Supersymmetric multi-charge AdS_5 black holes

0.22 MB·English

by  Hari K. Kunduri

AdS
Supersymmetric multi-charge black holes
5
Hari K. Kunduri
DAMTP, University of Cambridge, Wilberforce Road, Cambridge, CB3 0WA, UK
[email protected]
James Lucietti
6 St. John’s College, University of Oxford, Oxford, OX1 3JP, UK
0 [email protected]
0
2
Harvey S. Reall
n
a School of Physics and Astronomy, University of Nottingham, NG7 2RD, UK
J
[email protected]

0
2
DAMTP-2006-10

1
v
January 20, 2006
6
5
1
1
Abstract
0
6 A new supersymmetric, asymptotically anti-de Sitter, black hole solution of five-dimensional
0 U(1)3 gauged supergravity is presented. All known examples of such black holes arise as special
/
h cases of this solution, which is characterized by three charges and two angular momenta, with
t one constraint relating these five quantities. Analagous solutions of U(1)n gauged supergravity
-
p are also presented.
e
h
: 1 Introduction
v
i
X
Supersymmetric black hole solutions of five-dimensional gauged supergravity are of interest because
r
a theirexistenceraisesthequestionofwhetheritispossibletoprovideanexactcalculationofblackhole
entropy using four dimensional super Yang-Mills theory via the AdS/CFT correspondence [1, 2, 3].
Thefirstexamplesofsuchsolutionswereobtainedtwoyearsago[4,5]andfurtherexampleshavebeen
obtained morerecently [6,7]. Thepurposeof thispaperis topresentamoregeneralsupersymmetric
solution that contains all of these solutions as special limiting cases.
We shall beconcerned with = 1 gauged supergravity coupled to two abelian vector multiplets,
N
which has gauge group U(1)3. A solution of this theory with (electric) charges Q with respect to
i
the three U(1)’s can be oxidized to a solution of type IIB supergravity which, in CFT language has
SU(4) charge given by the weight vector (Q ,Q ,Q ) [8]. Sometimes we shall also refer to minimal
1 2 3
gauged supergravity. A charge Q solution of this theory oxidizes to a IIB solution with R-charge
(Q,Q,Q) [9].
The known supersymmetric black holes in five-dimensional gauged supergravity all have non-
vanishing angular momentum. Let J , J denote the angular momenta1. The mass is determined
1 2
by the BPS relation
M = g J +g J + Q + Q + Q , (1)
1 2 1 2 3
| | | | | | | | | |
1(J1,J2) is proportional to a weight vector of the rotation group SO(4). In CFT language, it is more common to
work with JL ≡J1+J2 and JR ≡J1−J2, which are proportional to weights with respect to the two SU(2) factors
in SO(4)∼SU(2)L×SU(2)R.
1
where the gauge coupling g is the reciprocal of the AdS radius of curvature.
The solutions of [4, 5, 6, 7] fall into several overlapping but distinct families. The 1-parameter
solution of [4] concerns minimal supergravity, and is a special case of the 3-parameter U(1)3 solution
of [5]. The latter is parameterized by its charges Q and has “self-dual” angular momenta J = J .
i 1 2
Supersymmetric black hole solutions with J = J were obtained in [7] and [6] for the minimal
1 2
6
and U(1)3 theories respectively. The former is not a special case of the latter. These are both 2-
parameter solutions: thetwo parameters can betaken to beJ andJ , which determinethe charges.
1 2
We have, then, three distinct solutions, namely those of [5, 6, 7]. In this paper, we shall present
a more general 4-parameter solution that contains all of these solutions as special cases. Just like
the known solutions, our solution preserves 1/4 of the supersymmetry in five dimensions, which
corresponds to 1/16 supersymmetry (i.e. 2 real supercharges) in ten dimensions.
The method that we shall use to construct the solution is based on the work of [10, 5, 11], which
reveals that supersymmetric solutions of five-dimensional = 1 gauged supergravity coupled to
N
abelian vector multiplets can always be written in a certain canonical form. This canonical form
involves a four-dimensional Ka¨hler “base space”. The choice of the base space is the difficult step
in constructing interesting solutions. The solutions of [4, 5] were discovered by requiring the base
space to have a certain highly symmetric form (cohomogeneity 1 with U(1) SU(2) isometry group)
×
which is the reason why these solutions have J = J . It turns out that the base space is singular.
1 2
One would therefore expect that solutions with J = J should also have a singular base space, with
1 2
6
less symmetry. Finding the appropriate space by guesswork would be very difficult.
Fortunately, guessworkisnotrequiredbecausewealreadyknowofsolutionswithJ = J ,namely
1 2
6
those of [6, 7]. These were obtained as limits of non-supersymmetric solutions rather than by using
theapproachof [10,5,11]. Sowestartby writingthesesolutions inthecanonical form. We findthat
both of these solutions have the same base space, which is indeed a less symmetric generalization
(cohomogeneity 2 with U(1)2 isometry group) of that of [4, 5]. In other words, we have a base space
that encompasses all of the known solutions. We take this base space as the starting point in our
generalization of these solutions. Applying the method of [10, 5, 11] together with a little more
guesswork based on the known solutions then leads to our new solution.
Rather than working directly with the U(1)3 theory, we have found it more convenient to study
gauged supergravity coupled to n 1 abelian vector multiplets (i.e. gauge group U(1)n) since this
−
wasthetheoryconsideredin[5,11]. Insection2, weshallobtaina(n+1)-parametersupersymmetric
black hole solution of this more general theory, and then specialize to the U(1)3 theory in section 3.
We have tried to make section 3 self-contained so that it can be read independently of section 2.
The supersymmetric solution that we present should arise as a limit of a non-supersymmetric
stationary black hole solution. It would be interesting to find this solution, which should generalize
the non-supersymmetric solutions in [6, 7]. A possible method for systematically attacking this
problem would be to use the Carter separability criterion, which is equivalent to the existence of a
second rank Killing tensor. Indeed, this is how the four dimensional Kerr-Newmann AdS solution
was found [12] and the non-supersymmetric solution of [7] also satisfies this criterion [13].
Finally, weshouldnotethattherehavebeenattemptstocalculatetheentropyofsupersymmetric,
asymptotically anti-de Sitter black holes using = 4 SU(N) super Yang-Mills theory. Free field
N
theory can reproduce certain qualitative aspects of these black holes for large R-charge, but, as one
might expect, always gives an overestimate of the entropy [14, 15]. An alternative approach is to
construct an index which only receives contributions from states in short superconformal multiplets
that cannot combine into long ones. All such indices where constructed in [15]. Unfortunately,
these indices count bosonic and fermionic states with opposite signs, which leads to a dramatic
cancellation: the index is (1) at large N whereas the black hole entropy is (N2) [15]. The
O O
authors of [15] speculated that other approaches, taking more account of the dynamics of the gauge
theory (e.g. by studying the chiral ring) might be more fruitful. We hope that our work will lead to
renewed interest in this problem.
2
2 The general solution
2.1 The theory
We shall consider the theory of five dimensional = 1 gauged supergravity coupled to n 1 abelian
N −
vector multiplets constructed in [16]. The bosonic sector this theory consists of the graviton, n
vectors AI and n 1 real scalars. The latter can be replaced with n real scalars XI subject to a
−
constraint
1
C XIXJXK = 1, (2)
IJK
6
where C are a set of real constants symmetric under permutations of (IJK). Indices I,J,K,...
IJK
run from 1 to n. It is convenient to define
1
X C XJXK. (3)
I IJK
≡ 6
The action is2
1 1
S = R ⋆1 Q FI ⋆FJ Q dXI ⋆dXJ C FI FJ AK +2g2 ⋆1 ,
5 IJ IJ IJK
16πG − ∧ − ∧ − 6 ∧ ∧ V
Z (cid:18) (cid:19)
(4)
where FI dAI and
≡ 9 1
Q X X C XK. (5)
IJ I J IJK
≡ 2 − 2
For simplicity, we shall assume that the scalars parametrize a symmetric space, which is equivalent
to the condition
CIJKCJ′(LMCPQ)K′δJJ′δKK′ = 4δI(LCMPQ). (6)
3
This condition ensures that the matrix Q is invertible, with inverse
IJ
QIJ = 2XIXJ 6CIJKX , (7)
K
−
where
CIJK C . (8)
IJK
≡
We then have
9
XI = CIJKX X . (9)
J K
2
The symmetric space condition is probably not essential – we expect that it should be possible to
use the results of [11] to relax it in what follows. However, we are mainly interested in a U(1)3
theory for which this condition is satisfied so we shall assume it for simplicity henceforth.
The scalar potential is3
= 27CIJKX¯ X¯ X , (10)
I J K
V
where X¯ are a set of constants. It was shown in [5] that the unique maximally supersymmetric
I
solution of this theory is AdS with radius g−1, vanishing vectors and constant scalars: XI = X¯I,
5
where
9
X¯I CIJKX¯ X¯ . (11)
J K
≡ 2
In section 3, we shall consider a particular U(1)3 gauged supergravity. In the above language, this
theory has n= 3, C = 1 if (IJK)is apermutation of (123) andC = 0 otherwise, andX¯I = 1
IJK IJK
(so X¯ = 1/3).
I
2Weuse a positive signature metric.
3This is related to thenotation of [5] via equations (2.23) and (2.29) of that paper.
3
2.2 Supersymmetric solutions
The general nature of supersymmetric solutions of this theory was deduced in [5] following closely
the corresponding analysis for the minimal theory given in [10]. Given a supercovariantly constant
spinor ǫ, one can construct a real scalar f ǫ¯ǫ and a real vector Vµ ǫ¯γµǫ. These obey V2 = f2,
∼ ∼ −
so V is timelike or null, and it turns out that V is always Killing. There are two cases: a “null”
case, in which V is globally null and a “timelike” case in which V is timelike in some region
U
of spacetime. The former case was treated in [10, 11] and does not concern us here because such
solutions cannot describe black holes.
In the timelike case, we can, without loss of generality, assume that f > 0 in , and introduce
U
local coordinates so that the metric takes the form
ds2 = f2(dt+ω)2+f−1h dxmdxn, (12)
mn
−
with V = ∂/∂t, h is a metric on a 4-dimensional Riemannian “base space” B and ω a 1-form
mn
on B. Supersymmetry implies that the base space is Ka¨hler [10, 5]. Let J denote the Ka¨hler form
and define the orientation of B so that J is anti-self-dual. If η denotes the volume form of B then
4
(dt+ω) η must then be positively oriented in space-time.
4
∧
Turning to the Maxwell fields, supersymmetry implies that [5]
FI = d XIf(dt+ω) +ΘI 9gf−1CIJKX¯ X J, (13)
J K
−
h i
and
2
X ΘI = G+, (14)
I
−3
where ΘI are self-dual 2-forms on B and
1
G+ = f (dω+⋆ dω), (15)
4
2
where ⋆ is the Hodge dual on B. Now define to be the Ricci form of B:
4
R
1
R Jpq, (16)
mn mnpq
R ≡ 2
where R is the Riemann tensor of B and indices are raised with hmn. Supersymmetry implies
mnpq
that [5]
= dP, (17)
R
where
P 3gX¯ AI fXIω . (18)
m ≡ I m− m
(cid:16) (cid:17)
These equations determine f [5] (R is the Ricci scalar of B):
108g2
f = CIJKX¯ X¯ X . (19)
I J K
− R
The above conditions are all necessary for supersymmetry. The analysis of [5] reveals that they
are also sufficient, i.e., a supercovariantly constant spinor will exist if the metric and maxwell field
are given by (12) and (13) for some Ka¨hler B, and equations (14) and (17) are satisfied. The field
equations of the theory are all satisfied once we impose the equations of motion for the Maxwell
fields [5], i.e., the Bianchi identities
dFI = 0, (20)
and the Maxwell equations,
1
d Q ⋆FJ = C FJ FK. (21)
IJ IJK
−4 ∧
(cid:16) (cid:17)
4
One can substitute the expression (13) into these equations to obtain a pair of equations on B [5]
but we shall not write them out here.
The analysis of [10, 5] reveals that the fraction of supersymmetry preserved by these solutions
is generically 1/4, i.e., such solutions are preserved by 2 real supercharges. This might be enhanced
in special cases. However, it has been shown [17] that this does not happen for the supersymmetric
blackholesof[4],evenwhenoxidizedtotendimensions. Sincetheseareaspecialcaseofoursolution,
we do not expect supersymmetry enhancement here either.
2.3 The base space
The first step in constructing a new solution using the above procedure is to choose the base space.
As discussed in the introduction, we shall use a base space obtained by writing the metrics of the
supersymmetric solutions of [6, 7] in the form (12). We find that the base space of both metrics
takes the form
dr2 dθ2 cos2θ
h dxmdxn = (r2 r2) + + Ξ +cos2θ ρ2g2+2(1+bg)(a+b)g dψ2
mn − 0 (∆r ∆θ Ξ2b b
h (cid:16) (cid:17)i
sin2θ
+ Ξ +sin2θ ρ2g2+2(1+ag)(a+b)g dφ2
Ξ2 a
a
h (cid:16) (cid:17)i
2sin2θcos2θ
+ ρ2g2+2(a+b)g+(a+b)2g2 dψdφ , (22)
ΞaΞb )
h i
where
∆ = (r2 r2)2[g2r2+(1+ag+bg)2]/r2, ∆ = 1 a2g2cos2θ b2g2sin2θ,
r − 0 θ − −
Ξ = 1 a2g2, Ξ =1 b2g2, ρ2 = r2+a2cos2θ+b2sin2θ. (23)
a b
− −
The coordinate ranges are r2 > r2, 0 θ π/2 and φ,ψ are angular coordinates with period
0 ≤ ≤
2π. Surfaces of constant r have topology S3. This metric is specified by three constants: a, b and
r2. The first two of these correspond roughly to angular momentum parameters for the φ and ψ
0
directions of the black holes of [6, 7]. We shall assume4 g−1 > a,b 0. The base space has a
≥
curvature singularity at r2 = r2 but the space-time metric (12) is smooth at r2 = r2, in fact this
0 0
corresponds to the event horizon. r2 is not an independent parameter for these solutions. For the
0
minimal gauged supergravity solution of [7], it takes the value r2 = r2 , where
0 m
r2 = g−1(a+b+abg). (24)
m
For the U(1)3 solution of [6], r02 = r∗2, where
ab
r∗2 = . (25)
1+ag+bg
For these values of r , supersymmetryimplies that the above base space metric must be Ka¨hler. We
0
find that the Ka¨hler form is
(r2 r2) cos2θdψ sin2θdφ
J = d − 0 + . (26)
− " 2 Ξb Ξa !#
4Itiseasy toseethata,b≥0can alwaysbearranged byacoordinate transformation. Forexample,ifa<0in the
generalnon-supersymmetricsolutionof[7]thentakingt→−t,ψ→−ψandq→−q (whereq isthechargeparameter
of [7]) effectively reverses thesign of a. The base space is determined after arranging a≥0,b≥0.
5
In fact, we find that the metric (22) is Ka¨hler for any r2, with Ka¨hler form given by (26). Anti-
0
self-duality of J determines the orientation of B: dr dθ dψ dφ is positively oriented. For later
∧ ∧ ∧
convenience, we also record the Ricci form:
cos2θ 4r2 r2 sin2θ 4r2 r2
= 3g2d ρ2+b2+ m 0 dψ+ ρ2+a2+ m 0 dφ . (27)
R " Ξb 3 − 3 ! Ξa 3 − 3 ! #
It appears, then, that we have a 3-parameter generalization of the 2-parameter base metrics of [6, 7].
However, this is not the case: it turns out one combination of these parameters can be eliminated
by a coordinate transformation. To see this, let
r2 r2 = α2sinh2(gσ), α2 r2+g−2(1+ag+bg)2. (28)
− 0 ≡ 0
The base metric now takes the simplified form5:
dψ dφ 2
h dxmdxn = dσ2+α2sinh2(gσ) α2g2cosh2(gσ) ∆ cos2θ +sin2θ
mn θ
− Ξ Ξ
(cid:16) (cid:17)(cid:18) b a(cid:19)
dθ2 dψ2 dφ2
+ α2sinh2(gσ) +∆ cos2θ +∆ sin2θ . (29)
∆θ θ Ξ2b θ Ξ2a !
Since∆ = Ξ cos2θ+Ξ sin2θ,itcannowbeseenthatthemetricdependsononlytwocombinations
θ a b
of the constants, namely A,B > 0, where
Ξ Ξ
A2 a , B2 b . (30)
≡ g2α2 ≡ g2α2
The redundancy in the metric (22) implies that r2 can be set to any convenient value by means of
0
a coordinate transformation. We shall choose r2 =0. Note that the parameters a, b also transform.
0
′ ′
If we start in the “gauge” r = 0 and transform to some new value of r then the new values (a,b)
0 0
can be determined by invariance of A,B. This is important when comparing our results with those
of [6, 7].
2.4 The solution
Having chosen a base space, it remains to solve the remaining conditions for supersymmetry. We
shall do this by making an Ansatz that is sufficiently general to encompass the known solutions of
[5, 6, 7]. We take the general supersymmetric metric (12) with the particular base space (22) with
r = 0. The Ansatz is
0
ω = ω dψ+ω dφ, (31)
ψ φ
gcos2θ gsin2θ
ω = ρ4+P ρ2+P , ω = ρ4+Q ρ2+Q , (32)
ψ − r2Ξ 2 0 φ − r2Ξ 2 0
b a
(cid:16) (cid:17) (cid:16) (cid:17)
1 X¯ ρ2+e
f−1X = H I I, (33)
I 3 I ≡ r2
gcos2θ gsin2θ
AI = fXI(dt+ω)+ X¯Iρ2+cI dψ+ X¯Iρ2+dI dφ, (34)
Ξ Ξ
b a
(cid:16) (cid:17) (cid:16) (cid:17)
where P ,P ,Q ,Q ,e ,cI and dI are constants. Equations (33) and (3) determine f:
2 0 2 0 I
f−3 = 1CIJKH H H , (35)
I J K
6
5Inthisform,itiseasytoseethatthebasemetricreducestothecohomogeneity1basemetricof[4,5]whena=b.
6
i.e.,
r2
f = , F(ρ2) ρ6+β ρ4+β ρ2+β , (36)
F(ρ2)1/3 ≡ 1 2 3
where
27 27 9
β = CIJKX¯ X¯ e = 3X¯Ie , β = CIJKX¯ e e , β = CIJKe e e . (37)
1 I J K I 2 I J K 3 I J K
2 2 2
For the solutions of [5, 6, 7], AI and AI decay as 1/r2 for large r so we impose the same condition
ψ φ
here, which determines
cI = X¯I (P β )+9CIJKX¯ e , dI = X¯I(Q β )+9CIJKX¯ e . (38)
2 1 J K 2 1 J K
− −
For orientation, we note that this Ansatz can be embedded in minimal gauged supergravity when
X X¯ and AI = X¯I where is the Maxwell potential of the minimal theory.
I I
≡ A A
Now we impose the conditions required for supersymmetry. First, equation (17) is satisfied for
the Ricci form (27) by taking P = (4r2 +β )/3+b2 and Q = (4r2 +β )/3+a2 (where r2 is
2 m 1 2 m 1 m
defined by (24)). Next, either from imposing self-duality of ΘI or from equation (19) one obtains
the constraint
β = 2r2 , (39)
1 m
and hence
P = 2r2 +b2, Q =2r2 +a2. (40)
2 m 2 m
The final equation required for supersymmetry is (14), which is satisfied if
1 1
P = β a2b2+g−2(a2 b2) , Q = β a2b2 g−2(a2 b2) . (41)
0 2 0 2
2 − − 2 − − −
h i h i
WemustnowimposetheequationsofmotionfortheMaxwellfields. TheBianchiidentitiesaretrivial
(we’ve specified potentials) and the Maxwell equations turn out to impose no further restrictions.
Hence we have a supersymmetric solution. It is specified by the constants a,b and e subject to the
I
constraint (39). Therefore the solution has n+1 independent parameters.
2.5 Asymptotics and charges
Now we show that oursupersymmetricsolution asymptotes to AdS . Totransform to a framewhich
5
is not rotating at infinity let t = t¯, ψ = ψ¯ gt¯, φ = φ¯ gt¯and y2 = r2 +2r2 /3. The spacetime
− − m
metric for large r (or y) then takes the form
∆ 1 1 1 1
ds2 = θ (1+g2y2)+ dt˜2+ dt˜dψ˜+ dt˜dφ˜+ dψ˜dφ˜
− Ξ Ξ O y2 O y2 O y2 O y2
(cid:20) a b (cid:18) (cid:19)(cid:21) (cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
cos2θ 1 sin2θ 1
+ (y2+b2)+ dψ˜2 + (y2+a2)+ dφ˜2
" Ξb O(cid:18)y2(cid:19)# " Ξa O(cid:18)y2(cid:19)#
ρ˜2+ (1/y2) ρ˜2+ (1/y2)
+ O dθ2+ O dy2 (42)
∆ ∆
θ y
ρ˜2 = y2+a2cos2θ+b2sin2θ,
2 2 4
∆ = y−2 y2 r2 1+g2y2+g2a2+g2b2+ g2r2 .
y − 3 m 3 m
(cid:18) (cid:19) (cid:18) (cid:19)
To bring the asymptotic metric into the familiar global AdS coordinates one needs to perform the
5
transformation:
Ξ Y2sin2Θ = (y2+a2)sin2θ, Ξ Y2cos2Θ = (y2+b2)cos2θ. (43)
a b
7
In the asymptotically static coordinates, the supersymmetric Killing field is
∂ ∂ ∂
V = +g +g . (44)
∂t¯ ∂φ¯ ∂ψ¯
The notation used above enabled the solution to be presented in a concise manner. However, in
calculating the charges it is useful to define new parameters q defined by
I
e = Ξ Ξ q g−2 1 Ξ Ξ X¯ . (45)
I a b I a b I
− −
p (cid:16) p (cid:17)
We shall see that q are closely related to the electric charges of our solution. Let
I
27 27 9
α = CIJKX¯ X¯ q = 3X¯Iq , α = CIJKX¯ q q , α = CIJKq q q . (46)
1 I J K I 2 I J K 3 I J K
2 2 2
These are related to the β via
i
β = Ξ Ξ α 3g−2(1 Ξ Ξ ),
1 a b 1 a b
− −
2
p 2√Ξ Ξ 1 p√Ξ Ξ 3 1 √Ξ Ξ
a b a b a b
β = Ξ Ξ α − α + − ,
2 a b 2− g2 1 g4
(cid:0) (cid:1) (cid:0) (cid:1)
3 ΞaΞb √ΞaΞb(1 √ΞaΞb)2 1 √ΞaΞb 3
β3 = (ΞaΞb)2α3− g2 1− ΞaΞb α2+ g−4 α1− − g6 . (47)
(cid:0) (cid:1)
(cid:16) p (cid:17)
The constraint (39) becomes
1 1 A B 1
α = + + 3 = 2r2 +3g−2 1 Ξ Ξ , (48)
1 g2 AB B A − √Ξ Ξ m − a b
(cid:18) (cid:19) a b h (cid:16) p (cid:17)i
where6
√Ξ √Ξ
a b
A , B . (49)
≡ (1+ag+bg) ≡ (1+ag+bg)
The electric charges are defined by:
1
Q = Q ⋆FJ, (50)
I IJ
8πG ZS3
where the integral is taken over a three-sphere at infinity. This gives:
3π g2α 3g2
Q = q 2X¯ + C X¯JCKLMq q . (51)
I I I IJK L M
4G − 2 2 !
Note that this is independent of a,b and hence coincides with the result of [5].
We will define the angular momenta via Komar integrals
1
J = ⋆dK , (52)
i i
16πG ZS3
where the Killing vector Kµ is either ∂ or ∂ and the S3 is at infinity. We find:
i φ ψ
π 1 (B A)
J = gα +g3α + − 1+g2α +g4α +g6α , (53)
φ 4G 2 2 3 Ag3 1 2 3
(cid:20) (cid:16) (cid:17)(cid:21)
π 1 (A B)
J = gα +g3α + − 1+g2α +g4α +g6α . (54)
ψ 4G 2 2 3 Bg3 1 2 3
(cid:20) (cid:16) (cid:17)(cid:21)
6NotethatA,Barethe“gauge-invariant”quantitiesdefinedin(30),herewritteninthegauger0 =0. Itistherefore
straightforward to convert our expressions to any other valuefor r0, which facilitates comparison with [6, 7].
8
The mass is fixed by the BPS condition
M = g J +g J + X¯IQ , (55)
φ ψ I
| | | | | |
which gives
π 3 (A B)2
M = α + g2α +2g4α + − 1+g2α +g4α +g6α . (56)
4G " 1 2 2 3 g2AB 1 2 3 #
(cid:16) (cid:17)
As written above, is trivial to see that these various charges reduce to those of the black holes of [5]
when a= b. We have have also checked that they are consistent with the results of [6, 7], for which
one needs to use (30) together with the appropriate value of r .
0
2.6 Absence of causal pathologies
Note that f is an invariant of the solution and should therefore remain finite on, and outside, the
horizon. Hence we demand
F(ρ2) > 0 for r 0. (57)
≥
Thisconstraint guarantees thatthescalars arefiniteforr 0. For A> B,evaluating this inequality
≥
at r = 0 gives
(A B) (A B)2
g6α > − − +g4α . (58)
3 B B2 2!
If A < B then the same expression holds with A and B interchanged.
Absence of closed causal curves requires that the φ ψ part of the metric be positive definite.
−
We find that this is the case as r 0 if, and only if,
→
g2α2 (A B)2
δ (1+g2α )α 2 − 1+g2α +g4α +g6α > 0. (59)
≡ 1 3− 4 − ABg6 1 2 3
(cid:16) (cid:17)
2.7 Extension through the horizon
We shall now show that our solution has an event horizon at r = 0. To this end, we transform to
′ ′
new coordinates (v,R,θ,φ,ψ ) where
R = gr2, (60)
A0 A1 ′ B0Ξb ′ C0Ξa
dv = dt + dR, dψ = dψ dR, dφ = dφ dR, (61)
− g2R2 gR − R − R
(cid:18) (cid:19)
where A ,A ,B ,C are constants to be determined. The exterior of the black hole corresponds to
0 1 0 0
′ ′
R > 0. Thescalars X can besmoothly extendedthrough R = 0, as can thev,φ andψ components
I
of the Maxwell potentials AI. The remaining non-zero components AI diverge as 1/R as R 0.
R →
However, this divergence is pure gauge if the coefficient of 1/R is independent of θ. This will be the
case if, and only if,
C Q a2Q β = B P b2P β , (62)
0 0 2 2 0 0 2 2
− − − −
(cid:16) (cid:17) (cid:16) (cid:17)
B b2P +β C a2Q +β +g−4(a2 b2)A =0. (63)
0 0 3 0 0 3 0
− −
(cid:16) (cid:17) (cid:16) (cid:17)
If A ,B ,C satisfy these equations then the Maxwell field strengths can be smoothly extended
0 0 0
through R = 0.
Now let’s consider the metric. It is convenient to work with independent variables R and ρ2
rather than R and θ. We then expand metric components as Laurent series in R (near R = 0). The
coefficients in this series are analytic functions of ρ2. We have
gvv = f2 = (R2), gvψ′ = f2ωψ = (R), gvφ′ = f2ωφ = (R),
− O − O − O
gφ′φ′ = gφφ = (1), gφ′ψ′ = gφψ = (1), gψ′ψ′ = gψψ = (1). (64)
O O O
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Using equations (62) and (63) we find
B C
gvR = a02−b20 F(ρ2)1/3+O(R), gRψ′ = O(1), gRφ′ = O(1). (65)
(cid:18) − (cid:19)
The RR component of the metric contains divergent 1/R2 and 1/R terms so we need the coefficients
of these terms to vanish. After using (62) and (63), the vanishing of the coefficient of the 1/R2 term
determines B ,C :
0 0
g2β /2+g2a4+g2a2b2/2+2g2a2r2 +(a2 b2)/2
B = 2 m − ,
0
± 2gΞ Ξ (1+ag+bg)2√δ
a b
g2β /2+g2b4+g2a2b2/2+2g2b2r2 (a2 b2)/2
C = 2 m− − , (66)
0
± 2gΞ Ξ (1+ag+bg)2√δ
a b
where δ is defined by (59). A is now determined by (63). Finally, vanishing of the coefficient of the
0
1/R term in g determines A :
RR 1
g2(C Q B P ) g2(a2 b2)
0 0 0 0
A = − + −
1 a2 b2 8(B C )(1+ag+bg)4
0 0
− −
(a+b)
B2(1+bg)(3b2g a+b+abg) 2B C (a2+b2)g(2+ag+bg)
− 2(a2 b2)(B C ) 0 − − 0 0
0 0
− − h
+ C2(1+ag)(3a2g+a b+abg) . (67)
0 −
i
We now have
g = (1). (68)
RR
O
Using the solutions for B ,C determines
0 0
F(ρ2)1/3
g = + (R). (69)
vR
±2gΞ Ξ √δ O
a b
The metric and its inverse are now analytic at R = 0 and can therefore be extended into a new
region R < 0. The surface R = 0 is a Killing horizon. To see this, consider the supersymmetric
Killing field V = ∂/∂v, which is null at R = 0. The above results imply
F(ρ2)1/3
V dxµ = dR . (70)
µ R=0
| ±"2gΞaΞb√δ #R=0
This reveals that the R = 0 is a null hypersurface with normal V, i.e., a Killing horizon of V. The
upper choice of sign corresponds to a future horizon and the lower choice to a past horizon.
Equation (44) determines the angular velocities of the horizon with respect to the static frame
at infinity:
Ω = Ω = g. (71)
φ ψ
The geometry of a spatial cross-section of the event horizon is determined by setting v = constant
and R = 0, or equivalently t =constant and r 0. This gives a deformed S3 with area
→
g2α2 (A B)2
= 2π2 (1+g2α )α 2 − (1+g2α +g4α +g6α ). (72)
AH s 1 3− 4 − ABg6 1 2 3
This reduces correctly to the results of [5, 6, 7] in the appropriate limits.
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