.1pt

Supersymmetric domain walls in maximal 6D gauged supergravity III

Parinya Karndumria and Patharadanai Nuchinob

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

E-mail<EMAIL_ADDRESS>
b Department of Physics, Faculty of Science, Ramkhamhaeng University, 282
Ramkhamhaeng Road, Bang Kapi, Bangkok 10240, Thailand

E-mail<EMAIL_ADDRESS>
###### Abstract

We continue our study of gaugings the maximal $N=(2,2)$ supergravity in six
dimensions with gauge groups obtained from decomposing the embedding tensor
under $\mathbb{R}^{+}\times SO(4,4)$ subgroup of the global symmetry
$SO(5,5)$. Supersymmetry requires the embedding tensor to transform in
$\mathbf{144}_{c}$ representation of $SO(5,5)$. Under $\mathbb{R}^{+}\times
SO(4,4)$ subgroup, this leads to the embedding tensor in $(\mathbf{8}^{\pm
3}$, $\mathbf{8}^{\pm 1},\mathbf{56}^{\pm 1})$ representations. Gaugings in
$\mathbf{8}^{\pm 3}$ representations lead to a translational gauge group
$\mathbb{R}^{8}$ while gaugings in $\mathbf{8}^{\pm 1}$ representations give
rise to gauge groups related to the scaling symmetry $\mathbb{R}^{+}$. On the
other hand, the embedding tensor in $\mathbf{56}^{\pm 1}$ representations
gives $CSO(4-p,p,1)\sim SO(4-p,p)\ltimes\mathbb{R}^{4}\subset SO(4,4)$ gauge
groups with $p=0,1,2$. More interesting gauge groups can be obtained by
turning on more than one representation of the embedding tensor subject to the
quadratic constraints. In particular, we consider gaugings in both
$\mathbf{56}^{-1}$ and $\mathbf{8}^{+3}$ representations giving rise to larger
$SO(5-p,p)$ and $SO(4-p,p+1)$ gauge groups for $p=0,1,2$. In this case, we
also give a number of half-supersymmetric domain wall solutions preserving
different residual symmetries. The solutions for gaugings obtained only from
$\mathbf{56}^{-1}$ representation are also included in these results when the
$\mathbf{8}^{+3}$ part is accordingly turned off.

##  1 Introduction

Supersymmetric domain walls are solutions to gauged supergravities that play
many important roles in various aspects of string/M-theory. These solutions
have provided a useful tool for studying different aspects of the AdS/CFT
correspondence since the beginning of the original proposal in [1], see also
[2, 3]. They are also vital in the so-called DW/QFT correspondence [4, 5, 6],
a generalization of the AdS/CFT correspondence to non-conformal field
theories. In particular, these solutions give holographic descriptions to RG
flows in strongly coupled dual field theories in various space-time
dimensions. Domain walls also appear in the study of cosmology via the domain
wall/cosmology correspondence, see for example [7, 8, 9]. A systematic
classification of supersymmetric domain walls from maximal gauged supergravity
in various space-time dimensions has been performed in [10], and many domain
wall solutions in gauged supergravities have been found in different space-
time dimensions, see [11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24,
25, 26, 27, 28] for an incomplete list.
In this paper, we are interested in domain wall solutions from maximal
$N=(2,2)$ six-dimensional gauged supergravity. The ungauged $N=(2,2)$
supergravity has been constructed in [29], and the first $N=(2,2)$ six-
dimensional gauged supergravity with $SO(5)$ gauge group has been found in
[30] by performing an $S^{1}$ reduction of the maximal $SO(5)$ gauged
supergravity in seven dimensions [31]. The most general $N=(2,2)$ gauged
supergravity has been constructed in [32] using the embedding tensor
formalism. The embedding tensor transforms in $\mathbf{144}_{c}$
representation of $SO(5,5)$ global symmetry as required by supersymmetry, and
some possible gaugings classified under $GL(5)$ and $SO(4,4)$ subgroups of
$SO(5,5)$ have also been identified in [32]. Many gaugings arising from
$GL(5)$ decomposition together with a large number of supersymmetric domain
wall solutions have been constructed recently in [17] and [18].
In this work, we will continue the study of the maximal $N=(2,2)$ gauged
supergravity and the corresponding supersymmetric domain walls by considering
gaugings arising from decomposing the embedding tensor under
$\mathbb{R}^{+}\times SO(4,4)\subset SO(5,5)$. Under $\mathbb{R}^{+}\times
SO(4,4)$, the embedding tensor in $\mathbf{144}_{c}$ representation of
$SO(5,5)$ decomposes into $\mathbf{8}^{\pm 1}$, $\mathbf{8}^{\pm 3}$, and
$\mathbf{56}^{\pm 1}$ representations of $\mathbb{R}^{+}\times SO(4,4)$. We
will determine explicit solutions of the embedding tensor giving rise to
consistent gauge groups of the $N=(2,2)$ gauged supergravity and look for
possible supersymmetric domain wall solutions.
According to the DW/QFT correspondence, the aforementioned domain wall
solutions should be dual to maximally supersymmetric Yang-Mills theory in five
dimensions. The latter plays an important role in defining the $N=(2,0)$
superconformal field theory in six dimensions compactified on $S^{1}$ and also
describing nonperturbative dynamics of $N=1,2$, class S, theories in four
dimensions, see for example [33, 34, 35, 36, 37, 38, 39]. We expect domain
wall solutions studied here could be useful in this context as well.
The paper is organized as follows. In section 2, we briefly review the
construction of six-dimensional maximal gauged supergravity in the embedding
tensor formalism. Possible gauge groups arising from decomposing the embedding
tensor under $SO(4,4)$ are determined in section 3. In section 4, we find
supersymmetric domain wall solutions from gaugings in $\mathbf{56}^{-1}$ and
$\mathbf{8}^{+3}$ representations. Conclusions and discussions are given in
section 5. Relevant branching rules for $SO(5,5)$ representations under
$SO(4,4)$ are given in the appendix.

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

We first give a brief review of six-dimensional $N=(2,2)$ gauged supergravity
in the embedding tensor formalism constructed in [32]. We will only collect
relevant formulae for determining possible gauge groups and finding
supersymmetric domain wall solutions. For more details, we refer the reader to
the original construction in [32].
The supergravity multiplet of the maximal $N=(2,2)$ supersymmetry in six
dimensions consists of the following component fields

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

In our convention, 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
$\overline{m},\overline{n},\ldots=1,\ldots,5$ indices label fundamental and
anti-fundamental representations of $GL(5)\subset SO(5,5)$, respectively.
Indices $A,B,\ldots=1,\ldots,16$ refer to Majorana-Weyl spinors of the
$SO(5,5)$ duality symmetry.
The electric two-form potentials $B_{\mu\nu\overline{m}}$, appearing in the
ungauged Lagrangian, transform as $\mathbf{5}$ under $GL(5)$ while the vector
fields $A^{A}_{\mu}$ transform as $\mathbf{16}_{c}$ under $SO(5,5)$. Together
with the magnetic duals ${B_{\mu\nu}}^{\overline{m}}$ transforming in
$\overline{\mathbf{5}}$ representation of $GL(5)$, the electric two-forms
$B_{\mu\nu\overline{m}}$ transform in a vector representation $\mathbf{10}$ of
the full global symmetry group $SO(5,5)$ denoted by $B_{\mu\nu
M}=(B_{\mu\nu\overline{m}},{B_{\mu\nu}}^{\overline{m}})$. Therefore, only the
subgroup $GL(5)\subset SO(5,5)$ is a manifest off-shell symmetry of the
theory. Indices $M,N,\ldots$ denote fundamental or vector representation of
$SO(5,5)$. Finally, there are $25$ scalar fields parametrizing the coset space
$SO(5,5)/SO(5)\times SO(5)$.
Fermionic fields, transforming under the local $SO(5)\times SO(5)$ symmetry,
are symplectic Majorana-Weyl (SMW) spinors. Indices $\alpha,\beta,\ldots$ and
$\dot{\alpha},\dot{\beta},\ldots$ 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 $a,b,\ldots$ and $\dot{a},\dot{b},\ldots$. We will use
$\pm$ to indicate the space-time chiralities of the spinors. Under the local
$SO(5)\times SO(5)$ symmetry, the two sets of gravitini $\psi_{+\mu\alpha}$
and $\psi_{-\mu\dot{\alpha}}$ transform as $(\mathbf{4},\mathbf{1})$ and
$(\mathbf{1},\mathbf{4})$ while the spin-$\frac{1}{2}$ fields
$\chi_{+a\dot{\alpha}}$ and $\chi_{-\dot{a}\alpha}$ transform as
$(\mathbf{5},\mathbf{4})$ and $(\mathbf{4},\mathbf{5})$, respectively.
In chiral spinor representation, the $SO(5,5)/SO(5)\times SO(5)$ coset is
described by a coset representative ${V_{A}}^{\alpha\dot{\beta}}$ transforming
under the global $SO(5,5)$ and local $SO(5)\times SO(5)$ symmetries by left
and right multiplications, respectively. The inverse elements
${(V^{-1})_{\alpha\dot{\beta}}}^{A}$ will be denoted by
${V^{A}}_{\alpha\dot{\beta}}$ satisfying the relations

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

On the other hand, in vector representation, the coset representative is given
by a $10\times 10$ matrix
${\mathcal{V}_{M}}^{\underline{A}}=({\mathcal{V}_{M}}^{a},{\mathcal{V}_{M}}^{\dot{a}})$
with $\underline{A}=(a,\dot{a})$ and 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. The inverse of
${\mathcal{V}_{M}}^{\underline{A}}$ 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 are
described by the embedding tensor in $\mathbf{144}_{c}$ representation of
$SO(5,5)$. This can be written in terms of a vector-spinor of $SO(5,5)$,
$\theta^{AM}$, subject to the linear constraint (LC)

$(\Gamma_{M})_{AB}\,\theta^{BM}\ =\ 0\,.$ (2.7)

The gauge covariant derivative is then given by

$D_{\mu}=\partial_{\mu}-gA^{A}_{\mu}\ {\Theta_{A}}^{MN}\boldsymbol{t}_{MN}$
(2.8)

with $g$ being a gauge coupling constant and ${\Theta_{A}}^{MN}$ defined by

${\Theta_{A}}^{MN}\ =\ -\theta^{B[M}(\Gamma^{N]})_{BA}\ \equiv\
\left(\Gamma^{[M}\theta^{N]}\right)_{A}\,.$ (2.9)

As usual, 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
$\boldsymbol{t}_{MN}$. Consistency also requires the gauge generators to form
a closed subalgebra of $SO(5,5)$ implying the quadratic constraint (QC)

$\left[X_{A},X_{B}\right]\ =\ -{(X_{A})_{B}}^{C}\,X_{C}\,.$ (2.10)

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

$\displaystyle\theta^{AM}\theta^{BN}\eta_{MN}$ $\displaystyle=$ $\displaystyle
0,$ (2.11) $\displaystyle\theta^{AM}\theta^{B[N}(\Gamma^{P]})_{AB}$
$\displaystyle=$ $\displaystyle 0\,.$ (2.12)

Any $\theta^{AM}\in\mathbf{144}_{c}$ satisfying these quadratic constraints
defines a consistent gauging.
In this work, we are only interested in classifying possible gauge groups and
finding supersymmetric domain wall solutions involving only the metric and
scalar fields. We have explicitly checked that the truncation of vector and
tensor fields is consistent in all the domain wall solutions given in section
4. This follows from the fact that the corresponding Yang-Mills currents
vanish for all of the solutions considered here. With all vector and tensor
fields set to zero, the bosonic Lagrangian of the maximal $N=(2,2)$ gauged
supergravity is given by

$e^{-1}\mathcal{L}=\frac{1}{4}R-\frac{1}{16}P_{\mu}^{a\dot{a}}P^{\mu}_{a\dot{a}}-\mathbf{V}$
(2.13)

while the supersymmetry transformations of fermionic fields read

$\displaystyle\delta\psi_{+\mu\alpha}$ $\displaystyle=$ $\displaystyle
D_{\mu}\epsilon_{+\alpha}+\frac{g}{4}\hat{\gamma}_{\mu}{T_{\alpha}}^{\dot{\beta}}\epsilon_{-\dot{\beta}},$
(2.14) $\displaystyle\delta\psi_{-\mu\dot{\alpha}}$ $\displaystyle=$
$\displaystyle
D_{\mu}\epsilon_{-\dot{\alpha}}-\frac{g}{4}\hat{\gamma}_{\mu}{T^{\beta}}_{\dot{\alpha}}\epsilon_{+\beta},$
(2.15) $\displaystyle\delta\chi_{+a\dot{\alpha}}$ $\displaystyle=$
$\displaystyle\frac{1}{4}P^{\mu}_{a\dot{a}}\hat{\gamma}_{\mu}{(\gamma^{\dot{a}})_{\dot{\alpha}}}^{\dot{\beta}}\epsilon_{-\dot{\beta}}+2g{(T_{a})^{\beta}}_{\dot{\alpha}}\epsilon_{+\beta}-\frac{g}{2}{T^{\alpha}}_{\dot{\alpha}}{(\gamma_{a})_{\alpha}}^{\beta}\epsilon_{+\beta},$
(2.16) $\displaystyle\delta\chi_{-\dot{a}\alpha}$ $\displaystyle=$
$\displaystyle\frac{1}{4}P^{\mu}_{a\dot{a}}\hat{\gamma}_{\mu}{(\gamma^{a})_{\alpha}}^{\beta}\epsilon_{+\beta}+2g{(T_{\dot{a}})_{\alpha}}^{\dot{\beta}}\epsilon_{-\dot{\beta}}+\frac{g}{2}{T_{\alpha}}^{\dot{\alpha}}{(\gamma_{\dot{a}})_{\dot{\alpha}}}^{\dot{\beta}}\epsilon_{-\dot{\beta}}.$
(2.17)

The covariant derivatives of supersymmetry parameters are defined as

$\displaystyle D_{\mu}\epsilon_{+\alpha}$ $\displaystyle=$
$\displaystyle\partial_{\mu}\epsilon_{+\alpha}+\frac{1}{4}{\omega_{\mu}}^{\nu\rho}\hat{\gamma}_{\nu\rho}\epsilon_{+\alpha}+\frac{1}{4}Q_{\mu}^{ab}{(\gamma_{ab})_{\alpha}}^{\beta}\epsilon_{+\beta},$
(2.18) $\displaystyle D_{\mu}\epsilon_{-\dot{\alpha}}$ $\displaystyle=$
$\displaystyle\partial_{\mu}\epsilon_{-\dot{\alpha}}+\frac{1}{4}{\omega_{\mu}}^{\nu\rho}\hat{\gamma}_{\nu\rho}\epsilon_{-\dot{\alpha}}+\frac{1}{4}Q_{\mu}^{\dot{a}\dot{b}}{(\gamma_{\dot{a}\dot{b}})_{\dot{\alpha}}}^{\dot{\beta}}\epsilon_{-\dot{\beta}}$
(2.19)

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

$\displaystyle P_{\mu}^{a\dot{a}}$ $\displaystyle=$
$\displaystyle\frac{1}{4}{(\gamma^{a})}^{\alpha\beta}{(\gamma^{\dot{a}})}^{\dot{\alpha}\dot{\beta}}{V^{A}}_{\alpha\dot{\alpha}}\partial_{\mu}V_{A\beta\dot{\beta}},$
(2.20) $\displaystyle Q_{\mu}^{ab}$ $\displaystyle=$
$\displaystyle\frac{1}{8}{(\gamma^{ab})}^{\alpha\beta}\Omega^{\dot{\alpha}\dot{\beta}}{V^{A}}_{\alpha\dot{\alpha}}\partial_{\mu}V_{A\beta\dot{\beta}},$
(2.21) $\displaystyle Q_{\mu}^{\dot{a}\dot{b}}$ $\displaystyle=$
$\displaystyle\frac{1}{8}\Omega^{\alpha\beta}{(\gamma^{\dot{a}\dot{b}})}^{\dot{\alpha}\dot{\beta}}{V^{A}}_{\alpha\dot{\alpha}}\partial_{\mu}V_{A\beta\dot{\beta}}\,.$
(2.22)

In these equations, $\Omega^{\alpha\beta}$ and
$\Omega^{\dot{\alpha}\dot{\beta}}$ are the $USp(4)\sim SO(5)$ symplectic forms
that satisfy the following relations

$\Omega_{\beta\alpha}\ =\ -\Omega_{\alpha\beta},\qquad\Omega^{\alpha\beta}\ =\
(\Omega_{\alpha\beta})^{*},\qquad\Omega_{\alpha\beta}\Omega^{\beta\gamma}\ =\
-\delta^{\gamma}_{\alpha}$ (2.23)

and similarly for $\Omega_{\dot{\alpha}\dot{\beta}}$. The scalar potential is
given by

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

with the T-tensors defined by

$(T^{a})^{\alpha\dot{\alpha}}={\mathcal{V}_{M}}^{a}\theta^{AM}{V_{A}}^{\alpha\dot{\alpha}},\qquad(T^{\dot{a}})^{\alpha\dot{\alpha}}=-{\mathcal{V}_{M}}^{\dot{a}}\theta^{AM}{V_{A}}^{\alpha\dot{\alpha}},$
(2.25)

and

$T^{\alpha\dot{\alpha}}\equiv(T^{a})^{\beta\dot{\alpha}}{(\gamma_{a})_{\beta}}^{\alpha}=-(T^{\dot{a}})^{\alpha\dot{\beta}}{(\gamma_{\dot{a}})_{\dot{\beta}}}^{\dot{\alpha}}\,.$
(2.26)

We also note useful identities involving various components of the T-tensors

$\displaystyle
D_{\mu}T^{a}=\frac{1}{4}P^{b\dot{b}}_{\mu}\left(\gamma^{b}T^{a}\gamma^{\dot{b}}-2\delta^{ab}T^{\dot{b}}\right),$
(2.27) $\displaystyle
D_{\mu}T^{\dot{a}}=\frac{1}{4}P^{b\dot{b}}_{\mu}\left(\gamma^{b}T^{\dot{a}}\gamma^{\dot{b}}-2\delta^{\dot{a}\dot{b}}T^{b}\right),$
(2.28) $\displaystyle
D_{\mu}T=\frac{1}{2}P^{a\dot{a}}_{\mu}\left(T^{a}\gamma^{\dot{a}}-\gamma^{a}T^{\dot{a}}-\frac{1}{2}\gamma^{a}T\gamma^{\dot{a}}\right).$
(2.29)

##  3 Gaugings of six-dimensional $N=(2,2)$ supergravity under $SO(4,4)$

In this section, we will determine explicit forms of the embedding tensor for
a number of possible gauge groups leading to consistent $N=(2,2)$ gauged
supergravities in six dimensions. To find gauge groups by decomposing the
embedding tensor under $\mathbb{R}^{+}\times SO(4,4)\subset SO(5,5)$, we
decompose the $SO(5,5)$ vector index as $M=(-,I,+)$ with $I=1,2,\ldots,8$
being the $SO(4,4)$ vector index. The $SO(5,5)$ generators
$\boldsymbol{t}_{MN}$ are decomposed accordingly as
$\boldsymbol{t}_{MN}=(\boldsymbol{t}_{-+}=\boldsymbol{d},\boldsymbol{t}_{+I}=\boldsymbol{p}_{I},\boldsymbol{t}_{-I}=\boldsymbol{k}_{I},\boldsymbol{t}_{IJ}=\boldsymbol{\tau}_{IJ})$
with $\boldsymbol{d}$ and $\boldsymbol{\tau}_{IJ}$ being $\mathbb{R}^{+}$ and
$SO(4,4)$ generators, respectively.
Similarly, the $SO(5,5)$ spinor index $A$ will also be split as
$A=(m,\dot{m})$ with $m=1,2,\ldots,8$ and
$\dot{m}=\dot{1},\dot{2},\ldots,\dot{8}$ being $SO(4,4)$ spinor indices, see
more detail in the appendix. As given in the appendix, the embedding tensor
transforming in $\mathbf{144}_{c}$ representation of $SO(5,5)$ will split into
the following representations under $\mathbb{R}^{+}\times SO(4,4)$

$\underbrace{\mathbf{144}_{c}}_{\theta^{AM}}\ \rightarrow\
\underbrace{\mathbf{56}^{-1}}_{\vartheta_{3}^{mI}}\,\oplus\,\underbrace{\mathbf{56}^{+1}}_{\vartheta_{4}^{\dot{m}I}}\,\oplus\,\underbrace{\mathbf{8}^{-1}}_{\theta_{6}^{\dot{m}+}}\,\oplus\,\underbrace{\mathbf{8}^{+1}}_{\theta_{1}^{m-}}\,\oplus\,\underbrace{\mathbf{8}^{+3}}_{\theta_{2}^{\dot{m}-}}\,\oplus\,\underbrace{\mathbf{8}^{-3}}_{\theta_{5}^{m+}}.$
(3.1)

For convenience, we also recall the identification of various components of
the embedding tensor of the form

$\theta^{AM}\ =\
\left(\begin{array}[]{c|c|c}\theta_{1}^{m-}&\theta_{3}^{mI}&\theta_{5}^{m+}\\\
\hline\cr\theta_{2}^{\dot{m}-}&\theta_{4}^{\dot{m}I}&\theta_{6}^{\dot{m}+\phantom{{}^{\prime}}}\end{array}\right).$
(3.2)

The components $\theta_{1}^{m-}$, $\theta_{4}^{\dot{m}I}$, and
$\theta_{5}^{m+}$ correspond to $\mathbf{8}^{+1}$, $\mathbf{56}^{+1}$, and
$\mathbf{8}^{-3}$ representations while $\theta_{2}^{\dot{m}-}$,
$\theta_{3}^{mI}$, and $\theta_{6}^{\dot{m}+}$ are respectively
$\mathbf{8}^{+3}$, $\mathbf{56}^{-1}$, and $\mathbf{8}^{-1}$ ones. The linear
combinations $\vartheta_{3}^{mI}$ and ${\vartheta_{4}^{\dot{m}I}}$ in terms of
$(\theta^{mI}_{3},\theta^{\dot{m}+}_{6})$ and
$(\theta^{n-}_{1},\theta^{\dot{m}I}_{4},)$, as required by the LC, are defined
in (A.43). For later convenience, we also repeat these relations here

$\vartheta_{3}^{mI}=\theta_{3}^{mI}+\frac{\sqrt{2}}{8}{(\gamma^{I})^{m}}_{\dot{n}}\theta_{6}^{\dot{n}+}\qquad\text{and
}\qquad\vartheta_{4}^{\dot{m}I}=\theta_{4}^{\dot{m}I}-\frac{\sqrt{2}}{8}{(\gamma^{I})^{\dot{m}}}_{n}\theta_{1}^{n-}\,.$
(3.3)

In this section, we will determine explicit forms of the embedding tensor by
imposing the quadratic constraint on the embedding tensor. With the above
decomposition, the first condition of QC given in (2.11) reduces to

$\theta^{A+}\theta^{B-}+\theta^{AI}\theta^{BJ}\eta_{IJ}+\theta^{A-}\theta^{B+}=0$
(3.4)

with $\eta_{IJ}$ being the $SO(4,4)$ invariant tensor defined in (A.3). On the
other hand, the second condition (2.12) splits into

$\displaystyle\theta^{mM}\theta_{1}^{n-}\mathbf{c}_{mn}+\theta^{\dot{m}M}\theta_{6}^{\dot{n}+}\mathbf{c}_{\dot{m}\dot{n}}$
$\displaystyle=$ $\displaystyle 0,$ (3.5)
$\displaystyle\theta^{mM}\theta_{2}^{\dot{n}-}(\gamma^{I})_{m\dot{n}}+\theta^{\dot{m}M}\theta_{1}^{n-}(\gamma^{I})_{\dot{m}n}-\sqrt{2}\theta^{\dot{m}M}\theta_{4}^{\dot{n}I}\mathbf{c}_{\dot{m}\dot{n}}$
$\displaystyle=$ $\displaystyle 0,$ (3.6)
$\displaystyle\theta^{mM}\theta_{6}^{\dot{n}+}(\gamma^{I})_{m\dot{n}}+\theta^{\dot{m}M}\theta_{5}^{n+}(\gamma^{I})_{\dot{m}n}+\sqrt{2}\theta^{mM}\theta_{3}^{nI}\mathbf{c}_{mn}$
$\displaystyle=$ $\displaystyle 0,$ (3.7)
$\displaystyle\theta^{mM}\theta_{4}^{\dot{n}[I}(\gamma^{J]})_{m\dot{n}}+\theta^{\dot{m}M}\theta_{3}^{n[I}(\gamma^{J]})_{\dot{m}n}$
$\displaystyle=$ $\displaystyle 0\,.$ (3.8)

In these equations, $\mathbf{c}_{mn}$ and $\mathbf{c}_{\dot{m}\dot{n}}$ are
elements of the $SO(4,4)$ charge conjugation matrix defined in (A.32), and
$(\gamma^{I})_{m\dot{n}}=(\gamma^{I})_{\dot{n}m}$ are chirally decomposed
$SO(4,4)$ gamma matrices given in (A.34)
Some possible gauge groups under $SO(4,4)$ have also been discussed in [40],
and it has been pointed out that turning on only $\theta_{1}^{m-}$ or
$\theta_{6}^{\dot{m}+}$ components leads to gaugings of the scaling symmetry
$\mathbb{R}^{+}$. Furthermore, with $\theta_{4}^{\dot{m}I}=0$ or
$\theta_{3}^{mI}=0$, we find from the LC given in (A.44) that
$\theta_{1}^{m-}$ or $\theta_{6}^{\dot{m}+}$ need to be zero, respectively.
Accordingly, we conclude that gaugings with only $\mathbf{8}^{+1}$ or
$\mathbf{8}^{-1}$ components non-vanishing are not consistent.

###  3.1 Gaugings in $\mathbf{8}^{+3}$ representation

We begin with gauge groups arising from the embedding tensor in
$\mathbf{8}^{+3}$ representation. In this case, we set all $\theta$’s
components to be zero except

$\theta_{2}^{\dot{m}-}=v^{\dot{m}}$ (3.9)

for a spinor $v^{\dot{m}}$. The $\theta^{AM}$ matrix of the form

$\theta^{AM}\ =\ \left(\begin{array}[]{c|c|c}&\phantom{\theta_{3}^{\alpha
I}}&\\\ \hline\cr
v^{\dot{m}}&&\phantom{\theta_{6}^{\dot{\alpha}}}\end{array}\right)$ (3.10)

makes the embedding tensor satisfy all the LC and QC. We have used the
notation that all vanishing elements are left as blank spaces.
For $A=(m,\dot{m})$, the corresponding gauge generators split into
$X_{A}=(X_{m},\,X_{\dot{m}})$. With the above embedding tensor, the last eight
generators vanish, $X_{\dot{m}}=0$, while the first eight generators are given
in terms of $\boldsymbol{k}_{I}$ as

$X_{m}=(\gamma^{I})_{m\dot{n}}v^{\dot{n}}\boldsymbol{k}_{I}\,.$ (3.11)

They are all linearly independent and commute with each other
$\left[X_{m},X_{n}\right]=0$. Thus, the resulting gauge group is an eight-
dimensional translational group $\mathbb{R}^{8}$ associated with the
$\boldsymbol{k}_{I}$ generators.

###  3.2 Gaugings in $\mathbf{8}^{-3}$ representation

As in the previous case, we set all $\theta$’s components to be zero except

$\theta_{5}^{m+}=w^{m}$ (3.12)

for any spinor $w^{m}$. All the LC and QC are satisfied by this embedding
tensor. In this case, there are also eight non-vanishing gauge generators, but
given in terms of the $\boldsymbol{p}_{I}$ generators, i.e.

$X_{\dot{m}}=(\gamma^{I})_{\dot{m}n}w^{n}\boldsymbol{p}_{I}\,.$ (3.13)

As in the previous case, they are all linearly independent and commute with
each other, $\left[X_{\dot{m}},X_{\dot{n}}\right]=0$. This implies again that
the resulting gauge group is an eight-dimensional translational group
$\mathbb{R}^{8}$ associated with the $\boldsymbol{p}_{I}$ generators.

###  3.3 Gaugings in $\mathbf{56}^{-1}$ representation

We now move to gaugings in $\mathbf{56}^{-1}$ representation by choosing only
$\theta_{3}^{mI}$ to be non-vanishing. The embedding tensor takes the form

$\theta^{AM}\ =\
\left(\begin{array}[]{c|c|c}&\theta_{3}^{mI}&\phantom{\theta_{6}^{\dot{\alpha}}}\\\
\hline\cr\phantom{\theta_{3}^{\alpha I}}&&\end{array}\right)$ (3.14)

subject to the LC

$(\gamma_{I})_{\dot{m}n}\theta_{3}^{nI}=0$ (3.15)

which is the same as the second condition in (A.45). For vanishing
$\theta_{6}^{\dot{m}+}$ component from $\mathbf{8}^{-1}$ representation, the
embedding tensor is simply given by $\theta_{3}^{mI}$. As in [40], all $56$
components in $\theta_{3}^{mI}$ can be parametrized by an antisymmetric tensor
$f_{\dot{m}\dot{n}\dot{p}}=f_{[\dot{m}\dot{n}\dot{p}]}$ by writing
$\theta_{3}^{mI}$ as

$\theta_{3}^{mI}=\frac{1}{48}f_{\dot{m}\dot{n}\dot{p}}(\gamma^{IJ})^{\dot{m}\dot{n}}(\gamma_{J})^{m\dot{p}}$
(3.16)

with
$(\gamma^{IJ})^{\dot{m}\dot{n}}=(\gamma^{[I})^{\dot{m}p}{(\gamma^{J]})_{p}}^{\dot{n}}$.
The LC given in (3.15) is now identically satisfied, and the corresponding
gauge generators are split into the following two sets

$\displaystyle X_{m}$ $\displaystyle=$
$\displaystyle\frac{1}{24\sqrt{2}}f_{\dot{m}\dot{n}\dot{p}}(\gamma^{IJ})^{\dot{m}\dot{n}}{(\gamma_{J})_{m}}^{\dot{p}}\boldsymbol{p}_{I},$
(3.17) $\displaystyle X_{\dot{m}}$ $\displaystyle=$
$\displaystyle\frac{1}{48}f_{\dot{n}\dot{p}\dot{q}}(\gamma^{IK})^{\dot{n}\dot{p}}(\gamma_{K})^{q\dot{q}}(\gamma^{J})_{q\dot{m}}\boldsymbol{\tau}_{IJ}.$
(3.18)

The first set contains eight nilpotent generators that commute with each
other, $\left[X_{m},\,X_{n}\right]=0$, so they generate a translational
subgroup associated with $\boldsymbol{p}_{I}$ generators. The other set gives
another subgroup embedded in the $SO(4,4)$ factor. According to [40], the QCs
in terms of the antisymmetric tensor $f_{\dot{m}\dot{n}\dot{p}}$ can be
written as

$f_{\dot{m}\dot{n}\dot{p}}f^{\dot{m}\dot{n}\dot{p}}=0\qquad\text{ and }\qquad
f_{\dot{r}[\dot{m}\dot{n}}{f_{\dot{p}\dot{q}]}}^{\dot{r}}=0\,.$ (3.19)

We will discuss some possible solutions to these conditions.

####  3.3.1 $CSO(4,0,1)\sim SO(4)\ltimes\mathbb{R}^{4}$ gauge group

We first consider a simple solution of the form

$f_{\dot{m}\dot{n}\dot{p}}=(\kappa_{1}\varepsilon_{ijk},\kappa_{2}\varepsilon_{rst})$
(3.20)

for $i,j,\ldots=\dot{1},\dot{2},\dot{3}$ and
$r,s,\ldots=\dot{5},\dot{6},\dot{7}$. To solve the first condition in (3.19),
we need to impose the relation $\kappa_{1}=\pm\kappa_{2}$. We will choose
$\kappa_{1}=\kappa_{2}=\kappa\in\mathbb{R}$ for definiteness.
With this form of the embedding tensor, we find that the gauge generators
$X_{\dot{4}}$ and $X_{\dot{8}}$ vanish. Commutation relations between $X_{i}$
and $X_{r}$ lead directly to $SO(3)\times SO(3)$ algebra

$\left[X_{i},\,X_{j}\right]=-\kappa\varepsilon_{ijk}X_{k},\qquad\left[X_{r},\,X_{s}\right]=\kappa\varepsilon_{rst}X_{t},\qquad\left[X_{i},\,X_{r}\right]=0\,.$
(3.21)

The remaining eight generators correspond to translational generators, but in
this case, there are four constraints among them

$X_{5}=X_{1},\qquad X_{6}=-X_{2},\qquad X_{7}=-X_{3},\qquad X_{8}=-X_{4}\,.$
(3.22)

Therefore, there are only four linearly independent translational generators.
To make the form of the resulting gauge group explicit, we redefine the gauge
generators as follows. We first introduce the $SO(4)\sim SO(3)\times SO(3)$
generators $M_{\tilde{\mu}\tilde{\nu}}=-M_{\tilde{\nu}\tilde{\mu}}$ for
$\tilde{\mu},\tilde{\nu}=1,2,3,4$. These satisfy the standard $SO(4)$ algebra
of the form

$\left[M_{\tilde{\mu}\tilde{\nu}},M_{\tilde{\rho}\tilde{\sigma}}\right]=2\kappa\left(\delta_{\tilde{\mu}[\tilde{\rho}}M_{\tilde{\sigma}]\tilde{\nu}}-\delta_{\tilde{\nu}[\tilde{\rho}}M_{\tilde{\sigma}]\tilde{\mu}}\right).$
(3.23)

In terms of $X_{i}$ and $X_{r}$ generators, we find, for $\tilde{\mu}=(i,4)$,

$M_{ij}=\varepsilon_{ijk}A_{k}\qquad\textrm{and}\qquad M_{4i}=-M_{i4}=B_{i}$
(3.24)

with

$\displaystyle A_{1}$ $\displaystyle=$ $\displaystyle
X_{\dot{1}}-X_{\dot{5}},\qquad A_{2}=X_{\dot{2}}-X_{\dot{6}},\qquad
A_{3}=X_{\dot{3}}-X_{\dot{7}},$ $\displaystyle B_{1}$ $\displaystyle=$
$\displaystyle X_{\dot{1}}+X_{\dot{5}},\qquad
B_{2}=X_{\dot{2}}+X_{\dot{6}},\qquad B_{3}=X_{\dot{3}}+X_{\dot{7}}\,.$ (3.25)

Furthermore, we redefine the four independent translational generators as

$K_{1}=X_{4},\qquad K_{2}=-X_{3},\qquad K_{3}=X_{2},\qquad K_{4}=X_{1}$ (3.26)

and obtain the following commutation relations

$\left[K_{\tilde{\mu}},M_{\tilde{\nu}\tilde{\rho}}\right]=2\kappa\delta_{\tilde{\mu}[\tilde{\nu}}\delta_{\tilde{\rho}]}^{\tilde{\sigma}}K_{\tilde{\sigma}}\,.$
(3.27)

This implies that the gauge group takes the form of

$SO(4)\ltimes\mathbb{R}^{4}\sim CSO(4,0,1).$ (3.28)

####  3.3.2 $CSO(3,1,1)\sim SO(3,1)\ltimes\mathbb{R}^{4}$ gauge group

There is another solution to the conditions (3.19) with the antisymmetric
tensor $f_{\dot{m}\dot{n}\dot{p}}$ of the form

$f_{\dot{m}\dot{n}\dot{p}}=\kappa(\varepsilon_{ijr},\varepsilon_{irs})$ (3.29)

for $i,j,\ldots=\dot{1},\dot{2},\dot{3}$ and
$r,s,\ldots=\dot{5},\dot{6},\dot{7}$. As in the previous case, there are eight
nilpotent generators $X_{m}$ subject to four constraints given in (3.22)
together with six non-vanishing gauge generators $X_{i}$ and $X_{r}$. The
latter satisfy the following commutation relations

$\left[X_{i},\,X_{j}\right]=\kappa\varepsilon_{ijr}X_{r},\quad\left[X_{r},\,X_{s}\right]=-\kappa\varepsilon_{rsi}X_{i},\quad\
\left[X_{i},\,X_{r}\right]=\kappa\varepsilon_{irk}(\delta_{ks}X_{s}-X_{k})$
(3.30)

with $\delta_{ir}=\text{diag}(1,1,1)$. These relations correspond to an
$SO(3,1)$ algebra which can be explicitly seen by defining

$M_{ij}=\varepsilon_{ijk}(\delta_{kr}X_{r}-X_{k})\qquad\textrm{and}\qquad
M_{4i}=-M_{i4}=-\frac{1}{\sqrt{3}}(\delta_{ir}X_{r}+X_{i}).$ (3.31)

These generators satisfy the $SO(3,1)$ algebra

$\left[M_{\tilde{\mu}\tilde{\nu}},M_{\tilde{\rho}\tilde{\sigma}}\right]=2\kappa\left(\eta_{\tilde{\mu}[\tilde{\rho}}M_{\tilde{\sigma}]\tilde{\nu}}-\eta_{\tilde{\nu}[\tilde{\rho}}M_{\tilde{\sigma}]\tilde{\mu}}\right)$
(3.32)

with $\eta_{\tilde{\mu}\tilde{\nu}}=\text{diag}(1,1,1,-1)$ for
$\tilde{\mu},\tilde{\nu}=1,2,3,4$.
Redefining the four linearly independent translational generators as

$K_{1}=X_{4},\qquad K_{2}=-X_{3},\qquad K_{3}=X_{2},\qquad
K_{4}=\frac{X_{1}}{\sqrt{3}},$ (3.33)

we find the commutation relations with the $SO(3,1)$ generators

$\left[K_{\tilde{\mu}},M_{\tilde{\nu}\tilde{\rho}}\right]=2\kappa\eta_{\tilde{\mu}[\tilde{\nu}}\delta_{\tilde{\rho}]}^{\tilde{\sigma}}K_{\tilde{\sigma}}\,.$
(3.34)

Accordingly, the resulting gauge group is given by

$SO(3,1)\ltimes\mathbb{R}^{4}\sim CSO(3,1,1).$ (3.35)

####  3.3.3 $CSO(2,2,1)\sim SO(2,2)\ltimes\mathbb{R}^{4}$ gauge group

As a final example for solutions to (3.19), we consider the embedding tensor
of the form

$f_{\dot{m}\dot{n}\dot{p}}=\kappa(\varepsilon_{\bar{i}\bar{j}\bar{k}},\varepsilon_{\bar{r}\bar{s}\bar{t}})$
(3.36)

with $\bar{i},\bar{j},...=\dot{1},\dot{6},\dot{7}$ and
$\bar{r},\bar{s},...=\dot{2},\dot{3},\dot{5}$ being two sets of indices that
can be raised and lowered by
$\eta^{\bar{i}\bar{j}}=\eta_{\bar{i}\bar{j}}=\text{diag}(-1,1,1)$ and
$\eta^{\bar{r}\bar{s}}=\eta_{\bar{r}\bar{s}}=\text{diag}(1,1,-1)$,
respectively. There are again eight nilpotent generators $X_{m}$ subject to
four constraints given in (3.22). We will choose the independent generators to
be

$K_{1}=-X_{3},\qquad K_{2}=X_{2},\qquad K_{3}=X_{4},\qquad K_{4}=X_{1}\,.$
(3.37)

Commutation relations between the remaining six non-vanishing gauge generators
$X_{\bar{i}}$ and $X_{\bar{r}}$ are given by

$\left[X_{\bar{i}},\,X_{\bar{j}}\right]=\kappa\varepsilon_{\bar{i}\bar{j}\bar{k}}\eta^{\bar{k}\bar{l}}X_{\bar{l}},\qquad\left[X_{\bar{r}},\,X_{\bar{s}}\right]=\kappa\varepsilon_{\bar{r}\bar{s}\bar{t}}\eta^{\bar{t}\bar{u}}X_{\bar{u}},\qquad\left[X_{\bar{i}},\,X_{\bar{r}}\right]=0\,.$
(3.38)

These lead to $SO(2,1)\times SO(2,1)\sim SO(2,2)$ algebra. As in the previous
cases, we can also redefine the generators as

$M_{i4}=-M_{4i}=\delta_{ij}G^{j}+H_{i}\qquad\text{ and }\qquad
M_{ij}=\varepsilon_{ijk}\eta^{kl}(\delta_{lm}G^{m}-H_{l})$ (3.39)

for $i,j,\ldots=1,2,3$ and $\eta^{ij}=\eta_{ij}=\text{diag}(1,1,-1)$ together
with

$\displaystyle G^{1}=X_{\dot{6}},\qquad G^{2}=X_{\dot{7}},\qquad
G^{3}=X_{\dot{1}},$ $\displaystyle H_{1}=X_{\dot{2}},\qquad
H_{2}=X_{\dot{3}},\qquad H_{3}=X_{\dot{5}}\,.$ (3.40)

These generators satisfy the algebra of the form given in (3.32) but with
$\eta_{\tilde{\mu}\tilde{\nu}}=\text{diag}(1,1,-1,-1)$.
Finally, the commutation relations between these $SO(2,2)$ generators and the
four nilpotent generators given in (3.37) are the same as in (3.34) for
$\eta_{\tilde{\mu}\tilde{\nu}}=\text{diag}(1,1,-1,-1)$. Consequently, the
resulting gauge group is given by

$SO(2,2)\ltimes\mathbb{R}^{4}\sim CSO(2,2,1).$ (3.41)

In summary, the embedding tensor in $\mathbf{56}^{-1}$ representation can lead
to $CSO(4-p,p,1)\sim SO(4-p,p)\ltimes\mathbb{R}^{4}$ gauge group for
$p=0,1,2$.

###  3.4 Gaugings in $\mathbf{56}^{+1}$ representation

Similar to the previous case, gaugings in $\mathbf{56}^{+1}$ representation
can be obtained by turning on only $\theta_{4}^{\dot{m}I}$ component with the
same parametrization as in (3.16)

$\theta_{4}^{\dot{m}I}=\frac{1}{48}f_{mnp}(\gamma^{IJ})^{mn}(\gamma_{J})^{\dot{m}p}$
(3.42)

where
$(\gamma^{IJ})^{mn}=(\gamma^{[I})^{m\dot{p}}{(\gamma^{J]})_{\dot{p}}}^{n}$.
The LC requires $f_{mnp}=f_{[mnp]}$ and the QCs reduce to

$f_{mnp}f^{mnp}=0\qquad\text{ and }\qquad f_{r[mn}{f_{pq]}}^{r}=0.$ (3.43)

We can repeat the same analysis as in the case of gaugings from
$\mathbf{56}^{-1}$ representation by solving these QCs and find the same
$CSO(4-p,p,1)\sim SO(4-p,p)\ltimes\mathbb{R}^{4}$ gauge group for $p=0,1,2$.
However, the gauge generators for the two sets of nilpotent and $SO(4-p,p)$
generators are interchanged as

$\displaystyle X_{m}$ $\displaystyle=$
$\displaystyle\frac{1}{48}f_{npq}(\gamma^{IK})^{np}(\gamma_{K})^{\dot{q}q}(\gamma^{J})_{m\dot{q}}\boldsymbol{\tau}_{IJ},$
(3.44) $\displaystyle X_{\dot{m}}$ $\displaystyle=$
$\displaystyle\frac{1}{24\sqrt{2}}f_{mnp}(\gamma^{IJ})^{mn}{(\gamma_{J})_{\dot{m}}}^{p}\boldsymbol{k}_{I}$
(3.45)

with the nilpotent generators given in terms of the $\boldsymbol{k}_{I}$
instead of $\boldsymbol{p}_{I}$.

###  3.5 Gaugings in $\mathbf{56}^{-1}$ and $\mathbf{8}^{+3}$ representations

As an example for gaugings from an embedding tensor with more than one
representation, we consider gaugings with both $\mathbf{56}^{-1}$ and
$\mathbf{8}^{+3}$ representations. The embedding tensor takes the form

$\theta^{AM}\ =\
\left(\begin{array}[]{c|c|c}&\theta_{3}^{mI}&\phantom{\theta_{5}^{m+}}\\\
\hline\cr\theta_{2}^{\dot{m}-}&&\end{array}\right).$ (3.46)

Only $\theta_{3}^{mI}$ is constrained by the LC given in equation (3.15). By a
similar analysis as in the previous cases, we can solve this condition by
parametrizing $\theta_{3}^{mI}$ as in equation (3.16). Denoting
$\theta_{2}^{\dot{m}-}=v^{\dot{m}}$, we find that the QCs lead to the
following four conditions, in accordance with the analysis of [40],

$f_{\dot{m}\dot{n}\dot{p}}f^{\dot{m}\dot{n}\dot{p}}=0,\qquad
f_{\dot{r}[\dot{m}\dot{n}}{f_{\dot{p}\dot{q}]}}^{\dot{r}}=0,\qquad
f_{\dot{m}\dot{n}\dot{p}}v^{\dot{p}}=0,\qquad{f_{[\dot{m}\dot{n}\dot{p}}v_{\dot{q}]}|}_{SD}=0$
(3.47)

in which $|_{SD}$ means the self-dual part of a four-form.
The first two conditions are independent of $v^{\dot{m}}$, and can be solved
by the antisymmetric tensor $f_{\dot{m}\dot{n}\dot{p}}$ given in the case of
gaugings from $\mathbf{56}^{-1}$ representation. We then consider the
following three possibilities:

$\displaystyle
f_{\dot{m}\dot{n}\dot{p}}=\kappa(\varepsilon_{ijk},\varepsilon_{rst})\quad\text{for
}i,j,...=\dot{1},\dot{2},\dot{3}\text{ and }r,s,...=\dot{5},\dot{6},\dot{7},$
(3.48) $\displaystyle
f_{\dot{m}\dot{n}\dot{p}}=\kappa(\varepsilon_{ijr},\varepsilon_{irs})\quad\text{for
}i,j,...=\dot{1},\dot{2},\dot{3}\text{ and }r,s,...=\dot{5},\dot{6},\dot{7},$
(3.49) $\displaystyle
f_{\dot{m}\dot{n}\dot{p}}=\kappa(\varepsilon_{\bar{i}\bar{j}\bar{k}},\varepsilon_{\bar{r}\bar{s}\bar{t}})\quad\text{for
}\bar{i},\bar{j},...=\dot{1},\dot{6},\dot{7}\text{ and
}\bar{r},\bar{s},...=\dot{2},\dot{3},\dot{5}$ (3.50)

for a real constant $\kappa$. The last two conditions in (3.47) can be solved
by taking $v^{\dot{m}}$ with the only non-vanishing components given by

$v^{\dot{4}}=v^{\dot{8}}=\lambda$ (3.51)

for a real constant $\lambda$. With all these, the resulting gauge generators
are given by

$\displaystyle X_{m}$ $\displaystyle=$
$\displaystyle(\gamma^{I})_{m\dot{n}}v^{\dot{n}}\boldsymbol{k}_{I}+\frac{1}{24\sqrt{2}}f_{\dot{m}\dot{n}\dot{p}}(\gamma^{IJ})^{\dot{m}\dot{n}}{(\gamma_{J})_{m}}^{\dot{p}}\boldsymbol{p}_{I},$
(3.52) $\displaystyle X_{\dot{m}}$ $\displaystyle=$
$\displaystyle\frac{1}{48}f_{\dot{n}\dot{p}\dot{q}}(\gamma^{IK})^{\dot{n}\dot{p}}(\gamma_{K})^{q\dot{q}}(\gamma^{J})_{q\dot{m}}\boldsymbol{\tau}_{IJ}\,.$
(3.53)

It turns out that by a suitable redefinition of $X_{\dot{m}}$ generators,
these generators can be shown to satisfy $SO(4-p,p)$ algebra

$\left[M_{\tilde{\mu}\tilde{\nu}},M_{\tilde{\rho}\tilde{\sigma}}\right]=2\kappa\left(\eta_{\tilde{\mu}[\tilde{\rho}}M_{\tilde{\sigma}]\tilde{\nu}}-\eta_{\tilde{\nu}[\tilde{\rho}}M_{\tilde{\sigma}]\tilde{\mu}}\right)$
(3.54)

with $\eta_{\tilde{\mu}\tilde{\nu}}=\text{diag}(1,1,1,1)$ for
$f_{\dot{m}\dot{n}\dot{p}}$ in (3.48),
$\eta_{\tilde{\mu}\tilde{\nu}}=\text{diag}(1,1,1,-1)$ for
$f_{\dot{m}\dot{n}\dot{p}}$ in (3.49), and
$\eta_{\tilde{\mu}\tilde{\nu}}=\text{diag}(1,1,-1,-1)$ for
$f_{\dot{m}\dot{n}\dot{p}}$ in (3.50). In addition, there are four constraints
among $X_{m}$ generators, given by (3.22), implying that only four generators
are linearly independent. Choosing these four generators as in section 3.3, we
find the following commutation relations

$\displaystyle\left[K_{\tilde{\mu}},M_{\tilde{\nu}\tilde{\rho}}\right]$
$\displaystyle=$ $\displaystyle
2\kappa\eta_{\tilde{\mu}[\tilde{\nu}}\delta_{\tilde{\rho}]}^{\tilde{\sigma}}K_{\tilde{\sigma}},$
(3.55) $\displaystyle\left[K_{\tilde{\mu}},K_{\tilde{\nu}}\right]$
$\displaystyle=$ $\displaystyle-\lambda\sqrt{2}M_{\tilde{\mu}\tilde{\nu}}\,.$
(3.56)

It should be noted that in this case with non-vanishing $\lambda$, generators
$K_{\tilde{\mu}}$ do not commute with each other but close onto the
$SO(4-p,p)$ part. These generators enlarge $SO(4-p,p)$ to a larger gauge
group. In particular, by setting $\lambda=\pm\frac{\kappa}{4\sqrt{2}}$ and
defining the generators

$M_{0\tilde{\mu}}=-M_{\tilde{\mu}0}=2K_{\tilde{\mu}},$ (3.57)

we obtain the algebra, with $\underline{\mu}=(0,\tilde{\mu})=0,1,2,3,4$,

$\left[M_{\underline{\mu}\underline{\nu}},M_{\underline{\rho}\underline{\sigma}}\right]=2\kappa\left(\eta_{\underline{\mu}[\underline{\rho}}M_{\underline{\sigma}]\underline{\nu}}-\eta_{\underline{\nu}[\underline{\rho}}M_{\underline{\sigma}]\underline{\mu}}\right)$
(3.58)

for $\eta_{\underline{\mu}\underline{\nu}}=\text{diag}(\pm
1,\eta_{\tilde{\mu}\tilde{\nu}})$. The two sign choices correspond to
$SO(5-p,p)$ or $SO(4-p,p+1)$ gauge groups. In particular, the $SO(4)$ group
with $\eta_{\tilde{\mu}\tilde{\nu}}=\delta_{\tilde{\mu}\tilde{\nu}}$ is
enlarged to $SO(5)$ or $SO(4,1)_{\textrm{I}}$. Similarly, the $SO(3,1)$ group
with $\eta_{\tilde{\mu}\tilde{\nu}}=\text{diag}(1,1,1,-1)$ is enlarged to
$SO(4,1)_{\textrm{II}}$ or $SO(3,2)_{\textrm{I}}$ while the $SO(2,2)$ group,
with $\eta_{\tilde{\mu}\tilde{\nu}}=\text{diag}(1,1,-1,-1)$, becomes
$SO(3,2)_{\textrm{II}}$. We have used subscripts I and II to distinguish the
gauge groups arising from different $SO(4-p,p)$ groups obtained from the
embedding tensor in $\mathbf{56}^{-1}$ representation.
We finally note that the analysis for gaugings from $\mathbf{56}^{+1}$ and
$\mathbf{8}^{-3}$ representations can be carried out in the same way leading
to the same gauge groups with the role of $X_{m}$ and $X_{\dot{m}}$
interchanged.

##  4 Supersymmetric domain wall solutions

In this section, we find supersymmetric domain walls which are half-
supersymmetric vacuum solutions of the maximal gauged supergravities
considered in the previous section. We take the space-time metric to be the
standard domain wall ansatz

$ds_{6}^{2}=e^{2A(r)}\eta_{\bar{\mu}\bar{\nu}}dx^{\bar{\mu}}dx^{\bar{\nu}}+dr^{2}$
(4.1)

where $\bar{\mu},\bar{\nu},\ldots$ are space-time indices of five-dimensional
Minkowski space, and $A(r)$ is a warp factor depending only on the radial
coordinate $r$.
Following [17] and [18], the coset representative of $SO(5,5)/SO(5)\times
SO(5)$, parametrized by 25 scalar fields, can be obtained by the following
non-compact generators of $SO(5,5)$ in diagonal basis

$\hat{\boldsymbol{t}}_{a\dot{b}}\,=\,{\mathbb{M}_{a}}^{M}{\mathbb{M}_{\dot{b}}}^{N}\boldsymbol{t}_{MN}$
(4.2)

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.24).
These non-compact generators are symmetric such that
$(\hat{\boldsymbol{t}}_{a\dot{b}})^{T}=\hat{\boldsymbol{t}}_{a\dot{b}}$.
Recall that an $SO(5,5)$ vector index in non-diagonal basis is decomposed as
$M=(-,I,+)$ under $SO(4,4)$, we further decompose vector indices of both
$SO(5)$ factors as $a=(0,i)$ and $\dot{a}=(\dot{i},\\#)$, with $i=1,2,3,4$ and
$\dot{i}=\dot{1},\dot{2},\dot{3},\dot{4}$. This leads to the following
decomposition of the non-compact generators

$\hat{\boldsymbol{t}}_{a\dot{b}}\,=\,(\hat{\boldsymbol{t}}_{0\dot{i}},\,\hat{\boldsymbol{t}}_{0\\#},\,\hat{\boldsymbol{t}}_{i\dot{j}},\,\hat{\boldsymbol{t}}_{i\\#}).$
(4.3)

Under the $SO(4,4)$ branching rule for $SO(5,5)$ adjoint representation given
in (A.15), these non-compact generators read

$\displaystyle\hat{\boldsymbol{t}}_{0\dot{i}}$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}{\mathbb{M}_{\dot{i}}}^{I}(\boldsymbol{p}_{I}+\boldsymbol{k}_{I}),\qquad\hat{\boldsymbol{t}}_{0\\#}=-\boldsymbol{d},$
$\displaystyle\hat{\boldsymbol{t}}_{i\\#}$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}{\mathbb{M}_{i}}^{I}(\boldsymbol{p}_{I}-\boldsymbol{k}_{I}),\qquad\hat{\boldsymbol{t}}_{i\dot{j}}={\mathbb{M}_{i}}^{I}{\mathbb{M}_{\dot{j}}}^{J}\boldsymbol{\tau}_{IJ}.$
(4.4)

Under the compact $SO(5)\times SO(5)\subset SO(5,5)$, the $25$ scalars
transform as $(\mathbf{5},\mathbf{5})$. The split of indices $a$ and $\dot{a}$
given above implies the branching $\mathbf{5}\rightarrow\mathbf{1}+\mathbf{4}$
of $SO(5)\rightarrow SO(4)$. Therefore, under $SO(4)\times SO(4)\subset
SO(5)\times SO(5)$, the scalars tranform as

$\underbrace{(\mathbf{5},\mathbf{5})}_{\hat{\boldsymbol{t}}_{a\dot{b}}}\
\rightarrow\
\underbrace{(\mathbf{1},\mathbf{1})}_{\boldsymbol{d}}\oplus\underbrace{(\mathbf{1},\mathbf{4})}_{\hat{\boldsymbol{t}}_{0\dot{i}}}\oplus\underbrace{(\mathbf{4},\mathbf{1})}_{\hat{\boldsymbol{t}}_{i\\#}}\,\oplus\underbrace{(\mathbf{4},\mathbf{4})}_{{\boldsymbol{t}}_{i\dot{j}}}\,.$
(4.5)

It is convenient to denote all $25$ scalar fields collectively as

$\Phi^{\mathbb{I}}=\\{\varphi,\zeta_{1},\ldots,\zeta_{4},\xi_{\dot{1}},\ldots,\xi_{\dot{4}},\phi_{1},\ldots,\phi_{16}\\}$
(4.6)

with $\mathbb{I}=1,\ldots,25$. The scalar $\varphi$ is the dilaton
corresponding to the $\mathbb{R}^{+}\sim SO(1,1)$ generator $\boldsymbol{d}$.
The two sets of four scalars $\\{\zeta_{1},...,\zeta_{4}\\}$ and
$\\{\xi_{\dot{1}},...,\xi_{\dot{4}}\\}$ respectively correspond to the
generators $\hat{\boldsymbol{t}}_{0\dot{i}}$ and
$\hat{\boldsymbol{t}}_{i\\#}$. The remaining sixteen scalar fields
$\\{\phi_{1},...,\phi_{16}\\}$ parametrize the submanifold
$SO(4,4)/SO(4)\times SO(4)$ of the $SO(5,5)/SO(5)\times SO(5)$ coset.
With this form of the scalar fields, we can rewrite the kinetic terms of the
scalar fields in (2.13) and obtain the following form of the bosonic
Lagrangian

$e^{-1}\mathcal{L}=\frac{1}{4}R-G_{{\mathbb{I}}{\mathbb{J}}}\partial_{\mu}\Phi^{\mathbb{I}}\partial^{\mu}\Phi^{\mathbb{J}}-\mathbf{V}$
(4.7)

with
$G_{{\mathbb{I}}{\mathbb{J}}}=\frac{1}{16}P^{a\dot{a}}_{\mathbb{I}}P_{a\dot{a}\mathbb{J}}$
being a symmetric scalar metric. The vielbein $P^{a\dot{a}}_{\mathbb{I}}$ on
the scalar manifold is related to $P^{a\dot{a}}_{\mu}$ via
$P^{a\dot{a}}_{\mu}=P^{a\dot{a}}_{\mathbb{I}}\partial_{\mu}\Phi^{\mathbb{I}}$.
We will find supersymmetric domain wall solutions from first-order
Bogomol’nyi-Prasad-Sommerfield (BPS) equations derived from the supersymmetry
transformations of fermionic fields. The procedure is essentially the same as
that given in [17] and [18], so we will mainly state the final results. The
variations of the gravitini in (2.14) and (2.15),
$\delta\psi_{+\bar{\mu}\alpha}$ and $\delta\psi_{-\bar{\mu}\dot{\alpha}}$
respectively gives

$\displaystyle
A^{\prime}\hat{\gamma}_{r}\epsilon_{+\alpha}+\frac{1}{2}\Omega_{\alpha\beta}T^{\beta\dot{\alpha}}\epsilon_{-\dot{\alpha}}=0,$
(4.8) $\displaystyle
A^{\prime}\hat{\gamma}_{r}\epsilon_{-\dot{\alpha}}-\frac{1}{2}\Omega_{\dot{\alpha}\dot{\beta}}T^{\alpha\dot{\beta}}\epsilon_{+\alpha}=0\,.$
(4.9)

Throughout the paper, we use the notation ′ to denote an $r$-derivative.
Multiply equation (4.8) by $A^{\prime}\hat{\gamma}_{r}$ and use equation (4.9)
or vice-versa, we find the following consistency conditions

$\displaystyle{A^{\prime}}^{2}{\delta_{\alpha}}^{\beta}$ $\displaystyle=$
$\displaystyle-\frac{1}{4}\Omega_{\alpha\gamma}T^{\gamma\dot{\alpha}}\Omega_{\dot{\alpha}\dot{\beta}}T^{\beta\dot{\beta}}=\mathcal{W}^{2}{\delta_{\alpha}}^{\beta},$
(4.10) $\displaystyle{A^{\prime}}^{2}{\delta_{\dot{\alpha}}}^{\dot{\beta}}$
$\displaystyle=$
$\displaystyle-\frac{1}{4}\Omega_{\dot{\alpha}\dot{\gamma}}T^{\alpha\dot{\gamma}}\Omega_{\alpha\beta}T^{\beta\dot{\beta}}=\mathcal{W}^{2}{\delta_{\dot{\alpha}}}^{\dot{\beta}}$
(4.11)

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

$A^{\prime}=\pm\mathcal{W}\,.$ (4.12)

With this result, equations (4.8) and (4.9) lead to the following (not
independent) projectors on the Killing spinors

$\hat{\gamma}_{r}\epsilon_{+\alpha}=-\frac{1}{2}\Omega_{\alpha\beta}\frac{T^{\beta\dot{\beta}}}{A^{\prime}}\epsilon_{-\dot{\beta}},\qquad\hat{\gamma}_{r}\epsilon_{-\dot{\alpha}}=\frac{1}{2}\Omega_{\dot{\alpha}\dot{\beta}}\frac{T^{\alpha\dot{\beta}}}{A^{\prime}}\epsilon_{+\alpha}\,.$
(4.13)

Using these projectors in the variations $\delta\chi_{+a\dot{\alpha}}$ and
$\delta\chi_{-\dot{a}\alpha}$ together with some identities involving the
T-tensors, in particular (2.29), we may rewrite the BPS equations for scalar
fields in the form

${\Phi^{\mathbb{I}}}^{\prime}=\mp
2G^{\mathbb{I}\mathbb{J}}\frac{\partial\mathcal{W}}{\partial\Phi^{\mathbb{J}}}$
(4.14)

in which $G^{\mathbb{I}\mathbb{J}}$ is the inverse of the scalar metric
$G_{\mathbb{I}\mathbb{J}}$. The remaining variations $\delta\psi_{+r\alpha}$
and $\delta\psi_{-r\dot{\alpha}}$ determine the $r$ dependence of the Killing
spinors.
In addition, we also note that the scalar potential can be written in terms of
$\mathcal{W}$ as

$\mathbf{V}=2G^{\mathbb{I}\mathbb{J}}\frac{\partial\mathcal{W}}{\partial\Phi^{\mathbb{I}}}\frac{\partial\mathcal{W}}{\partial\Phi^{\mathbb{J}}}-5\mathcal{W}^{2}\,.$
(4.15)

It is also straightforward to show that the BPS equations of the form (4.12)
and (4.14) satisfy the second-order field equations derived from the bosonic
Lagrangian (4.7) with the scalar potential given by (4.15), see [41, 42, 43,
44, 45, 46] for more detail. Finally, since we have imposed only one
independent projector on the Killing spinors, all solutions found in this work
are half-supersymmetric.

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

To deal with the $25$-dimensional scalar manifold, we will follow the approach
introduced in [47] by considering domain wall solutions that are invariant
under a particular subgroup of the gauge groups. These solutions involve only
a subset of all $25$ scalars. We also restrict ourselves to only gauge groups
obtained from the embedding tensor in $\mathbf{56}^{-1}$ and $\mathbf{8}^{+3}$
representations.
We first consider supersymmetric domain walls with an unbroken symmetry
$SO(4)$. Only $SO(5)$, $SO(4,1)_{\textrm{I}}$, and $CSO(4,0,1)$ gauge groups
contain a common $SO(4)$ subgroup which in turn lies within $SO(4,4)$. We
collectively describe them in a single framework by using
$f_{\dot{m}\dot{n}\dot{p}}$ given in (3.48) together with
$v^{\dot{4}}=v^{\dot{8}}=-\frac{\sigma\kappa}{4\sqrt{2}}$ with $\sigma=1,-1,$
and $0$, corresponding to $SO(5)$, $SO(4,1)_{\textrm{I}}$, and $CSO(4,0,1)$
gauge groups, respectively. The residual $SO(4)$ symmetry is embedded
diagonally in $SO(4,4)$ via the maximal compact subgroup $SO(4)\times SO(4)$.
From the decomposition of the scalars given in (4.5), we find two $SO(4)$
singlets according to the following decomposition

$\displaystyle(\mathbf{5},\mathbf{5})\rightarrow$
$\displaystyle(\mathbf{1}\otimes\mathbf{1})\oplus(\mathbf{1}\otimes\mathbf{4})\oplus(\mathbf{4}\otimes\mathbf{1})\oplus(\mathbf{4}\otimes\mathbf{4})$
(4.16) $\displaystyle\sim$
$\displaystyle\mathbf{1}\oplus\mathbf{4}\oplus\mathbf{4}\oplus\mathbf{1}\oplus\mathbf{6}\oplus\mathbf{9}\,.$

The first singlet corresponds to the $\mathbb{R}^{+}$ generator
$\boldsymbol{d}$ while the second one, arising from
$\mathbf{4}\otimes\mathbf{4}$, is given by the non-compact generator

$\mathcal{Y}=\hat{\boldsymbol{t}}_{1\dot{1}}-\hat{\boldsymbol{t}}_{2\dot{2}}-\hat{\boldsymbol{t}}_{3\dot{3}}-\hat{\boldsymbol{t}}_{4\dot{4}}$
(4.17)

from the $SO(4,4)/SO(4)\times SO(4)$ coset.
Using the coset representative of the form

$V=e^{\varphi\boldsymbol{d}+\phi\mathcal{Y}},$ (4.18)

we find the superpotential and the scalar potential of the form

$\displaystyle\mathcal{W}$ $\displaystyle=$
$\displaystyle\frac{g\kappa}{16\sqrt{2}}e^{-3\varphi-4\phi}\left(4e^{4(\varphi+\phi)}+\sigma\right),$
(4.19) $\displaystyle\mathbf{V}$ $\displaystyle=$
$\displaystyle-\frac{g^{2}\kappa^{2}}{64}e^{-6\varphi-8\phi}\left(8e^{8(\varphi+\phi)}+8\sigma
e^{4(\varphi+\phi)}-\sigma^{2}\right).\qquad$ (4.20)

It can be checked that the scalar potential can be written in terms of the
superpotential according to (4.15) using the scalar matrix
$G^{\mathbb{I}\mathbb{J}}=\text{diag}(\frac{1}{2},\frac{1}{8})$ for
$\Phi^{\mathbb{I}}=\\{\varphi,\phi\\}$ and $\mathbb{I}=1,2$. The general
analysis given above leads to the BPS equation for the warp factor

$A^{\prime}=\pm\frac{g\kappa}{16\sqrt{2}}e^{-3\varphi-4\phi}\left(4e^{4(\varphi+\phi)}+\sigma\right)$
(4.21)

together with the BPS equations for the scalar fields

$\displaystyle\varphi^{\prime}=\mp\frac{g\kappa}{16\sqrt{2}}e^{-3\varphi-4\phi}\left(4e^{4(\varphi+\phi)}-3\sigma\right)$
and $\displaystyle\phi^{\prime}=\pm\frac{\sigma
g\kappa}{16\sqrt{2}}e^{-3\varphi-4\phi}\,.$ (4.22)

For $\sigma=1,-1$ corresponding to $SO(5)$ and $SO(4,1)_{\textrm{I}}$ gauge
groups, the solutions for the warp factor $A$ and dilaton $\varphi$ can be
given in terms of $\phi$ as

$\displaystyle A$ $\displaystyle=$ $\displaystyle
C_{1}+3\phi-\frac{1}{4}\ln\left[e^{-4(4\phi+C_{2})}+\sigma\right],$ (4.23)
$\displaystyle\varphi$ $\displaystyle=$
$\displaystyle-\phi-\frac{1}{4}\ln\left[e^{-4(4\phi+C_{2})}+\sigma\right]$
(4.24)

in which $C_{1}$ and $C_{2}$ are integration constants. To obtain the solution
for $\phi$, we change the radial coordinate $r$ to $\rho$ defined by
$\frac{d\rho}{dr}=e^{-3\varphi-4\phi}$. The solution of $\phi$ is then readily
found to be

$\phi=\pm\frac{g\sigma\kappa\rho}{16\sqrt{2}}+C_{3}$ (4.25)

in which $\pm$ directly corresponds to the upper/lower signs in the BPS
equations. Thus, the two sign choices in the BPS equations can be absorbed by
flipping the sign of the radial coordinate. We will neglect these sign choices
by choosing the upper sign of the BPS equations from now on. Moreover, the
integration constants $C_{1}$ and $C_{3}$ can also be removed by rescaling the
coordinates $x^{\bar{\mu}}$ and shifting the radial coordinate $\rho$.
For $\sigma=0$ corresponding to $CSO(4,0,1)$ gauge group, the superpotential
and scalar potential are independent of $\phi$

$\mathcal{W}=\frac{g\kappa}{4\sqrt{2}}e^{\varphi}\qquad\text{ and
}\qquad\mathbf{V}=-\frac{g^{2}\kappa^{2}}{8}e^{2\varphi},$ (4.26)

and the BPS equations reduce to

$A^{\prime}=-\varphi^{\prime}=\frac{g\kappa}{4\sqrt{2}}e^{\varphi}\qquad\text{
and }\qquad\phi^{\prime}=0.$ (4.27)

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

$A=-\varphi=\ln\left(\frac{g\kappa r}{4\sqrt{2}}-C\right)\qquad\text{ and
}\qquad\phi=0\,.$ (4.28)

In this case, we can consistently truncate out the $SO(4)$ invariant scalar
$\phi$ since the scalar potential is independent of $\phi$.
For $SO(4,1)_{\textrm{II}}$ gauge group, the $SO(4)$ compact subgroup is not
embedded in $SO(4,4)$ since it involves $M_{0\tilde{i}}$ generators obtained
from the gauge generators $X_{m}$. However, a similar analysis can be carried
out by using $f_{\dot{m}\dot{n}\dot{p}}$ given in (3.49) together with
$v^{\dot{4}}=v^{\dot{8}}=\frac{\kappa}{4\sqrt{2}}$. In this case, there are
again two $SO(4)$ singlet scalars corresponding to the non-compact generators

$\overline{\mathbb{Y}}_{1}=\hat{\boldsymbol{t}}_{1\dot{1}}\qquad\text{ and
}\qquad\overline{\mathbb{Y}}_{2}=\boldsymbol{d}-\hat{\boldsymbol{t}}_{2\dot{2}}-\hat{\boldsymbol{t}}_{3\dot{3}}-\hat{\boldsymbol{t}}_{4\dot{4}}\,.$
(4.29)

Using the coset representative

$V=e^{\overline{\phi}_{1}\overline{\mathbb{Y}}_{1}+\overline{\phi}_{2}\overline{\mathbb{Y}}_{1}},$
(4.30)

we find the same form of the domain wall solution as given in (4.23)-(4.25)
with $\sigma=-3$, $\varphi=-\overline{\phi}_{1}$, and
$\phi=-\overline{\phi}_{2}$.

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

We next consider supersymmetric domain walls preserving a smaller residual
symmetry $SO(3)$ generated by $M_{ij}$ for $i,j,...=1,2,3$ from the
$SO(4-p,p)$ generators $M_{\tilde{\mu}\tilde{\nu}}$. There are many gauge
groups containing the $SO(3)\subset SO(4,4)$. These are given by $SO(5)$,
$SO(4,1)_{\textrm{I}}$, and $CSO(4,0,1)$ with $f_{\dot{m}\dot{n}\dot{p}}$
given in (3.48) and $v^{\dot{4}}=v^{\dot{8}}=-\frac{\sigma\kappa}{4\sqrt{2}}$
together with $SO(4,1)_{\textrm{II}}$, $SO(3,2)_{\textrm{I}}$, and
$CSO(3,1,1)$ with $f_{\dot{m}\dot{n}\dot{p}}$ given in (3.49) and
$v^{\dot{4}}=v^{\dot{8}}=\frac{\sigma\kappa}{4\sqrt{2}}$.
By further decomposing the residual symmetry of the previous section according
to $SO(4)\rightarrow SO(3)$ with the branching
$\mathbf{4}\rightarrow\mathbf{1}+\mathbf{3}$, we find the following
transformation of all $25$ scalars

$\displaystyle(\mathbf{5},\mathbf{5})\rightarrow$
$\displaystyle(\mathbf{1}\oplus\mathbf{1}\oplus\mathbf{3})\otimes(\mathbf{1}\oplus\mathbf{1}\oplus\mathbf{3})$
(4.31) $\displaystyle\sim$ $\displaystyle 4(\mathbf{1})\oplus
4(\mathbf{3})\oplus(\mathbf{1}\oplus\mathbf{3}\oplus\mathbf{5})$

with the last three representations arising from the product
$\mathbf{3}\otimes\mathbf{3}$. There are then five scalars invariant under the
residual symmetry $SO(3)$. One of them is again the dilaton corresponding to
the $\mathbb{R}^{+}$ generator. The other four singlets are associated with
the following non-compact generators

$\mathcal{Y}_{1}=\hat{\boldsymbol{t}}_{1\dot{1}},\qquad\mathcal{Y}_{2}=\hat{\boldsymbol{t}}_{2\dot{2}}+\hat{\boldsymbol{t}}_{3\dot{3}}+\hat{\boldsymbol{t}}_{4\dot{4}},\qquad\mathcal{Y}_{3}=\hat{\boldsymbol{t}}_{0\dot{1}},\qquad\mathcal{Y}_{4}=\hat{\boldsymbol{t}}_{1\\#}\,.$
(4.32)

We can use the coset representative of the form

$V=e^{\varphi\boldsymbol{d}+\phi_{1}\mathcal{Y}_{1}+\phi_{2}\mathcal{Y}_{2}+\zeta\mathcal{Y}_{3}+\xi\mathcal{Y}_{4}}\,.$
(4.33)

It turns out that consistency between the resulting BPS equations and the
second-order field equations requires $\xi=\zeta$. With this condition, the
superpotential and scalar potential become

$\displaystyle\mathcal{W}$ $\displaystyle=$
$\displaystyle\frac{g\kappa}{16\sqrt{2}}e^{-(3\varphi+\phi_{1}+\phi_{2})}\left[3e^{4\varphi}+e^{4\phi_{2}}\cosh{4\zeta}\left(ve^{4(\varphi+\phi_{1})}+\sigma\right)\right],$
(4.34) $\displaystyle\mathbf{V}$ $\displaystyle=$
$\displaystyle\frac{1}{64}e^{-2(3\varphi+\phi_{1}+\phi_{2})}g^{2}\kappa^{2}\left[\left[e^{4\phi_{2}}\cosh{4\zeta}\left(ve^{4(\varphi+\phi_{1})}+\sigma\right)-3e^{4\varphi}\right]^{2}\right.$
(4.35) $\displaystyle\left.-12e^{8\varphi}-4v\sigma
e^{4(\varphi+\phi_{1}+2\phi_{2})}\right].\qquad\quad$

We have used the parameter $v=2u-1$ with $u=\pm 1$ together with $\sigma=\pm
1,0$ to identify the gauge groups; $SO(5)$ ($u=\sigma=1$),
$SO(4,1)_{\textrm{I}}$ ($u=-\sigma=1$), $CSO(4,0,1)$ ($u=1,\sigma=0$),
$SO(4,1)_{\textrm{II}}$ ($u=-\sigma=-1$), $SO(3,2)_{\textrm{I}}$
($u=\sigma=-1$), and $CSO(3,1,1)$ ($u=-1,\sigma=0$).
Using

$G^{\mathbb{I}\mathbb{J}}=\frac{1}{8}\begin{pmatrix}(3+\cosh{8\zeta})\text{sech}^{2}{4\zeta}&-2\tanh^{2}{4\zeta}&0&0\\\
-2\tanh^{2}{4\zeta}&(3+\cosh{8\zeta})\text{sech}^{2}{4\zeta}&0&0\\\
0&0&\frac{4}{3}&0\\\ 0&0&0&2\end{pmatrix}$ (4.36)

for $\Phi^{\mathbb{I}}=\\{\varphi,\phi_{1},\phi_{2},\zeta\\}$ with
$\mathbb{I}=1,2,3,4$, we can rewrite the scalar potential in terms of the
superpotential as in (4.15).
With all these, we arrive at the BPS equations

$\displaystyle A^{\prime}$ $\displaystyle=$
$\displaystyle\frac{g\kappa}{16\sqrt{2}}e^{-(3\varphi+\phi_{1}+\phi_{2})}\left[3e^{4\varphi}+e^{4\phi_{2}}\cosh{4\zeta}\left(ve^{4(\varphi+\phi_{1})}+\sigma\right)\right],$
(4.37) $\displaystyle\varphi^{\prime}$ $\displaystyle=$
$\displaystyle-\frac{1}{32\sqrt{2}}e^{-3\varphi-\phi_{1}-\phi_{2}}g\kappa\left[6e^{4\varphi}\right.$
(4.38)
$\displaystyle\left.-e^{4\phi_{2}}\text{sech}{4\zeta}\left[ve^{4(\varphi+\phi_{1})}(\cosh{8\zeta}-3)+\sigma(\cosh{8\zeta}+5)\right]\right],\qquad\
$ $\displaystyle\phi^{\prime}_{1}$ $\displaystyle=$
$\displaystyle\frac{1}{32\sqrt{2}}e^{-3\varphi-\phi_{1}-\phi_{2}}g\kappa\left[6e^{4\varphi}\right.$
(4.39)
$\displaystyle\left.-e^{4\phi_{2}}\text{sech}{4\zeta}\left[ve^{4(\varphi+\phi_{1})}(\cosh{8\zeta}+5)+\sigma(\cosh{8\zeta}-3)\right]\right],$
$\displaystyle\phi^{\prime}_{2}$ $\displaystyle=$
$\displaystyle\frac{g\kappa}{16\sqrt{2}}e^{-(3\varphi+\phi_{1}+\phi_{2})}\left[e^{4\varphi}-e^{4\phi_{2}}\cosh{4\zeta}\left(ve^{4(\varphi+\phi_{1})}+\sigma\right)\right],$
(4.40) $\displaystyle\zeta^{\prime}$ $\displaystyle=$
$\displaystyle-\frac{g\kappa}{8\sqrt{2}}e^{-3\varphi-\phi_{1}+3\phi_{2}}\sinh{4\zeta}\left(ve^{4(\varphi+\phi_{1})}+\sigma\right).$
(4.41)

From these equations, the solutions for scalar fields $\phi_{2}$, $\phi_{1}$,
$\varphi$, and the warp factor $A$ can be obtained as functions of $\zeta$.
These are given by

$\displaystyle\phi_{2}$ $\displaystyle=$
$\displaystyle\frac{1}{16}\ln\left[\frac{\sinh{8\zeta}\sqrt{v\sigma+C_{1}^{2}-C_{1}^{2}\text{sech}^{2}{4\zeta}}-C_{2}v^{2}\sigma^{2}\sinh^{2}{4\zeta}}{2v\sigma}\right],$
(4.42) $\displaystyle\phi_{1}$ $\displaystyle=$
$\displaystyle\frac{1}{4}\ln\left[\frac{\sqrt{C_{1}^{2}+v\sigma-
C_{1}^{2}\text{sech}^{2}{4\zeta}}-C_{1}\tanh{4\zeta}}{v}\right]-\frac{1}{4}\ln{\sinh{4\zeta}}+3\phi_{2},\qquad$
(4.43) $\displaystyle\varphi$ $\displaystyle=$
$\displaystyle\frac{1}{4}\ln\left[\frac{\sqrt{C_{1}^{2}+v\sigma-
C_{1}^{2}\text{sech}^{2}{4\zeta}}-C_{1}\tanh{4\zeta}}{v}\right]+\frac{1}{4}\ln{\sinh{4\zeta}}-3\phi_{2},\qquad$
(4.44) $\displaystyle A$ $\displaystyle=$ $\displaystyle
3\phi_{2}-\frac{1}{2}\ln{\sinh{4\zeta}}$ (4.45)

in which we have chosen the integration constants for $\varphi$ and $A$ to be
zero for simplicity. To obtain the solution for $\zeta$, we change $r$ to a
new radial coordinate $\rho$ defined by
$\frac{d\rho}{dr}=e^{-4\zeta-\varphi-\phi_{1}+3\phi_{2}}(ve^{4(\varphi+\phi_{1})}+\sigma)$.
The solution for $\zeta$ is then given by

$e^{4\zeta}=\tanh\left(\frac{g\kappa\rho}{4\sqrt{2}}-C_{3}\right).$ (4.46)

It should be noted that for the case of $SO(3,2)_{\textrm{II}}$ gauge group,
the $SO(3)$ subgroup lies outside $SO(4,4)$. A similar solution with
$v=\sigma=-1$ can be found by using $f_{\dot{m}\dot{n}\dot{p}}$ given in
(3.50) together with $v^{\dot{4}}=v^{\dot{8}}=\frac{\kappa}{4\sqrt{2}}$, and
the coset representative

$V=e^{-\varphi\hat{\boldsymbol{t}}_{1\dot{1}}-\phi_{1}\hat{\boldsymbol{t}}_{4\dot{4}}+\phi_{2}(\boldsymbol{d}+\hat{\boldsymbol{t}}_{2\dot{2}}+\hat{\boldsymbol{t}}_{3\dot{3}})+\zeta(\hat{\boldsymbol{t}}_{1\dot{4}}-\hat{\boldsymbol{t}}_{4\dot{1}})}$
(4.47)

with the five $SO(3)$ singlets corresponding to the non-compact generators
$\hat{\boldsymbol{t}}_{1\dot{1}}$, $\hat{\boldsymbol{t}}_{4\dot{4}}$,
$\boldsymbol{d}+\hat{\boldsymbol{t}}_{2\dot{2}}+\hat{\boldsymbol{t}}_{3\dot{3}}$,
$\hat{\boldsymbol{t}}_{1\dot{4}}$, and $\hat{\boldsymbol{t}}_{4\dot{1}}$. We
also note the relation between scalar fields corresponding to
$\hat{\boldsymbol{t}}_{1\dot{4}}$ and $\hat{\boldsymbol{t}}_{4\dot{1}}$
generators arising from consistency between the BPS equations and the field
equations as in the above analysis.

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

We now move to domain wall solutions with $SO(2)\times SO(2)$ symmetry. Gauge
groups containing an $SO(2)\times SO(2)$ subgroup, embedded entirely in
$SO(4,4)$, are $SO(5)$, $SO(4,1)_{\textrm{I}}$, $CSO(4,0,1)$ with
$f_{\dot{m}\dot{n}\dot{p}}$ given in (3.48) and
$v^{\dot{4}}=v^{\dot{8}}=-\frac{\sigma\kappa}{4\sqrt{2}}$ together with
$SO(3,2)_{\textrm{II}}$ and $CSO(2,2,1)$ with $f_{\dot{m}\dot{n}\dot{p}}$
given in (3.50) and $v^{\dot{4}}=v^{\dot{8}}=\frac{\sigma\kappa}{4\sqrt{2}}$.
In order to incorporate the two sets of gauge groups within a single
framework, we will choose the $SO(2)\times SO(2)$ generators to be
$X_{\dot{1}}$ and $X_{\dot{5}}$.
The residual symmetry $SO(2)\times SO(2)$ is embedded in $SO(4)$ as
$\mathbf{4}\rightarrow(\mathbf{1},\mathbf{2})+(\mathbf{2},\mathbf{1})$ with
$\mathbf{2}$ denoting the fundamental or vector representation of $SO(2)$. As
in the previous cases, decomposing the transformation of all $25$ scalar
fields under $SO(2)\times SO(2)$ gives

$\displaystyle(\mathbf{5},\mathbf{5})\rightarrow$
$\displaystyle[(\mathbf{1},\mathbf{1})\oplus(\mathbf{1},\mathbf{2})\oplus(\mathbf{2},\mathbf{1})]\otimes[(\mathbf{1},\mathbf{1})\oplus(\mathbf{1},\mathbf{2})\oplus(\mathbf{2},\mathbf{1})]$
(4.48) $\displaystyle\sim$ $\displaystyle 5(\mathbf{1},\mathbf{1})\oplus
3(\mathbf{1},\mathbf{2})\oplus 3(\mathbf{2},\mathbf{1})\oplus
2(\mathbf{2},\mathbf{2}).$

Accordingly, there are five singles corresponding to the $\mathbb{R}^{+}$
generator $\boldsymbol{d}$ and the following four non-compact generators

$\mathds{Y}_{1}=\hat{\boldsymbol{t}}_{1\dot{1}}-\hat{\boldsymbol{t}}_{4\dot{4}},\qquad\mathds{Y}_{2}=\hat{\boldsymbol{t}}_{2\dot{2}}+\hat{\boldsymbol{t}}_{3\dot{3}},\qquad\mathds{Y}_{3}=\hat{\boldsymbol{t}}_{1\dot{4}}+\hat{\boldsymbol{t}}_{4\dot{1}},\qquad\mathds{Y}_{4}=\hat{\boldsymbol{t}}_{2\dot{3}}-\hat{\boldsymbol{t}}_{3\dot{2}}\,.$
(4.49)

leading to the coset representative of the form

$V=e^{\varphi\boldsymbol{d}+\phi_{1}\mathds{Y}_{1}+\phi_{2}\mathds{Y}_{2}+\phi_{3}\mathds{Y}_{3}+\phi_{4}\mathds{Y}_{4}}\,.$
(4.50)

In this case, consistency between the BPS equations and field equations
requires vanishing $\phi_{3}$ and $\phi_{4}$. With $\phi_{3}=\phi_{4}=0$, the
superpotential is given by

$\mathcal{W}=\frac{g\kappa}{16\sqrt{2}}e^{\varphi-2(\phi_{1}+\phi_{2})}\left(2e^{4(\phi_{1}+\phi_{2})}+2u+u\sigma
e^{4(\phi_{2}-\varphi)}\right),$ (4.51)

and the scalar potential takes the form

$\mathbf{V}=-\frac{ug^{2}\kappa^{2}}{64}e^{-6\varphi-4\phi_{1}}\left[8e^{8\varphi+4\phi_{1}}+4\sigma
e^{4\varphi}(e^{4(\phi_{1}+\phi_{2})}+u)-u\sigma^{2}e^{4\phi_{2}}\right].$
(4.52)

The parameters $u=\pm 1$ and $\sigma=\pm 1,0$ correspond to different gauge
groups; $SO(5)$ ($u=\sigma=1$), $SO(4,1)_{\textrm{I}}$ ($u=-\sigma=1$),
$CSO(4,0,1)$ ($u=1,\sigma=0$), $SO(3,2)_{\textrm{II}}$ ($u=-\sigma=-1$ or
$u=\sigma=-1$), and $CSO(2,2,1)$ ($u=-1,\sigma=0$).
The resulting BPS equations read

$\displaystyle A^{\prime}$ $\displaystyle=$
$\displaystyle\frac{g\kappa}{16\sqrt{2}}e^{\varphi-2(\phi_{1}+\phi_{2})}\left(2e^{4(\phi_{1}+\phi_{2})}+2u+u\sigma
e^{4(\phi_{2}-\varphi)}\right),$ (4.53) $\displaystyle\varphi^{\prime}$
$\displaystyle=$
$\displaystyle-\frac{g\kappa}{16\sqrt{2}}e^{\varphi-2(\phi_{1}+\phi_{2})}\left(2e^{4(\phi_{1}+\phi_{2})}+2u-3u\sigma
e^{4(\phi_{2}-\varphi)}\right),$ (4.54) $\displaystyle\phi^{\prime}_{1}$
$\displaystyle=$
$\displaystyle-\frac{g\kappa}{16\sqrt{2}}e^{\varphi-2(\phi_{1}+\phi_{2})}\left(2e^{4(\phi_{1}+\phi_{2})}-2u-u\sigma
e^{4(\phi_{2}-\varphi)}\right),$ (4.55) $\displaystyle\phi^{\prime}_{2}$
$\displaystyle=$
$\displaystyle-\frac{g\kappa}{16\sqrt{2}}e^{\varphi-2(\phi_{1}+\phi_{2})}\left(2e^{4(\phi_{1}+\phi_{2})}-2u+u\sigma
e^{4(\phi_{2}-\varphi)}\right).$ (4.56)

Defining a new radial coordinate $\rho$ by
$\frac{d\rho}{dr}=e^{\varphi-2(\phi_{1}+\phi_{2})}$, we find a domain wall
solution

$\displaystyle\phi_{2}$ $\displaystyle=$ $\displaystyle
C_{1}+\frac{1}{16}\ln\left[e^{C_{3}}+\sigma
e^{\frac{ug\kappa\rho}{\sqrt{2}}}\right]+\frac{1}{8}\ln\left[1-e^{u(C_{2}-\frac{g\kappa\rho}{\sqrt{2}})}\right],$
(4.57) $\displaystyle\phi_{1}$ $\displaystyle=$
$\displaystyle-\phi_{2}-\frac{1}{4}\ln\left[u(1-e^{u(C_{2}-\frac{g\kappa\rho}{\sqrt{2}})})\right],$
(4.58) $\displaystyle\varphi$ $\displaystyle=$
$\displaystyle\phi_{2}+\frac{1}{4}\ln\left[e^{C_{3}-\frac{ug\kappa\rho}{\sqrt{2}}}+\sigma\right],$
(4.59) $\displaystyle A$ $\displaystyle=$
$\displaystyle-5\phi_{2}-\frac{1}{2}\ln\left[u(1-e^{u(C_{2}-\frac{g\kappa\rho}{\sqrt{2}})})\right]-\frac{1}{4}\ln\left[e^{C_{3}-\frac{ug\kappa\rho}{\sqrt{2}}}+\sigma\right]$
(4.60)

with the integration constants $C_{1}$, $C_{2}$, and $C_{3}$. For simplicity,
we have chosen an integration constant for $A$ to be zero. As in the previous
cases, by suitably redefining $u$ and $v$ parameters together with
$SO(2)\times SO(2)$ singlet scalars, we can find a similar solution for
$SO(3,2)_{\textrm{I}}$ gauge group in which the $SO(2)\times SO(2)$ residual
symmetry lies outside the $SO(4,4)$.

###  4.4 $SO(2)_{\text{diag}}$ symmetric domain walls

We finally consider supersymmetric domain walls preserving an
$SO(2)_{\text{diag}}\subset SO(2)\times SO(2)$ symmetry. By taking the product
among the various representations in (4.48) to implement the
$SO(2)_{\text{diag}}$ subgroup, we find

$\displaystyle(\mathbf{5},\mathbf{5})\rightarrow 9(\mathbf{1})\oplus
8(\mathbf{2})$ (4.61)

leading to nine singlet scalars. Among these singlets, five of them correspond
to the $SO(2)\times SO(2)$ singlets given in the previous section. The
additional four singlets correspond to the following non-compact generators

$\mathds{Y}_{5}=\hat{\boldsymbol{t}}_{1\dot{2}}-\hat{\boldsymbol{t}}_{4\dot{3}},\qquad\mathds{Y}_{6}=\hat{\boldsymbol{t}}_{2\dot{1}}+\hat{\boldsymbol{t}}_{3\dot{4}},\qquad\mathds{Y}_{7}=\hat{\boldsymbol{t}}_{1\dot{3}}+\hat{\boldsymbol{t}}_{4\dot{2}},\qquad\mathds{Y}_{8}=\hat{\boldsymbol{t}}_{2\dot{4}}-\hat{\boldsymbol{t}}_{3\dot{1}}\,.$
(4.62)

The coset representative can be written as

$V=e^{\varphi\boldsymbol{d}+\phi_{1}\mathds{Y}_{1}+\phi_{2}\mathds{Y}_{2}+\phi_{3}\mathds{Y}_{3}+\phi_{4}\mathds{Y}_{4}+\phi_{5}\mathds{Y}_{5}+\phi_{6}\mathds{Y}_{6}+\phi_{7}\mathds{Y}_{7}+\phi_{8}\mathds{Y}_{8}}\,.$
(4.63)

It turns out that the resulting T-tensor, superpotential, and scalar potential
are highly complicated. Accordingly, we will look for some subtruncations to
simplify the analysis but still obtain interesting results. One possibility is
to impose the conditions $\phi_{3}=\phi_{4}=0$ together with
$\phi_{8}=\phi_{7}$ and $\phi_{6}=-\phi_{5}$. We have checked that these
indeed lead to a consistent subtruncation.
The truncated coset representative is now given by

$V=e^{\varphi\boldsymbol{d}+\phi_{1}\mathds{Y}_{1}+\phi_{2}\mathds{Y}_{2}+\tilde{\phi}_{3}(\mathds{Y}_{5}-\mathds{Y}_{6})+\tilde{\phi}_{4}(\mathds{Y}_{7}+\mathds{Y}_{8})}$
(4.64)

giving rise to the superpotential and scalar potential

$\displaystyle\mathcal{W}$ $\displaystyle=$
$\displaystyle\frac{g\kappa}{16\sqrt{2}}e^{\varphi-2(\phi_{1}+\phi_{2})}\left[(2e^{4(\phi_{1}+\phi_{2})}+2u)\cosh{4\tilde{\phi}_{3}}\cosh{4\tilde{\phi}_{4}}+u\sigma
e^{4(\phi_{2}-\varphi)}\right],\qquad$ (4.65) $\displaystyle\mathbf{V}$
$\displaystyle=$
$\displaystyle-\frac{1}{64}e^{-6\varphi-4\phi_{1}}ug^{2}\kappa^{2}\left[4\sigma
e^{4\varphi}(e^{4(\phi_{1}+\phi_{2})}+u)\cosh{4\tilde{\phi}_{3}}\cosh{4\tilde{\phi}_{4}}\right.$
$\displaystyle\left.+8e^{8\varphi
4\phi_{1}}-u\sigma^{2}e^{4\phi_{2}}\right].\quad$

The latter can be written in terms of the former according to (4.15) using

$G^{\mathbb{I}\mathbb{J}}=\frac{1}{8}\begin{pmatrix}4&0&0&0&0\\\
0&\text{sech}^{2}4\tilde{\phi_{3}}\,\text{sech}^{2}4\tilde{\phi_{4}}+1&\text{sech}^{2}4\tilde{\phi_{3}}\,\text{sech}^{2}4\tilde{\phi_{4}}-1&0&0\\\
0&\text{sech}^{2}4\tilde{\phi_{3}}\,\text{sech}^{2}4\tilde{\phi_{4}}-1&\text{sech}^{2}4\tilde{\phi_{3}}\,\text{sech}^{2}4\tilde{\phi_{4}}+1&0&0\\\
0&0&0&\text{sech}^{2}4\tilde{\phi_{4}}&0\\\ 0&0&0&0&1\\\ \end{pmatrix}$ (4.67)

for
$\Phi^{\mathbb{I}}=\\{\varphi,\phi_{1},\phi_{2},\tilde{\phi}_{3},\tilde{\phi}_{4}\\}$
with $\mathbb{I},\mathbb{J},\ldots=1,2,3,4,5$.
In this case, various gauge groups are identified by the parameters $u=\pm 1$
and $\sigma=\pm 1,0$ as $SO(5)$ ($u=\sigma=1$), $SO(4,1)_{\textrm{I}}$
($u=-\sigma=1$), $CSO(4,0,1)$ ($u=1,\sigma=0$), $SO(3,2)_{\textrm{II}}$
($u=-\sigma=-1$ or $u=\sigma=-1$), and $CSO(2,2,1)$ ($u=-1,\sigma=0$). With
all these, we find the BPS equations

$\displaystyle A^{\prime}$ $\displaystyle=$
$\displaystyle\frac{g\kappa}{16\sqrt{2}}e^{\varphi-2(\phi_{1}+\phi_{2})}\left[(2e^{4(\phi_{1}+\phi_{2})}+2u)\cosh{4\tilde{\phi}_{3}}\cosh{4\tilde{\phi}_{4}}+u\sigma
e^{4(\phi_{2}-\varphi)}\right],$ (4.68) $\displaystyle\varphi^{\prime}$
$\displaystyle=$
$\displaystyle-\frac{g\kappa}{16\sqrt{2}}e^{\varphi-2(\phi_{1}+\phi_{2})}\left[(2e^{4(\phi_{1}+\phi_{2})}+2u)\cosh{4\tilde{\phi}_{3}}\cosh{4\tilde{\phi}_{4}}-3u\sigma
e^{4(\phi_{2}-\varphi)}\right],\qquad\ $ (4.69)
$\displaystyle\phi^{\prime}_{1}$ $\displaystyle=$
$\displaystyle-\frac{g\kappa}{16\sqrt{2}}e^{\varphi-2(\phi_{1}+\phi_{2})}\left[(2e^{4(\phi_{1}+\phi_{2})}-2u)\text{sech}{4\tilde{\phi}_{3}}\,\text{sech}{4\tilde{\phi}_{4}}-u\sigma
e^{4(\phi_{2}-\varphi)}\right],$ (4.70) $\displaystyle\phi^{\prime}_{2}$
$\displaystyle=$
$\displaystyle-\frac{g\kappa}{16\sqrt{2}}e^{\varphi-2(\phi_{1}+\phi_{2})}\left[(2e^{4(\phi_{1}+\phi_{2})}-2u)\text{sech}{4\tilde{\phi}_{3}}\,\text{sech}{4\tilde{\phi}_{4}}+u\sigma
e^{4(\phi_{2}-\varphi)}\right],$ (4.71)
$\displaystyle\tilde{\phi}^{\prime}_{3}$ $\displaystyle=$
$\displaystyle-\frac{g\kappa}{16\sqrt{2}}e^{\varphi-2(\phi_{1}+\phi_{2})}(2e^{4(\phi_{1}+\phi_{2})}+2u)\sinh{4\tilde{\phi}_{3}}\,\text{sech}{4\tilde{\phi}_{4}},$
(4.72) $\displaystyle\tilde{\phi}^{\prime}_{4}$ $\displaystyle=$
$\displaystyle-\frac{g\kappa}{16\sqrt{2}}e^{\varphi-2(\phi_{1}+\phi_{2})}(2e^{4(\phi_{1}+\phi_{2})}+2u)\cosh{4\tilde{\phi}_{3}}\sinh{4\tilde{\phi}_{4}}.$
(4.73)

Changing the radial coordinate to $\rho$ defined by
$\frac{d\rho}{dr}=e^{\varphi-2(\phi_{1}+\phi_{2})}$, we can solve these
equations to find a domain wall solution

$\displaystyle\tilde{\phi}_{3}$ $\displaystyle=$
$\displaystyle\frac{1}{4}\sinh^{-1}\left[\frac{4C_{5}\sqrt{u}}{\sqrt{C_{0}^{6}g^{2}\kappa^{2}\left(C_{6}+2u\sigma
g^{3}\kappa^{3}\rho^{4}\right)^{2}-16C_{5}^{2}\left(C_{1}^{2}+u\right)}}\right],$
(4.74) $\displaystyle\tilde{\phi}_{4}$ $\displaystyle=$
$\displaystyle\frac{1}{4}\tanh^{-1}\left[\frac{4\sqrt{u}}{\sqrt{C_{0}^{6}g^{2}\kappa^{2}\left(C_{6}+2u\sigma
g^{3}\kappa^{3}\rho^{4}\right)^{2}-16C_{5}^{2}\left(C_{1}^{2}+u\right)}}\right],$
(4.75) $\displaystyle\phi_{1}$ $\displaystyle=$
$\displaystyle\frac{1}{4}\ln\left[C_{1}\tanh{4\tilde{\phi}_{3}}+\sqrt{C_{1}^{2}\tanh^{2}{4\tilde{\phi}_{3}}+u}\right]+\frac{1}{4}\ln\left[\frac{u\sigma
g\kappa\rho}{4\sqrt{2}}\right],$ (4.76) $\displaystyle\phi_{2}$
$\displaystyle=$
$\displaystyle\frac{1}{4}\ln\left[C_{1}\tanh{4\tilde{\phi}_{3}}+\sqrt{C_{1}^{2}\tanh^{2}{4\tilde{\phi}_{3}}+u}\right]-\frac{1}{4}\ln\left[\frac{u\sigma
g\kappa\rho}{4\sqrt{2}}\right],$ (4.77) $\displaystyle\varphi$
$\displaystyle=$ $\displaystyle\frac{3}{4}\ln\left[\sqrt{2}C_{0}u\sigma
g\kappa\rho\right]+\frac{1}{4}\ln{\sinh{4\tilde{\phi}_{4}}},$ (4.78)
$\displaystyle A$ $\displaystyle=$
$\displaystyle\frac{1}{3}\varphi-\frac{1}{3}\ln{\sinh{4\tilde{\phi}_{4}}}$
(4.79)

in which we have chosen the integration constants for $\phi_{2}$ and $A$ to be
zero.
We end this section by pointing out that domain wall solutions obtained from
gaugings in $\mathbf{56}^{+1}$ and $\mathbf{8}^{-3}$ representations can also
be found by a similar analysis. The resulting solutions take the same form as
the solutions given in this section with a sign change in some of the scalar
fields.

##  5 Conclusions and discussions

We have constructed the embedding tensors of six-dimensional maximal $N=(2,2)$
gauged supergravity for various gauge groups arising from the decomposition of
the embedding tensor under $\mathbb{R}^{+}\times SO(4,4)\subset SO(5,5)$
symmetry. Under this decomposition, viable gauge groups can be determined from
the embedding tensor in $\mathbf{8}^{\pm 1}$, $\mathbf{8}^{\pm 3}$, and
$\mathbf{56}^{\pm 1}$ representations. We have pointed out that gaugings in
$\mathbf{8}^{\pm 1}$ representation without $\mathbf{56}^{\pm 1}$ is not
consistent due to the linear constraint, and gaugings in $\mathbf{8}^{\pm 3}$
representation only lead to a translational gauge group $\mathbb{R}^{8}$. On
the other hand, gaugings in $\mathbf{56}^{\pm 1}$ representation give rise to
$CSO(4-p,p,1)\sim SO(4-p,p)\ltimes\mathbb{R}^{4}$ gauge groups with $p=0,1,2$.
Including $\mathbf{8}^{\pm 3}$ representation to $\mathbf{56}^{\mp 1}$ can
enlarge the gauge groups to $SO(4-p,p+1)$ or $SO(5-p,p)$. We have also found a
number of half-supersymmetric domain wall solutions from gaugings in
$\mathbf{56}^{-1}$ and $\mathbf{8}^{+3}$ representations with various residual
symmetries. The corresponding solutions for gaugings solely from
$\mathbf{56}^{-1}$ representation can be straightforwardly obtained from these
results by turning off the $\mathbf{8}^{+3}$ part.
As pointed out in [32], some of the gaugings under $SO(4,4)$ decomposition
could be truncated to gaugings of half-maximal $N=(1,1)$ supergravity coupled
to four vector multiplets in which supersymmetric AdS6 vacua are known to
exist [48, 49, 50]. It would be interesting to explicitly truncate the results
given here to half-maximal gauged supergravity and obtain new gaugings as well
as new supersymmetric $AdS_{6}$ vacua. Along this direction, a classification
of gauge groups with known eleven-dimensional origins, arising from truncating
the maximal theory to half-maximal one, has been given in [40]. It could also
be interesting to extend this analysis to many new gaugings identified in this
paper.
On the other hand, uplifting the six-dimensional gauged supergravity and the
corresponding domain wall solutions in this work and in [17, 18] to higher
dimensions could also be worth considering. This could be done by constructing
truncation ansatze of string/M-theory to six dimensions using $SO(5,5)$
exceptional field theory given in [51, 52] and would lead to interesting
holographic descriptions of maximal super Yang-Mills theory in five
dimensions. Finally, finding a holographic interpretation of the domain wall
solutions given in this paper could also be of particular interest. This could
be done along the line of [53] and [54] in which a holographic description
from a simple domain wall found in [30] with $SO(5)$ symmetry has been
studied.
Acknowledgement
This work is supported by the National Research Council of Thailand (NRCT),
Ramkhamhaeng University, and Chulalongkorn University under grants N42A660945
and N42A650263.

##  Appendix A $SO(4,4)$ branching rules

In this appendix, we collect the decompositions for various representations of
$SO(5,5)$ under $\mathbb{R}^{+}\times SO(4,4)$ in our convention. We note that
these branching rules have also been given in [40].

###  A.1 Vector

Under $\mathbb{R}^{+}\times SO(4,4)\subset SO(5,5)$, the $SO(5,5)$ vector
index is decomposed as $M=(-,I,+)$ where $I=1,...,8$ is an $SO(4,4)$ vector
index. Accordingly, an $SO(5,5)$ vector $V_{M}$ can be written as

$V_{M}=(V_{-},V_{I},V_{+})$ (A.1)

where $V_{I}$ is an $SO(4,4)$ vector. With the decomposition of the $SO(5,5)$
vector index $M=(-,I,+)$, the $SO(5,5)$ invariant metric together with its
inverse are given by

$\eta_{MN}\ =\ \left(\begin{array}[]{c|c|c}&&1\\\ \hline\cr&\
\eta_{IJ\phantom{K}}&\\\ \hline\cr 1&&\end{array}\right)\qquad\text{ and
}\qquad\eta^{MN}\ =\ \left(\begin{array}[]{c|c|c}&&1\\\ \hline\cr&\
\eta^{IJ\phantom{\hat{K}}}&\\\ \hline\cr 1&&\end{array}\right).$ (A.2)

We have also introduced the $SO(4,4)$ invariant metric

$\eta_{IJ}\ =\ \eta^{IJ}\ =\ \left(\begin{array}[]{c|c}&\mathds{1}_{4}\\\
\hline\cr\mathds{1}_{4}&\end{array}\right)$ (A.3)

where $\mathds{1}_{n}$ denotes an $n\times n$ identity matrix. The $SO(4,4)$
vector index can be raised and lowered as $V^{I}=\eta^{IJ}V_{J}$ and
$V_{I}=\eta_{IJ}V^{J}$.
The $SO(5,5)$ generators are defined in vector representation as

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

where $\delta^{M}_{N}=\mathds{1}_{10}$. These generators satisfy
$\boldsymbol{t}_{MN}=\boldsymbol{t}_{[MN]}$ as well as 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.5)

Under the $SO(4,4)$ decomposition, we define a generator corresponding to
$\mathbb{R}^{+}\sim SO(1,1)\subset\mathbb{R}^{+}\times SO(4,4)$ as
$\boldsymbol{d}\ =\ {\boldsymbol{t}}_{-+}$ whose explicit form in vector
representation is

${\left(\boldsymbol{d}\right)_{M}}^{N}\ =\ \left(\begin{array}[]{c|c|c}-2&&\\\
\hline\cr&\phantom{-2}&\\\ \hline\cr&&2\end{array}\right).$ (A.6)

With $\boldsymbol{d}$ of this form, we can assign the $\mathbb{R}^{+}$ weights
$\pm 1$ to the two singlets of the $SO(4,4)$ in (A.1). Therefore, the
branching rule for a vector representation reads

$\underbrace{\mathbf{10}}_{V_{M}}\ \rightarrow\
\underbrace{\mathbf{1}^{+1}}_{V_{+}}\,\oplus\,\underbrace{\mathbf{1}^{-1}}_{V_{-}}\,\oplus\,\underbrace{\mathbf{8}^{\,0}}_{V_{I}}\,.$
(A.7)

###  A.2 Adjoint

The decomposition of adjoint representation follows from the branching rule of
vector representations. Using $M=(-,I,+)$, we can decompose the $SO(5,5)$
generators as

$\boldsymbol{t}_{MN}\ =\
\left(\begin{array}[]{c|c|c}&\boldsymbol{k}_{J}&\boldsymbol{d}\\\
\hline\cr-\boldsymbol{k}_{I}&\boldsymbol{\tau}_{IJ}&-\boldsymbol{p}_{I}\\\
\hline\cr-\boldsymbol{d}&\boldsymbol{p}_{J}&\end{array}\right)$ (A.8)

where we have identified $\boldsymbol{t}_{+I}=\boldsymbol{p}_{I}$ and
$\boldsymbol{t}_{-I}=\boldsymbol{k}_{I}$. From the $SO(5,5)$ algebra (A.5), we
can derive the following commutation relations

$\displaystyle\left[\boldsymbol{d},\boldsymbol{d}\right]$ $\displaystyle=$
$\displaystyle
0,\qquad\qquad\quad\left[\boldsymbol{d},\boldsymbol{\tau}_{IJ}\right]\ =\ 0,$
(A.9) $\displaystyle\left[\boldsymbol{d},\boldsymbol{p}_{I}\right]$
$\displaystyle=$ $\displaystyle-2\boldsymbol{p}_{I},\quad\quad\quad\
\,\,\left[\boldsymbol{d},\boldsymbol{k}_{I}\right]\ =\ +2\boldsymbol{k}_{I},$
(A.10) $\displaystyle\left[\boldsymbol{p}_{I},\boldsymbol{p}_{J}\right]$
$\displaystyle=$ $\displaystyle
0,\qquad\quad\qquad\left[\boldsymbol{k}_{I},\boldsymbol{k}_{J}\right]\ =\ 0,$
(A.11) $\displaystyle\left[\boldsymbol{p}_{I},\boldsymbol{\tau}_{JK}\right]$
$\displaystyle=$ $\displaystyle 4\eta_{I[J}\boldsymbol{p}_{K]},\quad\
\left[\boldsymbol{k}_{I},\boldsymbol{\tau}_{JK}\right]\ =\
4\eta_{I[J}\boldsymbol{k}_{K]},$ (A.12)
$\displaystyle\left[\boldsymbol{p}_{I},\boldsymbol{k}_{J}\right]$
$\displaystyle=$ $\displaystyle
2(\boldsymbol{d}\eta_{IJ}-\boldsymbol{\tau}_{IJ}),$ (A.13)
$\displaystyle\left[\boldsymbol{\tau}_{IJ},\boldsymbol{\tau}_{KL}\right]$
$\displaystyle=$ $\displaystyle
4(\eta_{I[K}\boldsymbol{\tau}_{L]J}-\eta_{J[K}\boldsymbol{\tau}_{L]I}).$
(A.14)

The last commutation relation is the $SO(4,4)$ algebra implying that
$\boldsymbol{\tau}_{IJ}$ are $SO(4,4)$ generators. It follows that the
$SO(4,4)$ branching rule for the adjoint representation of $SO(5,5)$ is given
by

$\underbrace{\mathbf{45}}_{\boldsymbol{t}_{MN}}\ \rightarrow\
\underbrace{\mathbf{1}^{0}}_{\boldsymbol{d}}\,\oplus\,\underbrace{\mathbf{28}^{0}}_{\boldsymbol{\tau}_{IJ}}\,\oplus\,\underbrace{\mathbf{8}^{+2}}_{\boldsymbol{k}_{I}}\,\oplus\,\underbrace{\mathbf{8}^{-2}}_{\boldsymbol{p}_{I}}$
(A.15)

where the $\mathbb{R}^{+}$ weights are determined from the commutation
relations given in (A.9) and (A.10).

###  A.3 Spinor

In order to find the $SO(4,4)$ branching rule for spinor representation, we
start from the 32-dimensional $SO(5,5)$ gamma matrices given in [40]

$\displaystyle\tilde{\mathbf{\Gamma}}_{\underline{1}}$ $\displaystyle=$
$\displaystyle\sigma_{1}\otimes\sigma_{3}\otimes\mathds{1}_{2}\otimes\mathds{1}_{2}\otimes\mathds{1}_{2},\qquad\tilde{\mathbf{\Gamma}}_{\underline{2}}\
=\
\sigma_{1}\otimes\sigma_{1}\otimes\sigma_{2}\otimes\sigma_{1}\otimes\sigma_{2},$
$\displaystyle\tilde{\mathbf{\Gamma}}_{\underline{3}}$ $\displaystyle=$
$\displaystyle\sigma_{1}\otimes\sigma_{1}\otimes\sigma_{1}\otimes\sigma_{1}\otimes\mathds{1}_{2},\qquad\tilde{\mathbf{\Gamma}}_{\underline{4}}\
=\
\sigma_{1}\otimes\sigma_{1}\otimes\sigma_{1}\otimes\sigma_{3}\otimes\sigma_{1},$
$\displaystyle\tilde{\mathbf{\Gamma}}_{\underline{5}}$ $\displaystyle=$
$\displaystyle\sigma_{1}\otimes\sigma_{1}\otimes\sigma_{1}\otimes\sigma_{3}\otimes\sigma_{3},\qquad\tilde{\mathbf{\Gamma}}_{\underline{6}}\
=\
i\sigma_{1}\otimes\sigma_{1}\otimes\sigma_{3}\otimes\sigma_{1}\otimes\sigma_{2},$
$\displaystyle\tilde{\mathbf{\Gamma}}_{\underline{7}}$ $\displaystyle=$
$\displaystyle
i\sigma_{1}\otimes\sigma_{1}\otimes\mathds{1}_{2}\otimes\sigma_{2}\otimes\mathds{1}_{2},\quad\
\ \tilde{\mathbf{\Gamma}}_{\underline{8}}\ =\
i\sigma_{1}\otimes\sigma_{1}\otimes\mathds{1}_{2}\otimes\sigma_{3}\otimes\sigma_{2},$
$\displaystyle\tilde{\mathbf{\Gamma}}_{\underline{9}}$ $\displaystyle=$
$\displaystyle
i\sigma_{1}\otimes\sigma_{2}\otimes\mathds{1}_{2}\otimes\mathds{1}_{2}\otimes\mathds{1}_{2},\quad\
\ \tilde{\mathbf{\Gamma}}_{\underline{10}}=\
i\sigma_{2}\otimes\mathds{1}_{2}\otimes\mathds{1}_{2}\otimes\mathds{1}_{2}\otimes\mathds{1}_{2}\quad$
(A.16)

where $\\{\sigma_{1},\sigma_{2},\sigma_{3}\\}$ are the usual Pauli matrices

$\sigma_{1}\ =\ \begin{pmatrix}0&1\\\ 1&0\end{pmatrix},\qquad\sigma_{2}\ =\
\begin{pmatrix}0&-i\\\ i&0\end{pmatrix},\qquad\sigma_{3}\ =\
\begin{pmatrix}1&0\\\ 0&-1\end{pmatrix}.$ (A.17)

The $SO(5,5)$ gamma matrices $\tilde{\mathbf{\Gamma}}_{\underline{A}}$ with
$\underline{A}=\underline{1},...,\underline{10}$ satisfy the Clifford algebra

$\left\\{\tilde{\mathbf{\Gamma}}_{\underline{A}},\tilde{\mathbf{\Gamma}}_{\underline{B}}\right\\}\
=\ 2\eta_{\underline{A}\underline{B}}\mathds{1}_{32}$ (A.18)

in which
$\eta_{\underline{A}\underline{B}}=\textrm{diag}(\mathds{1}_{5},-\mathds{1}_{5})$
is the $SO(5,5)$ invariant metric in diagonal basis. In this explicit
representation, the $SO(5,5)$ chirality matrix is given by

$\tilde{\mathbf{\Gamma}}_{\ast}=\tilde{\mathbf{\Gamma}}_{\underline{1}...\underline{10}}=\textrm{diag}(-\mathds{1}_{16},\mathds{1}_{16})$
(A.19)

and the $SO(5,5)$ charge conjugation matrix is

$\tilde{\mathds{C}}\ =\
i\sigma_{2}\otimes\sigma_{3}\otimes\sigma_{3}\otimes\mathds{1}_{2}\otimes\mathds{1}_{2}$
(A.20)

satisfying

$\tilde{\mathbf{\Gamma}}_{\underline{A}}\tilde{\mathds{C}}=(\tilde{\mathbf{\Gamma}}_{\underline{A}}\tilde{\mathds{C}})^{T}\qquad\text{
and }\qquad\tilde{\mathds{C}}^{T}=-\tilde{\mathds{C}}.$ (A.21)

We now transform all these results to the basis with off-diagonal metric
$\eta_{MN}$ given in (A.2). In this basis, the Clifford algebra reads

$\left\\{\tilde{\mathbf{\Gamma}}_{M},\tilde{\mathbf{\Gamma}}_{N}\right\\}\ =\
2\eta_{MN}\mathds{1}_{32}\,.$ (A.22)

The 32-dimensional $SO(5,5)$ gamma matrices in the off-diagonal basis can be
obtained from

$\tilde{\mathbf{\Gamma}}_{M}\ =\
{\mathbb{M}_{M}}^{\underline{A}}\tilde{\mathbf{\Gamma}}_{\underline{A}}$
(A.23)

with the transformation matrix

$\mathbb{M}\ =\ \frac{1}{\sqrt{2}}(\eta_{\text{diag}}+\eta_{\text{off-diag}})\
=\ \frac{1}{\sqrt{2}}\left(\begin{array}[]{c|c|c|c}1&&&1\\\
\hline\cr&\mathds{1}_{4}&\mathds{1}_{4}&\\\
\hline\cr&\mathds{1}_{4}&-\mathds{1}_{4}&\\\ \hline\cr
1&&&-1\end{array}\right).$ (A.24)

Using this transformation matrix, the relation between diagonal and off-
diagonal $\eta$ is found to be

$\eta_{MN}\ =\
{\mathbb{M}_{M}}^{\underline{A}}\,{\mathbb{M}_{N}}^{\underline{B}}\,\eta_{\underline{A}\underline{B}}.$
(A.25)

The $SO(5,5)$ gamma matrices are chirally decomposed as

${(\tilde{\mathbf{\Gamma}}_{M})_{\mathcal{A}}}^{\mathcal{B}}\ =\
\left(\begin{array}[]{c|c}&{(\Gamma_{M})_{A}}^{B^{\prime}}\\\
\hline\cr{(\Gamma_{M})_{A^{\prime}}}^{B\phantom{\hat{i}}}&\end{array}\right).$
(A.26)

with spinor index $\mathcal{A}=1,2,...,32$ split according to
$\mathcal{A}=(A,A^{\prime})$ where $A=1,...,16$ and $A^{\prime}=17,...,32$. We
can raise and lower the chirally decomposed spinor indices with the charge
conjugation matrix given in (A.20) such that

$(\tilde{\mathbf{\Gamma}}_{M})_{\mathcal{A}\mathcal{B}}\ =\
{(\tilde{\mathbf{\Gamma}}_{M})_{\mathcal{A}}}^{\mathcal{C}}\tilde{\mathds{C}}_{\mathcal{C}\mathcal{B}}\
=\ \left(\begin{array}[]{c|c}(\Gamma_{M})_{AB}&\\\
\hline\cr&(\Gamma_{M})_{A^{\prime}B^{\prime}}\end{array}\right).$ (A.27)

The chirally decomposed $16\times 16$ gamma matrices $(\Gamma_{M})_{AB}$ play
an important role in determining explicit forms of the embedding tensor.
Following [40], $SO(4,4)$ gamma matrices can be extracted from the $SO(5,5)$
ones using the decomposition $M=(-,I,+)$. The ten $32$-dimensional $SO(5,5)$
gamma matrices are also decomposed as
$\tilde{\mathbf{\Gamma}}_{M}=(\tilde{\mathbf{\Gamma}}_{-},\tilde{\mathbf{\Gamma}}_{I},\tilde{\mathbf{\Gamma}}_{+})$.
In the explicit representation (A.16), only $\tilde{\mathbf{\Gamma}}_{I}$ can
be written as

$\tilde{\mathbf{\Gamma}}_{I}=\sigma_{1}\otimes\Gamma_{I}$ (A.28)

so that the chirally decomposed $16\times 16$ gamma matrices on the upper-
right block and the lower-left block in (A.26) are the same, i.e.,
${(\Gamma_{I})_{A}}^{B^{\prime}}={(\Gamma_{I})_{A^{\prime}}}^{B}$. We can
redefine these $16\times 16$ gamma matrices to be the $SO(4,4)$ gamma matrices
${(\Gamma_{I})_{A}}^{B}$ satisfying the Clifford algebra

$\\{\Gamma_{I},\Gamma_{J}\\}=2\eta_{IJ}\mathds{1}_{16}.$ (A.29)

In this case, the $SO(4,4)$ chirality matrix is

$\Gamma_{9}=\text{diag}(\mathds{1}_{8},-\mathds{1}_{8}),$ (A.30)

and the $SO(4,4)$ charge conjugation matrix, satisfying
$\Gamma_{I}\mathds{C}=(\Gamma_{I}\mathds{C})^{T}$ and
$\mathds{C}^{T}=\mathds{C}$, is given by

$\mathds{C}=\sigma_{3}\otimes\sigma_{3}\otimes\mathds{1}_{2}\otimes\mathds{1}_{2}.$
(A.31)

Decomposing the spinor index $A=(m,\dot{m})$ where $m=1,...,8$ and
$\dot{m}=\dot{1},...,\dot{8}$, we can write this charge conjugation matrix in
the form

$\mathds{C}_{AB}\ =\ \left(\begin{array}[]{c|c}\mathbf{c}_{mn}&\\\
\hline\cr&\mathbf{c}_{\dot{m}\dot{n}}\end{array}\right)\quad\text{ and
}\quad\mathds{C}^{AB}\ =\ \left(\begin{array}[]{c|c}\mathbf{c}^{mn}&\\\
\hline\cr&\mathbf{c}^{\dot{m}\dot{n}}\end{array}\right)$ (A.32)

where
$\mathbf{c}_{mn}=\mathbf{c}^{mn}=-\mathbf{c}_{\dot{m}\dot{n}}=-\mathbf{c}^{\dot{m}\dot{n}}=\text{diag}(\mathds{1}_{4},-\mathds{1}_{4})$.
Besides, we find that the $SO(4,4)$ gamma matrices are also chirally
decomposed as

${(\Gamma_{I})_{A}}^{B}\ =\
\left(\begin{array}[]{c|c}&{(\gamma_{I})_{m}}^{\dot{n}}\\\
\hline\cr{(\gamma_{I})_{\dot{m}}}^{n\phantom{{}^{\prime}}}&\end{array}\right)$
(A.33)

in which ${(\gamma_{I})_{m}}^{\dot{n}}$ and ${(\gamma_{I})_{\dot{m}}}^{n}$ are
in turn chirally decomposed $8\times 8$ gamma matrices. The spinor index $B$
can be lowered using the $SO(4,4)$ charge conjugation matrix as

$(\Gamma_{I})_{AB}\ =\ {(\Gamma_{I})_{A}}^{C}\mathds{C}_{CB}\ =\
\left(\begin{array}[]{c|c}&(\gamma_{I})_{m\dot{n}}\\\
\hline\cr(\gamma_{I})_{\dot{m}n}&\end{array}\right)$ (A.34)

where $(\gamma_{I})_{\dot{n}m}=(\gamma_{I})_{m\dot{n}}$ due to
$(\Gamma_{M})_{BA}=(\Gamma_{M})_{AB}$.
Following the same decomposition of spinor indices, we find the remaining two
$SO(5,5)$ gamma matrices of the forms

$(\Gamma_{-})_{AB}=-\sqrt{2}\left(\begin{array}[]{c|c}\mathbf{c}_{mn}&\phantom{\mathds{1}_{8}}\\\
\hline\cr&\end{array}\right)\qquad\text{ and
}\qquad(\Gamma_{+})_{AB}=\sqrt{2}\left(\begin{array}[]{c|c}\phantom{\mathds{1}_{8}}&\\\
\hline\cr&\mathbf{c}_{\dot{m}\dot{n}}\end{array}\right).$ (A.35)

The $SO(5,5)$ generators in spinor representation satisfying (A.5) are given
by

${(\boldsymbol{t}_{MN})_{A}}^{B}\ =\ {(\Gamma_{MN})_{A}}^{B}$ (A.36)

where

${(\Gamma_{MN})_{A}}^{B}=\frac{1}{2}\left[{(\Gamma_{M})_{A}}^{C^{\prime}}{(\Gamma_{N})_{C^{\prime}}}^{B}-{(\Gamma_{N})_{A}}^{C^{\prime}}{(\Gamma_{M})_{C^{\prime}}}^{B}\right].$
(A.37)

It should be noted that the $SO(5,5)$ generators in spinor representation
given in (A.36) also decompose according to (A.15) and satisfy the same
commutation relations given in (A.9) to (A.14). In particular, with the
explicit representation (A.16), the $\mathbb{R}^{+}$ generator in spinor
representation is

${\left(\boldsymbol{d}\right)_{A}}^{B}\ =\
\left(\begin{array}[]{c|c}\mathds{1}_{8}&\\\
\hline\cr&-\mathds{1}_{8}\end{array}\right).$ (A.38)

The $SO(4,4)$ branching rule for spinor representation is now straightforward.
For a given $SO(5,5)$ spinor in $\mathbf{16}_{s}$ representation $\Psi_{A}$,
we have the decomposition

$\Psi_{A}\ =\ \left(\begin{array}[]{c}\Psi_{m}\\\
\Psi_{\dot{m}}\end{array}\right).$ (A.39)

By assigning the $\mathbb{R}^{+}$ weights $+1$ and $-1$ to $\Psi_{m}$ and
$\Psi_{\dot{m}}$, we obtain the branching rule of the form

$\underbrace{\mathbf{16}_{s}}_{\Psi_{A}}\ \rightarrow\
\underbrace{\mathbf{8}^{+1}}_{\Psi_{m}}\,\oplus\,\underbrace{\mathbf{8}^{-1}}_{\Psi_{\dot{m}}}\,.$
(A.40)

###  A.4 Vector-spinor

The vector-spinor representation of $SO(5,5)$ we are interested in is given by
a $16\times 10$ matrix $\theta^{AM}\in\mathbf{144}_{c}$ subject to the linear
constraint (2.7), $(\Gamma_{M})_{AB}\,\theta^{BM}\ =\ 0\,$. This linear
constraint reduces $160$ components of $\theta^{AM}$ to $144$ in
$\mathbf{144}_{c}$ representation.
With the decomposition of $SO(5,5)$ vector and spinor indices as $M=(-,I,+)$
and $A=(m,\dot{m})$, we decompose the $\theta^{AM}$ matrix as

$\theta^{AM}\ =\
\left(\begin{array}[]{c|c|c}\theta_{1}^{m-}&\theta_{3}^{mI}&\theta_{5}^{m+}\\\
\hline\cr\theta_{2}^{\dot{m}-}&\theta_{4}^{\dot{m}I}&\theta_{6}^{\dot{m}+\phantom{{}^{\prime}}}\end{array}\right).$
(A.41)

It is straightforward to determine that $\mathbb{R}^{+}$ weights of
$(\theta_{1}^{m-},\theta_{2}^{\dot{m}-},\theta_{3}^{mI},\theta_{4}^{\dot{m}I},\theta_{5}^{m+},\theta_{6}^{\dot{m}})$
are given by $(+1,+3,-1,+1,-3,-1)$. Finally, as in [32], the $SO(4,4)$
branching rule for $\mathbf{144}_{c}$ representation is given by

$\underbrace{\mathbf{144}_{c}}_{\theta^{AM}}\ \rightarrow\
\underbrace{\mathbf{56}^{-1}}_{\vartheta_{3}^{mI}}\,\oplus\,\underbrace{\mathbf{56}^{+1}}_{\vartheta_{4}^{\dot{m}I}}\,\oplus\,\underbrace{\mathbf{8}^{-1}}_{\theta_{6}^{\dot{m}+}}\,\oplus\,\underbrace{\mathbf{8}^{+1}}_{\theta_{1}^{m-}}\,\oplus\,\underbrace{\mathbf{8}^{+3}}_{\theta_{2}^{\dot{m}-}}\,\oplus\,\underbrace{\mathbf{8}^{-3}}_{\theta_{5}^{m+}}$
(A.42)

with

$\vartheta_{3}^{mI}=\theta_{3}^{mI}+\frac{\sqrt{2}}{8}{(\gamma^{I})^{m}}_{\dot{n}}\theta_{6}^{\dot{n}+}\qquad\text{and
}\qquad\vartheta_{4}^{\dot{m}I}=\theta_{4}^{\dot{m}I}-\frac{\sqrt{2}}{8}{(\gamma^{I})^{\dot{m}}}_{n}\theta_{1}^{n-}\,.$
(A.43)

These two definitions follow from the linear constraint decomposed under
$SO(4,4)$ as

$0=\sqrt{2}\mathbf{c}_{mn}\theta_{1}^{n-}-(\gamma_{I})_{m\dot{n}}\theta_{4}^{\dot{n}I}\qquad\text{
and }\qquad
0=\sqrt{2}\mathbf{c}_{\dot{m}\dot{n}}\theta_{6}^{\dot{n}+}+(\gamma_{I})_{\dot{m}n}\theta_{3}^{nI}\,.$
(A.44)

It is also useful to note that $\vartheta_{3}^{mI}$ and
$\vartheta_{4}^{\dot{m}I}$ satisfy the following conditions

$(\gamma_{I})_{\dot{m}n}\vartheta_{3}^{nI}=0\qquad\text{ and
}\qquad(\gamma_{I})_{m\dot{n}}\vartheta_{4}^{\dot{n}I}=0\,.$ (A.45)

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