Supersymmetric domain walls in maximal 6D gauged supergravity II

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

In this paper, we are interested in supersymmetric domain walls from the maximal N = (2, 2) gauged supergravity in six dimensions constructed in [38], see [39] for an earlier result.The result of [38] describes the most general gaugings of N = (2, 2) supergravity in six dimensions using the embedding tensor formalism.The embedding tensor lives in representation 144 c of the global symmetry SO (5,5) and determines a viable gauge group G 0 ⊂ SO (5,5).In general, there are a large number of possible gauge groups.In this work, we will consider only gauge groups classified under GL(5) ⊂ SO (5,5) which is an off-shell symmetry of the N = (2, 2) supergravity Lagrangian in a particular symplectic frame.
We will study various possible gaugings and explicitly construct the corresponding embedding tensors for the resulting gauge groups.We will also look for supersymmetric domain wall solutions.According to the DW/QFT correspondence, these solutions are dual to maximally supersymmetric Yang-Mills theory (SYM) in five dimensions which recently plays an important role in studying dynamics of (conformal) field theories in higher and lower dimensions via a number of dualities, see for example [40][41][42][43][44][45].The five-dimensional SYM could be used to define N = (2, 0) superconformal field theory (SCFT) in six dimensions compactified on S 1 .Since the latter is well-known to describe dynamics of strongly coupled theory on M5-branes, we expect that supersymmetric domain walls of the maximal gauged supergravity in six dimensions could be useful in studying various aspects of the maximal SYM in five dimensions as well as six-dimensional SCFT and physics of M5-branes at strong coupling.
In addition, five-dimensional maximal SYM and compactifications on S 1 and S 2 can lead to some insights to S-duality of twisted gauge theories in four dimensions and monopoles in Aharony-Bergman-Jafferis-Maldacena (ABJM) theory.Therefore, the resulting domain walls could also be useful in this context as well.However, it should be pointed out that all gaugings classified here currently have no known higher dimensional origins.
Since N = 4 superconformal symmetry in five dimensions does not exist [46], see also a recent classification of maximally supersymmetric AdS vacua given in [47], there is no AdS 6 /CFT 5 duality with 32 supercharges.Accordingly, supersymmetric domain walls in N = (2, 2) gauged supergravity are expected to play a prominent role in holographic study in this case.A domain wall solution with SO(5) symmetry in SO(5) gauging has been found in [39] with the holographic interpretation given in [48] and [49].Moreover, a large number of domain wall solutions has been given recently in [25] with various gauge groups obtained from the embedding tensor in 15 −1 and 40 −1 representations of GL (5).We will extend this investigation by considering the embedding tensor in other representations of GL (5).These are given by 5 +7 , 5+3 , 10 −1 , 24 −5 , and 45 +3 representations.It turns out that most of the resulting gauge groups are of non-semisimple type without any compact subgroup.Due to the complexity of working with the full 25-dimensional scalar manifold SO(5, 5)/SO(5)×SO (5), we mainly follow the method introduced in [50] to reduce the number of scalar fields to make the analysis more traceable.However, this approach requires the existence of a non-trivial symmetry H 0 ⊂ G 0 .In many of the gaugings classified here, the residual H 0 is absent, so there are too many scalars to take into account.Accordingly, we will give domain wall solutions only for gauge groups with at least SO(2) subgroup.As in [25], there exist both 1 2 -and 1 4 -supersymmetric domain wall solutions in accord with the general classification of supersymmetric domain walls given in [37].
The paper is organized as follows.In section II, we briefly review the construction of six-dimensional maximal gauged supergravity in the embedding tensor formalism.Gaugings in 5 +7 , 5+3 , 10 −1 , 24 −5 , and 45 +3 representations of GL (5) are classified in section III.
In section IV, we explicitly construct a number of supersymmetric domain wall solutions.
Conclusions and discussions are given in section V, and useful formulae are given in the appendix.

II. N = (2, 2) GAUGED SUPERGRAVITY IN SIX DIMENSIONS
We first give a review of six-dimensional N = (2, 2) gauged supergravity in the embedding tensor formalism constructed in [38].We will mainly collect relevant formulae for constructing the embedding tensor in order to classify various gauge groups and find supersymmetric domain wall solutions.For more details, the reader is referred to the original construction in [38].
There is only one supermultiplet in N = (2, 2) supersymmetry in six dimensions, the graviton supermultiplet, with the following field content We note here all the conventions used throughout the paper.These mostly follow those used in [38].In chiral spinor representation, the coset manifold SO(5, 5)/SO(5)×SO( 5) is described by a coset representative V A α β transforming under the global SO(5, 5) and local SO(5) × SO (5) symmetries by left and right multiplications, respectively.The inverse elements (V −1 ) α β A will be denoted by V A α β satisfying the relations On the other hand, in vector representation, the coset representative is given by a 10 × 10 ).This is related to the coset representative in chiral spinor representation by the following relations In these equations, (Γ M ) AB and (Γ A ) α ) are respectively SO (5,5) gamma matrices in non-diagonal η M N and diagonal η AB bases.
The inverse will be denoted by V M A satisfying the following relations and In these equations, we have explicitly raised the SO(5) × SO( 5) resulting in a minus sign in equation ( 6).
The most general gaugings of six-dimensional N = (2, 2) supergravity are described by the embedding tensor Θ A M N leading to the following covariant derivative where g is a gauge coupling constant.The embedding tensor identifies generators X A = Θ A M N t M N of the gauge group G 0 ⊂ SO(5, 5) with particular linear combinations of the SO(5, 5) generators t M N .Supersymmetry requires the embedding tensor to transform as 144 c representation of SO (5,5).Accordingly, Θ A M N can be parametrized in terms of a vector-spinor θ AM of SO (5,5) as with θ AM subject to the constraint The SO(5, 5) generators in vector and spinor representations can be chosen as with η M N being the off-diagonal SO(5, 5) invariant tensor given by and We also note that the notation 1 n denotes an n × n identity matrix.(Γ M ) A B are chirally projected SO(5, 5) gamma matrices.
The corresponding gauge generators in these representations then take the forms Furthermore, consistency requires the gauge generators to form a closed subalgebra of SO(5, 5) implying the quadratic constraint In terms of θ AM , the quadratic constraint reduces to the following two conditions It follows that any θ AM ∈ 144 c satisfying this quadratic constraint defines a consistent gauging.
In this work, we are only interested in the classification of gauge groups under GL(5) ⊂ SO(5, 5) and domain wall solutions which only involve the metric and scalar fields.Therefore, we will, from now on, set all vector and tensor fields to zero with the bosonic Lagrangian of the maximal N = (2, 2) gauged supergravity given by We also need supersymmetry transformations of fermionic fields which, for vanishing fermions and vector/tensor fields, are given by The covariant derivatives of supersymmetry parameters, ǫ +α and ǫ − α, are defined by with γµ = e μ µ γμ .γμ are space-time gamma matrices, and for simplicity, we will suppress space-time spinor indices.
The scalar vielbein P a ȧ µ and SO(5) × SO(5) composite connections, Q ab µ and Q ȧḃ µ , are given by In these equations, Ω αβ and Ω α β are the USp(4) symplectic forms that satisfy the following relations and similarly for Ω α β .We will use the explicit form of Ω αβ and Ω α β given by We also note that the definitions ( 24), (25), and ( 26) can be derived from the following relation The scalar potential is given by with the T-tensors defined by and III. GAUGINGS OF N = (2, 2) SUPERGRAVITY UNDER GL (5) In this section, we consider gaugings under GL(5) ⊂ SO (5,5).The embedding tensor θ AM in 144 c representation of SO(5, 5) decomposes under GL(5) as The SO (5,5) spinor representation decomposes as Accordingly, the gauge generators can be written in terms of X m , X mn , and X * denoting respectively 5 +3 , 10 −1 , and 1 −5 as The decomposition matrices T Am , T mn A , and T A * are given in the appendix.By the decomposition of SO(5, 5) vector representation we can write the embedding tensor as The two components θ Am and θ A m contain the following irreducible GL(5) representations As pointed out in [38], gaugings triggered by θ Am are called electric gaugings in the sense that only electric two-forms participate in the gauged theory while gaugings triggered by θ A m are called magnetic gaugings involving magnetic two-forms together with additional three-form tensor fields.The decompositions in (38) and (39) imply that gaugings in 24 −5 ⊕ 40 −1 and 5 +7 ⊕ 15 −1 ⊕ 45 +3 representations are respectively purely electric and purely magnetic while those in 5 +3 ⊕ 10 −1 representations correspond to dyonic gaugings.Many possible dyonic gaugings can also arise from combinations of various electric and magnetic components.
Finally, we note that the quadratic constraint (15) is automatically satisfied for purely electric or purely magnetic gaugings that involve only θ Am or θ A m components.In accord with (38) and (39), we can parametrize the embedding tensor as, see [25] for more detail, Matrices T Am , T A mn , and T A * are inverses of the decomposition matrices T Am , T mn A , and T A * given by complex conjugations, T A = (T A ) −1 = (T A ) * .
We now look at various components in more detail.The first representation 24 −5  satisfying W nm m = 0. Finally, the last two 5 +3 and 5 +7 representations are parametrized by two GL(5) vectors J m and K m , respectively.The general structure of gaugings under GL(5) ⊂ SO(5, 5) is shown in Table I taken from [38].The left column represents the sixteen vector fields in GL( 5) representations while the top row corresponds to the decomposition of SO(5, 5) generators under GL (5).The table shows the couplings between SO(5, 5) generators and vector fields by various components of the embedding tensor.
In the present work, we will focus on the remaining representations and examples of possible combinations shown in Table I.In the following, we will determine an explicit form of the aforementioned GL(5) tensors by imposing the quadratic constraints ( 15) and ( 16) on the embedding tensor under consideration.We begin with gauge groups arising from the embedding tensor in 24 −5 representation.
Gaugings in this representation are purely electric and triggered by With θ A m = 0, the embedding tensor θ AM = ( T An S n m , 0 ) automatically satisfies the quadratic constraint.Therefore, every traceless 5 × 5 matrix S m n defines a viable gauge group generated by the following gauge generators τ m n are SL(5) generators defined as with τ m m = 0 and being the SO(1, 1) ∼ R + generator in GL( 5) ∼ R + × SL (5).We also use s mn = t mn to denote the generators corresponding to shift symmetries on the scalar fields, see more detail in the appendix.
Commutation relations between the gauge generators read where the generators with two antisymmetric pairs of GL(5) vector indices are defined as From equation ( 46), we readily see that the generators X mn form a translational group and transform non-trivially under a one-dimensional group generated by X * , a particular linear combination of SL(5) generators.
In vector or fundamental representation of SL( 5), X * generator takes the form If S m n is antisymmetric, X * will also be antisymmetric and generates a compact SO(2) group.On the other hand, for symmetric S m n , X * generates a non-compact SO(1, 1) group.
The resulting gauge groups then take the form of for 3 ≤ n ≤ 10.The values of n depend on the choices of S m n .It has been pointed out in [38] that this type of gaugings is related to Scherk-Schwarz reductions from seven-dimensional gauged supergravity, and gaugings in 24 −1 representation correspond to choosing a generator from the seven-dimensional symmetry group SL (5), see also [51] for a general discussion on Scherk-Schwarz reductions and gauged supergravities.
We end this case by giving an explicit example with X * antisymmetric.By choosing we find the following non-vanishing gauge generators in which i, j = 1, 2 and x, y = 4, 5.It can be straightforwardly verified that these gauge generators satisfy the commutation relations given in (46).The resulting gauge group is of the form in which SO(2) is generated by X * , and the shift translational group R 8 s is generated by X ix , X i3 , and X x3 .

B. Gaugings in 45
+3 representation We now consider the embedding tensor in 45 +3 representation.Gaugings in this case are purely magnetic and related to reductions of eleven-dimensional supergravity on twisted tori as pointed out in [38].The linear constraint (9) and the quadratic constraint (15) are both satisfied if we parametrize the embedding tensor as In terms of W np m , the constraint ( 16) reduces to As noted in [38], this constraint is the duality of a similar condition in gaugings from 40 −1 representation (U mn,r U pq,s ε mnpqt = 0), considered in [25].To solve this condition, we then follow the same procedure as in 40 −1 representation.We first write W np m in the form of and impose the condition u m m = 0 in order to satisfy the traceless condition W nm m = 0.This form of W np m is sufficient to solve the condition (54).To give an explicit example, we will choose a basis in which v m = δ m 5 and u 5 m = u m 5 = 0.
As a result, consistent gaugings satisfying the linear and quadratic constraints in 45

+3
representation are now parametrized by a traceless 4 × 4 matrix u i j with i, j = 1, ..., 4 and In this case, the gauge generators turn out to be X * = X i5 = 0 and the remaining non-vanishing generators given by We recall that h mn = t mn are SO(5, 5) generators associated with the hidden symmetries that do not constitute symmetries of the action.Commutation relations between the gauge generators read As seen from these relations, X ij generate a six-dimensional translational group R 6 h associated with the hidden symmetries while X i generate another translational group R 4 commuting with R 6 h .The X 5 generator takes the same form as X * in 24 −1 representation.We can similarly use the unbroken SL(4) symmetry to fix u i j in the form of With this explicit form of u i j , non-vanishing gauge generators are now given by where i = ( ī, x) with ī, j, ... = 1, 2 and x, ȳ, ... = 3, 4. Therefore, for arbitrary values of κ and λ, the corresponding gauge groups are given by or depending on the choices of the signs for κ and λ, making u i j in (58) respectively symmetric or antisymmetric.

C. Gaugings in 5 +7 representation
In this section, we look at the smallest representation giving rise to purely magnetic gaugings.With θ Am = 0, we parametrize the embedding tensor, in this case, by a GL( 5) Gaugings triggered by θ AM = ( 0 , T A * K m ) automatically satisfy the linear and quadratic constraints.As pointed out in [38], these gaugings might correspond to reductions from eleven dimensions with non-trivial four-form fluxes.
Only gauge generators X m are non-vanishing in this representation.They are explicitly given in terms of the generators of the hidden symmetry by These generators commute with each other However, due to the antisymmetric property of h nm , the gauge generators satisfy the condition K m X m = 0. Therefore, only four generators are linearly independent.The resulting gauge group in 5 +7 representation is then given by a four-dimensional translational group R 4 h associated with the hidden symmetries.
To explicitly parametrize this R 4 h gauge group, we will use the SL(5) symmetry to fix K m = κ δ 5 m with κ ∈ R. In this case, the gauge group is generated by for i = 1, 2, 3, 4.

D. Gaugings in 5 +3 representation
We now consider dyonic gaugings involving both electric and magnetic components of the embedding tensor.We begin with gaugings in 5 +3 representation.The embedding tensor in this representation, satisfying the linear constraint (9), is given by θ AM = (θ Am , θ A m ) with This form of the embedding tensor automatically satisfies the quadratic constraints given in (15) and (16).Therefore, any vectors J m define consistent gaugings in 5 +3 representation.
In this case, we find non-vanishing gauge generators given by with the following commutation relations To write down the explicit form of gauge generators, we can use SL( 5) symmetry to fix J m = κ δ m 5 as in the previous case.With this choice, non-vanishing gauge generators are given by with i, j, ... = 1, ..., 4. In vector representation of SL(5), the generator X 5 takes the form Commutation relations between these gauge generators become These imply the gauge group of the form in which the non-compact factor SO(1, 1) is generated by X 5 .

E. Gaugings in 10 −1 representation
In addition to gaugings from 5 +3 representation, gaugings in 10 −1 representation also require both θ Am and θ A m to be non-vanishing in order to satisfy the linear constraint.Thus, gaugings in this representation are dyonic and triggered by with . With this embedding tensor, the quadratic constraints in (15) and (16) reduce to As pointed in [38], this condition can be solved by Z mn of the form with u m and v n being arbitrary GL(5) vectors.
In this case, the corresponding gauge generators are given by with the commutation relations To determine the explicit form of possible gauge groups, we repeat the same procedure as in the previous cases by considering a particular parametrization of the two vectors u m and v m in the form of The gauge generators in this case become with m = (i, x) for i, j, ... = 1, 2, 3 and x, y, ... = 4, 5.The commutation relations read giving rise to the gauge group of the form The non-compact subgroup SO(1, 1) is generated by the gauge generator X 45 whose explicit form in GL(5) vector representation is given by The three commuting translational groups R 6 , R 3 s , and R h are respectively generated by X ix , X i , and X * .We now consider dyonic gaugings arising from combining two components of the embedding tensor in 5 +3 and 45 +3 representations.These gaugings are also dyonic since the embedding tensor in 5 +3 representation contains both electric and magnetic parts.The linear constraint (9) requires these components to take the form This is just a trivial combination between the embedding tensor from each representation given in ( 53) and (66).
With this form of the embedding tensor, the quadratic constraints ( 15) and ( 16) become To solve these conditions, we first fix the explicit form of the GL(5) vector J m = κδ m 5 with κ ∈ R, leading to a split of a GL(5) index m = (i, 5) with i, j, ... = 1, 2, 3, 4 as in section III D. With J m of this form, the condition (89) implies W mn 5 = 0, so only W ij k and W i5 j are non-vanishing.Note also that the condition W mn n = 0 imposes a traceless condition With only W ij k , W i5 j , and J 5 = κ non-vanishing, the condition (90) splits into The first condition takes the same form as the quadratic constraint (54) in 45 +3 representation and can be similarly solved by choosing with a GL(4) vector v i and a traceless 4 × 4 matrix w k j .
At this point, we can further use SL(4) ⊂ GL(5) symmetry to rotate v i such that v i = λδ i 4 .We will also split the index i = (x, 4) with x = 1, 2, 3.Moreover, the traceless condition W ij j = 0 requires that w 4 i = 0.For simplicity, we will also set w i 4 = 0 since these components do not appear in the resulting embedding tensor.The remaining components w x y can be described by a traceless 3 × 3 matrix satisfying w x x = 0.Among various components of x ), only the last components are non-vanishing and given by With all these, the condition (92) splits into The first condition can be solved by setting W x5 4 = 0.The remaining component W x5 y can be written in terms of a 3 × 3 matrix u y x as W 5x y = τ u y x with τ ∈ R. The condition (96) then gives rise to With the Schouten identity, w [x t ε yzt] = 0, this condition can be solved if and only if τ = − κ 2 and u x y = δ y x .In conclusion, the quadratic constraint determines the embedding tensor (88) in terms of J m and W mn p given by We note that W 45 x written in terms of a three-dimensional vector v x in the last relation do not appear in the linear and quadratic constraints.Therefore, this vector is unconstrained.
With all these, non-vanishing gauge generators are given by with the following commutation relations We find that the only possible compact subgroup is SO(2) generated by the gauge generator X 4 with w x y antisymmetric for any value of v x .If the matrix w x y is symmetric, the generator X 4 gives a non-compact SO(1, 1) group.For simplicity, we will set v x = 0 and restrict ourselves to the compact SO(2) case since we are mainly interested in domain wall solutions preserving some symmetry.For definiteness, we choose the matrix w x y to be with σ ∈ R. Together with v x = 0, the above gauge generators reduce to with x = (x, 3), x = 1, 2.
In vector representation, we can explicitly see that X 4 and X 5 are antisymmetric and symmetric generating SO(2) and SO(1, 1) subgroups, respectively.The gauge generators satisfy the following commutation relations From these relations, we see that (X 4 , X x5 ) and (X 3 , X 5 , X 12 ) form two commuting nonsemisimple groups ISO(2) and ISO(1, 1), respectively.The remaining generators (X x, X 3x ) generate a four-dimensional translational group transforming non-trivially under ISO(2) × ISO(1, 1).The gauge group is then given by For a simpler case of λ = 0, the non-vanishing components W 5x y still generate a non-trivial subgroup.In this case, we find X 4 = X x5 = 0 giving rise to the following gauge group The three factors SO(1, 1), R 3 , and R 3 h are respectively generated by X 5 , X xy , and X x .We also note that another possibility of setting κ = 0 is not possible since this choice leads to vanishing J m .G. Gaugings in (10 + 15) −1 representation For gaugings with the embedding tensor in (10 + 15) −1 representation, there are both electric and magnetic components θ Am and θ A m given by We recall that Y mn = Y (mn) and Z mn = Z [mn] are symmetric and antisymmetric tensors corresponding to 15 −1 and 10 −1 representations, respectively.In terms of Y mn and Z mn , the quadratic constraints (15) and ( 16) reduce to We can use the SL(5) ⊂ GL(5) symmetry to bring Y mn to a diagonal form where p + q + r = 5.We will split the GL(5) index m = (i, x) and Y mn into Y ij = diag(1, .., 1, −1, .., −1) with Y xy = 0. Solving the first condition in (112) using (113), we find only two possible solutions with both Y mn and Z mn non-vanishing for ranks of Y mn equal to 1 and 2.

RankY = 2
With Y ij = diag(1, ±1) for i, j, ... = 1, 2 and x, y, ... = 3, 4, 5, only Z 12 = κ with κ ∈ R is allowed to be non-vanishing.This automatically solves the other condition in (112) and gives rise to the following gauge generators Commutation relations between these generators are given by The generators X * , X x , and X ix generate three translational groups R h , R 3 s , and R 6 commuting with each other.For Y ij = diag(1, −1), the generator X 12 can only generate a non-compact SO(1, 1) group giving rise to the following gauge group For Y ij = δ ij , X 12 can become compact SO(2), non-compact SO(1, 1), or nilpotent generators depending on the values of κ.For particular values of κ = ± 3 5 , X 12 is nilpotent and will be denoted by T resulting in T ⋉ (R 6 × R 3 s × R h ) gauge group.For − 3 5 < κ < 3 5 and κ < − 3 5 or κ > 3 5 , X 12 is respectively compact and non-compact.Accordingly, the corresponding gauge group is given by or the gauge group given in (119).

RankY = 1
In this case, we will choose only Y 11 non-vanishing with Y 11 = κ = ±1.All components Z xy with x, y = 2, 3, 4, 5 need to be zero in order to solve the quadratic constraint (112).
Therefore, only the remaining four components Z 1x = z x give rise to the following gauge generators with the commutation relations To proceed further, we will choose z 2 = ζ and z 3 = z 4 = z 5 = 0.The above gauge generators simplify to with x, ȳ, ... = 3, 4, 5.The corresponding commutation relations read with ī, j, ... = 1, 2.
We also note that the generator X 12 does not commute with other generators.This generator generates an SO(1, 1) group while the remaining generators form three commuting translational groups.Accordingly, the resulting gauge group is given by in which the translational groups R 6 , R 3 s , and R h are respectively generated by X¯i x, X x, and X * .with U [mn,p] = 0.The embedding tensor solving the linear constraint (9) takes the form The quadratic constraint (15) imposes the following conditions U mn,p Z pq = 0, (127) while the other condition in (16) gives The condition ( 128) is similar to that given in the case of 10 −1 representation.Therefore, we will use the same ansatz for Z mn as given in (75).Moreover, as in section III E, we will further fix each GL( 5) vector in the ansatz Z mn = u [m v n] in order to see the corresponding gauge group explicitly.However, instead of using u m = (κ 1 , κ 2 , 0, 0, 0) and v m = (λ 1 , λ 2 , 0, 0, 0), we equivalently set Z 12 = κ with other components vanishing.The condition ( 128) is then automatically satisfied while the condition (127) can be solved if and only if We have split the GL( 5) index as m = (i, x) with i, j, ... = 1, 2 and x, y, ... = 3, 4, 5.
At this point, only U ix,y and U xy,z remain in terms of which the last condition in (129) leads to These conditions can be readily solved by setting U ix,y = 0.This is very similar to gaugings from (15 + 40) −1 representation with rankY = 2 studied in [25].Therefore, in order to solve the quadratic constraints for gaugings in (10 + 40) −1 representation, we are left with only the following non-vanishing components We have parametrized the components U xy,z in terms of a traceless 3 × 3 matrix u x y with The corresponding gauge generators are given by with the following commutation relations The gauge generator X 12 given in (137) generates either SO(1, 1) or SO(2) group for symmetric and antisymmetric u x y , respectively.The corresponding gauge group is given by or in which X ix , X x , and X * respectively generate the translational groups R 6 , R 3 s , and R h .
The quadratic constraints in ( 15) and ( 16) give together with To solve these conditions, we repeat the same procedure as in the previous case by setting all Z mn components to zero except Z 12 = κ together with Y mn = diag(1, −1, 0, 0, 0).With this choice together with the index splitting m = (i, x) for i, j, ... = 1, 2 and x, y, ... = 3, 4, 5, we can solve all the above conditions with all U mn,p components vanishing except This is the same ansatz (133) as used in the previous case.
These lead to the gauge generators with the same commutation relations as given in (138).The corresponding gauge group is again given by ( 139) or (140) depending on the values of κ and the form of u x y as in the previous case.It could be useful to look for more complicated solutions leading to new gauge groups.

IV. SUPERSYMMETRIC DOMAIN WALL SOLUTIONS
In this section, we will find supersymmetric domain wall solutions from six-dimensional gauged supergravities constructed from the embedding tensors given in the previous section.
We begin with a general procedure of finding supersymmetric domain walls.The analysis has already been performed in [25], so we will only review relevant formulae and refer to [25] for more detail.The metric ansatz for domain wall solutions take the general form of with μ, ν = 0, 1, . . ., 4.
In order to find supersymmetric domain wall solutions, we consider first-order Bogomol'nyi-Prasad-Sommerfield (BPS) equations derived from the supersymmetry transformations of fermionic fields.We begin with the variations of the gravitini from ( 18) and ( 19) which give Throughout the paper, we use the notation ′ to denote an r-derivative.Multiply equation (157) by A ′ γr and use equation ( 158) or vice-versa, we find the following consistency conditions in which we have introduced the "superpotential" W. We then obtain the BPS equations for the warped factor With this result, equations ( 157) and (158) lead to the following projectors on the Killing spinors It should be noted that these projectors are not independent.The conditions δψ +rα = 0 and δψ −r α = 0 will determine the Killing spinors as functions of the radial coordinate r as usual.We will give the corresponding expressions in explicit solutions obtained in subsequent analyses.
Using the γr projectors in δχ +a α = 0 and δχ − ȧα = 0 equations, we eventually obtain the BPS equations for scalar fields of the form in which G IJ is the inverse of the scalar metric G IJ .In addition, the scalar potential can also be written in terms of W as We are now in a position to find explicit domain wall solutions.As mentioned before, the complexity of working with all 25 scalars prohibits any traceable analysis.We will consider only solutions preserving a non-trivial residual symmetry.The solutions invariant under this symmetry are characterized by a smaller set of singlet scalars.Among the gauge groups classified in the previous section, only five gauge groups, SO(2) ⋉ R 8 , SO(2) ⋉ R 6 , CSO(2, 0, 2) ⋉ R 4 , CSO(2, 0, 2) ⋉ R 2 , and CSO(2, 0, 1) ⋉ R 4 , contain a compact SO(2) subgroup that can be used to reduce the number of scalar fields and result in consistent sets of BPS equations.Accordingly, in the following, we will consider only domain wall solutions for these gauge groups.We should also remark here that for (ISO(2) × ISO(1, 1)) ⋉ R 4   and CSO(2, 0, 3) ⋉ R It should be noted that the first non-compact generator corresponds to the SO(1, 1) factor defined in (155).This scalar is actually a singlet under the full compact subgroup SO(5) ⊂ GL (5).The scalar field corresponding to this generator will be called the dilaton ϕ.Additionally, there are two SL(5)/SO(5) scalars and two shift scalars invariant under the SO(2) symmetry.These are associated with Y 2,3 and Y 4,5 , respectively.
The coset representative can be written as Using this coset representative, we find that the scalar potential vanishes identically, and only the dilaton appears in the T-tensors which take the form of In these equations, we have written S b a = S ḃ ȧ = S m n .With S m n given in ( 50), the T-tensors become or explicitly There are two possible superpotentials Both of them give a valid superpotential in terms of which the scalar potential can be written as (164) using the matrix G IJ of the form for Φ I = {ϕ, φ 1 , φ 2 , ς 1 , ς 2 } with I = 1, 2, 3, 4, 5.It should be noted that for λ = 0, the two superpotentials are equal leading to half-supersymmetric domain walls.For λ = 0, one choice of the superpotentials gives rise to 1 4 -BPS domain walls since only the supersymmetry along the directions of this choice is unbroken.
We now consider the cases of λ = ±κ and λ = 0 corresponding to SO(2) ⋉ R 8 gauge group.In this case, choosing one of the two superpotentials corresponds to imposing an additional projector on the Killing spinors of the form for W = W 1 or W = W 2 , respectively.Together with the γr projector in (162), the resulting solutions will preserve only eight supercharges or 1 4 of the original supersymmetry.With all these, we obtain the following BPS equations It should be noted that although the superpotential does not depend on ς 1 and ς 2 , there are non-trivial BPS equations for these scalars due to the cross terms between these scalars and the dilaton in G IJ .
The plus/minus signs in the BPS equations are correlated with the plus/minus sign of the γ 3 projector (176).The solution is given by for constants C and ς (0) 1,2 .The remaining two scalars φ 1 and φ 2 are constant.It turns out that in this case, all the composite connections Q ab µ and Q ȧḃ µ vanish.The BPS equations from δψ +rα and δψ −r α then give the following Killing spinors with the constant SMW spinors ǫ 0 ± satisfying the projectors (162) and ( 176).In this solution, an integration constant for A has been neglected by rescaling the coordinates x μ.Note that the integration constant C and ς (0) 1,2 can also be removed by shifting the radial coordinate r and scalars ς 1,2 , but we keep it here for generality.We also note that from the BPS equations, we can consistently set φ 1 = φ 2 = ς 1 = ς 2 = 0. Indeed, it can be verified that redefining the shift scalars as ς 1,2 → ς1,2 = e −4ϕ ς 1,2 results in a set of BPS equations with ς′ 1,2 = 0.For λ = ±κ or λ = 0, the translational group R 8 s reduces to R 6 s .In the former case, X 14 = ±X 25 and X 15 = ∓X 24 while in the latter X 34 and X 35 vanish.In these two cases, there are more SO(2) singlet scalars.For λ = ±κ, there are four additional singlet scalars corresponding to non-compact generators The upper/lower signs in these generators are related to the upper/lower sign in λ = ±κ.
For λ = 0, we find six additional singlets given by Therefore, the solutions for these two special cases are given by the above solution with the substitution λ = ±κ and λ = 0.However, as mentioned before, solutions with λ = 0 is 1 2 -supersymmetric since in this case the two superpotentials W 1 and W 2 are equal.On the other hand, solutions with λ = ±κ preserve 1 4 of the original supersymmetry as in the more general case of λ = ±κ.Each choice of λ = ±κ makes one of the superpotentials vanish rendering the supersymmetry along the directions of this superpotential broken.
1. SO(2) symmetric domain walls from CSO(2, 0, 2) ⋉ R 4 gauge group With λ = ±κ, the gauge group is given by CSO(2, 0, 2) ⋉ R 4 as in (61).For convenience, we note the explicit form of the matrix u i j of the gauge group (61) where κ and λ are non-vanishing and λ = ±κ.There are five SO(2) singlet scalars containing the dilaton, two scalars from SL(5)/SO(5), and two shift scalars corresponding to the following non-compact generators The first non-compact generator corresponds to the dilaton ϕ as in the previous section.
With the coset representative given by the scalar potential vanishes identically as in the previous section.The T-tensor is more complicated due to the dependence on the shift scalars with the following form This gives rise to In this case, the two shift scalars ς 1 and ς 2 cannot be removed by a redefinition of scalar fields, and the composite connections Q ab µ and Q ȧḃ µ are non-vanishing.Therefore, as in [25], we need to modify the ansatz for the Killing spinors to in which B(r) is an r-dependent function.We now impose the γ 5 projector which implies γ 12 ǫ + = ∓γ 34 ǫ + and γ 1 2ǫ − = ∓γ 3 4ǫ − .The latter two conditions are sufficient to solve the conditions δψ +rα = 0 and δψ −r α = 0 with ǫ 0 ± being constant SMW spinors making ǫ ± satisfy the projectors (162) and (197).
2. SO(2) symmetric domain walls from CSO(2, 0, 2) ⋉ R 2 gauge group For λ = ±κ, we find that the gauge generators X 13 = ±X 24 and X 14 = ∓X 23 .This reduces the translational group R 4 h to R 2 h resulting in the gauge group of the form Using the coset representative we again find that the scalar potential vanishes identically.Using the γ 5 projector given in (197), we find that the superpotential is given by W = ge −3ϕ−8φ 0 κ 4ς 2 1 + 4ς 2 2 + 16ς 2 1 ς 2 2 + 32ς 1 ς 2 (ς 2 3 + ς Following [25], we will call the four scalars associated with t+ i 5, for i = 1, 2, 3, 4, nilpotent scalars.It turns out that the two nilpotent scalars associated with t+ 3 5 and t+ 4 5 need to vanish in order to find a consistent set of BPS equations in this case. As in all the previous cases, it turns out that the scalar potential vanishes.The superpotential is still given by (200) for λ = 0.This gives rise to the same set of BPS equations gauge groups.These solutions are either 1 2 -or 1 4 -supersymmetric preserving SO(2) symmetry.Some of the resulting gauge groups could have higher dimensional origins in terms of Scherk-Schwarz reductions from seven dimensions or truncations of eleven-dimensional supergravity on twisted tori (possibly) with fluxes.The consistent gaugings identified in this paper to some extent enlarge the known gauge groups pointed out in [38] and those constructed in [25].These would hopefully be useful in the context of DW/QFT correspondence and related aspects.The domain wall solutions given here are exhaustive for all the gaugings under consideration at least for domain walls with any residual symmetry.These could also be added to the known classification of supersymmetric domain walls.It would be useful to find more general and more complicated solutions without any residual symmetry.
Constructing truncation ansatze for embedding the domain wall solutions found here in string/M-theory using SO(5, 5) exceptional field theory given in [52] is of particular interest.This would provide a complete framework for a holographic study of five-dimensional maximal SYM.It is also interesting to perform a similar analysis for gaugings under SO(4, 4) ⊂ SO(5, 5) and construct the corresponding embedding tensors together with possible supersymmetric domain walls.These gaugings can be truncated to gaugings of half-maximal N = (1, 1) supergravity coupled to four vector multiplets in which supersymmetric AdS 6 vacua are known to exist [53][54][55].In this case, the results could be useful in the study of both DW 6 /QFT 5 duality and AdS 6 /CFT 5 correspondence as well.
Finally, generalizing the standard domain wall solutions in this paper and [25] to curved domain walls with non-vanishing vector and tensor fields could also be worth considering.This type of solutions describes conformal defects or holographic RG flows from fivedimensional N = 4 super Yang-Mills theories to lower-dimensional (conformal) field theories via twisted compactifications.A number of similar solutions in seven dimensions have been given in [56,57] while examples of holographic solutions dual to surface defects in fivedimensional N = 2 SCFTs from N = (1, 1) gauged supergravity have appeared recently in [58].
is described by a traceless 5 × 5 matrix S n m with S m m = 0.The tensor U np,m = U [np],m satisfying U [np,m] = 0 corresponds to 40 −1 representation.The symmetric tensor Y mn = Y (mn) and antisymmetric one Z mn = Z [mn] respectively denote 15 −1 and 10 −1 representations.The 45 +3 representation is written as the tensor W np m = W [np] m

H.
Gaugings in (10 + 40) −1 representation We now move to the next case with the embedding tensor in 10 −1 and 40 −1 representations.These two components are labelled respectively by Z mn = Z [mn] and U mn,p = U [mn],p

Y 10 =
s 34 , Y 11 = s 35 .(185) It turns out that even with all these extra SO(2) singlet scalars, the scalar potential still vanishes identically.Moreover, the T-tensors depend only on the dilaton and take the same form as given in (171).The resulting BPS equations for scalars from SL(5)/SO(5) associated with Y 6,7 and Y 6,7,8,9 give constant scalars.Similar to the previous case, the shift scalars associated with Y 8,9 and Y 10,11 can be redefined such that the corresponding BPS equations give constant scalars.All these constant scalars can in turn be consistently set to zero.

)Y 8 =
Apart from the five singlet scalars corresponding to non-compact generators in (187) to (191), there are additional four scalars invariant under the SO(2) subgroup generated by X 5 with λ = ±κ.These extra scalars correspond to the following non-compact generators s 13 ∓ s 24 , Y 9 = s 14 ± s 23 .

TABLE I :
Gauge couplings between the sixteen vector fields and SO(5, 5) generators from various GL(5) components of the embedding tensor.A. Gaugings in 24 −5 representation (50)ge groups given in (109), (120), and (140), we are not able to find consistent sets of BPS equations.The supersymmetry conditions from (δψ +µα , δψ −µ α) and .Supersymmetric domain walls from SO(2) ⋉ R 8 and SO(2) ⋉ R 6 gauge groupsWe now consider supersymmetric domain walls from SO(2) ⋉ R 8 and SO(2) ⋉ R 6 gauge groups with the embedding tensor given in(42)and the traceless matrix S m n in(50).We (δχ +a α, δχ − ȧα ) are not compatible with each other.Therefore, we argue that no supersymmetric domain walls with SO(2) symmetry exist in these gauge groups.A