Supersymmetric AdS5 solutions of type IIB supergravity without D3 branes
Received: November
Supersymmetric AdS5 solutions of type IIB
The Strand 0
London 0
WC 0
R 0
LS 0
U.K. 0
Open Access 0
c The Authors. 0
0 Department of Mathematics, King's College London
We analyse the most general bosonic supersymmetric solutions of type IIB supergravity whose metrics are warped products of vedimensional antide Sitter space try conditions in some special cases, and demonstrate how these are satis ed by a solution originally found in [13], utilising the method of nonAbelian Tduality.
AdSCFT Correspondence; Supergravity Models

(AdS5) with a
vedimensional Riemannian manifold M5, where the
ishes, while all remaining
uxes are allowed to be nonvanishing consistent with SO(4,2)
Contents
1 Introduction
4 Introducing local coordinates
Complex M4 and P = 0
An ansatz with P 6= 0
The solution of [13]
A Bilinear de nitions and the orthonormal frame
More details on the solution of [13]
Bilinear equations
Introduction
carried out in [8{10].
classi ed in [11] and later extended in [12].
IIB. The supersymmetric solutions that are obtained have seed solutions AdS5
Y p;q. Unfortunately these new solutions are singular and it was hoped that by
completing this classi cation we would be able to
nd new nonsingular solutions of this
T (1;1) and
supergravity with a warped metric of the form AdS5
M5, where M5 is an internal manifold
that admits a Riemannian metric. We set the selfdual veform
eld strength, F5, to be
vanishing but allow all other NeveuSchwarz NeveuSchwarz (NSNS) and RR
sider a simple ansatz and
nd a singular solution, in section 6 we present a less simpli ed
NATDTdual of AdS5
T (1;1) solution found in [13] satis es our equations. We conclude
section 7.
G = ie =2( dB
P =
Q =
DM = rM
supersymmetry variations through the following oneforms
U(1) gauge transformations, is de ned as
dimensional metric gMN , the bosonic elds comprise the axiondilaton
= C(0) + ie
complex threeform
ds120 = e2
where ds2AdS5 is the metric on AdS5 with Ricci tensor given by R
4m2(gAdS5 )
is the metric on a
vedimensional Riemannian internal space M5. In order
to preserve the SO(4; 2) symmetry of the metric we require the
elds to take values in;
3(M5; C). Notice that with this
the Killing spinor
Bianchi identities, and the supersymmetry variations for the gravitino
M and dilatino .
type IIB supergravity with SO(4; 2) symmetry and vanishing
ux. Namely we
require that
F5 = 0 ;
when it is not.
minimal supersymmetry in AdS5. This takes the form
= e =2(
Requiring supersymmetry to be preserved yields the following conditions
m m 1 = 0 ;
m m 2 = 0 ;
m1:::m3 2 = 0 ;
m1:::m3 1 = 0 ;
m1:::m3 Gm1:::m3 1 = 0 ;
m1:::m3 Gm1:::m3 2 = 0 :
ting f = 0.2
SU(2) structure.
tion (2.10) reads
Special cases.
The possible stabilizer groups of the Spin(5) spinors i are the
iden1 =
Following the use of Cli ord algebra identities one can show easily that @n
= 0, and
we conclude that there are no supersymmetric AdS5
M5 solutions with F5 = 0 in type
IIB supergravity with M5 admitting an SU(2) structure.3
implies 1 = 0 =
2 and hence no supersymmetry is preserved. We therefore conclude
uxes do not exist.
spinors i are not identically zero, thus giving a (local) identity structure on M5.
with M5 having an SU(2) structure either.
while the algebraic conditions are
G = 2P ^ K3
+ 4imK4) ^ K3
takes the form
ds2M5 =
This should be contrasted with the metric written in equation (3.53) of [2].
e 4 d(e4 S) = 3imK ;
e 6 D(e6 K3) = P ^ K3
e 4 d(e4 K4) =
e 8 d(e8 K5) =
Z = 0 = sin ;
A = 1 ;
= iK3 P ;
= 0 = iK5 P ;
we see that F5 = 0 implies that sin
from [2] where we set sin
0 and f
0. The resulting di erential conditions are5
LK5 G = 0 :
= LK5
= LK5 C(0) = 0 ;
4Following the argument in appendix C of [2], and imposing sin
= 0, we nd that it is not possible to
independent and admitting an identity structure.
5Here and in the rest of the paper
denotes the Hodge star operator with respect to the vedimensional
metric ds2M5 .
that applies to both cases. Using the supersymmetry equations, we nd
D(e6 X) = e6 (3im
e 6 D(e6 Y ) = 3im
X) =
and (3.13). As in [2], we conclude:
For the class of solutions with metric of the form (2.6), vanishing veform
uxes respecting SO(4; 2) symmetry, all the equations of motion and
Bianchi identities are implied by supersymmetry.
Introducing local coordinates
and RR twoform potentials.
K5. As a vector we have
and as a oneform
is a oneform with no d term. The factor of 3m is chosen for later convenience.
The Lie derivative of S with respect to K5# is
K5# = 3m
K5 = j j (d
LK5# S =
from which we nd
It is convenient to make the rede nitions
Then from (3.1) we have
and using the expression for K in appendix A we deduce that
not bered,
S =
= e 4 ;
= e4
K =
K5 =
K3 =
K4 =
tions read
2i P =
= L @@
supersymmetry.
the metric becomes
9m2 ds2 =
+ 2 2 + 2 d 2 + 2 2 d 2 :
is a real oneform and
is a complex oneform, and both have no leg along
equation (3.3) becomes
Equation (3.2) becomes
now the oneforms
have no d term, though they are still in general functions of
. We may then split the vedimensional exterior derivative as
almost product structure. Equation (4.10) now reads
d = d3 + d
whilst (4.11) reads6
where we have used (4.13).
ux we need to solve the four di erential equations (4.15){(4.18) subject to the
automatically satis ed upon using (4.12), (4.17) and (4.18).
We may now introduce the three remaining coordinates along
and , which we will
oneforms as
x dx + y1 dy1 + y2 dy2 ;
Re [ ] =
Im [ ] =
x dx + y1 dy1 + y2 dy2 ;
x dx + y1 dy1 + y2 dy2 :
example of a rather generic solution will be presented later in section 7.
can combine equations (3.11) and (3.12), to obtain
X )) =
+ X ) + e
6 P ^ (X
Y ) : (4.20)
!B =
e =2
the F5 6= 0 case were given in [14].
Complex M4 and P
= 0
Motivated by
P = 0 implies that
is a function of
necessary and su cient di erential equations to
only.8 Setting P = 0 and
( ) reduces the
d3Re [ ] =
d3Im [ ] =
phic twoform constructed from the orthonormal frame of appendix A as
This then de nes an almost complex structure on M4. In the second line we have expressed
in terms of
the twoform bilinears. Imposing that this is integrable implies
P = g(e4 + ie3) + f (e2 + ie5) + h(e4
equations to solve and we were unable to do so.
8To see this use (4.12) to note that d ln
that this is closed then implies that fK4 = 0.
@ Re [ ] =
@ Im [ ] =
= exp
Re [ ] = exp
Im [ ] = exp
We see immediately that we may solve (5.6){(5.8) as
may include arbitrary integration constants which we absorb into the
independent
oneforms. Upon substituting these expressions into (5.3){(5.5) one sees that the
in (5.3) cancels automatically as it should. However the
dependence in (5.4) and (5.5)
does not, we should have been suspicious if it cancelled as it would imply that
be any function of , requiring that this expression is
independent gives us the de ning
di erential equation for
@ ln exp
= 0 :
nonlinear di erential equation
(3 + 2 2) _ + 6 3 _ 2 + 3 (1
2 2) = 0 ;
and the three di erential equations
Where c is a constant satisfying
d3 ^ =
dR^ = c ^ ^ I^;
dI^ =
c =
@ ln exp
Notice that c is nonzero if
is nonconstant and we shall distinguish between these two
appendix B.
A singular solution.
constant . For simplicity we set
= (1
Re [ ] = (1
Im [ ] = (1
2)1=3 ^ ;
2)2=3R^ ;
2)2=3I^;
@ ^ = 0 ;
@ R^ = 0 ;
@ I^ = 0 :
d3R^ = 0 = d3I^;
d3 ^ =
R^ = dy2 ; I^ = dy1 :
(dx + y1dy2) :
^ =
We then need to solve
A solution to (5.22) is
The metric is
and we have
As R^ and I^ are closed we may de ne coordinates y1 and y2 such that
9m2 ds2 =
2)1=3( dy12 + dy22)
2)1=3 (y1 dy2 + dx)2 +
B =
C(2) =
2)2=3
2)2=3
Note that the range of
should be either
Scalar is given by R = 28m2, whilst R
2 [ 1; 0]. We nd that the Ricci
exhibits a singularity as
1 and therefore the solution is singular.
An ansatz with P 6= 0
The structure of the BPS equations suggests an ansatz in which the
coordinate plays a
rotation of
dependent phase . Namely, we consider
tations. Here the
dependent functions A; B; C are all real valued functions of
= 0, is
higher degree for one single function; as
appears \most often" in the system the simplest
_ =
=
(C2 + B2) =
2 (B2 + C2)
3 (9 + 47 2 2 + 31 4 4 + 9 6 6) _ 2
+36 4 (2 + 3 2 2 + 3 4 4) _ 3 + 9
+ (9 + 15 2 2 + 35 4 4 + 5 6 6) _ + 9 2(3 + 5 2 2 + 4 4
is discarded.
= A( ) 3 ;
@ log B =
@ log C =
@ log A =
A =
2 A
not su cient, notice that constant
solves (6.10) however it does not solve (6.7) as B
Dividing (6.3) by B2 and (6.4) by C2 and subtracting them we obtain
where k~ is an integration constant. Further using (6.6) we obtain
= k~
B2C2(C2
B2) = k
third order equation is found, the complete solution can be reconstructed.
A solution
of the third order equation depends generically on three integration
conand _ as known functions and solve for B and C. We have
B2 =
C2 =
we correctly have four integration constants, one for each function.
The remaining
dependent part of the system leads to the following equations
cos 2 @
+ sin 2 e @ C(0) =
= 0 ;
0), namely we have
where 0 and C(0) are two integration constants.
0
@ log cot 2 =
0) = sin 2 ;
C(0) =
equations. Once a solution is found, the phase
can be determined by integrating (6.17),
and nally the dilaton and axion are determined algebraically in terms of .
A_ =
B_ =
_ =
study of (6.10) as an open problem.
The solution of [13]
nal section we show that the supersymmetric NATDT dual of the AdS5
T(1;1) solution
in [2] for the PilchWarner solution.
We instead bypassed this problem by
nding the
rst begin this section by writing down the solution found in [13].
We use the coordinates x1 =
sin ; x2 =
and for simplicity set 0 = 1. The
d=10 metric in string frame10 is
ds2 = ds2(AdS5) + L
(L4 2 21 + x12) dx1 + x1x2 dx2
2(L4 41 + x22) cos 1 dx2)2
Q = L
x ; W =
21Q sin2 1 + 2 21x21 cos2 1 ; P = L
The constants
AdS5. The dilaton is
10Recall that the classi cation is in Einstein frame.
= L4W ;
whilst the NSNS twoform is given by11
F1 = 4L4
F3 =
B =
The nonzero RR uxes12 are
2x1 cos 1 d 1 + 2x2 sin2 1 dx1
x1( 2 cos2 1 + 12 sin2 1) dx2
m =
= L
= e ;
= ( x1 cos 1 dx1
x2 cos 1 dx2 + L
W (cos 1 dx2 + x2 sin 1 d 1
x1( 2 cos2 1 + 12 sin2 1) dx2 + 2x1 cos 1 d 1 ; (7.5)
elements are
the NSNS two form we nd
B =
e =2
F3 agrees with that derived from the general expressions (4.21) and (4.22).
11We correct a minor typographical error here by adding the cos 1 term in front of d 1.
= L16
21E[ dx2 ^ d 1 + x1 sin 1 d 1 ^ dx1 + x2 sin 1 d 1 ^ dx2] :
Upon substituting the values of the constants,
and 1 we nd that they are equal. The
equation for
follows similarly but is vastly more complicated than the one illustrated
above and for this reason we do not present it.
had no d term, we would like to verify this. To do so we must write the
oneforms in the form (4.19). To this end, we make the change of coordinates
x =
y1 =
y2 = ln
= L
(x12 + x22) + L
x = x2 ;
x =
x =
W 1=4
3 21W 1=4 ;
y1 =
y1 =
y1 = 0 ;
W 1=4
y2 = 0 ;
y2 =
W 1=4
y2 = L
3 21x2 cos 1W 1=4 :
this ansatz was swiftly dropped in favour of the ones we have presented.
whilst x1 ! 0. Furthermore the dilaton also blows up at these points.
Computing the invariants R
and R 1::: 4 R 1::: 4 we also
nd that these are singular
considered the type IIA nonAbelian T dual of AdS5
setup and a linear quiver to describe its dual SCFT.
S5 and propose a a D4/NS5 brane
namely the NATDT dual of the AdS5
Y p;q solution. This solution will also satisfy the
classi cation presented here however we have not checked the details.
One of the motivations for doing this work was to
nd new nonsingular
supersymhave been unable to nd new nonsingular solutions.
solution with at least SU(2)
U(1) and follows their procedure for applying the
Nonux has a leg over the
regularity is case dependent.
Acknowledgments
STFC studentship, number ST/N504361/1.
13We thank Daniel Waldram for clari cations on this point.
Bilinear de nitions and the orthonormal frame
We de ne all the bilinears appearing in the paper. The scalar bilinears are
The vector bilinears are
The twoform bilinears are
One nds that they satisfy the following algebraic relations
K5 = sin
K4 + Re [Z K3]
Re [S K] ;
0 = sin V
S X = (1 + sin )W
S Y = (1 sin )W
K ^ K + Re [iZ W ] ;
(K4 + K5) ^ K3 ;
Following [2] we take the basis of gamma matrices of Cli (5) to be
1 =
2 =
a =
where a =
a are the Pauli matrices. In this basis the charge conjugation
intertwiner is given by C = I
2. we label the corresponding basis by ei. We decompose
the spinors i as si
i where si are spinors of Cli (3) and i spinors of Cli (2). At the
K4 and K5 lie in the (e1
e2) plane and in particular K5 to be parallel with e1. We nd
s1 = p
cos cos
sin sin
s2 = p
sin cos
cos sin
now write the scalar and vector bilinears as functions of ; ; i. Requiring sin
= 0 implies
that cos 2 = 0 otherwise cos 2
(e3e4) plane one can choose:
1 =
2 =
from which we obtain the nal form of the vector bilinears
K5 = cos 2 e1 ;
K4 =
K3 = sin 2 (e4
K = e2i e1
and the one nontrivial scalar bilinear
The twoforms in terms of this orthonormal basis are
U =
V = e34
W = (i cos 2 e
X = e2i (sin 2 e1 + cos 2 e
S =
^ =
R^ =
I^ =
oneforms if c < 0.14 The metric becomes
9m2 ds2 = 2 2 d 2 +
wrong signature which can be clearly seen from the above.
and show that the metric is regular for all values of
in this range. To
do so we nd values of
for which the metric shrinks, equivalently some function of the
derivative in , we nd that the Ricci scalar is given in the two cases by
Rc>0 =
Rc<0 = Rc>0 +
3 + _ 2(9 + 171 2 2)) ;
c > 0 analysis.
We rst consider the case where the function 1
2 2 vanishes, let this
point be 0. Near to 0 we may write
for some constants
and . Making the change of coordinate,15
r =
(1=2)(1
2 2 = dr2 ;
oneforms satisfy d 1 = 2 ^
has a singularity as
= 1 one nds that the Ricciscalar
)2 = 1 ((1
solution, we nd
9m2 ds2 =
For regularity we require that the metric looks locally like S1
R4. For this to occur we
to be r2=4. Using the expression for
near 0 we
implicitly assumed that we are away from
is then necessarily unbounded in
order to satisfy 1
nd that for any 0 strictly negative with
and the metric will be
regular at this point.
2 2 = 0 away from
there is no point 3 2 ( 1; 2) such that 1
our range for
mind set 1 <
Without loss of generality and with the previous analysis in
2 < 0. Near to
a, a = 1; 2, we have _ ( ) =
2=(3 2). Therefore for
1=( 1 + 1) however near to 2
we have, for 2 a small positive number, ( 2
1=( 2
2). With the additional
and not unreasonable assumption that
is continuous we must have that at some point
3 existed. We conclude that no two such points exist.
Assume now that
takes a nite value at
if the last bracketed term in (B.4) is
nite in the limit as
goes to 0 or it goes to zero
as 2 and has the metric of a threesphere. If we consider these cases then
! 0. However one now
Ricci scalar as
! 0 in both cases. Moreover if we expand (5.13) about
incorrect signature. This suggests that
= 0 we
nonsingular metric.
( 0) = 0 or ( 0) = 0 at 0
. We can rule out both the rst and second choices from
nd that all the
derivatives of
vanish at this point by taking further derivatives of (5.13) and evaluating
. Assuming, not unreasonably, that
is analytic at this point we conclude that
a constant everywhere violating c 6= 0.
One can see immediately that this is not regular for any
> 0 and 0 as the Ricci scalar
diverges. For
diverging at 0 one still requires 1
2 2 > 0, for a metric with the correct
Finally we study the possibility that
1. It is best if we make the change of
9m2 ds2 =
We still require that 1
2=r2 > 0 and so for small r,
must take the form
= a1r + a2r2 + : : : ;
= ar. With a
= (
> 0 and the metric takes the form
9m2 ds2 =
with ja1j < 1. From looking at the last term in (B.4) we see that we need
9m2 ds2 = a2
The metric takes the form of S1
we have the form of
S3 where the S3 is squashed. Note however as r ! 0
solutions for c > 0 and we turn our attention to c < 0 in the following subsection.
c < 0 analysis.
We now consider the case of c < 0, recall that now i are the left invariant
we shall make use of these when possible. Note that the possibility of 1
2 2 = 0 at 0
scalar in equation (B.3).
Assume that
= 0 in
! 0 and this suggests
scalar diverges. Note that the nal possibility for a boundary value is
gives a singular point for the manifold we cannot have the range to be
1. As
= 0
1; 1) and
therefore there are no two points for
to take a value in. If one completes the analysis for
1 one again
nds that the manifold is singular at these points.
corresponding NATD and T dualities. It is this method that we present below.
of the Killing spinors under both NATD and Tduality for the ease of the reader.
and therefore de ne two di erent frame
elds. These two frames must be equivalent as
This Lorentz transformation induces an action on spinors by the matrix
which satis es
chirality. We shall denote these two spinors generically as
2, their chiralities are
NATD or Tduality
where for a Tduality along a Killing vector, @x,
takes the form
U(1) =
the at directions 1; 2 and 3,
takes the form
SU(2) =
where for our purposes
1 =
2 =
3 =
Note that both 's de ned above are unitary in our basis.
in the canonical vielbein basis for performing the NATD
e 1 = L 1 d 1 ;
e1;2 = L 1 1;2 ;
e 1 = L 1 sin 1 d 1 ;
e3 = L ( 3 + cos 1 d 1) ;
1 =
2 B@ i CA
2 =
1 B 0 C
2 B@ 1C
1 =
2 =
construct 1 and 2 as used in the classi cation
spinors become
whilst the vielbeins that change are16
e^1 =
e^2 =
e^3 =
LQ LQ
((L4 2 2 + x12) cos + L
1
+x1(x2 cos
+x1(L2 21 cos + x2 sin )( dx2 + L
[x1x2 dx1 + (L4 41 + x22) dx2
2 21x21( d + cos 1 d 1)]:
spinors become
also added some extra factors of
and 1 which we found to be missing.
One may now compute all the spinor bilinears. One nds for the scalar bilinears
A = 1 ;
= 0 ;
Z = 0 ;
S =
= L
x2 cos 1 dx2 + L4 41 sin 1 d 1 + x2 d 1) ;
1 [x1x2 sin2 1 dx1 + (L4 41 + x22) sin2 1 dx2 + x12 cos 1 d 1
From S one nds
frame and therefore we have the identi cation e2
1=2 = e
=2. From this we nd
spinors. Computing the oneform bilinears form the Killing spinors one nds
K =
K5 =
K4 =
K3 =
( x1 cos 1 dx1
(i(sin dx1 + x1 cos d )
W (cos 1 dx2 + x2 sin 1 d 1
d 1)]: (C.22)
(7.11). The change of coordinates (7.15){(7.18) follows from noticing that
can be
idenone recovers the change of coordinates presented.
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