Supersymmetric domain walls in maximal 6D gauged supergravity III

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 ${\textbf{144}}_c$ representation of SO(5, 5). Under ${\mathbb {R}}^+\times SO(4,4)$ subgroup, this leads to the embedding tensor in $({\textbf{8}}^{\pm 3},$ ${\textbf{8}}^{\pm 1},{\textbf{56}}^{\pm 1})$ representations. Gaugings in ${\textbf{8}}^{\pm 3}$ representations lead to a translational gauge group ${\mathbb {R}}^8$ while gaugings in ${\textbf{8}}^{\pm 1}$ representations give rise to gauge groups related to the scaling symmetry ${\mathbb {R}}^+.$ On the other hand, the embedding tensor in ${\textbf{56}}^{\pm 1}$ representations gives $CSO(4-p,p,1)\sim SO(4-p,p) < imes {\mathbb {R}}^4\subset SO(4,4)$ 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 ${\textbf{56}}^{-1}$ and ${\textbf{8}}^{+3}$ 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 ${\textbf{56}}^{-1}$ representation are also included in these results when the ${\textbf{8}}^{+3}$ 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 ${\textbf{144}}_c$ 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, 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 ${\textbf{144}}_c$ representation of SO(5, 5) decomposes into ${\textbf{8}}^{\pm 1},$ ${\textbf{8}}^{\pm 3},$ and ${\textbf{56}}^{\pm 1}$ 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 Sect. 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 Sect. 3. In Sect. 4, we find supersymmetric domain wall solutions from gaugings in ${\textbf{56}}^{-1}$ and ${\textbf{8}}^{+3}$ representations. Conclusions and discussions are given in Sect. 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

$$\begin{aligned} \left( e^{\hat{\mu }}_\mu , B_{\mu \nu \overline{m}}, A^{A}_\mu , {V_A}^{\alpha \dot{\alpha }}, \psi _{+\mu \alpha }, \psi _{-\mu \dot{\alpha }}, \chi _{+a\dot{\alpha }}, \chi _{-\dot{a}\alpha }\right) . \end{aligned}$$

(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 ${\textbf{5}}$ under GL(5) while the vector fields $A^{A}_\mu$ transform as ${\textbf{16}}_c$ under SO(5, 5). Together with the magnetic duals ${B_{\mu \nu }}^{\overline{m}}$ transforming in $\overline{{\textbf{5}}}$ representation of GL(5),  the electric two-forms $B_{\mu \nu \overline{m}}$ transform in a vector representation ${\textbf{10}}$ 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 ± 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 $({\textbf{4}},{\textbf{1}})$ and $({\textbf{1}},{\textbf{4}})$ while the spin-$\frac{1}{2}$ fields $\chi _{+a\dot{\alpha }}$ and $\chi _{-\dot{a}\alpha }$ transform as $({\textbf{5}},{\textbf{4}})$ and $({\textbf{4}},{\textbf{5}}),$ 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

$$\begin{aligned} {V_A}^{\alpha \dot{\beta }}{V^B}_{\alpha \dot{\beta }}=\delta ^B_A\quad \text {and}\quad {V_A}^{\alpha \dot{\beta }}{V^A}_{\gamma \dot{\delta }}=\delta ^{\alpha }_{\gamma } \delta ^{\dot{\beta }}_{\dot{\delta }}. \end{aligned}$$

(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

$$\begin{aligned} {{\mathcal {V}}_M}^a= & {} \frac{1}{16}V^{A\alpha \dot{\alpha }}(\Gamma _M)_{AB} {(\gamma ^a)_{\alpha \dot{\alpha }}}^{\beta \dot{\beta }}{V^B}_{\beta \dot{\beta }}, \end{aligned}$$

(2.3)

$$\begin{aligned} {{\mathcal {V}}_M}^{\dot{a}}= & {} -\frac{1}{16}V^{A\alpha \dot{\alpha }} (\Gamma _M)_{AB}{(\gamma ^{\dot{a}})_{\alpha \dot{\alpha }}}^{\beta \dot{\beta }}{V^B}_{\beta \dot{\beta }}. \end{aligned}$$

(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

$$\begin{aligned} {\mathcal {V}}^{Ma}{{\mathcal {V}}_M}^b=\delta ^{ab},\quad {\mathcal {V}}^{M\dot{a}} {{\mathcal {V}}_M}^{\dot{b}}=\delta ^{\dot{a}\dot{b}}, \quad {\mathcal {V}}^{Ma}{{\mathcal {V}}_M}^{\dot{a}}=0 \end{aligned}$$

(2.5)

and

$$\begin{aligned} {{\mathcal {V}}_M}^a{\mathcal {V}}^{Na}-{{\mathcal {V}}_M}^{\dot{a}} {\mathcal {V}}^{N\dot{a}}=\delta ^N_M. \end{aligned}$$

(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 Eq. (2.6).

The most general gaugings of six-dimensional $N=(2,2)$ supergravity are described by the embedding tensor in ${\textbf{144}}_c$ 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)

$$\begin{aligned} (\Gamma _M)_{AB}\,\theta ^{BM}=0. \end{aligned}$$

(2.7)

The gauge covariant derivative is then given by

$$\begin{aligned} D_\mu =\partial _\mu -gA^A_\mu \ {\Theta _A}^{MN}{\varvec{t}}_{MN} \end{aligned}$$

(2.8)

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

$$\begin{aligned} {\Theta _A}^{MN}=-\theta ^{B[M}(\Gamma ^{N]})_{BA}\equiv \left( \Gamma ^{[M}\theta ^{N]}\right) _A. \end{aligned}$$

(2.9)

As usual, the embedding tensor identifies generators $X_A={\Theta _A}^{MN}{\varvec{t}}_{MN}$ of the gauge group $G_0\subset SO(5,5)$ with particular linear combinations of the SO(5, 5) generators ${\varvec{t}}_{MN}.$ Consistency also requires the gauge generators to form a closed subalgebra of SO(5, 5) implying the quadratic constraint (QC)

$$\begin{aligned} \left[ X_A,X_B\right] =-{(X_A)_B}^C\,X_C. \end{aligned}$$

(2.10)

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

$$\begin{aligned} \theta ^{AM}\theta ^{BN}\eta _{MN}= & {} 0, \end{aligned}$$

(2.11)

$$\begin{aligned} \theta ^{AM}\theta ^{B[N}(\Gamma ^{P]})_{AB}= & {} 0. \end{aligned}$$

(2.12)

Any $\theta ^{AM}\in {\textbf{144}}_c$ 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 Sect. 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

$$\begin{aligned} e^{-1}{\mathcal {L}}=\frac{1}{4}R-\frac{1}{16}P_{\mu }^{a\dot{a}}P^\mu _{a\dot{a}} -{\textbf{V}} \end{aligned}$$

(2.13)

while the supersymmetry transformations of fermionic fields read

$$\begin{aligned} \delta \psi _{+\mu \alpha }= & {} D_\mu \epsilon _{+\alpha }+\frac{g}{4}\hat{\gamma }_\mu {T_\alpha }^{\dot{\beta }}\epsilon _{-\dot{\beta }}, \end{aligned}$$

(2.14)

$$\begin{aligned} \delta \psi _{-\mu \dot{\alpha }}= & {} D_\mu \epsilon _{-\dot{\alpha }}-\frac{g}{4}\hat{\gamma }_\mu {T^{\beta }}_{\dot{\alpha }}\epsilon _{+\beta }, \end{aligned}$$

(2.15)

$$\begin{aligned} \delta \chi _{+a\dot{\alpha }}= & {} \frac{1}{4}P^\mu _{a\dot{a}} \hat{\gamma }_\mu {(\gamma ^{\dot{a}})_{\dot{\alpha }}}^{\dot{\beta }}\epsilon _{-\dot{\beta }} +2g{(T_{a})^\beta }_{\dot{\alpha }}\epsilon _{+\beta }\nonumber \\{} & {} -\frac{g}{2}{T^{\alpha }}_{\dot{\alpha }}{(\gamma _a)_\alpha }^\beta \epsilon _{+\beta }, \end{aligned}$$

(2.16)

$$\begin{aligned} \delta \chi _{-\dot{a}\alpha }= & {} \frac{1}{4}P^\mu _{a\dot{a}}\hat{\gamma }_\mu {(\gamma ^a)_\alpha }^\beta \epsilon _{+\beta }+2g{(T_{\dot{a}})_{\alpha }}^{\dot{\beta }}\epsilon _{-\dot{\beta }}\nonumber \\{} & {} +\frac{g}{2}{T_{\alpha }}^{\dot{\alpha }}{(\gamma _{\dot{a}})_{\dot{\alpha }}}^{\dot{\beta }}\epsilon _{-\dot{\beta }}. \end{aligned}$$

(2.17)

The covariant derivatives of supersymmetry parameters are defined as

$$\begin{aligned} D_\mu \epsilon _{+\alpha }= & {} \partial _\mu \epsilon _{+\alpha }+\frac{1}{4}{\omega _\mu }^{\nu \rho }\hat{\gamma }_{\nu \rho }\epsilon _{+\alpha } +\frac{1}{4}Q_\mu ^{ab}{(\gamma _{ab})_\alpha }^\beta \epsilon _{+\beta },\nonumber \\ \end{aligned}$$

(2.18)

$$\begin{aligned} D_\mu \epsilon _{-\dot{\alpha }}= & {} \partial _\mu \epsilon _{-\dot{\alpha }}+\frac{1}{4}{\omega _\mu }^{\nu \rho } \hat{\gamma }_{\nu \rho }\epsilon _{-\dot{\alpha }}+\frac{1}{4}Q_\mu ^{\dot{a}\dot{b}} {(\gamma _{\dot{a}\dot{b}})_{\dot{\alpha }}}^{\dot{\beta }}\epsilon _{-\dot{\beta }}\nonumber \\ \end{aligned}$$

(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

$$\begin{aligned} P_{\mu }^{a\dot{a}}= & {} \frac{1}{4}{(\gamma ^a)}^{\alpha \beta }{(\gamma ^{\dot{a}})}^{\dot{\alpha } \dot{\beta }}{V^A}_{\alpha \dot{\alpha }}\partial _\mu V_{A\beta \dot{\beta }}, \end{aligned}$$

(2.20)

$$\begin{aligned} Q_{\mu }^{ab}= & {} \frac{1}{8}{(\gamma ^{ab})}^{\alpha \beta }\Omega ^{\dot{\alpha } \dot{\beta }} {V^A}_{\alpha \dot{\alpha }}\partial _\mu V_{A\beta \dot{\beta }}, \end{aligned}$$

(2.21)

$$\begin{aligned} Q_{\mu }^{\dot{a}\dot{b}}= & {} \frac{1}{8}\Omega ^{\alpha \beta }{(\gamma ^{\dot{a}\dot{b}})}^{\dot{\alpha } \dot{\beta }}{V^A}_{\alpha \dot{\alpha }}\partial _\mu V_{A\beta \dot{\beta }}. \end{aligned}$$

(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

$$\begin{aligned} \Omega _{\beta \alpha }= -\Omega _{\alpha \beta },\quad \Omega ^{\alpha \beta }= (\Omega _{\alpha \beta })^*,\quad \Omega _{\alpha \beta }\Omega ^{\beta \gamma }= -\delta ^\gamma _\alpha \end{aligned}$$

(2.23)

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

$$\begin{aligned} {\textbf{V}}= & {} \frac{g^2}{2}\theta ^{AM}\theta ^{BN}{{\mathcal {V}}_M}^a{{\mathcal {V}}_N}^b \left[ {V_A}^{\alpha \dot{\alpha }}{(\gamma _a)_\alpha }^\beta {(\gamma _b)_\beta }^\gamma V_{B\gamma \dot{\alpha }}\right] \nonumber \\= & {} -\frac{g^2}{2}\left[ T^{\alpha \dot{\alpha }}T_{\alpha \dot{\alpha }}-2(T^a)^{\alpha \dot{\alpha }} (T_a)_{\alpha \dot{\alpha }}\right] \end{aligned}$$

(2.24)

with the T-tensors defined by

$$\begin{aligned} (T^a)^{\alpha \dot{\alpha }}={{\mathcal {V}}_M}^a\theta ^{AM}{V_A}^{\alpha \dot{\alpha }},\quad (T^{\dot{a}})^{\alpha \dot{\alpha }}=-{{\mathcal {V}}_M}^{\dot{a}}\theta ^{AM} {V_A}^{\alpha \dot{\alpha }},\nonumber \\ \end{aligned}$$

(2.25)

and

$$\begin{aligned} T^{\alpha \dot{\alpha }}\equiv (T^a)^{\beta \dot{\alpha }}{(\gamma _a)_\beta }^\alpha =-(T^{\dot{a}})^{\alpha \dot{\beta }} {(\gamma _{\dot{a}})_{\dot{\beta }}}^{\dot{\alpha }}. \end{aligned}$$

(2.26)

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

$$\begin{aligned}{} & {} D_\mu T^a=\frac{1}{4}P^{b\dot{b}}_\mu \left( \gamma ^bT^a\gamma ^{\dot{b}}-2\delta ^{ab}T^{\dot{b}}\right) , \end{aligned}$$

(2.27)

$$\begin{aligned}{} & {} D_\mu T^{\dot{a}}=\frac{1}{4}P^{b\dot{b}}_\mu \left( \gamma ^bT^{\dot{a}}\gamma ^{\dot{b}}-2\delta ^{\dot{a}\dot{b}}T^{b}\right) , \end{aligned}$$

(2.28)

$$\begin{aligned}{} & {} D_\mu T=\frac{1}{2}P^{a\dot{a}}_\mu \left( T^a\gamma ^{\dot{a}}-\gamma ^aT^{\dot{a}}-\frac{1}{2}\gamma ^aT\gamma ^{\dot{a}}\right) . \end{aligned}$$

(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 ${\varvec{t}}_{MN}$ are decomposed accordingly as ${\varvec{t}}_{MN}=({\varvec{t}}_{-+}={\varvec{d}},{\varvec{t}}_{+I}={\varvec{p}}_{I},{\varvec{t}}_{-I}={\varvec{k}}_{I},{\varvec{t}}_{IJ}={\varvec{\tau }}_{IJ})$ with ${\varvec{d}}$ and ${\varvec{\tau }}_{IJ}$ 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 ${\textbf{144}}_c$ representation of SO(5, 5) will split into the following representations under ${\mathbb {R}}^+\times SO(4,4)$

$$\begin{aligned} \underbrace{{\textbf{144}}_c}_{\theta ^{AM}}\ \rightarrow \ \underbrace{{\textbf{56}}^{-1}}_{\vartheta _3^{m I}}\,\oplus \,\underbrace{{\textbf{56}}^{+1}}_{\vartheta _4^{\dot{m} I}}\,\oplus \,\underbrace{{\textbf{8}}^{-1}}_{\theta _6^{\dot{m} +}}\,\oplus \,\underbrace{{\textbf{8}}^{+1}}_{\theta _1^{m -}}\,\oplus \,\underbrace{{\textbf{8}}^{+3}}_{\theta _2^{\dot{m} -}}\,\oplus \,\underbrace{{\textbf{8}}^{-3}}_{\theta _5^{m +}}.\nonumber \\ \end{aligned}$$

(3.1)

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

(3.2)

The components $\theta _1^{m -},$ $\theta _4^{\dot{m} I},$ and $\theta _5^{m +}$ correspond to ${\textbf{8}}^{+1},$ ${\textbf{56}}^{+1},$ and ${\textbf{8}}^{-3}$ representations while $\theta _2^{\dot{m} -},$ $\theta _3^{m I},$ and $\theta _6^{\dot{m} +}$ are respectively ${\textbf{8}}^{+3},$ ${\textbf{56}}^{-1},$ and ${\textbf{8}}^{-1}$ ones. The linear combinations $\vartheta _3^{m I}$ 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

$$\begin{aligned}{} & {} \vartheta _3^{m I} = \theta _3^{m I}+\frac{\sqrt{2}}{8}{(\gamma ^I)^{m}}_{\dot{n}}\theta _6^{\dot{n} +}\nonumber \\ {}{} & {} \quad \text {and}\quad \vartheta _4^{\dot{m} I}=\theta _4^{\dot{m} I}-\frac{\sqrt{2}}{8}{(\gamma ^I)^{\dot{m}}}_n\theta _1^{n -}. \end{aligned}$$

(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

$$\begin{aligned} \theta ^{A+}\theta ^{B-}+\theta ^{AI}\theta ^{BJ}\eta _{IJ}+\theta ^{A-}\theta ^{B+}=0 \end{aligned}$$

(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

$$\begin{aligned}{} & {} \theta ^{m M}\theta _1^{n -}{\textbf{c}}_{mn}+\theta ^{\dot{m} M}\theta _6^{\dot{n} +}{\textbf{c}}_{\dot{m}\dot{n}}=0, \end{aligned}$$

(3.5)

$$\begin{aligned}{} & {} \theta ^{m M}\theta _2^{\dot{n} -}(\gamma ^I)_{m\dot{n}}+\theta ^{\dot{m} M}\theta _1^{n -}(\gamma ^I)_{\dot{m}n}-\sqrt{2}\theta ^{\dot{m} M}\theta _4^{\dot{n} I}{\textbf{c}}_{\dot{m}\dot{n}}=0, \end{aligned}$$

(3.6)

$$\begin{aligned}{} & {} \theta ^{m M}\theta _6^{\dot{n} +}(\gamma ^I)_{m\dot{n}}+\theta ^{\dot{m} M}\theta _5^{n +}(\gamma ^I)_{\dot{m}n}+\sqrt{2}\theta ^{m M}\theta _3^{n I}{\textbf{c}}_{mn}=0, \end{aligned}$$

(3.7)

$$\begin{aligned}{} & {} \theta ^{m M}\theta _4^{\dot{n} [I}(\gamma ^{J]})_{m\dot{n}}+\theta ^{\dot{m} M}\theta _3^{n [I}(\gamma ^{J]})_{\dot{m}n}=0. \end{aligned}$$

(3.8)

In these equations, ${\textbf{c}}_{mn}$ and ${\textbf{c}}_{\dot{m}\dot{n}}$ 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 $\theta _3^{m I}=0,$ 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 ${\textbf{8}}^{+1}$ or ${\textbf{8}}^{-1}$ components non-vanishing are not consistent.

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

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

$$\begin{aligned} \theta _2^{\dot{m} -}=v^{\dot{m}} \end{aligned}$$

(3.9)

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

(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 ${\varvec{k}}_{I}$ as

$$\begin{aligned} X_m=(\gamma ^I)_{m\dot{n}}v^{\dot{n}}{\varvec{k}}_{I}. \end{aligned}$$

(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 ${\varvec{k}}_{I}$ generators.

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

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

$$\begin{aligned} \theta _5^{m +}=w^{m} \end{aligned}$$

(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 ${\varvec{p}}_{I}$ generators, i.e.

$$\begin{aligned} X_{\dot{m}}=(\gamma ^I)_{\dot{m}n}w^{n}{\varvec{p}}_{I}. \end{aligned}$$

(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 ${\varvec{p}}_{I}$ generators.

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

We now move to gaugings in ${\textbf{56}}^{-1}$ representation by choosing only $\theta _3^{m I}$ to be non-vanishing. The embedding tensor takes the form

(3.14)

subject to the LC

$$\begin{aligned} (\gamma _I)_{\dot{m}n}\theta _3^{n I}=0 \end{aligned}$$

(3.15)

which is the same as the second condition in (A.45). For vanishing $\theta _6^{\dot{m} +}$ component from ${\textbf{8}}^{-1}$ representation, the embedding tensor is simply given by $\theta _3^{m I}.$ As in [40], all 56 components in $\theta _3^{m I}$ can be parametrized by an antisymmetric tensor $f_{\dot{m}\dot{n}\dot{p}}=f_{[\dot{m}\dot{n}\dot{p}]}$ by writing $\theta _3^{m I}$ as

$$\begin{aligned} \theta _3^{m I}=\frac{1}{48}f_{\dot{m}\dot{n}\dot{p}}(\gamma ^{IJ})^{\dot{m}\dot{n}}(\gamma _J)^{m\dot{p}} \end{aligned}$$

(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

$$\begin{aligned} X_m= & {} \frac{1}{24\sqrt{2}}f_{\dot{m}\dot{n}\dot{p}}(\gamma ^{IJ})^{\dot{m}\dot{n}} {(\gamma _J)_m}^{\dot{p}}{\varvec{p}}_{I}, \end{aligned}$$

(3.17)

$$\begin{aligned} X_{\dot{m}}= & {} \frac{1}{48}f_{\dot{n}\dot{p}\dot{q}}(\gamma ^{IK})^{\dot{n}\dot{p}} (\gamma _K)^{q\dot{q}}(\gamma ^J)_{q\dot{m}}{\varvec{\tau }}_{IJ}. \end{aligned}$$

(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 ${\varvec{p}}_{I}$ 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

$$\begin{aligned} f_{\dot{m}\dot{n}\dot{p}}f^{\dot{m}\dot{n}\dot{p}}=0\quad \text {and}\quad f_{\dot{r}[\dot{m}\dot{n}}{f_{\dot{p}\dot{q}]}}^{\dot{r}}=0. \end{aligned}$$

(3.19)

We will discuss some possible solutions to these conditions.

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

We first consider a simple solution of the form

$$\begin{aligned} f_{\dot{m}\dot{n}\dot{p}}=(\kappa _1\varepsilon _{ijk},\kappa _2\varepsilon _{rst}) \end{aligned}$$

(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

$$\begin{aligned}{} & {} \left[ X_i,\,X_j\right] =-\kappa \varepsilon _{ijk}X_k,\quad \left[ X_r,\,X_s\right] =\kappa \varepsilon _{rst}X_t,\nonumber \\{} & {} \left[ X_i,\,X_r\right] =0. \end{aligned}$$

(3.21)

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

$$\begin{aligned} X_5=X_1,\quad X_6=-X_2,\quad X_7=-X_3,\quad X_8=-X_4.\nonumber \\ \end{aligned}$$

(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

$$\begin{aligned} \left[ M_{\tilde{\mu }\tilde{\nu }},M_{\tilde{\rho }\tilde{\sigma }}\right] =2\kappa \left( \delta _{\tilde{\mu }[\tilde{\rho }}M_{\tilde{\sigma }]\tilde{\nu }} -\delta _{\tilde{\nu }[\tilde{\rho }}M_{\tilde{\sigma }]\tilde{\mu }}\right) . \end{aligned}$$

(3.23)

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

$$\begin{aligned} M_{ij}=\varepsilon _{ijk}A_k\quad \text {and}\quad M_{4i}=-M_{i4}=B_i \end{aligned}$$

(3.24)

with

$$\begin{aligned} A_1= & {} X_{\dot{1}}-X_{\dot{5}},\quad A_2=X_{\dot{2}}-X_{\dot{6}},\quad A_3=X_{\dot{3}}-X_{\dot{7}},\nonumber \\ B_1= & {} X_{\dot{1}}+X_{\dot{5}},\quad B_2=X_{\dot{2}}+X_{\dot{6}},\quad B_3=X_{\dot{3}}+X_{\dot{7}}. \end{aligned}$$

(3.25)

Furthermore, we redefine the four independent translational generators as

$$\begin{aligned} K_1=X_4,\quad K_2=-X_3,\quad K_3=X_2,\quad K_4=X_1 \end{aligned}$$

(3.26)

and obtain the following commutation relations

$$\begin{aligned} \left[ K_{\tilde{\mu }},M_{\tilde{\nu }\tilde{\rho }}\right] =2\kappa \delta _{\tilde{\mu } [\tilde{\nu }}\delta _{\tilde{\rho }]}^{\tilde{\sigma }}K_{\tilde{\sigma }}. \end{aligned}$$

(3.27)

This implies that the gauge group takes the form of

$$\begin{aligned} SO(4) < imes {\mathbb {R}}^4\sim CSO(4,0,1). \end{aligned}$$

(3.28)

3.3.2 $CSO(3,1,1)\sim SO(3,1) < imes {\mathbb {R}}^4$ 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

$$\begin{aligned} f_{\dot{m}\dot{n}\dot{p}}=\kappa (\varepsilon _{ijr},\varepsilon _{irs}) \end{aligned}$$

(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

$$\begin{aligned}{} & {} \left[ X_i,\,X_j\right] =\kappa \varepsilon _{ijr}X_r,\quad \left[ X_r,\,X_s\right] =-\kappa \varepsilon _{rsi}X_i,\nonumber \\{} & {} \left[ X_i,\,X_r\right] =\kappa \varepsilon _{irk}(\delta _{ks}X_s-X_k) \end{aligned}$$

(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

$$\begin{aligned}{} & {} M_{ij}=\varepsilon _{ijk}(\delta _{kr}X_r-X_k)\quad \text {and}\nonumber \\{} & {} M_{4i}=-M_{i4}=-\frac{1}{\sqrt{3}}(\delta _{ir}X_r+X_i). \end{aligned}$$

(3.31)

These generators satisfy the SO(3, 1) algebra

$$\begin{aligned} \left[ M_{\tilde{\mu }\tilde{\nu }},M_{\tilde{\rho }\tilde{\sigma }}\right] =2\kappa \left( \eta _{\tilde{\mu }[\tilde{\rho }}M_{\tilde{\sigma }]\tilde{\nu }} -\eta _{\tilde{\nu }[\tilde{\rho }}M_{\tilde{\sigma }]\tilde{\mu }}\right) \end{aligned}$$

(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

$$\begin{aligned} K_1=X_4,\quad K_2=-X_3,\quad K_3=X_2,\quad K_4=\frac{X_1}{\sqrt{3}}, \end{aligned}$$

(3.33)

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

$$\begin{aligned} \left[ K_{\tilde{\mu }},M_{\tilde{\nu }\tilde{\rho }}\right] =2\kappa \eta _{\tilde{\mu } [\tilde{\nu }}\delta _{\tilde{\rho }]}^{\tilde{\sigma }}K_{\tilde{\sigma }}. \end{aligned}$$

(3.34)

Accordingly, the resulting gauge group is given by

$$\begin{aligned} SO(3,1) < imes {\mathbb {R}}^4\sim CSO(3,1,1). \end{aligned}$$

(3.35)

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

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

$$\begin{aligned} f_{\dot{m}\dot{n}\dot{p}}=\kappa (\varepsilon _{\bar{i}\bar{j} \bar{k}},\varepsilon _{\bar{r}\bar{s}\bar{t}}) \end{aligned}$$

(3.36)

with $\bar{i},\bar{j},\ldots =\dot{1},\dot{6},\dot{7}$ and $\bar{r},\bar{s},\ldots =\dot{2},\dot{3},\dot{5}$ 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

$$\begin{aligned} K_1=-X_3,\quad K_2=X_2,\quad K_3=X_4,\quad K_4=X_1. \end{aligned}$$

(3.37)

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

$$\begin{aligned}{} & {} \left[ X_{\bar{i}},\,X_{\bar{j}}\right] =\kappa \varepsilon _{\bar{i}\bar{j}\bar{k}} \eta ^{\bar{k}\bar{l}}X_{\bar{l}},\quad \left[ X_{\bar{r}},\,X_{\bar{s}}\right] =\kappa \varepsilon _{\bar{r}\bar{s}\bar{t}}\eta ^{\bar{t}\bar{u}}X_{\bar{u}},\nonumber \\{} & {} \left[ X_{\bar{i}},\,X_{\bar{r}}\right] =0. \end{aligned}$$

(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

$$\begin{aligned}{} & {} M_{i4} =-M_{4i}=\delta _{ij}G^j+H_i\nonumber \\{} & {} \quad \text {and}\quad M_{ij}=\varepsilon _{ijk}\eta ^{kl}\quad (\delta _{lm}G^m-H_l) \end{aligned}$$

(3.39)

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

$$\begin{aligned}{} & {} G^1=X_{\dot{6}},\quad G^2=X_{\dot{7}},\quad G^3=X_{\dot{1}},\nonumber \\{} & {} H_1=X_{\dot{2}},\quad H_2=X_{\dot{3}},\quad H_3=X_{\dot{5}}. \end{aligned}$$

(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

$$\begin{aligned} SO(2,2) < imes {\mathbb {R}}^4\sim CSO(2,2,1). \end{aligned}$$

(3.41)

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

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

Similar to the previous case, gaugings in ${\textbf{56}}^{+1}$ representation can be obtained by turning on only $\theta _4^{\dot{m} I}$ component with the same parametrization as in (3.16)

$$\begin{aligned} \theta _4^{\dot{m} I}=\frac{1}{48}f_{mnp}(\gamma ^{IJ})^{mn}(\gamma _J)^{\dot{m}p} \end{aligned}$$

(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

$$\begin{aligned} f_{mnp}f^{mnp}=0\quad \text {and}\quad f_{r[mn}{f_{pq]}}^{r}=0. \end{aligned}$$

(3.43)

We can repeat the same analysis as in the case of gaugings from ${\textbf{56}}^{-1}$ representation by solving these QCs and find the same $CSO(4-p,p,1)\sim SO(4-p,p) < imes {\mathbb {R}}^4$ 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

$$\begin{aligned} X_m= & {} \frac{1}{48}f_{npq}(\gamma ^{IK})^{np}(\gamma _K)^{\dot{q}q} (\gamma ^J)_{m\dot{q}}{\varvec{\tau }}_{IJ}, \end{aligned}$$

(3.44)

$$\begin{aligned} X_{\dot{m}}= & {} \frac{1}{24\sqrt{2}}f_{mnp}(\gamma ^{IJ})^{mn} {(\gamma _J)_{\dot{m}}}^{p}{\varvec{k}}_{I} \end{aligned}$$

(3.45)

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

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

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

(3.46)

Only $\theta _3^{m I}$ 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 $\theta _3^{m I}$ as in Eq. (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],

$$\begin{aligned}{} & {} f_{\dot{m}\dot{n}\dot{p}}f^{\dot{m}\dot{n}\dot{p}}=0,\quad f_{\dot{r}[\dot{m}\dot{n}}{f_{\dot{p}\dot{q}]}}^{\dot{r}}=0,\quad f_{\dot{m}\dot{n}\dot{p}}v^{\dot{p}}=0,\quad \nonumber \\ {}{} & {} {f_{[\dot{m}\dot{n}\dot{p}}v_{\dot{q}]}|}_{SD}=0 \end{aligned}$$

(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 ${\textbf{56}}^{-1}$ representation. We then consider the following three possibilities:

$$\begin{aligned}{} & {} f_{\dot{m}\dot{n}\dot{p}} =\kappa (\varepsilon _{ijk},\varepsilon _{rst})\nonumber \\ {}{} & {} \quad \text {for } i,j,\ldots =\dot{1},\dot{2},\dot{3}\text { and }r,s,\ldots =\dot{5},\dot{6},\dot{7}, \end{aligned}$$

(3.48)

$$\begin{aligned}{} & {} f_{\dot{m}\dot{n}\dot{p}}=\kappa (\varepsilon _{ijr},\varepsilon _{irs})\nonumber \\ {}{} & {} \quad \text {for } i,j,\ldots =\dot{1},\dot{2},\dot{3}\text { and }r,s,\ldots =\dot{5},\dot{6},\dot{7}, \end{aligned}$$

(3.49)

$$\begin{aligned}{} & {} f_{\dot{m}\dot{n}\dot{p}} =\kappa (\varepsilon _{\bar{i}\bar{j} \bar{k}},\varepsilon _{\bar{r}\bar{s}\bar{t}})\nonumber \\ {}{} & {} \quad \text {for }\bar{i},\bar{j},\ldots =\dot{1},\dot{6},\dot{7}\text { and }\bar{r},\bar{s},\ldots =\dot{2},\dot{3},\dot{5} \end{aligned}$$

(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

$$\begin{aligned} v^{\dot{4}}=v^{\dot{8}}=\lambda \end{aligned}$$

(3.51)

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

$$\begin{aligned} X_m= & {} (\gamma ^I)_{m\dot{n}}v^{\dot{n}}{\varvec{k}}_{I}+\frac{1}{24\sqrt{2}}f_{\dot{m}\dot{n}\dot{p}} (\gamma ^{IJ})^{\dot{m}\dot{n}}{(\gamma _J)_m}^{\dot{p}}{\varvec{p}}_{I}, \end{aligned}$$

(3.52)

$$\begin{aligned} X_{\dot{m}}= & {} \frac{1}{48}f_{\dot{n}\dot{p}\dot{q}}(\gamma ^{IK})^{\dot{n}\dot{p}} (\gamma _K)^{q\dot{q}}(\gamma ^J)_{q\dot{m}}{\varvec{\tau }}_{IJ}. \end{aligned}$$

(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

$$\begin{aligned} \left[ M_{\tilde{\mu }\tilde{\nu }},M_{\tilde{\rho }\tilde{\sigma }}\right] =2\kappa \left( \eta _{\tilde{\mu }[\tilde{\rho }}M_{\tilde{\sigma }]\tilde{\nu }}-\eta _{\tilde{\nu } [\tilde{\rho }}M_{\tilde{\sigma }]\tilde{\mu }}\right) \end{aligned}$$

(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 Sect. 3.3, we find the following commutation relations

$$\begin{aligned}{} & {} \left[ K_{\tilde{\mu }},M_{\tilde{\nu }\tilde{\rho }}\right] = 2\kappa \eta _{\tilde{\mu } [\tilde{\nu }}\delta _{\tilde{\rho }]}^{\tilde{\sigma }}K_{\tilde{\sigma }}, \end{aligned}$$

(3.55)

$$\begin{aligned}{} & {} \left[ K_{\tilde{\mu }},K_{\tilde{\nu }}\right] = -\lambda \sqrt{2}M_{\tilde{\mu }\tilde{\nu }}. \end{aligned}$$

(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

$$\begin{aligned} M_{0\tilde{\mu }}=-M_{\tilde{\mu }0}=2K_{\tilde{\mu }}, \end{aligned}$$

(3.57)

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

$$\begin{aligned} \left[ M_{\underline{\mu }\underline{\nu }},M_{\underline{\rho }\underline{\sigma }}\right] =2 \kappa \left( \eta _{\underline{\mu }[\underline{\rho }}M_{\underline{\sigma }]\underline{\nu }} -\eta _{\underline{\nu }[\underline{\rho }}M_{\underline{\sigma }]\underline{\mu }}\right) \end{aligned}$$

(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 $SO(4,1)_{\text {I}}.$ Similarly, the SO(3, 1) group with $\eta _{\tilde{\mu }\tilde{\nu }}=\text {diag}(1,1,1,-1)$ is enlarged to $SO(4,1)_{\text {II}}$ or $SO(3,2)_{\text {I}}$ while the SO(2, 2) group, with $\eta _{\tilde{\mu }\tilde{\nu }}=\text {diag}(1,1,-1,-1),$ becomes $SO(3,2)_{\text {II}}.$ 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 ${\textbf{56}}^{-1}$ representation. We also note that the analysis for gaugings from ${\textbf{56}}^{+1}$ and ${\textbf{8}}^{-3}$ 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.

We end this section by some comments on the $SO(5-p,p)$ and $CSO(4-p,p,1)$ gauge groups identified in this section. The same gauge groups also arise in the classification of gauge groups under $GL(5)\subset SO(5,5)$ that has been extensively studied in [17]. In that case, both the $SO(5-p,p)$ and $CSO(4-p,p,1)$ gauge groups are embedded entirely in GL(5) and are described by purely magnetic gaugings in which only components of the embedding tensor that couple the magnetic two-form fields ${B_{\mu \nu }}^{\overline{m}}$ are non-vanishing. Unlike the electric two-form fields $B_{\mu \nu \overline{m}},$ these fields are also accompanied by the three-form fields. On the other hand, the $CSO(4-p,p,1)$ and $SO(5-p,p)$ gauge groups are embedded respectively in SO(4, 4) and $SO(4,4) < imes {\mathbb {R}}^8$ with the ${\mathbb {R}}^8$ factor generated by ${\varvec{p}}_I$ or ${\varvec{k}}_I$ generators. As can be seen from the structure of the deformed p-form hierarchy given in [32], the embedding tensor components ${\textbf{56}}^{\pm 1}$ couple both electric and magnetic two-form fields $B_{\mu \nu \overline{m}}$ and ${B_{\mu \nu }}^{\overline{m}}.$ Accordingly, the resulting gauged supergravities are not equivalent due to the different field contents among the tensor fields. In particular, gaugings obtained in [17] are known to arise from an $S^1$ reduction of $CSO(p,q,5-p-q)$ gauged supergravities in seven dimensions. However, higher dimensional origins of the gauge groups considered here are not clear at this stage.

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

$$\begin{aligned} ds_6^2=e^{2A(r)}\eta _{\bar{\mu } \bar{\nu }}dx^{\bar{\mu }} dx^{\bar{\nu }}+dr^2 \end{aligned}$$

(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, 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

$$\begin{aligned} \hat{{\varvec{t}}}_{a\dot{b}}= {\mathbb {M}_{a}}^M{\mathbb {M}_{\dot{b}}}^N{\varvec{t}}_{MN} \end{aligned}$$

(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 $(\hat{{\varvec{t}}}_{a\dot{b}})^T=\hat{{\varvec{t}}}_{a\dot{b}}.$ 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

$$\begin{aligned} \hat{{\varvec{t}}}_{a\dot{b}}=(\hat{{\varvec{t}}}_{0\dot{i}},\, \hat{{\varvec{t}}}_{0\#},\,\hat{{\varvec{t}}}_{i\dot{j}},\, \hat{{\varvec{t}}}_{i\#}). \end{aligned}$$

(4.3)

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

$$\begin{aligned} \hat{{\varvec{t}}}_{0\dot{i}}= & {} \frac{1}{\sqrt{2}}{\mathbb {M}_{\dot{i}}}^I({\varvec{p}}_{I}+{\varvec{k}}_{I}),\qquad \hat{{\varvec{t}}}_{0\#}=-{\varvec{d}},\nonumber \\ \hat{{\varvec{t}}}_{i\#}= & {} \frac{1}{\sqrt{2}}{\mathbb {M}_{i}}^I({\varvec{p}}_{I}-{\varvec{k}}_{I}),\qquad \hat{{\varvec{t}}}_{i\dot{j}}={\mathbb {M}_{i}}^I{\mathbb {M}_{\dot{j}}}^J{\varvec{\tau }}_{IJ}. \end{aligned}$$

(4.4)

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

$$\begin{aligned} \underbrace{({\textbf{5}},{\textbf{5}})}_{\hat{{\varvec{t}}}_{a\dot{b}}}\ \rightarrow \ \underbrace{({\textbf{1}},{\textbf{1}})}_{{\varvec{d}}}\oplus \underbrace{({\textbf{1}},{\textbf{4}})}_{\hat{{\varvec{t}}}_{0\dot{i}}} \oplus \underbrace{({\textbf{4}},{\textbf{1}})}_{\hat{{\varvec{t}}}_{i\#}}\, \oplus \underbrace{({\textbf{4}},{\textbf{4}})}_{{{\varvec{t}}}_{i\dot{j}}}. \end{aligned}$$

(4.5)

It is convenient to denote all 25 scalar fields collectively as

$$\begin{aligned} \Phi ^{{\mathbb {I}}}=\{\varphi ,\zeta _{\dot{1}},\ldots ,\zeta _{\dot{4}},\xi _{{1}},\ldots ,\xi _{{4}},\phi _1,\ldots ,\phi _{16}\} \end{aligned}$$

(4.6)

with ${\mathbb {I}}=1,\ldots ,25.$ The scalar $\varphi$ is the dilaton corresponding to the ${\mathbb {R}}^+\sim SO(1,1)$ generator ${\varvec{d}}.$ The two sets of four scalars $\{\zeta _{\dot{1}},\ldots ,\zeta _{\dot{4}}\}$ and $\{\xi _{{1}},\ldots ,\xi _{{4}}\}$ respectively correspond to the generators $\hat{{\varvec{t}}}_{0\dot{i}}$ and $\hat{{\varvec{t}}}_{i\#}.$ The remaining sixteen scalar fields $\{\phi _1,\ldots ,\phi _{16}\}$ 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

$$\begin{aligned} e^{-1}{\mathcal {L}}=\frac{1}{4}R-G_{{{\mathbb {I}}}{{\mathbb {J}}}}\partial _\mu \Phi ^{{\mathbb {I}}}\partial ^\mu \Phi ^{{\mathbb {J}}}-{\textbf{V}} \end{aligned}$$

(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, 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

$$\begin{aligned}&A'\hat{\gamma }_{r}\epsilon _{+\alpha }+\frac{1}{2}\Omega _{\alpha \beta } T^{\beta \dot{\alpha }}\epsilon _{-\dot{\alpha }}=0, \end{aligned}$$

(4.8)

$$\begin{aligned}&A'\hat{\gamma }_{r}\epsilon _{-\dot{\alpha }}-\frac{1}{2}\Omega _{\dot{\alpha } \dot{\beta }}T^{\alpha \dot{\beta }}\epsilon _{+\alpha }=0. \end{aligned}$$

(4.9)

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

$$\begin{aligned} {A'}^2{\delta _\alpha }^\beta= & {} -\frac{1}{4}\Omega _{\alpha \gamma }T^{\gamma \dot{\alpha }}\Omega _{\dot{\alpha } \dot{\beta }}T^{\beta \dot{\beta }}=\mathcal {W}^2{\delta _\alpha }^\beta , \end{aligned}$$

(4.10)

$$\begin{aligned} {A'}^2{\delta _{\dot{\alpha }}}^{\dot{\beta }}= & {} -\frac{1}{4}\Omega _{\dot{\alpha } \dot{\gamma }}T^{\alpha \dot{\gamma }}\Omega _{\alpha \beta }T^{\beta \dot{\beta }} =\mathcal {W}^2{\delta _{\dot{\alpha }}}^{\dot{\beta }} \end{aligned}$$

(4.11)

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

$$\begin{aligned} A'=\pm {\mathcal {W}}. \end{aligned}$$

(4.12)

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

$$\begin{aligned} \hat{\gamma }_r\epsilon _{+\alpha }=-\frac{1}{2}\Omega _{\alpha \beta } \frac{T^{\beta \dot{\beta }}}{A'}\epsilon _{-\dot{\beta }},\quad \hat{\gamma }_r\epsilon _{-\dot{\alpha }}=\frac{1}{2}\Omega _{\dot{\alpha } \dot{\beta }}\frac{T^{\alpha \dot{\beta }}}{A'}\epsilon _{+\alpha }.\nonumber \\ \end{aligned}$$

(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

$$\begin{aligned} {\Phi ^{{\mathbb {I}}}}'=\mp 2G^{{\mathbb {I}}{\mathbb {J}}}\frac{\partial {\mathcal {W}}}{\partial \Phi ^{{\mathbb {J}}}} \end{aligned}$$

(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

$$\begin{aligned} {\textbf{V}}=2G^{{\mathbb {I}}{\mathbb {J}}}\frac{\partial {\mathcal {W}}}{\partial \Phi ^{{\mathbb {I}}}} \frac{\partial {\mathcal {W}}}{\partial \Phi ^{{\mathbb {J}}}}-5{\mathcal {W}}^2. \end{aligned}$$

(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 ${\textbf{56}}^{-1}$ and ${\textbf{8}}^{+3}$ representations.

We first consider supersymmetric domain walls with an unbroken symmetry SO(4). Only SO(5),  $SO(4,1)_{\text {I}},$ 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),  $SO(4,1)_{\text {I}},$ 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

$$\begin{aligned}&({\textbf{5}},{\textbf{5}})\rightarrow ({\textbf{1}}\otimes {\textbf{1}})\oplus ({\textbf{1}}\otimes {\textbf{4}})\oplus ({\textbf{4}}\otimes {\textbf{1}})\oplus ({\textbf{4}}\otimes {\textbf{4}})\nonumber \\&\quad \sim {\textbf{1}}\oplus {\textbf{4}}\oplus {\textbf{4}}\oplus {\textbf{1}}\oplus {\textbf{6}}\oplus {\textbf{9}}. \end{aligned}$$

(4.16)

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

$$\begin{aligned} {\mathcal {Y}}=\hat{{\varvec{t}}}_{1\dot{1}}-\hat{{\varvec{t}}}_{2\dot{2}} -\hat{{\varvec{t}}}_{3\dot{3}}-\hat{{\varvec{t}}}_{4\dot{4}} \end{aligned}$$

(4.17)

from the $SO(4,4)/SO(4)\times SO(4)$ coset.

Using the coset representative of the form

$$\begin{aligned} V=e^{\varphi {\varvec{d}}+\phi {\mathcal {Y}}}, \end{aligned}$$

(4.18)

we find the superpotential and the scalar potential of the form

$$\begin{aligned} {\mathcal {W}}= & {} \frac{g \kappa }{16 \sqrt{2}} e^{-3 \varphi -4 \phi } \left( 4 e^{4 (\varphi +\phi )}+\sigma \right) , \end{aligned}$$

(4.19)

$$\begin{aligned} {\textbf{V}}= & {} -\frac{g^2 \kappa ^2}{64}e^{-6 \varphi -8 \phi } \left( 8 e^{8 (\varphi +\phi )}+8 \sigma e^{4 (\varphi +\phi )}-\sigma ^2\right) . \end{aligned}$$

(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

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

(4.21)

together with the BPS equations for the scalar fields

$$\begin{aligned} \varphi '{} & {} =\mp \frac{g \kappa }{16 \sqrt{2}} e^{-3 \varphi -4 \phi } \left( 4 e^{4 (\varphi +\phi )}-3\sigma \right) \nonumber \\{} & {} \quad \text {and}\quad \phi '=\pm \frac{\sigma g \kappa }{16 \sqrt{2}} e^{-3 \varphi -4 \phi }. \end{aligned}$$

(4.22)

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

$$\begin{aligned} A= & {} C_1+3\phi -\frac{1}{4} \ln \left[ e^{-4 (4 \phi +C_2)}+\sigma \right] , \end{aligned}$$

(4.23)

$$\begin{aligned} \varphi= & {} -\phi -\frac{1}{4} \ln \left[ e^{-4 (4 \phi +C_2)}+\sigma \right] \end{aligned}$$

(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

$$\begin{aligned} \phi =\pm \frac{g\sigma \kappa \rho }{16\sqrt{2}}+C_3 \end{aligned}$$

(4.25)

in which ± 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 $\phi$

$$\begin{aligned} {\mathcal {W}}=\frac{g \kappa }{4 \sqrt{2}} e^\varphi \quad \text {and} \quad {\textbf{V}}=-\frac{g^2 \kappa ^2}{8}e^{2\varphi }, \end{aligned}$$

(4.26)

and the BPS equations reduce to

$$\begin{aligned} A'=-\varphi '=\frac{g \kappa }{4 \sqrt{2}} e^\varphi \quad \text {and}\quad \phi '=0. \end{aligned}$$

(4.27)

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

$$\begin{aligned} A=-\varphi =\ln \left( \frac{g\kappa r}{4\sqrt{2}}-C\right) \quad \text {and}\quad \phi =0. \end{aligned}$$

(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 $SO(4,1)_{\text {II}}$ 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

$$\begin{aligned} \overline{\mathbb {Y}}_1=\hat{{\varvec{t}}}_{1\dot{1}}\quad \text {and}\quad \overline{\mathbb {Y}}_2={\varvec{d}}-\hat{{\varvec{t}}}_{2\dot{2}} -\hat{{\varvec{t}}}_{3\dot{3}} -\hat{{\varvec{t}}}_{4\dot{4}}. \end{aligned}$$

(4.29)

Using the coset representative

$$\begin{aligned} V=e^{\overline{\phi }_1\overline{\mathbb {Y}}_1+\overline{\phi }_2\overline{\mathbb {Y}}_1}, \end{aligned}$$

(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 $i,j,\ldots =1,2,3$ 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),  $SO(4,1)_{\text {I}},$ 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 $SO(4,1)_{\text {II}},$ $SO(3,2)_{\text {I}},$ 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 ${\textbf{4}}\rightarrow {\textbf{1}}+{\textbf{3}},$ we find the following transformation of all 25 scalars

$$\begin{aligned} ({\textbf{5}},{\textbf{5}}){} & {} \rightarrow ({\textbf{1}}\oplus {\textbf{1}}\oplus {\textbf{3}})\otimes ({\textbf{1}}\oplus {\textbf{1}}\oplus {\textbf{3}})\nonumber \\{} & {} \quad \sim 4({\textbf{1}})\oplus 4({\textbf{3}}) \oplus ({\textbf{1}}\oplus {\textbf{3}}\oplus {\textbf{5}}) \end{aligned}$$

(4.31)

with the last three representations arising from the product ${\textbf{3}}\otimes {\textbf{3}}.$ 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

$$\begin{aligned} \mathcal {Y}_1= & {} \hat{{\varvec{t}}}_{1\dot{1}},\quad \mathcal {Y}_2=\hat{{\varvec{t}}}_{2\dot{2}}+ \hat{{\varvec{t}}}_{3\dot{3}}+\hat{{\varvec{t}}}_{4\dot{4}},\nonumber \\ \mathcal {Y}_3= & {} \hat{{\varvec{t}}}_{0\dot{1}},\quad \mathcal {Y}_4=\hat{{\varvec{t}}}_{1\#}. \end{aligned}$$

(4.32)

We can use the coset representative of the form

$$\begin{aligned} V=e^{\varphi {\varvec{d}}+\phi _1\mathcal {Y}_1+\phi _2\mathcal {Y}_2+\zeta \mathcal {Y}_3+\xi \mathcal {Y}_4}. \end{aligned}$$

(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

$$\begin{aligned} {\mathcal {W}}= & {} \frac{g \kappa }{16 \sqrt{2}}e^{-(3\varphi +\phi _1+\phi _2)}\Big [3e^{4\varphi }+e^{4 \phi _2}\cosh {4\zeta }\nonumber \\ {}{} & {} \left( ve^{4 (\varphi +\phi _1)}+\sigma \right) \Big ],\nonumber \\ \end{aligned}$$

(4.34)

$$\begin{aligned} {\textbf{V}}= & {} \frac{1}{64}e^{-2(3\varphi +\phi _1+\phi _2)}g^2 \kappa ^2\left[ \left[ e^{4\phi _2} \cosh {4\zeta }\left( ve^{4 (\varphi +\phi _1)}+\sigma \right) \right. \right. \nonumber \\ {}{} & {} \left. \left. -3 e^{4\varphi }\right] ^2\right. \nonumber \\{} & {} \left. -12 e^{8 \varphi }-4v \sigma e^{4 (\varphi +\phi _1+2\phi _2)}\right] . \end{aligned}$$

(4.35)

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) ($u=\sigma =1$), $SO(4,1)_{\text {I}}$ ($u=-\sigma =1$), CSO(4, 0, 1) ($u=1, \sigma =0$), $SO(4,1)_{\text {II}}$ ($u=-\sigma =-1$), $SO(3,2)_{\text {I}}$ ($u=\sigma =-1$), and CSO(3, 1, 1) ($u=-1,\sigma =0$).

Using

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

(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

$$\begin{aligned} A'= & {} \frac{g \kappa }{16 \sqrt{2}}e^{-(3\varphi +\phi _1+\phi _2)}\left[ 3e^{4\varphi }+e^{4 \phi _2}\cosh {4\zeta }\left( ve^{4 (\varphi +\phi _1)}+\sigma \right) \right] ,\nonumber \\ \end{aligned}$$

(4.37)

$$\begin{aligned} \varphi '= & {} -\frac{1}{32 \sqrt{2}} e^{-3\varphi -\phi _1-\phi _2}g\kappa \left[ 6e^{4\varphi }\right. \nonumber \\{} & {} \left. -e^{4 \phi _2}\text {sech}{4\zeta }\left[ v e^{4 (\varphi +\phi _1)}(\cosh {8\zeta }-3)+\sigma (\cosh {8\zeta }+5) \right] \right] ,\nonumber \\ \end{aligned}$$

(4.38)

$$\begin{aligned} \phi '_1= & {} \frac{1}{32 \sqrt{2}} e^{-3\varphi -\phi _1-\phi _2}g\kappa \left[ 6e^{4\varphi }\right. \nonumber \\{} & {} \left. -e^{4 \phi _2}\text {sech}{4\zeta }\left[ v e^{4 (\varphi +\phi _1)}(\cosh {8\zeta }+5)+\sigma (\cosh {8\zeta }-3) \right] \right] ,\nonumber \\ \end{aligned}$$

(4.39)

$$\begin{aligned} \phi '_2= & {} \frac{g\kappa }{16 \sqrt{2}}e^{-(3\varphi +\phi _1+\phi _2)}\left[ e^{4\varphi }-e^{4 \phi _2}\cosh {4\zeta }\left( ve^{4 (\varphi +\phi _1)}+\sigma \right) \right] ,\nonumber \\ \end{aligned}$$

(4.40)

$$\begin{aligned} \zeta '= & {} -\frac{g \kappa }{8 \sqrt{2}}e^{-3\varphi -\phi _1+3\phi _2}\sinh {4\zeta } \left( ve^{4 (\varphi +\phi _1)}+\sigma \right) . \end{aligned}$$

(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

$$\begin{aligned} \phi _2= & {} \frac{1}{16} \ln \left[ \frac{\sinh {8\zeta }\sqrt{v \sigma +C_1^2-C_1^2 \text {sech}^2{4\zeta }}-C_2v^2\sigma ^2\sinh ^2{4\zeta }}{2 v \sigma }\right] ,\nonumber \\ \end{aligned}$$

(4.42)

$$\begin{aligned} \phi _1= & {} \frac{1}{4} \ln \left[ \frac{\sqrt{C_1^2+v\sigma -C_1^2\text {sech}^2{4\zeta }}-C_1 \tanh {4\zeta }}{v}\right] \nonumber \\{} & {} -\frac{1}{4}\ln {\sinh {4\zeta }}+3\phi _2,\end{aligned}$$

(4.43)

$$\begin{aligned} \varphi= & {} \frac{1}{4} \ln \left[ \frac{\sqrt{C_1^2+v\sigma -C_1^2\text {sech}^2{4\zeta }}-C_1 \tanh {4\zeta }}{v}\right] \nonumber \\{} & {} +\frac{1}{4}\ln {\sinh {4\zeta }}-3\phi _2,\end{aligned}$$

(4.44)

$$\begin{aligned} A= & {} 3\phi _2-\frac{1}{2}\ln {\sinh {4\zeta }} \end{aligned}$$

(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 $\frac{d\rho }{dr}=e^{-4\zeta -\varphi -\phi _1+3\phi _2}(v e^{4 (\varphi +\phi _1)}+\sigma ).$ The solution for $\zeta$ is then given by

$$\begin{aligned} e^{4\zeta }=\tanh \left( \frac{g\kappa \rho }{4\sqrt{2}}-C_3\right) . \end{aligned}$$

(4.46)

It should be noted that for the case of $SO(3,2)_{\text {II}}$ 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

$$\begin{aligned} V=e^{-\varphi \hat{{\varvec{t}}}_{1\dot{1}}-\phi _1\hat{{\varvec{t}}}_{4\dot{4}} +\phi _2({\varvec{d}}+\hat{{\varvec{t}}}_{2\dot{2}}+\hat{{\varvec{t}}}_{3\dot{3}}) +\zeta (\hat{{\varvec{t}}}_{1\dot{4}}-\hat{{\varvec{t}}}_{4\dot{1}})} \end{aligned}$$

(4.47)

with the five SO(3) singlets corresponding to the non-compact generators $\hat{{\varvec{t}}}_{1\dot{1}},$ $\hat{{\varvec{t}}}_{4\dot{4}},$ ${\varvec{d}}+\hat{{\varvec{t}}}_{2\dot{2}}+\hat{{\varvec{t}}}_{3\dot{3}},$ $\hat{{\varvec{t}}}_{1\dot{4}},$ and $\hat{{\varvec{t}}}_{4\dot{1}}.$ We also note the relation between scalar fields corresponding to $\hat{{\varvec{t}}}_{1\dot{4}}$ and $\hat{{\varvec{t}}}_{4\dot{1}}$ 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),  $SO(4,1)_{\text {I}},$ 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 $SO(3,2)_{\text {II}}$ 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 ${\textbf{4}}\rightarrow ({\textbf{1}},{\textbf{2}})+({\textbf{2}},{\textbf{1}})$ with ${\textbf{2}}$ 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

$$\begin{aligned} ({\textbf{5}},{\textbf{5}})&\rightarrow [({\textbf{1}},{\textbf{1}})\oplus ({\textbf{1}},{\textbf{2}})\oplus ({\textbf{2}},{\textbf{1}})]\otimes [({\textbf{1}},{\textbf{1}})\oplus ({\textbf{1}},{\textbf{2}})\oplus ({\textbf{2}},{\textbf{1}})]\nonumber \\&\quad \sim 5({\textbf{1}},{\textbf{1}})\oplus 3({\textbf{1}},{\textbf{2}})\oplus 3({\textbf{2}},{\textbf{1}}) \oplus 2({\textbf{2}},{\textbf{2}}). \end{aligned}$$

(4.48)

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

$$\begin{aligned}{} & {} {\mathbb {Y}}_1=\hat{{\varvec{t}}}_{1\dot{1}}-\hat{{\varvec{t}}}_{4\dot{4}},\quad {\mathbb {Y}}_2=\hat{{\varvec{t}}}_{2\dot{2}}+\hat{{\varvec{t}}}_{3\dot{3}},\quad {\mathbb {Y}}_3=\hat{{\varvec{t}}}_{1\dot{4}}+\hat{{\varvec{t}}}_{4\dot{1}},\nonumber \\ {}{} & {} \quad {\mathbb {Y}}_4=\hat{{\varvec{t}}}_{2\dot{3}}-\hat{{\varvec{t}}}_{3\dot{2}}. \end{aligned}$$

(4.49)

leading to the coset representative of the form

$$\begin{aligned} V=e^{\varphi {\varvec{d}}+\phi _1{\mathbb {Y}}_1+\phi _2{\mathbb {Y}}_2 +\phi _3{\mathbb {Y}}_3+\phi _4{\mathbb {Y}}_4}. \end{aligned}$$

(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

$$\begin{aligned} {\mathcal {W}}=\frac{g \kappa }{16\sqrt{2}}e^{\varphi -2 (\phi _1+\phi _2)} \left( 2e^{4 (\phi _1+\phi _2)}+2u+u\sigma e^{4(\phi _2-\varphi )}\right) ,\nonumber \\ \end{aligned}$$

(4.51)

and the scalar potential takes the form

$$\begin{aligned}{} & {} {\textbf{V}}=-\frac{ug^2 \kappa ^2}{64}e^{-6\varphi -4\phi _1} \left[ 8 e^{8 \varphi +4 \phi _1}+4 \sigma e^{4 \varphi }( e^{4 (\phi _1+\phi _2)}+u)\right. \nonumber \\ {}{} & {} \left. -u\sigma ^2 e^{4 \phi _2}\right] . \end{aligned}$$

(4.52)

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

The resulting BPS equations read

$$\begin{aligned} A'= & {} \frac{g \kappa }{16\sqrt{2}}e^{\varphi -2 (\phi _1+\phi _2)} \left( 2e^{4 (\phi _1+\phi _2)}+2u+u\sigma e^{4(\phi _2-\varphi )}\right) ,\nonumber \\ \end{aligned}$$

(4.53)

$$\begin{aligned} \varphi '= & {} -\frac{g \kappa }{16\sqrt{2}}e^{\varphi -2 (\phi _1+\phi _2)} \left( 2e^{4 (\phi _1+\phi _2)}+2u-3u\sigma e^{4(\phi _2-\varphi )}\right) ,\nonumber \\ \end{aligned}$$

(4.54)

$$\begin{aligned} \phi '_1= & {} -\frac{g \kappa }{16\sqrt{2}}e^{\varphi -2 (\phi _1+\phi _2)} \left( 2e^{4 (\phi _1+\phi _2)}-2u-u\sigma e^{4(\phi _2-\varphi )}\right) ,\nonumber \\ \end{aligned}$$

(4.55)

$$\begin{aligned} \phi '_2= & {} -\frac{g \kappa }{16\sqrt{2}}e^{\varphi -2 (\phi _1+\phi _2)} \left( 2e^{4 (\phi _1+\phi _2)}-2u+u\sigma e^{4(\phi _2-\varphi )}\right) .\nonumber \\ \end{aligned}$$

(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

$$\begin{aligned} \phi _2= & {} C_1+\frac{1}{16}\ln \left[ e^{C_3}+\sigma e^{\frac{ug \kappa \rho }{\sqrt{2}}}\right] +\frac{1}{8}\ln \left[ 1-e^{u(C_2-\frac{g \kappa \rho }{\sqrt{2}})}\right] ,\nonumber \\ \end{aligned}$$

(4.57)

$$\begin{aligned} \phi _1= & {} -\phi _2-\frac{1}{4}\ln \left[ u(1-e^{u(C_2-\frac{g \kappa \rho }{\sqrt{2}})})\right] ,\nonumber \\ \end{aligned}$$

(4.58)

$$\begin{aligned} \varphi= & {} \phi _2+\frac{1}{4}\ln \left[ e^{C_3-\frac{ug \kappa \rho }{\sqrt{2}}}+\sigma \right] ,\nonumber \\ \end{aligned}$$

(4.59)

$$\begin{aligned} A= & {} -5\phi _2-\frac{1}{2}\ln \left[ u(1-e^{u(C_2-\frac{g \kappa \rho }{\sqrt{2}})})\right] -\frac{1}{4}\ln \left[ e^{C_3-\frac{ug \kappa \rho }{\sqrt{2}}}+\sigma \right] \nonumber \\ \end{aligned}$$

(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 $SO(3,2)_{\text {I}}$ 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

$$\begin{aligned} ({\textbf{5}},{\textbf{5}})\rightarrow 9({\textbf{1}})\oplus 8({\textbf{2}}) \end{aligned}$$

(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

$$\begin{aligned}&{\mathbb {Y}}_5=\hat{{\varvec{t}}}_{1\dot{2}}-\hat{{\varvec{t}}}_{4\dot{3}},\quad {\mathbb {Y}}_6=\hat{{\varvec{t}}}_{2\dot{1}}+\hat{{\varvec{t}}}_{3\dot{4}},\nonumber \\&{\mathbb {Y}}_7=\hat{{\varvec{t}}}_{1\dot{3}}+\hat{{\varvec{t}}}_{4\dot{2}},\quad {\mathbb {Y}}_8=\hat{{\varvec{t}}}_{2\dot{4}}-\hat{{\varvec{t}}}_{3\dot{1}}. \end{aligned}$$

(4.62)

The coset representative can be written as

$$\begin{aligned} V=e^{\varphi {\varvec{d}}+\phi _1{\mathbb {Y}}_1+\phi _2{\mathbb {Y}}_2+\phi _3{\mathbb {Y}}_3 +\phi _4{\mathbb {Y}}_4+\phi _5{\mathbb {Y}}_5+\phi _6{\mathbb {Y}}_6+\phi _7{\mathbb {Y}}_7+\phi _8{\mathbb {Y}}_8}.\nonumber \\ \end{aligned}$$

(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

$$\begin{aligned} V=e^{\varphi {\varvec{d}}+\phi _1{\mathbb {Y}}_1+\phi _2{\mathbb {Y}}_2 +\tilde{\phi }_3({\mathbb {Y}}_5-{\mathbb {Y}}_6)+\tilde{\phi }_4({\mathbb {Y}}_7+{\mathbb {Y}}_8)} \end{aligned}$$

(4.64)

giving rise to the superpotential and scalar potential

$$\begin{aligned} {\mathcal {W}}= & {} \frac{g \kappa }{16\sqrt{2}}e^{\varphi -2 (\phi _1+\phi _2)}\left[ u\sigma e^{4(\phi _2-\varphi )}\right. \nonumber \\ {}{} & {} +(2e^{4 (\phi _1+\phi _2)}+2u)\left. \cosh {4\tilde{\phi }_3}\cosh {4\tilde{\phi }_4}+\right] ,\end{aligned}$$

(4.65)

$$\begin{aligned} {\textbf{V}}= & {} -\frac{1}{64}e^{-6\varphi -4\phi _1}ug^2 \kappa ^2 \left[ 8 e^{8 \varphi 4 \phi _1}-u\sigma ^2 e^{4 \phi _2}\right. \nonumber \\ {}{} & {} \left. +4 \sigma e^{4 \varphi }( e^{4 (\phi _1+\phi _2)}+u)\cosh {4\tilde{\phi }_3}\cosh {4\tilde{\phi }_4}\right] . \end{aligned}$$

(4.66)

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

$$\begin{aligned} G^{{\mathbb {I}}{\mathbb {J}}}=\frac{1}{8}\begin{pmatrix} 4 &{} 0 &{} 0 &{} 0 &{} 0 \\ 0 &{} \text {sech}^24\tilde{\phi _3}\,\text {sech}^24\tilde{\phi _4}+1 &{} \text {sech}^24\tilde{\phi _3}\,\text {sech}^24\tilde{\phi _4}-1 &{} 0 &{} 0 \\ 0 &{} \text {sech}^24\tilde{\phi _3}\,\text {sech}^24\tilde{\phi _4}-1 &{} \text {sech}^24\tilde{\phi _3}\,\text {sech}^24\tilde{\phi _4}+1 &{} 0 &{} 0 \\ 0 &{} 0 &{} 0 &{} \text {sech}^24\tilde{\phi _4} &{} 0 \\ 0 &{} 0 &{} 0 &{} 0 &{} 1 \\ \end{pmatrix}\nonumber \\ \end{aligned}$$

(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) ($u=\sigma =1$), $SO(4,1)_{\text {I}}$ ($u=-\sigma =1$), CSO(4, 0, 1) ($u=1, \sigma =0$), $SO(3,2)_{\text {II}}$ ($u=-\sigma =-1$ or $u=\sigma =-1$), and CSO(2, 2, 1) ($u=-1,\sigma =0$). With all these, we find the BPS equations

$$\begin{aligned} A'= & {} \frac{g \kappa }{16\sqrt{2}}e^{\varphi -2 (\phi _1+\phi _2)}\left[ (2e^{4 (\phi _1+\phi _2)}+2u)\right. \nonumber \\ {}{} & {} \left. \cosh {4\tilde{\phi }_3}\cosh {4\tilde{\phi }_4}+u\sigma e^{4(\phi _2-\varphi )}\right] , \end{aligned}$$

(4.68)

$$\begin{aligned} \varphi '= & {} -\frac{g \kappa }{16\sqrt{2}}e^{\varphi -2 (\phi _1+\phi _2)}\left[ (2e^{4 (\phi _1+\phi _2)}+2u)\right. \nonumber \\ {}{} & {} \left. \cosh {4\tilde{\phi }_3}\cosh {4\tilde{\phi }_4}-3u\sigma e^{4(\phi _2-\varphi )}\right] ,\end{aligned}$$

(4.69)

$$\begin{aligned} \phi '_1= & {} -\frac{g \kappa }{16\sqrt{2}}e^{\varphi -2 (\phi _1+\phi _2)}\left[ (2e^{4 (\phi _1+\phi _2)}-2u)\right. \nonumber \\ {}{} & {} \left. \text {sech}{4\tilde{\phi }_3}\,\text {sech}{4\tilde{\phi }_4}-u\sigma e^{4(\phi _2-\varphi )}\right] , \end{aligned}$$

(4.70)

$$\begin{aligned} \phi '_2= & {} -\frac{g \kappa }{16\sqrt{2}}e^{\varphi -2 (\phi _1+\phi _2)}\left[ (2e^{4 (\phi _1+\phi _2)}-2u)\right. \nonumber \\ {}{} & {} \left. \text {sech}{4\tilde{\phi }_3}\,\text {sech}{4\tilde{\phi }_4}+u\sigma e^{4(\phi _2-\varphi )}\right] , \end{aligned}$$

(4.71)

$$\begin{aligned} \tilde{\phi }'_3= & {} -\frac{g \kappa }{16\sqrt{2}}e^{\varphi -2 (\phi _1+\phi _2)}(2e^{4 (\phi _1+\phi _2)}+2u)\nonumber \\ {}{} & {} \sinh {4\tilde{\phi }_3}\,\text {sech}{4\tilde{\phi }_4}, \end{aligned}$$

(4.72)

$$\begin{aligned} \tilde{\phi }'_4= & {} -\frac{g \kappa }{16\sqrt{2}}e^{\varphi -2 (\phi _1+\phi _2)}(2e^{4 (\phi _1+\phi _2)}+2u)\nonumber \\ {}{} & {} \cosh {4\tilde{\phi }_3}\sinh {4\tilde{\phi }_4}. \end{aligned}$$

(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

$$\begin{aligned} \tilde{\phi }_3= & {} \frac{1}{4} \sinh ^{-1}\left[ \frac{4C_5 \sqrt{u}}{\sqrt{C_0^6 g^2 \kappa ^2 \left( C_6+2u\sigma g^3 \kappa ^3 \rho ^4\right) ^2-16 C_5^2 \left( C_1^2+u\right) }}\right] ,\nonumber \\ \end{aligned}$$

(4.74)

$$\begin{aligned} \tilde{\phi }_4= & {} \frac{1}{4} \tanh ^{-1}\left[ \frac{4 \sqrt{u}}{\sqrt{C_0^6 g^2 \kappa ^2 \left( C_6+2u\sigma g^3 \kappa ^3 \rho ^4\right) ^2-16 C_5^2 \left( C_1^2+u\right) }}\right] ,\nonumber \\ \end{aligned}$$

(4.75)

$$\begin{aligned} \phi _1= & {} \frac{1}{4} \ln \left[ C_1 \tanh {4\tilde{\phi }_3}+\sqrt{C_1^2\tanh ^2{4\tilde{\phi }_3}+u}\right] +\frac{1}{4} \ln \left[ \frac{u\sigma g\kappa \rho }{4 \sqrt{2}}\right] ,\nonumber \\ \end{aligned}$$

(4.76)

$$\begin{aligned} \phi _2= & {} \frac{1}{4} \ln \left[ C_1 \tanh {4\tilde{\phi }_3}+\sqrt{C_1^2\tanh ^2{4\tilde{\phi }_3}+u}\right] -\frac{1}{4} \ln \left[ \frac{u\sigma g\kappa \rho }{4 \sqrt{2}}\right] ,\nonumber \\ \end{aligned}$$

(4.77)

$$\begin{aligned} \varphi= & {} \frac{3}{4}\ln \left[ \sqrt{2} C_0 u\sigma g \kappa \rho \right] +\frac{1}{4}\ln {\sinh {4\tilde{\phi }_4}},\nonumber \\ \end{aligned}$$

(4.78)

$$\begin{aligned} A= & {} \frac{1}{3}\varphi -\frac{1}{3}\ln {\sinh {4\tilde{\phi }_4}} \end{aligned}$$

(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 ${\textbf{56}}^{+1}$ and ${\textbf{8}}^{-3}$ 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 ${\textbf{8}}^{\pm 1},$ ${\textbf{8}}^{\pm 3},$ and ${\textbf{56}}^{\pm 1}$ representations. We have pointed out that gaugings in ${\textbf{8}}^{\pm 1}$ representation without ${\textbf{56}}^{\pm 1}$ is not consistent due to the linear constraint, and gaugings in ${\textbf{8}}^{\pm 3}$ representation only lead to a translational gauge group ${\mathbb {R}}^8.$ On the other hand, gaugings in ${\textbf{56}}^{\pm 1}$ representation give rise to $CSO(4-p,p,1)\sim SO(4-p,p) < imes {\mathbb {R}}^4$ gauge groups with $p=0,1,2.$ Including ${\textbf{8}}^{\pm 3}$ representation to ${\textbf{56}}^{\mp 1}$ 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 ${\textbf{56}}^{-1}$ and ${\textbf{8}}^{+3}$ representations with various residual symmetries. The corresponding solutions for gaugings solely from ${\textbf{56}}^{-1}$ representation can be straightforwardly obtained from these results by turning off the ${\textbf{8}}^{+3}$ 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 AdS$_6$ 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, 54] in which a holographic description from a simple domain wall found in [30] with SO(5) symmetry has been studied.

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].

1.1 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 $I=1,\ldots ,8$ is an SO(4, 4) vector index. Accordingly, an SO(5, 5) vector $V_M$ can be written as

$$\begin{aligned} V_M=(V_-,V_I,V_+) \end{aligned}$$

(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

(A.2)

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

(A.3)

where ${\mathbb {1}}_n$ 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

$$\begin{aligned} {({\varvec{t}}_{MN})_P}^Q= 4\eta _{P[M}\delta ^Q_{N]} \end{aligned}$$

(A.4)

where $\delta ^M_N={\mathbb {1}}_{10}.$ These generators satisfy ${\varvec{t}}_{MN}={\varvec{t}}_{[MN]}$ as well as the SO(5, 5) algebra

$$\begin{aligned} \left[ {\varvec{t}}_{MN},{\varvec{t}}_{PQ}\right] = 4(\eta _{M[P}{\varvec{t}}_{Q]N}-\eta _{N[P}{\varvec{t}}_{Q]M}). \end{aligned}$$

(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 ${\varvec{d}}= {{\varvec{t}}}_{-+}$ whose explicit form in vector representation is

(A.6)

With ${\varvec{d}}$ 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

$$\begin{aligned} \underbrace{{\textbf{10}}}_{V_M}\ \rightarrow \ \underbrace{{\textbf{1}}^{+1}}_{V_+}\,\oplus \,\underbrace{{\textbf{1}}^{-1}}_{V_-}\,\oplus \,\underbrace{{\textbf{8}}^{\,0}}_{V_I}. \end{aligned}$$

(A.7)

1.2 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

(A.8)

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

$$\begin{aligned} \left[ {\varvec{d}},{\varvec{d}}\right]= & {} 0,\quad \left[ {\varvec{d}},{\varvec{\tau }}_{IJ}\right] = 0, \end{aligned}$$

(A.9)

$$\begin{aligned} \left[ {\varvec{d}},{\varvec{p}}_{I}\right]= & {} -2{\varvec{p}}_{I},\quad \left[ {\varvec{d}},{\varvec{k}}_{I}\right] = +2{\varvec{k}}_{I}, \end{aligned}$$

(A.10)

$$\begin{aligned} \left[ {\varvec{p}}_{I},{\varvec{p}}_{J}\right]= & {} 0,\qquad \left[ {\varvec{k}}_{I},{\varvec{k}}_{J}\right] =0, \end{aligned}$$

(A.11)

$$\begin{aligned} \left[ {\varvec{p}}_{I},{\varvec{\tau }}_{JK}\right]= & {} 4\eta _{I[J}{\varvec{p}}_{K]},\quad \left[ {\varvec{k}}_{I},{\varvec{\tau }}_{JK}\right] = 4\eta _{I[J}{\varvec{k}}_{K]}, \end{aligned}$$

(A.12)

$$\begin{aligned} \left[ {\varvec{p}}_{I},{\varvec{k}}_{J}\right]= & {} 2({\varvec{d}}\eta _{IJ}-{\varvec{\tau }}_{IJ}), \end{aligned}$$

(A.13)

$$\begin{aligned} \left[ {\varvec{\tau }}_{IJ},{\varvec{\tau }}_{KL}\right]= & {} 4(\eta _{I[K} {\varvec{\tau }}_{L]J}-\eta _{J[K}{\varvec{\tau }}_{L]I}). \end{aligned}$$

(A.14)

The last commutation relation is the SO(4, 4) algebra implying that ${\varvec{\tau }}_{IJ}$ 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

$$\begin{aligned} \underbrace{\textbf{45}}_{{\varvec{t}}_{MN}}\ \rightarrow \ \underbrace{{\textbf{1}}^{0}}_{{\varvec{d}}}\,\oplus \, \underbrace{\textbf{28}^{0}}_{{\varvec{\tau }}_{IJ}}\, \oplus \,\underbrace{{\textbf{8}}^{+2}}_{{\varvec{k}}_{I}} \,\oplus \,\underbrace{{\textbf{8}}^{-2}}_{{\varvec{p}}_I} \end{aligned}$$

(A.15)

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

1.3 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]

$$\begin{aligned} \tilde{{\varvec{\Gamma }}}_{\underline{1}}= & {} \sigma _1\otimes \sigma _3\otimes {\mathbb {1}}_2\otimes {\mathbb {1}}_2\otimes {\mathbb {1}}_2,\qquad \tilde{{\varvec{\Gamma }}}_{\underline{2}}= \sigma _1\otimes \sigma _1\otimes \sigma _2\otimes \sigma _1\otimes \sigma _2,\nonumber \\ \tilde{{\varvec{\Gamma }}}_{\underline{3}}= & {} \sigma _1\otimes \sigma _1\otimes \sigma _1\otimes \sigma _1\otimes {\mathbb {1}}_2,\qquad \tilde{{\varvec{\Gamma }}}_{\underline{4}}= \sigma _1\otimes \sigma _1\otimes \sigma _1\otimes \sigma _3\otimes \sigma _1,\nonumber \\ \tilde{{\varvec{\Gamma }}}_{\underline{5}}= & {} \sigma _1\otimes \sigma _1\otimes \sigma _1\otimes \sigma _3\otimes \sigma _3,\qquad \tilde{{\varvec{\Gamma }}}_{\underline{6}}= i \sigma _1\otimes \sigma _1\otimes \sigma _3\otimes \sigma _1\otimes \sigma _2,\nonumber \\ \tilde{{\varvec{\Gamma }}}_{\underline{7}}= & {} i\sigma _1\otimes \sigma _1\otimes {\mathbb {1}}_2 \otimes \sigma _2\otimes {\mathbb {1}}_2,\quad \ \ \tilde{{\varvec{\Gamma }}}_{\underline{8}}= i\sigma _1\otimes \sigma _1\otimes {\mathbb {1}}_2\otimes \sigma _3\otimes \sigma _2,\nonumber \\ \tilde{{\varvec{\Gamma }}}_{\underline{9}}= & {} i\sigma _1\otimes \sigma _2\otimes {\mathbb {1}}_2\otimes {\mathbb {1}}_2\otimes {\mathbb {1}}_2,\quad \tilde{{\varvec{\Gamma }}}_{\underline{10}}= i\sigma _2\otimes {\mathbb {1}}_2\otimes {\mathbb {1}}_2\otimes {\mathbb {1}}_2\otimes {\mathbb {1}}_2\quad \nonumber \\ \end{aligned}$$

(A.16)

where $\{\sigma _1,\sigma _2,\sigma _3\}$ are the usual Pauli matrices

$$\begin{aligned} \sigma _1= \begin{pmatrix} 0 &{} 1\\ 1 &{} 0 \end{pmatrix},\quad \sigma _2= \begin{pmatrix} 0 &{} -i\\ i &{} 0 \end{pmatrix},\quad \sigma _3 = \begin{pmatrix} 1 &{} 0\\ 0 &{} -1 \end{pmatrix}. \end{aligned}$$

(A.17)

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

$$\begin{aligned} \left\{ \tilde{{\varvec{\Gamma }}}_{\underline{A}},\tilde{{\varvec{\Gamma }}}_{\underline{B}}\right\} \ =\ 2 \eta _{\underline{A}\underline{B}}{\mathbb {1}}_{32} \end{aligned}$$

(A.18)

in which $\eta _{\underline{A}\underline{B}} = \text {diag}({\mathbb {1}}_5,-{\mathbb {1}}_5)$ is the SO(5, 5) invariant metric in diagonal basis. In this explicit representation, the SO(5, 5) chirality matrix is given by

$$\begin{aligned} \tilde{{\varvec{\Gamma }}}_*=\tilde{{\varvec{\Gamma }}}_{\underline{1}\ldots } \tilde{{\varvec{\Gamma }}}\underline{10} =\text {diag}(-{\mathbb {1}}_{16},{\mathbb {1}}_{16}) \end{aligned}$$

(A.19)

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

$$\begin{aligned} \tilde{{\mathbb {C}}}= i \sigma _2\otimes \sigma _3\otimes \sigma _3\otimes {\mathbb {1}}_2\otimes {\mathbb {1}}_2 \end{aligned}$$

(A.20)

satisfying

$$\begin{aligned} \tilde{{\varvec{\Gamma }}}_{\underline{A}}\tilde{{\mathbb {C}}} =(\tilde{{\varvec{\Gamma }}}_{\underline{A}}\tilde{{\mathbb {C}}})^T\qquad \text { and }\qquad \tilde{{\mathbb {C}}}^T=-\tilde{{\mathbb {C}}}. \end{aligned}$$

(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

$$\begin{aligned} \left\{ \tilde{{\varvec{\Gamma }}}_M,\tilde{{\varvec{\Gamma }}}_N\right\} = 2 \eta _{MN}{\mathbb {1}}_{32}. \end{aligned}$$

(A.22)

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

$$\begin{aligned} \tilde{{\varvec{\Gamma }}}_M= {\mathbb {M}_M}^{\underline{A}}\tilde{{\varvec{\Gamma }}}_{\underline{A}} \end{aligned}$$

(A.23)

with the transformation matrix

(A.24)

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

$$\begin{aligned} \eta _{MN}={\mathbb {M}_M}^{\underline{A}}\,{\mathbb {M}_N}^{\underline{B}} \,\eta _{\underline{A}\underline{B}}. \end{aligned}$$

(A.25)

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

(A.26)

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

(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 $\tilde{{\varvec{\Gamma }}}_M=(\tilde{{\varvec{\Gamma }}}_-,\tilde{{\varvec{\Gamma }}}_I,\tilde{{\varvec{\Gamma }}}_+).$ In the explicit representation (A.16), only $\tilde{{\varvec{\Gamma }}}_I$ can be written as

$$\begin{aligned} \tilde{{\varvec{\Gamma }}}_I=\sigma _1\otimes \Gamma _I \end{aligned}$$

(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., ${(\Gamma _I)_{A}}^{B'}={(\Gamma _I)_{A'}}^{B}.$ 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

$$\begin{aligned} \{\Gamma _I,\Gamma _J\}=2\eta _{IJ}{\mathbb {1}}_{16}. \end{aligned}$$

(A.29)

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

$$\begin{aligned} \Gamma _9=\Gamma _{1\ldots }\Gamma _{8}=\text {diag}({\mathbb {1}}_8,-{\mathbb {1}}_8), \end{aligned}$$

(A.30)

and the SO(4, 4) charge conjugation matrix, satisfying $\Gamma _I{\mathbb {C}}=(\Gamma _I{\mathbb {C}})^T$ and ${\mathbb {C}}^T={\mathbb {C}},$ is given by

$$\begin{aligned} {\mathbb {C}}=\sigma _3\otimes \sigma _3\otimes {\mathbb {1}}_2\otimes {\mathbb {1}}_2. \end{aligned}$$

(A.31)

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

(A.32)

where ${\textbf{c}}_{mn}={\textbf{c}}^{mn}=-{\textbf{c}}_{\dot{m}\dot{n}}=-{\textbf{c}}^{\dot{m}\dot{n}}=\text {diag}({\mathbb {1}}_4,-{\mathbb {1}}_4).$ Besides, we find that the SO(4, 4) gamma matrices are also chirally decomposed as

(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

(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

(A.35)

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

$$\begin{aligned} {({\varvec{t}}_{MN})_A}^B = {(\Gamma _{MN})_A}^B \end{aligned}$$

(A.36)

where

$$\begin{aligned} {(\Gamma _{MN})_A}^B=\frac{1}{2}\left[ {(\Gamma _M)_A}^{C'}{(\Gamma _N)_{C'}}^B -{(\Gamma _N)_A}^{C'}{(\Gamma _M)_{C'}}^B\right] .\nonumber \\ \end{aligned}$$

(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

(A.38)

The SO(4, 4) branching rule for spinor representation is now straightforward. For a given SO(5, 5) spinor in ${\textbf{16}}_s$ representation $\Psi _A,$ we have the decomposition

$$\begin{aligned} \Psi _A = \left( \begin{array}{c} \Psi _m \\ \Psi _{\dot{m}} \end{array}\right) . \end{aligned}$$

(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

$$\begin{aligned} \underbrace{{\textbf{16}}_s}_{\Psi _A}\ \rightarrow \ \underbrace{{\textbf{8}}^{+1}}_{\Psi _m}\,\oplus \,\underbrace{{\textbf{8}}^{-1}}_{\Psi _{\dot{m}}}. \end{aligned}$$

(A.40)

1.4 A.4 Vector-spinor

The vector-spinor representation of SO(5, 5) we are interested in is given by a $16\times 10$ matrix $\theta ^{AM}\in {\textbf{144}}_c$ 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 ${\textbf{144}}_c$ 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

(A.41)

It is straightforward to determine that ${\mathbb {R}}^+$ weights of $(\theta _1^{m -},\theta _2^{\dot{m} -},\theta _3^{m I},\theta _4^{\dot{m} I},\theta _5^{m +},\theta _6^{\dot{m}})$ are given by $(+1,+3,-1,+1,-3,-1).$ Finally, as in [32], the SO(4, 4) branching rule for ${\textbf{144}}_c$ representation is given by

$$\begin{aligned} \underbrace{{\textbf{144}}_c}_{\theta ^{AM}}\ \rightarrow \ \underbrace{{\textbf{56}}^{-1}}_{\vartheta _3^{m I}}\,\oplus \,\underbrace{{\textbf{56}}^{+1}}_{\vartheta _4^{\dot{m} I}}\,\oplus \,\underbrace{{\textbf{8}}^{-1}}_{\theta _6^{\dot{m} +}}\,\oplus \,\underbrace{{\textbf{8}}^{+1}}_{\theta _1^{m -}}\,\oplus \,\underbrace{{\textbf{8}}^{+3}}_{\theta _2^{\dot{m} -}}\,\oplus \,\underbrace{{\textbf{8}}^{-3}}_{\theta _5^{m +}}\nonumber \\ \end{aligned}$$

(A.42)

with

$$\begin{aligned}{} & {} \vartheta _3^{m I}=\theta _3^{m I}+\frac{\sqrt{2}}{8}{(\gamma ^I)^{m}}_{\dot{n}}\theta _6^{\dot{n} +}\quad \text {and}\nonumber \\{} & {} \vartheta _4^{\dot{m} I}=\theta _4^{\dot{m} I}-\frac{\sqrt{2}}{8}{(\gamma ^I)^{\dot{m}}}_n\theta _1^{n -}. \end{aligned}$$

(A.43)

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

$$\begin{aligned}{} & {} 0=\sqrt{2}{\textbf{c}}_{mn}\theta _1^{n -}-(\gamma _I)_{m\dot{n}}\theta _4^{\dot{n} I}\quad \text {and}\nonumber \\{} & {} 0=\sqrt{2}{\textbf{c}}_{\dot{m}\dot{n}}\theta _6^{\dot{n} +}+(\gamma _I)_{\dot{m}n}\theta _3^{n I}. \end{aligned}$$

(A.44)

It is also useful to note that $\vartheta _3^{m I}$ and $\vartheta _4^{\dot{m} I}$ satisfy the following conditions

$$\begin{aligned} (\gamma _I)_{\dot{m}n}\vartheta _3^{n I}=0\quad \text {and} \quad (\gamma _I)_{m\dot{n}}\vartheta _4^{\dot{n} I}=0. \end{aligned}$$

(A.45)