# Five-branes wrapped on topological disks from 7D $N=2$ gauged supergravity

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

###### Abstract

We study supersymmetric $AdS_{5}\times\Sigma$ solutions, for $\Sigma$ being a
topological disk with non-trivial $U(1)$ holonomy at the boundary or “half-
spindle”, in seven-dimensional $N=2$ gauged supergravity coupled to three
vector multiplets. We consider compact and non-compact gauge groups,
$SO(4-p,p)$, for $p=0,1,2$, and find a number of $AdS_{5}\times\Sigma$
solutions with $SO(2)\times SO(2)$ and $SO(2)_{\text{diag}}$ residual symmetry
from $SO(4)$ and $SO(2,2)$ gauge groups. We also find an $SO(2)_{R}\subset
SO(3)_{R}\sim SU(2)_{R}$ symmetric solution which can be regarded as a
solution of pure $N=2$ gauged supergravity with $SO(3)_{R}$ gauge group and
all the fields from vector multiplets vanishing. The solutions preserve
$\frac{1}{2}$ of the original $N=2$ supersymmetry and could be interpreted as
supergravity duals of $N=1$ superconformal field theories in four dimensions.
In particular, some of these solutions can be embedded in ten or eleven
dimensions in which a description in terms of five-branes wrapped on a
topological disk can be given.

##  I Introduction

The AdS/CFT correspondence maldacena ; Gubser_AdS_CFT ; Witten_AdS_CFT  leads
to holographic descriptions of strongly coupled superconformal field theories
(SCFTs). Various aspects of these SCFTs including their non-conformal phases
can be studied by the corresponding dual gravity solutions in string/M-theory
or, at low energy, supergravity theories in ten or eleven dimensions. In this
framework, the conformal field theories can be considered as world-volume
theories on the branes in the near horizon limit.
A class of solutions that describes branes wrapping on particular manifolds is
of particular interest since these can lead to supersymmetric field theories
in lower dimensions arising from world-volume theories of wrapped branes.
These configurations describe RG flows across dimensions from higher-
dimensional SCFTs to lower-dimensional ones and provide a useful holographic
description of the less-known higher-dimensional SCFTs such as $N=(2,0),(1,0)$
SCFTs in six dimensions via four-dimensional SCFTs of which many aspects are
better understood.

At low-energy, these wrapped branes can be described by supersymmetric
$AdS_{m}\times M^{n}$ solutions of gauged supergravity in $m+n$ dimensions
Maldacena_Twist . $M^{n}$ is an $n$-dimensional compact manifold with constant
curvature on which the wrapped $(m+n-2)$-branes leading to an
$(m-1)$-dimensional SCFT from a compactification of the dual
$(m+n-1)$-dimensional SCFT on $M^{n}$. The corresponding supergravity
solutions preserve some amount of the original supersymmetry by means of a
topological twist Witten_twist , implemented by turning on some gauge fields
to cancel the spin connections on the compact manifold $M^{n}$. A large number
of these solutions have previously been found in various dimensions, see
flow_across_Gauntlett1 ; flow_across_Gauntlett2 ; flow_across_Gauntlett3 ;
Kim_AdS_factor ; Cucu_AdSD-2 ; BB ; Wraped_D3 ; 3D_CFT_from_LS_point ;
flow_acrossD_bobev ; BBC ; 4D_SCFT_from_M5 ; Wraped_M5 ; N3_AdS2 ;
flow_across_Betti ; AdS2_trisasakian ; 7D_twist ; 6D_twist ; 5D_N4_flow ;
5Dtwist ; BH_microstate_6D2 ; Minwoo_6D_BH2 ; Calos_6D_flow1 ; AdS3_7D_N2 ;
7D_Max_twist , for an incomplete list.
Recently, new classes of $AdS\times\Sigma$ solutions, in which unbroken
supersymmetry is not realized by a topological twist, have been found for
$\Sigma$ being a two-dimensional space with non-constant curvature. These
solutions describe supersymmetric branes wrapped on a spindle, which is
topologically a two-sphere with orbifold singularities at the poles, Ferro_D3
; Hosseini_D3 ; Boido_D3 ; Ferrero_M2 ; Cassani_M2 ; Ferrero_M5 ; D4_spindle ,
see also Gaunlett_super_spindle  for a more recent result, or on a topological
disk with non-trivial $U(1)$ holonomy on the boundary or “half-spindle” Bah_AD
; Bah_M5 ; Suh_D3 ; Suh_D4 ; Suh_M2 ; D3_disk . These lead to new
supersymmetric AdS geometries from gauged supergravities which are dual to
lower-dimensional SCFTs arising from compactifications of higher-dimensional
SCFTs on a spindle or a half-spindle. In particular, supersymmetric
$AdS_{5}\times\Sigma$ solutions from the $U(1)^{2}$ truncation of the maximal
$SO(5)$ gauged supergravity in seven dimensions obtained in Bah_AD ; Bah_M5
have been shown to be dual to four-dimensional $N=2$ SCFTs of Argyres-Douglas
(AD) type AD_origin . Furthermore, it should be remarked that both the spindle
and half-spindle can be obtained from different global extensions of the same
local solutions as pointed out recently in M2-spindle .
In this work, we are interested in supersymmetric $AdS_{5}\times\Sigma$
solutions from matter-coupled $N=2$ gauged supergravity in seven dimensions
constructed in Eric_N2_7Dmassive , see Pure_N2_7D1 ; Pure_N2_7D2 ; Eric_N2_7D
; Park_7D ; Salam_7DN2  for earlier constructions. We mainly consider $N=2$
gauged supergravity coupled to three vector multiplets with possible gauge
groups given by $SO(4)\sim SO(3)\times SO(3)$, $SO(2,2)\sim SO(2,1)\times
SO(2,1)$ and $SO(3,1)$. It is well-known that only $SO(4)$ and $SO(3,1)$ gauge
groups admit supersymmetric $AdS_{7}$ vacua dual to $N=(1,0)$ SCFTs in six
dimensions with $SO(3)_{R}$ R-symmetry 7D_flow ; 7D_noncompact ; AdS_7_N2_Jan
. Furthermore, a number of interesting holographic solutions have been found
in 7D_twist ; AdS3_7D_N2 ; 7D_flow ; 7D_noncompact ; 7D_defect . We will give
more solutions to this list by finding supersymmetric $AdS_{5}\times\Sigma$
solutions within this $N=2$ gauged supergravity for $\Sigma$ being a half-
spindle.
It has been pointed out in Ferrero_M5  that supersymmetric
$AdS_{5}\times\Sigma$ solutions with $\Sigma$ being a spindle do not exist in
minimal or pure $N=2$ gauged supergravity with $SO(3)$ gauge group, see also
Gaunlett_super_spindle . It turns out that solutions with $\Sigma$ being a
half-spindle do exist in pure $N=2$ gauged supergravity. These solutions
preserve $\frac{1}{2}$ of the supersymmetry and $SO(2)_{R}\subset SO(3)_{R}$
and can be obtained from a truncation of $AdS_{5}\times\Sigma$ solutions with
$SO(2)\times SO(2)$ symmetry in $SO(4)$ gauged supergravity considered in this
work. Furthermore, we also find $SO(2)_{\textrm{diag}}$ symmetric solutions
that can be mapped to the solution found in Bah_M5  from $U(1)^{2}$ truncation
of the maximal $SO(5)$ gauged supergravity.

However, unlike the solutions in Bah_M5  dual to $N=2$ SCFTs of AD type, these
solutions preserve only eight supercharges and should be dual to $N=1$ SCFTs
in four dimensions. In addition, most of the solutions found in this paper
currently have no known higher dimensional origin. In particular, it has been
shown in Henning_Malek_AdS7_6  that uplifting seven-dimensional $N=2$ gauged
supergravity with $AdS_{7}$ vacua to ten dimensions can be achieved only if
there is no vector multiplet or one vector multiplet. On the other hand, the
uplift to eleven dimensions can be performed via an $S^{4}$ truncation if the
$N=2$ theories are truncations of the maximal $N=4$ gauged supergravity. The
embedding in this case can be obtained from the results of Pure_Red_Ans  and
7D_from_11D . Moreover, pure $N=2$ gauged supergravity with $SO(3)$ gauge
group can also be uplifted to type IIA supergravity uni_6D_7D ;
minimalAdS7_AdS6_Henning .
Finally, we will find $AdS_{5}\times\Sigma$ solutions in $N=2$ gauged
supergravity with a non-compact $SO(2,2)\sim SO(2,1)\times SO(2,1)$ gauge
group. Since this gauged supergravity does not admit any supersymmetric
$AdS_{7}$ vacua, the maximally supersymmetric vacua are given by half-
supersymmetric domain walls dual to $N=(1,0)$ non-conformal field theories in
six dimensions according to the DW/QFT correspondence DW_QFT1 ; DW_QFT2 ;
DW_QFT3 ; DW_QFT4 . In this case, the resulting $AdS_{5}\times\Sigma$
solutions are expected to describe four-dimensional $N=1$ SCFTs arising from
six-dimensional $N=(1,0)$ field theories compactified on a topological disk or
half-spindle. To the best of our knowledge, these are the first example of
$AdS_{5}\times\Sigma$ solutions involving half-spindles with domain wall
asymptotics.
The paper is organized as follows. In Section II, we give a brief review of
seven-dimensional $N=2$ gauged supergravity coupled to an arbitrary number of
vector multiplets. Supersymmetric $AdS_{5}\times\Sigma$ solutions in $SO(4)$,
$SO(2,2)$ and $SO(3,1)$ gauge groups will be considered in Sections III, IV,
and V, respectively. Some conclusions and comments will be given in Section
VI.

##  II Matter-coupled $N=2$ gauged supergravity in seven dimensions

In this section, we give relevant formulae involving bosonic Lagrangian and
supersymmetry transformations of fermions to find supersymmetric solutions of
matter-coupled $N=2$ gauged supergravity in seven dimensions. We follow most
of the conventions and notations in Eric_N2_7D  in which the detailed
construction can be found, see also Dibietto_7D_embedding_tensor  for gaugings
using the embedding tensor formalism in the case of three vector multiplets.

In seven dimensions, the half-maximal $N=2$ supergravity multiplet contains
the following component fields

$(e^{\hat{\mu}}_{\mu},\psi^{a}_{\mu},A^{i}_{\mu},\chi^{a},B_{\mu\nu},\sigma)\\\
$

given by the graviton $e^{\hat{\mu}}_{\mu}$, two gravitini $\psi^{a}_{\mu}$,
three vectors $A^{i}_{\mu}$, two spin-$\frac{1}{2}$ fields $\chi^{a}$, a two-
form field $B_{\mu\nu}$ and the scalar field or dilaton $\sigma$. We
respectively denote curved and flat space-time indices by $\mu,\nu,\ldots$ and
$\hat{\mu},\hat{\nu},\ldots$. Indices $i,j=1,2,3$ and $a,b=1,2$ respectively
label triplet and doublet of $SO(3)_{R}\sim SU(2)_{R}$ $R$-symmetry.

The supergravity multiplet can couple to an arbitrary number $n$ of vector
multiplets with the field content

$(A_{\mu},\lambda^{a},\phi^{i})^{r}\,.$ (1)

Each vector multiplet, labelled by an index $r=1,\ldots,n$, consists of a
vector field $A_{\mu}$, two gaugini $\lambda^{a}$, and three scalars
$\phi^{i}$. Together with the supergravity multiplet, there are $3+n$ vector
fields transforming in a fundamental representation of the global symmetry
$SO(3,n)$, collectively denoted by $A^{I}_{\mu}=(A^{i}_{\mu},A^{r}_{\mu})$.
The $SO(3,n)$ fundamental indices $I,J=1,\ldots,3+n$ are lowered and raised
respectively by the $SO(3,n)$ invariant tensor and its inverse

$\eta_{IJ}=\eta^{IJ}=\text{diag}(-1,-1,-1,\underbrace{1,\ldots,1}_{n}).$ (2)

Similar to the dilaton $\sigma$ described by a coset manifold
$SO(1,1)\sim\mathbb{R}^{+}$, the $3n$ scalars $\phi^{ir}$ from the $n$ vector
multiplets are parametrized by $SO(3,n)/SO(3)\times SO(n)$ coset manifold.
With $A=(i,r)$ being an $SO(3)\times SO(n)$ index, the associated coset
representative can be written as

${L_{I}}^{A}=({L_{I}}^{i},{L_{I}}^{r}).$ (3)

${L_{I}}^{A}$ transforms under the global $SO(3,n)$ and the local $SO(3)\times
SO(n)$ by left and right multiplications, respectively. The inverse of
${L_{I}}^{A}$ will be denoted by ${L_{A}}^{I}=({L_{i}}^{I},{L_{r}}^{I})$ and
satisfies the following relations

${L_{j}}^{I}{L_{I}}^{i}=\delta^{i}_{j},\qquad{L_{s}}^{I}{L_{I}}^{r}=\delta^{r}_{s}\,,\qquad\eta_{IJ}=-{L_{I}}^{i}{L_{J}}^{i}+{L_{I}}^{r}{L_{J}}^{r}\,.$
(4)

Note also that indices $i,j$ and $r,s$ are raised and lowered by $\delta_{ij}$
and $\delta_{rs}$, respectively.

Gaugings of the matter-coupled $N=2$ supergravity can be obtained by promoting
a subgroup $G\subset SO(3,n)$ to be a local symmetry. The embedding of $G$ in
$SO(3,n)$ is described by the $SO(3,n)$ tensor ${f_{IJ}}^{K}$ identified with
structure constants of the gauge group $G$ via the gauge algebra

$[T_{I},T_{J}]={f_{IJ}}^{K}T_{K}$ (5)

where $T_{I}$ are the gauge generators. In the embedding tensor formalism,
${f_{IJ}}^{K}$ is one of the components of the embedding tensor, see
Dibietto_7D_embedding_tensor  for more detail.
In order for the gauging to be consistent with supersymmetry, ${f_{IJ}}^{K}$
must satisfy the conditions

$f_{IJK}=\eta_{KL}{f_{IJ}}^{L}=f_{[IJK]}\qquad\textrm{and}\qquad{f_{[IJ}}^{L}{f_{K]L}}^{M}=0\,.$
(6)

Since $\eta_{IJ}$ has only three negative eigenvalues, any gauge group can
have at most three compact or three non-compact generators. Therefore, the
allowed semi-simple gauge groups are of the form $G\sim G_{0}\times H\subset
SO(3,n)$ with $H$ being a compact group of dimension
$\text{dim}H\leq(n+3-\text{dim}G_{0})$ Eric_N2_7D . On the other hand, $G_{0}$
can only be one of the six possibilities: $SO(3)$, $SO(3,1)$,
$SL(3,\mathbb{R})$, $SO(2,1)$, $SO(2,2)$, and $SO(2,2)\times SO(2,1)$.

Apart from the usual gaugings, there is also a massive deformation given by
adding a topological mass term to the three-form field $C_{\mu\nu\rho}$ dual
to the two-form field $B_{\mu\nu}$. This additional deformation is crucial for
the gauged supergravity to admit $AdS_{7}$ vacua. With both of these
deformations, the bosonic Lagrangian of the matter-coupled $N=2$ gauged
supergravity are given in differential form language by

$\displaystyle\mathcal{L}$ $\displaystyle=$ $\displaystyle\
\frac{1}{2}R\ast\mathbf{1}-\frac{1}{2}e^{\sigma}a_{IJ}\ast F^{I}_{(2)}\wedge
F^{J}_{(2)}-\frac{1}{2}e^{-2\sigma}\ast H_{(4)}\wedge H_{(4)}-\frac{5}{8}\ast
d\sigma\wedge d\sigma$ (7) $\displaystyle-\frac{1}{2}\ast P^{ir}_{(1)}\wedge
P^{ir}_{(1)}+\frac{1}{\sqrt{2}}H_{(4)}\wedge\omega_{(3)}-4hH_{(4)}\wedge
C_{(3)}-\mathbf{V}\ast\mathbf{1}\,.$

The constant $h$ describes the topological mass term for the three-form
$C_{(3)}$ with the field strength $H_{(4)}=dC_{(3)}$. The gauge field strength
is defined by

$F^{I}_{(2)}=dA^{I}_{(1)}+\frac{1}{2}{f_{JK}}^{I}A^{J}_{(1)}\wedge
A^{K}_{(1)}.$ (8)

The scalar matrix $a_{IJ}$ appearing in the kinetic term of vector fields is
given in terms of the coset representative as follows

$a_{IJ}={L_{I}}^{i}{L_{J}}^{i}+{L_{I}}^{r}{L_{J}}^{r}\,.$ (9)

The Chern-Simons three-form satisfying $d\omega_{(3)}=F^{I}_{(2)}\wedge
F^{I}_{(2)}$ is defined by

$\omega_{(3)}=F^{I}_{(2)}\wedge
A^{I}_{(1)}-\frac{1}{6}{f_{IJ}}^{K}A^{I}_{(1)}\wedge A^{J}_{(1)}\wedge
A_{(1)K}\,.$ (10)

The scalar potential is given by

$\mathbf{V}=\frac{1}{4}e^{-\sigma}\left(C^{ir}C_{ir}-\frac{1}{9}C^{2}\right)+16h^{2}e^{4\sigma}-\frac{4\sqrt{2}}{3}he^{\frac{3\sigma}{2}}C$
(11)

where $C$-functions, or fermion-shift matrices, are defined as

$\displaystyle C$ $\displaystyle=$
$\displaystyle-\frac{1}{\sqrt{2}}{f_{IJ}}^{K}{L_{i}}^{I}{L_{j}}^{J}L_{Kk}\varepsilon^{ijk},$
(12) $\displaystyle C^{ir}$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}{f_{IJ}}^{K}{L_{j}}^{I}{L_{k}}^{J}{L_{K}}^{r}\varepsilon^{ijk},$
(13) $\displaystyle C_{rsi}$ $\displaystyle=$
$\displaystyle{f_{IJ}}^{K}{L_{r}}^{I}{L_{s}}^{J}L_{Ki}\,.$ (14)

The scalar kinetic term is defined in terms of the vielbein on the
$SO(3,n)/SO(3)\times SO(n)$ coset as

$P^{ir}_{(1)}=L^{rI}\left(\delta^{K}_{I}d+{f_{IJ}}^{K}A^{J}_{(1)}\right){L_{K}}^{i}.$
(15)

Supersymmetry transformations of fermionic fields read

$\displaystyle\delta\psi_{\mu}^{a}$ $\displaystyle=$ $\displaystyle
2D_{\mu}\epsilon^{a}-\frac{\sqrt{2}}{30}e^{-\frac{\sigma}{2}}C\Gamma_{\mu}\epsilon^{a}-\frac{4}{5}he^{2\sigma}\Gamma_{\mu}\epsilon^{a}-\frac{i}{20}e^{\frac{\sigma}{2}}F^{i}_{\rho\sigma}{(\sigma^{i})^{a}}_{b}(3\Gamma_{\mu}\Gamma^{\rho\sigma}-5\Gamma^{\rho\sigma}\Gamma_{\mu})\epsilon^{b}$
(16)
$\displaystyle-\frac{1}{240\sqrt{2}}e^{-\sigma}H_{\rho\sigma\lambda\tau}(\Gamma_{\mu}\Gamma^{\rho\sigma\lambda\tau}+5\Gamma^{\rho\sigma\lambda\tau}\Gamma_{\mu})\epsilon^{a},$
$\displaystyle\delta\chi^{a}$ $\displaystyle=$
$\displaystyle-\frac{1}{2}\Gamma^{\mu}\partial_{\mu}\sigma\epsilon^{a}+\frac{\sqrt{2}}{30}e^{-\frac{\sigma}{2}}C\epsilon^{a}-\frac{16}{5}e^{2\sigma}h\epsilon^{a}-\frac{i}{10}e^{\frac{\sigma}{2}}F^{i}_{\mu\nu}{(\sigma^{i})^{a}}_{b}\Gamma^{\mu\nu}\epsilon^{b}$
(17)
$\displaystyle-\frac{1}{60\sqrt{2}}e^{-\sigma}H_{\mu\nu\rho\sigma}\Gamma^{\mu\nu\rho\sigma}\epsilon^{a},$
$\displaystyle\delta\lambda^{ar}$ $\displaystyle=$ $\displaystyle
i\Gamma^{\mu}P^{ir}_{\mu}{(\sigma^{i})^{a}}_{b}\epsilon^{b}-\frac{1}{2}e^{\frac{\sigma}{2}}F^{r}_{\mu\nu}\Gamma^{\mu\nu}\epsilon^{a}-\frac{i}{\sqrt{2}}e^{-\frac{\sigma}{2}}C^{ir}{(\sigma^{i})^{a}}_{b}\epsilon^{b}.$
(18)

In these equations, ${(\sigma^{i})^{a}}_{b}$ are the usual Pauli matrices and
$\Gamma_{\mu}=e^{\hat{\mu}}_{\mu}\Gamma_{\hat{\mu}}$ in which
$\Gamma_{\hat{\mu}}$ are seven-dimensional space-time gamma matrices
satisfying the $SO(1,6)$ Clifford algebra

$\\{\Gamma_{\hat{\mu}},\Gamma_{\hat{\nu}}\\}=2\eta_{\hat{\mu}\hat{\nu}},\qquad\eta_{\hat{\mu}\hat{\nu}}=\text{diag}(-++++++).$
(19)

The dressed field strengths $F^{i}_{(2)}$ and $F^{r}_{(2)}$ are defined by

$F^{i}_{(2)}={L_{I}}^{i}F^{I}_{(2)}\qquad\textrm{and}\qquad
F^{r}_{(2)}={L_{I}}^{r}F^{I}_{(2)}\,.$ (20)

The covariant derivative of the supersymmetry parameter $\epsilon^{a}$ is
given by

$D_{\mu}\epsilon^{a}=\partial_{\mu}\epsilon^{a}+\frac{1}{4}{\omega_{\mu}}^{\hat{\nu}\hat{\rho}}\Gamma_{\hat{\nu}\hat{\rho}}\epsilon^{a}+\frac{1}{2\sqrt{2}}Q^{i}_{\mu}{(\sigma^{i})^{a}}_{b}\epsilon^{b}$
(21)

where $Q^{i}_{\mu}$ is defined in terms of the $SO(3)_{R}$ composite
connection $Q^{ij}_{\mu}$ as

$Q^{i}_{\mu}=\frac{i}{\sqrt{2}}\varepsilon^{ijk}Q^{jk}_{\mu}$ (22)

with

$Q^{ij}_{\mu}=L^{jI}\left(\delta^{K}_{I}\partial_{\mu}+{f_{IJ}}^{K}A^{J}_{\mu}\right){L_{K}}^{i}\,.$
(23)

For convenience, we also give all the bosonic field equations derived from the
Lagrangian (7)

$\displaystyle 0$ $\displaystyle=$ $\displaystyle d(e^{-2\sigma}\ast
H_{(4)})+8hH_{(4)}-\frac{1}{\sqrt{2}}F^{I}_{(2)}\wedge F^{I}_{(2)},$ (24)
$\displaystyle 0$ $\displaystyle=$ $\displaystyle D(e^{\sigma}a_{IJ}\ast
F^{I}_{(2)})-\sqrt{2}H_{(4)}\wedge F^{J}_{(2)}+\ast
P^{ir}_{(1)}{f_{IJ}}^{K}{L_{r}}^{I}L_{Ki},\phantom{\frac{1}{\sqrt{2}}}$ (25)
$\displaystyle 0$ $\displaystyle=$ $\displaystyle D(\ast
P^{ir}_{(1)})-2e^{\sigma}{L_{I}}^{i}{L_{J}}^{r}\ast F^{I}_{(2)}\wedge
F^{J}_{(2)}$ (26)
$\displaystyle-\left(\frac{1}{\sqrt{2}}e^{-\sigma}C^{js}C_{rsk}\varepsilon^{ijk}+4\sqrt{2}he^{\frac{3\sigma}{2}}C^{ir}\right)\varepsilon_{(7)},$
$\displaystyle 0$ $\displaystyle=$ $\displaystyle\frac{5}{4}d(\ast
d\sigma)-\frac{1}{2}e^{\sigma}a_{IJ}\ast F^{I}_{(2)}\wedge
F^{J}_{(2)}+e^{-2\sigma}\ast H_{(4)}\wedge H_{(4)}$ (27)
$\displaystyle+\left[\frac{1}{4}e^{-\sigma}\left(C^{ir}C_{ir}-\frac{1}{9}C^{2}\right)+2\sqrt{2}he^{\frac{3\sigma}{2}}C-64h^{2}e^{4\sigma}\right]\varepsilon_{(7)},$
$\displaystyle 0$ $\displaystyle=$ $\displaystyle
R_{\mu\nu}-\frac{5}{4}\partial_{\mu}\sigma\partial_{\nu}\sigma-
a_{IJ}e^{\sigma}\left(F^{I}_{\mu\rho}{F^{J}_{\nu}}^{\rho}-\frac{1}{10}g_{\mu\nu}F^{I}_{\rho\sigma}F^{J\
\rho\sigma}\right)$ (28)
$\displaystyle-P^{ir}_{\mu}P^{ir}_{\nu}-\frac{2}{5}g_{\mu\nu}\mathbf{V}-\frac{1}{6}e^{-2\sigma}\left(H_{\mu\rho\sigma\lambda}{H_{\nu}}^{\rho\sigma\lambda}-\frac{3}{20}g_{\mu\nu}H_{\rho\sigma\lambda\tau}H^{\rho\sigma\lambda\tau}\right).\quad$

We finally give a general parametrization of the $SO(3,n)/SO(3)\times SO(n)$
coset which is useful for finding explicit solutions. We first introduce
$(n+3)^{2}$ basis elements of a general $(n+3)\times(n+3)$ matrix as follow

$(e_{IJ})_{KL}=\delta_{IK}\delta_{JL}\,.$ (29)

The composite $SO(3)\times SO(n)$ generators are given by

$\displaystyle SO(3):\qquad J^{(1)}_{ij}$ $\displaystyle=$ $\displaystyle
e_{ji}-e_{ij},\qquad\qquad\qquad\ i,j=1,2,3,$ $\displaystyle SO(n):\qquad
J^{(2)}_{rs}$ $\displaystyle=$ $\displaystyle e_{s+3,r+3}-e_{r+3,s+3},\qquad
r,s=1,\ldots,n\,.$ (30)

The non-compact generators corresponding to the $3n$ scalars are given by

$Y_{ir}=e_{i,r+3}+e_{r+3,i}\,.$ (31)

##  III $SO(4)$ gauge group

We first consider $N=2$ gauged supergravity with $SO(4)\sim SO(3)\times SO(3)$
gauge group obtained by coupling the gravity multiplet to $n=3$ vector
multiplets. The first $SO(3)$ factor is identified with the $SO(3)_{R}\sim
SU(2)_{R}$ R-symmetry. The corresponding structure constants are given by

$f_{IJK}=(\widetilde{g}_{1}\varepsilon_{ijk},-\widetilde{g}_{2}\varepsilon_{rst}),\qquad
r,s,\ldots=1,2,3$ (32)

in which $\widetilde{g}_{1}$ and $\widetilde{g}_{2}$ are coupling constants of
$SO(3)_{R}$ and $SO(3)$ generated by $J^{(1)}_{ij}$ and $J^{(2)}_{rs}$,
respectively.

We are interested in supersymmetric solutions in the form of a product space
between an $AdS_{5}$ and a topological disk $\Sigma$ with a non-trivial $U(1)$
holonomy at the boundary. Following Bah_M5 , we take the ansatz for the seven-
dimensional metric to be

$ds^{2}_{7}=f(r)ds^{2}_{AdS_{5}}+g_{1}(r)dr^{2}+g_{2}(r)dz^{2}$ (33)

where the metric on $AdS_{5}$ with unit radius is given by

$ds^{2}_{AdS_{5}}=\frac{1}{\rho^{2}}(dx^{2}_{1,3}+d\rho^{2})$ (34)

with $dx^{2}_{1,3}=\eta_{mn}dx^{m}dx^{n}$, $m,n=0,\ldots,3$ being the flat
metric on the four-dimensional Minkowski space $Mkw_{4}$. $r$ and $z$ are
respectively radial and angular coordinates on the topological disk $\Sigma$
whose ranges will be determined later on. The seven-dimensional curved and
flat space-time indices will be split into $\mu=(m,\rho,r,z)$ and
$\hat{\mu}=(\hat{m},\hat{\rho},\hat{r},\hat{z})$, respectively.

With the following vielbein one-forms

$\displaystyle e^{\hat{m}}_{(1)}=\frac{\sqrt{f(r)}}{\rho}dx^{m},\qquad
e^{\hat{\rho}}_{(1)}=\frac{\sqrt{f(r)}}{\rho}d\rho,$ $\displaystyle
e^{\hat{r}}_{(1)}=\sqrt{g_{1}(r)}dr,\qquad
e^{\hat{z}}_{(1)}=\sqrt{g_{2}(r)}dz,$ (35)

we can straightforwardly compute all non-vanishing components of the spin
connections

$\displaystyle\omega_{(1)}^{\hat{m}\hat{\rho}}=-\frac{e^{\hat{m}}}{\sqrt{f}},\qquad\omega_{(1)}^{\hat{m}\hat{r}}=\frac{f^{\prime}e^{\hat{m}}}{2f\sqrt{g_{1}}},$
$\displaystyle\omega_{(1)}^{\hat{\rho}\hat{r}}=\frac{f^{\prime}e^{\hat{\rho}}}{2f\sqrt{g_{1}}},\qquad\omega_{(1)}^{\hat{z}\hat{r}}=\frac{g^{\prime}_{2}e^{\hat{z}}}{2g_{2}\sqrt{g_{1}}}\,.$
(36)

From now on, we will use primes to denote $r$-derivatives and mostly suppress
arguments of the $r$-dependent functions for convenience.

###  III.1 $SO(2)_{R}$ and $SO(2)\times SO(2)$ symmetric solutions

We now move to the ansatz for gauge fields in the case of $SO(2)_{R}\subset
SO(3)_{R}$ and $SO(2)\times SO(2)$ symmetry. Since the former can be obtained
as a truncation of the latter, we will firstly consider the case of
$SO(2)\times SO(2)$ symmetry and perform a suitable truncation to obtain
$SO(2)_{R}$ symmetric solutions. For $SO(2)\times SO(2)$ symmetric solutions,
we will choose the $SO(2)\times SO(2)$ subgroup generated by $J^{(1)}_{12}$
and $J^{(2)}_{12}$ and turn on the following gauge fields

$A^{I}_{(1)}=\left[A_{1}(r)\delta^{I}_{3}+A_{2}(r)\delta^{I}_{6}\right]dz\,.$
(37)

The corresponding two-form field strengths are given by

$F^{I}_{(2)}=(A^{\prime}_{1}\delta^{I}_{3}+A^{\prime}_{2}\delta^{I}_{6})\,dr\wedge
dz\,.$ (38)

This ansatz leads to $F^{I}_{(2)}\wedge F^{I}_{(2)}=0$. According to the field
equation of the three-form field given in (24), we can consistently set
$C_{(3)}=0$.

Among the nine scalars from the $SO(3,3)/SO(3)\times SO(3)$ coset, there is
only one $SO(2)\times SO(2)$ singlet scalar. Following 7D_flow , this scalar
field corresponds to the non-compact generator $Y_{33}$, and the coset
representative can be written as

$L=e^{\phi Y_{33}}\,.$ (39)

This singlet scalar $\phi$ and the dilaton $\sigma$ depend only on the radial
coordinate. It is now straightforward to compute the $C$-functions appearing
in the supersymmetry transformations

$C=3\sqrt{2}\,\widetilde{g}_{1}\cosh\phi,\qquad
C^{ir}=-\sqrt{2}\,\widetilde{g}_{1}\sinh{\phi}\,\delta^{i}_{3}\delta^{r}_{3}\,.$
(40)

The scalar vielbein and $SO(2)_{R}$ composite connection have the following
non-vanishing components

$P^{ir}_{(1)}=\phi^{\prime}\,\delta^{i}_{3}\delta^{r}_{3}dr\qquad\text{ and
}\qquad Q^{ij}_{(1)}=\widetilde{g}_{1}A_{1}\varepsilon^{ij3}dz\,.$ (41)

It should be noted here that only $A_{1}$ appears in the composite connection
because $A^{3}_{(1)}$ is the vector field that gauges $SO(2)_{R}\subset
SO(3)_{R}$ under which the gravitini and supersymmetry parameters are charged.
With all these and a similar analysis as in Bah_M5 , we can determine all the
BPS equations from the supersymmetry transformations of fermionic fields. The
detailed analysis and relevant results can be found in the appendix. In the
following, we will separately consider solutions with $SO(2)_{R}$ and
$SO(2)\times SO(2)$ symmetries.

####  III.1.1 $SO(2)_{R}$ symmetric solution in pure $N=2$ gauged supergravity

We first consider a simple case of $SO(2)_{R}$ symmetric solutions which can
be obtained by setting $A^{6}_{(1)}=0$. Equation (291) then gives

$\mathbf{F}_{1}=be^{-\sigma}\cosh{\phi}f^{-\frac{5}{2}}\qquad\text{ and
}\qquad\mathbf{F}_{2}=-be^{-\sigma}\sinh{\phi}f^{-\frac{5}{2}}$ (42)

in which we have written $a_{1}=a_{2}=b$.

With this explicit form of $\mathbf{F}_{1}$ and $\mathbf{F}_{2}$, the BPS
condition in (282) implies that

$bh\sinh\phi=0\,.$ (43)

For $h=0$, the gauged supergravity does not admit any supersymmetric $AdS_{7}$
vacua, so we will keep $h\neq 0$. With $b=0$, all the gauge fields vanish.
This clearly does not lead to any solutions of the form $AdS_{5}\times\Sigma$.
Therefore, to find possible $AdS_{5}\times\Sigma$ solutions with $AdS_{7}$
asymptotics, we need to set $\phi=0$. Effectively, all the fields from vector
multiplets are truncated out. The resulting solutions can be then considered
as solutions of the minimal or pure $N=2$ gauged supergravity with $SO(3)_{R}$
gauge group.

With $\phi=0$, equation (291) implies that $A^{\prime}_{2}=0$. All the BPS
conditions from (257) and (276) to (288) as well as the field equation (26)
for scalars from the vector multiplets are automatically satisfied. With all
these, we are left with the algebraic conditions (261) to (275). Firstly, we
solve for $f$ from equation (271) with the solution given by

$f=\frac{2\sqrt{b\,h}\,e^{\sigma}}{\sqrt{s(16he^{\frac{5\sigma}{2}}-\widetilde{g}_{1})}}\,.$
(44)

With this result, equation (263) can be solved for $g_{1}$ giving rise to

$g_{1}=\frac{200\sqrt{b\,h^{5}}\,e^{\sigma}\,(\sigma^{\prime})^{2}}{(16he^{\frac{5\sigma}{2}}-\widetilde{g}_{1})^{2}\left[32\sqrt{b\,h^{5}}-e^{-5\sigma}\sqrt{s(16he^{\frac{5\sigma}{2}}-\widetilde{g}_{1})}\right]}.$
(45)

The condition (272) together with the solution for
$A^{\prime}_{1}=\widehat{A}^{\prime}_{1}$ in (291) give an ordinary
differential equation of the form

$\widehat{A}^{\prime}_{1}=\frac{5\widetilde{g}_{1}\,\sigma^{\prime}}{2(12he^{\frac{5\sigma}{2}}-\widetilde{g}_{1})}\widehat{A}_{1}$
(46)

which can be readily solved by

$\widehat{A}_{1}=c\left(12h-\widetilde{g}_{1}e^{-\frac{5\sigma}{2}}\right)$
(47)

for an integration constant $c$. Substituting all these results in (267) or
(272) leads to the following solution for $g_{2}$ of the form

$g_{2}=\frac{c^{2}\widetilde{g}^{2}_{1}e^{\sigma}\left[32\sqrt{b\,h^{5}}-e^{-5\sigma}\sqrt{s(16he^{\frac{5\sigma}{2}}-\widetilde{g}_{1})}\right]}{\sqrt{s(16he^{\frac{5\sigma}{2}}-\widetilde{g}_{1})}}.$
(48)

Finally, it can be verified that all the BPS conditions (261) to (275) as well
as the dilaton’s and Einstein’s field equations, (27) and (28), are satisfied
by these solutions provided that

$\text{sign}(c\widetilde{g}_{1}\sigma^{\prime})=+1\,.$ (49)

Furthermore, the BPS equations (254) to (256) are satisfied by the following
form of the two-component spinor

$\eta=e^{iqz}Ye^{\frac{5\sigma}{4}}\begin{pmatrix}\sqrt{8(h^{5}b)^{\frac{1}{4}}+\sqrt{2}se^{-\frac{5\sigma}{2}}\left[s(16he^{\frac{5\sigma}{2}}-\widetilde{g}_{1})\right]^{\frac{1}{4}}}\\\
-\sqrt{8(h^{5}b)^{\frac{1}{4}}-\sqrt{2}se^{-\frac{5\sigma}{2}}\left[s(16he^{\frac{5\sigma}{2}}-\widetilde{g}_{1})\right]^{\frac{1}{4}}}\end{pmatrix}$
(50)

with the function $Y$ being the solution of an ordinary differential equation
given in (247). The explicit form of the solution for $Y$ can be written as

$Y=\frac{Y_{0}\,e^{-\sigma}}{\left[s(16he^{\frac{5\sigma}{2}}-\widetilde{g}_{1})\right]^{\frac{1}{8}}}$
(51)

in which $Y_{0}$ is an integration constant. It should be noted that the
solution is characterized by a set of functions that are determined in terms
of the dilaton $\sigma$ together with its derivative. However, the
$r$-dependent function $\sigma(r)$ is not determined by the BPS equations.
This is very similar to the solutions obtained in Bah_M5 ; Suh_D3 ; Suh_D4 ;
Suh_M2 .

To further analyse the solution, we first define the following parameters

$B=8h^{2}\sqrt{b},\qquad
m=\frac{\widetilde{g}_{1}}{16h},\qquad\mathcal{C}=2\,\widetilde{g}_{1}hc$ (52)

together with the function

$W=B-e^{-5\sigma}\sqrt{s(e^{\frac{5\sigma}{2}}-m)}\,.$ (53)

In terms of these quantities, the seven-dimensional metric reads

$ds^{2}_{7}=\frac{B\,e^{\sigma}}{16h^{2}\sqrt{s(e^{\frac{5\sigma}{2}}-m)}}ds^{2}_{AdS_{5}}+\frac{25B\,e^{\sigma}\,(\sigma^{\prime})^{2}}{4^{5}h^{2}W(e^{\frac{5\sigma}{2}}-m)^{2}}dr^{2}+\frac{\mathcal{C}^{2}We^{\sigma}}{4h^{2}\sqrt{s(e^{\frac{5\sigma}{2}}-m)}}dz^{2}$
(54)

which is singular when $W\rightarrow 0$. It turns out that the analysis near
$W=0$ is simpler if we fix the solution of $\sigma$ to

$\sigma=-\frac{2}{5}\ln r\,.$ (55)

This choice implies $r>0$, and the sign condition (49) requires
$\text{sign}(c\widetilde{g}_{1})=-1$. Accordingly, the constant $\mathcal{C}$
must be negative for $h>0$.

In terms of the radial coordinate, the seven-dimensional metric is given by

$\displaystyle ds^{2}_{7}$ $\displaystyle=$
$\displaystyle\frac{B\,r^{1/10}}{16h^{2}\sqrt{s(1-mr)}}\left[ds^{2}_{AdS_{5}}+ds^{2}_{\Sigma}\right]$
$\displaystyle\textrm{with}\qquad ds^{2}_{\Sigma}$ $\displaystyle=$
$\displaystyle\frac{r^{-1/2}}{16W\left[s(1-mr)\right]^{3/2}}dr^{2}+\frac{4\mathcal{C}^{2}W}{B}dz^{2}$
(56)

and

$W=B-r^{3/2}\sqrt{s(1-mr)}\,.$ (57)

The equation $W=0$ admits four roots given by

$r_{(\pm_{1},\pm_{2})}=\frac{1}{4m}\left[1\pm_{1}2m\sqrt{X}\pm_{2}2m\sqrt{\frac{3}{4m^{2}}-X\pm_{1}\frac{1}{4m^{3}\sqrt{X}}}\,\right]$
(58)

with

$X=\frac{1}{4m^{2}}+\frac{4(\frac{2}{3})^{1/3}sB^{4/3}}{\left[9s+\sqrt{81-768sB^{2}m^{3}}\right]^{\frac{1}{3}}}+\frac{B^{2/3}\left[9s+\sqrt{81-768sB^{2}m^{3}}\right]^{\frac{1}{3}}}{18^{1/3}m}\,.$
(59)

These roots are all distinct due to different sign choices $\pm_{1}$ and
$\pm_{2}$ appearing in (58).

Using (253), we find the explicit form of the $SO(2)_{R}$ vector field

$A_{1}=\frac{1}{8mh}\left[\,|\mathcal{C}|(4mr-3)-q\,\right].$ (60)

In terms of the radial coordinate, the two-component spinor $\eta$ is given by

$\eta=Y_{0}e^{iqz}\frac{2^{1/4}r^{1/40}}{\left[s(1-mr)\right]^{\frac{1}{8}}}\begin{pmatrix}\sqrt{\sqrt{B}+sr^{3/4}\left[s(1-mr)\right]^{\frac{1}{4}}}\\\
-\sqrt{\sqrt{B}-sr^{3/4}\left[s(1-mr)\right]^{\frac{1}{4}}}\end{pmatrix}.$
(61)

The allowed ranges of the radial coordinate $r$ for a regular solution are
constrained by requiring that the dilaton scalar is real ($r>0$) and all the
seven-dimensional metric functions are positive. There are seven possibilities
depending on the values of the parameters $s$, $m$, and $B$. It should also be
noted that these ranges together with the corresponding behaviours of the
solution are very similar to those considered in Bah_M5 . We now discuss these
possibilities in detail.
Case I: $s=1$, $m>0$, $0<B<\frac{3\sqrt{3}}{16m\sqrt{m}}$, $0<r<r_{(+,-)}$
For clarity, we plot a representative solution of the warp factors with $s=1$,
$m=\frac{3}{4}$, $B=h=\frac{1}{4}$, and $\mathcal{C}=-1$ in Figure 1. As seen
from the figure, as $r\rightarrow 0$, both $f$ and $g_{2}$ approach zero while
$g_{1}$ diverges to $+\infty$. This is a curvature singularity of the seven-
dimensional metric as pointed out in Bah_M5 . Setting $r=R^{4/3}$, we find
that the seven-dimensional metric becomes conformal to a product of $AdS_{5}$
and a cylinder near $R=0$

$ds_{7}^{2}\approx\frac{B\,R^{2/15}}{16h^{2}}\left[ds^{2}_{AdS_{5}}+\frac{1}{9B}dR^{2}+4\mathcal{C}^{2}dz^{2}\right].$
(62)

(a) $f$ solution

(b) $g_{1}$ solution

(c) $g_{2}$ solution

Figure 1: Numerical plots of the warp factors for the $SO(2)$ symmetric
solution in pure $N=2$ gauged supergravity with $s=1$, $m=\frac{3}{4}$,
$B=h=\frac{1}{4}$, and $\mathcal{C}=-1$. All the warp factors are positive in
the range $0<r<r_{(+,-)}=0.456$ with the two vertical red dashed lines
representing the two boundaries.

As $r\rightarrow r_{(+,-)}$, the $AdS_{5}$ warp factor is smooth and the $z$
circle shrinks due to $W(r_{(+,-)})=B-r^{3/2}_{(+,-)}\sqrt{1-mr_{(+,-)}}=0$.
By introducing a new radial coordinate $R=\sqrt{r_{(+,-)}-r}$, we find that
$g_{2}$ is finite as $r\rightarrow r_{(+,-)}$. The seven-dimensional metric
near this endpoint takes the form of

$ds_{7}^{2}\approx\frac{B\,r_{(+,-)}^{1/10}}{16h^{2}\sqrt{1-mr_{(+,-)}}}\left[ds^{2}_{AdS_{5}}+\frac{dR^{2}+4\mathcal{C}^{2}\left[3-4mr_{(+,-)}\right]^{2}R^{2}dz^{2}}{-4W^{\prime}(r_{(+,-)})\sqrt{r_{(+,-)}}(1-mr_{(+,-)})^{3/2}}\right].$
(63)

The $R-z$ surface becomes locally an $\mathbb{R}^{2}/\mathbb{Z}_{l}$ orbifold
near the point $r=r_{(+,-)}$ if we set

$|\mathcal{C}|=\frac{1}{2l\left[3-4mr_{(+,-)}\right]},\qquad
l=1,2,3,\ldots\,.$ (64)

In this case, the function $3-4mr_{(+,-)}$ depends on two constants, $m$ and
$B$. However, its explicit form obtained from equation (58) is highly
complicated, so we will only show that $3-4mr_{(+,-)}$ is strictly positive in
the range of the $r$ coordinate under consideration by giving a numerical plot
of $3-4mr_{(+,-)}$ in Figure 2.

Figure 2: A numerical plot of the function $3-4mr_{(+,-)}$ appearing in (64).
Note that $3-4mr_{(+,-)}\rightarrow 3$ and $3-4mr_{(+,-)}\rightarrow 0$ as
$B\rightarrow 0$ or $m\rightarrow 0$ and
$B\rightarrow\frac{3\sqrt{3}}{16m\sqrt{m}}$, respectively.

By using the Gauss-Bonnet theorem and the same computation as in Bah_M5 , we
can calculate the Euler characteristic of $\Sigma$

$\displaystyle\chi(\Sigma)$ $\displaystyle=$
$\displaystyle\frac{1}{4\pi}\int_{\Sigma}R_{\Sigma}\text{vol}_{\Sigma}\ =\
\frac{2|\mathcal{C}|\left[3-4mr_{(+.-)}\right]r_{(+.-)}^{3/4}\left[1-mr_{(+,-)}\right]^{1/4}}{\sqrt{B}}$
(65) $\displaystyle=$ $\displaystyle 2|\mathcal{C}|\left[3-4mr_{(+.-)}\right]\
=\ \frac{1}{l}$

which is a natural result for a disk with an $\mathbb{R}^{2}/\mathbb{Z}_{l}$
orbifold singularity at $r=r_{(+,-)}$. Noted that the integration has been
performed on the interval $0<r<r_{(+.-)}$ and $0<z<2\pi$, and we have used
$B=r^{3/2}_{(+,-)}\sqrt{1-mr_{(+,-)}}$ obtained from $W(r_{(+,-)})=0$ with
$s=+1$.

In order for the $SO(2)_{R}$ gauge field to vanish at $r=r_{(+,-)}$, at which
the $z$ circle shrinks, we fix the constant $q$ to be

$q=-|\mathcal{C}|\left[3-4mr_{(+,-)}\right]=-\frac{1}{2l}$ (66)

giving rise to the $SO(2)_{R}$ gauge field of the form

$A_{1}=\frac{|\mathcal{C}|}{2h}[r-r_{(+,-)}].$ (67)

The explicit form of the Killing spinor at $r=r_{(+,-)}$ is given by

$\eta=2^{3/4}Y_{0}e^{-\frac{iz}{2l}}r^{2/5}_{(+,-)}\footnotesize\begin{pmatrix}1\\\
0\end{pmatrix}.$ (68)

Since only the upper component is non-vanishing, only $\frac{1}{4}$ of the
original supersymmetry or $4$ supercharges are preserved at $r=r_{(+,-)}$.
Moreover, $\eta$ is also well-behaved near $r=r_{(+,-)}$ and hence globally
defined on the disk $\Sigma$ as in Bah_M5 .
Case II: $s=1$, $m<0$, $B>0$, $0<r<r_{(-,+)}$
An example of numerical solutions with $s=1$, $m=-1$, $B=1$, $h=\frac{1}{8}$,
and $\mathcal{C}=-\frac{1}{2}$ is given in Figure 3. This case is very similar
to Case I. The previous analysis at both $r=0$ and $r=r_{(-,+)}$ endpoints can
be repeated by formally replacing $m\rightarrow-|m|$ and $r_{(+,-)}\rightarrow
r_{(-,+)}$. However, in order for the $z$ circle to shrink smoothly at
$r=r_{(-,+)}$, instead of (64), we have to impose

$|\mathcal{C}|=\frac{1}{2l\left[3+4|m|r_{(-,+)}\right]},\qquad
l=1,2,3,\ldots\,,\,.$ (69)

A numerical plot of $3+4|m|r_{(-,+)}$ is shown in Figure 4. The function is
very different from $3-4mr_{(+,-)}$ in Case I since the condition on the
constant $B$ in this case is less stringent.

(a) $f$ solution

(b) $g_{1}$ solution

(c) $g_{2}$ solution

Figure 3: Numerical plots of the warp factors for the $SO(2)$ symmetric
solution in pure $N=2$ gauged supergravity with $s=1$, $m=-1$, $B=1$,
$h=\frac{1}{8}$, and $\mathcal{C}=-\frac{1}{2}$. The warp factors are positive
in the range $0<r<r_{(-,+)}=0.819$ with the two vertical red dashed lines
representing the two boundaries. Figure 4: A numerical plot of the function
$3+4|m|r_{(-,+)}$ appearing in the condition (69). We also note that
$3+4|m|r_{(-,+)}\rightarrow 3$ as $B\rightarrow 0$ or $m\rightarrow 0$.

Case III: $s=1$, $m>0$, $0<B<\frac{3\sqrt{3}}{16m\sqrt{m}}$,
$r_{(+,+)}<r<\frac{1}{m}$
As $r\rightarrow\frac{1}{m}$, the $AdS_{5}$ warp factor goes to $+\infty$ as
can be seen from Figure 5. Setting
$r=\frac{1}{m}-16\mathcal{C}^{4}m^{2}B^{2}R^{4}$ , we find the seven
dimensional-metric of the form

$ds^{2}_{7}\approx\frac{1}{16h^{2}m^{8/5}R^{2}}\left[\frac{1}{4\mathcal{C}^{2}}ds^{2}_{AdS_{5}}+dR^{2}+dz^{2}\right],\qquad
R\rightarrow 0\,.$ (70)

This metric is again conformally related to a product of $AdS_{5}$ and a
cylinder.

(a) $f$ solution

(b) $g_{1}$ solution

(c) $g_{2}$ solution

Figure 5: Numerical plots of the warp factors for the $SO(2)$ symmetric
solution in pure $N=2$ gauged supergravity with $s=1$, $m=\frac{1}{2}$,
$B=\frac{1}{2}$, $h=\frac{1}{4}$, and $\mathcal{C}=-1$. The warp factors are
all positive in the range $r_{(+,+)}=1.93<r<\frac{1}{m}=2$ with the two
vertical red dashed lines representing the two boundaries.

By changing the radial coordinate to $\rho=\sqrt{r_{(+,+)}-r}$, we find the
seven-dimensional metric, as $r\rightarrow r_{(+,+)}$, of the form

$ds_{7}^{2}\approx\frac{B\,r_{(+,+)}^{1/10}}{16h^{2}\sqrt{1-mr_{(+,+)}}}\left[ds^{2}_{AdS_{5}}+\frac{dR^{2}+4\mathcal{C}^{2}\left[3-4mr_{(+,+)}\right]^{2}R^{2}dz^{2}}{-4W^{\prime}(r_{(+,+)})\sqrt{r_{(+,+)}}(1-mr_{(+,+)})^{3/2}}\right]$
(71)

which is similar to (63) with $r_{(+,-)}$ replaced by $r_{(+,+)}$. The $z$
circle also shrinks smoothly near the endpoint at $r=r_{(+,+)}$, and the $R-z$
surface is locally an $\mathbb{R}^{2}/\mathbb{Z}_{l}$ if we choose

$\mathcal{C}=\frac{1}{2l\left[3-4mr_{(+,+)}\right]},\qquad l=1,2,3,\ldots\,.$
(72)

Unlike all previous cases, $3-4mr_{(+,+)}$ is negative in the regularity
ranges of $B$ and $m$ as can be seen from the numerical plot given in Figure
6. In Figure 7, we also plot $3-4mr_{(+,+)}$ (blue surface) and
$3-4mr_{(+,-)}$ (orange surface) in Case I. The two joint smoothly at
$B=\frac{3\sqrt{3}}{16m\sqrt{m}}$ where $3-4mr_{(+,-)}=0=3-4mr_{(+,+)}$.

Figure 6: A numerical plot of the function $3-4mr_{(+,+)}$ appearing in the
condition (72). The function approaches $-1$ as $B\rightarrow 0$ or
$m\rightarrow 0$. As $B\rightarrow\frac{3\sqrt{3}}{16m\sqrt{m}}$, we find that
$3-4mr_{(+,-)}\rightarrow 0$. Figure 7: A numerical plot of $3-4mr_{(+,-)}$
(orange surface) and $3-4mr_{(+,+)}$ (blue surface) which are connected to
each other at $B=\frac{3\sqrt{3}}{16m\sqrt{m}}$.

To obtain the $SO(2)_{R}$ gauge field that vanishes at $r=r_{(+,+)}$, we
choose

$q=\mathcal{C}\left[3-4mr_{(+,+)}\right]=\frac{1}{2l}$ (73)

resulting in $A_{1}$ of the form

$A_{1}=\frac{|\mathcal{C}|}{2h}[r-r_{(+,+)}].$ (74)

Since $q=\frac{1}{2l}$ in this case, the Killing spinor $\eta$ near
$r=r_{(+,+)}$ is the same (68) with $r_{(+,-)}$ and $z$ replaced by
$r_{(+,+)}$ and $-z$.
Case IV: $s=1$, $m>0$, $B=\frac{3\sqrt{3}}{16m\sqrt{m}}$, $0<r<r_{(+,-)}$
With $B=\frac{3\sqrt{3}}{16m\sqrt{m}}$, the quantity $X$ in the four roots of
$W=0$ reduces to

$X=\frac{1}{m^{2}}$ (75)

giving rise to a much simple form of $r_{(\pm_{1},\pm_{2})}$ solution

$r_{(\pm_{1},\pm_{2})}=\frac{1}{4m}\left[1\pm_{1}2\pm_{2}\sqrt{-1\pm_{1}1}\right].$
(76)

In this case, there is only one real root of $W=0$ given by
$r_{\ast}=r_{(+,\pm_{2})}=\frac{3}{4m}$.

However, as pointed out in Bah_M5 , the $z$ circle does not shrink smoothly at
$r=r_{\ast}$ for any value of $\mathcal{C}$, due to the function
$W(r=r_{\ast})$ having a double zero. This can also be seen explicitly in the
present work by considering Case I with
$B\rightarrow\frac{3\sqrt{3}}{16m\sqrt{m}}$. In this limit, we cannot impose
the condition (64) since $|\mathcal{C}|$ diverges as $3-4mr_{(+,-)}\rightarrow
0$. If we set $r=\frac{3}{4m}-\frac{1}{R}$, the seven-dimensional metric is
approximately given by

$ds^{2}_{7}\approx\frac{3^{3/5}}{128(2^{1/5})h^{2}m^{8/5}}\left[3ds^{2}_{AdS_{5}}+\frac{dR^{2}+64\mathcal{C}^{2}m^{2}dz^{2}}{R^{2}}\right]$
(77)

as $R\rightarrow+\infty$. On the other hand, for $r\rightarrow 0$, we find the
seven-dimensional metric given in (62) as in Case I.
Case V: $s=1$, $m>0$, $B=\frac{3\sqrt{3}}{16m\sqrt{m}}$,
$r_{(+,+)}<r<\frac{1}{m}$
Since $r_{(+,+)}=r_{(+,-)}=r_{\ast}=\frac{3}{4m}$ for
$B=\frac{3\sqrt{3}}{16m\sqrt{m}}$, Case IV and Case V are connected at
$r=r_{\ast}$. Therefore, we will give a representative numerical solution for
these two cases collectively in Figure 8. For $r\rightarrow r_{\ast}$ both
from the left and from the right, the behaviour of the metric is given by
(77). On the other side, as $r\rightarrow\frac{1}{m}$, the metric becomes (70)
as in Case III.

(a) $f$ solution

(b) $g_{1}$ solution

(c) $g_{2}$ solution

Figure 8: Numerical plots of the warp factors for the $SO(2)$ symmetric
solution in pure $N=2$ gauged supergravity with $s=1$, $m=\frac{3}{4}$,
$B=\frac{1}{2}$, $h=\frac{1}{4}$, and $\mathcal{C}=-\frac{1}{2}$. The warp
factors are positive in the ranges $0<r<r_{\ast}=1$ (Case IV) and
$r_{\ast}=1<r<\frac{1}{m}=\frac{4}{3}$ (Case V). The three vertical red dashed
lines represent the three boundaries.

For the $SO(2)_{R}$ vector, we find that taking $r=r_{\ast}=\frac{3}{4m}$ in
(60) results in the $SO(2)_{R}$ gauge field

$A_{1}(r_{\ast})=-\frac{q}{8mh}\,.$ (78)

Therefore, we will set $q=0$ in order to make the $SO(2)_{R}$ gauge field
vanish at $r=r_{\ast}$.
Case VI: $s=1$, $m>0$, $B>\frac{3\sqrt{3}}{16m\sqrt{m}}$, $0<r<\frac{1}{m}$
This case combines the behaviour near $r=0$ in Case I and the feature near
$r=\frac{1}{m}$ in Case III. An example of numerical solution with
$m=\frac{3}{4}$, $B=1$, $h=\frac{1}{4}$, and $\mathcal{C}=-\frac{1}{2}$ is
given in Figure 9.

(a) $f$ solution

(b) $g_{1}$ solution

(c) $g_{2}$ solution

Figure 9: Numerical plots of the warp factors for the $SO(2)$ symmetric
solution in pure $N=2$ gauged supergravity with $s=1$, $m=\frac{3}{4}$, $B=1$,
$h=\frac{1}{4}$, and $\mathcal{C}=-\frac{1}{2}$. The warp factors are positive
in the range $0<r<\frac{1}{m}=\frac{4}{3}$ with the two vertical red dashed
lines representing the two boundaries.

Case VII: $s=-1$, $m>0$, $B>0$, $\frac{1}{m}<r<r_{(+,+)}$
This is the only case with $s=-1$ with a numerical solution given in Figure
10. In comparison with other cases, we find that this case is very similar to
Case III with the two boundaries interchanged. As $r\rightarrow\frac{1}{m}$,
setting $r=\frac{1}{m}+16\mathcal{C}^{4}m^{2}B^{2}R^{4}$ , we find the seven-
dimensional metric given in (70). Thus, the seven-dimensional space-time is
conformal to a product of $AdS_{5}$ and a cylinder as in Case III.

(a) $f$ solution

(b) $g_{1}$ solution

(c) $g_{2}$ solution

Figure 10: Numerical plots of the warp factors for the $SO(2)$ symmetric
solution in pure $N=2$ gauged supergravity with $s=-1$, $m=2$, $B=1$,
$h=\frac{1}{4}$, and $\mathcal{C}=-\frac{1}{2}$. The warp factors are positive
in the range $\frac{1}{m}=\frac{1}{2}<r<r_{(+,+)}=1$ with the two vertical red
dashed lines representing the two boundaries.

On the other side, as $r\rightarrow r_{(+,+)}$, we find the following form of
the seven-dimensional metric

$ds_{7}^{2}\approx\frac{B\,r_{(+,+)}^{1/10}}{16h^{2}\sqrt{mr_{(+,+)}-1}}\left[ds^{2}_{AdS_{5}}+\frac{dR^{2}+4\mathcal{C}^{2}\left[4mr_{(+,+)}-3\right]^{2}R^{2}dz^{2}}{-4W^{\prime}(r_{(+,+)})\sqrt{r_{(+,+)}}(mr_{(+,+)}-1)^{3/2}}\right]$
(79)

after changing the radial coordinate to $\rho=\sqrt{r_{(+,+)}-r}$ as in Case
III. The $z$ circle shrinks smoothly if we impose

$|\mathcal{C}|=\frac{1}{2l\left[4mr_{(+,+)}-3\right]},\qquad
l=1,2,3,\ldots\,.$ (80)

The $SO(2)_{R}$ gauge field is given by (74) with the constant $q$ chosen to
be

$q=|\mathcal{C}|\left[4mr_{(+,+)}-3\right]=\frac{1}{2l}\,.$ (81)

Unlike all the previous cases with $s=1$, only the lower component of the
Killing spinor $\eta$ is non-vanishing as $r\rightarrow r_{(+,+)}$

$\eta=-2^{3/4}Y_{0}e^{\frac{iz}{2l}}r^{2/5}_{(+,+)}\footnotesize\begin{pmatrix}0\\\
1\end{pmatrix}.$ (82)

We end this section by pointing out that in the matter-coupled $N=2$ gauged
supergravity with $SO(4)$ gauge group, there are three $SO(2)_{R}$ singlet
scalars corresponding to $SO(3,3)$ non-compact generators $Y_{31}$, $Y_{32}$,
and $Y_{33}$. We have also looked for $SO(2)_{R}$ symmetric solutions in this
case. However, in order to satisfy the vector field equation (25), it turns
out that either the $SO(2)_{R}$ gauge field needs to be constant or all three
vector-multiplet scalars must vanish. This implies that the $SO(2)_{R}$
symmetric $AdS_{5}\times\Sigma$ solution can only exist in pure $N=2$ gauged
supergravity.

####  III.1.2 $SO(2)\times SO(2)$ symmetric solution

We now repeat the same procedure for more complicated solutions with
$SO(2)\times SO(2)$ symmetry. With the explicit form of $A^{\prime}_{1}$ and
$A^{\prime}_{2}$ given in (291), non-vanishing components of the dressed field
strength tensors read

$\mathbf{F}_{1}=\frac{e^{-\sigma-\phi}}{2}(a_{1}+a_{2}e^{2\phi})f^{-\frac{5}{2}},\qquad\mathbf{F}_{2}=\frac{e^{-\sigma-\phi}}{2}(a_{1}-a_{2}e^{2\phi})f^{-\frac{5}{2}}\,.$
(83)

We can now solve equation (282) and find the solution for the $SO(2)\times
SO(2)$ singlet scalar $\phi$ of the form

$\phi=\ln\left[\frac{e^{-\frac{5\sigma}{2}}}{32a_{2}h}\left[(a_{2}-\lambda
a_{1})\widetilde{g}_{1}+K\right]\right]$ (84)

in which

$K=\kappa\sqrt{1024a_{1}a_{2}h^{2}e^{5\sigma}+(a_{2}-\lambda
a_{1})^{2}\widetilde{g}_{1}^{2}}$ (85)

and $\kappa=\pm 1$. We have included a sign factor $\lambda=+1,-1$ for later
convenience. In the present case of $SO(4)$ gauge group, we can set the
constant $\lambda=1$. In the next section, we will consider non-compact
$SO(2,2)$ gauge group and find that the solution takes the same form with
$\lambda=-1$.

With the solution for $\phi$ given above, we can solve equations (271) and
(263) resulting in the following solutions for $f$ and $g_{1}$ of the form

$\displaystyle f$ $\displaystyle=$
$\displaystyle\sqrt{2}\sqrt{\frac{he^{2\sigma}\left[(a_{2}+\lambda
a_{1})\widetilde{g}_{1}+K\right]}{s(256h^{2}e^{5\sigma}-\lambda\widetilde{g}_{1}^{2})}},$
(86) $\displaystyle g_{1}$ $\displaystyle=$
$\displaystyle\frac{25,600\sqrt{2}h^{9/2}e^{6\sigma}K^{-2}(256h^{2}e^{5\sigma}-\lambda\widetilde{g}_{1}^{2})^{-2}\left[(a_{2}+\lambda
a_{1})\widetilde{g}_{1}+K\right]^{5/2}(\sigma^{\prime})^{2}}{\left[16\sqrt{2}h^{9/2}\sqrt{(a_{2}+\lambda
a_{1})\widetilde{g}_{1}+K}+\lambda
e^{-5\sigma}\sqrt{s(256h^{2}e^{5\sigma}-\lambda\widetilde{g}_{1}^{2})}\right]}.$

We now determine the function $\widehat{A}_{1}$ defined in (253). With all the
previous results, the first equation in (291) becomes

$\widehat{A}^{\prime}_{1}=\frac{10\widetilde{g}_{1}\widehat{A}\sigma^{\prime}\left[K^{2}(a_{2}+\lambda
a_{1})-2K\widetilde{g}_{1}(a_{2}-\lambda
a_{1})^{2}+\widetilde{g}_{1}^{2}(a_{2}+\lambda a_{1})(a_{2}-\lambda
a_{1})^{2}\right]}{K\left[3K^{2}-4K\widetilde{g}_{1}(a_{2}+\lambda
a_{1})+\widetilde{g}_{1}^{2}(a_{2}-\lambda a_{1})^{2}\right]}\,.$ (88)

Solving this equation leads to the following solution

$\displaystyle\widehat{A}_{1}$ $\displaystyle=$
$\displaystyle\frac{c}{8}e^{-5\sigma}\sqrt{\frac{9K^{4}-2K^{2}\widetilde{g}_{1}^{2}\left(5a_{1}^{2}+22\lambda
a_{1}a_{2}+5a_{2}^{2}\right)+\widetilde{g}_{1}^{4}(a_{2}-\lambda
a_{1})^{4}}{a_{1}a_{2}}}$ (89)
$\displaystyle\times\exp\left[-\frac{\left(a_{1}^{2}+2a_{1}(7\lambda
a_{2}-y)+a_{2}(a_{2}-2\lambda
y)\right)\tan^{-1}\left(\frac{3K}{u\widetilde{g}_{1}}\right)}{\lambda
uy}\right.$
$\displaystyle\qquad\quad\left.-\frac{\left(a_{1}^{2}+2a_{1}(7\lambda
a_{2}+y)+a_{2}(a_{2}+2\lambda
y)\right)\tanh^{-1}\left(\frac{3K}{v\widetilde{g}_{1}}\right)}{\lambda
vy}\right]\quad$

with an integration constant $c$ and

$\displaystyle u$ $\displaystyle=$ $\displaystyle\sqrt{4a_{1}y+4\lambda
a_{2}y-5a_{1}^{2}-22\lambda a_{1}a_{2}-5a_{2}^{2}},$ (90) $\displaystyle v$
$\displaystyle=$ $\displaystyle\sqrt{4a_{1}y+4\lambda
a_{2}y+5a_{1}^{2}+22\lambda a_{1}a_{2}+5a_{2}^{2}},$ (91) $\displaystyle y$
$\displaystyle=$ $\displaystyle\sqrt{a_{1}^{2}+14\lambda
a_{1}a_{2}+a_{2}^{2}}\,.$ (92)

Similarly, we can find the solution for $g_{2}$ as

$\displaystyle g_{2}$ $\displaystyle=$
$\displaystyle\frac{400\widetilde{g}_{1}^{2}f^{5}\widehat{A}_{1}^{2}e^{2\sigma}\left[K^{2}-(a_{2}-\lambda
a_{1})^{2}\widetilde{g}_{1}^{2}\right](\sigma^{\prime})^{2}}{K^{2}g_{1}\left[3K^{2}-4K\widetilde{g}_{1}(a_{2}+\lambda
a_{1})+\widetilde{g}_{1}^{2}(a_{2}-\lambda a_{1})^{2}\right]^{2}}\,.$ (93)

Finding the solution for $A_{2}$ in the other $SO(2)$ gauge field is much more
difficult since, with all the previous results, the second equation in (291)
leads to a highly complicated differential equation. It should also be noted
that $A_{2}$ does not appear in all of the BPS equations which depend only on
$A_{2}^{\prime}$ and $A_{1}^{\prime}$. In particular, we have verified that
all the BPS conditions are satisfied by the above solution without the
explicit solution for $A_{2}$.
To make the subsequent analysis of the solutions more traceable, we will
further simplify the ansatz for vector fields by setting

$a_{1}=-a_{2}=b\,.$ (94)

In this case, we find the non-vanishing components of the dressed field
strength tensors given by

$\mathbf{F}_{1}=-be^{-\sigma}\sinh{\phi}f^{-\frac{5}{2}}\qquad\text{ and
}\qquad\mathbf{F}_{2}=be^{-\sigma}\cosh{\phi}f^{-\frac{5}{2}}\,.$ (95)

The solution found above now takes a much simpler form, with $\lambda=1$,

$\displaystyle\phi$ $\displaystyle=$
$\displaystyle\kappa\cosh^{-1}\left[\frac{\widetilde{g}_{1}e^{-\frac{5\sigma}{2}}}{16h}\right],$
(96) $\displaystyle f$ $\displaystyle=$
$\displaystyle\frac{2\sqrt{\kappa\,b\,h}e^{\sigma}}{\sqrt{s}(-256h^{2}e^{5\sigma}+\widetilde{g}_{1}^{2})^{1/4}},$
(97) $\displaystyle g_{1}$ $\displaystyle=$
$\displaystyle\frac{51,200\sqrt{\kappa\,b\,h^{9}}\,e^{6\sigma}\,(\sigma^{\prime})^{2}}{(-256h^{2}e^{5\sigma}+\widetilde{g}_{1}^{2})^{2}\left[32\sqrt{\kappa\,b\,h^{5}}-\sqrt{s}e^{-5\sigma}(-256h^{2}e^{5\sigma}+\widetilde{g}_{1}^{2})^{1/4}\right]},\qquad$
(98) $\displaystyle g_{2}$ $\displaystyle=$
$\displaystyle\frac{256c_{1}^{2}h^{2}\widetilde{g}_{1}^{2}e^{\sigma}\left[32\sqrt{\kappa\,b\,h^{5}}-\sqrt{s}e^{-5\sigma}(-256h^{2}e^{5\sigma}+\widetilde{g}_{1}^{2})^{1/4}\right]}{\sqrt{s}(-256h^{2}e^{5\sigma}+\widetilde{g}_{1}^{2})^{1/4}},$
(99) $\displaystyle\widehat{A}_{1}$ $\displaystyle=$ $\displaystyle
c_{1}\left(192h^{2}-\widetilde{g}_{1}^{2}e^{-5\sigma}\right).$ (100)

It can also be verified that all of the BPS conditions and the field equations
are satisfied if

$\text{sign}(c_{1}\widetilde{g}_{1}\sigma^{\prime})=-1\,.$ (101)

In addition, the second equation in (291) gives the following differential
equation

$A^{\prime}_{2}=\frac{5c_{1}\widetilde{g}_{1}(128h^{2}-\widetilde{g}_{1}^{2}e^{-5\sigma})\sigma^{\prime}}{\sqrt{-256e^{5\sigma}h^{2}+\widetilde{g}_{1}^{2}}}$
(102)

which can be readily solved by

$A_{2}=c_{1}\widetilde{g}_{1}e^{-5\sigma}\sqrt{-256h^{2}e^{5\sigma}+\widetilde{g}_{1}^{2}}+c_{2}$
(103)

with an integration constant $c_{2}$. We also have an explicit form of the
$A_{1}$ solution

$A_{1}=c_{1}\left(192h^{2}-\widetilde{g}_{1}^{2}e^{-5\sigma}\right)-\frac{2q}{\widetilde{g}_{1}}$
(104)

together with the Killing spinor

$\eta=\frac{Y_{0}\,e^{iqz-\frac{\sigma}{4}}}{s^{1/8}(-256h^{2}e^{5\sigma}+\widetilde{g}_{1}^{2})^{\frac{1}{16}}}\begin{pmatrix}\sqrt{4\sqrt{2}(\kappa\,b\,h^{5})^{\frac{1}{4}}+s^{\frac{1}{4}}e^{-\frac{5\sigma}{2}}(-256h^{2}e^{5\sigma}+\widetilde{g}_{1}^{2})^{\frac{1}{8}}}\\\
-\sqrt{4\sqrt{2}(\kappa\,b\,h^{5})^{\frac{1}{4}}-s^{\frac{1}{4}}e^{-\frac{5\sigma}{2}}(-256h^{2}e^{5\sigma}+\widetilde{g}_{1}^{2})^{\frac{1}{8}}}\end{pmatrix}$
(105)

with $Y_{0}$ being a constant.

We now turn to the regularity of the solution. From the $f$ solution in (97),
it is immediately seen that we have to impose the conditions
$-256e^{5\sigma}h^{2}+\widetilde{g}_{1}^{2}>0$ and $\frac{\kappa\,b}{s}>0$.
The latter is clearly consistent with the condition sign$(\kappa bs)=+1$. As
in the previous section, we define the following parameters

$B=\frac{8h^{2}\sqrt{\kappa\,b}}{\sqrt{s}},\qquad
m=\frac{\widetilde{g}_{1}}{16h},\qquad\mathcal{C}=32\,\widetilde{g}_{1}h^{2}c_{1}\,.$
(106)

The complete solution then reads

$\displaystyle f$ $\displaystyle=$
$\displaystyle\frac{Be^{\sigma}}{16h^{2}(-e^{5\sigma}+m^{2})^{1/4}},$ (107)
$\displaystyle g_{1}$ $\displaystyle=$
$\displaystyle\frac{25Be^{6\sigma}\,(\sigma^{\prime})^{2}}{1024h^{2}(-e^{5\sigma}+m^{2})^{2}\left[B-e^{-5\sigma}(-e^{5\sigma}+m^{2})^{1/4}\right]},\qquad$
(108) $\displaystyle g_{2}$ $\displaystyle=$
$\displaystyle\frac{\mathcal{C}^{2}e^{\sigma}\left[B-e^{-5\sigma}(-e^{5\sigma}+m^{2})^{1/4}\right]}{4h^{2}(-e^{5\sigma}+m^{2})^{1/4}},$
(109) $\displaystyle A_{1}$ $\displaystyle=$
$\displaystyle\frac{1}{8mh}\left[\mathcal{C}(3-4m^{2}e^{-5\sigma})-q\right],$
(110) $\displaystyle A_{2}$ $\displaystyle=$
$\displaystyle\frac{\mathcal{C}e^{-5\sigma}}{2h}\sqrt{-e^{5\sigma}+m^{2}}+c_{2},$
(111) $\displaystyle\phi$ $\displaystyle=$
$\displaystyle\kappa\cosh^{-1}\left[me^{-\frac{5\sigma}{2}}\right].$ (112)

We will also fix the form of $\sigma$ solution to

$\sigma=-\frac{1}{5}\ln r$ (113)

for $r>0$. This form of $\sigma$ implies that the constant $\mathcal{C}$ can
only be a positive integer due to (101).

In terms of the radial coordinate, the seven-dimensional metric can be written
as

$ds^{2}_{7}=\frac{Br^{1/20}}{16h^{2}(m^{2}r-1)^{1/4}}\left[ds^{2}_{AdS_{5}}+\frac{r^{-5/4}}{64W(m^{2}r-1)^{7/4}}dr^{2}+\frac{4\mathcal{C}^{2}W}{B}dz^{2}\right]$
(114)

with

$W=B-r^{3/4}(m^{2}r-1)^{1/4}$ (115)

while the solutions for $\phi$, $A_{1}$, and $A_{2}$ are respectively given by

$\displaystyle\phi$ $\displaystyle=$
$\displaystyle\kappa\cosh^{-1}\left[m\sqrt{r}\right],$ (116) $\displaystyle
A_{1}$ $\displaystyle=$
$\displaystyle\frac{1}{8mh}\left[\mathcal{C}(3-4m^{2}r)-q\right],$ (117)
$\displaystyle A_{2}$ $\displaystyle=$
$\displaystyle\frac{\mathcal{C}}{2h}\sqrt{r(m^{2}r-1)}+c_{2}\,.$ (118)

Finally, the Killing spinor $\eta$ takes the form

$\eta=Y_{0}e^{iqz}\frac{r^{1/80}}{(m^{2}r-1)^{\frac{1}{16}}}\begin{pmatrix}\sqrt{\sqrt{B}+r^{3/8}(m^{2}r-1)^{\frac{1}{8}}}\\\
-\sqrt{\sqrt{B}-r^{3/8}(m^{2}r-1)^{\frac{1}{8}}}\end{pmatrix}.$ (119)

In the present case, we find only one possible range of the radial coordinate
$r$ that leads to a regular solution

$m\neq 0,\qquad B>0,\qquad\frac{1}{m^{2}}<r<r_{1}$ (120)

with $r_{1}$ determined from $W(r_{1})=0$. The explicit form of $r_{1}$ is
given by

$r_{1}=\frac{1}{4m^{2}}+\frac{1}{2}\sqrt{X_{1}}+\frac{1}{2}\sqrt{\frac{3}{4m^{4}}-X_{1}+\frac{1}{4m^{6}\sqrt{X_{1}}}}$
(121)

in which

$X_{1}=\frac{1}{4m^{4}}-\frac{4(\frac{2}{3})^{1/3}B^{8/3}}{(-9+\sqrt{3(27+256B^{4}m^{6})})^{1/3}}+\frac{B^{4/3}(-9+\sqrt{3(27+256B^{4}m^{6})})^{1/3}}{18^{1/3}m^{2}}\,.$
(122)

Note also that the two possibilities with $m>0$ and $m<0$ are equivalent. For
definiteness, we will only consider $m>0$ case. An example of numerical plots
of the three warp factors for $m=1$, $B=1$, $h=\frac{1}{8}$, and
$\mathcal{C}=1$ is given in Figure 11.

(a) $f$ solution

(b) $g_{1}$ solution

(c) $g_{2}$ solution

Figure 11: Numerical plots of the warp factors for $SO(2)\times SO(2)$
symmetric solution in $SO(4)$ gauged supergravity with $m=1$, $B=1$,
$h=\frac{1}{8}$, and $\mathcal{C}=1$. The solution is regular in the range
$\frac{1}{m^{2}}=1<r<r_{1}=1.38$ with the two vertical red dashed lines
representing the two boundaries.

The result is very similar to Case VII in the previous section. As
$r\rightarrow\frac{1}{m^{2}}$, the seven-dimensional metric becomes conformal
to a product of $AdS_{5}$ and a cylinder. With the new radial coordinate $R$
given by $r=\frac{1}{m^{2}}+256C^{8}m^{4}B^{4}R^{8}$, the metric near $R=0$ is
given by (70). On the other side, as $r\rightarrow r_{1}$, the seven-
dimensional metric is approximately given by

$ds_{7}^{2}\approx\frac{B\,r_{1}^{1/20}}{16h^{2}(m^{2}r_{1}-1)^{1/4}}\left[ds^{2}_{AdS_{5}}+\frac{dR^{2}+4\mathcal{C}^{2}\left[3-4m^{2}r_{1}\right]^{2}R^{2}dz^{2}}{-16W^{\prime}(r_{1})r_{1}^{5/4}(m^{2}r_{1}-1)^{7/4}}\right]$
(123)

in which $R$ is the new radial coordinate defined by $R=\sqrt{r_{1}-r}$. The
$z$ circle shrinks smoothly giving rise to an $\mathbb{R}^{2}/\mathbb{Z}_{l}$
orbifold at $r=r_{1}$ if we impose

$\mathcal{C}=-\frac{1}{2l\left[3-4m^{2}r_{1}\right]},\qquad l=1,2,3,\ldots\,.$
(124)

As in the previous section, the explicit form of $3-4m^{2}r_{1}$ is very
complicated, so we will not give it here but only show that this function is
negative and less than -1 in the regularity range by a numerical plot in
Figure 12.

Figure 12: A numerical plot of the function $3-4m^{2}r_{1}$ appearing in the
condition (124). Note also that $3-4m^{2}r_{1}\leq-1$ in the regularity range
(120).

In order for the $SO(2)\times SO(2)$ gauge fields to vanish at $r=r_{1}$, we
fix the constants $q$ and $c_{2}$ to be

$\displaystyle q=-\mathcal{C}\left[3-4m^{2}r_{1}\right]=\frac{1}{2l}$ and
$\displaystyle
c_{2}=-\frac{\mathcal{C}}{2h}\sqrt{r_{1}(1+m^{2}r_{1})}=-\frac{\mathcal{C}B}{2h}$
(125)

leading to

$\displaystyle A_{1}=\frac{m\,\mathcal{C}}{2h}(r_{1}-r)$ and $\displaystyle
A_{2}=\frac{\mathcal{C}}{2h}\left[\sqrt{r(m^{2}r-1)}-\sqrt{r_{1}(m^{2}r_{1}-1)}\right].$
(126)

We also note that $A_{1}$ is well-defined for all values of $r$ while $A_{2}$
is complex for $r<\frac{1}{m^{2}}$.

The Killing spinor $\eta$ at the endpoint $r=r_{1}$ is given by

$\eta=\sqrt{2}Y_{0}e^{\frac{iz}{2l}}r_{1}^{1/5}\footnotesize\begin{pmatrix}1\\\
0\end{pmatrix}.$ (127)

###  III.2 $SO(2)_{\text{diag}}$ symmetric solution

We now consider $AdS_{5}\times\Sigma$ solutions with
$SO(2)_{\text{diag}}\subset SO(2)\times SO(2)$ symmetry generated by
$J^{(1)}_{12}+J^{(2)}_{12}$. In this case, the two $SO(2)$ gauge fields are
related by

$\widetilde{g}_{2}A_{2}=\widetilde{g}_{1}A_{1}$ (128)

resulting in the following ansatz for the gauge fields

$A^{I}_{(1)}=A_{1}\left(\delta^{I}_{3}+\frac{\widetilde{g}_{1}}{\widetilde{g}_{2}}\delta^{I}_{6}\right)dz\,.$
(129)

The corresponding two-form field strength is given by

$F^{I}_{(2)}=A^{\prime}\left(\delta^{I}_{3}+\frac{\widetilde{g}_{1}}{\widetilde{g}_{2}}\delta^{I}_{6}\right)\,dr\wedge
dz\,.$ (130)

We can also consistently set $C_{(3)}=0$ as in the previous cases.

Apart from the $SO(2)\times SO(2)$ singlet scalar $\phi$ corresponding to the
non-compact-generator $Y_{33}$, there are two additional scalars from
$SO(3,3)/SO(3)\times SO(3)$ coset that are invariant under
$SO(2)_{\text{diag}}$. These scalars correspond to the non-compact generators

$\hat{Y}_{1}=Y_{11}+Y_{22}\qquad\text{ and
}\qquad\hat{Y}_{2}=Y_{12}-Y_{21}\,.$ (131)

The coset representative then takes the form of

$L=e^{\varphi_{1}\hat{Y}_{1}}e^{\phi Y_{33}}e^{\varphi_{2}\hat{Y}_{2}}\,.$
(132)

This leads to the following non-vanishing $C$-functions

$\displaystyle C$ $\displaystyle=$
$\displaystyle\frac{3}{\sqrt{2}}\left[\widetilde{g}_{1}\cosh{\phi}-\widetilde{g}_{2}\sinh{\phi}+\cosh{2\varphi_{1}}\cosh{2\varphi_{2}}\left(\widetilde{g}_{1}\cosh{\phi}+\widetilde{g}_{2}\sinh{\phi}\right)\right],$
$\displaystyle C^{11}$ $\displaystyle=$ $\displaystyle
C^{22}=-\frac{1}{\sqrt{2}}\left(\widetilde{g}_{1}\cosh{\phi}+\widetilde{g}_{2}\sinh{\phi}\right)\sinh{2\varphi_{1}},$
$\displaystyle C^{12}$ $\displaystyle=$
$\displaystyle-C^{21}=-\frac{1}{\sqrt{2}}\left(\widetilde{g}_{1}\cosh{\phi}+\widetilde{g}_{2}\sinh{\phi}\right)\cosh{2\varphi_{1}}\sinh{2\varphi_{2}},$
$\displaystyle C^{33}$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}\left[\widetilde{g}_{2}\cosh{\phi}-\widetilde{g}_{1}\sinh{\phi}-\cosh{2\varphi_{1}}\cosh{2\varphi_{2}}\left(\widetilde{g}_{2}\cosh{\phi}+\widetilde{g}_{1}\sinh{\phi}\right)\right].$

The scalar vielbein is given by

$P^{ir}_{(1)}=\begin{pmatrix}\varphi^{\prime}_{1}\cosh{2\varphi_{2}}&\varphi^{\prime}_{2}&0\\\
-\varphi^{\prime}_{2}&\varphi^{\prime}_{1}\cosh{2\varphi_{2}}&0\\\
0&0&\phi^{\prime}\end{pmatrix}dr$ (134)

while the composite connection takes the form of

$Q^{ij}_{(1)}=-\varepsilon^{ij3}(\varphi^{\prime}_{1}\sinh{2\varphi_{2}}dr-\widetilde{g}_{1}A_{1}dz).$
(135)

With all these, the vector field equation (25) gives rise to the following
equations

$\displaystyle\varphi^{\prime}_{2}$ $\displaystyle=$
$\displaystyle\frac{\varphi^{\prime}_{1}}{2}\text{coth}{2\varphi_{1}}\sinh{4\varphi_{2}},$
(136) $\displaystyle A^{\prime\prime}_{1}$ $\displaystyle=$
$\displaystyle-\left[\frac{5f^{\prime}}{2f}-\frac{g^{\prime}_{1}}{2g_{1}}-\frac{g^{\prime}_{2}}{2g_{2}}+\sigma^{\prime}+\frac{2(\widetilde{g}_{1}\cosh{2\phi}+\widetilde{g}_{2}\sinh{2\phi})\phi^{\prime}}{\widetilde{g}_{2}\cosh{2\phi}+\widetilde{g}_{1}\sinh{2\phi}}\right]A^{\prime}_{1},$
(137) $\displaystyle A^{\prime\prime}_{1}$ $\displaystyle=$
$\displaystyle-\left[\frac{5f^{\prime}}{2f}-\frac{g^{\prime}_{1}}{2g_{1}}-\frac{g^{\prime}_{2}}{2g_{2}}+\sigma^{\prime}+\frac{2(\widetilde{g}_{2}\cosh{2\phi}+\widetilde{g}_{1}\sinh{2\phi})\phi^{\prime}}{\widetilde{g}_{1}\cosh{2\phi}+\widetilde{g}_{2}\sinh{2\phi}}\right]A^{\prime}_{1}.\qquad$
(138)

Consistency between the last two equations for $\phi^{\prime}\neq 0$ requires

$\widetilde{g}_{2}=\pm\widetilde{g}_{1}$ (139)

leading to a differential equation for the $SO(2)_{\text{diag}}$ gauge field

$A^{\prime\prime}_{1}=-\left(\frac{5f^{\prime}}{2f}-\frac{g^{\prime}_{1}}{2g_{1}}-\frac{g^{\prime}_{2}}{2g_{2}}+\sigma^{\prime}\pm
2\phi^{\prime}\right)A^{\prime}_{1}\,.$ (140)

The plus/minus sign arises from that in the relation
$\widetilde{g}_{2}=\pm\widetilde{g}_{1}$. From now on, we will choose
$\widetilde{g}_{2}=\widetilde{g}_{1}$ for definiteness and find

$A^{\prime}_{1}=A^{\prime}_{2}=be^{-\sigma-2\phi}\sqrt{g_{1}g_{2}}f^{-\frac{5}{2}}\,.$
(141)

Non-vanishing components of the dressed field strength tensors are given by

$\mathbf{F}_{1}=\mathbf{F}_{2}=be^{-\sigma-\phi}f^{-\frac{5}{2}}\,.$ (142)

Unlike the previous cases, there are additional BPS conditions arising from
the transformations $\delta\lambda^{a1}=0$ and $\delta\lambda^{a2}=0$
resulting in

$0=\left[\frac{\varphi^{\prime}_{1}\cosh{2\varphi_{2}}}{\sqrt{g_{1}}}\Gamma^{\hat{r}}-\frac{1}{\sqrt{2}}e^{-\frac{\sigma}{2}}C^{11}\right]{(\sigma^{1})^{a}}_{b}\epsilon^{b}+\left[\frac{\varphi^{\prime}_{2}}{\sqrt{g_{1}}}\Gamma^{\hat{r}}-\frac{1}{\sqrt{2}}e^{-\frac{\sigma}{2}}C^{12}\right]{(\sigma^{2})^{a}}_{b}\epsilon^{b}\,.$
(143)

We note here that the projector in (245) implies

${(\sigma^{2})^{a}}_{b}\epsilon^{b}=i{(\sigma^{1})^{a}}_{b}\epsilon^{b}\,.$
(144)

Therefore, with the supersymmetry parameter (240), we can rewrite the above
condition as

$0=\frac{1}{\sqrt{g_{1}}}\left(\varphi^{\prime}_{1}\cosh{2\varphi_{2}}+i\varphi^{\prime}_{2}\right)\sigma^{3}\eta-\frac{1}{\sqrt{2}}e^{-\frac{\sigma}{2}}\left(C^{11}+iC^{12}\right)(i\sigma^{2}\eta).$
(145)

There are seventeen BPS conditions obtained from the vanishing of the matrices
$\mathcal{A}$, $\mathcal{B}$, and $\mathcal{C}$ derived from this BPS
equation. Two of these are given by

$\displaystyle 0$ $\displaystyle=$
$\displaystyle\frac{s}{\sqrt{f\,g_{1}}}(\varphi^{\prime}_{1}\cosh{2\varphi_{2}}+i\varphi^{\prime}_{2}),$
(146) $\displaystyle 0$ $\displaystyle=$
$\displaystyle-\frac{\widetilde{g}_{1}}{\sqrt{2}}e^{\phi}A^{\prime}_{1}(\sinh{2\varphi_{1}}+i\cosh{2\varphi_{1}}\sinh{2\varphi_{2}}).\quad$
(147)

For non-vanishing $\widetilde{g}_{1}$ and $A^{\prime}_{1}$, the second
condition is satisfied only by setting $\varphi_{1}=\varphi_{2}=0$. All the
remaining conditions are also satisfied by this result.

In addition, repeating the same procedure as in the previous sections with
$\varphi_{1}=\varphi_{2}=0$, we find the same set of the BPS conditions (261)
to (288). The resulting $SO(2)_{\text{diag}}$ symmetric solution is then given
by

$\displaystyle f$ $\displaystyle=$
$\displaystyle\frac{2\sqrt{2}\sqrt{\widetilde{g}_{1}b\,h}e^{\sigma}}{\sqrt{s(256h^{2}e^{5\sigma}-\widetilde{g}_{1}^{2})}},$
(148) $\displaystyle g_{1}$ $\displaystyle=$
$\displaystyle\frac{204,800\sqrt{2}\sqrt{\widetilde{g}_{1}b\,h^{9}}\,e^{6\sigma}\,(\sigma^{\prime})^{2}}{(256h^{2}e^{5\sigma}-\widetilde{g}_{1}^{2})^{2}\left[32\sqrt{2}\sqrt{b\,h^{5}}-e^{-5\sigma}\sqrt{s(256h^{2}e^{5\sigma}-\widetilde{g}_{1}^{2})}\right]},$
(149) $\displaystyle g_{2}$ $\displaystyle=$
$\displaystyle\frac{1,024c^{2}\widetilde{g}_{1}^{2}h^{2}e^{\sigma}\left[32\sqrt{2}\sqrt{\widetilde{g}_{1}b\,h^{5}}-e^{-5\sigma}\sqrt{s(256h^{2}e^{5\sigma}-\widetilde{g}_{1}^{2})}\right]}{\sqrt{s(256h^{2}e^{5\sigma}-\widetilde{g}_{1}^{2})}},$
(150) $\displaystyle\phi$ $\displaystyle=$
$\displaystyle\frac{5\sigma}{2}+\ln\left[\frac{16h}{\widetilde{g}_{1}}\right],$
(151) $\displaystyle A_{1}$ $\displaystyle=$ $\displaystyle
c\left(128h^{2}-\widetilde{g}_{1}^{2}e^{-5\sigma}\right)-\frac{2q}{\widetilde{g}_{1}}$
(152)

with the usual sign condition
$\text{sign}(c\widetilde{g}_{1}\sigma^{\prime})=+1$. The Killing spinor is
given by

$\eta=\frac{Y_{0}\,e^{iqz+\frac{\sigma}{4}}}{\left[s(256h^{2}e^{5\sigma}-\widetilde{g}_{1}^{2})\right]^{\frac{1}{8}}}\begin{pmatrix}\sqrt{8(\widetilde{g}_{1}b\,h^{5})^{\frac{1}{4}}+se^{-\frac{5\sigma}{2}}\left[2s(256h^{2}e^{5\sigma}-\widetilde{g}_{1}^{2})\right]^{\frac{1}{4}}}\\\
-\sqrt{8(\widetilde{g}_{1}b\,h^{5})^{\frac{1}{4}}-se^{-\frac{5\sigma}{2}}\left[2s(256h^{2}e^{5\sigma}-\widetilde{g}_{1}^{2})\right]^{\frac{1}{4}}}\end{pmatrix}.$
(153)

It should be remarked that the half-spindle solution found in Bah_M5  from the
$U(1)^{2}$ truncation of the maximal seven-dimensional gauged supergravity can
also be recovered from the above solution by the following identification

$\displaystyle\sigma$ $\displaystyle=$
$\displaystyle-\frac{2}{3}\lambda,\qquad\widetilde{g}_{1}=2m,\qquad
b=\frac{8B^{2}}{m^{4}},$ $\displaystyle c$ $\displaystyle=$
$\displaystyle\frac{\mathcal{C}^{2}}{2m^{2}},\qquad h=\frac{m}{8}$ (154)

where $\lambda$, $B$, $m$, and $\mathcal{C}$ are the scalar field and
constants used in Bah_M5 . In more detail, the non-vanishing gauge field in
the solution of Bah_M5  is given by

$A^{(1)}=A^{12}=\frac{1}{2}(A^{3}+A^{6})=A^{3}$ (155)

with $A^{6}=A^{3}$. The scalar fields are related by the following form of the
unimodular matrix

$T_{ab}=\text{diag}(-\frac{\sigma}{2}-\phi,-\frac{\sigma}{2}-\phi,-\frac{\sigma}{2}+\phi,-\frac{\sigma}{2}+\phi,\,2\sigma).$
(156)

with $\sigma=-2(\lambda_{1}+\lambda_{2})$ and $\phi=\lambda_{2}-\lambda_{1}$.
In this equation, $\lambda_{1}$ and $\lambda_{2}$ are the two scalars of the
$U(1)^{2}$ truncation of $N=4$ gauged supergravity used in Bah_M5 . Therefore,
in a sense, the $SO(2)_{\text{diag}}$ symmetric solution found in this work
contains the solution of Bah_M5  for a special value of the gauge coupling
constant $\widetilde{g}_{1}=16h$. However, the present solution preserves only
eight supercharges corresponding to $N=1$ superconformal symmetry in four
dimensions.

To further analyze the solution, we introduce the following constants, as in
the previous cases,

$B=2\sqrt{2}h^{3/2}\sqrt{\widetilde{g}_{1}b},\qquad
m=\frac{\widetilde{g}_{1}}{16h},\qquad\mathcal{C}=32\widetilde{g}_{1}h^{2}c\,.$
(157)

We will also take the solution for $\sigma$ as given in (113). The seven-
dimensional metric reads

$ds^{2}_{7}=\frac{Br^{3/10}}{16h^{2}\sqrt{s(1-m^{2}r)}}\left[ds^{2}_{AdS_{5}}+\frac{r^{-3/2}}{16W\left[s(1-m^{2}r)\right]^{3/2}}dr^{2}+\frac{16\mathcal{C}^{2}W}{B}dz^{2}\right]$
(158)

with

$W=B-\sqrt{sr(1-m^{2}r)}\,.$ (159)

The scalar field $\phi$ and the $SO(2)_{\text{diag}}$ gauge field are
respectively given by

$\phi=-\frac{1}{2}\ln r-\ln m\qquad\text{ and }\qquad
A_{1}=-\frac{1}{8mh}\left[2|\mathcal{C}|(1-2m^{2}r)+q\right].$ (160)

Since we have used $\sigma<0$, it follows that the constant $\mathcal{C}$ must
be negative due to the condition (49). The Killing spinor takes the form of

$\eta=Y_{0}e^{iqz}\frac{2^{1/8}r^{3/40}}{\left[s(1-m^{2}r)\right]^{\frac{1}{8}}}\begin{pmatrix}\sqrt{\sqrt{B}+sr^{1/4}\left[s(1-m^{2}r)\right]^{\frac{1}{4}}}\\\
-\sqrt{\sqrt{B}-sr^{1/4}\left[s(1-m^{2}r)\right]^{\frac{1}{4}}}\end{pmatrix}.$
(161)

The behaviours of the solution are very similar to those given in Section
III.1.1 and in Bah_M5 . However, the first equation in (160) implies that $r$
and $m$ must be positive in order to find a real solution for $\phi$.
Therefore, there is no regular solution with $m<0$ as in Case II of Section
III.1.1. For convenience, the remaining six possibilities with $m>0$ will be
combined into four different cases as follows

$\displaystyle i)\qquad s$ $\displaystyle=$ $\displaystyle+1,\quad
0<B<\frac{1}{2m},\quad r\in(0,r_{-})\cup(r_{+},\frac{1}{m^{2}}),\quad$
$\displaystyle ii)\qquad s$ $\displaystyle=$ $\displaystyle+1,\quad
B=\frac{1}{2m},\quad r\in(0,r_{\ast})\cup(r_{\ast},\frac{1}{m^{2}}),$
$\displaystyle iii)\qquad s$ $\displaystyle=$ $\displaystyle+1,\quad
B>\frac{1}{2m},\quad r\in(0,\frac{1}{m^{2}}),$ $\displaystyle iv)\qquad s$
$\displaystyle=$ $\displaystyle-1,\quad B>0,\quad r\in(\frac{1}{m^{2}},r_{+})$
(162)

in which

$r_{\pm}=\frac{1\pm\sqrt{1-4sB^{2}m^{2}}}{2m^{2}}$ (163)

are the two roots of $W=B-\sqrt{sr(1-m^{2}r)}=0$. For $B=\frac{1}{2m}$ with
$s=+1$, the two roots are equal, $r_{+}=r_{-}=r_{\ast}=\frac{1}{2m^{2}}$.

Examples of numerical solutions for the warp factors in these four cases are
given in Figures 13, 14, 15, and 16, respectively.

(a) $f$ solution

(b) $g_{1}$ solution

(c) $g_{2}$ solution

Figure 13: Numerical plots of the warp factors for $SO(2)_{\text{diag}}$
symmetric solution in $SO(4)$ gauge group in Case $i$ with $s=1$,
$m=\frac{1}{2}$, $B=\frac{4}{5}$, $h=\frac{1}{2}$, and $\mathcal{C}=-1$. All
the warp factors are positive in the ranges $0<r<r_{-}=0.8$ and
$r_{+}=3.2<r<\frac{1}{m^{2}}=4$ with the four vertical red dashed lines
representing the four boundaries.

(a) $f$ solution

(b) $g_{1}$ solution

(c) $g_{2}$ solution

Figure 14: Numerical plots of the warp factors for $SO(2)_{\text{diag}}$
symmetric solution in $SO(4)$ gauge group in Case $ii$ with $s=1$,
$m=\frac{1}{2}$, $B=1$, $h=\frac{1}{2}$, and $\mathcal{C}=-1$. The warp
factors are positive in the ranges $0<r<r_{\ast}=2$ and
$r_{\ast}=2<r<\frac{1}{m^{2}}=4$ with the three vertical red dashed lines
representing the three boundaries.

(a) $f$ solution

(b) $g_{1}$ solution

(c) $g_{2}$ solution

Figure 15: Numerical plots of the warp factors for $SO(2)_{\text{diag}}$
symmetric solution in $SO(4)$ gauge group in Case $iii$ with $s=1$,
$m=\frac{1}{2}$, $B=2$, $h=\frac{1}{2}$, and $\mathcal{C}=-1$. The warp
factors are positive in the range $0<r<\frac{1}{m^{2}}=4$ with the two
vertical red dashed lines representing the two boundaries.

(a) $f$ solution

(b) $g_{1}$ solution

(c) $g_{2}$ solution

Figure 16: Numerical plots of the warp factors for $SO(2)_{\text{diag}}$
symmetric solution in $SO(4)$ gauge group in Case $iv$ with $s=-1$, $m=4$,
$B=4$, $h=\frac{1}{2}$, and $\mathcal{C}=-1$. The warp factors are positive in
the range $\frac{1}{m^{2}}=0.0625<r<r_{+}=1.03$ with the two vertical red
dashed lines representing the two boundaries.

The endpoint behaviours are also similar to the $SO(2)_{R}$ symmetric
solution. As $r\rightarrow 0$ or $r\rightarrow\frac{1}{m^{2}}$, the seven-
dimensional metric is conformal to a product of $AdS_{5}$ and a cylinder. The
explicit form of the metric in these limits can be found in the same way as in
Case I with $r\rightarrow 0$ and Case III with $r\rightarrow\frac{1}{m^{2}}$.

The more interesting behaviours as $r\rightarrow r_{\pm}$ are in Case $i$ and
Case $iv$. In terms of the new radial coordinate $R=\sqrt{r_{\pm}-r}$, the
seven-dimensional metric, as $r\rightarrow r_{\pm}$, is approximately given by

$ds_{7}^{2}\approx\frac{B\,r_{\pm}^{3/10}}{16h^{2}\sqrt{s(1-m^{2}r_{\pm})}}\left[ds^{2}_{AdS_{5}}+\frac{dR^{2}+16\mathcal{C}^{2}\left[1-2m^{2}r_{\pm}\right]^{2}R^{2}dz^{2}}{-4W^{\prime}(r_{\pm})r_{\pm}^{3/2}(1-m^{2}r_{\pm})^{3/2}}\right].$
(164)

The $z$ circle shrinks smoothly near these endpoints $r=r_{\pm}$ if we impose

$|\mathcal{C}|=\frac{1}{4l\left|1-2m^{2}r_{\pm}\right|}=\frac{1}{4l\sqrt{1-4sB^{2}m^{2}}},\qquad
l=1,2,3,\ldots\,.$ (165)

Unlike in the previous solutions, this condition can be written explicitly in
terms of $B$ and $m$ since $r_{\pm}$ takes a much simpler form than
$r_{(\pm_{1},\pm_{2})}$ and $r_{1}$.

In order for the $SO(2)_{\text{diag}}$ gauge field to vanish at $r=r_{\pm}$,
we fix the constant $q$ such that

$q=\mp 2|\mathcal{C}||1-2m^{2}r_{\pm}|=\mp\frac{1}{2l}\qquad\textrm{and}\qquad
A_{1}=\frac{m|\mathcal{C}|}{2h}(r-r_{\pm})$ (166)

which gives rise to the Killing spinor $\eta$ at $r=r_{\pm}$ of the form

$\eta=2^{-3/8}Y_{0}e^{\mp\frac{iz}{2l}}r^{1/5}_{\pm}\footnotesize\begin{pmatrix}s+1\\\
s-1\end{pmatrix}.$ (167)

Finally, it should be pointed out that if $B=\frac{1}{2m}$, we find that
$|\mathcal{C}|$ diverges due to $\sqrt{1-4B^{2}m^{2}}=0$. This implies that
the $z$ circle does not shrink smoothly at $r=r_{\ast}=\frac{1}{2m^{2}}$ in
Case $ii$ as in Cases IV and V of the $SO(2)_{R}$ symmetric solution. As
$r\rightarrow r_{\ast}$, the seven-dimensional metric is approximately given
by

$ds^{2}_{7}\approx\frac{1}{32(2^{4/5})h^{2}m^{8/5}}\left[2ds^{2}_{AdS_{5}}+\frac{dR^{2}+64\mathcal{C}^{2}m^{4}dz^{2}}{R^{2}}\right]$
(168)

with the new radial coordinate $R$ given by $r=\frac{3}{4m}-\frac{1}{R}$ and
$R\rightarrow+\infty$. In order for the $SO(2)_{\text{diag}}$ gauge field to
vanish at $r=r_{\ast}$, we have to set $q=0$ resulting in the $z$-independent
Killing spinor as in Case IV and Case V of the pure supergravity solution.

##  IV $SO(2,2)$ gauge group

In this section, we consider $N=2$ gauged supergravity with non-compact
$SO(2,2)\sim SO(2,1)\times SO(2,1)$ gauge group. The embedding of this gauge
group in $SO(3,n)$ requires at least $n=3$ vector multiplets. We will work
with the minimal number of $n=3$ and choose the following $SO(2,2)$ structure
constants

$f_{IJK}=(\widetilde{g}_{1}\varepsilon_{\bar{i}\bar{j}\bar{k}},-\widetilde{g}_{2}\varepsilon_{\bar{r}\bar{s}\bar{t}}).$
(169)

Indices $\bar{i},\bar{j},...=1,2,6$ and $\bar{r},\bar{s},...=3,4,5$ are two
sets of $SO(2,1)$ indices raised and lowered by
$\eta_{\bar{i}\bar{j}}=\text{diag}(-1,-1,1)$ and
$\eta_{\bar{r}\bar{s}}=\text{diag}(-1,1,1)$, respectively. Moreover,
$\widetilde{g}_{1}$ and $\widetilde{g}_{2}$ are coupling constants for the two
$SO(2,1)$ subgroups respectively generated by the generators

$\displaystyle T_{1}$ $\displaystyle=$ $\displaystyle Y_{23},\qquad\ T_{2}\ =\
-Y_{13},\qquad T_{6}\ =\ J^{(1)}_{12},$ (170) $\displaystyle T_{3}$
$\displaystyle=$ $\displaystyle J^{(2)}_{12},\qquad T_{4}\ =\
Y_{32},\qquad\quad T_{5}\ =\ -Y_{31}\,.$ (171)

These satisfy $SO(2,1)\times SO(2,1)$ algebra

$[T_{\bar{i}},T_{\bar{j}}]=-{\varepsilon_{\bar{i}\bar{j}}}^{\bar{k}}T_{\bar{k}},\qquad[T_{\bar{r}},T_{\bar{s}}]={\varepsilon_{\bar{r}\bar{s}}}^{\bar{t}}T_{\bar{t}},\qquad[T_{\bar{i}},T_{\bar{r}}]=0\,.$
(172)

The compact $SO(2)\times SO(2)$ subgroup is generated by $T_{3}$ and $T_{6}$.
In this case, $T_{6}$ generates $SO(2)_{R}$ subgroup of the $SO(3)_{R}$
R-symmetry while $T_{3}$ generates the other $SO(2)$ in the $SO(3)$ symmetry
of the vector multiplets. We now repeat the same procedure of finding
$AdS_{5}\times\Sigma$ solutions as in the previous section. Unlike the $SO(4)$
gauge group, this gauged supergravity does not admit any supersymmetric
$AdS_{7}$ vacua. The maximally supersymmetric vacua are given by half-
supersymmetric domain walls dual to non-conformal $N=(1,0)$ field theories in
six dimensions 7D_noncompact . In this case, $AdS_{5}\times\Sigma$ solutions
would describe conformal fixed points in four dimensions of six-dimensional
$N=(1,0)$ field theories on a half-spindle.

###  IV.1 $SO(2)\times SO(2)$ symmetric solution

As in Section III, supersymmetric $AdS_{5}\times\Sigma$ solutions preserving
$SO(2)\times SO(2)\subset SO(2,1)\times SO(2,1)\sim SO(2,2)$ can be found by
using the same metric ansatz (33), $SO(2)\times SO(2)$ gauge fields (37), and
$SO(2)\times SO(2)$ singlet scalar $\phi$ corresponding to the non-compact
generator $Y_{33}$. As in $SO(4)$ gauge group, we can consistently set
$C_{(3)}=0$ due to $F^{I}_{(2)}\wedge F^{I}_{(2)}=0$.

With the coset representative (39), the scalar vielbein is still given by
(41). However, the composite connection for $SO(2)_{R}$ is now given by

$Q^{ij}_{(1)}=\widetilde{g}_{1}A_{2}\varepsilon^{ij3}dz\,.$ (173)

The $C$-functions take the form of

$C=-3\sqrt{2}\,\widetilde{g}_{1}\sinh\phi,\qquad
C^{ir}=\sqrt{2}\,\widetilde{g}_{1}\cosh{\phi}\,\delta^{i}_{3}\delta^{r}_{3}.$
(174)

By the same analysis, we can derive a similar set of BPS equations with
$A_{1}$ and $A_{2}$ interchanged. General solutions for $\phi$, $f$, $g_{1}$,
$\widehat{A}_{2}$, and $g_{2}$ are given in (84), (86), (LABEL:g1_gen_sol),
(89), and (93), respectively with $\lambda=-1$. As in the case of $SO(4)$
gauge group, we are not able to determine an analytic solution for the second
$SO(2)$ gauge field in this case given by $A_{1}$. Therefore, to explicitly
write down a complete solution and discuss some properties of the solution, we
further simplify the BPS equations by choosing $a_{1}=-a_{2}=b$ as in the
$SO(4)$ case. It should be noted here that setting $a_{1}=a_{2}=b$ is also
possible in this case but does not lead to any new solutions.

Repeating the same analysis as in the $SO(4)$ case, we can eventually solve
all the BPS conditions and find the $SO(2)\times SO(2)$ symmetric solution

$\displaystyle\phi$ $\displaystyle=$
$\displaystyle\sinh^{-1}\left[\frac{\widetilde{g}_{1}e^{-\frac{5\sigma}{2}}}{16h}\right],$
(175) $\displaystyle f$ $\displaystyle=$
$\displaystyle\frac{2\sqrt{bh}e^{\sigma}}{\sqrt{s}(256h^{2}e^{5\sigma}+\widetilde{g}_{1}^{2})^{1/4}},$
(176) $\displaystyle g_{1}$ $\displaystyle=$
$\displaystyle\frac{51,200\sqrt{b\,h^{9}}\,e^{6\sigma}\,(\sigma^{\prime})^{2}}{(256h^{2}e^{5\sigma}+\widetilde{g}_{1}^{2})^{2}\left[32\sqrt{b\,h^{5}}-\sqrt{s}e^{-5\sigma}(256h^{2}e^{5\sigma}+\widetilde{g}_{1}^{2})^{1/4}\right]},$
(177) $\displaystyle g_{2}$ $\displaystyle=$
$\displaystyle\frac{256c_{1}^{2}h^{2}\widetilde{g}_{1}^{2}e^{\sigma}\left[32\sqrt{b\,h^{5}}-\sqrt{s}e^{-5\sigma}(256h^{2}e^{5\sigma}+\widetilde{g}_{1}^{2})^{1/4}\right]}{\sqrt{s}(256h^{2}e^{5\sigma}+\widetilde{g}_{1}^{2})^{1/4}},$
(178) $\displaystyle A_{1}$ $\displaystyle=$ $\displaystyle-
c_{1}\widetilde{g}_{1}e^{-5\sigma}\sqrt{256h^{2}e^{5\sigma}+\widetilde{g}_{1}^{2}}+c_{2},$
(179) $\displaystyle A_{2}$ $\displaystyle=$ $\displaystyle
c_{1}\left(192h^{2}+\widetilde{g}_{1}^{2}e^{-5\sigma}\right)-\frac{2q}{\widetilde{g}_{1}}$
(180)

together with the sign condition
$\text{sign}(c\widetilde{g}_{1}\sigma^{\prime})=+1$. The explicit form of the
Killing spinor is given by

$\eta=\frac{Y_{0}\,e^{iqz-\frac{\sigma}{4}}}{s^{1/8}(256h^{2}e^{5\sigma}+\widetilde{g}_{1}^{2})^{\frac{1}{16}}}\begin{pmatrix}\sqrt{4\sqrt{2}(h^{5}b)^{\frac{1}{4}}+s^{1/4}e^{-\frac{5\sigma}{2}}(256h^{2}e^{5\sigma}+\widetilde{g}_{1}^{2})^{\frac{1}{8}}}\\\
-\sqrt{4\sqrt{2}(h^{5}b)^{\frac{1}{4}}-s^{1/4}e^{-\frac{5\sigma}{2}}(256h^{2}e^{5\sigma}+\widetilde{g}_{1}^{2})^{\frac{1}{8}}}\end{pmatrix}$
(181)

with $Y_{0}$ being a constant.

The analysis of the regularity of the solution proceeds as in the previous
cases. We first take the solution for the dilaton as in (113) and introduce
the following parameters

$B=\frac{8h^{2}\sqrt{b}}{\sqrt{s}},\qquad
m=\frac{\widetilde{g}_{1}}{16h},\qquad\mathcal{C}=32\,\widetilde{g}_{1}h^{2}c_{1}\,.$
(182)

The seven-dimensional metric reads

$ds^{2}_{7}=\frac{Br^{1/20}}{16h^{2}(1+m^{2}r)^{1/4}}\left[ds^{2}_{AdS_{5}}+\frac{r^{-5/4}}{64W(1+m^{2}r)^{7/4}}dr^{2}+\frac{4\mathcal{C}^{2}W}{B}dz^{2}\right]$
(183)

with

$W=B-r^{3/4}(1+m^{2}r)^{1/3}$ (184)

while the solutions for $\phi$, $A_{1}$, and $A_{2}$ are given by

$\displaystyle\phi$ $\displaystyle=$
$\displaystyle\sinh^{-1}\left[m\sqrt{r}\right],$ (185) $\displaystyle A_{1}$
$\displaystyle=$
$\displaystyle\frac{|\mathcal{C}|}{2h}\sqrt{r(1+m^{2}r)}+c_{2},$ (186)
$\displaystyle A_{2}$ $\displaystyle=$
$\displaystyle-\frac{1}{8mh}\left[|\mathcal{C}|(4m^{2}r+3)-q\right].$ (187)

The Killing spinor $\eta$ takes the form

$\eta=Y_{0}e^{iqz}\frac{r^{1/80}}{(1+m^{2}r)^{\frac{1}{16}}}\begin{pmatrix}\sqrt{\sqrt{B}+r^{3/8}(1+m^{2}r)^{\frac{1}{8}}}\\\
-\sqrt{\sqrt{B}-r^{3/8}(1+m^{2}r)^{\frac{1}{8}}}\end{pmatrix}.$ (188)

It should be noted that since we have chosen sign$(\sigma)=-1$, we need to
impose the condition sign$(c_{1}\widetilde{g}_{1})=-1$ resulting in
sign$(\mathcal{C})=-1$.

There is only one possible range for the radial coordinate $r$ in order to
obtain a regular solution

$m\neq 0,\qquad B>0,\qquad 0<r<r_{2}$ (189)

where $r_{2}$ is determined from $W(r_{2})=0$ as

$r_{2}=-\frac{1}{4m^{2}}-\frac{1}{2}\sqrt{X_{2}}+\frac{1}{2}\sqrt{\frac{3}{4m^{4}}-X_{2}+\frac{1}{4m^{6}\sqrt{X_{2}}}}$
(190)

with

$X_{2}=\frac{1}{4m^{4}}-\frac{4(2/3)^{1/3}B^{8/3}}{(\sqrt{81+768B^{4}m^{6}}-9)^{1/3}}+\frac{B^{4/3}(\sqrt{81+768B^{4}m^{6}}-9)^{1/3}}{18^{1/3}m^{2}}\,.$
(191)

There are two possibilities with $m>0$ and $m<0$. However, these are related
to each other by a sign change in the $\phi$ solution. In the following, we
will choose $m>0$ for definiteness. An example of numerical plots for the
three warp factors is given in Figure 17.

(a) $f$ solution

(b) $g_{1}$ solution

(c) $g_{2}$ solution

Figure 17: Numerical plots of the warp factors for $SO(2)\times SO(2)$
symmetric solution in $SO(2,2)$ gauge group with $m=1$, $B=1$,
$h=\frac{1}{8}$, and $\mathcal{C}=-\frac{1}{4}$. The solution is regular in
the range $0<r<r_{2}=0.819$ with the two vertical red dashed lines
representing the two boundaries.

This is very similar to Case II in Section III.1.1. As $r\rightarrow 0$, the
seven-dimensional metric is again conformal to a product of $AdS_{5}$ and a
cylinder. With the new radial coordinate $R=r^{3/8}$, the metric near $R=0$ is
given in (62). As $r\rightarrow r_{2}$, the seven-dimensional metric is
approximately given by

$ds_{7}^{2}\approx\frac{B\,r_{2}^{1/20}}{16h^{2}(1+m^{2}r_{2})^{1/4}}\left[ds^{2}_{AdS_{5}}+\frac{dR^{2}+4\mathcal{C}^{2}\left[3+4m^{2}r_{2}\right]^{2}R^{2}dz^{2}}{-16W^{\prime}(r_{2})r_{2}^{5/4}(1+m^{2}r_{2})^{7/4}}\right]$
(192)

with the coordinate $R$ defined by $R=\sqrt{r_{2}-r}$. The $z$ circle shrinks
smoothly giving rise to an $\mathbb{R}^{2}/\mathbb{Z}_{l}$ orbifold at
$r=r_{2}$ after imposing the condition

$|\mathcal{C}|=\frac{1}{2l\left[3+4m^{2}r_{2}\right]},\qquad
l=1,2,3,\ldots\,.$ (193)

Since the explicit form of $3+4m^{2}r_{2}$ derived from (190) is rather
complicated, we only numerically show that this function is always greater
than $3$ in the regularity range in Figure 18.

Figure 18: A numerical plot of the function $3+4m^{2}r_{2}$ appearing in the
condition (193).

Fixing the constants

$q=|\mathcal{C}|\left[3+4m^{2}r_{2}\right]=\frac{1}{2l},\qquad
c_{2}=-\frac{|\mathcal{C}|}{2h}\sqrt{r_{2}(1+m^{2}r_{2})}=-\frac{|\mathcal{C}|B}{2h},$
(194)

we obtain the $SO(2)\times SO(2)$ gauge fields

$A_{1}=\frac{|\mathcal{C}|}{2h}\left[\sqrt{r(1+m^{2}r)}-\sqrt{r_{2}(1+m^{2}r_{2})}\right],\qquad
A_{2}=\frac{m|\mathcal{C}|}{2h}(r_{2}-r)$ (195)

which vanish at $r=r_{2}$. The Killing spinor at the endpoint $r=r_{2}$ is
explicitly given by

$\eta=\sqrt{2}Y_{0}e^{\frac{iz}{2l}}r_{2}^{1/5}\footnotesize\begin{pmatrix}1\\\
0\end{pmatrix}.$ (196)

###  IV.2 $SO(2)_{\text{diag}}$ symmetric solution

To find solutions preserving $SO(2)_{\text{diag}}$ symmetry generated by
$T_{3}+T_{6}$ in $SO(2,2)$ gauge group, we use the following ansatz for vector
fields

$A^{I}_{(1)}=A_{2}\left(\frac{\widetilde{g}_{1}}{\widetilde{g}_{2}}\delta^{I}_{3}+\delta^{I}_{6}\right)dz\,.$
(197)

There are three $SO(2)_{\text{diag}}$ singlets corresponding to non-compact
generators $Y_{33}$ together with $\hat{Y}_{1}$ and $\hat{Y}_{2}$ given in
(131). As in $SO(4)$ gauge group, solving the vector field equations (25)
requires $\widetilde{g}_{2}=\pm\widetilde{g}_{1}$. We again choose
$\widetilde{g}_{2}=\widetilde{g}_{1}$ giving rise to non-vanishing components
of the dressed field strength tensors in (142).

Repeating the same procedure as in Section III.2, we can solve all the BPS
conditions and find the solution

$\displaystyle\phi$ $\displaystyle=$
$\displaystyle\frac{5\sigma}{2}+\ln\left[-\frac{16h}{\widetilde{g}_{1}}\right],\qquad\varphi_{1}=\varphi_{2}=0,$
(198) $\displaystyle f$ $\displaystyle=$
$\displaystyle-\frac{2\sqrt{2}\sqrt{-\widetilde{g}_{1}b\,h}e^{\sigma}}{\sqrt{s(256h^{2}e^{5\sigma}+\widetilde{g}_{1}^{2})}},$
(199) $\displaystyle g_{1}$ $\displaystyle=$
$\displaystyle\frac{204,800\sqrt{2}\sqrt{-\widetilde{g}_{1}b\,h^{9}}\,e^{6\sigma}\,(\sigma^{\prime})^{2}}{(256h^{2}e^{5\sigma}+\widetilde{g}_{1}^{2})^{2}\left[32\sqrt{2}\sqrt{-\widetilde{g}_{1}b\,h^{5}}+e^{-5\sigma}\sqrt{s(256h^{2}e^{5\sigma}+\widetilde{g}_{1}^{2})}\right]},$
(200) $\displaystyle g_{2}$ $\displaystyle=$
$\displaystyle-\frac{1,024c^{2}h^{2}e^{\sigma}\left[32\sqrt{2}\sqrt{-\widetilde{g}_{1}b\,h^{5}}+e^{-5\sigma}\sqrt{s(256h^{2}e^{5\sigma}+\widetilde{g}_{1}^{2})}\right]}{\sqrt{s(256h^{2}e^{5\sigma}+\widetilde{g}_{1}^{2})}},$
(201) $\displaystyle A_{2}$ $\displaystyle=$ $\displaystyle
c\left(128h^{2}+\widetilde{g}_{1}^{2}e^{-5\sigma}\right)-\frac{2q}{\widetilde{g}_{1}}$
(202)

together with the Killing spinor

$\eta=\frac{Y_{0}\,e^{iqz+\frac{\sigma}{4}}}{\left[2s(256h^{2}e^{5\sigma}+\widetilde{g}_{1}^{2})\right]^{\frac{1}{8}}}\begin{pmatrix}\sqrt{8(-\widetilde{g}_{1}b\,h^{5})^{\frac{1}{4}}+e^{-\frac{5\sigma}{2}}\left[2s(256h^{2}e^{5\sigma}+\widetilde{g}_{1}^{2})\right]^{\frac{1}{4}}}\\\
-\sqrt{8(-\widetilde{g}_{1}b\,h^{5})^{\frac{1}{4}}-e^{-\frac{5\sigma}{2}}\left[2s(256h^{2}e^{5\sigma}+\widetilde{g}_{1}^{2})\right]^{\frac{1}{4}}}\end{pmatrix}$
(203)

and the sign condition $\text{sign}(c\widetilde{g}_{1}\sigma^{\prime})=+1$. We
also note that the $\phi$ solution in (198) requires
$\frac{16h}{\widetilde{g}_{1}}<0$. Defining the following parameters

$B=-\frac{2\sqrt{2}h\sqrt{-\widetilde{g}_{1}b\,h}}{\sqrt{s}},\qquad
m=\frac{\widetilde{g}_{1}}{16h},\qquad\mathcal{C}=32\widetilde{g}_{1}h^{2}c$
(204)

and using the solution for $\sigma$ from (113), we find the seven-dimensional
metric given by

$ds^{2}_{7}=\frac{Br^{3/10}}{16h^{2}\sqrt{1+m^{2}r}}\left[ds^{2}_{AdS_{5}}+\frac{r^{-3/2}}{16W(1+m^{2}r)^{3/2}}dr^{2}+\frac{16\mathcal{C}^{2}W}{B}dz^{2}\right]$
(205)

with

$W=B-\sqrt{r(1+m^{2}r)}\,.$ (206)

The scalar $\phi_{2}$ and the gauge field can be written as

$\phi_{2}=-\frac{1}{2}\ln r-\ln\left[-m\right]\qquad\text{ and }\qquad
A_{2}=-\frac{1}{8mh}\left[2|\mathcal{C}|(1+2m^{2}r)+q\right].$ (207)

Furthermore, since we have chosen $\sigma<0$, the constant $\mathcal{C}$ must
be negative. Finally, the Killing spinor reads

$\eta=Y_{0}e^{iqz}\frac{2^{1/8}r^{3/40}}{(1+m^{2}r)^{\frac{1}{8}}}\begin{pmatrix}\sqrt{\sqrt{B}+\left[r(1+m^{2}r)\right]^{\frac{1}{4}}}\\\
-\sqrt{\sqrt{B}-\left[r(1+m^{2}r)\right]^{\frac{1}{4}}}\end{pmatrix}.$ (208)

As in the $SO(2)\times SO(2)$ symmetric solution, there is only one possible
range of the radial coordinate $r$ in order to obtain regular solutions. This
is given by

$m<0,\qquad B>0,\qquad 0<r<r_{3},\qquad
r_{3}=\frac{-1+\sqrt{1+4B^{2}m^{2}}}{2m^{2}}$ (209)

with $r_{3}$ determined from $W(r_{3})=0$. An example of numerical plots for
the three warp factors is given in Figure 19.

(a) $f$ solution

(b) $g_{1}$ solution

(c) $g_{2}$ solution

Figure 19: Numerical plots of the warp factors for $SO(2)_{\text{diag}}$
symmetric solution in $SO(2,2)$ gauge group with $m=-1$, $B=1$,
$h=\frac{1}{2}$, and $\mathcal{C}=-1$. The solution is regular in the range
$0<r<r_{3}=0.618$ with the two vertical red dashed lines representing the two
boundaries.

As $r\rightarrow 0$, the seven-dimensional metric becomes a conformal
rescaling of a product between $AdS_{5}$ and a cylinder as in Case I of the
$SO(4)$ gauge group. As $r\rightarrow r_{3}$, the metric is approximately
given by

$ds_{7}^{2}\approx\frac{B\,r_{3}^{3/10}}{16h^{2}\sqrt{1+m^{2}r_{3}}}\left[ds^{2}_{AdS_{5}}+\frac{dR^{2}+16\mathcal{C}^{2}\left[1+2m^{2}r_{3}\right]^{2}R^{2}dz^{2}}{-4W^{\prime}(r_{3})r_{3}^{3/2}(1+m^{2}r_{3})^{3/2}}\right]$
(210)

with the new radial coordinate $R$ defined by $R=\sqrt{r_{3}-r}$. The $z$
circle shrinks smoothly giving rise to an $\mathbb{R}^{2}/\mathbb{Z}_{l}$
orbifold at $r=r_{3}$ for

$|\mathcal{C}|=\frac{1}{4l\left[1+2m^{2}r_{3}\right]}=\frac{1}{4l\sqrt{1+4B^{2}m^{2}}},\qquad
l=1,2,3,\ldots\,.$ (211)

With the constant $q$ chosen to be

$q=-2|\mathcal{C}|\left[1+2m^{2}r_{2}\right]=-\frac{1}{2l},$ (212)

the $SO(2)_{\textrm{diag}}$ gauge field takes the form

$A_{2}=\frac{m|\mathcal{C}|}{2h}(r_{3}-r).$ (213)

The Killing spinor at the endpoint $r=r_{2}$ is given by

$\eta=2^{5/8}Y_{0}e^{\frac{-iz}{2l}}r_{2}^{1/5}\footnotesize\begin{pmatrix}1\\\
0\end{pmatrix}.$ (214)

##  V $SO(3,1)$ gauge group

In this section, we consider non-compact $SO(3,1)$ gauge group which can also
be embedded in $SO(3,3)$. The gauge structure constants are chosen to be

$f_{IJK}=-\widetilde{g}(\varepsilon_{ijk},\varepsilon_{rsi}),\qquad
r,s,\ldots=1,2,3$ (215)

with a coupling constant $\widetilde{g}$. The $SO(3,1)$ generators can be
written as $T_{I}=(T_{i},\hat{T}_{r})$ with the explicit form in terms of
$SO(3,3)$ generators given by

$\displaystyle T_{1}$ $\displaystyle=$ $\displaystyle
J^{(1)}_{23}-J^{(2)}_{23},\quad T_{2}\ =\ J^{(1)}_{31}+J^{(2)}_{31},\quad
T_{3}\ =\ J^{(1)}_{12}-J^{(2)}_{12},$ $\displaystyle\hat{T}_{1}$
$\displaystyle=$ $\displaystyle Y_{23}+Y_{32},\quad\ \,\hat{T}_{2}\ =\
Y_{13}-Y_{31},\quad\ \,\hat{T}_{3}\ =\ -Y_{12}-Y_{21}$ (216)

satisfying $SO(3,1)$ algebra

$[T_{i},T_{j}]={\varepsilon_{ij}}^{k}T_{k},\qquad[T_{i},\hat{T}_{r}]={\varepsilon_{ir}}^{s}\hat{T}_{s},\qquad[\hat{T}_{r},\hat{T}_{s}]={\varepsilon_{rs}}^{i}T_{i}\,.$
(217)

$T_{i}$ are compact generators of the maximal compact subgroup $SO(3)\subset
SO(3,1)$ while $\hat{T}_{r}$ are non-compact generators.

We now look for supersymmetric $AdS_{5}\times\Sigma$ solutions preserving
$SO(2)\subset SO(3)$ symmetry by using the same analysis as in the previous
two gauge groups. The metric is still given by (33), and the non-vanishing
$SO(2)$ gauge field takes the form of

$A^{3}_{(1)}=A(r)dz\,.$ (218)

There are three $SO(2)$ singlet scalars corresponding to the non-compact
generators

$\bar{Y}_{1}=Y_{11}-Y_{22},\qquad\bar{Y}_{2}=Y_{33},\qquad\bar{Y}_{3}=Y_{12}+Y_{21},$
(219)

and the coset representative is given by

$L=e^{\phi_{1}\bar{Y}_{1}}e^{\phi_{2}\bar{Y}_{2}}e^{\phi_{3}\bar{Y}_{3}}\,.$
(220)

With all these, the vector field equation (25) gives rise to

$\displaystyle\phi^{\prime}_{3}$ $\displaystyle=$
$\displaystyle\frac{\phi^{\prime}_{1}}{2}\tanh{2\phi_{1}}\sinh{4\phi_{3}},$
(221) $\displaystyle A^{\prime\prime}$ $\displaystyle=$
$\displaystyle-\left[\frac{5f^{\prime}}{2f}-\frac{g^{\prime}_{1}}{2g_{1}}-\frac{g^{\prime}_{2}}{2g_{2}}+\sigma^{\prime}+2\phi^{\prime}_{2}\tanh{2\phi_{2}}\right]A^{\prime},$
(222) $\displaystyle A^{\prime\prime}$ $\displaystyle=$
$\displaystyle-\left[\frac{5f^{\prime}}{2f}-\frac{g^{\prime}_{1}}{2g_{1}}-\frac{g^{\prime}_{2}}{2g_{2}}+\sigma^{\prime}+2\phi^{\prime}_{2}\,\text{coth}\,{2\phi_{2}}\right]A^{\prime}.$
(223)

Consistency between equations (222) and (223) requires $\phi^{\prime}_{2}=0$
for $A^{\prime}\neq 0$. We will write the constant $\phi_{2}$ as $c$. This
leads to a single differential equation for $A$

$A^{\prime\prime}=-\left[\frac{5f^{\prime}}{2f}-\frac{g^{\prime}_{1}}{2g_{1}}-\frac{g^{\prime}_{2}}{2g_{2}}+\sigma^{\prime}\right]A$
(224)

with the solution

$A^{\prime}=be^{-\sigma}\sqrt{g_{1}g_{2}}f^{-\frac{5}{2}}$ (225)

for an integration constant $b$. The dressed field strength tensors are now
given by

$\mathbf{F}_{1}=b\cosh{c}e^{-\sigma}f^{-\frac{5}{2}}\qquad\text{ and
}\qquad\mathbf{F}_{2}=b\sinh{c}e^{-\sigma}f^{-\frac{5}{2}}.$ (226)

In this case, non-vanishing conponenets of the $C$-functions read

$\displaystyle C$ $\displaystyle=$
$\displaystyle-3\sqrt{2}\,\widetilde{g}(\cosh{c}-\sinh{c}\sinh{2\phi_{1}}\cosh{2\phi_{3}}),$
$\displaystyle C^{11}$ $\displaystyle=$
$\displaystyle-C^{22}=-\sqrt{2}\,\widetilde{g}\sinh{c}\cosh{2\phi_{1}},$
$\displaystyle C^{12}$ $\displaystyle=$ $\displaystyle
C^{21}=-\sqrt{2}\,\widetilde{g}\sinh{c}\sinh{2\phi_{1}}\sinh{2\phi_{3}},$
$\displaystyle C^{33}$ $\displaystyle=$
$\displaystyle\sqrt{2}\,\widetilde{g}(\sinh{c}-\cosh{c}\sinh{2\phi_{1}}\cosh{2\phi_{3}}).$
(227)

The scalar vielbein and the $SO(2)$ composite connection are given by

$P^{ir}_{(1)}=\begin{pmatrix}\phi^{\prime}_{1}\cosh{2\phi_{3}}&\phi^{\prime}_{3}&0\\\
\phi^{\prime}_{3}&-\phi^{\prime}_{1}\cosh{2\phi_{3}}&0\\\
0&0&0\end{pmatrix}dr$ (228)

and

$Q^{ij}_{(1)}=\varepsilon^{ij3}(\phi^{\prime}_{1}\sinh{2\phi_{3}}dr-\widetilde{g}Adz).$
(229)

With the supersymmetry parameter (240) subject to the projector (245), the
supersymmetry transformations $\delta\lambda^{a1}=0$ and
$\delta\lambda^{a2}=0$ give the same BPS condition while
$\delta\lambda^{a3}=0$ gives another equation. These equations take the form

$\displaystyle 0$ $\displaystyle=$
$\displaystyle\widetilde{g}\sinh{c}e^{-\frac{\sigma}{2}}\left[\cosh{2\phi_{1}}+i\sinh{2\phi_{1}}\sinh{2\phi_{3}}\right](i\sigma^{2}\eta)$
(230)
$\displaystyle+\frac{1}{\sqrt{g_{1}}}\left[\phi^{\prime}_{1}\cosh{2\phi_{3}}-i\phi^{\prime}_{3}\right]\sigma^{3}\eta,$
$\displaystyle 0$ $\displaystyle=$
$\displaystyle\widetilde{g}e^{-\frac{\sigma}{2}}[\sinh{c}-\cosh{c}\sinh{2\phi_{1}}\cosh{2\phi_{3}}]\eta-e^{\frac{\sigma}{2}}be^{-\sigma}\sinh{c}f^{-\frac{5}{2}}\sigma^{3}\eta\,.\qquad$
(231)

Solving these conditions gives $\phi_{1}=0$, $c=0$ and $\phi^{\prime}_{3}=0$.
Since the constant $\phi_{3}$ does not appear in other equations, we can
simply set $\phi_{3}=0$ without losing any generality. However, with
$\phi_{1}=\phi_{2}=\phi_{3}=0$, all matter fields from the vector multiplets
vanish. The resulting solution turns out to be the same as the $SO(2)_{R}$
symmetric solution given in pure $N=2$ gauged supergravity discussed in
Section III.1.1.

##  VI Conclusions

We have found supersymmetric $AdS_{5}\times\Sigma$ solutions in which $\Sigma$
is a topological disk with a non-trivial $U(1)$ holonomy at the boundary or a
half-spindle from matter-coupled $N=2$ gauged supergravity in seven dimensions
with $SO(4)$ and $SO(2,2)$ gauge groups. These solutions preserve eight
supercharges and $SO(2)_{R}$, $SO(2)\times SO(2)$ and $SO(2)_{\text{diag}}$
symmetries. These solutions represent a new class of supersymmetric solutions
of $N=2$ gauged supergravity in seven dimensions and might be useful in
holographic studies. We have also extensively discussed various possible
ranges of the radial coordinate in which the resulting solutions are regular.
The $SO(2)_{R}$ symmetric solution requires vanishing of all fields from the
vector multiplets and can be considered as a solution of pure $N=2$ gauged
supergravity with $SO(3)$ gauge group.

For $SO(3,1)$ gauge group, we have found only solutions with $SO(2)_{R}\subset
SO(3)_{R}$ symmetry. Furthermore, we have also considered other possible gauge
groups namely $SL(3,\mathbb{R})$, $SO(2,1)$, and $SO(2,2)\times SO(2,1)\sim
SO(2,1)\times SO(2,1)\times SO(2,1)$ gauge groups. However, all of these gauge
groups do not lead to $AdS_{5}\times\Sigma$ solutions with non-vanishing
fields from vector multiplets. On the other hand, these gauge groups do admit
a solution with $SO(2)_{R}$ symmetry which appears to be a universal solution
to all gauge groups.

Similar to the result of Ferrero_M5 , the solutions should be dual to $N=1$
SCFTs in four dimensions obtained from compactifications on a half-spindle of
six-dimensional $N=(1,0)$ SCFTs in the case of $SO(4)$ gauge group or
$N=(1,0)$ non-conformal field theory in the case of $SO(2,2)$ gauge group.
Solutions of pure $N=2$ and $SO(4)$ gauged supergravity with equal $SO(3)$
coupling constants, $\widetilde{g}_{1}=\widetilde{g}_{2}$, can be embedded in
eleven-dimensional supergravity via consistent truncations constructed in
Pure_Red_Ans  and 7D_from_11D . In this case, the uplifted solutions would
describe M5-branes wrapped on a half-spindle as in Ferrero_M5 , but the brane
configurations preserve only $\frac{1}{4}$ of the maximal supersymmetry rather
than $\frac{1}{2}$. For $SO(2,2)$ gauge group, an embedding in ten-dimensional
type I or heterotic theories via a truncation on a hyperbolic space $H^{2,2}$
might be obtained by extending the result of Pope_SO2_2_7D  to $SO(2,2)$
gauged supergravity with a non-vanishing topological mass for the three-form
field. In this case, the uplifted solutions would describe D5-branes or
NS5-branes wrapped on half-spindles.

It would be interesting to explicitly identify the four-dimensional $N=1$
SCFTs that are dual to the supergravity solutions found in this paper.
Uplifting the solutions in $SO(2,2)$ gauge group and in $SO(4)$ gauge group
with different $SO(3)$ coupling constants to ten or eleven dimensions is of
particular interest in the holographic context and could lead to new
configurations of D5/NS5-branes or M5-branes wrapped on half-spindles. This
could be done along the line of Henning_dualize_IIA_IIB  in which the
embedding of maximal $N=4$ gauged supergravity with various gauge groups has
been given by using $SL(5)$ exceptional field theory. In the present case of
half-maximal gauged supergravity, the results on $SO(3,n)$ double field theory
and non-geometric fluxes in Dibitetto_non_Geometric ; 7D_N2_Malek ;
EFT_Heterotic_Malek ; Henning_Malek_AdS7_6  would be very useful. We hope to
come back to these issues in future works.

###### Acknowledgements.

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

##  Appendix A Derivation of BPS equations in $SO(4)$ gauge group

In this appendix, we analyze first-order BPS conditions derived from fermionic
supersymmetry transformations. We begin with the first condition
$\delta\psi_{\mu}^{a}=0$ along $m$, $\rho$, $r$, and $z$ directions. These are
respectively given by

$\displaystyle 0$ $\displaystyle=$ $\displaystyle
2\partial_{m}\epsilon^{a}-\frac{1}{\sqrt{f}}\Gamma_{m}\Gamma^{\hat{\rho}}\epsilon^{a}+\frac{f^{\prime}}{2f\sqrt{g_{1}}}\Gamma_{m}\Gamma^{\hat{r}}\epsilon^{a}-\left[\frac{\sqrt{2}}{30}e^{-\frac{\sigma}{2}}C+\frac{4}{5}he^{2\sigma}\right]\Gamma_{m}\epsilon^{a}$
(232)
$\displaystyle+\frac{i}{5}e^{\frac{\sigma}{2}}\mathbf{F}_{1}{(\sigma^{3})^{a}}_{b}\Gamma_{m}\Gamma^{\hat{r}\hat{z}}\epsilon^{b},$
$\displaystyle 0$ $\displaystyle=$ $\displaystyle
2\partial_{\rho}\epsilon^{a}+\frac{f^{\prime}}{2f\sqrt{g_{1}}}\Gamma_{\rho}\Gamma^{\hat{r}}\epsilon^{a}-\left[\frac{\sqrt{2}}{30}e^{-\frac{\sigma}{2}}C+\frac{4}{5}he^{2\sigma}\right]\Gamma_{\rho}\epsilon^{a}$
(233)
$\displaystyle+\frac{i}{5}e^{\frac{\sigma}{2}}\mathbf{F}_{1}{(\sigma^{3})^{a}}_{b}\Gamma_{\rho}\Gamma^{\hat{r}\hat{z}}\epsilon^{b},$
$\displaystyle 0$ $\displaystyle=$ $\displaystyle
2\partial_{r}\epsilon^{a}-\sqrt{g_{1}}\left[\frac{\sqrt{2}}{30}e^{-\frac{\sigma}{2}}C+\frac{4}{5}he^{2\sigma}\right]\Gamma^{\hat{r}}\epsilon^{a}-\frac{4i}{5}\sqrt{g_{1}}e^{\frac{\sigma}{2}}\mathbf{F}_{1}{(\sigma^{3})^{a}}_{b}\Gamma^{\hat{z}}\epsilon^{b},$
(234) $\displaystyle 0$ $\displaystyle=$ $\displaystyle
2\partial_{z}\epsilon^{a}-\frac{g^{\prime}_{2}}{2\sqrt{g_{1}g_{2}}}\Gamma^{\hat{r}\hat{z}}\epsilon^{a}+i\widetilde{g}_{1}A_{1}{(\sigma^{3})^{a}}_{b}\epsilon^{b}-\sqrt{g_{2}}\left[\frac{\sqrt{2}}{30}e^{-\frac{\sigma}{2}}C+\frac{4}{5}he^{2\sigma}\right]\Gamma^{\hat{z}}\epsilon^{a}$
(235)
$\displaystyle+\frac{4i}{5}\sqrt{g_{2}}e^{\frac{\sigma}{2}}\mathbf{F}_{1}{(\sigma^{3})^{a}}_{b}\Gamma^{\hat{r}}\epsilon^{b}\,.$

The next condition $\delta\chi^{a}=0$ yields

$0\ =\
-\frac{\sigma^{\prime}}{2\sqrt{g_{1}}}\Gamma^{\hat{r}}\epsilon^{a}+\left[\frac{\sqrt{2}}{30}e^{-\frac{\sigma}{2}}C-\frac{16}{5}he^{2\sigma}\right]\epsilon^{a}-\frac{i}{5}e^{\frac{\sigma}{2}}\mathbf{F}_{1}{(\sigma^{3})^{a}}_{b}\Gamma^{\hat{r}\hat{z}}\epsilon^{b}\,.$
(236)

From the supersymmetry transformation of the gaugini, only
$\delta\lambda^{a3}=0$ gives rise to non-trivial conditions given by

$0\ =\
\frac{\phi^{\prime}}{\sqrt{g_{1}}}{(\sigma^{3})^{a}}_{b}\Gamma^{\hat{r}}\epsilon^{b}-\frac{1}{\sqrt{2}}e^{-\frac{\sigma}{2}}C^{33}{(\sigma^{3})^{a}}_{b}\epsilon^{b}+ie^{\frac{\sigma}{2}}\mathbf{F}_{2}\Gamma^{\hat{r}\hat{z}}\epsilon^{a}\,.$
(237)

In these equations, for convenience, we have introduced the notations
$\mathbf{F}_{1}$ and $\mathbf{F}_{2}$ for non-vanishing components of the
dressed field strength tensors via

$F^{i}_{(2)}={L_{I}}^{i}F^{I}_{(2)}=\mathbf{F}_{1}\delta^{i}_{3}\,e^{\hat{r}}\wedge
e^{\hat{z}}\qquad\text{and}\qquad
F^{r}_{(2)}={L_{I}}^{r}F^{I}_{(2)}=\mathbf{F}_{2}\delta^{r}_{3}\,e^{\hat{r}}\wedge
e^{\hat{z}}\,.$ (238)

Explicitly, in the case of $SO(2)\times SO(2)$ symmetric solutions, these are
given by

$\displaystyle\mathbf{F}_{1}=\frac{1}{\sqrt{g_{1}g_{2}}}\left(A^{\prime}_{1}\cosh{\phi}+A^{\prime}_{2}\sinh{\phi}\right)$
and
$\displaystyle\mathbf{F}_{2}=\frac{1}{\sqrt{g_{1}g_{2}}}\left(A^{\prime}_{1}\sinh{\phi}+A^{\prime}_{2}\cosh{\phi}\right).$
(239)

To solve the resulting BPS equations, we follow Bah_M5  and use the
supersymmetry parameters of the form

$\epsilon^{a}=n^{a}\,\vartheta\otimes\eta\,.$ (240)

$n^{a}$ are two components of a constant object in the doublet representation
of $SO(3)_{R}$ while $\eta$ is a two-component spinor depending on $r$ and $z$
coordinates. On the other hand, $\vartheta$ is a four-component Killing spinor
on $AdS_{5}$ satisfying

$\nabla_{\hat{\alpha}}^{AdS_{5}}\vartheta=\frac{s}{2}\gamma_{\hat{\alpha}}\vartheta$
(241)

where $\hat{\alpha}=(\hat{m},\hat{\rho})=0,1,...,4$ is a flat space-time index
on $AdS_{5}$, $s=\pm 1$ is an arbitrary sign, and $\gamma_{\hat{\alpha}}$ are
five-dimensional $4\times 4$ gamma matrices satisfying the Clifford algebra

$\\{\gamma_{\hat{\alpha}},\gamma_{\hat{\beta}}\\}=2\eta_{\hat{\alpha}\hat{\beta}},\qquad\eta_{\hat{\alpha}\hat{\beta}}=\text{diag}(-++++).$
(242)

In terms of $\gamma_{\hat{\alpha}}$, we further decompose the seven-
dimensional gamma matrices as

$\Gamma_{\hat{\alpha}}=\gamma_{\hat{\alpha}}\otimes\sigma^{3},\qquad\quad\Gamma_{\hat{r}}=\mathds{1}_{4}\otimes\sigma^{1},\qquad\quad\Gamma_{\hat{z}}=\mathds{1}_{4}\otimes\sigma^{2}$
(243)

in which $\mathds{1}_{n}$ is an $n\times n$ identity matrix.

With the supersymmetry parameter (240), the first two BPS equations given in
(232) and (233) reduce to a single equation

$\displaystyle 0$ $\displaystyle=$
$\displaystyle\frac{s}{\sqrt{f}}(\gamma_{\hat{\alpha}}\otimes\mathds{1}_{2})\epsilon^{a}+\frac{f^{\prime}\Gamma_{\hat{\alpha}}\Gamma^{\hat{r}}\epsilon^{a}}{2f\sqrt{g_{1}}}-\left[\frac{\sqrt{2}}{30}e^{-\frac{\sigma}{2}}C+\frac{4}{5}he^{2\sigma}\right]\Gamma_{\hat{\alpha}}\epsilon^{a}+\frac{i}{5}e^{\frac{\sigma}{2}}\mathbf{F}_{1}{(\sigma^{3})^{a}}_{b}\Gamma_{\hat{\alpha}}\Gamma^{\hat{r}\hat{z}}\epsilon^{b}$
$\displaystyle=$ $\displaystyle
n^{a}(\gamma_{\hat{\alpha}}\vartheta)\otimes\left\\{\frac{s}{\sqrt{f}}\eta+\frac{f^{\prime}}{2f\sqrt{g_{1}}}(i\sigma^{2}\eta)-\left[\frac{\sqrt{2}}{30}e^{-\frac{\sigma}{2}}C+\frac{4}{5}he^{2\sigma}\right](\sigma^{3}\eta)-\frac{1}{5}e^{\frac{\sigma}{2}}\mathbf{F}_{1}\eta\right\\}$

in which we have expressed all the supersymmetry parameters in terms of two-
and four-component spinors and imposed the projector

${(\sigma^{3})^{a}}_{b}n^{b}=n^{a}\,.$ (245)

By using the same procedure in the remaining conditions, we find the following
set of BPS equations on the two-component spinor $\eta$

$\displaystyle 0$ $\displaystyle=$
$\displaystyle\frac{s}{\sqrt{f}}\eta+\frac{f^{\prime}}{2f\sqrt{g_{1}}}(i\sigma^{2}\eta)-\left[\frac{\sqrt{2}}{30}e^{-\frac{\sigma}{2}}C+\frac{4}{5}he^{2\sigma}\right](\sigma^{3}\eta)-\frac{1}{5}e^{\frac{\sigma}{2}}\mathbf{F}_{1}\eta,\quad$
(246) $\displaystyle 0$ $\displaystyle=$ $\displaystyle
2\partial_{r}\eta-\sqrt{g_{1}}\left[\frac{\sqrt{2}}{30}e^{-\frac{\sigma}{2}}C+\frac{4}{5}he^{2\sigma}\right]\sigma^{1}\eta-\frac{4}{5}\sqrt{g_{1}}e^{\frac{\sigma}{2}}\mathbf{F}_{1}(i\sigma^{2}\eta),\quad$
(247) $\displaystyle 0$ $\displaystyle=$
$\displaystyle-2i\partial_{z}\eta+\widetilde{g}_{1}A_{1}\eta-\frac{g^{\prime}_{2}}{2\sqrt{g_{1}g_{2}}}\sigma^{3}\eta+\sqrt{g_{2}}\left[\frac{\sqrt{2}}{30}e^{-\frac{\sigma}{2}}C+\frac{4}{5}he^{2\sigma}\right](i\sigma^{2}\eta)$
(248)
$\displaystyle+\frac{4}{5}\sqrt{g_{2}}e^{\frac{\sigma}{2}}\mathbf{F}_{1}\sigma^{1}\eta,$
$\displaystyle 0$ $\displaystyle=$
$\displaystyle-\frac{\sigma^{\prime}}{2\sqrt{g_{1}}}\sigma^{1}\eta+\left[\frac{\sqrt{2}}{30}e^{-\frac{\sigma}{2}}C-\frac{16}{5}he^{2\sigma}\right]\eta+\frac{1}{5}e^{\frac{\sigma}{2}}\mathbf{F}_{1}\sigma^{3}\eta,$
(249) $\displaystyle 0$ $\displaystyle=$
$\displaystyle\frac{\phi^{\prime}}{\sqrt{g_{1}}}\sigma^{1}\eta-\frac{1}{\sqrt{2}}e^{-\frac{\sigma}{2}}C^{33}\eta-e^{\frac{\sigma}{2}}\mathbf{F}_{2}\sigma^{3}\eta\,.$
(250)

As in Bah_M5 , we assume a definite charge under the $U(1)_{z}$ isometry for
the two-component spinor leading to the ansatz for $\eta$ of the form

$\eta(r,z)=e^{iqz}\widehat{\eta}(r)$ (251)

with a constant $q$. With this explicit form of $\eta$, it follows that the
combination
$(-2i\partial_{z}+\widetilde{g}_{1}A_{1})\eta=(2q+\widetilde{g}_{1}A_{1})\eta$
in (248) is invariant under the transformations

$A_{1}\rightarrow
A_{1}-\frac{2\alpha_{0}}{\widetilde{g}_{1}},\qquad\eta\rightarrow
e^{i\alpha_{0}z}\eta$ (252)

where $\alpha_{0}$ is an arbitrary constant. It is then convenient to define

$\widetilde{g}_{1}\widehat{A}_{1}=2q+\widetilde{g}_{1}A_{1}$ (253)

with $\widehat{A}^{\prime}_{1}=A^{\prime}_{1}$.
With suitable left-multiplications by Pauli matrices and additions of (249) to
(246) and (248), we find the following set of algebraic equations

$\displaystyle 0$ $\displaystyle=$
$\displaystyle\frac{s}{\sqrt{f}}\eta+\frac{1}{2\sqrt{g_{1}}}\left[\frac{f^{\prime}}{f}-\sigma^{\prime}\right](i\sigma^{2}\eta)-4he^{2\sigma}\sigma^{3}\eta,$
(254) $\displaystyle 0$ $\displaystyle=$
$\displaystyle\frac{\widetilde{g}_{1}\widehat{A}_{1}}{\sqrt{g_{2}}}\sigma^{1}\eta+\frac{1}{2\sqrt{g_{1}}}\left[\frac{g^{\prime}_{2}}{g_{2}}-\sigma^{\prime}\right](i\sigma^{2}\eta)-4he^{2\sigma}\sigma^{3}\eta+e^{\frac{\sigma}{2}}\mathbf{F}_{1}\eta,\qquad$
(255) $\displaystyle 0$ $\displaystyle=$
$\displaystyle\frac{5\sigma^{\prime}}{2\sqrt{g_{1}}}\sigma^{3}\eta-e^{-\frac{\sigma}{2}}\left[\frac{C}{3\sqrt{2}}-16he^{\frac{5\sigma}{2}}\right](i\sigma^{2}\eta)+e^{\frac{\sigma}{2}}\mathbf{F}_{1}\sigma^{1}\eta,$
(256) $\displaystyle 0$ $\displaystyle=$
$\displaystyle\frac{\phi^{\prime}}{\sqrt{g_{1}}}\sigma^{3}\eta-\frac{1}{\sqrt{2}}e^{-\frac{\sigma}{2}}C^{33}(i\sigma^{2}\eta)+e^{\frac{\sigma}{2}}\mathbf{F}_{2}\sigma^{1}\eta\,.$
(257)

These equations are of the form $M^{(x)}\eta=0$ where $x=1,2,3,4$ respectively
labels the BPS equations in (254) to (257). The four $2\times 2$ matrices
$M^{(x)}$ can be parametrized by

$M^{(x)}=X^{(x)}_{0}\mathds{1}_{2}+X^{(x)}_{1}\sigma^{1}+X^{(x)}_{2}(i\sigma^{2})+X^{(x)}_{3}\sigma^{3}\,.$
(258)

Following Bah_M5 , for each matrix $M^{(x)}$, we define the two-component
vectors

$v^{(x)}=\begin{pmatrix}X^{(x)}_{1}+X^{(x)}_{2}\\\
-X^{(x)}_{0}-X^{(x)}_{3}\end{pmatrix},\qquad
w^{(x)}=\begin{pmatrix}X^{(x)}_{0}-X^{(x)}_{3}\\\
-X^{(x)}_{1}+X^{(x)}_{2}\end{pmatrix}$ (259)

together with

$\mathcal{A}^{xy}=\text{det}(v^{(x)}|w^{(y)}),\qquad\mathcal{B}^{xy}=\text{det}(v^{(x)}|v^{(y)}),\qquad\mathcal{C}^{xy}=\text{det}(w^{(x)}|w^{(y)}).$
(260)

The notation $(a|b)$ denotes a $2\times 2$ matrix obtained from a
juxtaposition of the two-column vectors $a$ and $b$. As pointed out in Bah_M5
, the vanishing of $\mathcal{A}^{xy}$, $\mathcal{B}^{xy}$, and
$\mathcal{C}^{xy}$ gives a number of necessary conditions for the existence of
a non-trivial solution for $\eta$. We will separately determine the conditions
from the supergravity and vector multiplets by splitting the index
$x=(\bar{x},4)$ with $\bar{x}=1,2,3$.

Starting from the first matrix $\mathcal{A}^{xy}$, the vanishing of the
diagonal components $\mathcal{A}^{\bar{x}\bar{x}}$ gives the following
conditions

$\displaystyle 0$ $\displaystyle=$
$\displaystyle\frac{1}{f}-16h^{2}e^{4\sigma}+\frac{1}{4g_{1}}\left[\frac{f^{\prime}}{f}-\sigma^{\prime}\right]^{2},$
(261) $\displaystyle 0$ $\displaystyle=$ $\displaystyle
e^{\sigma}(\mathbf{F}_{1})^{2}+\frac{1}{4g_{1}}\left[\frac{g^{\prime}_{2}}{g_{2}}-\sigma^{\prime}\right]^{2}-16h^{2}e^{4\sigma}-\frac{(\widetilde{g}_{1}\widehat{A}_{1})^{2}}{g_{2}},$
(262) $\displaystyle 0$ $\displaystyle=$ $\displaystyle\frac{25\sigma^{\prime
2}}{4g_{1}}+e^{\sigma}(\mathbf{F}_{1})^{2}-e^{-\sigma}\left[\frac{C}{3\sqrt{2}}-16he^{\frac{5\sigma}{2}}\right]^{2}.$
(263)

The vanishing of the off-diagonal symmetric components
$\mathcal{A}^{\bar{x}\bar{y}}+\mathcal{A}^{\bar{y}\bar{x}}$
($\bar{x}\neq\bar{y}$) yields

$\displaystyle 0$ $\displaystyle=$
$\displaystyle\frac{1}{4g_{1}}\left[\frac{f^{\prime}}{f}-\sigma^{\prime}\right]\left[\frac{g^{\prime}_{2}}{g_{2}}-\sigma^{\prime}\right]+\frac{se^{\frac{\sigma}{2}}}{\sqrt{f}}\mathbf{F}_{1}-16h^{2}e^{4\sigma},$
(264) $\displaystyle 0$ $\displaystyle=$ $\displaystyle
20he^{2\sigma}\sigma^{\prime}-e^{-\frac{\sigma}{2}}\left[\frac{C}{3\sqrt{2}}-16he^{\frac{5\sigma}{2}}\right]\left[\frac{f^{\prime}}{f}-\sigma^{\prime}\right],$
(265) $\displaystyle 0$ $\displaystyle=$
$\displaystyle\frac{20he^{2\sigma}\sigma^{\prime}}{\sqrt{g_{1}}}-\frac{e^{-\frac{\sigma}{2}}}{\sqrt{g_{1}}}\left[\frac{C}{3\sqrt{2}}-16he^{\frac{5\sigma}{2}}\right]\left[\frac{g^{\prime}_{2}}{g_{2}}-\sigma^{\prime}\right]-\frac{2\widetilde{g}_{1}e^{\frac{\sigma}{2}}}{\sqrt{g_{2}}}\widehat{A}_{1}\mathbf{F}_{1}\qquad$
(266)

while the off-diagonal antisymmetric components
$\mathcal{A}^{\bar{x}\bar{y}}-\mathcal{A}^{\bar{y}\bar{x}}=0$ gives

$\displaystyle 0$ $\displaystyle=$ $\displaystyle
8he^{\frac{5\sigma}{2}}\mathbf{F}_{1}-\frac{8she^{2\sigma}}{\sqrt{f}}+\frac{\widetilde{g}_{1}}{\sqrt{g_{1}g_{2}}}\left[\frac{f^{\prime}}{f}-\sigma^{\prime}\right]\widehat{A}_{1},$
(267) $\displaystyle 0$ $\displaystyle=$ $\displaystyle
e^{\frac{\sigma}{2}}\left[\frac{f^{\prime}}{f}-\sigma^{\prime}\right]\mathbf{F}_{1}+\frac{5s}{\sqrt{f}}\sigma^{\prime},$
(268) $\displaystyle 0$ $\displaystyle=$
$\displaystyle\frac{e^{\sigma}}{\sqrt{g_{1}}}\left[\frac{g^{\prime}_{2}}{g_{2}}+4\sigma^{\prime}\right]\mathbf{F}_{1}+\frac{2\widetilde{g}_{1}}{\sqrt{g_{2}}}\left[\frac{C}{3\sqrt{2}}-16he^{\frac{5\phi}{2}}\right]\widehat{A}_{1}\,.$
(269)

Matrices $\mathcal{B}^{xy}$ and $\mathcal{C}^{xy}$ are both antisymmetric in
$xy$ indices. It turns out that the resulting BPS conditions arising from the
combinations $\mathcal{B}^{xy}\pm\mathcal{C}^{xy}=0$ take a simpler form. The
combinations $\mathcal{B}^{\bar{x}\bar{y}}+\mathcal{C}^{\bar{x}\bar{y}}=0$
give the following conditions

$\displaystyle 0$ $\displaystyle=$
$\displaystyle\frac{e^{\frac{\sigma}{2}}}{\sqrt{g_{1}}}\left[\frac{f^{\prime}}{f}-\sigma^{\prime}\right]\mathbf{F}_{1}-\frac{s}{\sqrt{f\,g_{1}}}\left[\frac{g^{\prime}_{2}}{g_{2}}-\sigma^{\prime}\right]+\frac{8\widetilde{g}_{1}he^{2\sigma}}{\sqrt{g_{2}}}\widehat{A}_{1},$
(270) $\displaystyle 0$ $\displaystyle=$ $\displaystyle
4he^{3\sigma}\mathbf{F}_{1}+\frac{s}{\sqrt{f}}\left[\frac{C}{3\sqrt{2}}-16he^{\frac{5\phi}{2}}\right],$
(271) $\displaystyle 0$ $\displaystyle=$
$\displaystyle\frac{5\widetilde{g}_{1}\sigma^{\prime}}{\sqrt{g_{1}g_{2}}}\widehat{A}_{1}+2\left[\frac{C}{3\sqrt{2}}-12he^{\frac{5\phi}{2}}\right]\mathbf{F}_{1}$
(272)

while the combinations
$\mathcal{B}^{\bar{x}\bar{y}}-\mathcal{C}^{\bar{x}\bar{y}}=0$ lead to

$\displaystyle 0$ $\displaystyle=$
$\displaystyle\frac{2he^{2\sigma}}{\sqrt{g_{1}}}\left[\frac{f^{\prime}}{f}-\frac{g^{\prime}_{2}}{g_{2}}\right]+\frac{s\widetilde{g}_{1}}{\sqrt{f\,g_{2}}}\widehat{A}_{1},$
(273) $\displaystyle 0$ $\displaystyle=$
$\displaystyle\frac{2se^{\frac{\sigma}{2}}}{\sqrt{f}}\mathbf{F}_{1}-\frac{5\sigma^{\prime}}{2g_{1}}\left[\frac{f^{\prime}}{f}-\sigma^{\prime}\right]+8he^{\frac{3\sigma}{2}}\left[\frac{C}{3\sqrt{2}}-16he^{\frac{5\phi}{2}}\right],$
(274) $\displaystyle 0$ $\displaystyle=$ $\displaystyle
2e^{\sigma}(\mathbf{F}_{1})^{2}-\frac{5\sigma^{\prime}}{2g_{1}}\left[\frac{g^{\prime}_{2}}{g_{2}}-\sigma^{\prime}\right]+8he^{\frac{3\sigma}{2}}\left[\frac{C}{3\sqrt{2}}-16he^{\frac{5\phi}{2}}\right].$
(275)

There are, in total, fifteen algebraic conditions obtained from the BPS
equations of the supergravity multiplet in (254) to (256).

Extending this procedure to the BPS equation (257) from vector multiplets
gives additional BPS conditions derived from the vanishing of
$\mathcal{A}^{x4}$, $\mathcal{B}^{\bar{x}4}$, and $\mathcal{C}^{\bar{x}4}$.
The first condition obtained from $\mathcal{A}^{44}=0$ takes the form

$0=\frac{(\phi^{\prime})^{2}}{g_{1}}+e^{\sigma}(\mathbf{F}_{2})^{2}-\frac{e^{-\sigma}}{2}(C^{33})^{2}.$
(276)

The symmetric part $\mathcal{A}^{\bar{x}4}+\mathcal{A}^{4\bar{x}}=0$ gives

$\displaystyle 0$ $\displaystyle=$ $\displaystyle
16he^{\frac{5\sigma}{2}}\phi^{\prime}-\sqrt{2}C^{33}\left[\frac{f^{\prime}}{f}-\sigma^{\prime}\right],$
(277) $\displaystyle 0$ $\displaystyle=$
$\displaystyle\frac{16he^{\frac{5\sigma}{2}}\phi^{\prime}}{\sqrt{g_{1}}}-\frac{\sqrt{2}C^{33}}{\sqrt{g_{1}}}\left[\frac{g^{\prime}_{2}}{g_{2}}-\sigma^{\prime}\right]-\frac{4\widetilde{g}_{1}e^{\sigma}}{\sqrt{g_{2}}}\mathbf{F}_{2}\widehat{A}_{1},$
(278) $\displaystyle 0$ $\displaystyle=$
$\displaystyle\frac{5\sigma^{\prime}\phi^{\prime}}{g_{1}}-\sqrt{2}e^{-\sigma}C^{33}\left[\frac{C}{3\sqrt{2}}-16he^{\frac{5\phi}{2}}\right]+2e^{\sigma}\mathbf{F}_{1}\mathbf{F}_{2}$
(279)

while the antisymmetric part $\mathcal{A}^{\bar{x}4}-\mathcal{A}^{4\bar{x}}=0$
results in

$\displaystyle 0$ $\displaystyle=$
$\displaystyle\frac{2s\phi^{\prime}}{\sqrt{f}}+e^{\frac{\sigma}{2}}\mathbf{F}_{2}\left[\frac{f^{\prime}}{f}-\sigma^{\prime}\right],$
(280) $\displaystyle 0$ $\displaystyle=$
$\displaystyle\frac{2e^{\sigma}\phi^{\prime}}{\sqrt{g_{1}}}\mathbf{F}_{1}+\frac{e^{\sigma}}{\sqrt{g_{1}}}\left[\frac{g^{\prime}_{2}}{g_{2}}-\sigma^{\prime}\right]\mathbf{F}_{2}+\frac{\sqrt{2}\widetilde{g}_{1}C^{33}}{\sqrt{g_{2}}}\widehat{A}_{1},$
(281) $\displaystyle 0$ $\displaystyle=$ $\displaystyle
C^{33}\mathbf{F}_{1}-\sqrt{2}\left[\frac{C}{3\sqrt{2}}-16he^{\frac{5\phi}{2}}\right]\mathbf{F}_{2}\,.$
(282)

Moreover, the combinations $\mathcal{B}^{\bar{x}4}+\mathcal{C}^{\bar{x}4}=0$
and $\mathcal{B}^{\bar{x}4}-\mathcal{C}^{\bar{x}4}=0$ respectively lead to the
following sets of extra conditions

$\displaystyle 0$ $\displaystyle=$
$\displaystyle\frac{2\widetilde{g}_{1}\phi^{\prime}}{\sqrt{g_{1}g_{2}}}\widehat{A}_{1}+\sqrt{2}C^{33}\mathbf{F}_{1}+8he^{\frac{5\sigma}{2}}\mathbf{F}_{2},$
(283) $\displaystyle 0$ $\displaystyle=$ $\displaystyle
2\phi^{\prime}\mathbf{F}_{1}-5\sigma^{\prime}\mathbf{F}_{2},$ (284)
$\displaystyle 0$ $\displaystyle=$
$\displaystyle\frac{\sqrt{2}sC^{33}}{\sqrt{f}}+8he^{3\sigma}\mathbf{F}_{2}$
(285)

and

$\displaystyle 0$ $\displaystyle=$
$\displaystyle\frac{\phi^{\prime}}{g_{1}}\left[\frac{f^{\prime}}{f}-\sigma^{\prime}\right]-\frac{2se^{\frac{\sigma}{2}}}{\sqrt{f}}\mathbf{F}_{2}-4\sqrt{2}he^{\frac{3\sigma}{2}}C^{33},$
(286) $\displaystyle 0$ $\displaystyle=$
$\displaystyle\frac{\phi^{\prime}}{g_{1}}\left[\frac{g^{\prime}_{2}}{g_{2}}-\sigma^{\prime}\right]-2e^{\sigma}\mathbf{F}_{1}\mathbf{F}_{2}-4\sqrt{2}he^{\frac{3\sigma}{2}}C^{33},$
(287) $\displaystyle 0$ $\displaystyle=$ $\displaystyle
5\sqrt{2}\sigma^{\prime}C^{33}-4\phi^{\prime}\left[\frac{C}{3\sqrt{2}}-16he^{\frac{5\phi}{2}}\right].$
(288)

Finally, the BPS equation (247) will be used to determine the explicit form of
$\eta(r,z)$.

We now turn to the vector fields and determine the explicit forms of
$A^{\prime}_{1}$ and $A^{\prime}_{2}$ which are relevant for solving the BPS
conditions obtained previously. With only the $SO(2)\times SO(2)$ singlet
scalar non-vanishing, we find that $\ast
P^{ir}_{(1)}{f_{IJ}}^{K}{L_{r}}^{I}L_{Ki}=0$. Together with $C_{(3)}=0$, the
field equation (25) reduces to

$\displaystyle A^{\prime\prime}_{1}$ $\displaystyle=$
$\displaystyle-2\phi^{\prime}A^{\prime}_{2}-\left(\frac{5f^{\prime}}{2f}-\frac{g^{\prime}_{1}}{2g_{1}}-\frac{g^{\prime}_{2}}{2g_{2}}+\sigma^{\prime}\right)A^{\prime}_{1},$
(289) $\displaystyle A^{\prime\prime}_{2}$ $\displaystyle=$
$\displaystyle-2\phi^{\prime}A^{\prime}_{1}-\left(\frac{5f^{\prime}}{2f}-\frac{g^{\prime}_{1}}{2g_{1}}-\frac{g^{\prime}_{2}}{2g_{2}}+\sigma^{\prime}\right)A^{\prime}_{2}\,.$
(290)

The most general solution to these equations is given by

$\displaystyle
A^{\prime}_{1}=\frac{e^{-\sigma-2\phi}}{2}(a_{1}+a_{2}e^{4\phi})\sqrt{g_{1}g_{2}}f^{-\frac{5}{2}}$
and $\displaystyle
A^{\prime}_{2}=\frac{e^{-\sigma-2\phi}}{2}(a_{1}-a_{2}e^{4\phi})\sqrt{g_{1}g_{2}}f^{-\frac{5}{2}}$
(291)

where $a_{1}$ and $a_{2}$ are constants.

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