Supersymmetric AdS5 solutions of type IIB supergravity without D3 branes

Journal of High Energy Physics, Jan 2017

We analyse the most general bosonic supersymmetric solutions of type IIB supergravity whose metrics are warped products of five-dimensional anti-de Sitter space (AdS5) with a five-dimensional Riemannian manifold M 5, where the five-form flux vanishes, while all remaining fluxes are allowed to be non-vanishing consistent with SO(4,2) symmetry. This completes the program of classifying all supersymmetric solutions of ten and eleven-dimensional supergravity with an AdS5 factor. We investigate the supersymmetry conditions in some special cases, and demonstrate how these are satisfied by a solution originally found in [13], utilising the method of non-Abelian T-duality.

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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 ve-dimensional anti-de 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 non-Abelian T-duality. AdS-CFT Correspondence; Supergravity Models - (AdS5) with a ve-dimensional Riemannian manifold M5, where the ishes, while all remaining uxes are allowed to be non-vanishing 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 non-singular 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 self-dual ve-form eld strength, F5, to be vanishing but allow all other Neveu-Schwarz Neveu-Schwarz (NS-NS) and R-R sider a simple ansatz and nd a singular solution, in section 6 we present a less simpli ed NATD-T-dual 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 one-forms U(1) gauge transformations, is de ned as dimensional metric gMN , the bosonic elds comprise the axion-dilaton = C(0) + ie complex three-form ds120 = e2 where ds2AdS5 is the metric on AdS5 with Ricci tensor given by R 4m2(gAdS5 ) is the metric on a ve-dimensional 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 ve-dimensional 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 ve-form uxes respecting SO(4; 2) symmetry, all the equations of motion and Bianchi identities are implied by supersymmetry. Introducing local coordinates and R-R two-form potentials. K5. As a vector we have and as a one-form is a one-form 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 one-form and is a complex one-form, and both have no leg along equation (3.3) becomes Equation (3.2) becomes now the one-forms have no d term, though they are still in general functions of . We may then split the ve-dimensional 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 one-forms 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 two-form 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 two-form 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 : non-linear 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 non-zero if is non-constant 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 NATD-T dual of the AdS5 T(1;1) solution in [2] for the Pilch-Warner 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 NS-NS two-form is given by11 F1 = 4L4 F3 = B = The non-zero 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 NS-NS 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 one-forms 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 non-Abelian T dual of AdS5 set-up and a linear quiver to describe its dual SCFT. S5 and propose a a D4/NS5 brane namely the NATD-T 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 non-singular supersymhave been unable to nd new non-singular 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 two-form 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 (e3-e4) 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 non-trivial scalar bilinear The two-forms 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^ = one-forms 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 ; one-forms satisfy d 1 = 2 ^ has a singularity as = 1 one nds that the Ricci-scalar )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 three-sphere. 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 non-singular 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 T-duality 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 T-duality where for a T-duality 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. 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Christopher Couzens. Supersymmetric AdS5 solutions of type IIB supergravity without D3 branes, Journal of High Energy Physics, 2017, DOI: 10.1007/JHEP01(2017)041