.1pt

Supersymmetric domain walls in maximal 6D gauged supergravity I

Parinya Karndumria and Patharadanai Nuchinob

String Theory and Supergravity Group, Department of Physics, Faculty of
Science, Chulalongkorn University, 254 Phayathai Road, Pathumwan, Bangkok
10330, Thailand

E-mail<EMAIL_ADDRESS>
E-mail<EMAIL_ADDRESS>

###### Abstract

We find a large class of supersymmetric domain wall solutions from six-
dimensional $N=(2,2)$ gauged supergravity with various gauge groups. In
general, the embedding tensor lives in $\mathbf{144}_{c}$ representation of
the global symmetry $SO(5,5)$. We explicitly construct the embedding tensors
in $\mathbf{15}^{-1}$ and $\overline{\mathbf{40}}^{-1}$ representations of
$GL(5)\sim\mathbb{R}^{+}\times SL(5)\subset SO(5,5)$ leading to
$CSO(p,q,5-p-q)$ and $CSO(p,q,4-p-q)\ltimes\mathbb{R}^{4}_{\boldsymbol{s}}$
gauge groups, respectively. These gaugings can be obtained from $S^{1}$
reductions of seven-dimensional gauged supergravity with $CSO(p,q,5-p-q)$ and
$CSO(p,q,4-p-q)$ gauge groups. As in seven dimensions, we find half-
supersymmetric domain walls for purely magnetic or purely electric gaugings
with the embedding tensors in $\mathbf{15}^{-1}$ or
$\overline{\mathbf{40}}^{-1}$ representations, respectively. In addition, for
dyonic gauge groups with the embedding tensors in both $\mathbf{15}^{-1}$ and
$\overline{\mathbf{40}}^{-1}$ representations, the domain walls turn out to be
$\frac{1}{4}$-supersymmetric as in the seven-dimensional analogue. By the
DW/QFT duality, these solutions are dual to maximal and half-maximal super
Yang-Mills theories in five dimensions. All of the solutions can be uplifted
to seven dimensions and further embedded in type IIB or M-theories by the
well-known consistent truncation of the seven-dimensional $N=4$ gauged
supergravity.

##  1 Introduction

Supersymmetric domain walls in gauged supergravities in various space-time
dimensions have provided a useful tool for studying various aspects of the
AdS/CFT correspondence since the original proposal in [1], see also [2, 3]. In
particular, these solutions play an important role in the so-called DW/QFT
correspondence [4, 5, 6], a generalization of the AdS/CFT correspondence to
non-conformal field theories. They are also useful in studying some aspects of
cosmology, see for example [7, 8, 9]. Due to their importance in many areas of
applications, many domain wall solutions in gauged supergravities have been
found in different space-time dimensions [10, 11, 12, 13, 14, 15, 16, 17, 18,
19, 20, 21, 22, 23, 24, 25]. A systematic classification of supersymmetric
domain walls from maximal gauged supergravity in various space-time dimensions
can also be found in [26].

In this paper, we are interested in maximal $N=(2,2)$ six-dimensional gauged
supergravity with $SO(5,5)$ global symmetry. Compared to other dimensions,
supersymmetric solutions to this six-dimensional gauged supergravity have not
been systematically studied since the original construction of the ungauged
$N=(2,2)$ supergravity long ago in [27]. The first $N=(2,2)$ six-dimensional
gauged supergravity with $SO(5)$ gauge group has been constructed in [28] by
performing an $S^{1}$ reduction of the $SO(5)$ maximal gauged supergravity in
seven dimensions [29]. More recently, the most general gaugings have been
constructed and classified in [30] using the embedding tensor formalism. From
the results of [30], there are two particularly interesting classes of
gaugings under $GL(5)$ and $SO(4,4)$ subgroups of $SO(5,5)$. The former
contains gaugings obtained from an $S^{1}$ reduction of seven-dimensional
maximal gauged supergravity while the latter can be truncated to half-maximal
$N=(1,1)$ gauged supergravity.

We will consider only gaugings in the first class with the embedding tensor in
$\mathbf{15}^{-1}$ and $\overline{\mathbf{40}}^{-1}$ representations of
$GL(5)$. These gaugings have known seven-dimensional origins via an $S^{1}$
reduction and can also be embedded in string/M-theory using the truncations to
maximal gauged supergravity in seven dimensions. The fact that there does not
exist an $N=4$ superconformal symmetry in five dimensions [31] is in agreement
with the recent classification of maximally supersymmetric $AdS$ vacua given
in [32]. This implies that there is no AdS6/CFT5 duality in the case of $32$
supercharges. Therefore, maximally supersymmetric vacuum solutions of the
$N=(2,2)$ gauged supergravity are expected to be half-supersymmetric domain
walls. In this work, we will systematically study this type of solutions and
give a large number of them including $\frac{1}{4}$-supersymmetric solutions.

It has been shown recently that maximally supersymmetric Yang-Mills theory in
five dimensions plays an important role in the dynamics of (conformal) field
theories in both higher and lower dimensions via a number of dualities, see
for example [33, 34, 35, 36, 37, 38]. In particular, this theory could even be
used to define the less known $N=(2,0)$ superconformal field theory in six
dimensions compactified on $S^{1}$. The latter is well-known to describe the
dynamics of strongly coupled theory on M5-branes. Accordingly, we expect that
supersymmetric domain walls of the maximal gauged supergravity in six
dimensions could be useful in studying various aspects of the maximal super
Yang-Mills theory in five dimensions via the DW/QFT correspondence. A simple
domain wall solution with $SO(5)$ symmetry has already been given in [28] for
$SO(5)$ gauging, see [39] and [40] for the holographic interpretation of this
solution. In this paper, we extend this study by including a large class of
supersymmetric domain walls with different unbroken symmetries in $N=(2,2)$
gauged supergravity with various gauge groups.

The paper is organized as follows. In section 2, the construction of six-
dimensional maximal gauged supergravity in the embedding tensor formalism is
reviewed. Supersymmetric domain wall solutions from gaugings in
$\mathbf{15}^{-1}$, $\overline{\mathbf{40}}^{-1}$, and
$(\mathbf{15}+\overline{\mathbf{40}})^{-1}$ representations are respectively
given in sections 3, 4, and 5. Conclusions and discussions are given in
section 6. Branching rules for relevant $SO(5,5)$ representations under
$GL(5)$ are given in appendix A. The conventions on symplectic-Majorana-Weyl
Spinors in six-dimensional space-time used throughout this work are collected
in appendix B. Finally, consistent truncation ansatze for seven-dimensional
$SO(5)$ gauged supergravity on $S^{1}$ giving rise to $SO(5)$ maximal gauged
supergravity in six dimensions are reviewed in appendix C.

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

We begin by giving a brief review of six-dimensional $N=(2,2)$ gauged
supergravity in the embedding tensor formalism constructed in [30]. We will
mainly collect relevant formulae for constructing the embedding tensor and
finding supersymmetric domain wall solutions. For more details, we refer the
reader to the original construction in [30].

As in other dimensions, $N=(2,2)$ maximal supersymmetry in six dimensions
allows only a unique graviton supermultiplet with the following field content

$\left(e^{\hat{\mu}}_{\mu},B_{\mu\nu
m},A^{A}_{\mu},{V_{A}}^{\alpha\dot{\alpha}},\psi_{+\mu\alpha},\psi_{-\mu\dot{\alpha}},\chi_{+a\dot{\alpha}},\chi_{-\dot{a}\alpha}\right).$
(2.1)

Most of the conventions are the same as in [30]. 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 according to this convention, the electric
two-form potentials $B_{\mu\nu m}$ transform as $\mathbf{5}$ under $GL(5)$
while the vector fields $A^{A}_{\mu}$ transform as $\mathbf{16}_{c}$ under
$SO(5,5)$.

Fermionic fields, transforming under the local $SO(5)\times SO(5)$ symmetry,
are symplectic-Majorana-Weyl (SMW) spinors, see appendix B for more detail on
the convention. 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 $(\mathbf{4},\mathbf{5})$.

In ungauged supergravity, only the electric two-forms $B_{\mu\nu m}$ appear in
the Lagrangian while the magnetic duals ${B_{\mu\nu}}^{m}$ transforming in
$\overline{\mathbf{5}}$ representation of $GL(5)$ are introduced on-shell. The
electric and magnetic two-forms are combined into 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. On the
other hand, the full $SO(5,5)$ duality symmetry is the on-shell symmetry
interchanging field equations and Bianchi identities of the two-form
potentials. However, the most general gaugings of the ungauged supergravity
can involve a symmetry that is not a subgroup of the off-shell $GL(5)$
symmetry. Moreover, the magnetic two-forms can also appear in the gauged
Lagrangian via topological terms.

In $N=(2,2)$ supergravity, there are $25$ scalar fields parametrizing the
coset space $SO(5,5)/\left(SO(5)\times SO(5)\right)$. In chiral spinor
representation, we can describe the coset manifold by a coset representative
${V_{A}}^{\alpha\dot{\beta}}$ transforming under the global $SO(5,5)$ and
local $SO(5)\times SO(5)$ 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

${V_{A}}^{\alpha\dot{\beta}}{V^{B}}_{\alpha\dot{\beta}}=\delta^{B}_{A}\qquad\textrm{and}\qquad{V_{A}}^{\alpha\dot{\beta}}{V^{A}}_{\gamma\dot{\delta}}=\delta^{\alpha}_{\gamma}\delta^{\dot{\beta}}_{\dot{\delta}}\,.$
(2.2)

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

$\displaystyle{\mathcal{V}_{M}}^{a}$ $\displaystyle=$
$\displaystyle\frac{1}{16}V^{A\alpha\dot{\alpha}}(\Gamma_{M})_{AB}{(\gamma^{a})_{\alpha\dot{\alpha}}}^{\beta\dot{\beta}}{V^{B}}_{\beta\dot{\beta}},$
(2.3) $\displaystyle{\mathcal{V}_{M}}^{\dot{a}}$ $\displaystyle=$
$\displaystyle-\frac{1}{16}V^{A\alpha\dot{\alpha}}(\Gamma_{M})_{AB}{(\gamma^{\dot{a}})_{\alpha\dot{\alpha}}}^{\beta\dot{\beta}}{V^{B}}_{\beta\dot{\beta}}\,.$
(2.4)

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, see appendix A.3 for more
detail.
The inverse will be denoted by $\mathcal{V}^{M\underline{A}}$ satisfying the
following relations

$\mathcal{V}^{Ma}{\mathcal{V}_{M}}^{b}=\delta^{ab},\qquad\mathcal{V}^{M\dot{a}}{\mathcal{V}_{M}}^{\dot{b}}=\delta^{\dot{a}\dot{b}},\qquad\mathcal{V}^{Ma}{\mathcal{V}_{M}}^{\dot{a}}=0$
(2.5)

and

${\mathcal{V}_{M}}^{a}\mathcal{V}^{Na}-{\mathcal{V}_{M}}^{\dot{a}}\mathcal{V}^{N\dot{a}}=\delta^{N}_{M}\,.$
(2.6)

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 (2.6).

The most general gaugings of six-dimensional $N=(2,2)$ supergravity can be
efficiently described by using the embedding tensor ${\Theta_{A}}^{MN}$. This
tensor introduces the minimal coupling of various fields via the covariant
derivative

$D_{\mu}=\partial_{\mu}-gA^{A}_{\mu}\ {\Theta_{A}}^{MN}\boldsymbol{t}_{MN}$
(2.7)

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 term of a vector-spinor
$\theta^{AM}$ of $SO(5,5)$ as

${\Theta_{A}}^{MN}\ =\ -\theta^{B[M}(\Gamma^{N]})_{BA}\ \equiv\
\left(\Gamma^{[M}\theta^{N]}\right)_{A}$ (2.8)

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

$(\Gamma_{M})_{AB}\,\theta^{BM}\ =\ 0\,.$ (2.9)

With the $SO(5,5)$ generators in vector and spinor representations given by

${(\boldsymbol{t}_{MN})_{P}}^{Q}=4\eta_{P[M}\delta^{Q}_{N]}\qquad\text{and}\qquad{(\boldsymbol{t}_{MN})_{A}}^{B}\
=\ {(\Gamma_{MN})_{A}}^{B}$ (2.10)

in which $\eta_{MN}$ is the off-diagonal $SO(5,5)$ invariant tensor given in
(A.1), the corresponding gauge generators take the forms

${(X_{A})_{M}}^{N}=2\left(\Gamma_{M}\theta^{N}\right)_{A}+2\left(\Gamma^{N}\theta_{M}\right)_{A}\qquad\text{and}\qquad{(X_{A})_{B}}^{C}=\left(\Gamma^{M}\theta^{N}\right)_{A}{(\Gamma_{MN})_{B}}^{C}\,.$
(2.11)

For consistency, the gauge generators must form a closed subalgebra of
$SO(5,5)$, so the embedding tensor needs to satisfy the quadratic constraint

$\left[X_{A},X_{B}\right]\ =\ -{(X_{A})_{B}}^{C}\,X_{C}\,.$ (2.12)

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

$\theta^{AM}\theta^{BN}\eta_{MN}\ =\
0,\qquad\theta^{AM}\theta^{B[N}(\Gamma^{P]})_{AB}=0\,.$ (2.13)

It follows that any $\theta^{AM}\in\mathbf{144}_{c}$ satisfying this quadratic
constraint defines a consistent gauging of the theory.
To identify possible gaugings, we first decompose $\theta^{AM}$ under a given
subgroup of $SO(5,5)$. As pointed out before, the $GL(5)$ subgroup of
$SO(5,5)$ is of particular interest since this is the symmetry of the ungauged
Lagrangian. As given in [30], $\theta^{AM}\in\mathbf{144}_{c}$ decomposes
under $GL(5)\subset SO(5,5)$ as

$\mathbf{144}_{c}\ \rightarrow\
\overline{\mathbf{5}}^{+3}\,\oplus\,\mathbf{5}^{+7}\,\oplus\,\mathbf{10}^{-1}\,\oplus\,\mathbf{15}^{-1}\,\oplus\,\mathbf{24}^{-5}\,\oplus\,\overline{\mathbf{40}}^{-1}\,\oplus\,\overline{\mathbf{45}}^{+3}\,.$
(2.14)

The explicit form of all the seven irreducible components can be found in
appendix A.4. In this case, determining consistent gaugings is to find the
irreducible components satisfying the quadratic constraint (2.13).

By decomposing the $SO(5,5)$ vector index under $GL(5)$, we can write
$\theta^{AM}=(\theta^{Am},\theta^{A}_{m})$ with $\theta^{Am}$ and
$\theta^{A}_{m}$ containing the following irreducible components

$\displaystyle\theta^{Am}$ $\displaystyle=$
$\displaystyle\overline{\mathbf{5}}^{+3}\,\oplus\,\mathbf{10}^{-1}\,\oplus\,\mathbf{24}^{-5}\,\oplus\,\overline{\mathbf{40}}^{-1},$
(2.15) $\displaystyle\theta^{A}_{m}$ $\displaystyle=$
$\displaystyle\overline{\mathbf{5}}^{+3}\,\oplus\,\mathbf{5}^{+7}\,\oplus\,\mathbf{10}^{-1}\,\oplus\,\mathbf{15}^{-1}\,\oplus\,\overline{\mathbf{45}}^{+3}\,.$
(2.16)

It is easily seen that the first equation in (2.13) is automatically satisfied
for purely electric or purely magnetic gaugings that involve only
$\theta^{Am}$ or $\theta^{A}_{m}$ components. We note that as pointed out in
[30], gaugings triggered by $\theta^{Am}$ are electric in the sense that only
electric two-forms participate in the resulting gauged theory while magnetic
gaugings triggered by $\theta^{A}_{m}$ involve magnetic two-forms together
with additional three-form tensor fields. Comparing (2.15) and (2.16) to
(2.14), we immediately see 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 whereas
those in $\overline{\mathbf{5}}^{+3}\,\oplus\,\mathbf{10}^{-1}$ representation
correspond to dyonic gaugings involving both electric and magnetic two-forms.
Other dyonic gaugings can also arise from combinations of various electric and
magnetic components leading to many possible gauge groups.

Apart from the minimal coupling implemented by the covariant derivative (2.7),
gaugings also lead to hierarchies of non-abelian vector and tensor fields of
various ranks. However, since we are only interested in domain wall solutions
which only involve the metric and scalar fields, we will, from now on, set all
vector and tensor fields to zero. It is straightforward to verify that this is
indeed a consistent truncation. With only the metric and scalars non-
vanishing, the bosonic Lagrangian of the maximal $N=(2,2)$ gauged supergravity
takes the form

$e^{-1}\mathcal{L}=\frac{1}{4}R-\frac{1}{16}P_{\mu}^{a\dot{a}}P^{\mu}_{a\dot{a}}-\mathbf{V},$
(2.17)

and supersymmetry transformations of fermionic fields are given by

$\displaystyle\delta\psi_{+\mu\alpha}$ $\displaystyle=$ $\displaystyle
D_{\mu}\epsilon_{+\alpha}+\frac{g}{4}\hat{\gamma}_{\mu}{T_{\alpha}}^{\dot{\beta}}\epsilon_{-\dot{\beta}},$
(2.18) $\displaystyle\delta\psi_{-\mu\dot{\alpha}}$ $\displaystyle=$
$\displaystyle
D_{\mu}\epsilon_{-\dot{\alpha}}-\frac{g}{4}\hat{\gamma}_{\mu}{T^{\beta}}_{\dot{\alpha}}\epsilon_{+\beta},$
(2.19) $\displaystyle\delta\chi_{+a\dot{\alpha}}$ $\displaystyle=$
$\displaystyle\frac{1}{4}P^{\mu}_{a\dot{a}}\hat{\gamma}_{\mu}{(\gamma^{\dot{a}})_{\dot{\alpha}}}^{\dot{\beta}}\epsilon_{-\dot{\beta}}+2g{(T_{a})^{\beta}}_{\dot{\alpha}}\epsilon_{+\beta}-\frac{g}{2}{T^{\alpha}}_{\dot{\alpha}}{(\gamma_{a})_{\alpha}}^{\beta}\epsilon_{+\beta},$
(2.20) $\displaystyle\delta\chi_{-\dot{a}\alpha}$ $\displaystyle=$
$\displaystyle\frac{1}{4}P^{\mu}_{a\dot{a}}\hat{\gamma}_{\mu}{(\gamma^{a})_{\alpha}}^{\beta}\epsilon_{+\beta}+2g{(T_{\dot{a}})_{\alpha}}^{\dot{\beta}}\epsilon_{-\dot{\beta}}+\frac{g}{2}{T_{\alpha}}^{\dot{\alpha}}{(\gamma_{\dot{a}})_{\dot{\alpha}}}^{\dot{\beta}}\epsilon_{-\dot{\beta}}.$
(2.21)

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

$\displaystyle D_{\mu}\epsilon_{+\alpha}$ $\displaystyle=$
$\displaystyle\partial_{\mu}\epsilon_{+\alpha}+\frac{1}{4}{\omega_{\mu}}^{\nu\rho}\hat{\gamma}_{\nu\rho}\epsilon_{+\alpha}+\frac{1}{4}Q_{\mu}^{ab}{(\gamma_{ab})_{\alpha}}^{\beta}\epsilon_{+\beta},$
(2.22) $\displaystyle D_{\mu}\epsilon_{-\dot{\alpha}}$ $\displaystyle=$
$\displaystyle\partial_{\mu}\epsilon_{-\dot{\alpha}}+\frac{1}{4}{\omega_{\mu}}^{\nu\rho}\hat{\gamma}_{\nu\rho}\epsilon_{-\dot{\alpha}}+\frac{1}{4}Q_{\mu}^{\dot{a}\dot{b}}{(\gamma_{\dot{a}\dot{b}})_{\dot{\alpha}}}^{\dot{\beta}}\epsilon_{-\dot{\beta}}$
(2.23)

with $\hat{\gamma}_{\mu}=e_{\mu}^{\hat{\mu}}\hat{\gamma}_{\hat{\mu}}$.
Matrices $\hat{\gamma}_{\hat{\mu}}$ are space-time gamma matrices, see the
convention in appendix B. For simplicity, we will suppress all 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

$\displaystyle P_{\mu}^{a\dot{a}}$ $\displaystyle=$
$\displaystyle\frac{1}{4}{(\gamma^{a})}^{\alpha\beta}{(\gamma^{\dot{a}})}^{\dot{\alpha}\dot{\beta}}{V^{A}}_{\alpha\dot{\alpha}}\partial_{\mu}V_{A\beta\dot{\beta}},$
(2.24) $\displaystyle Q_{\mu}^{ab}$ $\displaystyle=$
$\displaystyle\frac{1}{8}{(\gamma^{ab})}^{\alpha\beta}\Omega^{\dot{\alpha}\dot{\beta}}{V^{A}}_{\alpha\dot{\alpha}}\partial_{\mu}V_{A\beta\dot{\beta}},$
(2.25) $\displaystyle Q_{\mu}^{\dot{a}\dot{b}}$ $\displaystyle=$
$\displaystyle\frac{1}{8}\Omega^{\alpha\beta}{(\gamma^{\dot{a}\dot{b}})}^{\dot{\alpha}\dot{\beta}}{V^{A}}_{\alpha\dot{\alpha}}\partial_{\mu}V_{A\beta\dot{\beta}}$
(2.26)

in which $\Omega^{\alpha\beta}$ and $\Omega^{\dot{\alpha}\dot{\beta}}$ are the
two $USp(4)$ symplectic forms whose explicit forms can be found in (A.23).
These definitions can be derived from the following relation

${V^{A}}_{\alpha\dot{\alpha}}\partial_{\mu}V_{A\beta\dot{\beta}}=\frac{1}{4}P_{\mu}^{a\dot{a}}(\gamma_{a})_{\alpha\beta}(\gamma_{\dot{a}})_{\dot{\alpha}\dot{\beta}}+\frac{1}{4}Q_{\mu}^{ab}(\gamma_{ab})_{\alpha\beta}\Omega_{\dot{\alpha}\dot{\beta}}+\frac{1}{4}Q_{\mu}^{\dot{a}\dot{b}}\Omega_{\alpha\beta}(\gamma_{\dot{a}\dot{b}})_{\dot{\alpha}\dot{\beta}}.$
(2.27)

The scalar potential is given by

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

where we have introduced the T-tensors defined by

$(T^{a})^{\alpha\dot{\alpha}}={\mathcal{V}_{M}}^{a}\theta^{AM}{V_{A}}^{\alpha\dot{\alpha}},\qquad(T^{\dot{a}})^{\alpha\dot{\alpha}}=-{\mathcal{V}_{M}}^{\dot{a}}\theta^{AM}{V_{A}}^{\alpha\dot{\alpha}}$
(2.29)

with

$T^{\alpha\dot{\alpha}}\equiv(T^{a})^{\beta\dot{\alpha}}{(\gamma_{a})_{\beta}}^{\alpha}=-(T^{\dot{a}})^{\alpha\dot{\beta}}{(\gamma_{\dot{a}})_{\dot{\beta}}}^{\dot{\alpha}}.$
(2.30)

##  3 Supersymmetric domain walls from gaugings in $\mathbf{15}^{-1}$
representation

In this section, we consider gauge groups arising from the embedding tensor in
$\mathbf{15}^{-1}$ representation. These are purely magnetic gaugings with the
corresponding embedding tensor given by

$\theta^{A}_{m}\ =\ \mathbb{T}^{An}Y_{nm}\,.$ (3.1)

The matrix $\mathbb{T}^{An}$ is the inverse of the transformation matrix
$\mathbb{T}_{An}$ given in (A.59) and $Y_{mn}$ is a symmetric $5\times 5$
matrix.

As previously mentioned, for $\theta^{Am}=0$, the embedding tensor
$\theta^{AM}=(\,0\,,\,\mathbb{T}^{An}Y_{nm}\,)$ automatically satisfies the
quadratic constraint (2.13). Therefore, every symmetric tensor $Y_{mn}$
defines a viable gauging in $\mathbf{15}^{-1}$ representation. As in [41], we
can use $SL(5)\subset GL(5)$ symmetry to bring $Y_{mn}$ to the form

$Y_{mn}\ =\
\text{diag}(\underbrace{1,..,1}_{p},\underbrace{-1,..,-1}_{q},\underbrace{0,..,0}_{r})$
(3.2)

where $p+q+r=5$.

Under $GL(5)$, the gauge generators transforming as a spinor $\mathbf{16}_{s}$
of $SO(5,5)$ decompose as follows

$X_{A}\ =\
\mathbb{T}_{Am}X^{m}+\mathbb{T}_{A}^{mn}X_{mn}+\mathbb{T}_{A\ast}X_{\ast}\,.$
(3.3)

For the embedding tensor in $\mathbf{15}^{-1}$ representation, the only non-
vanishing gauge generators are given by

$X_{mn}=2Y_{p[m}{\boldsymbol{t}^{p}}_{n]}$ (3.4)

with ${\boldsymbol{t}^{m}}_{n}$ being $GL(5)$ generators. In vector
representation, the explicit form of $X_{mn}$ is given by

${(X_{mn})_{P}}^{Q}\ =\
-4(\delta^{Q}_{[m}Y_{n]p}\delta^{p}_{P}+\eta_{P[m}Y_{n]q}\eta^{qQ}).$ (3.5)

These generators satisfy the commutation relations

$[X_{mn},X_{pq}]={(X_{mn})_{pq}}^{rs}X_{rs}$ (3.6)

in which ${(X_{mn})_{pq}}^{rs}=2{(X_{mn})_{[p}}^{[r}\delta_{q]}^{s]}$.
Therefore, the corresponding gauge group is determined to be

$G_{0}\ =\ CSO(p,q,r)\ =\ SO(p,q)\ltimes\mathbb{R}^{(p+q)\cdot r}.$ (3.7)

These gaugings arise from an $S^{1}$ reduction of seven-dimensional maximal
gauged supergravity with the same gauge groups. In the case of $SO(5)$ gauge
group ($p=5$ and $q=r=0$), the complete reduction ansatz has already been
constructed in [28].

###  3.1 Supersymmetric domain walls

In order to find supersymmetric domain wall solutions, we take the space-time
metric to be the standard domain wall ansatz

$ds_{6}^{2}=e^{2A(r)}\eta_{\bar{\mu}\bar{\nu}}dx^{\bar{\mu}}dx^{\bar{\nu}}+dr^{2}$
(3.8)

where $\bar{\mu},\bar{\nu}=0,1,\ldots,4$, and $A(r)$ is a warped factor
depending only on the radial coordinate $r$. To parametrize the coset
representative of $SO(5,5)/(SO(5)\times SO(5))$, we first identify the
corresponding non-compact generators of $SO(5,5)$ in the basis with diagonal
$SO(5,5)$ metric $\eta_{\underline{A}\underline{B}}$. These are given by

$\hat{\boldsymbol{t}}_{a\dot{b}}\,=\,{\mathbb{M}_{a}}^{M}{\mathbb{M}_{\dot{b}}}^{N}\boldsymbol{t}_{MN}$
(3.9)

where
${\mathbb{M}_{\underline{A}}}^{M}=({\mathbb{M}_{a}}^{M},{\mathbb{M}_{\dot{a}}}^{M})$
is the inverse of the transformation matrix $\mathbb{M}$ given in (A.50).
We then split these generators into two parts that are symmetric and
antisymmetric in $a$ and $\dot{b}$ indices as follows

$\hat{\boldsymbol{t}}_{a\dot{b}}\,=\,\hat{\boldsymbol{t}}^{+}_{a\dot{b}}+\hat{\boldsymbol{t}}^{-}_{a\dot{b}}$
(3.10)

with

$\hat{\boldsymbol{t}}^{+}_{a\dot{b}}\,=\,\frac{1}{2}\left(\hat{\boldsymbol{t}}_{a\dot{b}}+\hat{\boldsymbol{t}}_{b\dot{a}}\right)\qquad\text{
and
}\qquad\hat{\boldsymbol{t}}^{-}_{a\dot{b}}\,=\,\frac{1}{2}\left(\hat{\boldsymbol{t}}_{a\dot{b}}-\hat{\boldsymbol{t}}_{b\dot{a}}\right).$
(3.11)

It is now straightforward to check that symmetric generators
$\hat{\boldsymbol{t}}^{+}_{a\dot{b}}$ are given by
$\frac{1}{2}\left({\boldsymbol{t}^{m}}_{n}+{\boldsymbol{t}^{n}}_{m}\right)$
which are non-compact generators of $GL(5)$. Accordingly, the scalars
corresponding to these generators parametrize the submanifold $GL(5)/SO(5)$.
The antisymmetric generators $\hat{\boldsymbol{t}}^{-}_{a\dot{b}}$ correspond
to the shift generators $\boldsymbol{s}_{mn}$. Therefore, the $25$ non-compact
generators decompose into

$\underbrace{25}_{\hat{\boldsymbol{t}}_{a\dot{b}}}\ \rightarrow\
\underbrace{1+14}_{\hat{\boldsymbol{t}}^{+}_{a\dot{b}}}\,+\underbrace{10}_{\boldsymbol{s}_{mn}}.$
(3.12)

We can also separate the trace part of $\hat{\boldsymbol{t}}^{+}_{a\dot{b}}$,
corresponding to the dilaton scalar field $\varphi$ in
$GL(5)/SO(5)\sim\mathbb{R}^{+}\times SL(5)/SO(5)$ scalar coset. This generator
is the $\mathbb{R}^{+}\sim SO(1,1)$ generator defined in (A.4). In terms of
$\hat{\boldsymbol{t}}^{+}_{a\dot{b}}$, this is given by

$\boldsymbol{d}\,=\,\hat{\boldsymbol{t}}^{+}_{1\dot{1}}+\hat{\boldsymbol{t}}^{+}_{2\dot{2}}+\hat{\boldsymbol{t}}^{+}_{3\dot{3}}+\hat{\boldsymbol{t}}^{+}_{4\dot{4}}+\hat{\boldsymbol{t}}^{+}_{5\dot{5}}.$
(3.13)

The remaining generators can be identified as the fourteen non-compact
generators corresponding to scalar fields $\\{\phi_{1},...,\phi_{14}\\}$ in
the $SL(5)/SO(5)$ coset. These generators are given by the symmetric traceless
part

$\tilde{\boldsymbol{t}}_{a\dot{b}}\,=\,\hat{\boldsymbol{t}}^{+}_{a\dot{b}}-\frac{1}{5}\boldsymbol{d}\,\delta_{a\dot{b}}$
(3.14)

satisfying $\delta^{a\dot{b}}\tilde{\boldsymbol{t}}_{a\dot{b}}=0$.
The other ten scalars denoted by $\\{\varsigma_{1},...,\varsigma_{10}\\}$
correspond to the shift generators $\boldsymbol{s}_{mn}$. These will be called
the axions or shift scalars in this work. The decomposition in equation (3.12)
is in agreement with that in [28] in which the consistent circle reduction of
seven-dimensional $SO(5)$ gauged supergravity giving rise to $SO(5)$ gauged
theory in six dimensions is performed. From a higher-dimensional perspective,
the fourteen scalars are the seven-dimensional scalars parameterizing the
$SL(5)/SO(5)$ coset in seven dimensions while the dilaton and shift scalars
descend from the reduction of seven-dimensional metric and vector fields,
respectively, see appendix C for more detail.
By this decomposition of the scalar fields, we can rewrite the kinetic terms
of the scalars in (2.17) and obtain the following bosonic Lagrangian

$e^{-1}\mathcal{L}=\frac{1}{4}R-G_{IJ}\partial_{\mu}\Phi^{I}\partial^{\mu}\Phi^{J}-\mathbf{V}$
(3.15)

in which $G_{IJ}$ is a symmetric matrix depending on scalar fields denoted by
$\Phi^{I}=\\{\varphi,\phi_{1},\ldots,\phi_{14},\varsigma_{1},\ldots,\varsigma_{10}\\}$
with $I,J=1,\ldots,25$.
In order to find supersymmetric solutions, we consider first-order BPS
equations derived from the supersymmetry transformations of fermionic fields
in the background with vanishing fermionic fields. In this section, we only
discuss a general structure of the procedure leaving a more detailed analysis
and explicit results in subsequent sections. We begin with the variations of
the gravitini which are given by

$\displaystyle\delta\psi_{+\bar{\mu}\alpha}$ $\displaystyle:$
$\displaystyle\qquad
A^{\prime}\hat{\gamma}_{r}\epsilon_{+\alpha}+\frac{1}{2}gT_{\alpha\dot{\alpha}}\epsilon_{-}^{\dot{\alpha}}=0,$
(3.16) $\displaystyle\delta\psi_{-\bar{\mu}\dot{\alpha}}$ $\displaystyle:$
$\displaystyle\qquad
A^{\prime}\hat{\gamma}_{r}\epsilon_{-\dot{\alpha}}-\frac{1}{2}gT_{\dot{\alpha}\alpha}\epsilon_{+}^{\alpha}=0\,.$
(3.17)

In these equations, we have used the notation $A^{\prime}=\frac{dA}{dr}$. We
will use a prime to denote an $r$-derivative throughout the paper.
Multiply the first equation by $A^{\prime}\hat{\gamma}_{r}$ and use the second
equation or vice-versa, we find the following consistency conditions

$\displaystyle{A^{\prime}}^{2}{\delta_{\alpha}}^{\beta}$ $\displaystyle=$
$\displaystyle\frac{1}{4}g^{2}T_{\alpha\dot{\alpha}}\Omega^{\dot{\alpha}\dot{\beta}}T_{\gamma\dot{\beta}}\Omega^{\beta\gamma}=\mathcal{W}^{2}{\delta_{\alpha}}^{\beta},$
(3.18) $\displaystyle{A^{\prime}}^{2}{\delta_{\dot{\alpha}}}^{\dot{\beta}}$
$\displaystyle=$
$\displaystyle\frac{1}{4}g^{2}T_{\dot{\alpha}\alpha}\Omega^{\alpha\beta}T_{\beta\dot{\gamma}}\Omega^{\dot{\beta}\dot{\gamma}}=\mathcal{W}^{2}{\delta_{\dot{\alpha}}}^{\dot{\beta}}$
(3.19)

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

$A^{\prime}=\pm\mathcal{W}\,.$ (3.20)

Using this result in equations (3.16) and (3.17) leads to the following
projectors on the Killing spinors

$\hat{\gamma}_{r}\epsilon_{+\alpha}=P_{\alpha\dot{\alpha}}\epsilon^{\dot{\alpha}}_{-}\qquad\textrm{and}\qquad\hat{\gamma}_{r}\epsilon_{-\dot{\alpha}}=P_{\dot{\alpha}\alpha}\epsilon^{\alpha}_{+}$
(3.21)

with

$P_{\alpha\dot{\alpha}}=-\frac{1}{2}g\frac{T_{\alpha\dot{\alpha}}}{A^{\prime}}\qquad\textrm{and}\qquad
P_{\dot{\alpha}\alpha}=\frac{1}{2}g\frac{T_{\dot{\alpha}\alpha}}{A^{\prime}}$
(3.22)

satisfying
${P_{\alpha}}^{\dot{\alpha}}{P_{\dot{\alpha}}}^{\beta}={\delta_{\alpha}}^{\beta}$
and
${P_{\dot{\alpha}}}^{\alpha}{P_{\alpha}}^{\dot{\beta}}={\delta_{\dot{\alpha}}}^{\dot{\beta}}$.
The conditions $\delta\psi_{+r\alpha}=0$ and $\delta\psi_{-r\dot{\alpha}}=0$
determine the Killing spinors as functions of the radial coordinate $r$ as
usual.
Using these projectors in $\delta\chi_{+a\dot{\alpha}}=0$ and
$\delta\chi_{-\dot{a}\alpha}=0$ equations, we eventually obtain the BPS
equations for scalars. These equations are of the form

${\Phi^{I}}^{\prime}=\mp 2G^{IJ}\frac{\partial\mathcal{W}}{\partial\Phi^{J}}$
(3.23)

in which $G^{IJ}$ is the inverse of the scalar metric $G_{IJ}$ defined in
(3.15).
In addition, the scalar potential can also be written in term of $\mathcal{W}$
as

$\mathbf{V}=2G^{IJ}\frac{\partial\mathcal{W}}{\partial\Phi^{I}}\frac{\partial\mathcal{W}}{\partial\Phi^{J}}-5\mathcal{W}^{2}\,.$
(3.24)

It is well-known that the BPS equations of the form (3.20) and (3.23) satisfy
the second-order field equations derived from the bosonic Lagrangian (3.15)
with the scalar potential given by (3.24), see [42, 43, 44, 45, 46, 47] for
more detail.
As in other dimensions, we will follow the approach introduced in [48] to
explicitly find supersymmetric domain wall solutions involving only a subset
of the $25$ scalars that is invariant under a particular subgroup
$H_{0}\subset G_{0}$ to make the analysis more traceable.

###  3.2 $SO(5)$ symmetric domain walls

We first consider supersymmetric domain walls with the maximal unbroken
symmetry $SO(5)\subset CSO(p,q,5-p-q)$. The only gauge group containing
$SO(5)$ as a subgroup is $SO(5)$ with $Y_{mn}=\delta_{mn}$. In this case, only
the dilaton $\varphi$ corresponding to the non-compact generator (3.13) is
invariant under $SO(5)$. Thus, the coset representative can be written as

$V=e^{\varphi\boldsymbol{d}}\,.$ (3.25)

We recall that this coset representative is a $16\times 16$ matrix with an
index structure ${V_{A}}^{B}$. To compute the T-tensor, we need to write the
$SO(5)\times SO(5)$ index as a pair of $SO(5)$ spinor indices resulting in the
coset representative of the form ${V_{A}}^{\alpha\dot{\alpha}}$. To achieve
this, we use the transformation matrices $\boldsymbol{p}$ introduced in (A.3)
so that ${V_{A}}^{\alpha\dot{\alpha}}$ and its inverse
${V^{A}}_{\alpha\dot{\alpha}}$ are given by

${V_{A}}^{\alpha\dot{\alpha}}={V_{A}}^{B}{\boldsymbol{p}_{B}}^{\alpha\dot{\alpha}}\qquad\text{and}\qquad{V^{A}}_{\alpha\dot{\alpha}}={(V^{-1})_{B}}^{A}{\boldsymbol{p}^{B}}_{\alpha\dot{\alpha}}\,.$
(3.26)

With all these, it is now straightforward to find the T-tensor

$T^{\alpha\dot{\beta}}=\frac{5}{2\sqrt{2}}e^{\varphi}\,\Omega^{\alpha\beta}\delta^{\dot{\beta}}_{\beta}=\frac{2}{g}\mathcal{W}\,\Omega^{\alpha\beta}\delta^{\dot{\beta}}_{\beta}$
(3.27)

from which the superpotential is given by

$\mathcal{W}=\frac{5g}{4\sqrt{2}}e^{\varphi}.$ (3.28)

The resulting scalar potential reads

$\mathbf{V}=-\frac{15g^{2}}{4}e^{2\varphi}$ (3.29)

which does not admit any stationary points.
The general analysis given above leads to the BPS equation for the warped
factor

$A^{\prime}=\frac{5g}{4\sqrt{2}}e^{\varphi}$ (3.30)

and the following projector

$\hat{\gamma}_{r}\epsilon_{\pm}=\epsilon_{\mp}\,.$ (3.31)

For definiteness, we have chosen a particular sign choice in the $A^{\prime}$
equation and the $\hat{\gamma}_{r}$ projector. The condition $\delta\psi_{\pm
r}=0$ gives the standard solution for the Killing spinors

$\epsilon_{\pm}=e^{\frac{A(r)}{2}}\epsilon_{\pm}^{0}$ (3.32)

with the constant spinors $\epsilon_{\pm}^{0}$ satisfying
$\hat{\gamma}_{r}\epsilon_{\pm}^{0}=\epsilon_{\mp}^{0}$. Accordingly, the
solution is half-supersymmetric.
The BPS equation for the dilaton can be found from the condition
$\delta\chi_{\pm}=0$ with the projector (3.31). This results in a simple
equation

$\varphi^{\prime}=-\frac{g}{4\sqrt{2}}e^{\varphi}\,.$ (3.33)

All of these equations can be readily solved to obtain the solution

$A=5\ln\left(\frac{gr}{4\sqrt{2}}-C\right)\qquad\textrm{and}\qquad\varphi=-\ln\left(\frac{gr}{4\sqrt{2}}-C\right).$
(3.34)

The integration constant $C$ can be removed by shifting the radial coordinate
$r$. We have also neglected an additive integration constant for $A$ since it
can be absorbed by rescaling the coordinates $x^{\bar{\mu}}$. This is the
$SO(5)$ domain wall originally found in [28]. In order to recover the same
form of the solution, we redefine the radial coordinate as
$r\rightarrow\frac{4\sqrt{2}}{g}\left[C+(3\sqrt{2}gr+C)^{-\frac{1}{24}}\right]$
and set $\varphi=\frac{1}{2\sqrt{10}}\sigma$.

###  3.3 $SO(4)$ symmetric domain walls

We now look for more complicated solutions with $SO(4)$ symmetry. The gauge
groups that contain $SO(4)$ as a subgroup are $SO(5)$, $SO(4,1)$, and
$CSO(4,0,1)$. To incorporate all of these gauge groups within a single
framework, we write the embedding tensor in the form

$Y_{mn}=\text{diag}(1,1,1,1,\kappa)$ (3.35)

with $\kappa=1,0,-1$ corresponding to $SO(5)$, $CSO(4,0,1)$, and $SO(4,1)$
gauge groups, respectively.

There are two $SO(4)$ singlet scalars. The first one is the dilaton
corresponding to the non-compact generator (3.13), and the other one comes
from the $SL(5)/SO(5)$ coset corresponding to the non-compact generator

$\mathcal{Y}=\hat{\boldsymbol{t}}^{+}_{1\dot{1}}+\hat{\boldsymbol{t}}^{+}_{2\dot{2}}+\hat{\boldsymbol{t}}^{+}_{3\dot{3}}+\hat{\boldsymbol{t}}^{+}_{4\dot{4}}-4\,\hat{\boldsymbol{t}}^{+}_{5\dot{5}}.$
(3.36)

Using the coset representative

$V=e^{\varphi\boldsymbol{d}+\phi\mathcal{Y}}\,,$ (3.37)

we find that the T-tensor is given by

$T^{\alpha\dot{\beta}}=\frac{1}{2\sqrt{2}}e^{\varphi-4\phi}(4+\kappa
e^{20\phi})\,\Omega^{\alpha\beta}\delta^{\dot{\beta}}_{\beta}=\frac{2}{g}\mathcal{W}\,\Omega^{\alpha\beta}\delta^{\dot{\beta}}_{\beta}\,.$
(3.38)

This leads to the superpotential and the scalar potential of the form

$\displaystyle\mathcal{W}$ $\displaystyle=$
$\displaystyle\frac{g}{4\sqrt{2}}e^{\varphi-4\phi}(4+\kappa e^{20\phi}),$
(3.39) $\displaystyle\mathbf{V}$ $\displaystyle=$
$\displaystyle-\frac{g^{2}}{4}e^{2\varphi-8\phi}\left(8+8\kappa
e^{20\phi}-\kappa^{2}e^{40\phi}\right).$ (3.40)

Using the projector (3.31), we find the BPS equations

$\displaystyle A^{\prime}$ $\displaystyle=$
$\displaystyle\frac{g}{4\sqrt{2}}e^{\varphi-4\phi}(4+\kappa e^{20\phi}),$
(3.41) $\displaystyle\varphi^{\prime}$ $\displaystyle=$
$\displaystyle-\frac{g}{20\sqrt{2}}e^{\varphi-4\phi}(4+\kappa e^{20\phi}),$
(3.42) $\displaystyle\phi^{\prime}$ $\displaystyle=$
$\displaystyle\frac{g}{5\sqrt{2}}e^{\varphi-4\phi}(1-\kappa e^{20\phi}).$
(3.43)

The resulting solutions for the dilation $\varphi$ and the warped factor $A$
as functions of $\phi$ are given by

$\displaystyle\varphi$ $\displaystyle=$
$\displaystyle-\phi+C+\frac{1}{16}\ln\left(1-\kappa e^{20\phi}\right),$ (3.44)
$\displaystyle A$ $\displaystyle=$
$\displaystyle-5\varphi\,=\,5\phi-5C-\frac{5}{16}\ln\left(1-\kappa
e^{20\phi}\right).$ (3.45)

To obtain the solution for $\phi$, we change $r$ to a new radial coordinate
$\rho$ defined by $\frac{d\rho}{dr}=e^{\varphi+6\phi}$. The solution for
$\phi$ is then given by

$e^{10\phi}=\frac{1}{\sqrt{\kappa}}\tanh\left[\sqrt{\kappa}(\sqrt{2}g\rho+C_{1})\right]$
(3.46)

for an integration constant $C_{1}$. It is useful to note that for
$\kappa=-1$, the solution for $\phi$ can be written as

$e^{10\phi}=\tan\left[\sqrt{2}g\rho+C_{1}\right].$ (3.47)

For $\kappa=0$, the solution is simply given by

$e^{10\phi}=\sqrt{2}g\rho+C_{1}\,.$ (3.48)

###  3.4 $SO(3)\times SO(2)$ symmetric domain walls

We now consider $SO(3)\times SO(2)$ residual symmetry, which is possible only
for $SO(5)$ and $SO(3,2)$ gauge groups. In this case, we write the embedding
tensor as

$Y_{mn}=\textrm{diag}(1,1,1,\kappa,\kappa)$ (3.49)

with $\kappa=1$ and $\kappa=-1$ corresponding to $SO(5)$ and $SO(3,2)$,
respectively. The $SO(3)\times SO(2)$ symmetry is generated by $X_{ij}$,
$i,j=1,2,3$, and $X_{45}$. There are three singlet scalars corresponding to
the dilaton and the following non-compact generators

$\mathcal{Y}_{1}=2\,\hat{\boldsymbol{t}}^{+}_{1\dot{1}}+2\,\hat{\boldsymbol{t}}^{+}_{2\dot{2}}+2\,\hat{\boldsymbol{t}}^{+}_{3\dot{3}}-3\,\hat{\boldsymbol{t}}^{+}_{4\dot{4}}-3\,\hat{\boldsymbol{t}}^{+}_{5\dot{5}},\qquad\mathcal{Y}_{2}=\boldsymbol{s}_{45}\,.$
(3.50)

With the coset representative

$V=e^{\varphi\boldsymbol{d}+\phi\mathcal{Y}_{1}+\varsigma\mathcal{Y}_{2}},$
(3.51)

we find the scalar potential

$\mathbf{V}=-\frac{3g^{2}}{4}e^{2(\varphi-8\phi)}(1+4\kappa e^{20\phi}).$
(3.52)

The superpotential reads

$\mathcal{W}=\frac{g}{4\sqrt{2}}e^{\varphi-8\phi}\sqrt{(3+2\kappa
e^{20\phi})^{2}+8\kappa^{2}\varsigma^{2}e^{40\phi}}$ (3.53)

which can be found from the T-tensor given by

$T^{\alpha\dot{\beta}}=\frac{1}{2\sqrt{2}}e^{\varphi-8\phi}(3+2\kappa
e^{20\phi})\,\Omega^{\alpha\beta}\delta^{\dot{\beta}}_{\beta}-\sqrt{2}\kappa\varsigma
e^{\varphi+12\phi}\delta^{\alpha\dot{\beta}}.$ (3.54)

In this case, it turns out that consistency of the supersymmetry conditions
from $\delta\chi_{\pm}$ requires $\varsigma=0$. Therefore, in order to find a
consistent set of BPS equations, we need to truncate the axion out. With
$\varsigma=0$, the superpotential is given by

$\mathcal{W}=\frac{g}{4\sqrt{2}}e^{\varphi-8\phi}(3+2\kappa e^{20\phi}).$
(3.55)

With the projector (3.31), we find the following BPS equations

$\displaystyle A^{\prime}$ $\displaystyle=$
$\displaystyle\frac{g}{4\sqrt{2}}e^{\varphi-8\phi}(3+2\kappa e^{20\phi}),$
(3.56) $\displaystyle\varphi^{\prime}$ $\displaystyle=$
$\displaystyle-\frac{g}{20\sqrt{2}}e^{\varphi-8\phi}(3+2\kappa e^{20\phi}),$
(3.57) $\displaystyle\phi^{\prime}$ $\displaystyle=$
$\displaystyle\frac{g}{5\sqrt{2}}e^{\varphi-8\phi}(1-\kappa e^{20\phi}).$
(3.58)

It can be verified that all these equations satisfy the corresponding field
equations as expected.

With a new radial coordinate $\rho$ given by
$\frac{d\rho}{dr}=e^{\varphi+2\phi}$, we obtain the domain wall solution

$\displaystyle\varphi$ $\displaystyle=$
$\displaystyle-\frac{3\phi}{4}+C+\frac{1}{16}\ln\left(1-\kappa
e^{20\phi}\right),$ (3.59) $\displaystyle A$ $\displaystyle=$
$\displaystyle-5\varphi\,=\,\frac{15\phi}{4}-5C-\frac{5}{16}\ln\left(1-\kappa
e^{20\phi}\right),$ (3.60) $\displaystyle e^{10\phi}$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{\kappa}}\tanh\left[\sqrt{\kappa}(\sqrt{2}g\rho+C_{1})\right].$
(3.61)

###  3.5 $SO(3)$ symmetric domain walls

We now move to domain wall solutions with $SO(3)$ symmetry. Many gauge groups
contain $SO(3)$ as a subgroup with the embedding tensor parameterized by

$Y_{mn}=\text{diag}(1,1,1,\kappa,\lambda)$ (3.62)

for $\kappa,\lambda=0,\pm 1$. With this embedding tensor, the $SO(3)$ symmetry
is generated by $X_{mn}$, $m,n=1,2,3$. In addition to the dilaton, there are
four singlet scalars corresponding to the following non-compact generators

$\displaystyle\mathcal{Y}_{1}$ $\displaystyle=$ $\displaystyle
2\,\hat{\boldsymbol{t}}^{+}_{1\dot{1}}+2\,\hat{\boldsymbol{t}}^{+}_{2\dot{2}}+2\,\hat{\boldsymbol{t}}^{+}_{3\dot{3}}-3\,\hat{\boldsymbol{t}}^{+}_{4\dot{4}}-3\,\hat{\boldsymbol{t}}^{+}_{5\dot{5}},\qquad\mathcal{Y}_{2}\,=\,\hat{\boldsymbol{t}}^{+}_{4\dot{5}},$
$\displaystyle\mathcal{Y}_{3}$ $\displaystyle=$
$\displaystyle\hat{\boldsymbol{t}}^{+}_{4\dot{4}}-\hat{\boldsymbol{t}}^{+}_{5\dot{5}},\qquad\quad\mathcal{Y}_{4}\,=\,\boldsymbol{s}_{45}\,.$
(3.63)

With the only exception for $\kappa=\lambda=0$ corresponding to $CSO(3,0,2)$
gauge group, we need to truncate out the scalar corresponding to
$\boldsymbol{s}_{45}$ generator in order to find a consistent set of BPS
equations as in the previous case. For the moment, we will set this shift
scalar to zero and consider the particular case of $\kappa=\lambda=0$
afterward.

For vanishing shift scalars, the coset representative is given by

$V=e^{\varphi\boldsymbol{d}+\phi_{1}\mathcal{Y}_{1}+\phi_{2}\mathcal{Y}_{2}+\phi_{3}\mathcal{Y}_{3}}$
(3.64)

giving rise to the superpotential and the scalar potential of the form

$\displaystyle\mathcal{W}$ $\displaystyle=$
$\displaystyle\frac{ge^{\varphi-8\phi_{1}}}{4\sqrt{2}}\left[3+e^{20\phi_{1}}\left((\kappa+\lambda)\cosh{2\phi_{2}}\cosh{4\phi_{3}}-(\kappa-\lambda)\sinh{4\phi_{3}}\right)\right],\qquad$
(3.65) $\displaystyle\mathbf{V}$ $\displaystyle=$
$\displaystyle-\frac{g^{2}e^{2(\varphi-8\phi_{1})}}{4}\left[3+6e^{20\phi_{1}}\left((\kappa+\lambda)\cosh{2\phi_{2}}\cosh{4\phi_{3}}-(\kappa-\lambda)\sinh{4\phi_{3}}\right)\phantom{\frac{1}{2}}\right.$
(3.66)
$\displaystyle\left.+\frac{e^{40\phi_{1}}}{4}\left(\kappa^{2}+10\kappa\lambda+\lambda^{2}-(3\kappa^{2}-2\kappa\lambda+3\lambda^{2})\cosh{8\phi_{3}}\phantom{\frac{1}{2}}\right.\right.$
$\displaystyle\left.\left.\phantom{\frac{1}{2}}-2(\kappa+\lambda)^{2}\cosh{4\phi_{2}}\cosh^{2}{4\phi_{3}}+4(\kappa^{2}-\lambda^{2})\cosh{2\phi_{2}}\sinh{8\phi_{3}}\right)\right].$

We also note the matrix $G^{IJ}$ in this case

$G^{IJ}=\frac{1}{60}\text{diag}(6,1,60\text{sech}^{2}{4\phi_{3}},15)$ (3.67)

for $\Phi_{I}=\\{\varphi,\phi_{1},\phi_{2},\phi_{3}\\}$ with $I,J=1,2,3,4$.

In this case, the Killing spinors are different from the ansatz given in
(3.32) due to the non-vanishing composite connections $Q_{r}^{45}$ and
$Q_{r}^{\dot{4}\dot{5}}$ appearing in $\delta\psi_{\pm r}=0$ conditions. In
more detail, there are additional terms involving
${(\gamma_{45})_{\alpha}}^{\beta}\epsilon_{+\beta}$ and
${(\gamma_{\dot{4}\dot{5}})_{\dot{\alpha}}}^{\dot{\beta}}\epsilon_{-\dot{\beta}}$
in the covariant derivative of the supersymmetry parameters, see equations
(2.22) and (2.23). According to this, we modify the ansatz for the Killing
spinors to

$\epsilon_{+}=e^{\frac{A(r)}{2}+B(r)\gamma_{45}}\epsilon_{+}^{0}\qquad\text{
and
}\qquad\epsilon_{-}=e^{\frac{A(r)}{2}+B(r)\gamma_{\dot{4}\dot{5}}}\epsilon_{-}^{0}$
(3.68)

where $B(r)$ is an $r$-dependent function, and $\epsilon_{\pm}^{0}$ are
constant symplectic-Majorana-Weyl spinors satisfying
$\hat{\gamma}_{r}\epsilon_{\pm}^{0}=\epsilon_{\mp}^{0}$.

Using this ansatz for the Killing spinors satisfying the projector (3.31), we
find the following set of BPS equations from the supersymmetry transformations
of fermions

$\displaystyle A^{\prime}$ $\displaystyle=$
$\displaystyle\frac{ge^{\varphi-8\phi_{1}}}{4\sqrt{2}}\left[3+e^{20\phi_{1}}\left((\kappa+\lambda)\cosh{2\phi_{2}}\cosh{4\phi_{3}}-(\kappa-\lambda)\sinh{4\phi_{3}}\right)\right],$
(3.69) $\displaystyle\varphi^{\prime}$ $\displaystyle=$
$\displaystyle-\frac{ge^{\varphi-8\phi_{1}}}{20\sqrt{2}}\left[3+e^{20\phi_{1}}\left((\kappa+\lambda)\cosh{2\phi_{2}}\cosh{4\phi_{3}}-(\kappa-\lambda)\sinh{4\phi_{3}}\right)\right],\qquad\
\,$ (3.70) $\displaystyle\phi^{\prime}_{1}$ $\displaystyle=$
$\displaystyle\frac{ge^{\varphi-8\phi_{1}}}{10\sqrt{2}}\left[2-e^{20\phi_{1}}\left((\kappa+\lambda)\cosh{2\phi_{2}}\cosh{4\phi_{3}}-(\kappa-\lambda)\sinh{4\phi_{3}}\right)\right],$
(3.71) $\displaystyle\phi^{\prime}_{2}$ $\displaystyle=$
$\displaystyle-\frac{g}{\sqrt{2}}e^{\varphi+12\phi_{1}}(\kappa+\lambda)\sinh{2\phi_{2}}\,\text{sech
}{4\phi_{3}},$ (3.72) $\displaystyle\phi^{\prime}_{3}$ $\displaystyle=$
$\displaystyle\frac{ge^{\varphi+12\phi_{1}}}{2\sqrt{2}}\left((\kappa-\lambda)\cosh{4\phi_{3}}-(\kappa+\lambda)\cosh{2\phi_{2}}\sinh{4\phi_{3}}\right)$
(3.73)

together with

$B^{\prime}=-\frac{g}{2\sqrt{2}}e^{\varphi+12\phi_{1}}(\kappa+\lambda)\sinh{2\phi_{2}}\tanh{4\phi_{3}}.$
(3.74)

To find explicit solutions, we will separately discuss various possible values
of $\kappa$ and $\lambda$.

####  3.5.1 Domain walls in $CSO(4,0,1)$ and $CSO(3,1,1)$ gauge groups

For $\lambda=0$ and $\kappa\neq 0$, the gauge groups are given by $CSO(4,0,1)$
and $CSO(3,1,1)$ for $\kappa=1$ and $\kappa=-1$, respectively. Using a new
radial coordinate $\rho$ defined by $\frac{d\rho}{dr}=e^{\varphi+12\phi_{1}}$,
a domain wall solution to the BPS equations can be obtained

$\displaystyle\phi_{2}$ $\displaystyle=$
$\displaystyle\frac{1}{4}\ln\left[\frac{g^{2}\rho^{2}(1+2C_{3})^{2}+2(1+C_{3})^{2}}{g^{2}\rho^{2}(1+2C_{3})^{2}+2C_{3}^{2}}\right],$
(3.75) $\displaystyle\phi_{3}$ $\displaystyle=$
$\displaystyle\frac{1}{8}\ln\left[\frac{e^{2\phi_{2}}-C_{3}e^{4\phi_{2}}+C_{3}+1}{e^{2\phi_{2}}+C_{3}e^{4\phi_{2}}-C_{3}-1}\right],$
(3.76) $\displaystyle\phi_{1}$ $\displaystyle=$
$\displaystyle\frac{1}{20}\ln\left[\frac{\kappa\left(2+C_{1}(e^{4\phi_{2}}-1)\right)}{\sqrt{(1-e^{4\phi_{2}})\left(C_{3}^{2}e^{4\phi_{2}}-(C_{3}+1)^{2}\right)}}\right],$
(3.77) $\displaystyle\varphi$ $\displaystyle=$
$\displaystyle\frac{\phi_{1}}{2}+C-\frac{1}{16}\ln\left[\frac{C_{1}(e^{4\phi_{2}}-1)+2}{e^{4\phi_{2}}-1}\right],$
(3.78) $\displaystyle A$ $\displaystyle=$
$\displaystyle-5\varphi\,=\,-\frac{5\phi_{1}}{2}-5C+\frac{5}{16}\ln\left[\frac{C_{1}(e^{4\phi_{2}}-1)+2}{e^{4\phi_{2}}-1}\right]$
(3.79)

together with

$B=\frac{1}{4}\sin^{-1}\left[C_{3}\sqrt{\frac{e^{4\phi_{2}}-1}{2C_{3}+1}}\,\right]+\frac{1}{4}\tan^{-1}\left[\sqrt{\frac{(1-e^{4\phi_{2}})(C_{3}+1)^{2}}{C_{3}^{2}e^{4\phi_{2}}-(C_{3}+1)^{2}}}\,\right].$
(3.80)

We have chosen integration constants for $\phi_{2}$ and $B$ to be zero for
simplicity.

####  3.5.2 Domain walls in $SO(4,1)$ gauge group

In $SO(4,1)$ gauge group with $\kappa=-\lambda=1$, the BPS equations give
$\phi_{2}^{\prime}=B^{\prime}=0$. Accordingly, we can set $B=0$ and
$\phi_{2}=0$. We can readily verify that this is a consistent truncation.
Taking $\phi_{2}=0$ and redefining the radial coordinate $r$ to $\rho$ as
given in the $CSO(4,0,1)$ and $CSO(3,1,1)$ gauge groups, we obtain a domain
wall solution

$\displaystyle\phi_{3}$ $\displaystyle=$
$\displaystyle\frac{1}{2}\tanh^{-1}\left[\tan\left[\sqrt{2}g\rho+C_{3}\right]\right],$
(3.81) $\displaystyle\phi_{1}$ $\displaystyle=$
$\displaystyle\frac{1}{20}\ln\left[e^{4\phi_{3}}(C_{1}+1)+C_{1}e^{-4\phi_{3}}\right],$
(3.82) $\displaystyle\varphi$ $\displaystyle=$
$\displaystyle\frac{\phi_{1}}{2}+C+\frac{1}{16}\ln\left[\frac{e^{8\phi_{3}}+1}{2\left(C_{1}+e^{8\phi_{3}}(C_{1}+1)\right)}\right],$
(3.83) $\displaystyle A$ $\displaystyle=$
$\displaystyle-5\varphi\,=\,-\frac{5\phi_{1}}{2}-5C-\frac{5}{16}\ln\left[\frac{e^{8\phi_{3}}+1}{2\left(C_{1}+e^{8\phi_{3}}(C_{1}+1)\right)}\right].$
(3.84)

####  3.5.3 Domain walls in $SO(5)$ and $CSO(3,2)$ gauge groups

For $\kappa=\lambda=\pm 1$ corresponding to $SO(5)$ and $SO(3,2)$ gauge
groups. we find the following domain wall solution

$\displaystyle\phi_{2}$ $\displaystyle=$
$\displaystyle\frac{1}{4}\ln\left[\frac{1-2e^{2\sqrt{2}g\kappa\rho}+e^{4\sqrt{2}g\kappa\rho}+e^{4\sqrt{2}g\kappa\rho+2C_{3}}}{1+2e^{2\sqrt{2}g\kappa\rho}+e^{4\sqrt{2}g\kappa\rho}+e^{4\sqrt{2}g\kappa\rho+2C_{3}}}\right],$
(3.85) $\displaystyle\phi_{3}$ $\displaystyle=$
$\displaystyle\frac{1}{8}\ln\left[\frac{2e^{2\phi_{2}}+e^{4\phi_{2}+C_{3}}-e^{C_{3}}}{2e^{2\phi_{2}}-e^{4\phi_{2}+C_{3}}+e^{C_{3}}}\right],$
(3.86) $\displaystyle\phi_{1}$ $\displaystyle=$
$\displaystyle\frac{\phi_{2}}{10}+\frac{1}{20}\ln\left[\frac{\kappa
e^{-2\phi_{2}}\left(C_{1}(e^{4\phi_{2}}-1)-2\right)}{\sqrt{4e^{4\phi_{2}}-e^{2C_{3}}(e^{4\phi_{2}}-1)^{2}}}\right],$
(3.87) $\displaystyle\varphi$ $\displaystyle=$
$\displaystyle\frac{\phi_{1}}{2}+C-\frac{1}{16}\ln\left[e^{4\phi_{2}}-1\right]+\frac{1}{16}\ln[C_{1}(e^{4\phi_{2}}-1)-2],$
(3.88) $\displaystyle A$ $\displaystyle=$
$\displaystyle-5\varphi\,=\,-\frac{5\phi_{1}}{2}-5C+\frac{5}{16}\ln\left[e^{4\phi_{2}}-1\right]+\frac{1}{16}\ln[C_{1}(e^{4\phi_{2}}-1)-2]\qquad$
(3.89)

in terms of the new radial coordinate $\rho$ defined previously. The function
$B(r)$ appearing in the Killing spinors is given in term of $\phi_{2}$ as

$B=-\frac{1}{8}\tan^{-1}\left[\frac{1-e^{4\phi_{2}}+2e^{-2C_{3}}}{\sqrt{4e^{4\phi_{2}-2C_{3}}-(e^{4\phi_{2}}-1)^{2}}}\right]-\frac{1}{8}\tan^{-1}\left[\frac{e^{4\phi_{2}}(1+2e^{-2C_{3}})-1}{\sqrt{4e^{4\phi_{2}-2C_{3}}-(e^{4\phi_{2}}-1)^{2}}}\right]$
(3.90)

in which the integration constant has been set to zero.

####  3.5.4 Domain walls in $CSO(3,0,2)$ gauge group

In the case of $CSO(3,0,2)$ gauge group with $\kappa=\lambda=0$, supersymmetry
allows a non-vanishing axion corresponding to $\mathcal{Y}_{4}$ generator. We
write the coset representative as

$V=e^{\varphi\boldsymbol{d}+\phi_{1}\mathcal{Y}_{1}+\phi_{2}\mathcal{Y}_{2}+\phi_{3}\mathcal{Y}_{3}+\varsigma\mathcal{Y}_{4}}$
(3.91)

and find a simple scalar potential

$\mathbf{V}=-\frac{3g^{2}}{4}e^{2(\varphi-8\phi_{1})}\,.$ (3.92)

We also note that this potential does not depend on $\varsigma$ and can be
obtained from (3.66) by setting $\kappa=\lambda=0$. This potential can also be
written in the form (3.24) using the superpotential

$\mathcal{W}=\frac{3g}{4\sqrt{2}}e^{\varphi-8\phi_{1}}$ (3.93)

and the symmetric matrix

$G^{IJ}=\begin{pmatrix}\frac{1}{10}&0&0&0&\frac{2\varsigma}{5}\\\
0&\frac{1}{60}&0&0&-\frac{\varsigma}{5}\\\
0&0&\text{sech}^{2}{4\phi_{3}}&0&0\\\ 0&0&0&\frac{1}{4}&0\\\
\frac{2\varsigma}{5}&-\frac{\varsigma}{5}&0&0&1+4\varsigma^{2}\end{pmatrix}$
(3.94)

for $\Phi^{I}=\\{\varphi,\phi_{1},\phi_{2},\phi_{3},\varsigma\\}$,
$I,J=1,2,3,4,5$.

With all these and the usual ansatz for the Killing spinors (3.32) together
with the projector (3.31), we find the BPS equations

$\displaystyle A^{\prime}$ $\displaystyle=$
$\displaystyle\frac{3g}{4\sqrt{2}}e^{\varphi-8\phi_{1}},\quad\varphi^{\prime}=-\frac{3g}{20\sqrt{2}}e^{\varphi-8\phi_{1}},\quad\phi^{\prime}_{1}=\frac{g}{5\sqrt{2}}e^{\varphi-8\phi_{1}},$
$\displaystyle\phi^{\prime}_{2}$ $\displaystyle=$
$\displaystyle\phi^{\prime}_{3}\ =\ 0,\qquad\
\,\varsigma^{\prime}=-\frac{3g}{\sqrt{2}}e^{\varphi-8\phi_{1}}\varsigma\,.$
(3.95)

Except for an additional equation for $\varsigma$, these are the BPS equations
obtained from (3.69) to (3.73) by setting $\kappa=\lambda=0$. Furthermore,
$\phi_{2}$ and $\phi_{3}$ can be consistently truncated out since the scalar
potential (3.92) is independent of $\phi_{2}$ and $\phi_{3}$.
With all these, we find a domain wall solution

$\displaystyle\phi_{1}$ $\displaystyle=$
$\displaystyle\frac{1}{4}\ln\left[\frac{2}{5}(\sqrt{2}g\rho+C_{1})\right],$
(3.96) $\displaystyle\varphi$ $\displaystyle=$ $\displaystyle
C-\frac{1}{8}\ln\left[\frac{2}{5}(\sqrt{2}g\rho+C_{1})\right],$ (3.97)
$\displaystyle\varsigma$ $\displaystyle=$ $\displaystyle
C_{4}e^{-15\phi_{1}}=\frac{C_{4}}{\left(\frac{2}{5}(\sqrt{2}g\rho+C_{1})\right)^{\frac{15}{4}}},$
(3.98) $\displaystyle A$ $\displaystyle=$
$\displaystyle-5\varphi\,=\,-5C+\frac{5}{8}\ln\left[\frac{2}{5}(\sqrt{2}g\rho+C_{1})\right]$
(3.99)

in which $\rho$ is a new radial coordinate defined by
$\frac{d\rho}{dr}=e^{\varphi-4\phi_{1}}$. It should also be noted that the
axion $\varsigma$ can also be truncated out.

###  3.6 $SO(2)\times SO(2)$ symmetric domain walls

As a final example of domain wall solutions in $\mathbf{15}^{-1}$
representation, we consider an $SO(2)\times SO(2)$ unbroken symmetry. In this
case, the embedding tensor for all possible gauge groups takes the form

$Y_{mn}=\textrm{diag}(1,1,\kappa,\kappa,\lambda)$ (3.100)

for $\lambda=0,\pm 1$ and $\kappa=\pm 1$. These gauge groups are $SO(5)$
($\kappa=\lambda=1$), $SO(4,1)$ ($\kappa=-\lambda=1$), $SO(3,2)$
($\kappa=-\lambda=-1$), $CSO(4,0,1)$ ($\kappa=1,\lambda=0$), and $CSO(2,2,1)$
($\kappa=-1,\lambda=0$).

There are five scalars invariant under $SO(2)\times SO(2)$ generated by
$X_{12}$ and $X_{34}$. As usual, one of these is the dilaton and the other
four are associated with the following non-compact generators

$\overline{\mathcal{Y}}_{1}=\hat{\boldsymbol{t}}^{+}_{1\dot{1}}+\hat{\boldsymbol{t}}^{+}_{2\dot{2}}-2\,\hat{\boldsymbol{t}}^{+}_{5\dot{5}},\qquad\overline{\mathcal{Y}}_{2}=\hat{\boldsymbol{t}}^{+}_{3\dot{3}}+\hat{\boldsymbol{t}}^{+}_{4\dot{4}}-2\,\hat{\boldsymbol{t}}^{+}_{5\dot{5}},\qquad\overline{\mathcal{Y}}_{3}=\boldsymbol{s}_{12},\qquad\overline{\mathcal{Y}}_{4}=\boldsymbol{s}_{34}\,.$
(3.101)

As in many previous cases, we need to truncate out the axions corresponding to
the shift generators $\boldsymbol{s}_{12}$ and $\boldsymbol{s}_{34}$ in order
to find a consistent set of BPS equations that are compatible with the field
equations. We then take the coset representative of the form

$V=e^{\varphi\boldsymbol{d}+\phi_{1}\overline{\mathcal{Y}}_{1}+\phi_{2}\overline{\mathcal{Y}}_{2}}\,.$
(3.102)

The resulting scalar potential reads

$\mathbf{V}=-\frac{g^{2}}{4}e^{2\left(\varphi-2(\phi_{1}+\phi_{2})\right)}\left[4\kappa(2+\lambda
e^{12\phi_{1}+8\phi_{2}})+\lambda e^{8\phi_{1}+12\phi_{2}}(4-\lambda
e^{12\phi_{1}+8\phi_{2}})\right]$ (3.103)

which can be written in terms of the superpotential

$\mathcal{W}=\frac{g}{4\sqrt{2}}e^{\varphi}(2e^{-4\phi_{1}}+2\kappa
e^{-4\phi_{2}}+\lambda e^{8(\phi_{1}+\phi_{2})})$ (3.104)

using

$G^{IJ}=\begin{pmatrix}\frac{1}{10}&0&0\\\ 0&\frac{3}{20}&-\frac{1}{10}\\\
0&-\frac{1}{10}&\frac{3}{20}\end{pmatrix}$ (3.105)

for $\Phi^{I}=\\{\varphi,\phi_{1},\phi_{2}\\}$, $I,J=1,2,3$.

Using the projector (3.31) together with the Killing spinors (3.32), we find
the following BPS equations

$\displaystyle A^{\prime}$ $\displaystyle=$
$\displaystyle\frac{g}{4\sqrt{2}}e^{\varphi}(2e^{-4\phi_{1}}+2\kappa
e^{-4\phi_{2}}+\lambda e^{8(\phi_{1}+\phi_{2})}),$ (3.106)
$\displaystyle\varphi^{\prime}$ $\displaystyle=$
$\displaystyle-\frac{g}{20\sqrt{2}}e^{\varphi}(2e^{-4\phi_{1}}+2\kappa
e^{-4\phi_{2}}+\lambda e^{8(\phi_{1}+\phi_{2})}),$ (3.107)
$\displaystyle\phi^{\prime}_{1}$ $\displaystyle=$
$\displaystyle\frac{g}{5\sqrt{2}}e^{\varphi}(3e^{-4\phi_{1}}-2\kappa
e^{-4\phi_{2}}-\lambda e^{8(\phi_{1}+\phi_{2})}),$ (3.108)
$\displaystyle\phi^{\prime}_{2}$ $\displaystyle=$
$\displaystyle\frac{g}{5\sqrt{2}}e^{\varphi}(3\kappa
e^{-4\phi_{2}}-2e^{-4\phi_{1}}-\lambda e^{8(\phi_{1}+\phi_{2})}).$ (3.109)

Solving these BPS equations gives a domain wall solution

$\displaystyle\phi_{2}$ $\displaystyle=$
$\displaystyle-\frac{3\sqrt{2}g\rho}{10}+\frac{3}{20}\ln\left[\kappa
e^{2\sqrt{2}g\rho}+C_{2}\right]-\frac{1}{20}\ln\left[C_{1}e^{-2\sqrt{2}g\rho}+\lambda\right],$
(3.110) $\displaystyle\phi_{1}$ $\displaystyle=$
$\displaystyle-\frac{2\phi_{2}}{3}-\frac{1}{12}\ln\left[C_{1}e^{-2\sqrt{2}g\rho}+\lambda\right],$
(3.111) $\displaystyle\varphi$ $\displaystyle=$
$\displaystyle-\frac{\phi_{2}}{6}-\frac{3\sqrt{2}g\rho}{24}+C-\frac{1}{48}\ln\left[C_{1}e^{-2\sqrt{2}g\rho}+\lambda\right],$
(3.112) $\displaystyle A$ $\displaystyle=$
$\displaystyle-5\varphi\,=\,\frac{5\phi_{2}}{6}+\frac{15\sqrt{2}g\rho}{24}-5C+\frac{5}{48}\ln\left[C_{1}e^{-2\sqrt{2}g\rho}+\lambda\right]$
(3.113)

in which $\rho$ is the new radial coordinate defined by the relation
$\frac{d\rho}{dr}=e^{\varphi-4\phi_{1}}$.

For domain walls preserving smaller residual symmetries such as
$SO(2)_{\text{diag}}\subset SO(2)\times SO(2)$ and $SO(2)$, there are many
more scalars, and the analysis is much more involved without any possibility
for complete analytic solutions. We will not consider these cases in this
work.

##  4 Domain walls from gaugings in $\overline{\mathbf{40}}^{-1}$
representation

In this section, we consider gaugings in which the irreducible part of the
embedding tensor transforms in $\overline{\mathbf{40}}^{-1}$ representation.
These gauged theories are obtained from a consistent circle reduction of the
maximal seven-dimensional $CSO(p,q,4-p-q)$ gauged supergravity constructed in
[41].

In six dimensions, gaugings in $\overline{\mathbf{40}}^{-1}$ representation
are purely electric and triggered by

$\theta^{Am}\ =\ \mathbb{T}^{A}_{np}U^{np,m}$ (4.1)

where $U^{mn,p}=U^{[mn],p}$ satisfying $U^{[mn,p]}=0$. With
$\theta^{AM}=(\,\mathbb{T}^{A}_{np}U^{np,m}\,,\,0\,)$, the second condition
from the quadratic constraint (2.13) reduces to

$U^{mn,r}U^{pq,s}\varepsilon_{mnpqt}\ =\ 0\,.$ (4.2)

This condition can be solved by setting

$U^{mn,p}\ =\ v^{[m}w^{n]p}$ (4.3)

in which $v^{m}$ is a $GL(5)$ vector and $w^{mn}$ is a symmetric tensor,
$w^{mn}=w^{(mn)}$.

To classify possible gauge groups, we follow [41] by using the $SL(5)$
symmetry to further fix $v^{m}=\delta^{m}_{5}$ and split the index $m=(i,5)$,
$i=1,..,4$. For simplicity, we also restrict to cases with $w^{i5}=w^{55}=0$.
The remaining $SL(4)$ residual symmetry can be used to diagonalize the
$4\times 4$ block corresponding to $w^{ij}$ as

$w^{ij}\ =\
\text{diag}(\underbrace{1,..,1}_{p},\underbrace{-1,..,-1}_{q},\underbrace{0,..,0}_{r})$
(4.4)

with $p+q+r=4$. From the decomposition in (3.3), we find that in this case,
only $X_{ij}$ and $X^{i}$ gauge generators are non-vanishing. The generators
$X_{ij}$ are given in terms of the $GL(5)$ generators while $X^{i}$ only
involve the shift generators. Explicitly, these generators are given by

$X_{ij}=\frac{1}{\sqrt{2}}\varepsilon_{ijkm}w^{kl}{\boldsymbol{t}^{m}}_{l}\qquad\textrm{and}\qquad
X^{i}=w^{ij}\boldsymbol{s}_{5j}\,.$ (4.5)

It is now straightforward to show that the gauge generators satisfy the
following commutation relations

$[X^{i},X^{j}]=0,\qquad[X_{ij},X^{k}]={(X_{ij})_{l}}^{k}X^{l},\qquad[X_{ij},X_{kl}]={(X_{ij})_{kl}}^{mn}X_{mn}$
(4.6)

in which ${(X_{ij})_{kl}}^{mn}=2{(X_{ij})_{[k}}^{[m}\delta_{l]}^{n]}$. This
implies that the corresponding gauge group is of the form

$G_{0}=CSO(p,q,4-p-q)\ltimes\mathbb{R}^{4}_{\boldsymbol{s}}=SO(p,q)\ltimes\left(\mathbb{R}^{(p+q)(4-p-q)}\times\mathbb{R}^{4}_{\boldsymbol{s}}\right).$
(4.7)

The $CSO(p,q,4-p-q)$ factor and the four-dimensional translation group from
the shift symmetries $\mathbb{R}^{4}_{\boldsymbol{s}}$ are respectively
generated by $X_{ij}$ and $X^{i}$.

We should note here that the corresponding gauge group in seven dimensions is
just $CSO(p,q,4-p-q)$. After an $S^{1}$ reduction, this gauge group is
accompanied by a translation group $\mathbb{R}^{4}_{\boldsymbol{s}}$. As
pointed out in [30], the complete off-shell symmetry group of the maximal six-
dimensional gauged supergravity is $GL(5)\ltimes\mathbf{10}^{-4}$, with
$\mathbf{10}^{-4}$ being shift symmetries of scalar fields. The gauge group
given in (4.7) is embedded in $GL(5)\ltimes\mathbf{10}^{-4}$ as
$CSO(p,q,4-p-q)\subset GL(5)$ and
$\mathbb{R}^{4}_{\boldsymbol{s}}\subset\mathbf{10}^{-4}$. We also note that in
vector representation of $SO(5,5)$, the gauge generators are given by

${(X^{i})_{P}}^{Q}=4\eta_{P[j}\delta^{Q}_{k]}\delta^{j}_{5}w^{ki},\qquad{(X_{ij})_{P}}^{Q}=\sqrt{2}\varepsilon_{ijkm}w^{kl}(\delta_{P}^{m}\delta^{Q}_{l}-\eta^{mP}\eta_{lP}).$
(4.8)

By splitting the $SO(5)\times SO(5)$ vector indices as $a=(i,5)$ and
$\dot{a}=(\dot{i},\dot{5})$, we find the following decomposition for non-
compact generators of $SL(5)\subset GL(5)\subset SO(5,5)$ under
$SL(5)\rightarrow SL(4)\times SO(1,1)$

$\tilde{\boldsymbol{t}}_{a\dot{b}}\rightarrow\left(\tilde{\boldsymbol{t}}_{i\dot{j}},\,\tilde{\boldsymbol{t}}_{i\dot{5}},\,\tilde{\boldsymbol{t}}_{5\dot{5}}\right).$
(4.9)

Since the $SL(5)$ generators $\tilde{\boldsymbol{t}}_{a\dot{b}}$ are
traceless, the generator $\tilde{\boldsymbol{t}}_{5\dot{5}}$ is related to the
trace part of $\tilde{\boldsymbol{t}}_{i\dot{j}}$ according to
$\tilde{\boldsymbol{t}}_{1\dot{1}}+\tilde{\boldsymbol{t}}_{2\dot{2}}+\tilde{\boldsymbol{t}}_{3\dot{3}}+\tilde{\boldsymbol{t}}_{4\dot{4}}=-\tilde{\boldsymbol{t}}_{5\dot{5}}$.
It is then convenience to define new non-compact generators
$\overline{\boldsymbol{t}}_{i\dot{j}}$ as

$\overline{\boldsymbol{t}}_{i\dot{j}}=\tilde{\boldsymbol{t}}_{i\dot{j}}+\frac{1}{4}\tilde{\boldsymbol{t}}_{5\dot{5}}\delta_{i\dot{j}}$
(4.10)

which are symmetric traceless. The nine scalar fields corresponding to these
generators then parametrize an $SL(4)/SO(4)$ coset. The other four scalars
associated with
$\tilde{\boldsymbol{t}}_{i\dot{5}}=\hat{\boldsymbol{t}}^{+}_{i\dot{5}}$ are
nilpotent scalars and will be denoted by $b_{i}$ as in seven dimensions. In
addition, there are also ten axions corresponding to the antisymmetric shift
generators as in the previous section.

As in the previous section, we will systematically find supersymmetric domain
walls invariant under some residual symmetries of the $CSO(p,q,4-p-q)$ factor
in the gauge group.

###  4.1 $SO(4)$ symmetric domain walls

We first consider domain walls with the largest possible unbroken symmetry
namely $SO(4)$. The only gauge group containing $SO(4)$ as a subgroup is
$SO(4)\ltimes\mathbb{R}^{4}_{\boldsymbol{s}}$ with the embedding tensor
parametrized by $w^{ij}=\delta^{ij}$. The $SO(4)$ symmetry is generated by
$X_{ij}$, $i,j=1,2,3,4$, generators.

There are two $SO(4)$ singlet scalars given by the dilaton $\varphi$ and
another dilatonic scalar corresponding to the $SO(1,1)$ factor in $SL(4)\times
SO(1,1)\subset SL(5)$. The latter is given by the non-compact generator

$\widetilde{\mathcal{Y}}_{0}=\hat{\boldsymbol{t}}^{+}_{1\dot{1}}+\hat{\boldsymbol{t}}^{+}_{2\dot{2}}+\hat{\boldsymbol{t}}^{+}_{3\dot{3}}+\hat{\boldsymbol{t}}^{+}_{4\dot{4}}-4\,\hat{\boldsymbol{t}}^{+}_{5\dot{5}}$
(4.11)

and will be denoted by $\phi_{0}$.

The coset representative can be written as

$V=e^{\varphi\boldsymbol{d}+\phi_{0}\widetilde{\mathcal{Y}}_{0}}$ (4.12)

leading to the T-tensor given by

$T^{\alpha\dot{\beta}}=e^{\varphi-4\phi_{0}}\,(\gamma^{5})^{\alpha\beta}\delta^{\dot{\beta}}_{\beta}=\frac{2}{g}\mathcal{W}\,(\gamma^{5})^{\alpha\beta}\delta^{\dot{\beta}}_{\beta}\,$
(4.13)

with the superpotential

$\mathcal{W}=\frac{g}{2}e^{\varphi-4\phi_{0}}\,.$ (4.14)

The appearance of $\gamma^{5}$ rather than other $SO(5)$ gamma matrices is due
to the specific choice of $v^{m}=\delta^{m}_{5}$ for the tensor $U^{mn,p}$.
The scalar potential can also be directly computed and is given by

$\mathbf{V}=-g^{2}e^{2\varphi-8\phi_{0}}\,.$ (4.15)

The Killing spinors are given by the same ansatz as in (3.32) but in this case
subject to the following projector

$\hat{\gamma}_{r}\epsilon_{\pm}=\gamma^{5}\epsilon_{\mp}.$ (4.16)

because of the appearance of $\gamma^{5}$ in the T-tensor. With this new
projector, it is now straightforward to derive the following BPS equations

$A^{\prime}=\frac{g}{2}e^{\varphi-4\phi_{0}},\qquad\varphi^{\prime}=-\frac{g}{10}e^{\varphi-4\phi_{0}},\qquad\phi^{\prime}_{0}=\frac{g}{10}e^{\varphi-4\phi_{0}}\,.$
(4.17)

These equations are solved by the solution

$\displaystyle\varphi$ $\displaystyle=$ $\displaystyle
4C-\frac{1}{5}\ln\left[\frac{gr}{2}+C_{0}\right],$ (4.18)
$\displaystyle\phi_{0}$ $\displaystyle=$ $\displaystyle
C+\frac{1}{5}\ln\left[\frac{gr}{2}+C_{0}\right],$ (4.19) $\displaystyle A$
$\displaystyle=$ $\displaystyle-5\varphi\ =\
\ln\left[\frac{gr}{2}+C_{0}\right]-20C\,.$ (4.20)

###  4.2 $SO(3)$ symmetric domain walls

We now look for more complicated solutions with $SO(3)$ symmetry. Gauge groups
with an $SO(3)$ subgroup are $SO(4)\ltimes\mathbb{R}^{4}_{\boldsymbol{s}}$,
$SO(3,1)\ltimes\mathbb{R}^{4}_{\boldsymbol{s}}$, and
$CSO(3,0,1)\ltimes\mathbb{R}^{4}_{\boldsymbol{s}}$ which are collectively
described by the symmetric tensor

$w^{ij}=\textrm{diag}(1,1,1,\kappa)$ (4.21)

for $\kappa=1,-1,0$, respectively.

The residual symmetry $SO(3)$ is generated by the generators $X_{\hat{i}4}$
with $\hat{i}=1,2,3$. Apart from the two dilatons, there are three additional
$SO(3)$ singlet scalars, one from the $SL(4)/SO(4)$ coset and the other two
from symmetric and antisymmetric axions denoted by $b$ and $\varsigma$. These
three singlets correspond to the following $SO(5,5)$ non-compact generators

$\widetilde{\mathcal{Y}}_{1}=\hat{\boldsymbol{t}}^{+}_{1\dot{1}}+\hat{\boldsymbol{t}}^{+}_{2\dot{2}}+\hat{\boldsymbol{t}}^{+}_{3\dot{3}}-3\,\hat{\boldsymbol{t}}^{+}_{4\dot{4}},\qquad\widetilde{\mathcal{Y}}_{2}=\hat{\boldsymbol{t}}^{+}_{4\dot{5}},\qquad\widetilde{\mathcal{Y}}_{3}=\boldsymbol{s}_{45}\,.$
(4.22)

Using the coset representative of the form

$V=e^{\varphi\boldsymbol{d}+\phi_{0}\widetilde{\mathcal{Y}}_{0}+\phi\widetilde{\mathcal{Y}}_{1}+b\widetilde{\mathcal{Y}}_{2}+\varsigma\widetilde{\mathcal{Y}}_{3}},$
(4.23)

we find the scalar potential and the T-tensor given by

$\mathbf{V}=-\frac{g^{2}}{16}e^{2(\varphi-4(\phi_{0}+3\phi))}\left(6\kappa
e^{16\phi}+(9e^{32\phi}+\kappa^{2})\cosh{2b}\right)$ (4.24)

and

$T^{\alpha\dot{\beta}}=\frac{1}{4}e^{\varphi-4(\phi_{0}+3\phi)}\left[(3e^{16\phi}+\kappa)\cosh{b}\,(\gamma^{5})^{\alpha\beta}\delta^{\dot{\beta}}_{\beta}+(3e^{16\phi}-\kappa)\sinh{b}\,(\gamma^{4})^{\alpha\beta}\delta^{\dot{\beta}}_{\beta}\right].$
(4.25)

It turns out that consistency of the BPS equations from $\delta\chi_{\pm}$
conditions requires vanishing symmetric axion $b$ unless $\kappa=0$
corresponding to $CSO(3,0,1)\ltimes\mathbb{R}^{4}_{\boldsymbol{s}}$ gauge
group.

####  4.2.1 Domain walls without the symmetric axion

With $b=0$, the scalar potential and superpotential read

$\displaystyle\mathbf{V}$ $\displaystyle=$
$\displaystyle-\frac{g^{2}}{8}e^{2(\varphi-4(\phi_{0}+3\phi))}(3e^{32\phi}+6\kappa
e^{16\phi}-\kappa^{2}),$ (4.26) $\displaystyle\mathcal{W}$ $\displaystyle=$
$\displaystyle\frac{g}{8}e^{\varphi-4(\phi_{0}+3\phi)}(3e^{16\phi}+\kappa).$
(4.27)

Imposing the projector (4.16) on the Killing spinors (3.32), we can derive the
following set of BPS equations

$\displaystyle A^{\prime}$ $\displaystyle=$
$\displaystyle\frac{g}{8}e^{\varphi-4(\phi_{0}+3\phi)}(3e^{16\phi}+\kappa),$
(4.28) $\displaystyle\varphi^{\prime}$ $\displaystyle=$
$\displaystyle-\frac{g}{40}e^{\varphi-4(\phi_{0}+3\phi)}(3e^{16\phi}+\kappa),$
(4.29) $\displaystyle\phi^{\prime}_{0}$ $\displaystyle=$
$\displaystyle\frac{g}{40}e^{\varphi-4(\phi_{0}+3\phi)}(3e^{16\phi}+\kappa),$
(4.30) $\displaystyle\phi^{\prime}$ $\displaystyle=$
$\displaystyle-\frac{g}{8}e^{\varphi-4(\phi_{0}+3\phi)}(3e^{16\phi}-\kappa),$
(4.31) $\displaystyle\varsigma^{\prime}$ $\displaystyle=$ $\displaystyle-
ge^{\varphi-4(\phi_{0}+3\phi)}\varsigma\,.$ (4.32)

From these equations, we can find the solutions for $A$, $\varphi$, and
$\phi_{0}$ as functions of $\phi$ of the form

$\displaystyle\phi_{0}$ $\displaystyle=$
$\displaystyle\frac{\phi}{5}+C_{0}-\frac{1}{20}\ln(e^{16\phi}-\kappa),$ (4.33)
$\displaystyle\varphi$ $\displaystyle=$
$\displaystyle-\frac{\phi}{5}+C-C_{0}+\frac{1}{20}\ln(e^{16\phi}-\kappa),$
(4.34) $\displaystyle A$ $\displaystyle=$
$\displaystyle-5\varphi\,=\,\phi-5C+5C_{0}-\frac{1}{4}\ln(e^{16\phi}-\kappa).$
(4.35)

With the new radial coordinate $\rho$ defined by
$\frac{d\rho}{dr}=e^{\varphi-4(\phi_{0}+\phi)}$, the solutions for $\phi$ and
$\varsigma$ are given by

$e^{8\phi}=\sqrt{\kappa}\tanh\left[\sqrt{\kappa}(g\rho+C_{1})\right]\qquad\textrm{and}\qquad\varsigma=C_{2}\,\text{csch}\left[\sqrt{\kappa}(g\rho+C_{1})\right].$
(4.36)

In particular, for $\kappa=-1$ and $\kappa=0$, we find respectively

$e^{8\phi}=-\tan\left[\sqrt{\kappa}(g\rho+C_{1})\right],\qquad\varsigma=C_{2}\,\text{csc}\left(g\rho+C_{1}\right)$
(4.37)

and

$\displaystyle e^{8\phi}=\frac{1}{(g\rho+C_{1})},\qquad\varsigma=C_{2}\,.$
(4.38)

####  4.2.2 Domain walls with the symmetric axion

For $\kappa=0$ corresponding to
$CSO(3,0,1)\ltimes\mathbb{R}^{4}_{\boldsymbol{s}}$ gauge group, it is possible
to find solutions with the symmetric axion $b$ non-vanishing. With $\kappa=0$,
the scalar potential and the T-tensor are given by

$\mathbf{V}=-\frac{3g^{2}}{8}e^{2(\varphi-4\phi_{0}+4\phi)}\cosh{2b}$ (4.39)

and

$T^{\alpha\dot{\beta}}=\frac{3}{4}e^{\varphi-4(\phi_{0}-\phi)}\left[\cosh{b}\,(\gamma^{5})^{\alpha\beta}\delta^{\dot{\beta}}_{\beta}+\sinh{b}\,(\gamma^{4})^{\alpha\beta}\delta^{\dot{\beta}}_{\beta}\right].$
(4.40)

By the general procedure given in section 3.1, we find the superpotential and
$\hat{\gamma}_{r}$ projectors on the Killing spinors

$\mathcal{W}=\frac{3g}{8}e^{\varphi-4\phi_{0}+4\phi}\sqrt{\cosh{2b}}$ (4.41)

and

$\displaystyle\hat{\gamma}_{r}\epsilon_{+\alpha}$ $\displaystyle=$
$\displaystyle\frac{\Omega_{\alpha\beta}}{\sqrt{\cosh{2b}}}\left[\cosh{b}\,(\gamma^{5})^{\beta\gamma}\delta^{\dot{\alpha}}_{\gamma}+\sinh{b}\,(\gamma^{4})^{\beta\gamma}\delta^{\dot{\alpha}}_{\gamma}\right]\epsilon_{-\dot{\alpha}},$
(4.42) $\displaystyle\hat{\gamma}_{r}\epsilon_{-\dot{\alpha}}$
$\displaystyle=$
$\displaystyle-\frac{\Omega_{\dot{\alpha}\dot{\beta}}}{\sqrt{\cosh{2b}}}\left[\cosh{b}\,(\gamma^{5})^{\alpha\beta}\delta^{\dot{\beta}}_{\beta}+\sinh{b}\,(\gamma^{4})^{\alpha\beta}\delta^{\dot{\beta}}_{\beta}\right]\epsilon_{+\alpha}\,.$
(4.43)

It should be noted that these projectors are not independent. Therefore, the
resulting solutions will preserve half of the supersymmetry. Moreover, we can
easily see that these projectors reduce to that given in (4.16) for $b=0$.
With all these, we find the following set of BPS equations

$\displaystyle A^{\prime}$ $\displaystyle=$
$\displaystyle\frac{3g}{8}e^{\varphi-4\phi_{0}+4\phi}\sqrt{\cosh{2b}},\qquad\quad\
\ B^{\prime}\ =\
-\frac{3ge^{\varphi-4\phi_{0}+4\phi}\tanh{2b}}{8\sqrt{\cosh{2b}}},$ (4.44)
$\displaystyle\varphi^{\prime}$ $\displaystyle=$
$\displaystyle-\frac{3g}{40}e^{\varphi-4\phi_{0}+4\phi}\sqrt{\cosh{2b}},\qquad\
\ \,\phi^{\prime}_{0}\ =\
-\frac{3ge^{\varphi-4\phi_{0}+4\phi}(\cosh{4b}-9)}{320\cosh^{3/2}{2b}},\qquad$
(4.45) $\displaystyle\phi^{\prime}$ $\displaystyle=$
$\displaystyle-\frac{ge^{\varphi-4\phi_{0}+4\phi}(\cosh{4b}+7)}{64\cosh^{3/2}{2b}},\qquad
b^{\prime}\ =\
-\frac{3ge^{\varphi-4\phi_{0}+4\phi}\sinh{2b}}{4\sqrt{\cosh{2b}}}$ (4.46)

together with $\varsigma^{\prime}=0$. Since the scalar potential does not
depend on $\varsigma$, we can consistently truncate $\varsigma$ out by setting
$\varsigma=0$. The domain wall solution to the above BPS equations is then
given by

$\displaystyle\sqrt{\sinh{2b}}(\frac{3g\rho}{4}+C_{b})$ $\displaystyle=$
$\,{}_{2}F_{1}(\frac{1}{4},\frac{1}{4},\frac{5}{4},-\frac{1}{\sinh^{2}{2b}}),$
(4.47) $\displaystyle\varphi$ $\displaystyle=$ $\displaystyle
C-\frac{b}{10}+\frac{1}{20}\ln(1-e^{4b}),$ (4.48) $\displaystyle\phi_{0}$
$\displaystyle=$ $\displaystyle
C_{0}-\frac{b}{40}-\frac{1}{20}\ln(1-e^{4b})+\frac{1}{16}\ln(1+e^{4b}),$
(4.49) $\displaystyle\phi$ $\displaystyle=$ $\displaystyle
C_{1}-\frac{b}{24}+\frac{1}{12}\ln(1-e^{4b})-\frac{1}{16}\ln(1+e^{4b}),$
(4.50) $\displaystyle A$ $\displaystyle=$
$\displaystyle-5\varphi\,=\,-5C+\frac{b}{2}-\frac{1}{4}\ln(1-e^{4b}),$ (4.51)
$\displaystyle B$ $\displaystyle=$
$\displaystyle\frac{1}{2}\tan^{-1}\left(e^{2b}\right)+C_{B}$ (4.52)

in which the new radial coordinate $\rho$ is defined by
$\frac{d\rho}{dr}=e^{\varphi-4\phi_{0}+4\phi}$, and ${}_{2}F_{1}$ is the
hypergeometric function.

###  4.3 $SO(2)\times SO(2)$ symmetric domain walls

Domain walls preserving $SO(2)\times SO(2)$ symmetry can be found in
$SO(4)\ltimes\mathbb{R}^{4}_{\boldsymbol{s}}$ and
$SO(2,2)\ltimes\mathbb{R}^{4}_{\boldsymbol{s}}$ gauge groups described by the
embedding tensor with

$w^{ij}=\textrm{diag}(1,1,\kappa,\kappa),\qquad\kappa=1,-1\,.$ (4.53)

In addition to the two dilatons, there are three $SO(2)\times SO(2)$ singlet
scalars corresponding to the following $SO(5,5)$ non-compact generators

$\widehat{\mathcal{Y}}_{1}=\hat{\boldsymbol{t}}^{+}_{1\dot{1}}+\hat{\boldsymbol{t}}^{+}_{2\dot{2}}-\hat{\boldsymbol{t}}^{+}_{3\dot{3}}-\hat{\boldsymbol{t}}^{+}_{4\dot{4}},\qquad\widehat{\mathcal{Y}}_{2}=\boldsymbol{s}_{12},\qquad\widehat{\mathcal{Y}}_{3}=\boldsymbol{s}_{34}\,.$
(4.54)

In this case, a consistent set of BPS equations can be found only when the
scalars corresponding to $\widehat{\mathcal{Y}}_{2}$ and
$\widehat{\mathcal{Y}}_{3}$ generators vanish.

With the coset representative

$V=e^{\varphi\boldsymbol{d}+\phi_{0}\widetilde{\mathcal{Y}}_{0}+\phi\widehat{\mathcal{Y}}_{1}},$
(4.55)

the scalar potential and superpotential are given by

$\mathbf{V}=-g^{2}\kappa
e^{2\varphi-8\phi_{0}}\qquad\textrm{and}\qquad\mathcal{W}=\frac{g}{4}e^{\varphi-4(\phi_{0}+\phi)}(e^{8\phi}+\kappa).$
(4.56)

With all these and the usual Killing spinors (3.32) subject to the projector
(4.16), the resulting BPS equations read

$\displaystyle A^{\prime}$ $\displaystyle=$
$\displaystyle\frac{g}{4}e^{\varphi-4(\phi_{0}+\phi)}(e^{8\phi}+\kappa),$
(4.57) $\displaystyle\varphi^{\prime}$ $\displaystyle=$
$\displaystyle-\frac{g}{20}e^{\varphi-4(\phi_{0}+\phi)}(e^{8\phi}+\kappa),$
(4.58) $\displaystyle\phi^{\prime}_{0}$ $\displaystyle=$
$\displaystyle\frac{g}{20}e^{\varphi-4(\phi_{0}+\phi)}(e^{8\phi}+\kappa),$
(4.59) $\displaystyle\phi^{\prime}$ $\displaystyle=$
$\displaystyle-\frac{g}{4}e^{\varphi-4(\phi_{0}+\phi)}(e^{8\phi}-\kappa).$
(4.60)

Using a new radial coordinate $\rho$ defined by
$\frac{d\rho}{dr}=e^{\varphi-2\phi_{0}}$, we find a domain wall solution

$\displaystyle\phi_{0}$ $\displaystyle=$
$\displaystyle\frac{\phi}{5}+C_{0}-\frac{1}{20}\ln\left(e^{8\phi}-\kappa\right),$
(4.61) $\displaystyle\varphi$ $\displaystyle=$
$\displaystyle-\frac{\phi}{5}+C-C_{0}+\frac{1}{20}\ln\left(e^{8\phi}-\kappa\right),$
(4.62) $\displaystyle A$ $\displaystyle=$
$\displaystyle-5\varphi\,=\,\phi-5C+5C_{0}-\frac{1}{4}\ln\left(e^{8\phi}-\kappa\right),$
(4.63) $\displaystyle e^{4\phi}$ $\displaystyle=$
$\displaystyle\sqrt{\kappa}\tanh\left[\sqrt{\kappa}(g\rho+C_{1})\right].$
(4.64)

###  4.4 $SO(2)$ symmetric domain walls

As a final example in this case, we consider $SO(2)$ symmetric domain walls.
There are many gauge groups admitting an $SO(2)$ subgroup. They are
collectively characterized by the following component of the embedding tensor

$w^{ij}=\text{diag}(1,1,\kappa,\lambda).$ (4.65)

Together with the two dilatons, there are additional nine $SO(2)$ singlet
scalars. Three of them are in the $SL(4)/SO(4)$ coset corresponding to non-
compact generators

$\overline{\mathcal{Y}}_{1}=\hat{\boldsymbol{t}}^{+}_{1\dot{1}}+\hat{\boldsymbol{t}}^{+}_{2\dot{2}}-\hat{\boldsymbol{t}}^{+}_{3\dot{3}}-\hat{\boldsymbol{t}}^{+}_{4\dot{4}},\qquad\overline{\mathcal{Y}}_{2}=\hat{\boldsymbol{t}}^{+}_{3\dot{4}},\qquad\overline{\mathcal{Y}}_{3}=\hat{\boldsymbol{t}}^{+}_{3\dot{3}}-\hat{\boldsymbol{t}}^{+}_{4\dot{4}}\,.$
(4.66)

The remaining ones consist of two nilpotent scalars associated with

$\overline{\mathcal{Y}}_{4}=\hat{\boldsymbol{t}}^{+}_{3\dot{5}},\qquad\overline{\mathcal{Y}}_{5}=\hat{\boldsymbol{t}}^{+}_{4\dot{5}}$
(4.67)

and four shift scalars corresponding to

$\overline{\mathcal{Y}}_{6}=\boldsymbol{s}_{12},\qquad\overline{\mathcal{Y}}_{7}=\boldsymbol{s}_{35},\qquad\overline{\mathcal{Y}}_{8}=\boldsymbol{s}_{45},\qquad\overline{\mathcal{Y}}_{9}=\boldsymbol{s}_{34}\,.$
(4.68)

However, dealing with all eleven scalars turns out to be highly complicated,
so we perform a subtruncation by setting the shift scalar corresponding to
$\boldsymbol{s}_{12}$ and the two nilpotent scalars to zero. It is
straightforward to verify that this is a consistent truncation and still gives
interesting solutions. We now end up with eight singlet scalars with the coset
representative

$V=e^{\varphi\boldsymbol{d}+\phi_{0}\widetilde{\mathcal{Y}}_{0}+\phi_{1}\overline{\mathcal{Y}}_{1}+\phi_{2}\overline{\mathcal{Y}}_{2}+\phi_{3}\overline{\mathcal{Y}}_{3}+\varsigma_{1}\overline{\mathcal{Y}}_{7}+\varsigma_{2}\overline{\mathcal{Y}}_{8}+\varsigma_{3}\overline{\mathcal{Y}}_{9}}\,.$
(4.69)

Consistency of the resulting BPS equations requires vanishing of the shift
scalar $\varsigma_{3}$ unless $\kappa=\lambda=0$ corresponding to
$CSO(2,0,2)\ltimes\mathbb{R}^{4}_{\boldsymbol{s}}$ gauge group. In what
follows, we will for the moment set $\varsigma_{3}=0$ and separately consider
the $CSO(2,0,2)\ltimes\mathbb{R}^{4}_{\boldsymbol{s}}$ gauge group with
$\varsigma_{3}\neq 0$.

With $\varsigma_{3}=0$, we can compute the scalar potential and the
superpotential of the form

$\displaystyle\mathbf{V}$ $\displaystyle=$
$\displaystyle-\frac{g^{2}}{32}e^{2\left(\varphi-4(\phi_{0}+\phi_{1})\right)}\left[\kappa^{2}+10\kappa\lambda+\lambda^{2}-2(\kappa+\lambda)^{2}\cosh{4\phi_{2}}\cosh^{2}{4\phi_{3}}\phantom{\frac{1}{\sqrt{2}}}\right.$
(4.70)
$\displaystyle\left.\phantom{+\frac{1}{\sqrt{2}}}-(3\kappa^{2}-2\kappa\lambda+3\lambda^{2})\cosh{8\phi_{3}}+16(\kappa-\lambda)e^{8\phi_{1}}\sinh{4\phi_{3}}\right.$
$\displaystyle\left.\phantom{+\frac{1}{\sqrt{2}}}+4(\kappa+\lambda)\cosh{2\phi_{2}}\left(4e^{8\phi_{1}}\cosh{4\phi_{3}}-(\kappa-\lambda)\sinh{8\phi_{3}}\right)\right],$
$\displaystyle\mathcal{W}$ $\displaystyle=$
$\displaystyle\frac{g}{8}e^{\varphi-4(\phi_{0}+\phi_{1})}\left[2e^{8\phi_{1}}+(\kappa+\lambda)\cosh{2\phi_{2}}\cosh{4\phi_{3}}+(\kappa-\lambda)\sinh{4\phi_{3}}\right].$

This scalar potential can be written in term of the superpotential according
to (3.24) using

$G^{IJ}=\begin{pmatrix}G^{ij}&G^{iy}\\\ G^{xj}&G^{xy}\end{pmatrix}$ (4.72)

where

$G^{ij}=\frac{1}{40}\text{diag}(4,1,5,40\text{sech}^{2}{4\phi_{3}},10),\qquad
G^{xj}=\begin{pmatrix}\frac{2\varsigma_{1}}{5}&-\frac{3\varsigma_{1}}{20}&-\frac{\varsigma_{1}}{4}&\frac{4\varsigma_{2}e^{12\phi_{3}}}{(1+e^{8\phi_{3}})^{2}}&\frac{\varsigma_{1}}{2}\\\
\frac{2\varsigma_{2}}{5}&-\frac{3\varsigma_{2}}{20}&-\frac{\varsigma_{2}}{4}&\frac{4\varsigma_{1}e^{12\phi_{3}}}{(1+e^{8\phi_{3}})^{2}}&\frac{\varsigma_{2}}{2}\end{pmatrix},$
(4.73)

and

$G^{xy}=\begin{pmatrix}1+4\varsigma_{1}^{2}+\varsigma_{2}^{2}e^{8\phi_{3}}\,\text{sech}^{2}{4\phi_{3}}&\frac{2\varsigma_{1}\varsigma_{2}(1+4e^{8\phi_{3}}+e^{16\phi_{3}})}{(1+e^{8\phi_{3}})^{2}}\\\
\frac{2\varsigma_{1}\varsigma_{2}(1+4e^{8\phi_{3}}+e^{16\phi_{3}})}{(1+e^{8\phi_{3}})^{2}}&1+4\varsigma_{2}^{2}+\varsigma_{1}^{2}e^{8\phi_{3}}\,\text{sech}^{2}{4\phi_{3}}\end{pmatrix}.$
(4.74)

Here, we have denoted
$\Phi^{I}=\\{\varphi,\phi_{0},\phi_{1},\phi_{2},\phi_{3},\varsigma_{1},\varsigma_{2}\\}=\\{\Phi^{i},\Phi^{x}\\}$
for $i,j=1,2,\ldots,5$ and $x,y=6,7$. Note also that the scalar potential for
$CSO(2,0,2)\ltimes\mathbb{R}^{4}_{\boldsymbol{s}}$ gauge group with
$\kappa=\lambda=0$ vanishes identically leading to a family of Minkowski
vacua.

Imposing the projector (4.16) on the Killing spinors of the form

$\epsilon_{+}=e^{\frac{A(r)}{2}+B(r)\gamma_{34}}\epsilon_{+}^{0}\qquad\text{
and
}\qquad\epsilon_{-}=e^{\frac{A(r)}{2}+B(r)\gamma_{\dot{3}\dot{4}}}\epsilon_{-}^{0},$
(4.75)

we obtain the following set of BPS equations

$\displaystyle A^{\prime}$ $\displaystyle=$
$\displaystyle\frac{g}{8}e^{\varphi-4(\phi_{0}+\phi_{1})}\left[2e^{8\phi_{1}}+(\kappa+\lambda)\cosh{2\phi_{2}}\cosh{4\phi_{3}}+(\kappa-\lambda)\sinh{4\phi_{3}}\right],$
(4.76) $\displaystyle\varphi^{\prime}$ $\displaystyle=$
$\displaystyle-\frac{g}{40}e^{\varphi-4(\phi_{0}+\phi_{1})}\left[2e^{8\phi_{1}}+(\kappa+\lambda)\cosh{2\phi_{2}}\cosh{4\phi_{3}}+(\kappa-\lambda)\sinh{4\phi_{3}}\right],\qquad\
$ (4.77) $\displaystyle\phi^{\prime}_{0}$ $\displaystyle=$
$\displaystyle\frac{g}{40}e^{\varphi-4(\phi_{0}+\phi_{1})}\left[2e^{8\phi_{1}}+(\kappa+\lambda)\cosh{2\phi_{2}}\cosh{4\phi_{3}}+(\kappa-\lambda)\sinh{4\phi_{3}}\right],$
(4.78) $\displaystyle\phi_{1}^{\prime}$ $\displaystyle=$
$\displaystyle-\frac{g}{8}e^{\varphi-4(\phi_{0}+\phi_{1})}\left[2e^{8\phi_{1}}-(\kappa+\lambda)\cosh{2\phi_{2}}\cosh{4\phi_{3}}-(\kappa-\lambda)\sinh{4\phi_{3}}\right],$
(4.79) $\displaystyle\phi_{2}^{\prime}$ $\displaystyle=$
$\displaystyle-\frac{g}{2}e^{\phi_{0}-4(\phi_{0}+\phi_{1})}(\kappa+\lambda)\sinh{2\phi_{2}}\,\text{sech}\,{4\phi_{3}},$
(4.80) $\displaystyle\phi_{3}^{\prime}$ $\displaystyle=$
$\displaystyle-\frac{g}{4}e^{\phi_{0}-4(\phi_{0}+\phi_{1})}\left((\kappa+\lambda)\cosh{2\phi_{2}}\sinh{3\phi_{3}}+(\kappa-\lambda)\cosh{4\phi_{3}}\right)$
(4.81)

together with

$\displaystyle B^{\prime}$ $\displaystyle=$
$\displaystyle-\frac{g}{4}e^{\phi_{0}-4(\phi_{0}+\phi_{1})}(\kappa+\lambda)\sinh{2\phi_{2}}\tanh{4\phi_{3}},$
(4.82) $\displaystyle\varsigma^{\prime}_{1}$ $\displaystyle=$
$\displaystyle-\frac{ge^{\varphi+4\phi_{3}}}{2e^{4(\phi_{0}+\phi_{1})}}\left[\varsigma_{1}\left(\kappa-\lambda+(\kappa+\lambda)\cosh{2\phi_{2}}\right)+\varsigma_{2}(\kappa+\lambda)\sinh{2\phi_{2}}\,\text{sech}\,{4\phi_{3}}\right],$
$\displaystyle\varsigma^{\prime}_{2}$ $\displaystyle=$
$\displaystyle-\frac{ge^{\varphi-4\phi_{3}}}{2e^{4(\phi_{0}+\phi_{1})}}\left[\varsigma_{1}(\kappa+\lambda)\sinh{2\phi_{2}}\,\text{sech}\,{4\phi_{3}}-\varsigma_{2}\left(\kappa-\lambda-(\kappa+\lambda)\cosh{2\phi_{2}}\right)\right].$

We are unable to completely solve these equations for arbitrary values of the
parameters $\kappa$ and $\lambda$. However, the solutions can be separately
found for each value of $\kappa$ and $\lambda$.

####  4.4.1 Domain walls in $SO(3,1)\ltimes\mathbb{R}^{4}_{\boldsymbol{s}}$
gauge group

In this case, we set $\kappa=-\lambda=1$, and the BPS equations give
$B^{\prime}=\phi_{2}^{\prime}=0$. We can again truncate $\phi_{2}$ out and set
the constant $B=0$. As a result, we find a domain wall solution

$\displaystyle\phi_{1}$ $\displaystyle=$
$\displaystyle\frac{1}{2}\phi_{3}-\frac{1}{8}\ln\left[1+C_{1}(1+e^{8\phi_{3}})\right],$
(4.85) $\displaystyle\phi_{0}$ $\displaystyle=$ $\displaystyle
C_{0}+\frac{1}{10}\phi_{3}-\frac{1}{20}\ln(1+e^{8\phi_{3}})+\frac{1}{40}\ln\left[1+C_{1}(1+e^{8\phi_{3}})\right],$
(4.86) $\displaystyle\varphi$ $\displaystyle=$ $\displaystyle
C-C_{0}-\frac{1}{10}\phi_{3}+\frac{1}{20}\ln(1+e^{8\phi_{3}})-\frac{1}{40}\ln\left[1+C_{1}(1+e^{8\phi_{3}})\right],$
(4.87) $\displaystyle A$ $\displaystyle=$
$\displaystyle-5\varphi\,=\,5(C_{0}-C)+\frac{1}{2}\phi_{3}-\frac{1}{4}\ln(1+e^{8\phi_{3}})+\frac{1}{8}\ln\left[1+C_{1}(1+e^{8\phi_{3}})\right],\qquad\quad$
(4.88) $\displaystyle\phi_{3}$ $\displaystyle=$
$\displaystyle\frac{1}{4}\ln\tan(C_{3}-g\rho),$ (4.89)
$\displaystyle\varsigma_{1}$ $\displaystyle=$ $\displaystyle
C_{4}\sec(C_{3}-g\rho),$ (4.90) $\displaystyle\varsigma_{2}$ $\displaystyle=$
$\displaystyle C_{5}\csc(C_{3}-g\rho)$ (4.91)

with $\rho$ defined by $\frac{d\rho}{dr}=e^{\varphi-4(\phi_{0}+\phi_{1})}$.

####  4.4.2 Domain walls in $CSO(3,0,1)\ltimes\mathbb{R}^{4}_{\boldsymbol{s}}$
and $CSO(2,1,1)\ltimes\mathbb{R}^{4}_{\boldsymbol{s}}$ gauge groups

For $\lambda=0$ and $\kappa=\pm 1$ corresponding to
$CSO(3,0,1)\ltimes\mathbb{R}^{4}_{\boldsymbol{s}}$ and
$CSO(2,1,1)\ltimes\mathbb{R}^{4}_{\boldsymbol{s}}$ gauge groups, the domain
wall solution is given by

$\displaystyle\phi_{1}$ $\displaystyle=$
$\displaystyle\frac{1}{16}\ln\left[(e^{4\phi_{2}}-1)(1+2e^{C_{3}}+e^{2C_{3}}-e^{2C_{3}+4\phi_{2}})\right]-\frac{1}{8}\ln(4-2e^{4\phi_{2}}),\qquad\,$
(4.92) $\displaystyle\phi_{2}$ $\displaystyle=$
$\displaystyle\frac{1}{4}\ln\left[\frac{4(1+e^{C_{3}})^{2}+(1+2e^{C_{3}})^{2}g^{2}\rho^{2}}{4e^{2C_{3}}+(1+2e^{C_{3}})^{2}g^{2}\rho^{2}}\right],$
(4.93) $\displaystyle\phi_{3}$ $\displaystyle=$
$\displaystyle\frac{1}{8}\ln\left[\frac{(e^{2\phi_{2}}-1)(1+e^{C_{3}}+e^{C_{3}+2\phi_{2}})}{1+e^{C_{3}}+e^{2\phi_{2}}-e^{C_{3}+4\phi_{2}}}\right],$
(4.94) $\displaystyle\phi_{0}$ $\displaystyle=$ $\displaystyle
C_{0}-\frac{1}{5}\phi_{1}+\frac{1}{40}\ln(1-e^{4\phi_{2}})-\frac{1}{40}\ln\left[1+2e^{C_{3}}+e^{2C_{3}}-e^{2C_{3}+4\phi_{2}}\right],\quad$
(4.95) $\displaystyle\varphi$ $\displaystyle=$ $\displaystyle C-\phi_{0},$
(4.96) $\displaystyle A$ $\displaystyle=$ $\displaystyle-5\varphi$ (4.97)

together with

$B=C_{B}+\frac{1}{4}\sin^{-1}\left[e^{C_{3}}\sqrt{\frac{e^{4\phi_{2}}-1}{1+2e^{C_{3}}}}\right]+\frac{1}{4}\tan^{-1}\left[\sqrt{\frac{(e^{4\phi_{2}}-1)(1+e^{C_{3}})^{2}}{1+2e^{C_{3}}+e^{2C_{3}}-e^{2C_{3}+4\phi_{2}}}}\right].$
(4.98)

In this solution, we have defined the coordinate $\rho$ by
$\frac{d\rho}{dr}=e^{-2\phi_{0}-2\phi_{1}}$ and set the integration constant
for $\phi_{2}$ solution to be $C_{2}=\frac{1}{16(1+2e^{C_{3}})^{2}}$ in order
to simplify the expression for the solution. We also note that the two gauge
groups have exactly the same domain wall solution since the parameter $\kappa$
does not appear anywhere in the solution. In more detail, $\kappa^{2}$ appears
in $\phi_{2}$ solution as $g^{2}\kappa^{2}\rho^{2}$, but this term is simply
given by $g^{2}\rho^{2}$ for $\kappa=\pm 1$.
For the remaining scalars $\varsigma_{1}$ and $\varsigma_{2}$, we are not able
to analytically find their solutions. We can instead perform a numerical
analysis to find these solutions, but we will not pursue any further along
this direction. In any case, these scalars can be consistently truncated out
since they do not appear in the scalar potential.

####  4.4.3 Domain walls in $SO(4)\ltimes\mathbb{R}^{4}_{\boldsymbol{s}}$ and
$SO(2,2)\ltimes\mathbb{R}^{4}_{\boldsymbol{s}}$ gauge groups

In this case, we set $\kappa=\lambda=\pm 1$ corresponding to
$SO(4)\ltimes\mathbb{R}^{4}_{\boldsymbol{s}}$ and
$SO(2,2)\ltimes\mathbb{R}^{4}_{\boldsymbol{s}}$ gauge groups. As in the
previous case, the resulting BPS equations are very complicated to find
explicit solutions. Therefore, we will set $\varsigma_{1}=\varsigma_{2}=0$ and
find the domain wall solution for the remaining fields as follows

$\displaystyle\phi_{1}$ $\displaystyle=$
$\displaystyle\frac{1}{16}\ln\left[e^{4\phi_{2}}-e^{2C_{3}}(e^{4\phi_{2}}-1)^{2}\right]-\frac{1}{8}\ln(2-e^{4\phi_{2}}),$
(4.99) $\displaystyle\phi_{2}$ $\displaystyle=$
$\displaystyle\frac{1}{4}\ln\left[\frac{1-2e^{2g\kappa\rho}+e^{4g\kappa\rho}+4e^{2C_{3}}}{1+2e^{2g\kappa\rho}+e^{4g\kappa\rho}+4e^{2C_{3}}}\right],$
(4.100) $\displaystyle\phi_{3}$ $\displaystyle=$
$\displaystyle\frac{1}{8}\ln\left[\frac{e^{2\phi_{2}}-e^{C_{3}}+e^{C_{3}+4\phi_{2}}}{e^{2\phi_{2}}+e^{C_{3}}-e^{C_{3}+4\phi_{2}}}\right],$
(4.101) $\displaystyle\phi_{0}$ $\displaystyle=$ $\displaystyle
C_{0}-\frac{\phi_{1}}{5}-\frac{1}{20}\ln(e^{4\phi_{2}}-1)+\frac{1}{4}\ln\left[e^{4\phi_{2}}-e^{2C_{3}}+e^{4\phi_{2}+2C_{3}}(2-e^{4\phi_{2}})\right],\qquad\
\ $ (4.102) $\displaystyle\varphi$ $\displaystyle=$ $\displaystyle
C-\phi_{0},$ (4.103) $\displaystyle B$ $\displaystyle=$ $\displaystyle
C_{B}-\frac{1}{8}\tan^{-1}\left[\frac{e^{-C_{3}}(e^{4\phi_{2}}-2e^{2C_{3}}+2e^{2C_{3}+4\phi_{2}})}{2\sqrt{e^{4\phi_{2}}-e^{2C_{3}}+2e^{2C_{3}+4\phi_{2}}-e^{2C_{3}+8\phi_{2}}}}\right]$
(4.104)
$\displaystyle-\frac{1}{8}\tan^{-1}\left[\frac{e^{-C_{3}}(1+2e^{2C_{3}}+2e^{2C_{3}+4\phi_{2}})}{2\sqrt{e^{4\phi_{2}}(1+2e^{2C_{3}})-2e^{2C_{3}}-e^{2C_{3}+8\phi_{2}}}}\right],$
$\displaystyle A$ $\displaystyle=$ $\displaystyle-5\varphi$ (4.105)

with $\frac{d\rho}{dr}=e^{\varphi-4(\phi_{0}+\phi_{1})}$.

####  4.4.4 Domain walls in $CSO(2,0,2)\ltimes\mathbb{R}^{4}_{\boldsymbol{s}}$
gauge group

Finally, we consider the case of $\kappa=\lambda=0$ corresponding to
$CSO(2,0,2)\ltimes\mathbb{R}^{4}_{\boldsymbol{s}}$ gauge group. Using the
coset representative (4.69), we find the T-tensor given by

$T^{\alpha\dot{\beta}}=\frac{1}{2}e^{\varphi-4(\phi_{0}-\phi_{1})}\left[\,(\gamma^{5})^{\alpha\beta}\delta^{\dot{\beta}}_{\beta}+2\varsigma_{3}\,(\gamma^{12})^{\alpha\beta}\delta^{\dot{\beta}}_{\beta}\right].$
(4.106)

By the general procedure given in section 3.1, we find the superpotential

$\mathcal{W}=\frac{g}{4}e^{\varphi-4(\phi_{0}-\phi_{1})}\sqrt{\varsigma_{3}^{2}+1}$
(4.107)

and the following projectors

$\displaystyle\hat{\gamma}_{r}\epsilon_{+\alpha}$ $\displaystyle=$
$\displaystyle\Omega_{\alpha\beta}\frac{\left[\,(\gamma^{5})^{\beta\gamma}\delta^{\dot{\beta}}_{\gamma}+2\varsigma_{3}\,(\gamma^{12})^{\beta\gamma}\delta^{\dot{\beta}}_{\gamma}\right]}{\sqrt{\varsigma_{3}^{2}+1}}\epsilon_{-\dot{\beta}},$
(4.108) $\displaystyle\hat{\gamma}_{r}\epsilon_{-\dot{\alpha}}$
$\displaystyle=$
$\displaystyle-\Omega_{\dot{\alpha}\dot{\beta}}\frac{\left[\,(\gamma^{5})^{\alpha\gamma}\delta^{\dot{\beta}}_{\gamma}+2\varsigma_{3}\,(\gamma^{12})^{\alpha\gamma}\delta_{\dot{\gamma}}^{\beta}\right]}{\sqrt{\varsigma_{3}^{2}+1}}\epsilon_{+\alpha}\,.$
(4.109)

As expected for half-supersymmetric solutions, these projectors are not
independent. In addition, for $\varsigma_{3}=0$, they reduce to a simpler
projector given in (4.16). At this point, it is useful to note that for this
gauge group, the scalar potential vanishes as previously mentioned, so there
exists a six-dimensional Minkowski vacuum for this gauge group. However, the
superpotential (4.107) does not have any stationary points, so this Minkowski
vacuum is not supersymmetric.

With the following ansatz for the Killing spinors

$\epsilon_{+}=e^{\frac{A(r)}{2}+B(r)\gamma_{34}}\epsilon_{+}^{0}\quad\text{
and
}\quad\epsilon_{-}=e^{\frac{A(r)}{2}-B(r)\gamma_{\dot{3}\dot{4}}}\epsilon_{-}^{0},$
(4.110)

we obtain the BPS equations

$\displaystyle A^{\prime}$ $\displaystyle=$
$\displaystyle\frac{g}{4}e^{\varphi-4(\phi_{0}-\phi_{1})}\sqrt{\varsigma_{3}^{2}+1},\qquad\varphi^{\prime}\
=\
-\frac{ge^{\varphi-4(\phi_{0}-\phi_{1})}(1+20\varsigma_{3}^{2})}{20\sqrt{\varsigma_{3}^{2}+1}},$
$\displaystyle\phi^{\prime}_{0}$ $\displaystyle=$
$\displaystyle\frac{ge^{\varphi-4(\phi_{0}-\phi_{1})}}{20\sqrt{\varsigma_{3}^{2}+1}},\qquad\phi_{1}^{\prime}\
=\
-\frac{ge^{\varphi-4(\phi_{0}-\phi_{1})}}{4\sqrt{\varsigma_{3}^{2}+1}},\qquad\phi_{2}^{\prime}\
=\ \phi_{3}^{\prime}\ =\ 0,$ $\displaystyle\varsigma_{1}^{\prime}$
$\displaystyle=$
$\displaystyle-\frac{4ge^{\varphi-4(\phi_{0}-\phi_{1})}\varsigma_{3}^{2}\varsigma_{1}}{\sqrt{\varsigma_{3}^{2}+1}},\qquad\varsigma_{2}^{\prime}\
=\
-\frac{4ge^{\varphi-4(\phi_{0}-\phi_{1})}\varsigma_{3}^{2}\varsigma_{2}}{\sqrt{\varsigma_{3}^{2}+1}},$
$\displaystyle\varsigma_{3}^{\prime}$ $\displaystyle=$ $\displaystyle-
ge^{\varphi-4(\phi_{0}-\phi_{1})}\varsigma_{3}\sqrt{\varsigma_{3}^{2}+1},\qquad
B^{\prime}\ =\
\frac{ge^{\varphi-4(\phi_{0}-\phi_{1})}}{20\sqrt{\varsigma_{3}^{2}+1}}\,.$
(4.111)

With a new radial coordinate $\rho$ defined by
$\frac{d\rho}{dr}=e^{\varphi-4\phi_{0}}$, the corresponding solution is given
by

$\displaystyle A$ $\displaystyle=$
$\displaystyle-\frac{1}{4}\ln\varsigma_{3},\qquad\quad\ \ B\ =\
C_{B}-\frac{1}{2}\tan^{-1}{2\varsigma_{3}},$ (4.112) $\displaystyle\varphi$
$\displaystyle=$ $\displaystyle
C+\frac{1}{20}\ln\varsigma_{3}+\frac{1}{10}\ln(1+4\varsigma_{3}^{2}),$ (4.113)
$\displaystyle\phi_{0}$ $\displaystyle=$ $\displaystyle
C_{0}+\frac{1}{20}\ln\varsigma_{3}-\frac{1}{40}\ln(1+4\varsigma_{3}^{2}),$
(4.114) $\displaystyle\phi_{1}$ $\displaystyle=$ $\displaystyle
C_{1}+\frac{1}{4}\ln\varsigma_{3}-\frac{1}{8}\ln(1+4\varsigma_{3}^{2}),$
(4.115) $\displaystyle\varsigma_{1}$ $\displaystyle=$ $\displaystyle
C_{4}\sqrt{1+4\varsigma_{3}^{2}},\qquad\varsigma_{2}\ =\
C_{5}\sqrt{1+4\varsigma_{3}^{2}},$ (4.116) $\displaystyle\varsigma_{3}$
$\displaystyle=$ $\displaystyle\frac{1}{g\rho e^{4C_{1}}+C_{6}}\,.$ (4.117)

##  5 Domain walls from gaugings in
$(\mathbf{15}+\overline{\mathbf{40}})^{-1}$ representation

We now consider gaugings with non-vanishing components of the embedding tensor
in both $\mathbf{15}^{-1}$ and $\overline{\mathbf{40}}^{-1}$ representations.
These gaugings are dyonic with the embedding tensor containing both electric
and magnetic parts. The full embedding tensor is given by
$\theta^{AM}=(\theta^{Am},\theta^{A}_{m})$ with

$\theta^{Am}\ =\
\mathbb{T}^{A}_{np}U^{np,m}\qquad\textrm{and}\qquad\theta^{A}_{m}\ =\
\mathbb{T}^{An}Y_{nm}$ (5.1)

for $Y_{mn}=Y_{(mn)}$ and $U^{mn,p}=U^{[mn],p}$ satisfying $U^{[mn,p]}=0$.

However, for dyonic gaugings, the first condition in the quadratic constraint
(2.13) is not automatically satisfied. For the embedding tensor given in
(5.1), we find that this constraint imposes the following condition

$U^{np,m}Y_{qm}=0\,.$ (5.2)

To solve this condition, we follow [41] and split the $GL(5)$ index as
$m=(i,x)$. By choosing a suitable basis, we can take $Y_{mn}$ to be

$Y_{ij}=\text{diag}(+1,...,+1,-1,...,-1)\qquad\text{and}\qquad Y_{xy}=0\,.$
(5.3)

The constraint (5.2) then implies that only the components $U^{xy,z}$ and
$U^{ix,y}=U^{i(x,y)}$ are non-vanishing. As a result, the embedding tensor is
parametrized by the following tensors

$Y_{ij},\qquad U^{i(x,y)},\qquad U^{xy,z}\,.$ (5.4)

We now consider different possible gauge groups with $\text{rank}Y=0,1,...,5$.
There are two trivial cases for $\text{rank}Y=5$ with $U^{mn,p}=0$ and
$\text{rank}Y=0$ with all $Y_{mn}=0$. These correspond respectively to
gaugings in $\mathbf{15}^{-1}$ and $\overline{\mathbf{40}}^{-1}$
representations and have already been considered in the previous two sections.

For $\text{rank}Y=4$, only $U^{i5,5}$ can be non-vanishing, but another
condition from the quadratic constraint (2.13) requires $U^{i5,5}=0$.
Accordingly, the corresponding gauge groups are given by $CSO(4,0,1)$,
$CSO(3,1,1)$ and $CSO(2,2,1)$ which again have been considered in section 3.

In the following, we will study supersymmetric domain walls in the two non-
trivial cases with $\text{rank}Y=3$ and $\text{rank}Y=2$. Gaugings in these
cases are expected to arise from a circle reduction of seven-dimensional
maximal gauged supergravity with the embedding tensor in both $\mathbf{15}$
and $\overline{\mathbf{40}}$ representations of $SL(5)$. Similar to the seven-
dimensional solutions given in [15], we will find that in these gaugings, the
domain walls are $\frac{1}{4}$-BPS preserving eight supercharges. For the case
of $\text{rank}Y=1$, the second condition from the quadratic constraint (2.13)
is much more complicated to find a non-trivial solution for $U^{i(x,y)}$ and
$U^{xy,z}$. We refrain from discussing this case here.

###  5.1 $\frac{1}{4}$-BPS domain walls for $\text{rank}Y=3$

We first consider the case of $\text{rank}Y=3$ with $i,j=1,2,3$. The second
condition from the quadratic constraint (2.13) becomes

$\varepsilon_{ijk}U^{jx,z}\varepsilon_{zw}U^{kw,y}=\frac{1}{\sqrt{2}}Y_{ij}U^{jx,y}$
(5.5)

which can be solved by $U^{ix,y}$ of the form

$U^{ix,y}=-\frac{1}{2\sqrt{2}}\varepsilon^{xz}{(\Sigma^{i})_{z}}^{y}$ (5.6)

where ${(\Sigma^{i})_{x}}^{y}$ are $2\times 2$ matrices. In terms of these
$\Sigma^{i}$, the quadratic constraint (5.5) can be rewritten as

$[\Sigma^{i},\Sigma^{j}]=2\varepsilon^{ijk}Y_{kl}\Sigma^{l}.$ (5.7)

As pointed out in [41], a real, non-vanishing solution for $U^{ix,y}$ is
possible only for

$Y_{ij}=\textrm{diag}(1,1,-1)$ (5.8)

with the explicit form of $\Sigma^{i}$ given in terms of Pauli matrices as

$\Sigma^{1}=\sigma_{1},\qquad\Sigma^{2}=\sigma_{3},\qquad\Sigma^{3}=i\sigma_{2}\,.$
(5.9)

The constraint (5.7) is then the Lie algebra of a non-compact group $SO(2,1)$.
It should also be noted that the tensor $U^{xy,z}$ is not constrained by this
condition, so it can be parametrized by an arbitrary two-component vector
$u^{x}$ as

$U^{xy,z}=\varepsilon^{xy}u^{z}\,.$ (5.10)

We now consider the corresponding gauge algebra spanned by the following gauge
generators

$\displaystyle X^{x}$ $\displaystyle=$
$\displaystyle-\frac{1}{2\sqrt{2}}\varepsilon^{yz}{(\Sigma^{i})_{z}}^{x}\boldsymbol{s}_{iy}+\varepsilon^{yz}u^{x}\boldsymbol{s}_{yz},$
(5.11) $\displaystyle X_{ij}$ $\displaystyle=$ $\displaystyle
2Y_{k[i}{\boldsymbol{t}^{k}}_{j]}+2\sqrt{2}\varepsilon_{ijk}u^{x}{\boldsymbol{t}^{k}}_{x}-\frac{1}{2}\varepsilon_{ijk}{(\Sigma^{k})_{z}}^{x}{\boldsymbol{t}^{z}}_{x},$
(5.12) $\displaystyle X_{ix}$ $\displaystyle=$ $\displaystyle
Y_{ik}{\boldsymbol{t}^{k}}_{x}+\frac{1}{2}\varepsilon_{ijk}{(\Sigma^{j})_{x}}^{z}{\boldsymbol{t}^{k}}_{z}.$
(5.13)

To determine the form of the corresponding gauge group, we explicitly evaluate
these generators in vector representation and find the following commutation
relations

$\displaystyle\left[X^{x},X^{y}\right]$ $\displaystyle=$ $\displaystyle
0,\qquad\,\left[X_{ij},X^{x}\right]={(X_{ij})_{y}}^{x}X^{y},\qquad\left[X_{ix},X^{y}\right]=0,$
(5.14) $\displaystyle\left[X_{ix},X_{jy}\right]$ $\displaystyle=$
$\displaystyle
0,\qquad\left[X_{ij},X_{kx}\right]=-2{(X_{ij})_{kx}}^{ly}X_{ly},$ (5.15)
$\displaystyle\left[X_{ij},X_{kl}\right]$ $\displaystyle=$
$\displaystyle-{(X_{ij})_{kl}}^{pq}X_{pq}-2{(X_{ij})_{kl}}^{px}X_{px}.$ (5.16)

Redefining the $X_{ij}$ generators as

$\widetilde{X}_{ij}=X_{ij}-\frac{2\sqrt{2}}{3}\varepsilon_{ijk}\eta^{kl}X_{lx}u^{x}$
(5.17)

with $\eta^{ij}=\text{diag}(+1,+1,-1)$, we find that $\widetilde{X}_{ij}$
generate an $SO(2,1)$ subgroup with the Lie algebra

$\left[\widetilde{X}_{ij},\widetilde{X}_{kl}\right]=-{(\widetilde{X}_{ij})_{kl}}^{pq}\widetilde{X}_{pq}\,.$
(5.18)

The remaining generators $X_{ix}$ and $X_{x}$, which transform non-trivially
under $SO(2,1)$, generate two translation groups. Note also that there are
only four independent $X_{ix}$ generators. With all these, the resulting gauge
group is then given by

$G_{0}=SO(2,1)\ltimes\left(\mathbb{R}^{4}\times\mathbb{R}^{2}_{\boldsymbol{s}}\right)$
(5.19)

in which $\mathbb{R}^{2}_{\boldsymbol{s}}$ is the translation group from the
shift symmetries generated by $X^{x}$. As also pointed out in [41], we see
that the vector $u^{x}$ does not change the gauge algebra, so we can set
$u^{x}=0$ for simplicity.

We now look for supersymmetric domain wall solutions invariant under
$SO(2)\subset SO(2,1)$ generated by $X_{12}$. There are five $SO(2)$ singlet
scalars corresponding to the non-compact generators

$\displaystyle\mathbf{Y}_{\boldsymbol{d}}$ $\displaystyle=$
$\displaystyle\hat{\boldsymbol{t}}^{+}_{1\dot{1}}+\hat{\boldsymbol{t}}^{+}_{2\dot{2}}+\hat{\boldsymbol{t}}^{+}_{3\dot{3}}+\hat{\boldsymbol{t}}^{+}_{4\dot{4}}+\hat{\boldsymbol{t}}^{+}_{5\dot{5}},$
(5.20) $\displaystyle\mathbf{Y}_{1}$ $\displaystyle=$ $\displaystyle
2\,\hat{\boldsymbol{t}}^{+}_{1\dot{1}}+2\,\hat{\boldsymbol{t}}^{+}_{2\dot{2}}+2\,\hat{\boldsymbol{t}}^{+}_{3\dot{3}}-3\,\hat{\boldsymbol{t}}^{+}_{4\dot{4}}-3\,\hat{\boldsymbol{t}}^{+}_{5\dot{5}},$
(5.21) $\displaystyle\mathbf{Y}_{2}$ $\displaystyle=$
$\displaystyle\hat{\boldsymbol{t}}^{+}_{1\dot{1}}+\hat{\boldsymbol{t}}^{+}_{2\dot{2}}-2\,\hat{\boldsymbol{t}}^{+}_{3\dot{3}},$
(5.22) $\displaystyle\mathbf{Y}_{3}$ $\displaystyle=$
$\displaystyle\boldsymbol{s}_{12},$ (5.23) $\displaystyle\mathbf{Y}_{4}$
$\displaystyle=$ $\displaystyle\boldsymbol{s}_{45}.$ (5.24)

Using the coset representative of the form

$V=e^{\varphi\mathbf{Y}_{\boldsymbol{d}}+\phi_{1}\mathbf{Y}_{1}+\phi_{2}\mathbf{Y}_{2}+\varsigma_{1}\mathbf{Y}_{3}+\varsigma_{2}\mathbf{Y}_{4}},$
(5.25)

we find the scalar potential

$\mathbf{V}=-\frac{g^{2}}{4}e^{2(\varphi-8\phi_{1}+2\phi_{2})}(e^{12\phi_{2}}+6)\,.$
(5.26)

Consistency of the BPS equations from $\delta\chi_{\pm}$ conditions requires
$\varsigma_{1}=0$. After truncating out $\varsigma_{1}$, we find the T-tensor

$T^{\alpha\dot{\beta}}=\frac{2}{g}e^{\varphi-8\phi_{1}-4\phi_{2}}\left[\mathcal{W}_{1}(\delta^{\alpha}_{1}\delta^{\dot{\beta}}_{3}-\delta^{\alpha}_{3}\delta^{\dot{\beta}}_{1})+\mathcal{W}_{2}(\delta^{\alpha}_{2}\delta^{\dot{\beta}}_{4}-\delta^{\alpha}_{4}\delta^{\dot{\beta}}_{2})\right]$
(5.27)

with

$\mathcal{W}_{1}=\frac{g}{4\sqrt{2}}e^{\varphi-8\phi_{1}-4\phi_{2}}(3-e^{12\phi_{2}}),\qquad\mathcal{W}_{2}=\frac{g}{4\sqrt{2}}e^{\varphi-8\phi_{1}-4\phi_{2}}(1-e^{12\phi_{2}}).$
(5.28)

It turns out that only $\mathcal{W}_{1}$ gives rise to the superpotential in
term of which the scalar potential can be written.
With the superpotential given by $\mathcal{W}_{1}$, the unbroken supersymmetry
corresponds to $\epsilon^{1}_{\pm}$ and $\epsilon^{3}_{\pm}$. Therefore, we
set $\epsilon^{2}_{\pm}=\epsilon^{4}_{\pm}=0$ in the following analysis.
Alternatively, we can implement this by imposing an additional projector of
the form

$\gamma^{3}\epsilon_{\mp}=\epsilon_{\mp}\,.$ (5.29)

By the same procedure as in the previous cases together with the projector
(3.31), we obtain the BPS equations, with $\varsigma_{2}=\varsigma$,

$\displaystyle A^{\prime}$ $\displaystyle=$
$\displaystyle\frac{g}{4\sqrt{2}}e^{\varphi-8\phi_{1}-4\phi_{2}}(3-e^{12\phi_{2}}),$
(5.30) $\displaystyle\varphi^{\prime}$ $\displaystyle=$
$\displaystyle-\frac{g}{20\sqrt{2}}e^{\varphi-8\phi_{1}-4\phi_{2}}(3-e^{12\phi_{2}}),$
(5.31) $\displaystyle\phi_{1}^{\prime}$ $\displaystyle=$
$\displaystyle\frac{g}{15\sqrt{2}}e^{\varphi-8\phi_{1}-4\phi_{2}}(3-e^{12\phi_{2}}),$
(5.32) $\displaystyle\phi_{2}^{\prime}$ $\displaystyle=$
$\displaystyle\frac{g}{6\sqrt{2}}e^{\varphi-8\phi_{1}-4\phi_{2}}(3+e^{12\phi_{2}}),$
(5.33) $\displaystyle\varsigma$ $\displaystyle=$
$\displaystyle-\frac{g}{\sqrt{2}}e^{\varphi-8\phi_{1}-4\phi_{2}}(3-e^{12\phi_{2}})\varsigma\,.$
(5.34)

Introducing a new radial coordinate $\rho$ via
$\frac{d\rho}{dr}=e^{\varphi-8\phi_{1}+2\phi_{2}}$, we find a domain wall
solution

$\displaystyle e^{6\phi_{2}}$ $\displaystyle=$
$\displaystyle\sqrt{\frac{3}{2}}\tan(\sqrt{3}g\rho+C_{2}),$ (5.35)
$\displaystyle\phi_{1}$ $\displaystyle=$ $\displaystyle
C_{1}+\frac{2}{5}\phi_{2}-\frac{1}{20}\ln(3+2e^{12\phi_{2}}),$ (5.36)
$\displaystyle\varsigma$ $\displaystyle=$ $\displaystyle
C_{3}e^{-6\phi_{2}}(3+2e^{12\phi_{2}})^{\frac{3}{4}},$ (5.37)
$\displaystyle\varphi$ $\displaystyle=$ $\displaystyle
C-\frac{3}{4}C_{1}-\frac{3}{10}\phi_{2}-\frac{3}{80}\ln(3+2e^{12\phi_{2}}),$
(5.38) $\displaystyle A$ $\displaystyle=$ $\displaystyle-5\varphi\ =\
-5C+\frac{15}{4}C_{1}+\frac{3}{2}\phi_{2}+\frac{3}{16}\ln(3+2e^{12\phi_{2}}).$
(5.39)

###  5.2 $\frac{1}{4}$-BPS domain walls for $\text{rank}Y=2$

In this case, $i,j=1,2$, we have $Y_{ij}=\text{diag}(1,\pm 1)$. The second
condition from the quadratic constraint (2.13) allows only the components
$U^{xy,z}$, $x,y,\ldots=3,4,5$, which can be parametrized by a $3\times 3$
traceless matrix ${u_{x}}^{y}$ as

$U^{xy,z}=\frac{1}{2\sqrt{2}}\varepsilon^{xyt}{u_{t}}^{z}$ (5.40)

with ${u_{x}}^{x}=0$. The non-vanishing gauge generators read

$\displaystyle X^{x}$ $\displaystyle=$
$\displaystyle\frac{1}{2\sqrt{2}}\epsilon^{yz,t}{u_{t}}^{x}\boldsymbol{s}_{yz},$
(5.41) $\displaystyle X_{12}$ $\displaystyle=$ $\displaystyle
2Y_{k[1}{\boldsymbol{t}^{k}}_{2]}+\frac{1}{2}{u_{x}}^{y}{\boldsymbol{t}^{x}}_{y},$
(5.42) $\displaystyle X_{ix}$ $\displaystyle=$ $\displaystyle
Y_{ij}{\boldsymbol{t}^{j}}_{x}-\frac{1}{2}\varepsilon_{ij}{u_{x}}^{y}{\boldsymbol{t}^{j}}_{y}$
(5.43)

with the commutation relations given by

$\displaystyle\left[X^{x},X^{y}\right]$ $\displaystyle=$ $\displaystyle
0,\qquad\left[X^{x},X_{iy}\right]=0,\qquad\left[X_{ix},X_{jy}\right]=0,$
(5.44) $\displaystyle\left[X_{12},X^{x}\right]$ $\displaystyle=$
$\displaystyle{(X_{12})_{y}}^{x}X^{y},\qquad\left[X_{12},X_{ix}\right]=-2{(X_{12})_{ix}}^{jy}X_{jy}.$
(5.45)

$X^{x}$ and $X_{ix}$ commute with each other and separately generate two
translation groups $\mathbb{R}^{3}_{\boldsymbol{s}}$ and $\mathbb{R}^{6}$
which transform non-trivially under $X_{12}$. The single $X_{12}$ generator in
turn leads to a compact $SO(2)$ or a non-compact $SO(1,1)$ group for
$Y_{ij}=\text{diag}(1,1)$ or $Y_{ij}=\text{diag}(1,-1)$, respectively. The
corresponding gauge groups are then given by
$SO(2)\ltimes\left(\mathbb{R}^{6}\times\mathbb{R}^{3}_{\boldsymbol{s}}\right)$
or
$SO(1,1)\ltimes\left(\mathbb{R}^{6}\times\mathbb{R}^{3}_{\boldsymbol{s}}\right)$.

####  5.2.1 Domain walls in
$SO(2)\ltimes\left(\mathbb{R}^{6}\times\mathbb{R}^{2}_{\boldsymbol{s}}\right)$
gauge group

To find solutions with a non-trivial residual symmetry, we will consider
$SO(2)\ltimes\left(\mathbb{R}^{6}\times\mathbb{R}^{2}_{\boldsymbol{s}}\right)$
gauge group with $Y_{ij}=\delta_{ij}$. In vector representation, the $X_{12}$
generator is given by

${(X_{12})_{m}}^{n}=\begin{pmatrix}2i{(\sigma_{2})_{i}}^{j}&0_{2\times 6}\\\
0_{2\times 6}&{u_{x}}^{y}\end{pmatrix}.$ (5.46)

Accordingly, we choose the matrix ${u_{x}}^{y}$ to be

${u_{x}}^{y}=\begin{pmatrix}0&0&0\\\ 0&0&-\lambda\\\ 0&\lambda&0\end{pmatrix}$
(5.47)

with $\lambda\in\mathbb{R}$. The $SO(2)$ subgroup is then embedded diagonally
with only $X^{4}$ and $X^{5}$ non-vanishing. Thus, the corresponding gauge
group, in this case, is given by
$SO(2)\ltimes\left(\mathbb{R}^{6}\times\mathbb{R}^{2}_{\boldsymbol{s}}\right)$.

There are five $SO(2)$ singlets corresponding to the following non-compact
generators commuting with $X_{12}$

$\displaystyle\overline{\mathbf{Y}}_{\boldsymbol{d}}$ $\displaystyle=$
$\displaystyle\hat{\boldsymbol{t}}^{+}_{1\dot{1}}+\hat{\boldsymbol{t}}^{+}_{2\dot{2}}+\hat{\boldsymbol{t}}^{+}_{3\dot{3}}+\hat{\boldsymbol{t}}^{+}_{4\dot{4}}+\hat{\boldsymbol{t}}^{+}_{5\dot{5}},$
(5.48) $\displaystyle\overline{\mathbf{Y}}_{1}$ $\displaystyle=$
$\displaystyle
3\,\hat{\boldsymbol{t}}^{+}_{1\dot{1}}+3\,\hat{\boldsymbol{t}}^{+}_{2\dot{2}}-2\,\hat{\boldsymbol{t}}^{+}_{3\dot{3}}-2\,\hat{\boldsymbol{t}}^{+}_{4\dot{4}}-2\,\hat{\boldsymbol{t}}^{+}_{5\dot{5}},$
(5.49) $\displaystyle\overline{\mathbf{Y}}_{2}$ $\displaystyle=$
$\displaystyle-2\,\hat{\boldsymbol{t}}^{+}_{3\dot{3}}+\hat{\boldsymbol{t}}^{+}_{4\dot{4}}+\,\hat{\boldsymbol{t}}^{+}_{5\dot{5}},$
(5.50) $\displaystyle\overline{\mathbf{Y}}_{3}$ $\displaystyle=$
$\displaystyle\boldsymbol{s}_{12},$ (5.51)
$\displaystyle\overline{\mathbf{Y}}_{4}$ $\displaystyle=$
$\displaystyle\boldsymbol{s}_{45}\,.$ (5.52)

With the coset representative

$V=e^{\varphi\overline{\mathbf{Y}}_{\boldsymbol{d}}+\phi_{1}\overline{\mathbf{Y}}_{1}+\phi_{2}\overline{\mathbf{Y}}_{2}+\varsigma_{1}\overline{\mathbf{Y}}_{3}+\varsigma_{2}\overline{\mathbf{Y}}_{4}},$
(5.53)

it turns out that the scalar potential vanishes identically. On the other
hand, the T-tensor is given by

$T^{\alpha\dot{\beta}}=\frac{e^{\varphi-12\phi_{1}}}{2\sqrt{2}}\left[\lambda\,(\gamma^{3})^{\alpha\beta}+2\,\Omega^{\alpha\beta}+2\varsigma_{1}\left[\lambda\,(\gamma^{45})^{\alpha\beta}-2\,(\gamma^{12})^{\alpha\beta}\right]\right]\delta^{\dot{\beta}}_{\beta}$
(5.54)

or explicitly

$T^{\alpha\dot{\beta}}=\frac{e^{\varphi-12\phi_{1}}}{2\sqrt{2}}\begin{pmatrix}2(\lambda+2)\varsigma_{1}&0&(\lambda+2)&0\\\
0&2(\lambda-2)\varsigma_{1}&0&-(\lambda-2)\\\
-(\lambda+2)&0&2(\lambda+2)\varsigma_{1}&0\\\
0&(\lambda-2)&0&2(\lambda-2)\varsigma_{1}\end{pmatrix}.$ (5.55)

This leads to two superpotentials

$\displaystyle\mathcal{W}_{1}$ $\displaystyle=$
$\displaystyle\frac{g}{4\sqrt{2}}e^{\varphi-12\phi_{1}}(\lambda+2)\sqrt{1+4\varsigma_{1}^{2}},$
(5.56) $\displaystyle\mathcal{W}_{2}$ $\displaystyle=$
$\displaystyle\frac{g}{4\sqrt{2}}e^{\varphi-12\phi_{1}}(\lambda-2)\sqrt{1+4\varsigma_{1}^{2}}\,.$
(5.57)

Unlike the previous $\text{rank}Y=3$ case, both of these give a valid
superpotential in term of which the scalar potential can be written. As in the
previous case, half of the supersymmetry is broken by choosing any one of
these two possibilities which again corresponds to imposing an additional
$\gamma^{3}$ projector of the form

$\gamma^{3}\epsilon_{\pm}=\epsilon_{\pm}\qquad\text{ or
}\qquad\gamma^{3}\epsilon_{\pm}=-\epsilon_{\pm}$ (5.58)

for $\mathcal{W}=\mathcal{W}_{1}$ or $\mathcal{W}=\mathcal{W}_{2}$,
respectively. Together with the usual $\hat{\gamma}_{r}$ projectors

$\hat{\gamma}_{r}\epsilon_{+\alpha}=\Omega_{\alpha\beta}\frac{T^{\beta\dot{\beta}}}{A^{\prime}}\epsilon_{-\dot{\beta}},\qquad\hat{\gamma}_{r}\epsilon_{-\dot{\alpha}}=-\Omega_{\dot{\alpha}\dot{\beta}}\frac{T^{\alpha\dot{\beta}}}{A^{\prime}}\epsilon_{+\alpha},$
(5.59)

the resulting solutions will preserve only eight supercharges or $\frac{1}{4}$
of the original supersymmetry.

With the following ansatz for the Killing spinors

$\epsilon_{+}=e^{\frac{A(r)}{2}+B(r)\gamma_{12}}\epsilon_{+}^{0}\qquad\text{
and
}\qquad\epsilon_{-}=e^{\frac{A(r)}{2}-B(r)\gamma_{\dot{1}\dot{2}}}\epsilon_{-}^{0},$
(5.60)

for $\epsilon^{0}_{\pm}$ satisfying the projectors (5.58) and (5.59), we
obtain the following BPS equations

$\displaystyle A^{\prime}$ $\displaystyle=$
$\displaystyle\frac{g}{4\sqrt{2}}e^{\varphi-12\phi_{1}}(\lambda\pm
2)\sqrt{1+4\varsigma_{1}^{2}},\qquad B^{\prime}\ =\
\frac{g\varsigma_{1}e^{\varphi-12\phi_{1}}(\lambda\pm
2)}{\sqrt{2+8\varsigma_{1}^{2}}},$ (5.61) $\displaystyle\varphi^{\prime}$
$\displaystyle=$ $\displaystyle-\frac{ge^{\varphi-12\phi_{1}}(\lambda\pm
2)(1+20\varsigma_{1}^{2})}{20\sqrt{2+8\varsigma_{1}^{2}}},\qquad\
\phi^{\prime}_{1}\ =\ \frac{ge^{\varphi-12\phi_{1}}(\lambda\pm
2)}{10\sqrt{2+8\varsigma_{1}^{2}}},$ (5.62) $\displaystyle\phi_{2}^{\prime}$
$\displaystyle=$ $\displaystyle
0,\qquad\varsigma^{\prime}_{1}=\frac{g}{\sqrt{2}}\varsigma_{1}e^{\varphi-12\phi_{1}}(\lambda\pm
2)\sqrt{1+4\varsigma_{1}^{2}},$ (5.63) $\displaystyle\varsigma^{\prime}_{2}$
$\displaystyle=$
$\displaystyle\frac{g}{\sqrt{2}}\varsigma_{2}e^{\varphi-12\phi_{1}}(\lambda\pm
2)\sqrt{1+4\varsigma_{1}^{2}}\,.$ (5.64)

The choices of plus or minus signs in these equations are correlated with the
plus or minus signs of the two projectors given in (5.58).

We can consistently set $\phi_{2}=0$ and find a domain wall solution

$\displaystyle A$ $\displaystyle=$
$\displaystyle-\frac{1}{4}\ln\varsigma_{1},\quad B\ =\
C_{B}-\frac{1}{2}\tan^{-1}{2\varsigma_{1}},$ (5.65) $\displaystyle\varphi$
$\displaystyle=$ $\displaystyle
C+\frac{1}{20}\ln\varsigma_{1}+\frac{1}{10}\ln(1+4\varsigma_{1}^{2}),$ (5.66)
$\displaystyle\phi_{1}$ $\displaystyle=$ $\displaystyle
C_{1}+\frac{1}{10}\ln\varsigma_{1}-\frac{1}{20}\ln(1+4\varsigma_{1}^{2}),$
(5.67) $\displaystyle\varsigma_{1}$ $\displaystyle=$
$\displaystyle\frac{1}{2}\tan\left[\sqrt{2}e^{-10C_{1}}(\lambda\pm
2)g\rho+C_{3}\right],$ (5.68) $\displaystyle\varsigma_{2}$ $\displaystyle=$
$\displaystyle C_{4}\varsigma_{1}$ (5.69)

where $\rho$ is the new radial coordinate defined by
$\frac{d\rho}{dr}=e^{\varphi-2\phi_{1}}$.

####  5.2.2 Domain walls in $CSO(2,0,2)\ltimes\mathbb{R}^{2}_{\boldsymbol{s}}$
gauge group

From the previous result, there are special values of $\lambda=\pm 2$ at which
the
$SO(2)\ltimes\left(\mathbb{R}^{6}\times\mathbb{R}^{2}_{\boldsymbol{s}}\right)$
gauge group reduces to
$SO(2)\ltimes\left(\mathbb{R}^{4}\times\mathbb{R}^{2}_{\boldsymbol{s}}\right)\sim
CSO(2,0,2)\ltimes\mathbb{R}^{2}_{\boldsymbol{s}}$. The two choices are
equivalent, so we will choose $\lambda=2$ for definiteness.

In this case, there are nine scalars invariant under the residual $SO(2)$
symmetry generated by $X_{12}$. They are given by the five scalars associated
with the non-compact generators given in (5.48) to (5.52) together with
additional two symmetric and two shift scalars respectively corresponding to

$\displaystyle\overline{\mathbf{Y}}_{6}$ $\displaystyle=$
$\displaystyle\hat{\boldsymbol{t}}^{+}_{1\dot{4}}+\hat{\boldsymbol{t}}^{+}_{2\dot{5}},\qquad\,\overline{\mathbf{Y}}_{7}\
=\ \hat{\boldsymbol{t}}^{+}_{1\dot{5}}-\hat{\boldsymbol{t}}^{+}_{2\dot{4}},$
(5.70) $\displaystyle\overline{\mathbf{Y}}_{8}$ $\displaystyle=$
$\displaystyle\boldsymbol{s}_{14}+\boldsymbol{s}_{25},\qquad\overline{\mathbf{Y}}_{9}\
=\ \boldsymbol{s}_{15}-\boldsymbol{s}_{24}\,.$ (5.71)

However, with this large number of scalar fields, the analysis is highly
complicated. To make things more manageable, we will further truncate the nine
scalars to the previous five singlets together with each of the two sets of
axionic scalars separately.

Turning on two shift scalars, denoted by $\varsigma_{3}$ and $\varsigma_{4}$,
corresponding to $\overline{\mathbf{Y}}_{8}$ and $\overline{\mathbf{Y}}_{9}$
generators, we find the solution given in equations (5.65) to (5.69) together
with the solutions for $\varsigma_{3}$ and $\varsigma_{4}$ of the form

$\varsigma_{3}=C_{5}\sqrt{1+4\varsigma_{1}^{2}}\qquad\text{ and
}\qquad\varsigma_{4}=C_{6}\sqrt{1+4\varsigma_{1}^{2}}\,.$ (5.72)

More interesting solutions are obtained by including the scalars corresponding
to $\overline{\mathbf{Y}}_{6}$ and $\overline{\mathbf{Y}}_{7}$ generators.
With the coset representative

$V=e^{\varphi\overline{\mathbf{Y}}_{\boldsymbol{d}}+\phi_{1}\overline{\mathbf{Y}}_{1}+\phi_{2}\overline{\mathbf{Y}}_{2}+\phi_{3}\overline{\mathbf{Y}}_{6}+\phi_{4}\overline{\mathbf{Y}}_{7}+\varsigma_{1}\overline{\mathbf{Y}}_{3}+\varsigma_{2}\overline{\mathbf{Y}}_{4}},$
(5.73)

we find that the scalar potential vanishes as in the previous case. There are
also two superpotentials. One of them vanishes identically while the non-
trivial one is given by

$\mathcal{W}=\frac{g}{\sqrt{2}}e^{\varphi-12\phi_{1}}\sqrt{\cosh^{2}{2\phi_{3}}\cosh^{2}{2\phi_{4}}+\left(\varsigma_{1}-\varsigma_{2}+\cosh{2\phi_{3}}\cosh{2\phi_{4}}(\varsigma_{1}+\varsigma_{2})\right)^{2}}\,.$
(5.74)

Unlike the previous case, the Minkowski vacuum in this case is half-
supersymmetric with the unbroken supersymmetry corresponding to the vanishing
superpotential. This is very similar to $CSO(2,0,2)$ gauged supergravity in
seven dimensions [41].
Only the supersymmetry corresponding to the superpotential (5.74) is preserved
by the domain wall. This again amounts to imposing a $\gamma^{3}$ projector of
the form (5.58). Furthermore, consistency of the BPS equations from
$\delta\chi_{\pm}$ requires $\varsigma_{1}=\varsigma_{2}=\varsigma$. It is
useful to note the explicit form of the T-tensor for
$\varsigma_{1}=\varsigma_{2}=\varsigma$ which is given by

$T^{\alpha\dot{\beta}}=\frac{e^{\varphi-12\phi_{1}}}{\sqrt{2}}\cosh{2\phi_{3}}\cosh{2\phi_{4}}\left[(\gamma^{3})^{\alpha\beta}+\Omega^{\alpha\beta}+2\varsigma\left((\gamma^{45})^{\alpha\beta}-(\gamma^{12})^{\alpha\beta}\right)\right]\delta^{\dot{\beta}}_{\beta}\,.$
(5.75)

Using the Killing spinors (5.60) subject to the projectors in (5.59) and the
first projector in (5.58), we can derive the following BPS equations

$\displaystyle A^{\prime}$ $\displaystyle=$
$\displaystyle\frac{g}{\sqrt{2}}e^{\varphi-12\phi_{1}}\cosh{2\phi_{3}}\cosh{2\phi_{4}}\sqrt{1+4\varsigma^{2}},$
(5.76) $\displaystyle B^{\prime}$ $\displaystyle=$
$\displaystyle\frac{2ge^{\varphi-12\phi_{1}}\cosh{2\phi_{3}}\cosh{2\phi_{4}}\varsigma}{\sqrt{2+8\varsigma^{2}}},$
(5.77) $\displaystyle\varphi^{\prime}$ $\displaystyle=$
$\displaystyle-\frac{ge^{\varphi-12\phi_{1}}\cosh{2\phi_{3}}\cosh{2\phi_{4}}(1+20\varsigma^{2})}{5\sqrt{2+8\varsigma^{2}}},$
(5.78) $\displaystyle\phi_{1}^{\prime}$ $\displaystyle=$
$\displaystyle\frac{ge^{\varphi-12\phi_{1}}(\cosh^{2}{2\phi_{3}}\cosh^{2}{2\phi_{4}}+5)\,\text{sech}\,{2\phi_{3}}\,\text{sech}\,{2\phi_{4}}}{15\sqrt{2+8\varsigma^{2}}},$
(5.79) $\displaystyle\phi_{2}^{\prime}$ $\displaystyle=$
$\displaystyle\frac{ge^{\varphi-12\phi_{1}}(\cosh^{2}{2\phi_{3}}\cosh^{2}{2\phi_{4}}-1)\,\text{sech}\,{2\phi_{3}}\,\text{sech}\,{2\phi_{4}}}{3\sqrt{2+8\varsigma^{2}}},$
(5.80) $\displaystyle\phi_{3}^{\prime}$ $\displaystyle=$
$\displaystyle-\frac{\sqrt{2}ge^{\varphi-12\phi_{1}}\sinh{2\phi_{3}}\,\text{sech}\,{2\phi_{4}}}{\sqrt{1+4\varsigma^{2}}},$
(5.81) $\displaystyle\phi_{4}^{\prime}$ $\displaystyle=$
$\displaystyle-\frac{\sqrt{2}ge^{\varphi-12\phi_{1}}\cosh{2\phi_{3}}\sinh{2\phi_{4}}}{\sqrt{1+4\varsigma^{2}}},$
(5.82) $\displaystyle\varsigma^{\prime}$ $\displaystyle=$
$\displaystyle-2ge^{\varphi-12\phi_{1}}\cosh{2\phi_{3}}\cosh{2\phi_{4}}\varsigma\sqrt{2+8\varsigma^{2}}\,.$
(5.83)

Introducing a new radial coordinate $\rho$ via
$\frac{d\rho}{dr}=\frac{e^{\varphi-12\phi_{1}}}{\sqrt{1+4\varsigma^{2}}}$, we
eventually find a domain wall solution

$\displaystyle\phi_{1}$ $\displaystyle=$ $\displaystyle
C_{1}+\frac{\phi_{2}}{5}-\frac{1}{10}\ln(e^{4\phi_{3}}-1)+\frac{1}{10}\ln(e^{4\phi_{3}}+1),$
(5.84) $\displaystyle\phi_{2}$ $\displaystyle=$ $\displaystyle
C_{2}-\frac{1}{12}\ln\left(e^{4\phi_{3}}+1\right)+\frac{1}{24}\ln\left[e^{2C_{4}}(1-2e^{4\phi_{3}}+e^{8\phi_{3}})-e^{4\phi_{3}}\right],$
(5.85) $\displaystyle\phi_{3}$ $\displaystyle=$
$\displaystyle\frac{1}{4}\ln\left[\frac{1+2e^{2\sqrt{2}g\rho}+e^{4\sqrt{2}g\rho}+4e^{2C_{4}}}{1-2e^{2\sqrt{2}g\rho}+e^{4\sqrt{2}g\rho}+4e^{2C_{4}}}\right],$
(5.86) $\displaystyle\phi_{4}$ $\displaystyle=$
$\displaystyle\frac{1}{4}\ln\left[\frac{e^{2\phi_{3}}-e^{C_{4}}+e^{C_{4}+4\phi_{3}}}{e^{2\phi_{3}}+e^{C_{3}}-e^{C_{3}+4\phi_{3}}}\right],$
(5.87) $\displaystyle\varphi$ $\displaystyle=$ $\displaystyle
C+\frac{1}{20}\ln(e^{4\phi_{3}}-1)+\frac{1}{10}\ln\left[e^{2C_{4}}(1-2e^{4\phi_{3}}+e^{8\phi_{3}})-e^{4\phi_{3}}\right]$
(5.89)
$\displaystyle-\frac{1}{8}\ln\left[e^{4\phi_{3}}-e^{2C_{4}}(1-2e^{4\phi_{3}}+e^{8\phi_{3}})+4e^{2C_{5}}(1-2e^{4\phi_{3}}+e^{8\phi_{3}})\right],\quad$
$\displaystyle A$ $\displaystyle=$
$\displaystyle\frac{1}{8}\ln\left[\frac{e^{4\phi_{3}}-e^{2C_{4}}(1-2e^{4\phi_{3}}+e^{8\phi_{3}})+4e^{2C_{5}}(1-2e^{4\phi_{3}}+e^{8\phi_{3}})}{(e^{4\phi_{3}}-1)^{2}}\right],$
(5.90) $\displaystyle\varsigma$ $\displaystyle=$
$\displaystyle\frac{e^{C_{5}}(e^{4\phi_{3}}-1)}{\sqrt{e^{2C_{4}}(1-2e^{4\phi_{3}}+e^{8\phi_{3}})-4e^{2C_{5}}(1-2e^{4\phi_{3}}+e^{8\phi_{3}})-e^{4\phi_{3}}}}.$
(5.91)

We end this section by noting that a domain wall solution with $\varsigma=0$
can similarly be obtained with the coordinate $\rho$ defined by
$\frac{d\rho}{dr}=e^{\varphi-12\phi_{1}}$. In this case, the solutions for the
dilaton and warped factor are given by

$\displaystyle\varphi$ $\displaystyle=$ $\displaystyle
C+\frac{1}{20}\ln(e^{4\phi_{3}}-1)-\frac{1}{40}\ln\left[e^{4\phi_{3}}-e^{2C_{4}}(1-2e^{4\phi_{3}}+e^{8\phi_{3}})\right],\quad$
(5.92) $\displaystyle A$ $\displaystyle=$ $\displaystyle-5\varphi$ (5.93)

while solutions for the remaining scalars are the same as given above.

##  6 Conclusions and discussions

We have constructed the embedding tensors of six-dimensional maximal $N=(2,2)$
gauged supergravity for various gauge groups with known seven-dimensional
origins via an $S^{1}$ reduction. These gaugings are triggered by the
embedding tensor in $\mathbf{15}^{-1}$ and $\overline{\mathbf{40}}^{-1}$
representations of $GL(5)\subset SO(5,5)$ duality symmetry. In
$\mathbf{15}^{-1}$ representation, the corresponding gauge group is
$CSO(p,q,5-p-q)$ which is the same as its seven-dimensional counterpart. On
the other hand, for gaugings in $\overline{\mathbf{40}}^{-1}$ representation,
additional translation groups $\mathbb{R}^{n}_{\boldsymbol{s}}$ associated
with the shift symmetries on the scalar fields appear in the gaugings
resulting in $CSO(p,q,4-p-q)\ltimes\mathbb{R}^{4}_{\boldsymbol{s}}$ gauge
group. This is also the case for gaugings in
$(\mathbf{15}+\overline{\mathbf{40}})^{-1}$ representation with gauge groups
$SO(2,1)\ltimes\left(\mathbb{R}^{4}\times\mathbb{R}^{2}_{\boldsymbol{s}}\right)$,
$SO(2)\ltimes\left(\mathbb{R}^{6}\times\mathbb{R}^{2}_{\boldsymbol{s}}\right)$,
and $CSO(2,0,2)\ltimes\mathbb{R}^{2}_{\boldsymbol{s}}$.

We have also studied supersymmetric domain wall solutions and found a large
number of half-supersymmetric domain walls from purely magnetic and purely
electric gaugings in $\mathbf{15}^{-1}$ and $\overline{\mathbf{40}}^{-1}$
representations, respectively. In addition, we have given
$\frac{1}{4}$-supersymmetric domain walls for dyonic gaugings involving the
embedding tensor in both $\mathbf{15}^{-1}$ and $\overline{\mathbf{40}}^{-1}$
representations. These are similar to the seven-dimensional solutions and in
agreement with the general classification of supersymmetric domain walls in
[26] in which the existence of $\frac{1}{4}$-BPS domain walls has been pointed
out.

Apart from solutions with seven-dimensional analogues, we have also found
solutions that are not uplifted to seven-dimensional domain walls due to the
presence of axionic scalars leading to non-vanishing vector fields in seven
dimensions. This can be explicitly seen from the truncation ansatz collected
in appendix C. Although this ansatz has originally been given only for $SO(5)$
gauge group, a similar ansatz with possibly suitable modifications in the
tensor field content is also applicable for other gauge groups. In particular,
the fact that a truncation of seven-dimensional vectors leads to axionic
scalars in six dimensions is still true. Therefore, domain wall solutions with
non-vanishing axionic scalars obtained in this work cannot be obtained from an
$S^{1}$ reduction of any domain wall solutions in seven dimensions.
Accordingly, these solutions are genuine six-dimensional domain walls without
seven-dimensional analogues. As a final comment, we note that there is no
$SO(5)$ symmetric domain wall in seven dimensions since there is no $SO(5)$
singlet scalar in $SL(5)/SO(5)$ coset. The six-dimensional $SO(5)$ symmetric
domain wall, on the other hand, arises form an $S^{1}$ reduction of the
supersymmetric $AdS_{7}$ vacuum by the general result of [49].

The seven-dimensional origin of all the gaugings considered in this work can
also be embedded in ten or eleven dimensions, so the six-dimensional domain
wall solutions can be embedded in string/M-theory via the corresponding seven-
dimensional truncations. Accordingly, the solutions given here are hopefully
useful in the study of DW6/QFT5 duality for maximal supersymmetric Yang-Mills
theory in five dimensions from both six-dimensional framework and
string/M-theory context. It is interesting to explicitly uplift the domain
wall solutions to seven dimensions and subsequently to ten or eleven
dimensions using the truncation ansatze given in [50, 51, 52, 53, 54].

Constructing truncation ansatze of string/M-theory to six dimensions using
$SO(5,5)$ exceptional field theory given in [55] is also of particular
interest. This would allow uplifting the six-dimensional solutions directly to
ten or eleven dimensions. In this paper, we have considered only gaugings with
the embedding tensor in $\mathbf{15}^{-1}$ and $\overline{\mathbf{40}}^{-1}$
representations. It is natural to extend this study by performing a similar
analysis for the embedding tensors in other $GL(5)$ representations as well as
finding supersymmetric domain walls. Unlike the solutions obtained in this
paper, these solutions will not have seven-dimensional counterparts via an
$S^{1}$ reduction.

It is also interesting to construct the embedding tensors for various gaugings
under $SO(4,4)\subset SO(5,5)$. These gaugings can be truncated to gaugings in
half-maximal $N=(1,1)$ supergravity coupled to four vector multiplets in which
supersymmetric $AdS_{6}$ vacua are known to exist in the presence of both
conventional gaugings and massive deformations [56, 57, 58]. Finding
supersymmetric solutions from these gauge groups could be useful in the study
of AdS6/CFT5 correspondence. Finally, finding supersymmetric curved domain
walls with non-vanishing vector and tensor fields as in seven-dimensional
maximal gauged supergravity in [59, 60] is worth considering. This type of
solutions can describe conformal defects or holographic RG flows from five-
dimensional $N=4$ super Yang-Mills theories to lower dimensions. Along this
line, examples of solutions dual to surface defects from $N=(1,1)$ gauged
supergravity have appeared recently in [61].
Acknowledgement
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 $GL(5)$ Branching rules

In this appendix, we collect all of the $SO(5,5)\rightarrow GL(5)$ branching
rules used throughout the paper. Relevant decompositions have already been
given in [30], but in order to construct the embedding tensor, we need a
concrete realization. Therefore, we will determine the decompositions for
various representations of $SO(5,5)$ in terms of $GL(5)$ representations using
explicit matrix forms.

###  A.1 Vector

A vector or fundamental representation of $SO(5,5)$ decomposes under
$GL(5)\subset SO(5,5)$ as $\mathbf{5}$ and $\overline{\mathbf{5}}$, i.e.,
$V_{M}=(V_{m},V^{m})$. The $SO(5,5)$ vector index $M=1,..,10$ can be raised
and lowered through the following $SO(5,5)$ invariant metric in the light cone
or off-diagonal basis

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

in which $\mathds{1}_{n}$ is an ($n\times n$) identity matrix. For example,
$V^{M}=\eta^{MN}V_{N}=(V^{m},V_{m})$.

In vector representation, the $SO(5,5)$ algebra

$\left[\boldsymbol{t}_{MN},\boldsymbol{t}_{PQ}\right]\ =\
4(\eta_{M[P}\boldsymbol{t}_{Q]N}-\eta_{N[P}\boldsymbol{t}_{Q]M})$ (A.2)

is realized by $SO(5,5)$ generators ,
$\boldsymbol{t}_{MN}=\boldsymbol{t}_{[MN]}$, of the form

${(\boldsymbol{t}_{MN})_{P}}^{Q}\ =\ 4\eta_{P[M}\delta^{Q}_{N]}$ (A.3)

where $\delta^{M}_{N}=\mathds{1}_{10}$. Defining an $\mathbb{R}^{+}\sim
SO(1,1)\subset GL(5)\sim\mathbb{R}^{+}\times SL(5)$ generator by

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

we find the explicit form of the $\mathbb{R}^{+}$ generator in vector
representation given by

$\boldsymbol{d}\ =\ 2\begin{pmatrix}\mathds{1}_{5}&0\\\
0&-\mathds{1}_{5}\end{pmatrix}.$ (A.5)

With an $SO(5,5)$ vector decomposing as $V_{M}=(V_{m},V^{m})$, we obtain the
commutation relations

$\left[\boldsymbol{d},V_{m}\right]\ =\
+2V_{m}\qquad\text{and}\qquad\left[\boldsymbol{d},V^{m}\right]\ =\ -2V^{m}\,.$
(A.6)

These imply that we can assign the $\mathbb{R}^{+}$ weights $\pm 2$ to the
$\mathbf{5}$ and $\overline{\mathbf{5}}$ representations of $SL(5)\subset
GL(5)$. Therefore, the branching rule for a vector representation reads

$\underbrace{\mathbf{10}}_{V_{M}}\ \rightarrow\
\underbrace{\mathbf{5}^{+2}}_{V_{m}}\,\oplus\,\underbrace{\overline{\mathbf{5}}^{-2}}_{V^{m}}\,.$
(A.7)

###  A.2 Adjoint

The decomposition of adjoint representation follows from the branching rule of
vector representations. Using (A.7), we can decompose the $SO(5,5)$ generators
as

$\boldsymbol{t}_{MN}\ \rightarrow\
(\boldsymbol{t}_{mn},{\boldsymbol{t}^{m}}_{n},\boldsymbol{t}^{mn})$ (A.8)

with ${\boldsymbol{t}_{m}}^{n}=-{\boldsymbol{t}^{n}}_{m}$. The 25 generators
${\boldsymbol{t}^{m}}_{n}$ of $GL(5)$ consist of the $\mathbb{R}^{+}$
generator defined in (A.4) and the $SL(5)$ generators given by

${\boldsymbol{\tau}^{m}}_{n}\ =\
{\boldsymbol{t}^{m}}_{n}-\frac{1}{5}\,\boldsymbol{d}\,\delta^{m}_{n}$ (A.9)

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

We denote the shift and hidden generators by
$\boldsymbol{s}_{mn}=\boldsymbol{t}_{mn}$ and
$\boldsymbol{h}^{mn}=\boldsymbol{t}^{mn}$, respectively. In vector
representation, the $SO(5,5)$ generators can be written as

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