{2}-B_{1}(r)\gamma_{\dot{1}\dot{2}}-B_{2}(r)\gamma_{\dot{3}\dot{4}}}\epsilon_{-}^{0}.$
(232)

The BPS equations for $B_{1}$ and $B_{2}$ are given by

$\displaystyle B^{\prime}_{1}$ $\displaystyle=$
$\displaystyle\frac{2ge^{-3\varphi-8\phi_{0}}\kappa\varsigma_{1}\sqrt{1+4\varsigma_{2}^{2}}}{\sqrt{1+4\varsigma_{1}^{2}}}$
(233) $\displaystyle\textrm{and}\qquad B^{\prime}_{2}$ $\displaystyle=$
$\displaystyle\frac{2ge^{-3\varphi-8\phi_{0}}\kappa\varsigma_{2}\sqrt{1+4\varsigma_{1}^{2}}}{\sqrt{1+4\varsigma_{2}^{2}}}\,.$
(234)

All these equations lead to the same solution given in (208) to (211) and
(213) for $\lambda=0$ together with

$\displaystyle B_{1}$ $\displaystyle=$ $\displaystyle
C_{6}-\frac{1}{2}\tan^{-1}(2\varsigma_{1})$ (235)
$\displaystyle\textrm{and}\qquad B_{2}$ $\displaystyle=$ $\displaystyle
C_{7}-\frac{1}{2}\tan^{-1}(2C_{1}\varsigma_{1})$ (236)

in which $C_{6}$ and $C_{7}$ are integration constants. For a particular case
of $\varsigma_{1}=\varsigma_{2}=0$, we find the domain wall solution given in
(217) with $\lambda=0$ and the Killing spinors (183).

##  V Conclusions and discussions

We have classified a number of gauge groups in six-dimensional maximal gauged
supergravity with the embedding tensor in $\mathbf{5}^{+7}$,
$\bar{\mathbf{5}}^{+3}$, $\mathbf{10}^{-1}$, $\mathbf{24}^{-5}$, and
$\overline{\mathbf{45}}^{+3}$ representations of $GL(5)\subset SO(5,5)$. All
these gauge groups are non-semisimple and consist mostly translational groups
of various dimensions. We have also found supersymmetric domain wall solutions
in $SO(2)\ltimes\mathbb{R}^{8}$, $SO(2)\ltimes\mathbb{R}^{6}$,
$CSO(2,0,2)\ltimes\mathbb{R}^{4}$, $CSO(2,0,2)\ltimes\mathbb{R}^{2}$, and
$CSO(2,0,1)\ltimes\mathbb{R}^{4}$ gauge groups. These solutions are either
$\frac{1}{2}$\- or $\frac{1}{4}$-supersymmetric preserving $SO(2)$ symmetry.
Some of the resulting gauge groups could have higher dimensional origins in
terms of Scherk-Schwarz reductions from seven dimensions or truncations of
eleven-dimensional supergravity on twisted tori (possibly) with fluxes. The
consistent gaugings identified in this paper to some extent enlarge the known
gauge groups pointed out in 6D_Max_Gauging  and those constructed in 6D_DW_I .
These would hopefully be useful in the context of DW/QFT correspondence and
related aspects. The domain wall solutions given here are exhaustive for all
the gaugings under consideration at least for domain walls with any residual
symmetry. These could also be added to the known classification of
supersymmetric domain walls. It would be useful to find more general and more
complicated solutions without any residual symmetry.

Constructing truncation ansatze for embedding the domain wall solutions found
here in string/M-theory using $SO(5,5)$ exceptional field theory given in
SO5_5_EFT  is of particular interest. This would provide a complete framework
for a holographic study of five-dimensional maximal SYM. It is also
interesting to perform a similar analysis for gaugings under $SO(4,4)\subset
SO(5,5)$ and construct the corresponding embedding tensors together with
possible supersymmetric domain walls. These gaugings can be truncated to
gaugings of half-maximal $N=(1,1)$ supergravity coupled to four vector
multiplets in which supersymmetric $AdS_{6}$ vacua are known to exist D4D8 ;
F4_flow ; AdS6_Jan . In this case, the results could be useful in the study of
both DW6/QFT5 duality and AdS6/CFT5 correspondence as well.

Finally, generalizing the standard domain wall solutions in this paper and
6D_DW_I  to curved domain walls with non-vanishing vector and tensor fields
could also be worth considering. This type of solutions describes conformal
defects or holographic RG flows from five-dimensional $N=4$ super Yang-Mills
theories to lower-dimensional (conformal) field theories via twisted
compactifications. A number of similar solutions in seven dimensions have been
given in 7D_Defect ; 7D_twist  while examples of holographic solutions dual to
surface defects in five-dimensional $N=2$ SCFTs from $N=(1,1)$ gauged
supergravity have appeared recently in 6D_surface_defect .

###### Acknowledgements.

This work is supported by the Second Century Fund (C2F), Chulalongkorn
University. P. K. is supported by The Thailand Research Fund (TRF) under grant
RSA6280022.

##  Appendix A Useful formulae

In this appendix, we collect some formulae and relations used in the main
text, see 6D_DW_I  for more detail. We begin with $SO(5,5)\rightarrow GL(5)$
branching rules used throughout the paper. With the branching rule for an
$SO(5,5)$ vector representation

$\mathbf{10}\ \rightarrow\
\mathbf{5}^{+2}\,\oplus\,\overline{\mathbf{5}}^{-2},$ (237)

we can decompose the $SO(5,5)$ generators as

$\boldsymbol{t}_{MN}\ \rightarrow\
(\boldsymbol{t}_{mn},{\boldsymbol{t}^{m}}_{n},\boldsymbol{t}^{mn})$ (238)

with ${\boldsymbol{t}_{m}}^{n}=-{\boldsymbol{t}^{n}}_{m}$.

In vector representation, $SO(5,5)$ generators
$\boldsymbol{t}_{MN}=\boldsymbol{t}_{[MN]}$ can be chosen as

${(\boldsymbol{t}_{MN})_{P}}^{Q}\ =\ 4\eta_{P[M}\delta^{Q}_{N]}$ (239)

satisfying the Lie algebra

$\left[\boldsymbol{t}_{MN},\boldsymbol{t}_{PQ}\right]\ =\
4(\eta_{M[P}\boldsymbol{t}_{Q]N}-\eta_{N[P}\boldsymbol{t}_{Q]M}).$ (240)

With the generators of the shift and hidden symmetries respectively denoted by
$\boldsymbol{s}_{mn}=\boldsymbol{t}_{mn}$ and
$\boldsymbol{h}^{mn}=\boldsymbol{t}^{mn}$, the $SO(5,5)$ generators can be
written as

${(\boldsymbol{t}_{MN})_{P}}^{Q}=\begin{pmatrix}{\boldsymbol{t}^{m}}_{n}&\boldsymbol{h}^{mn}\\\
\boldsymbol{s}_{mn}&-{\boldsymbol{t}^{n}}_{m}\end{pmatrix}.$ (241)

In this form, we can see that the $GL(5)$ subgroup generated by
${\boldsymbol{t}_{m}}^{n}$ is embedded diagonally.
The branching rules for spinor and conjugate spinor representations of
$SO(5,5)$

$\mathbf{16}_{s}\ \rightarrow\
\overline{\mathbf{5}}^{+3}\,\oplus\,\mathbf{10}^{-1}\,\oplus\,\mathbf{1}^{-5}\qquad\text{and}\qquad\mathbf{16}_{c}\
\rightarrow\
\mathbf{5}^{-3}\,\oplus\,\overline{\mathbf{10}}^{+1}\,\oplus\,\mathbf{1}^{+5}$
(242)

are realized respectively by the following relations

$\Psi_{A}\ =\
\mathbb{T}_{Am}\Psi^{m}+\mathbb{T}_{A}^{mn}\Psi_{mn}+\mathbb{T}_{A\ast}\Psi_{\ast}$
(243)

with $\Psi_{mn}=\Psi_{[mn]}$ and

$\Psi^{A}\ =\
\mathbb{T}^{Am}\Psi_{m}+\mathbb{T}^{A}_{mn}\Psi^{mn}+\mathbb{T}^{A}_{\ast}\Psi_{\ast}\,.$
(244)

The transformation matrices are defined by

$\displaystyle\mathbb{T}_{Am}$ $\displaystyle=$
$\displaystyle\frac{1}{2\sqrt{2}}(\Gamma_{m})_{AB}\boldsymbol{p}^{B}_{\alpha\beta}\Omega^{\alpha\beta},$
(245) $\displaystyle\mathbb{T}_{A}^{mn}$ $\displaystyle=$
$\displaystyle\frac{1}{4\sqrt{2}}{(\Gamma^{mn})_{A}}^{B}\boldsymbol{p}_{B}^{\alpha\beta}\Omega_{\alpha\beta},$
(246) $\displaystyle\mathbb{T}_{A\ast}$ $\displaystyle=$
$\displaystyle\frac{1}{10}{({\Gamma^{m}}_{m})_{A}}^{B}\boldsymbol{p}_{B}^{\alpha\beta}\Omega_{\alpha\beta}$
(247)

with the inverse matrices of $\mathbb{T}_{A}$ are simply given by their
complex conjugation
$\mathbb{T}^{A}=(\mathbb{T}_{A})^{-1}=(\mathbb{T}_{A})^{*}$ satisfying the
relations

$\displaystyle\mathbb{T}^{Am}\mathbb{T}_{An}\ =\
\delta^{m}_{n},\qquad\quad\mathbb{T}^{A}_{mn}\mathbb{T}_{A}^{pq}$
$\displaystyle=$
$\displaystyle\delta^{[p}_{m}\delta^{q]}_{n},\qquad\mathbb{T}^{A}_{\ast}\mathbb{T}_{A\ast}\
=\ 1,$ $\displaystyle\mathbb{T}^{Am}\mathbb{T}_{Anp}\ =\
0,\qquad\quad\,\mathbb{T}^{Am}\mathbb{T}_{A\ast}$ $\displaystyle=$
$\displaystyle 0,\qquad\quad\mathbb{T}^{A}_{mn}\mathbb{T}_{A\ast}\ =\ 0$ (248)

together with

$\mathbb{T}^{Am}\mathbb{T}_{Bm}+\mathbb{T}^{A}_{mn}\mathbb{T}_{B}^{mn}+\mathbb{T}^{A}_{\ast}\mathbb{T}_{B\ast}=\delta^{A}_{B}\,.$
(249)

The transformation matrices $\boldsymbol{p}_{A}^{\alpha\beta}$ and inverse
matrices $\boldsymbol{p}^{A}_{\alpha\beta}$ are given by

$\displaystyle\boldsymbol{p}_{A}^{\alpha\beta}$ $\displaystyle=$
$\displaystyle\delta_{A}^{\alpha}\delta_{1}^{\beta}+\delta_{A}^{\alpha+4}\delta_{2}^{\beta}+\delta_{A}^{\alpha+8}\delta_{3}^{\beta}+\delta_{A}^{\alpha+12}\delta_{4}^{\beta},$
$\displaystyle\boldsymbol{p}^{A}_{\alpha\beta}$ $\displaystyle=$
$\displaystyle\delta^{A}_{\alpha}\delta^{1}_{\beta}+\delta^{A}_{\alpha+4}\delta^{2}_{\beta}+\delta^{A}_{\alpha+8}\delta^{3}_{\beta}+\delta^{A}_{\alpha+12}\delta^{4}_{\beta}\,.$
(250)

These matrices satisfy the relations

$\boldsymbol{p}_{A}^{\alpha\beta}\boldsymbol{p}^{B}_{\alpha\beta}\ =\
\delta_{A}^{B}\qquad\text{and}\qquad\boldsymbol{p}_{A}^{\alpha\delta}\boldsymbol{p}^{A}_{\beta\gamma}\
=\ \delta^{\alpha}_{\beta}\delta^{\delta}_{\gamma}$ (251)

and can be used to express a spinor index of $SO(5,5)$, $A,B,\ldots$, in terms
of a pair of $USp(4)$ fundamental or $SO(5)$ spinor indices $(\alpha\beta)$.

The matrices $(\Gamma_{m})_{AB}$ appearing in the $SO(5,5)$ gamma matrices
$\Gamma_{M}=\left(\Gamma_{m},\Gamma^{m}\right)$ are related to $SO(5)\times
SO(5)$ gamma matrices $\Gamma_{\underline{A}}=(\Gamma_{a},\Gamma_{\dot{a}})$
as

$(\Gamma_{M})_{AB}={{\mathbb{M}}_{M}}^{\underline{A}}(\Gamma_{\underline{A}})_{AB}$
(252)

with

$(\Gamma_{\underline{A}})_{AB}={(\Gamma_{\underline{A}})_{A}}^{C}\boldsymbol{c}_{CB}\,.$
(253)

The charge conjugation matrix $\boldsymbol{c}_{AB}$ is given by

$\boldsymbol{c}_{AB}=\boldsymbol{p}_{A}^{\alpha\dot{\alpha}}\boldsymbol{p}_{B}^{\beta\dot{\beta}}\Omega_{\alpha\beta}\Omega_{\dot{\alpha}\dot{\beta}}$
(254)

in which $\boldsymbol{p}_{A}^{\alpha\dot{\alpha}}$ and
$\boldsymbol{p}^{A}_{\alpha\dot{\alpha}}$ are defined in parallel with (A).
For convenience, we also note the transformation matrix $\mathbb{M}$ here

$\mathbb{M}\ =\
\frac{1}{\sqrt{2}}\begin{pmatrix}\mathds{1}_{5}&\mathds{1}_{5}\\\
\mathds{1}_{5}&-\mathds{1}_{5}\end{pmatrix}.$ (255)

The $SO(5)\times SO(5)$ gamma matrices can be written as

${(\Gamma_{\underline{A}})_{A}}^{B}=\left({(\gamma_{a})_{A}}^{B},{(\gamma_{\dot{a}})_{A}}^{B}\right)$
(256)

in which ${(\gamma_{a})_{A}}^{B}$ and ${(\gamma_{\dot{a}})_{A}}^{B}$ are
explicitly given by

$\displaystyle{(\gamma_{a})_{A}}^{B}$ $\displaystyle=$
$\displaystyle\boldsymbol{p}_{A}^{\alpha\dot{\alpha}}{(\gamma_{a})_{\alpha\dot{\alpha}}}^{\beta\dot{\beta}}\boldsymbol{p}^{B}_{\beta\dot{\beta}}\
=\
\boldsymbol{p}_{A}^{\alpha\dot{\alpha}}{(\gamma_{a})_{\alpha}}^{\beta}\delta_{\dot{\alpha}}^{\dot{\beta}}\boldsymbol{p}^{B}_{\beta\dot{\beta}},$
(257) $\displaystyle{(\gamma_{\dot{a}})_{A}}^{B}$ $\displaystyle=$
$\displaystyle\boldsymbol{p}_{A}^{\alpha\dot{\alpha}}{(\gamma_{\dot{a}})_{\alpha\dot{\alpha}}}^{\beta\dot{\beta}}\boldsymbol{p}^{B}_{\beta\dot{\beta}}\
=\
\boldsymbol{p}_{A}^{\alpha\dot{\alpha}}\delta_{\alpha}^{\beta}{(\gamma_{\dot{a}})_{\dot{\alpha}}}^{\dot{\beta}}\boldsymbol{p}^{B}_{\beta\dot{\beta}}\,.$
(258)

With all these, the matrices $(\Gamma_{m})_{AB}$ can be explicitly defined in
terms of the $SO(5)$ gamma matrices as

$(\Gamma_{m})_{AB}=\frac{1}{\sqrt{2}}\boldsymbol{p}_{A}^{\alpha\dot{\alpha}}\left[{(\gamma_{m})_{\alpha}}^{\beta}\delta_{\dot{\alpha}}^{\dot{\beta}}+\delta_{\alpha}^{\beta}{(\gamma_{m})_{\dot{\alpha}}}^{\dot{\beta}}\right]\Omega_{\beta\gamma}\Omega_{\dot{\beta}\dot{\gamma}}\boldsymbol{p}_{B}^{\gamma\dot{\gamma}}\,.$
(259)

Finally, ${(\Gamma^{mn})_{A}}^{B}$ and ${({\Gamma^{m}}_{m})_{A}}^{B}$ matrices
are given by

${(\Gamma^{mn})_{A}}^{B}=\frac{1}{2}\left[{(\Gamma^{m})_{A}}^{C}{(\Gamma^{n})_{C}}^{B}-{(\Gamma^{n})_{A}}^{C}{(\Gamma^{m})_{C}}^{B}\right]$
(260)

and

${({\Gamma^{m}}_{m})_{A}}^{B}=\frac{1}{2}\left[{(\Gamma^{m})_{A}}^{C}{(\Gamma_{m})_{C}}^{B}-{(\Gamma_{m})_{A}}^{C}{(\Gamma^{m})_{C}}^{B}\right]$
(261)

with

$\displaystyle{(\Gamma_{m})_{A}}^{B}$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}\boldsymbol{p}_{A}^{\alpha\dot{\alpha}}\left[{(\gamma_{m})_{\alpha}}^{\beta}\delta_{\dot{\alpha}}^{\dot{\beta}}+\delta_{\alpha}^{\beta}{(\gamma_{m})_{\dot{\alpha}}}^{\dot{\beta}}\right]\boldsymbol{p}^{B}_{\beta\dot{\beta}},$
(262) $\displaystyle{(\Gamma^{m})_{A}}^{B}$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}\boldsymbol{p}_{A}^{\alpha\dot{\alpha}}\left[{(\gamma_{m})_{\alpha}}^{\beta}\delta_{\dot{\alpha}}^{\dot{\beta}}-\delta_{\alpha}^{\beta}{(\gamma_{m})_{\dot{\alpha}}}^{\dot{\beta}}\right]\boldsymbol{p}^{B}_{\beta\dot{\beta}}\,.$
(263)

In this paper, we use the following representation for $SO(5)$ gamma matrices

$\displaystyle\gamma_{1}$ $\displaystyle=$
$\displaystyle-\sigma_{2}\otimes\sigma_{2},\quad\,\gamma_{2}\ =\
\mathds{1}_{2}\otimes\sigma_{1},\quad\gamma_{3}\ =\
\mathds{1}_{2}\otimes\sigma_{3},$ $\displaystyle\gamma_{4}$ $\displaystyle=$
$\displaystyle\sigma_{1}\otimes\sigma_{2},\qquad\gamma_{5}\ =\
\sigma_{3}\otimes\sigma_{2}$ (264)

in which $\sigma_{1}$, $\sigma_{2}$ and $\sigma_{3}$ are the usual Pauli
matrices given by

$\sigma_{1}\ =\ \begin{pmatrix}0&1\\\ 1&0\end{pmatrix},\qquad\sigma_{2}\ =\
\begin{pmatrix}0&-i\\\ i&0\end{pmatrix},\qquad\sigma_{3}\ =\
\begin{pmatrix}1&0\\\ 0&-1\end{pmatrix}\,.$ (265)

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