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Exact Three Dimensional Black Holes in String Theory

arXiv:hep-th/9302126v1 26 Feb 1993NSF-ITP-93-21hep-th/9302126Exact Three Dimensional Black Holes in String TheoryGary T. Horowitz∗Institute for Theoretical PhysicsUniversity of CaliforniaSanta Barbara, CA 93106-9530gary@cosmic.physics.ucsb.eduDean L. WelchDepartment of PhysicsUniversity of CaliforniaSanta Barbara, CA 93106-9530dean@cosmic.physics.ucsb.eduAbstractA black hole solution to three dimensional general relativity with a negative cosmolog-ical constant has recently been found. We show that a slight modification of this solutionyields an exact solution to string theory.

This black hole is equivalent (under duality) tothe previously discussed three dimensional black string solution. Since the black string isasymptotically flat and the black hole is asymptotically anti-de Sitter, this suggests thatstrings are not affected by a negative cosmological constant in three dimensions.2/93∗On leave from the Physics Department, University of California, Santa Barbara, CA.

In a recent paper [1], Banados et. al.

showed that there is a black hole solution tothree dimensional general relativity with a negative cosmological constant. At first sightthis is surprising, since the field equation for this theory requires that, locally, the curvatureis constant.

However they showed that by identifying certain points of three dimensionalanti-de Sitter space, one obtains a solution with almost all of the usual features of a blackhole. In fact, there are a two parameter family of inequivalent identifications leading toblack holes with mass M and angular momentum J.

Even though the curvature is constant,the solutions have trapped surfaces, an event horizon, and nonzero Hawking temperature.When J ̸= 0, they also have an ergosphere, and inner horizon. The solutions all approachanti-de Sitter space (without identifications) asymptotically.This solution is easily modified to obtain an exact solution to string theory.

Onesimply adds an antisymmetric tensor field Hµνρ proportional to the volume form ǫµνρ.The reason is the following. There is a well known construction (the Wess-Zumino-Wittenmodel) for obtaining a conformal field theory describing string propagation on a Lie group.The natural metric on the group SL(2, R) is precisely the three dimensional anti-de Sittermetric.

So the WZW model based on SL(2, R) is an exact conformal field theory describingstring propagation on anti-de Sitter space [2]. The Hµνρ field is required by the Wess-Zumino term and must be chosen so that the connection with torsion Hµνρ is flat.

Toobtain the black hole, one applies the orbifold procedure [3] to obtain a conformal fieldtheory describing string propagation on the quotient space. This projects onto the stateswhich are invariant under the discrete group, and adds the winding states.This solution is of interest for several reasons.

An exact four dimensional black holein string theory has not yet been found. A few years ago, Witten showed [4] that an exacttwo dimensional black hole could be obtained by gauging a one dimensional subgroup ofSL(2, R).

The three dimensional black hole has a number of advantages over the twodimensional one. First, strings in three dimensions resemble higher dimensional solutionsin that there are an infinite number of propagating modes.

One can thus examine theireffect on Hawking evaporation. Second, the construction is even simpler than the twodimensional black hole.

One merely quotients by a discrete subgroup rather than gauginga continuous one. In three dimensions, there will presumably be a tachyon, which can beremoved by considering the supersymmetric WZW model.One of the most interesting properties of string theory is that different spacetimegeometries can correspond to equivalent classical solutions.

We will show that the threedimensional black hole is equivalent to the charged black string solution discussed earlier1

[5]. Some implications of this equivalence for spacetime singularities and the cosmologicalconstant problem will be considered.We begin by reviewing the black hole solution discovered by Banados et.al.

[1].Anti-de Sitter space can be represented as the surface−x20 −x21 + x22 + x23 = −l2(1)in the flat space of signature (– – + +)ds2 = −dx20 −dx21 + dx22 + dx23(2)It has curvature Rµν = −(2/l2)gµν. This space is clearly invariant under SO(2, 2).

Thesix independent Killing vectors consist of two rotations (in the (0, 1) and (2, 3) planes) andfour boosts. A convenient way to parameterize the surface is to choose two commutingKilling vectors and let two of the coordinates be the parameters along these symmetrydirections.

If t and ϕ are parameters along the two rotations, the metric takes the familiarformds2 = −1 + r2l2dt2 +1 + r2l2−1dr2 + r2dϕ2(3)To obtain the black hole, we let ˆt and ˆϕ be parameters along the boosts in the (0,3) and(1,2) planes. Explicitly, setx1 =ˆr cosh ˆϕx0 =pl2 −ˆr2 cosh (ˆt/l)x2 =ˆr sinh ˆϕx3 =pl2 −ˆr2 sinh (ˆt/l)(4)Notice that ˆr2 > 0 only covers the region x21 −x22 > 0.

Since this is not the entire space, itmight be more natural to use the radial coordinate ρ = ˆr2 which takes both positive andnegative values. For now, we will continue to use ˆr to maintain agreement with [1].

Interms of these coordinates, anti-de Sitter space becomesds2 =1 −ˆr2l2dˆt2 + ˆr2l2 −1−1dˆr2 + ˆr2d ˆϕ2(5)Since ˆt and ˆϕ are both parameters along a boost, they can take any real value. If weidentify ˆϕ = ˆϕ + 2π, (5) describes a black hole.

The surfaces of constant ˆt and ˆr < l arenow compact trapped surfaces. One could also identify ˆϕ with a period other than 2π.

Itturns out that this corresponds to changing the mass of the black hole. This is analogous2

to the fact that in the absence of a cosmological constant, the mass (in three dimensions)is related to the deficit angle at infinity [6]. To add angular momentum, one periodicallyidentifies a linear combination of ˆϕ and ˆt, rather than ˆϕ itself.To be more explicit, choose two constants r+, r−and introduce new coordinates ˆt =(r+t/l) −r−ϕ,ˆϕ = (r+ϕ/l) −(r−t/l2),ˆr2 = l2(r2 −r2−)/(r2+ −r2−).

Then the metric(5) becomes [1]ds2 =M −r2l2dt2 −Jdtdϕ + r2dϕ2 +r2l2 −M + J24r2−1dr2(6)where the constants M and J are related to r± byM = r2+ + r2−l2J = 2r+r−l(7)Identifying ϕ with ϕ + 2π, yields a two parameter family of black holes. By paying carefulattention to the surface terms in a Hamiltonian analysis [1], one finds that M is themass and J is the angular momentum of the solution.

In general, there are two horizonswhere ∇µr becomes null, which are located at r = r±. These two horizons coincide when|J| = Ml which is the extremal limit.

The extremal black hole, as well as the masslesssolution M = J = 0, cannot be obtained by the above identifications. Instead, one mustuse null boosts.

The original anti-de Sitter space (3) is recovered when M = −1 and J = 0.The Killing vector ∂/∂t becomes null at r2 = Ml2 which lies outside the event horizonr = r+ when J ̸= 0. This is similar to the ergosphere in the Kerr solution.

Physically, itmeans that an observer cannot remain at rest with respect to infinity when she is close tothe horizon.What is the spacetime like near r = 0? Since the curvature is constant, there cannotbe a curvature singularity.

When J = 0 and M > 0, the ϕ translation symmetry has afixed point in the (1,2) plane. This causes the solution, near r = 0, to resemble the Taub-NUT solution and have incomplete null geodesics.

However, when J ̸= 0, the symmetryhas no fixed points and the spacetime is completely nonsingular. This is consistent withthe singularity theorems [7] even though the spacetime has trapped surfaces and satisfiesthe strong energy condition, because there are closed timelike curves.

(Recall that thecontinuation past r = 0 consists of r2 becoming negative, so ϕ becomes timelike.) Banadoset.

al. argue that one should end the spacetime at r = 0, which avoids the causalityproblem but creates incomplete geodesics.

However, this appears very unnatural. Thefour dimensional Kerr solution also has closed timelike curves inside the inner horizon.3

(Although the vector ∂/∂ϕ is timelike only for r < 0 in Kerr, there are closed timelikecurves through every point with r < r−. It is likely that a similar result holds here.

)The closed timelike curves are not expected to produce a physical violation of causalitybecause of the instability of the inner horizon. String theory provides another reason fornot ending the spacetime at r = 0.

The WZW construction clearly includes all regions ofthe spacetime, including r2 < 0.The Hawking temperature for this black hole isT = r2+ −r2−2πr+l2(8)(The factor of l2 was omitted in [1].) These black holes do not evaporate completely ina finite time.

To see this, notice that since the temperature is reduced by rotation, wecan obtain a lower limit on the lifetime by setting r−= 0. Then the temperature and thehorizon size are both proportional to r+ ∼√M.

In three dimensions, the energy flux inthermal radiation is proportional to T 3, so˙M ∝M 2 which implies M(t) ∝1/t. Sincetheir lifetime is infinite, small black holes will act like stable remnants.We now turn to string theory.

We first consider these black holes in the context of thelow energy approximation, and then consider the exact conformal field theory. In threedimensions, the low energy string action isS =Zd3x√−g e−2φ4k + R + 4(∇φ)2 −112HµνρHµνρ(9)The equations of motion which follow from this action areRµν + 2∇µ∇νφ −14HµλσHνλσ = 0(10a)∇µ(e−2φHµνρ) = 0(10b)4∇2φ −4(∇φ)2 + 4k + R −112H2 = 0(10c)A special property of three dimensions is that Hµνρ must be proportional to the volumeform ǫµνρ.

If we assume φ = 0, then (10b) yields Hµνρ = (2/l)ǫµνρ where l is a constantwith dimensions of length. Substituting this form of H into (10a) yieldsRµν = −2l2 gµν(11)which is exactly Einstein’s equation with a negative cosmological constant.

The dilatonequation (10c) will also be satisfied provided k = l2. Thus every solution to three dimen-sional general relativity with negative cosmological constant, is a solution to low energy4

string theory with φ = 0, Hµνρ = (2/l)ǫµνρ and k = l2. In particular, the two parameterfamily of black holes (6) is a solution withBϕt = r2l ,φ = 0(12)where H = dB.

An earlier argument [8] claiming that three dimensional black hole solu-tions to (10) do not exist assumed that Hµνρ = 0 [9].We now consider the dual of this solution. Duality is a well known symmetry of stringtheory that maps any solution of the low energy string equations (10) with a translationalsymmetry to another solution.

(Under certain conditions, the two solutions correspond toequivalent conformal field theories [10][11].) Given a solution (gµν, Bµν, φ) that is inde-pendent of one coordinate, say x, then (˜gµν, ˜Bµν, ˜φ) is also a solution where [12]˜gxx = 1/gxx,˜gxα = Bxα/gxx˜gαβ = gαβ −(gxαgxβ −BxαBxβ)/gxx˜Bxα = gxα/gxx,˜Bαβ = Bαβ −2gx[αBβ]x/gxx˜φ = φ −12 ln gxx(13)and α, β run over all directions except x.Applying this transformation to the ϕ translational symmetry of the black hole solu-tion (6)(12) yieldseds2 =M −J24r2dt2 + 2l dtdϕ+ 1r2 dϕ2 +r2l2 −M + J24r2−1dr2˜Bϕt = −J2r2φ = −ln r(14)To better understand this solution, we diagonalize the metric.

Lett =l(ˆx −ˆt)(r2+ −r2−)1/2 ,ϕ = r2+ ˆt −r2−ˆx(r2+ −r2−)1/2 ,r2 = lˆr(15)Then the solution becomeseds2 = −1 −Mˆrdˆt2 +1 −Q2Mˆrdˆx2+1 −Mˆr−1 1 −Q2Mˆr−1 l2 dˆr24ˆr2φ = −12 ln ˆrl ,Bˆxˆt = Qˆr(16)5

where M = r2+/l and Q = J/2. This is precisely the previously studied three dimensionalcharged black string solution [5].Notice that the charge of the black string is simplyproportional to the angular momentum of the black hole.

The horizons of the black stringare at the same location as the black hole r2 = r2±. Since ϕ is periodic, both ˆt and ˆx willin general be periodic.

To avoid closed timelike curves, one must go to the covering space.Since the dual of the black hole is the black string, it must be possible to dualize theblack string and recover the black hole. This is a little puzzling since it has been shown[13] that if you dualize (16) on ˆx, you obtain a boosted uncharged black string.

The chargeQ is dual to the momentum in the symmetry direction Pˆx. However, one can apply dualityto any translational symmetry ∂/∂ˆx + α∂/∂ˆt.

If α < 1, then the dual is again a chargedblack string. If α = 1 the result is different.

The Killing vector ∂/∂ˆx + ∂/∂ˆt has norm(M2 −Q2)/Mˆr, so it is spacelike everywhere but asymptotically null. One can easilyverify that the dual of the black string (16) with respect to this symmetry is precisely thethree dimensional black hole.We now consider a few special cases.

The dual of the nonrotating black hole, J = 0and M > 0, is the uncharged black string. This is simply the two dimensional black holecross S1.

For the zero mass solution (M = J = 0) and the extremal limit (|J| = Ml)the dual is still given by (14), but the transformation to the black string breaks down.The duals are not the zero mass and extremal black string, but rather these solutionssuperposed with a plane fronted wave. Setting M = J = 0 in (14), and introducing newcoordinates t = −v, ϕ = ul/2, r = eˆr/l yieldseds2 = −dudv + dˆr2 + l24 e−2ˆr/ldu2(17)This is a plane fronted wave in the presence of a dilaton [13].Setting J2 = M 2l2 in(14), and introducing new coordinates t = −v/M, ϕ = l(v + Mu)/2, r2 = lˆr, the metricbecomeseds2 = −1 −Ml2ˆrdudv +l2dˆr24ˆr −Ml22 + M 2l4ˆr du2(18)This describes a wave of constant amplitude traveling along an extremal black string [14].Finally, recall that the full anti-de Sitter space corresponds to J = 0 and M = −1.Inserting these values into (14) and setting t = ˆt + ϕ/l yieldseds2 = −dˆt2 +1 + r2l2−1dr2 +r2 + l2r2l2dϕ2(19)6

which is the product of time and the dual of the two dimensional Euclidean black hole.We now turn to the exact conformal field theory description of the black hole. Asmentioned earlier, this is in terms of the SL(2, R) WZW model.

One can parameterizeSL(2, R) byg =x2 + x1x0 + x3x0 −x3x2 −x1(20)The statement that this matrix has unit determinant is just equation (1) with −l2 replacedby 1. The difference in sign causes the induced metric to have signature (+ – –).

Onecan correct for this and reintroduce the scale by multiplying the WZW action by −k. Inother words, one considers the level k WZW model.

For the noncompact group SL(2, R),k is not required to be an integer. The central charge is c = 3k/(k −2), so c = 26 whenk = 52/23.

One can also consider larger values of k and take the product of this blackhole with an internal conformal field theory. In terms of the group, translations of ˆϕ in(5) correspond to the axial symmetryδg = ǫ100−1g + g100−1(21)while translations of ˆt correspond to the vector symmetryδg = ǫ100−1g −g100−1(22)The general black hole is obtained by quotienting under a discrete subgroup of a linearcombination of these symmetries.

This can be carried out by the standard orbifold con-struction [3].There is a close connection between the two and three dimensional black holes instring theory. Witten has shown [4] that the two dimensional black hole can be obtainedby starting with the SL(2, R) WZW model and gauging the axial symmetry (21).

If onegauges the vector symmetry, one obtains the dual of the black hole, which turns out tohave the same geometry. One cannot gauge a general linear combination of the symmetriesbecause of an anomaly.Rocek and Verlinde have shown [10] that for a positive definite target space havinga spacelike symmetry with compact orbits, the low energy duality (13) corresponds to anequivalence between exact conformal field theories.

For Lorentzian target spaces, equiv-alence has not yet been rigorously established. In addition to the obvious difficulty ofconvergence of the functional integral, there are other issues involving potential closed7

timelike curves. Nevertheless, since the equivalence does not explicitly depend on the sig-nature of the target space, one expects it to hold in this case also.

The effect of dualityon a WZW model has been investigated [10][15]. If one dualizes with respect to an axialor vector symmetry, the result (after a simple shift of coordinate) is just the product ofU(1) and the WZW model with this symmetry gauged [10].

This explains why the dualof the nonrotating black hole is just the product of the two dimensional black hole andU(1). It also explains why the dual of anti-de Sitter is the product of time and the dualof the Euclidean black hole.

However, we have seen that the dual of the general rotatingblack hole is the charged black string which is not a simple product. The exact conformalfield theory associated with this solution is also known [5].

One starts with the groupSL(2, R) × U(1) and gauges the axial symmetry (21) of SL(2, R) together with rotationsof U(1).Solutions to the low energy field equations cannot be trusted in regions of large cur-vature. But since the three dimensional black hole has constant curvature (which is smallfor large k), it should be a good approximation everywhere.

In fact, for a WZW model,the exact metric differs from the low energy approximation only by an overall rescaling[16][17]. So the exact three dimensional black hole metric is simply proportional to (6),and (for J ̸= 0) is nonsingular.

(Although the stability of the inner horizon and the effectof the closed timelike curves remain to be investigated.) A candidate for the exact twodimensional black hole metric has been found [18][19][16].

It has recently been shown [20]that this metric is also free of curvature singularities (although the dilaton diverges insidethe event horizon). This can be viewed as increasing evidence that exact black holes instring theory do not have curvature singularities.

But the evidence is far from conclusive.A candidate for the exact three dimensional black string metric has also been found[21][17], which does have a curvature singularity. However, the fact that the solution isequivalent to one without a curvature singularity, suggests that it may also correspond toa nonsingular conformal field theory.Perhaps the most remarkable consequence of the equivalence between the black holeand black string comes from the fact that the black hole is asymptotically anti-de Sitterwhile the black string is asymptotically flat.

Since they are equivalent, it suggests that anegative cosmological constant has no effect on strings in three dimensions. The reason thatthe asymptotic structure of the spacetime changes under duality is that the length of thecircles parameterized by ϕ does not approach a constant at infinity.

This phenomenon hasbeen noticed before, but in previous examples the asymptotic behavior of the dual space8

did not have a simple physical interpretation.For example, consider four dimensionalMinkowski spacetime ds2 = −dt2 +dr2 +r2(dθ2 +sin2 θdϕ2). If we dualize on ϕ the metricis identical except that gϕϕ is changed to (r2 sin2 θ)−1 which is singular along the axisθ = 0, π.

The three dimensional black hole seems to be the first example in which twodifferent asymptotic behaviors each have a simple physical interpretation.This suggests a novel resolution of the cosmological constant problem. Perhaps a so-lution with a cosmological constant in string theory is equivalent to one without.

Stringsmay not be affected by a cosmological constant. Unfortunately, a straightforward general-ization of our results to higher dimensions, does not relate a solution with a cosmologicalconstant to one without.

One can start with the charged black string in D dimensions[22] which is asymptotically flat, and dualize with respect to a symmetry that is spacelikebut asymptotically null. The result is a metric which is neither asymptotically flat norasymptotically anti-de Sitter.

However, given our elementary understanding of duality instring theory, one cannot rule out the possibility that this effect will play a role in resolvingthe cosmological constant problem.AcknowledgmentsIt is a pleasure to thank T. Banks, A. Giveon, J. Horne, M. Rocek, and A. Stromingerfor discussions. This work was supported in part by NSF Grants PHY-8904035 and PHY-9008502.9

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