Synthesis, characterization and anticancer activity of two Ru(II) polypyridyl complexes [Ru(dpq)2L](PF6)2 (L = maip, paip)

Available online at www.sciencedirect.com ScienceDirect Nuclear Physics B 933 (2018) 185–233 www.elsevier.com/locate/nuclphysb Mink3 × S 3 solutions of type II supergravity Niall T. Macpherson a,b,c , Jesús Montero a,d , Daniël Prins a,e,∗ a Dipartimento di Fisica, Università di Milano-Bicocca, I-20126 Milano, Italy b INFN, sezione di Milano-Bicocca, I-20126 Milano, Italy c SISSA International School for Advanced Studies and INFN, sezione di Trieste, 34136, Trieste, Italy d Department of Physics, University of Oviedo, Avda. Calvo Sotelo 18, 33007 Oviedo, Spain e Institut de physique théorique, Université Paris Saclay, CNRS, CEA F-91191 Gif-sur-Yvette, France Received 5 March 2018; accepted 29 May 2018 Available online 6 June 2018 Editor: Stephan Stieberger Abstract We initiate the classification of supersymmetric solutions of type II supergravity on R1,2 × S 3 × M4 . We find explicit local expressions for all backgrounds with either a single Killing spinor or two of equal norm, up to PDE’s. We show that the only type II AdS4 × S 3 solution is the known N = 4 AdS4 background obtained from the near-horizon limit of intersecting D2–D6 branes. Various known branes and intersecting brane systems are recovered, and we obtain a novel class of R1,2 × S 2 × S 3 solutions in IIA. © 2018 The Author(s). Published by Elsevier B.V. This is an open access article under the CC BY license (http://creativecommons.org/licenses/by/4.0/). Funded by SCOAP3 . 1. Introduction The advent of the AdS-CFT correspondence has led to significant interest in the construction of Anti-de Sitter string backgrounds in various dimensions and with various amounts of supersymmetry. One of the most famous AdS4 backgrounds is the AdS4 × CP3 solution. The discovery of this solution far pre-dates the correspondence [1], however it was not realised how it fit into the holographic paradigm until the works of [2] and [3]. A plethora of other such AdS4 classes and explicit examples have been found using (in some cases vastly) different methods * Corresponding author. E-mail addresses: nmacpher@sissa.it (N.T. Macpherson), monteroaragon@uniovi.es (J. Montero), daniel.prins@cea.fr (D. Prins). https://doi.org/10.1016/j.nuclphysb.2018.05.021 0550-3213/© 2018 The Author(s). Published by Elsevier B.V. This is an open access article under the CC BY license (http://creativecommons.org/licenses/by/4.0/). Funded by SCOAP3 . 186 N.T. Macpherson et al. / Nuclear Physics B 933 (2018) 185–233 and exhibiting different amounts of supersymmetry: consider the very incomplete list of [4–14] for N = 1, [15–21] for N = 2, [22] for N = 3 and [23–26] for N = 4. Solutions with N > 4 where recently classified in [27], they are very restricted. One of the more prominent methods of finding AdSd backgrounds is to find bosonic solutions with an AdSd factor to the supersymmetry constraints, which also satisfy the Bianchi identities. As a consequence of various integrability theorems, such solutions automatically solve the equations of motions. The Killing spinor equations reduce to constraints on the internal manifold, which can then be solved by means of G-structure and generalised geometrical techniques. The literature usually approaches this problem by assuming an AdSd from the start. However we are also interested in solutions of relevance to flux compactifications and the broader definition of holography that includes non-conformal solutions. As such we shall consider assume Minkowski factor, in this case Mink3 , so that our results are more broadly applicable. Finding Minkowski solutions using G-structure techniques [28–31] or otherwise is by now quite a mature program, see [32–35] for some recent examples. Usually the aim is to preserve minimal or even no supersymmetry for phenomenological reason which makes the problem in general quite hard. We shall take inspiration from [36] and assume the existence of an S 3 factor in the metric. This will necessarily mean that we are dealing with at least N = 2 which is of less phenomenological interest, however with these solutions classified it should then be possible to systematically break some (or even all) of this symmetry by deforming the S 3 . In this paper we classify all supersymmetric solutions of Type II supergravity on R1,2 × S 3 × M4 , under the assumption that the seven-dimensional internal Killing spinors have equal norms and that the physical fields of the solution respect the I SO(1, 2) ×SO(4) isometry subgroup. Our classification is quite detailed, going as far as to give explicit local expressions the metric, fluxes and dilaton in terms of simple (Laplace-like) PDE’s. As we shall see, solutions in this class are generically N = 2, from the Minkowski perspective, and support a SU (2) R-symmetry realised geometrically as one factor of the SO(4) ≃ SU (2)+ × SU (2)− isometry group manifold of S 3 – the remaining SU (2) factor is a “flavour” under which the Killing spinors are uncharged.1 This may sound strange as there is no 3d superconformal algebra with SU (2)R , but this only matters for solutions where R1,2 is part of a AdS4 factor so that SO(2, 3) is realised. Ultimately our results end up side stepping this issue as in general there is either an enhancement of the R-symmetry to SO(4) via the emergence of an additional S 2,3 factor, or an enhancement of the Minkowski factor to dimensions where SU (2)R is a necessary part of the superconformal algebra. The classification recovers various well-known intersecting brane systems listed in [40] and some of their U-duals and some of their S-duals. New classes we find include a pure NS R1,2 × S 3 × S 2 × R vacuum, its U-dual in IIB, the cone over R1,2 × S 3 × S 3 , and a novel class of R1,2 × S 2 × S 3 × 2 solutions in massless and massive IIA. One of our main results is that the only compact AdS4 × S 3 × M3 solution of type II supergravity is the known N = 4 solution of type IIA on a foliation of AdS4 × S 3 × S 2 over an interval which is the near-horizon limit of the D2–D6 brane system. The required SO(4)R is realised with one SU (2) from each sphere, and not the S 3 alone. Indeed this is to be expected as if the 3-sphere does realise two SU (2) R-symmetries there would be two sets of N = 2 spinors transforming in the (2, 1) and (1, 2) of SO(4) – there is no N = 4 super-conformal algebra in 3d 1 As such all classes we present are compatible with performing non-Abelian T-duality [45–47] on this SU (2) whilst preserving SU (2)R [48]. N.T. Macpherson et al. / Nuclear Physics B 933 (2018) 185–233 187 with Q-generators that transform in this fashion. So it seems likely that the only avenue left open for holographic duals of N = 4 is to seek AdS4 × S 2 × S 2 solution like [23,24], but in massive IIA. Our other main result is the discovery of a new class of N = 4 solutions on R1,2 × S 2 × S 3 × 2 preserving an SO(4) R-symmetry but no AdS4 . These generically have all possible IIA fluxes turned on and can be divided into cases either in massless or massive IIA at which point solutions are in one to one correspondence with a single PDE on σ2 . In particular the massless solutions are governed by a 3d cylindrical Laplace equation with axial symmetry. These classes look very promising both for finding compact Mink3 solutions, but also possibly solutions that asymptote to AdS. Let us now describe the outline of the paper: in order to solve the supersymmetry constraints, we will make use of the reformulation of the Killing spinor equations in terms of so-called pure spinor equations. Such pure spinor equations were first used for backgrounds of the form M10 = R1,3 × M6 , where it was shown that they are related to integrability constraints of generalised almost complex structures on the internal space M6 [30]. For backgrounds of the form M10 = R1,2 × M7 , the pure spinor equations were constructed in [31] (see also [37]). Next, we decompose M7 = S 3 × M4 , leading to pure spinor equations on the internal M4 . We explain this setup in detail in section 2. The resulting supersymmetry constraints vary significantly, depending on whether the theory at hand is type IIA or type IIB. We will solve the supersymmetry constraints as well as the Bianchi identities for IIB backgrounds in section 3 and for IIA backgrounds in section 4. In section 5, we then show that there is a unique solution with a warped AdS4 factor, obtained from the D2–D6 system. In addition to the case where the internal Killing spinors have equivalent norm, in section 6 we examine all backgrounds in the case where one of the Killing spinors vanishes, i.e., 2 = 0. In this case, there is no need to distinguish between IIA and IIB; we demonstrate that all such backgrounds are pure NSNS and give the solutions. In the appendix, we discuss conventions and identities used, a mild extension of the 3 + 7 pure spinor equation construction (including the non-equivalent norm case), and a discussion on similar backgrounds from an M-theory perspective. 2. Mink3 with an S 3 factor We are interested in solutions to type II with at least a three-dimensional external Minkowski component, with the fluxes respecting the three-dimensional Poincaré invariance: ds 2 = e2A ds 2 (R1,2 ) + ds 2 (M7 ) , F = f + e3A Vol3 ∧ 7 λ(f ) , (2.1) where the RR flux f is a polyform on M7 and the warp factor A and the dilaton  are functions on M7 .2 Moreover, we take the NSNS 3-form H to be internal as well. The Killing spinors for N = 1 supersymmetric solutions decompose as     1 1 1 = ⊗ ζ ⊗ χ1 , 2 = ⊗ ζ ⊗ χ2 , (2.2) −i ±i where ζ is a Majorana spinor of Spin(1, 2) and χ1,2 are Majorana spinors of Spin(7) and where the upper (lower) signs are taken in IIA (IIB). Following [31], we define two real sevendimensional bispinors ± in terms of χ1,2 : 2 We work in the democratic formalism. Other conventions can be found in appendix A. 188 N.T. Macpherson et al. / Nuclear Physics B 933 (2018) 185–233 + + i− = 8e−A χ1 ⊗ χ2† , (2.3) where the subscript +/− refers to the even/odd forms in the decomposition of the polyform. The conditions for unbroken N = 1 supersymmetry are equivalent to dH (e2A− ± ) = 0 , (2.4a) dH (e3A− ∓ ) + e3A 7 λ(f ) = 0 ,  (± ∧ λ(f ))Top = 0 , (2.4b) (2.4c) as long as the norms of spinors χ1,2 are equal,3 which leads to |χ 1 |2 = |χ 1 |2 = eA (2.5) The assumption of equal norm is a global requirement for AdS4 (see footnote 7 in section 5), and a local requirement for the existence of calibrated D-branes or O-planes (see section 6), however this is not a requirement in general – rather we view this as a well-motivated simplifying assumption. Next, we require that the internal space can be decomposed locally as M7 = S 3 × M4 , and in order to ensure that compactification leads to an SO(4) global symmetry we insist that the fluxes respect the SO(4) isometry. As a result, the metric and fluxes decompose further as ds 2 (M7 ) = e2C ds 2 (S 3 ) + ds 2 (M4 ) , f = G∓ + e3C Vol(S 3 ) ∧ G± , H = H3 + H0 e3C Vol(S 3 ) . (2.6) We decompose the 7d spinors in the same fashion in terms of a single4 pseudoreal (i.e., (ξ c )c = −ξ ) Killing spinor ξ on S 3 , and two pseudoreal spinors η1,2 on M4 : A A χi = e 2 (ξ ⊗ ηi + ξ c ⊗ ηic ) = e 2 ξ a ⊗ ηia , i = 1, 2 (2.7) which is the most general parameterisation consistent with an S 3 × M4 product and the Majorana condition.5 Note that we do not restrict the Spin(4) spinors ηi to be chiral and we normalise † η1,2 = 1. The Killing spinors on S 3 satisfy the Killing spinor equation η1,2 1 (2.8) ∇α ξ = iνσα ξ , ν = ±1 , 2 which preserves two supercharges for each of ν = ±1. We will not make a choice of ν so we can establish whether any solutions are independent of this choice – the S 3 of such a solution would preserve 4 supercharges. As explained in Appendix C a spinor on S 3 defines a doublet 3 In [31], [37] an additional constraint that was imposed in order to derive (2.4) was that the external component of the NSNS 3-form flux is trivial; unlike in four dimensions, this is not enforced by Poincaré invariance. It turns out that this second assumption is redundant though, as is shown in appendix D: if |χ1 |2 = |χ2 |2 and spacetime does not admit a cosmological constant, then supersymmetry enforces that the external NSNS flux vanishes. 4 As explained in Appendix C, there are two independent types of Killing spinors on S 3 , ξ and ξ – however they + − cannot be mapped to each other using the SO(4) invariants of the fluxes or the Killing spinor equations. This is all that appears when one decompose M7 = S 3 × M4 , so if one were to include terms like ξ+ ⊗ η+ and ξ− ⊗ η− then reduced the 7d spinor conditions to 4d ones you would find that η± never mix. So setting one of η± to zero excludes no solutions in our analysis. 5 One might imagine it was possible to construct a more general 7d spinor from two 4d spinors like ξ ⊗ η + ξ c ⊗ η̃. But if one then adds the Majorana conjugate to this the resulting spinor can be put in the form of (2.7) by redefining η, η̃. N.T. Macpherson et al. / Nuclear Physics B 933 (2018) 185–233  ξa = ξ ξc 189  (2.9) which is charged under one SU (2) factor of SO(4) = SU (2)+ × SU (2)− , depending on the sign of ν – ξ a is a singlet under the action of the second SU (2). As such, a generic solution with Mink3 × S 3 will have an R-symmetry SU (2)R and an additional global flavour symmetry SU (2)F . Such solutions preserve at least N = 2 supersymmetry from the 3d perspective, so 4 real supercharges – indeed, the 10d Killing spinors may be written as     1 1 a a 1 ac 1c ⊗ ζ ⊗ (ξ ⊗ η +ξ ⊗ η ) , 2 = ⊗ ζ a ⊗ (ξ a ⊗ η2 +ξ ac ⊗ η2c ) , 1 = −i ±i where ζ a is a doublet of Killing spinors on R1,2 , that allow the 10d spinors to be invariant under SU (2)R transformations. However we only need to solve an N = 1 sub-sector, because the part of the Killing spinor which couples to ζ 1 is mapped to the part coupling to ζ 2 under the action of SU (2)R – so if you solve one part, the other is guaranteed. If a solution ends up being independent of ν then there is a copy of (2.2) for each sign and supersymmetry is doubled to N = 4 – there are two SU (2) R-symmetries, but they do not appear as a product so do not form SO(4)R – as we shall see, this only happen in a small number of special cases. Using the gamma matrix decomposition (A.2), the seven-dimensional bispinor (2.3) decomposes as χ1 ⊗ χ2† = (ξ a ⊗ ξ b )+ ∧ (ηa ⊗ ηb ) + (ξ a ⊗ ξ b )+ ∧ (γ̂ ηa ⊗ ηb ). (2.10) Here, γ̂ is the four-dimensional chirality matrix and the ± subscripts again refer to even and odd form components. We see that the components are in fact matrices and that the seven-dimensional bispinor is constructed as the trace of the product of the components. The S 3 component leads to the bispinor matrix    i 1 1 C ν 2C a b† 3C 3 ξ ⊗ξ = (1 − ie Vol(S )) + e Ki − ie dKi (σ )ab , (2.11) 2 2 2 where Ki is a vielbein defining a trivial structure on S 3 (see appendix A). The M4 component leads to the bispinor matrix    1 2 ψγ̂1 ± ψ± ψ± b† b† a a (η1 ⊗ η2 )± = ⊗ η ) = , ( γ̂ η ± 2 )∗ ±(ψ 1 )∗ 1 2 ∓(ψ± ∓(ψγ̂2 ± )∗ ± ψγ̂2 ± ±(ψγ̂1 ± )∗  , (2.12) where ψ 1 = 4η1 ⊗ η2† , ψ 2 = 4η1 ⊗ η2c† , ψγ̂1 = 4γ̂ η1 ⊗ η2† , ψγ̂2 = 4γ̂ η1 ⊗ η2c† (2.13) Since the matrix entries are somewhat involved, we refer to appendix B for details. Plugging both components (2.11), (2.12) into the seven-dimensional bispinors (2.10), it follows that 1 + = Reψ+ − e3C Vol(S 3 ) ∧ Imψγ̂1 − + − eC (K1 ∧ Reψγ̂2− + K2 ∧ Imψγ̂2 − + K3 ∧ Reψγ̂1− ), 2 e2C 1 2 2 + K1 ∧ K3 ∧ Reψ+ + K2 ∧ K3 ∧ Imψ+ ), (K1 ∧ K2 ∧ Imψ+ 4 (2.14) 190 N.T. Macpherson et al. / Nuclear Physics B 933 (2018) 185–233 1 − = Imψ− − e3C Vol(S 3 ) ∧ Reψγ̂1 + + eC (K1 ∧ Imψγ̂2+ − K2 ∧ Reψγ̂2 + + K3 ∧ Imψγ̂1+ ), 2 e2C 1 2 2 − K1 ∧ K3 ∧ Imψ− + K2 ∧ K3 ∧ Reψ− ). (K1 ∧ K2 ∧ Reψ− 4 At this point, the IIA and IIB supersymmetry equations diverge, and we shall relegate their explicit form to the relevant sections. With our set up, a solution to the supersymmetry equations is a solution to the equation of motion if and only if it satisfies the Bianchi identities [4] [38] [39]. These are given by dH F = dH = 0 away from localised sources. By definition, a localised (magnetic) source manifests itself in the Bianchi identity of some field strength F as dF = Qδ n (x) and hence in such cases F is discontinuous. Loosely speaking, a localised source corresponds physically to an extended object (such as a brane) located at a submanifold of the ten-dimensional spacetime S ⊂ M10 which is pointlike in some of the local coordinates. The standard approach to obtaining backgrounds, which we follow as well, is to first solve the supersymmetry equations by introducing local coordinates, and then afterwards determine the physically sensible range of these local coordinates by examining the obtained geometry and fluxes. The presence of localised sources is signified by discontinuities of not just the fluxes, but of the spacetime geometry as well, precisely at the location of the sources. Therefore, it is possible to obtain solutions with localised sources even when making use of the Bianchi identities with no sources: one examines possible discontinuities in the geometry and fluxes and determines whether or not such discontinuities are associated with localised sources or not by comparing them with the divergent behaviour of known extended objects. Making use of the flux decomposition (2.1), (2.6), the Bianchi identities thus reduce to     dH3 e3A+3C 4 λ(G± ) = dH3 e3C 4 λ(G∓ ) = 0 ,   (2.15) dH3 (G± ) = dH3 e3C G∓ = 0 , + dH3 = 0 . This is after imposing H0 = 0, which turns out to be a requirement for every solution to the supersymmetry equations that we obtain. 2.1. Summary of obtained backgrounds As the rest of the paper is somewhat technical, let us summarise our results here. We find a number of well-known backgrounds, as well as some new ones. In type IIB with internal Killing spinors of equal norm, we find: 1. The intersecting D3–D7 system with metric (3.27), fluxes (3.25) and scalar field constraints (3.26). 2. The D5-brane with metric (3.37), fluxes (3.35) and scalar field constraints (3.36). 3. A generalization of the D5-brane generated by U-duality. The metric is given by (3.43), the fluxes by (3.41), scalar field constraints by (3.42). 3 , with S 3 a generically squashed three4. A new background on the cone over R1,2 × S 3 × Ssq sq sphere admitting an SU (2) × U (1) isometry group. For the unsquashed limit, the metric is given by (3.59), the fluxes by (3.57), the scalar field constraints by (3.58). In the generic squashed case, the metric and dilaton are given by (3.75), the fluxes by (3.74). We note that the more general squashed case can be obtained from the unsquashed case by a duality chain. N.T. Macpherson et al. / Nuclear Physics B 933 (2018) 185–233 191 In type IIA with internal Killing spinors of equal norm, we find: 1. The intersecting D4–D8 system with metric (4.14), fluxes (4.12) and scalars constraints (4.11), (4.13). 2. The intersecting D2–D6 system with metric (4.24), fluxes (4.22) and scalar constraints (4.21), (4.23). 3. A generalization of the D4–D8 system generated by U-duality. The metric is given by (4.30), the fluxes by (4.29), and scalar constraints by (4.27). 4. A class of new backgrounds. The metric contains an R1,2 × S 3 × S 2 factor, with various warpings, and is given by (4.42). The warp factors are constrained by various PDE, given in (4.43). In general, all fluxes are turned on and are given by (4.44). This new class of backgrounds contains a subset with a U (1) isometry. In this case, T-dualising along the isometry direction leads to the new IIB backgrounds outlined above, with generic squashing. In addition, we find two more backgrounds when setting 2 = 0. These backgrounds are pure NS, and as such, can be found in both type IIA and type IIB. We find: 1. The NS5-brane, with metric (6.18), flux (6.16) and the scalar constraints (6.17). 2. A pure NS background on R1,2 × R × S 3 × S 3 , dual to the new (unsquashed) IIB background. The metric is given by (6.22), the flux by (6.21). All scalars are determined up to constant factors. 3. Mink3 with an S 3 factor in IIB The type IIB supersymmetry equations are obtained by plugging the decomposed sevendimensional bispinors (2.14) into the seven-dimensional supersymmetry constraints (2.4). This leads to the following constraints on the four-dimensional bispinors 1 dH3 (e2A− Reψ+ )=0, (3.1a) 2 dH3 (e3A+2C− ψ− ) + 2iνe3A+C− ψγ̂2 + = 0 , (3.1b) 2 dH3 (e2A+2C− ψ+ ) + 2iνe2A+C− ψγ̂2 − = 0 , (3.1c) 1 ) − 2νe3A+C− Imψγ̂1 + = 0 , dH3 (e3A+2C− Reψ− (3.1d) 1 dH3 (e2A+2C− Imψ+ ) + 2νe2A+C− Reψγ̂1 − = 0 , (3.1e) 1 dH3 (e2A+3C− Imψγ̂1 − ) + e2A+3C− H0 Reψ+ = 0, (3.1f) while the fluxes are determined by 1 dH3 (e3A− Imψ− ) + e3A 4 λ(G+ ) = 0 , (3.2a) 1 dH3 (e3A+3C− Reψγ̂1+ ) − e3A+3C− H0 Imψ− + νe3A+3C 4 λ(G− ) = 0 (3.2b) and must additionally satisfy the pairing equation    1 1 Imψγ̂ − ∧ λ(G− ) − Reψ+ ∧ λ(G+ )  = 0 . 4 (3.3) 192 N.T. Macpherson et al. / Nuclear Physics B 933 (2018) 185–233 In order to solve these, we will first examine the 0-form conditions. These are given by (ψγ̂2 )0 = (Imψγ̂1 )0 = H0 (Reψγ̂1 )0 = 0 . (3.4) We solve the first two of these in Appendix B, which leads to a spinor ansatz depending on 6 real functions with support on M4 α, a1 , b1 , λ1 , λ2 , λ3 (3.5) subject to the constraint a12 + b12 + λ21 + λ22 + λ23 = 1. (3.6) The third 0-form constraint, which is unique to IIB, still needs to dealt with. After making use of (B.12), it reduces to α α H0 (a1 cos + b1 sin ) = 0 . (3.7) 2 2 Here, as well as in IIA, the solutions depend drastically on the behaviour of α. We can distinguish between three different cases: α = 0, α = 12 π , and generic α ∈ (0, π), α = 12 π . Let us reiterate that we introduced α in (B.5) by defining α α η1 = cos η + sin γ̂ η , (3.8) 2 2 where η is a locally defined non-chiral spinor, where the chiral components are normalised. Note that the non-chirality is crucial: it ensures that η can be used to define the local trivial structure (i.e., the vielbein) via (B.3). In the case that α = 0, the 4d internal Killing spinors η1 = η are such that the chiral components of η1 have equal norm. In the case that α = π/2, we see that η1 becomes chiral. It turns out that we can treat this case together with α = 0, but find no such solutions. Thus we separate our solutions into two branches. Branch I: Here α = 0. The only non-trivial zero form is a1 H0 = 0, which a priori can be solved in two ways. However, we shall see in the next section that only H0 = 0 is consistent with the higher form conditions. In order to solve (B.6) we parametrise a1 = sin β, b1 = cos β sin δ, λ1 = y1 cos β cos δ, λ1 = y2 cos β cos δ, λ3 = y3 cos β cos δ, (3.9) with y1 = sin θ cos φ, y2 = sin θ sin φ, y3 = cos θ. (3.10) Branch II: Here 0 < α < π . Note that α = π is equivalent to α = 0, which is easiest to see by sending η → γ̂ η in (B.5). We choose to parametrise α α a1 = cos β sin(δ − ) , b1 = cos β cos(δ − ) 2 2 λ1 = − cos β cos δy1 , λ2 = − sin βy3 λ3 = − sin βy2 where yi is defined in (3.10). This ensures (B.6) and all but the first equation of (3.7) are solved, which becomes H0 cos β sin δ = 0 . (3.12) N.T. Macpherson et al. / Nuclear Physics B 933 (2018) 185–233 193 3.1. Branch I: solutions with α = 0 In order to solve branch I, it is convenient to first examine a number of lower form conditions that follow from (3.1). To do this it is useful to first rotate the canonical frame of (B.3) such that v1 → sin φw2 + cos φ(cos θ v1 + sin θ (cos δw1 − sin δv2 )) v2 → cos δv2 + sin δw1 w1 → cos θ (cos δw1 − sin δv2 ) − sin θ v1 (3.13) w2 → cos φw2 − sin φ(cos θ v1 + sin θ (cos δw1 − sin δv2 )) . Making use of these, one finds that the supersymmetry equations imply sin βH0 = 0 , (3.14a) d(e2A+2C− cos β cos δ) − 2νe2A+C− cos βv2 = 0 , (3.14b) d(e2A+3C− sin β(cos δv2 + sin δw1 )) − e2A+3C− H0 cos β cos δv1 ∧ w2 = 0, (3.14c) cos β(eC cos δdθ + 2ν sin δv1 ) = cos β(eC cos δ sin θ dφ − 2ν sin δw2 ) = 0 , (3.14d) d(e3A+2C− sin β(cos δw1 − sin δv2 )) + 2νe3A+C− (sin βv2 ∧ w1 + sin δ cos βv1 ∧ w2 ) + e3A+2C− cos β cos δ(dθ ∧ w2 + sin θ dφ ∧ v1 ) = 0, (3.14e) which is not a compete list. The first thing to establish is how to solve (3.14a) – if we set sin β = 0, one needs to set cos δ = 0 to solve (3.14c), but since ν = ±1, (3.14b) leads to a contradiction. The next conditions we consider are (3.14d). For cos β = 0 we see that either sin δ = dθ = dφ = 0, or 0 < sin δ < π2 in which case (θ, φ) define local coordinates on a 2-sphere. We are ignoring cos β = 0 because, as should be clear from, (3.9), this is a subcase of sin δ = dθ = dφ = 0. Let us now prove that 0 < sin δ < π2 is not possible: Since H0 = 0 we can solve (3.14b)–(3.14c) by introducing local coordinates x and ρ = e2A+2C− cos β cos δ such that √ cos δ −A+  cos β cos3/2 δ A−  1 2 νv2 = dρ, w1 = e e 2 (2ν cot βdx −sec βe−2A+ ρdρ). 2 ρ cos β 2νρ 3/2 sin δ We can also rewrite (3.14e) as d(e3A+2C− sin β(cos δw1 −sin δv2 ))+2νe3A+C− (sin βv2 ∧w1−sin δ cos βv1 ∧w2 ) = 0, using (3.14d). The key point here is that v2 , w1 only have legs in (ρ, x) while v1 , w2 sit orthogonal to this with legs in (θ, φ) only. This means that the equation above, cannot be solved as there is a Vol(S 2 ) term whose coefficient is non-vanishing. Thus we can conclude in general that sin δ = dθ = dφ = 0. Plugging this back into (4.1), one finds that nothing depends on the specific values these parameters take so we can set H0 = θ = φ = δ = 0 , (3.15) without loss of generality, leaving one undetermined function β. We are now ready to write the supersymmetry conditions that follow when α = 0, however we find it helpful to perform a second rotation of the canonical vielbein by considering 194 N.T. Macpherson et al. / Nuclear Physics B 933 (2018) 185–233 v1 + iw2 → e−iβ w, v2 − iw1 → −iv, (3.16) to ease presentation. The necessary and sufficient conditions for supersymmetry in the α = 0 branch are 1  H0 = d(e2A− sin β) = d(eA+C− 2  cos β) − νeA− 2 = d(e 2A+3C− cos βv2 sin βv2 ) = 0 , d(e3A+2C− w) + 2νe3A+C− w ∧ v2 = d(e3A+2C− sin βv1 ) + 2νe3A+C− sin βv1 ∧ v2 = 0 , d(e2A+2C− v1 ∧ w) − 2νe2A+C− v1 ∧ w ∧ v2 = 0 , d(e2A+2C− sin βw1 ∧ w2 ) − 2νe2A+C− sin βw1 ∧ w2 ∧ v2 + e2A+2C− cos βH3 = 0 , H3 + 2β ∧ w1 ∧ w2 = d(e4A− cos β) ∧ w1 ∧ w2 ∧ v2 = 0 , d(e3A− cos βv1 ) − d(e3A− sin βv1 ∧ w1 ∧ w2 ) + e3A− cos βv1 ∧ H3 − e3A 4 λ(G+ ) =0, d(e3A+3C− cos βv1 ∧ v2 ) − e3C+3A 4 λ(G− ) = 0 ,    (sin β + cos βw1 ∧ w2 ) ∧ v2 ∧ λ(G− ) − (sin β + cos βw1 ∧ w2 ) ∧ λ(G+ )  = 0. 4 (3.17) We can simplify this system further, but not without making assumptions about β. We now proceed to study the systems that follow from different values of β, we find that the physical interpretation is quite different in each case. 3.1.1. Subcase: β = 0 Upon setting β = 0 in (3.17) one can show that the supersymmetry conditions reduce to H3 = H0 = d(e− ) ∧ w1 ∧ w2 = d(e−4A ) ∧ v2 ∧ w1 ∧ w2 = 0 , 1 1 (3.18a) d(e−A v1 ) ∧ w = d(eA+C− 2  ) − νeA− 2  v2 = d(eA w) = 0 , (3.18b) e3A 4 λ(G+ ) = d(e3A− v1 ), (3.18c) e3A+3C 4 λ(G− ) = d(e3A+3C− v1 ∧ v2 ), λ(G− ) ∧ v2 ∧ w1 ∧ w2 − λ(G+ ) ∧ w1 ∧ w2 = 0. (3.18d) We can solve (3.18b) by using it to define a vielbein in terms of local coordinates ψ, x1, x2 and 1 ρ = eA+C− 2  (3.19) such that v1 = eA (dψ + V ) , 1 v2 = νe−A+ 2  dρ , w = e−A (dx1 + idx2 ) , V = f1 (x1 , x2 )dx1 + f2 (x1 , x2 )dx2 . (3.20) N.T. Macpherson et al. / Nuclear Physics B 933 (2018) 185–233 195 From (3.18a) we see there is no NSNS flux, ∂ψ is an isometry and A = A(ρ, x1 , x2 ),  = (x1 , x2 ). We now have enough information to calculate the fluxes. First we find     4 G− = −eA ∂x1 (e− )w1 + ∂x2 (e− )w2 ∧ v1 ∧ v2 − e3A− ∂x1 f2 − ∂x2 f1 v1 1 4 G+ = −e3A ∂x1 (e4A− )w1 ∧ v2 + ∂x2 (e4A− )w2 ∧ v2 + νe− 2  ∂ρ e4A− v2 ∧ v1 − e3A− (∂1 f2 − ∂2 f1 ) w1 ∧ w2 . (3.21) We can then use coordinate dependence of the physical fields and local expression for the vielbein (3.20) to take the Hodge dual in (3.18c) arriving at G− = (∂x2 f1 − ∂x1 f2 )e4A− (dψ + V ) + ∂x2 (e− )dx1 − ∂x1 (e− )dx2 ,  3A− 32  G+ = −νe ∂x2 (e−4A+ )dx1 ∧ dρ − ∂x1 (e−4A+ )dx2 ∧ dρ −e − ∂ρ (e −4A+  )dx1 ∧ dx2 − νe (∂x2 f1 − ∂x1 f2 )(dψ + V ) ∧ dρ .  (3.22) (3.23) Plugging this into (3.18d) we find (∂x2 f1 − ∂x1 f2 ) = 0 which mean that V is closed and so we can locally fix V =0, (3.24) with a shift ψ → ψ − η for dη = V , without loss of generality. Taking this into account the ten-dimensional fluxes are F1 = ∂x2 (e− )dx1 − ∂x1 (e− )dx2 , F5 = dψ ∧ d(e4A− ) ∧ Vol3 (3.25)   3 −4A+ −4A+ − −4A+ − νρ ∂x2 (e )dx1 ∧ dρ − ∂x1 (e )dx2 ∧ dρ − e ∂ρ (e )dx1 ∧ dx2 ∧ Vol(S 3 ). The final thing we need to do is impose the Bianchi identities, which away from localised sources rise to the PDEs (∂x21 + ∂x22 )e− = 0 , e− ∂ρ (ρ 3 ∂ρ e−4A+ ) + (∂x21 + ∂x22 )(e−4A+ ) = 0 . ρ3 (3.26) The local form of the metric is then 1 ds 2 (R1,3 ) + ds 2 = √ fH H =e −4A+ , f =e   H dρ 2 + ρ 2 ds 2 (S 3 ) + f f H dx12 + dx22 , (3.27) − This corresponds to the intersecting D3–D7 brane system, where the D3-branes are embedded in the D7-branes [40]. 196 N.T. Macpherson et al. / Nuclear Physics B 933 (2018) 185–233 3.1.2. Subcase: β = π2 Setting β = π2 in (3.17) leads to the following necessary and sufficient conditions for unbroken supersymmetry d(e2A− ) = d(e2A+3C− v2 ) = H3 = H0 = G− = 0 , (3.28a) d(e3A+2C− ui ) + 2νe3A+C− ui ∧ v2 = 0 , (3.28b) d(e2A+2C− ij k uj ∧ uk ) − 2νe2A+C− ij k uj ∧ uk ∧ v2 = 0 , (3.28c) e3A 4 λ(G+ ) = d(e3A− v1 ∧ w1 ∧ w2 ). (3.28d) Here we have introduced the notation u = (v1 , w1 , w2 ), (3.29) both to ease notation and to stress that the vielbeine ui obey a cyclic property. Exploiting this property will be very helpful in solving this system and other systems we shall encounter which mirror this behaviour, so we will be very explicit in our derivation here, but less so elsewhere. The first thing to note is that the combination (3.28c)i + e−A ij k uj ∧(3.28b)k leads to   1 1 ij k d(e2A+C− 2  ) − νe2A− 2  v2 ∧ uj ∧ uk = 0 . (3.30) This implies that the 1-form in large brackets is zero. This can be seen by writing it as Xj uj + X0 v2 for some functions Xj , noting that the vielbeine u1,2,3 are independent, and then considering the resulting constraints for i = 1, 2, 3. Next by examining (3.28c)i + e−A uj ∧(3.28b)k and (3.28c)i − e−A ∧(3.28b)j ∧ uk for cyclic permutations of (i, j, k) = (1, 2, 3) one realises that d(e−A ui ) ∧ uj = 0 , i = j, (3.31) which implies that d(e−A ui ) has no leg in ui , it is then not hard to see that since (3.28b) has no ij k uj ∧ uk term it is in fact zero. So we can conclude without loss of generality that 1 1 d(e−A ui ) = d(e2A+C− 2  ) − νe2A− 2  v2 = 0 , (3.32) which imply (3.28b)–(3.28c) without further constants. We can then solve these conditions by using them to define the vielbeine in terms of local coordinates as 1 1 v1 = eA dx1 , w1 = eA dx2 , w2 = eA dx3 , v2 = νe−2A+ 2  dρ , ρ = e2A+C− 2  . (3.33) We can now solve (3.28a), which in fact just tells us that e = e2A , (3.34) up to rescaling gs and that e2A is a function of ρ only, making ∂xi all isometry directions parameterising either R3 or T 3 locally. The only non-trivial flux is the RR 3-form F3 = −νρ 3 ∂ρ (e−4A )Vol(S 3 ) and its Bianchi identity, dF3 = 0, imposes that (3.35) N.T. Macpherson et al. / Nuclear Physics B 933 (2018) 185–233 e−4A = c1 + c2 , dci = 0 . ρ2 197 (3.36) This is the warp factor of a D5-brane or O5-hole, depending on the sign of c2 (see for example [41]). Indeed the metric locally takes the form   ds 2 = e2A ds 2 (R1,5 ) + e−2A dρ 2 + ρ 2 ds 2 (S 3 ) (3.37) As we will see, this is a subcase of the solution in the next section. 3.1.3. Subcase: generic β For generic 0 < β < π2 we are free to divide by the trigonometric functions in (3.17). Using sin β = 0 it is possible to show that supersymmetry requires d(e2A− sin β) = d(e− cos β) = deA ∧ v2 = 0,  d(e−A v1 ) = d(e2A+C− 2  sin β) − νe2A− 2 sin βv2 = d(e−A w csc β) = 0 , H3 + 2dβ ∧ w1 ∧ w2 = dβ ∧ v2 ∧ w1 ∧ w2 = 0 , (3.38a) (3.38b) (3.38c) e3A 4 λ(G+ ) = d(e3A− cos βv1 ) + d(e3A− sin βv1 ∧ w1 ∧ w2 ) − e3A− cos βH3 ∧ v1 , e3A+3C 4 λ(G− ) = d(e3A+3C− cos βv1 ∧ v2 ), (3.38d) λ(G− ) ∧ v2 ∧ (sin α + cos αw1 ∧ w2 ) − λ(G+ ) ∧ (sin α + cos αw1 ∧ w2 ) = 0 , (3.38e) by following the same line of reasoning as in the previous subsection. First we solve (3.38b) by using it to define the vielbeine on M4 locally v1 = eA dx1 , w = eA sin β(dx2 + idx3 ), v2 = νe−A dρ, ρ = eA+C , (3.39) where we have used the first of (3.38a) to simplify these somewhat. Next (3.38a) is solved when e = e2A sin β, cot β = ce2A , dc = 0 , (3.40) with A = A(ρ), β = β(ρ). As a result, ∂xi are isometries. We then use (3.39) to take the Hodge dual of (3.38d), (3.38e) arriving at the fluxes F3 = −νρ 3 ∂ρ (e−4A )Vol(S 3 ), H = 2e2A ∂ρ β sin2 βdρ ∧ dx2 ∧ dx3 , (3.41)   F5 = Vol3 ∧ dx1 ∧ d(e2A cot β) + νe−2A ρ 3 sin(2β)∂ρ A − ∂ρ β dx2 ∧ dx3 ∧ Vol(S 3 ), which solve (3.38e) without restriction. The Bianchi identities impose that c2 e−4A = c1 + 2 , dci = 0 , ρ (3.42) which is again the warp factor of a D5-brane or O5-hole. However, in this case the metric takes the local form   2 2A 2 1,3 2A 2 2 2 −2A 2 2 2 3 ds = e ds (R ) + e sin β ds (T ) + e (3.43) dρ + ρ ds (S ) where T 2 is spanned by (x2 , x3 ). This generalises the solution in the previous section by introducing an additional warping factor for a T 2 submanifold, thus breaking SO(1, 5) Lorentz 198 N.T. Macpherson et al. / Nuclear Physics B 933 (2018) 185–233 symmetry and leading to more general fluxes. In fact, this solution can be generated from the D5-brane solution of the previous section via “G-structure rotation” [42] which is formally a U-duality [43]. 3.2. Branch II: α non-zero solutions For the second branch with 0 < α < π , we begin by studying the lower form conditions that follow from (3.1). Here we find it useful to rotate the canonical frame of (B.3) as v → − sin w1 + cos φ(cos θ v1 − sin θ w2 ) + iv2 , (3.44) w → cos φw1 + sin φ(cos θ v1 − sin θ w2 ) + i(cos θ w2 + sin θ v1 ). (3.45) We then find the following necessary, but not sufficient, conditions for supersymmetry cos β sin δH0 = 0 , (3.46a) eC cos α sin βdθ − 2ν(cos β cos δv1 + sin β sin αw1 ) = 0 , (3.46b) eC cos α sin β sin θ dφ − 2ν(cos β cos δw1 − sin α sin βv1 ) = 0 , (3.46c) d(e2A+2C− cos α sin β) − 2νe2A+C− (cos β cos δw2 − sin βv2 ), (3.46d) d(e2A+3C− (sin α sin βw2 + sin(α − δ) cos βv2 )) + e2A+3C− H0 sin β cos αv1 ∧ w1 = 0, (3.46e) d(e3A+2C− (sin β sin αv2 − cos β sin(α − δ)w2 )) + 2νe3A+C− cos β(sin δw2 ∧ v2 + cos(α − δ)v1 ∧ w1 )  − e3A+2C− dθ ∧ (cos β sin(α − δ)v1 + sin βw1 )  + sin θ dφ ∧ (cos β sin(α − δ)w1 − sin βv1 ) = 0 . (3.46f) First we note that if either θ or φ become constant or if cos α = 0 then (3.46b), (3.46c) require that sin β = cos δ = 0 which makes (θ, φ) drop out of (3.11) entirely and the final line of (3.46f) vanishes (setting sin θ = 0 leads to the same conclusion). In this case we can conclude that we can set π H0 = θ = φ = (δ − ) = 0 (3.47) 2 without loss of generality, which we study in section 3.2.1. If we assume sin θ and cos α don’t vanish, then (θ, φ) are local coordinates on a 2-sphere and we can take ρ = e2A+2C− cos α sin β as a local coordinate. We can then use (3.46b)–(3.46d) to rewrite (3.46f) as d(e3A+2C− (sin β sin αv2 − cos β sin(α − δ)w2 )) + 2νe3A+C− cos β(sin δw2 ∧ v2 − cos(α − δ)v1 ∧ w1 ) = 0, (3.48) which we can then use to fix some of the free functions. First we note that if we solve (3.46a) with sin δ = 0, then (3.48) fixes cos β = 0 – this is because sin β sin αv2 − cos β sin αw2 is parallel to dρ in this limit and so cannot generate the Vol(S 2 ) factor that comes from v1 ∧w1 . Next, for H0 = N.T. Macpherson et al. / Nuclear Physics B 933 (2018) 185–233 199 0 and for generic values of (α, β, δ) we can use (3.46b)–(3.46e) to locally define the vielbein on M4 by introducing another local coordinate such that dx = e2A+3C− (sin α sin βw2 + sin α − δ cos βv2 ), but then we must once more set the v1 ∧ w1 term in (3.48) to zero which fixes either cos β = 0 or cos(α − δ) = 0. Thus, for H0 = 0 and a priori generic (θ, φ, α, β, δ), we end up with just two cases. Firstly cos β = 0, which solves (3.46a) and makes the δ dependence of (3.11) drop out such that we can set without loss of generality π δ=0, β = . (3.49) 2 We shall examine this case in detail in section 3.2.2 where we find that it contains no solution. Secondly cos(α − δ) = 0, such that we can set without loss of generality π δ=α+ , (3.50) 2 which we shall study in section 3.2.3, finding a new class of solution. There is one final option one can consider for H0 = 0, by taking both cos α and sin θ nonvanishing – one can tune the values of (α, δ, β) such that (sin α sin βw2 + sin(α − δ) cos βv2 ) becomes parallel to dρ and so can no longer be used to introduce a local coordinate. This requires fixing cos δ sin(δ − α) . (3.51) sin α It will then be (3.48) that will be used to define the final vielbein direction, which will necessarily be fibred over S 2 . We shall examine this possibility in section 3.2.4. tan β = 3.2.1. Subcase β = 0 For β = 0 the supersymmetry conditions reduce to d(e2A− ) = d(e2A+3C− cos αv2 ) = H3 = H0 = 0 , d(e3A+2C− cos αui ) + 2νe3A+C− (ui ∧ v2 + 1 sin αij k uj ∧ uk ) = 0 , 2 (3.52a) (3.52b) 1 d(e2A+2C− (sin αui ∧ v2 + ij k uj ∧ uk )) − νe2A+C− cos αij k uj ∧ uk ∧ v2 = 0 , 2 (3.52c) e3A 4 λ(G+ ) = d(e3A− cos αv1 ∧ w1 ∧ w2 ), e3A+3C 4 λ(G− ) = −d(e3A+3C− sin α), (3.52d)  (3.52e) (cos αv2 ∧ λ(G− ) + sin αv1 ∧ v2 ∧ w1 ∧ w2 λ(G+ ))4 = 0, where as usual u = (v1 , w1 , w2 ). Note that this system reduces to that of section 3.1.2 when sin α = 0, and that (3.52b) imposes that cos α = 0, so we can take 0 < 2α < π . By adding linear combinations of wedge products of (3.52a), (3.52b) and the vielbein to (3.52b), it is then possible to derive enough independent 2-form conditions to establish that d(eA−C sin α)∧v2 = dα ∧v2 = d(eA+C cos α)−νeA v2 = d(e−A sec αui )∧uj = 0, i = j. (3.53) This is sufficient to establish that e2A , e2C and α are functions of a single local coordinate ρ, which v2 is parallel to – specifically 200 N.T. Macpherson et al. / Nuclear Physics B 933 (2018) 185–233 ρ = eA+C cos α , v2 = νe−A dρ . (3.54) The final condition in (3.53) implies that d(e−A sec αu1 ) ∝ u2 ∧ u3 and cyclic permutations, however plugging this into (3.52b) and (3.52b) we realise we can without loss of generality take ui = c1 A ν e−A+C e cos α K̃i , K̃i + ij k K̃ j ∧ K̃ k , c1 = , dc1 = 0 , 2 2 sin α (3.55) where K̃ i are necessarily SU (2) invariant forms which furnish a frame for a round S 3 . We have now without loss of generality determined the vielbein on M4 , which is a foliation of S 3 over an interval, and (3.52a)–(3.52c) are solved when e−2A = c1 sin α cos α , e2C = e2A c12 sin2 α , e2A− = c2 , dci = 0 . ρ (3.56) The only non-trivial 10d flux can be extracted from (3.52d) and is given by     F3 = 2c12 c2 ν sin2 α − ρ tan α∂ρ α Vol(S 3 ) + 2c12 c2 ν cos2 α + ρ cot α∂ρ α Vol(S̃ 3 ) . (3.57) The pairing equation (3.52e) is equivalent to the Bianchi identity at this point; either one implies dα = 0 . The metric is of the form  ds 2 = e2A ds 2 (R1,2 ) + e−2A dρ 2 + (3.58)  ρ2 ρ2 2 3 2 3 (S ) + (S ) ds ds cos2 α sin2 α (3.59) This solution has both an SO(4) R-symmetry and SO(4) flavour symmetry, and is S-dual to the one that we find in section 6, as will be explained in that section. 3.2.2. Subcase β = π2 Here one can show that supersymmetry implies d(e2A+2C− cos α) − 2νe2A+C− v2 = d(eC cot αw2 ) = 0 , (3.60a) eC dθ − 2ν tan αw1 = eC sin θ dφ + 2ν tan αv1 = 0 , (3.60b) d(e−A−C sin α) = d(e sin α tan α) ∧ v2 = d(e−A−C+ tan α) ∧ w2 ∧ v2 = 0 , (3.60c) √   1√ cos αeA+C− 2 H0 v1 ∧ w1 = 0 , d(eA+C− 2 tan α cos α) ∧ w2 + 2 (3.60d) d(e2A+2C− cos α cot2 α) ∧ Vol(S 2 ) = 0 , (3.60e) ν H3 + eC cot αw2 ∧ Vol(S 2 ) = d(e−2C tan α) ∧ v2 ∧ Vol(S 2 ) = 0 , 2 (3.60f) e3A 4 λ(G+ ) − d(e3A− w2 ) + d(e3A− sin αv1 ∧ v2 ∧ w1 ) = 0 , (3.60g) e3A+3C 4 λ(G− ) + e3A+3C− H0 w2 + d(e3A+3C− cos αv2 ∧ w2 ) = 0 ,    (sin αw2 − v1 ∧ v2 ∧ w1 ) ∧ λ(G− ) − cos αv1 ∧ v2 ∧ λ(G+ )  = 0 . (3.60h) 4 (3.60i) N.T. Macpherson et al. / Nuclear Physics B 933 (2018) 185–233 201 There is no solution to this set of constraints, as we will now show. First, we note that H0 = 0 is imposed by (3.60d). Let ρ = e2A+2C− cos α. Due to (3.60a), we have w2 ∼ dx, v2 ∼ dρ. It is then possible to rewrite (3.60d), (3.60e) as √  d ρ tan α ∧ dx = 0 , d ρ cot2 α ∧ dθ ∧ dφ = 0 . (3.61) The first equation implies tan α = ρ −1/2 f (x), which is incompatible with the second equation – thus this putative class contains no solutions. 3.2.3. Subcase δ = α + π2 , H0 = 0 As explained below (3.46f), here we necessarily have β > 0 and 0 < α < π2 – for this reason it will turn out the case contains no solutions. As the proof is similar to that of the previous section we shall be brief, this time only quoting sufficient supersymmetry conditions to prove this. In addition to the rotation of (B.3) we find it useful to send v1 + iw1 → e−iβ (v1 + iw1 ), then a set necessary (but insufficient) conditions for supersymmetry are d(e2A− cos α cos β) = 0, (3.62a) d(eC cot αw2 ) = 0, (3.62b) (v1 + iw1 ) tan α + eC sin β (dθ + i sin θ dφ), d(e2A+3C− (sin α sin βw2 − cos βv2 )) = 0, 2ν (3.62c) d(e2A+2C− cos α sin β) − 2νe2A+C− (cos β sin αw2 + sin βv2 ) = 0, (3.62d) e2A− cos α cos βH3 + d(e2A− cos α sin βv1 ∧ w1 ) = 0, (3.62e) e2A+3C− (sin α sin βw2 − cos βv2 ) ∧ H3 + d(e2A+3C− v1 ∧ w1 ∧ (sin βv2 − cos β sin αw2 )) = 0. (3.62f) As elsewhere we can take (3.62c)–(3.62d) as a local definition of the vielbein without loss of generality – v1 , w1 are clearly the local vielbeine of a round S 2 in terms of the local coordinates 1 √ (θ, φ). For v2 , w2 we introduce local coordinates x and ρ = eA+C− 2  cos α sin β such that e2A+3C− (sin α sin βw2 − cos βv2 ) = dx. (3.63) We can then use (3.62e) to define H3 without loss of generality, which leaves (3.62a), (3.62b) and (3.62f) to solve – this  turns out to be impossible.  To see this, one needs to consider the combination 4(3.62f) + f1 (3.62a) ∧ dρ + f2 (3.62a) ∧ Vol(S 2 ). When one tunes 3  f1 = e−A+5C+ 2 ν 5 cos 2 α sin 2 β sin2 α cos β , f2 = e2A+4C− cos α sin β tan β, (3.64) this leads to cot α csc α sin β tan βdx ∧ dρ ∧ Vol(S 2 ) = 0 (3.65) which cannot be solved without violating the initial assumptions that lead to this case. We conclude that there exist no solutions. 202 N.T. Macpherson et al. / Nuclear Physics B 933 (2018) 185–233 3.2.4. Subcase: special value of β, H0 = 0 The final case in IIB requires us to tune tan β to a specific value. After redefining δ → δ + α, this value is cos(α + δ) sin δ . sin α In addition, we rotate the vielbein (with respect to (3.44)) as tan β = v1 + iw1 → −i  v2 + iw2 → −i  cos β cos(α + δ) + i sin α sin β cos2 β cos2 (α + δ) + sin2 α sin2 β sin α sin β + i cos β sin δ sin2 α sin2 β + cos2 β sin2 δ (3.66) (v1 + iw1 ), (v2 + iw2 ) (3.67) In what follows we assume that the undefined functions of the spinor ansatz are bounded as 0 < 2α < π and 0 < δ + α < π2 , as the upper and lower limits have been dealt with in the preceding sections. It is then possible to show that the necessary and sufficient conditions for supersymmetry for this case are     A−C 2A− sin α sin(α + δ) − sin 2(α + δ) d(e sin α) = d e =d e = 0, (3.68a) cos δ sin 2δ √ 2νe−C tan α cot δ(w1 − iv1 ) − (dθ + i sin θ dφ) = 0,    cos α sin(α + δ) −C 2 2νd e tan α w2 − Vol(S ) = d eA+C sin δ (3.68b) cos α sin δ sin(α + δ)  − νeA v2 = 0, (3.68c)   cos2 α sin δ 1 H3 + d e2C 2 cos(α + δ) sin α sin δ ∧ Vol(S 2 ) = dα ∧ v2 = 0 4 sin α sin(α + δ) cos δ (3.68d)   cos α cos(α + δ) e3A 4 λ(G+ ) − dH3 e3A− cos δ  −d e 3A−  sin α cos α sin δ v1 ∧ w1 ∧ w2 = 0, cos δ  e 3A+3C 4 λ(G− ) + dH3 e  −d e 3A+3C− 3A+3C− cos δ sin α sin(α + δ) (3.68e)   sin δ cos(α + δ) cos α v2 ∧ w2 = 0, cos δ sin(α + δ)   sin α cos α sin δ v2 − cos δ  cos α cos(α + δ) v1 ∧ w1 ∧ v2 ∧ λ(G− ) cos δ (3.68f) N.T. Macpherson et al. / Nuclear Physics B 933 (2018) 185–233 −  sin α sin(α + δ) sin δ cos(α + δ) − cos α v 1 ∧ w1 cos δ cos δ sin(α + δ) −    sin α cos δ v1 ∧ v2 ∧ w1 ∧ w2 ∧ λ(G+ )  = 0, sin(α + δ) 4 203 where Vol(S 2 ) is the volume form on the S 2 spanned by (θ, φ). We solve these conditions by first using (3.68b)–(3.68c) to define the vielbein locally without loss of generality as ν w1 = e C 2 cos α sin δ ν dθ , v1 = − eC sin α cos δ 2 cos α sin δ sin θ dφ , sin α cos δ sin δ ν w2 = − eC cot α (dψ + cos θ dφ) , v2 = νe−A dρ , 2 sin(α + δ) cos α (3.69) (3.70) where we have taken θ, φ as local coordinates and introduced the additional coordinates ψ and ρ = eA+C cos α sin δ . sin(α + δ) (3.71) We can invert this conditions then use (3.68a) to define A, C, , δ in terms of α, ρ and some integration constants ci as c12 cos2 α sin2 α c3 sin2 α − , e2C = c12 sin2 αe2A , ρ2 c22   c3 ρ 2 c3 ρ 2 −2 −4A 2 =e e c2 + 2 , cot(α + δ) = 2 2 . 2 c1 sin α c1 c2 sin α cos α e−4A = (3.72) The second equality in (3.68d) implies that α is itself a function of ρ only, so we realise that ∂ψ is an isometry of the solution and that M4 is foliation of a (SU (2) × U (1) preserving) squashed 3-sphere over an interval. We now turn our attention to the fluxes. We have that (3.68d) simply defines the NSNS flux in such a way that it is automatically closed, while the RR fluxes are defined through the 4d fluxes that follow from (3.68e). We could use the definitions of the functions in (3.72) and vielbein in (3.69) to calculate the 10d fluxes immediately, however we already have enough information to first fix α. The 3-form component of 4 λ(G− ) is necessarily parallel to v1 ∧ w1 ∧ v2 from which it follows that the 10d flux F1 is parallel to w2 . As this vielbein is fibred over the S 2 we have that dF1 = 0 iff (4 λ(G− ))3 = 0. For cos(α + δ) = 0 this imposes d(c12 c22 sin2 α + c3 ρ 2 ) = 0, which implies that sin2 α = c4 − c3 ρ 2 , c12 c22 dci = 0, (3.73) and as a result, every function has been solved in terms of ρ and the four integration constants ci . We are now ready to calculate the fluxes. The non-trivial ones take the form B= √ 1 − c4 2 1 c3 ρ Vol(S 2 ) , F3 = νc12 c2 (1 − c4 )dψ ∧ Vol(S 2 ) + 2νc12 c2 c4 Vol(S 3 ) , 4c4 c2 4 (3.74) 204 N.T. Macpherson et al. / Nuclear Physics B 933 (2018) 185–233 √ F5 = B ∧ F3 + c1 c22 ν c3 d   ρ2 (dψ + cos θ dφ) 2(c3 ρ 2 − c12 c22 c4 ) √  νc12 c3 c4  2 d ρ (dψ + cos θ dφ) ∧ Vol(S 3 ) , + 2 where dB = H . Clearly the Bianchi identities of the fluxes are implied automatically and one can show that this is true of (3.68f) also. So this case contains a single example, expressed in terms of 4 integration constants. The 10d metric, warp factor and dilation then take the form  1 ρ2 2 2A 2 1,2 −2A ds = e ds (R ) + e ρ 2 ds 2 (S 3 ) + (dψ + cos θ dφ)2 dρ 2 + 1 − c4 4(c4 − 2c3 2 ρ 2 ) +  ρ2 2 2 ds (S ) , 4c4  e−4A = (1 − c4 ) c 1 c4 c3 − 2 ρ2 c2  c1 c2 , e−4A+2 = 1 c3 − ρ2 . c22 c12 c24 c4 (3.75) This solution preserves an SO(4) R-symmetry realised by one SU (2) factor of the round S 3 and the SU (2) of the squashed sphere – the residual symmetries of the spheres make up an SU (2) × U (1) flavour symmetry. Despite our assumption that α + δ = π2 (which is when c3 = 0) when deriving (3.68a)–(3.68f) there is in fact no issue with taking this limit, which merely collapses this solution to that of section 3.2.1. There is good reason for this, as one can actually generate this solution from section 3.2.1 by first T-dualising on ∂ψ then performing a formal U-duality6 on the Mink3 followed by another T-duality on ∂ψ . Additionally, this solution is also contained in section 4.2.2: it can be obtained by imposing that the coordinate x there (which should be identified with ψ in this section) is an isometry and then T-dualising it. This concludes our IIB classification, we shall now turn our attention towards IIA. 4. Mink3 with an S 3 factor in IIA The type IIA supersymmetry conditions obtained from plugging (2.14) into the sevendimensional supersymmetry constraints (2.4) lead to the following constraints on the fourdimensional bispinors 1 dH3 (e2A− Imψ− )=0, (4.1a) 2 dH3 (e2A+2C− ψ− ) + 2iνe2A+C− ψγ̂2 + = 0 , (4.1b) 2 ) + 2iνe3A+C− ψγ̂2 − = 0 , dH3 (e3A+2C− ψ+ (4.1c) 1 ) − 2νe2A+C− Imψγ̂1 + = 0 , dH3 (e2A+2C− Reψ− (4.1d) 1 =0, dH3 (e2A+3C− Reψγ̂1+ ) − e2A+3C− H0 Imψ− (4.1e) 6 T-dualities along the spatial components of Mink , then a lift to M-theory followed by a boost along the M-theory 3 U (1) before reducing to IIA, and finally undoing the spatial T-dualities. This process needs to be supplemented by rescaling the coordinates along the way, which is why we refer to this as a formal U-duality. N.T. Macpherson et al. / Nuclear Physics B 933 (2018) 185–233 1 dH3 (e3A+2C− Imψ+ ) + 2νe3A+C− Reψγ̂1 − = 0 , 205 (4.1f) while the fluxes are defined through 1 dH3 (e3A− Reψ+ ) + e3A 4 λ(G− ) = 0 , (4.2a) 1 + νe3A+3C 4 λ(G+ ) = 0 , dH3 (e3A+3C− Imψγ̂1 − ) + e3A+3C− H0 Reψ+ (4.2b) and must additionally satisfy    1 ∧ λ(G− )  = 0 Reψγ̂1 + ∧ λ(G+ ) + Imψ− (4.3) 4 As before, we will first examine the 0-form constraints. These are given by two of the three constraints that were found for type IIB: (ψγ̂2 )0 = (Imψγ̂1 )0 = 0 . (4.4) Again, the solutions branch off similar to type IIB, with an α = 0 and an α = 0 branch. We parameterise Branch I as in (3.9) and Branch II as in (3.11). 4.1. Branch I: solutions with α = 0 As was the case in IIB we first study the lower form conditions that follow from (3.1). After once more rotating the canonical frame of (B.3) by (3.13) we extract the necessary, but not sufficient, supersymmetry conditions d(e2A+3C− cos β sin δ) + e2A+3C− H0 cos βw1 = 0 , (4.5a) e2A+3C− H0 sin β(cos δw1 − sin δv2 ) ∧ v1 ∧ w2 + cos β(...) = 0 , (4.5b) d(e2A− cos βw1 ) = d(e3A+2C− cos β cos δ) − 2νe3A+C− cos βv2 = 0 , (4.5c) ν ν cos β(sin δv1 + eC cos δdθ) = cos β(sin δw2 − eC cos δ sin θ dφ) = 0 , 2 2 (4.5d) d(e2A+2C− sin β(cos δw1 − sin δv2 )) + 2νe2A+C− (sin βw1 ∧ v2 − cos β sin δv1 ∧ w2 ) − cos β cos δe2A+2C− (dθ ∧ w2 + sin θ dφ ∧ v1 ) = 0, (4.5e) where cos β(...) represents further terms which vanish when cos β = 0. These are sufficient to truncate the ansatz considerably. We note from (4.5a) that if sin δ = 0 then either H0 = 0 or cos β = 0, however the latter also leads to H0 = 0 because of (4.5b) – so sin δ = 0 implies H0 = 0. We also observe that if sin δ = 0 then (4.5c) requires dθ = dφ = 0, or naively cos θ = 0 but this is a subcase of the former (see (3.9)), so sin δ = 0 also implies dθ = dφ = 0. Our task now is to show that, as in IIB, sin δ = 0 is a necessary condition: first we note that if we set cos δ = 0 then there is no solution as (4.5c) sets the vielbein to zero, thus we can restrict our considerations to 0 < sin δ < π2 where (w1 , v2 ) must span an S 2 . However, as was the case in IIB, (4.5e) can be rewritten as d(e2A+2C− sin β(cos δw1 −sin δv2 ))+2νe2A+C− (sin βw1 ∧v2 +cos β sin δv1 ∧w2 )=0, (4.6) 206 N.T. Macpherson et al. / Nuclear Physics B 933 (2018) 185–233 using (4.5d) which excludes this because w1 ∧ v2 gives rise to a Vol(S 2 ) term that can not be cancelled by the parts involving (v1 , w2 ). Thus we can once more conclude that H0 = δ = θ = φ = 0 . (4.7) Given this, we can write the necessary and sufficient solutions for supersymmetry in the α = 0 limit in a relatively simple way. After rotating the canonical frame, this time as in (3.16), we find 3  H0 = d(e2A− cos βv1 ) = d(e 2 A+C− 2 3  cos β) − νe 2 A− 2 cos βv2 = 0 , d(e2A+2C− w) + 2νe2A+C− w ∧ v2 = d(e2A+2C− sin βv1 ) + 2νe2A+C− sin βv1 ∧ v2 = 0 , d(e3A+2C− v1 ∧ w) − 2νe3A+2C− v1 ∧ w ∧ v2 = d(e3A− cos β) ∧ v1 ∧ v2 = 0 , d(e3A+2C− sin βw1 ∧ w2 ) − 2νe3A+2C− sin βw1 ∧ w2 ∧ v2 + e3A+2C− cos βH3 = 0 , e2A H3 ∧ v1 + cos2 βd(e2A tan β) ∧ v1 ∧ w1 ∧ w2 = d(e−A+ sin β) ∧ v1 ∧ w1 ∧ w2 = w ∧ H3 = 0 , d(e3A− sin β) + d(e3A− cos βw1 ∧ w2 ) − e3A− sin βH3 − e3A 4 λ(G− ) = 0 , d(e3A+3C− sin βv2 ) + d(e3A+3C− cos βw1 ∧ w2 ∧ v2 ) − e3A+3C− sin βH3 ∧ v2 − e3A+3C 4 λ(G+ ) = 0 ,    cos βv1 ∧ v2 ∧ λ(G+ ) + (cos β − sin βw1 ∧ w2 ) ∧ v1 ∧ λ(G− )  = 0 . (4.8) 4 This is as far as we can go without making assumptions about β, which we now proceed to do. 4.1.1. Subcase: β = 0 Setting β = 0 in (4.8) immediately leads to H3 = 0, the rest of the conditions are implied by 3  3  d(e 2 A+C− 2 ) − νe 2 A− 2 v2 = d(e2A− v1 ) = d(e−A w) = 0 , (4.9a) d(e2A ) ∧ v1 ∧ v2 = d(e−A+ ) ∧ w ∧ v1 = 0 , (4.9b) e3A 4 λ(G− ) − d(e3A− w1 ∧ w2 ) = 0 , (4.9c) e3A+3C 4 λ(G+ ) − d(e3A+3C− w1 ∧ w2 ∧ v2 ) = 0 ,    v1 ∧ v2 ∧ λ(G+ ) + v1 ∧ λ(G− )  = 0 . (4.9d) (4.9e) 4 The first thing we note is that given (4.9c)–(4.9d), G+ must be a 0-form and G− a 1-form which means (4.9e) is solved automatically. Next, we solve (4.9a) by using it to define the vielbein 3  v1 = e−2A+ g(x)dx, w = eA (dψ1 + idψ2 ) , v2 = ν e−3A/2+/2 dρ, ρ = e 2 A+C− 2 , (4.10) N.T. Macpherson et al. / Nuclear Physics B 933 (2018) 185–233 207 where g is a function parametrising a potential coordinate transformation in x. From eq. (4.9c), we see that the combination eA− only depends on x and that eA , e and eC are functions of x, ρ only, so that ∂ψ1 and ∂ψ2 are isometries. We thus choose to parametrise eA− = f (x) , g = −f , (4.11) the latter of which is a convenient choice we make without loss of generality. For the fluxes, we use the v1 , v2 , w1 , w2 vielbein on M4 to compute the Hodge duals from eq. (4.9c) and (4.9d), arriving at the ten-dimensional fluxes   3 −1 −4A −1 −4A F0 = ∂x f , F4 = νρ f ∂ρ (f e )dx − ∂x (f e )dρ ∧ Vol(S 3 ) . (4.12) The Bianchi identities reduce to dF0 = dF4 = 0 which leads to ∂x2 f = 0 , ∂x2 (f −1 e−4A ) + f 1 ∂ρ (ρ 3 ∂ρ (f −1 e−4A )) = 0 , ρ3 (4.13) the former of which can be immediately integrated as f = (c + F0 x), dc = 0. The metric takes the form   H 1 ds 2 (R1,4 ) + ds 2 = √ dρ 2 + ρ 2 ds 2 (S 3 ) + f H dx 2 , H = f −1 e−4A . (4.14) f fH This solution corresponds to an intersecting D4–D8 brane system, where the localised D4-branes are embedded in the D8-branes [40]. 4.1.2. Subcase: β = π2 The β = π2 limit of (4.8) leads to H3 ∝ v1 ∧ w1 ∧ w2 with the remaining conditions implied by d(e2A+2C− ui ) + 2νe2A+C− ui ∧ v2 = 0 , (4.15a) d(e3A+2C− ij k uj ∧ uk ) − 2νe3A+C− ij k uj ∧ uk ∧ v2 = 0 , (4.15b) d(e3A− ) − e3A− H3 + e3A 4 λ(G− ) = 0 , (4.15c) d(e3A+3C− v2 ) − e3A+3C− H3 ∧ v2 + e3A+3C 4 λ(G+ ) = 0 ,   −A− ) ∧ v1 ∧ w1 ∧ w2 = v1 ∧ w1 ∧ w2 ∧ λ(G− ) = 0 , d(e (4.15d) (4.15e) 4 where we introduce u = (v1 , w1 , w2 ), (4.16) to ease notation, and to make clear the cyclic property of these vielbein. The first thing we note is that, given (4.15c), the second of (4.15e) reads 4 H3 ∧ v1 ∧ w1 ∧ w2 = 0, but since H3 ∝ v1 ∧ w1 ∧ w2 we must set H3 = 0 . Next one can show that both (4.15a) and (4.15b) together imply the useful identities   d(eA ui ) ∧ uj = d(eA/2+C−/2 ) − νeA/2−/2 v2 ∧ ui ∧ uj = 0 , i = j, (4.17) (4.18) 208 N.T. Macpherson et al. / Nuclear Physics B 933 (2018) 185–233 so we must have ci (4.19) ij k uj ∧ uk = d(eA/2+C−/2 ) − νeA/2−/2 v2 = dci = 0 . 2 Consistency of the first of these with (4.15a) implies that ci = 0, and with this fixed (4.15b) also follows from (4.19). We can use the standard trick of taking (4.19) to define a vielbein in terms of local coordinates, namely d(eA ui ) + ui = e−A dxi , v2 = νe−A/2+/2 dρ, ρ = eA/2+C−/2 . (4.20) Having defined the vielbein, it is then a simple matter to solve the first of (4.15e) by introducing a free function e−A− = f (x1 , x2 , x3 ). (4.21) All that remains is to calculate the fluxes, and impose their Bianchi identities. Using (4.20) to take the Hodge duals of 4.15c and 4.15d we find the 10d fluxes 1 F2 = ij k ∂xi f dx j ∧ dx k , (4.22) 2   1 F6 = −νρ 3 ij k ∂xi (f −1 e−4A )dx j ∧ dx k ∧ dρ − f ∂ρ (f −1 e−4A )dx1 ∧ dx2 ∧ dx3 2 ∧ Vol(S 3 ), which clearly means the Bianchi identities, away from localised sources, follow from ∂x2i (f −1 e−4A ) + f ∂x2i f = 0 , 1 ∂ρ (ρ 3 ∂ρ (f −1 e−4A )) = 0 . ρ3 (4.23) The metric takes the form 1 ds 2 = √ ds 2 (R1,2 ) + fH H =f −1 −4A e H dρ 2 + ρ 2 ds 2 (S 3 ) + f f H (dx12 + dx22 + dx33 ) , (4.24) . The solution corresponds to an intersecting D2–D6 brane system [44]. 4.1.3. Subcase: generic β For 0 < β < π2 one is able to divide by sin β, cos β freely when simplifying (4.8). Assuming that cos β = 0 the result is d(e2A− cos βv1 ) = d(e−A sec βw) = 0 , 3 1 3 1 (4.25a) d(e 2 A+C− 2  cos β) − νe 2 A− 2  cos βv2 = 0 , (4.25b) d(e−A+ sec β) ∧ v1 ∧ w = d(e2A ) ∧ v1 ∧ v2 = d(e−2A tan β) ∧ v1 = 0 , (4.25c) e2A cos2 βH3 + d(e2A sin β cos β) ∧ w1 ∧ w2 = 0 , (4.25d) d(e3A+3C− sin βv2 ) + d(e3A+3C− cos βw1 ∧ w2 ∧ v2 ) − e3A+3C− sin βH3 ∧ v2 − e3A+3C 4 λ(G+ ) = 0, N.T. Macpherson et al. / Nuclear Physics B 933 (2018) 185–233 209 d(e3A− sin β) + d(e3A− cos βw1 ∧ w2 ) − e3A− sin βH3 − e3A 4 λ(G− ) = 0 , (4.25e)    cos βv1 ∧ v2 ∧ λ(G+ ) + (cos β − sin βw1 ∧ w2 ) ∧ v1 ∧ λ(G− )  = 0 , (4.25f) 4 where H3 is closed given (4.25a) and (4.25c). As usual we solve (4.25a) and (4.25b) by using them to define a vielbein in terms of local coordinates v1 = g(x)e−2A+ sec βdx, w = eA cos β(dψ1 + idψ2 ), 3 1 3 1 v2 = νe− 2 A+ 2  sec βdρ, ρ = e 2 A+C− 2  cos β, (4.26) where g(x1 ) is a function parametrising a potential diffeomorphism in x1 . With local coordinates introduced we can solve (4.25c) in terms of them as eA− cos β = f (x) , A = A(ρ, x) , tan β = c(x1 )e2A , g = −f , (4.27) so that ∂ψi are necessarily isometry directions. We can then calculate the ten-dimensional fluxes as before – first we note that ∂x cf tan2 β , (4.28) c should be constant. We shall restrict ourselves to the case ∂x c = 0. For generic β then the fluxes may be expressed as F0 = ∂x f + sin2 β dψ1 ∧ dψ2 , dc = 0 c   3 −1 −4A −1 −4A F4 = B2 ∧ F2 + νρ f ∂ρ (f e )dx − ∂x (f e )dρ ∧ Vol(S 3 ). F0 = ∂x f, F2 = F0 B2 , B2 = − (4.29) We note that the Bianchi identities follow when anything not coupled to B2 is closed, and since these terms reproduce (4.12) the Bianchi identities imply the PDEs of (4.13) once more. This is because the class of solutions in this section can be generated via U-duality, from the intersecting D4–D8 system in section 4.1.1. For completeness the metric takes the form 1 cos2 β 2 2 ds 2 = √ ds 2 (R1,2 ) + √ ds (T ) + fH fH   H dρ 2 + ρ 2 ds 2 (S 3 ) + f f H dx 2 , (4.30) H = f −1 e−4A , where T 2 is spanned by (ψ1 , ψ2 ). 4.2. Branch II: α non-zero For the second branch with 0 < α < π , we begin by studying the lower form conditions that follow from (4.1). As in IIB we find it useful to rotate the canonical frame of (B.3) as (3.44). Necessary but insufficient conditions for supersymmetry are 210 N.T. Macpherson et al. / Nuclear Physics B 933 (2018) 185–233 d(e2A+3C− cos β cos(α − δ)) + e2A+3C− H0 (cos β cos δv2 + sin βw2 ) = 0, (4.31a) d(e2A− (cos β cos δv2 + sin βw2 )) = 0, (4.31b) d(e3A+2C− cos α sin β) + 2νe3A+C− (cos β cos δw2 − sin βv2 ) = 0, (4.31c) eC cos α sin β sin θ dφ − 2ν(cos β cos δw1 − sin α sin βv1 ) = 0, (4.31d) eC cos α sin βdθ − 2ν(cos β cos δv1 + sin α sin βw1 ) = 0, (4.31e) d(e2A+2C− (sin α sin βv2 − sin(α − δ) cos βw2 )) + 2e2A+C− ν cos β(sin δw2 ∧ v2 + cos(α − δ)v1 ∧ w1 ) = 0, (4.31f) First from (4.31a) we observe that when cos(α − δ) = 0 (cos β = 0 is a subcase of this) we necessary have H0 = 0. Next we observe that generically (4.31b)–(4.31e) can be used to locally define the vielbein on M4 , the only exception is when sin β = 0 (cos α = 0 is a subcase of this). Setting sin β = 0 means that in order to solve (4.31d)–(4.31e) we must take δ = π2 , additionally (θ, φ) drop out of the definition of the spinors so we can fix β = θ = φ = (δ − π )=0 2 (4.32) without loss of generality. Interestingly one doesn’t need to set H0 = 0, however as we shall see in section 4.2.1 this case actually contains no solutions. If one assumes sin β = 0 we see that v1 , w1 must span S 2 while v2 , w2 can be expressed in terms of local coordinates in such a way that they have no legs in S 2 . This is a problem for (4.31f) which generically has an Vol(S 2 ) factor, due to the v1 ∧ w1 term which sits orthogonal to everything else. Thus the only resolution is to fix cos(α − δ) = 0 which leads to H0 = 0 also. This actually leads to a novel class of solutions that we shall derive in section 4.2.2. 4.2.1. Subcase: β = 0 Upon setting β = 0 we are led to the following conditions for supersymmetry d(e2A+3C− sin α) = d(e−A− cos4 α) ∧ v1 ∧ w1 ∧ w2 = 0 , (4.33a) d(e2A+2C− cos αui ) + νe2A+C− (sin αij k uj ∧ uk + 2ui ∧ v2 ) = 0 , (4.33b) d(e3A+2C− (ij k uj ∧ uk + 2 sin αui ∧ v2 )) − 2νe3A+C− cos αij k uj ∧ uk ∧ v2 = 0 , (4.33c) sin αH3 − cos αH0 v1 ∧ w1 ∧ w2 = 0 , (4.33d) e3A 4 λ(G− ) + d(e3A− ) − e3A− H0 cot αv1 ∧ w1 ∧ w2 = 0 , (4.33e) N.T. Macpherson et al. / Nuclear Physics B 933 (2018) 185–233 211 e3A+3C 4 λ(G+ ) + e3A+3C− H0 + d(e3A+3C− cos αv2 ) − e3A+3C− H0 csc αv1 ∧ v2 ∧ w1 ∧ w2 = 0,    cos αv1 ∧ w1 ∧ w2 ∧ λ(G− ) − (sin α − v1 ∧ v2 ∧ w1 ∧ w2 ) ∧ λ(G+ )  = 0 , (4.33f) 4 where here as elsewhere u = (v1 , w1 , w2 ). Using the same techniques as are spelled out in section 3.1.2, it is possible to establish that 1 1 1 1 d(e 2 A+C− 2  ) − νe 2 A− 2  cos αv2 = d(eA ui ) ∧ uj = 0, for i = j (4.34) which we can use as in section 3.2.1 to define the vielbein in terms of the local coordinate 1 1 ρ = e 2 A+C− 2  and a set of left invariant 1-forms such that 1 1 1 v2 = νe− 2 A+ 2  dρ, ui = e−A cos αci K̃i , d K̃i = K̃j ∧ K̃k , 2 under the assumption that α = 0. Plugging this back into (4.33b) fixes e2A = c4 e−2 sin4 α , ρ4 sin α = c1 , ci = c ρ (4.35) (4.36) however plugging this back into (4.33a) leads to dρ ∧ K̃1 ∧ K̃2 ∧ K̃3 = 0 (4.37) which cannot be solved. 4.2.2. Subcase: δ = α + π2 The final case we consider is when δ = α + π2 and contains β = π2 as a subcase. In addition to the rotating the canonical frame (B.3) by (3.44) we find it useful to send v1 + iw1 → e−iβ (v1 + iw1 ), then the necessary and sufficient conditions for supersymmetry are d(eA−C sin α) = H0 = 0 , (4.38a) d(e3A+2C− cos α sin β) − 2νe3A+C− k1 = d(e2A− k2 ) = 0 ., (4.38b) eC cos α sin βdθ + 2ν sin αv1 = eC cos α sin β sin θ dφ + 2ν sin αw1 = 0 , (4.38c) 1 (4.38d) d(e2C cot2 α cos β sin β) − 2eC ν cos αw2 ∧ Vol(S 2 ), 4  −A+C  −3A+   e sin α cos α cos β e d ∧ k1 − e2A+C− cos α sin βd ∧ k2 = 0 ,   (4.38e)  3  √ 1 3 e− 2 A+C+ 2  cos 2 α sin β cos β d ∧ k1  H3 = −e 1 1 2 A− 2   −2A+C+  cot α sin β e cos α sin βd ∧ k2 = 0 ,  212 N.T. Macpherson et al. / Nuclear Physics B 933 (2018) 185–233 e3A 4 λ(G− ) + d(e3A− cos α cos β) − d(e3A− cos α sin βv1 ∧ w1 ) − e3A− cos α cos βH3 = 0 , e3A+3C 4 λ(G+ ) + d(e3A+3C− k3 ) − d(e3A+3C− v1 ∧ w1 ∧ k1 ) + e3A+3C− k2 ∧ H3 , (4.38f)    cos α sin β(1+ cot βv1 ∧ w1 ) ∧ v2 ∧ w2 ∧ λ(G+ ) − (k2 − k4 ∧ v1 ∧ w1 ) ∧ λ(G− )  =0, 4 (4.38g) where we introduce k1 = (sin α cos βw2 + sin βv2 ), k2 = (sin α cos βv2 − sin βw2 ), k3 = (cos αv2 − sin α sin βw2 ), k4 = (cos αw2 + sin α sin βw1 ),  = sin2 β + cos2 β sin2 α, Vol(S 2 ) = sin θ dθ ∧ dφ, (4.39) to ease presentation. We can use (4.38b)–(4.38c) to locally define the vielbein through 3 1 k1 = ν cos α sin βe− 2 A+ 2  dρ, k2 = e−2A+ dx, v1 = − (4.40) 3 1 1 νe− 2 A+ 2  ρ cos α sin βdθ, 2 sin α 3 1 1 νe− 2 A+ 2  ρ cos α sin β sin θ dφ, 2 sin α where (θ, φ, x) and w1 = − 3 1 ρ = e 2 A+C− 2  cos α sin β (4.41) are local coordinates on M4 . The ten-dimensional metric then takes the form ds 2 = e2A ds 2 (R1,2 ) + (4.42) e−3A+ e−4A+2 dx 2 ρ 2 ds 2 (S 3 ) + cos α sin β    e−3A+ cos α sin β 4 sin2 αdρ 2 2 2 2 ds (S ) . + ρ +  4 sin2 α We can now turn our attention to (4.38a) and (4.38e) which lead to the PDEs ∂ρ (eA−C sin α) = ∂x (eA−C sin α) = 0 ,    −3A+  −A+C cos α cos β sin α e e νρ∂ρ + ∂x =0,     3 √   −2A+C+ 1 3 cot α sin β e e− 2 A+C+ 2  cos 2 α sin β cos β + ∂x =0, ν∂ρ   (4.43a) (4.43b) (4.43c) and tell us that e2A , e2C , e , α, β are functions of (ρ, x) only, which means these solutions support an additional SU (2) isometry due to round S 2 spanned by (θ, φ). Actually this SU (2) N.T. Macpherson et al. / Nuclear Physics B 933 (2018) 185–233 213 is an additional part of an enhanced R-symmetry which together with the SU (2)R of S 3 gives SO(4)R – there is also an SU (2) flavour symmetry. It is now a simple matter to calculate the Hodge dual of the fluxes from (4.38f) and then the fluxes themselves given (4.38d) and our local vielbein (4.40). At first G± take a complicated form that we will not quote here, however, we are yet to impose also (4.38g): doing so and making extensive use of (4.43a)–(4.43c) one can express the ten-dimensional fluxes as:  1 H= d(e2C cot2 α cos β sin β) 4  2 −3A+ −2A+C+ cos α cos βdρ − νe cot α sin βdx) ∧ Vol(S 2 ) , − (ρe   −2 ∂x (e2A sin α) , e sin α √ 1 √ 3 3 e−3A− 2  ρ cos α F2 = F0 e 2  ρ cos α cos β − 2νe 2 A sin α sin β Vol(S 2 ) , 2 4 sin α  5 1 e− 2 A+ 2  ρ 2 3A− F4 = −Vol3 ∧ d(e d(e−2A cot β) cos α sin β) + √ cos α sin β sin α F0 = 2  5  1 2e− 2 A sin α  1  sin β A − νe 2 dρ − e 2 sin α cos βdx ∧ Vol(S 3 ) , sin β cos α F6 = −Vol3 ∧ (d(e3A− cos α sin βv1 ∧ w1 ) + e3A− cos α cos βH3 )  νe−10A+2 ρ 5 d(e3A− cos α cos β) ∧ dρ + 4 sin2 α  e−A+ 3A− − cos α cos β) ∧ dx ∧ Vol(S 2 ) ∧ Vol(S 3 ) d(e cos α sin β (4.44) where (4.38g) can be expressed in terms of F0 as √ 1 3 ν e 2 A+ 2  sin2 α cos α cos β ρ sin β∂ρ (ρe2A sin α) + F0 = 0. (4.45) 2  Imposing that F0 is constant together with (4.43a)–(4.43c) and (4.45) then implies dH = 0 and the Bianchi identities of the remaining fluxes. This system is quite complicated, however taking inspiration from section 4.1 of [36] (which re-derives [49]) we anticipate that the system can be further simplified if we treat the cases F0 = 0 and F0 = 0 separately. Subcase F0 = 0 If we set F0 = 0 then (4.45) and (4.44) impose that ρ e2A sin α = L2 , dL = 0 (4.46) This leaves (4.43a)–(4.43c) to solve. We first integrate (4.43a) as eA−C sin α = c, dc = 0, (4.47) 214 N.T. Macpherson et al. / Nuclear Physics B 933 (2018) 185–233 then use it to express (4.43b) as     −3A+ cos α cos β 1 e e−5A+ c∂x + ν ∂ρ = 0. cos α sin β ρ  (4.48) We note that this defines an integrability condition that we can solve by introducing an auxiliary function h(ρ, x) such that c e−5A+ ν = ∂ρ h, cos α sin β ρ e−3A+ cos α cos β = −∂x h.  (4.49) Plugging these definitions into (4.43c) and making use of (4.46)–(4.47) we arrive at c2 ∂x2 h + 1 ∂ρ (ρ∂ρ h) = 0. ρ (4.50) This is a 3d Laplacian expressed in axially symmetric cylindrical polar coordinates (up to rescaling x). Solution in this class are in one to one correspondence with solution to this Laplace equation. The physical data can be expressed in terms of h and the 2 constants (c, L), as L4 e−4A = ρ 2 sin2 α, e−+5A = c3 ρ(∂ρ h)2 νcρ , =1− 4 , cos α sin β∂ρ h L ∂ρ h(1 − cρ∂ρ h) L4 (1 − cρ∂ρ h) − 1, 4 2 c ρ (∂x h)2 + L4 (1 − cρ∂ρ h)2 tan α =  ν 1 − cρ∂ρ h L4 ∂ρ h(1 − cρ∂ρ h) − c3 ρ(∂x h)2 tan β = . 3√ c 2 ρ∂x h (4.51) It is interesting that this class depends on axially symmetric Laplacian, indeed the same is true of the class of AdS5 × S 2 solutions in IIA [50] one obtains by dimensionally reducing the M-theory class of Lin–Lunin–Maldecena [51]. The M-theory class actually depend on a 3d Toda equation, which is equivalent to the Laplacian only when one imposes an additional U (1) isometry, which one then uses to reduce to IIA. As the class of this section is in massless IIA it can be lifted to M-theory, so an obvious question poses itself: Is there a class in M-theory governed by a 3d Toda from which the backgrounds in this section descend? It would be interesting to look into this and what connection, if any, this class has to AdS5 × S 2 or indeed any AdS class. Subcase: F0 = 0 We expect to be able to perform a similar simplification of the system of PDE’s for F0 = 0 case, however up to this point we have failed to do so in general. However there is a special case which is far more simple, namely β = π2 . Here (4.43a)–(4.43c) and (4.45) force e2A = ρ 1 g(x) 4 sin α , e−5A = cos α sin2 α , dα = dc = 0, c2 ρ 2 (4.52) all that remains is to ensure that dF0 = 0 which is ensured as long as g = c̃ − 2F0 x cos α . c14 (4.53) N.T. Macpherson et al. / Nuclear Physics B 933 (2018) 185–233 215 This solution has all the generic fluxes except the internal part of F6 turned on, it bears some resemblance to D8-branes on some sort of cone, but the precise picture depends on what values the free constant α takes. We shall come back to study the solutions that follow from these massless and massive systems in [71]. 5. The unique type II AdS4 × S 3 background We have classified all Mink3 × S 3 with internal Killing spinors of equal norm, up to certain PDE determining various warp factors. As equal norms is a requirement7 for the existence of AdS4 it is reasonable to ask whether such solutions are contained within our classification. Any AdS4 solution can be expressed as a Mink3 solution, one needs only parametrise AdS as the Poincaré patch. This comes about quite naturally in terms of the Mink3 × M7 set up by imposing that   dr 2 ds 2 = e2A ds 2 (R1,2 ) + ds 2 (M7 ) = e2à r 2 ds 2 (R1,2 ) + 2 + ds 2 (M6 ) , (5.2) r with e2à and M6 independent of r. As we have local expressions on M7 = S 3 × M4 , this makes our task relatively easy. A quick scan through the various cases in section 3 and 4 indicates that the only class that are potentially compatible with AdS4 are in sections 4.1.2 and 4.2.2 – the others manifestly cannot be put in the form (5.2). These are both in IIA, but closer inspection leads one to realise that the class of section 4.2.2 cannot work as (4.38a) would break the putative AdS isometry. This leaves only sections 4.1.2. We will now show that there is a unique compact8 AdS4 solution, at least locally, for the class of solutions of section 4.1.2. This background corresponds to a foliation of AdS4 × S 3 × S 2 over an interval and is the near-horizon of a D2–D6 brane intersection, and can also be obtained by dimensionally reducing a certain Zk orbifold of AdS4 × S 7 . Starting from M-theory one first parameterises S 7 as a foliation of S 3 × S 3 over a closed interval, then performs both the orbifolding and reduction to IIA on the Hopf fibre of one of the S 3 ’s (see e.g. [52]) – there by preserving 16 supercharges. For this purpose, we take the metric (4.24), expressed in the form ds 2 = e2A ds 2 (R1,2 ) + e−A+ dρ 2 + ρ 2 ds 2 (S 3 ) + e−2A (dx12 + dx22 + dx33 ) , and assume an AdS4 factor, which requires e2A = r 2 e2à , with the rescaled warp factor à as well as the dilaton  undetermined functions independent of the AdS4 radial coordinate r. Further7 It is established in Appendix D that the 7d spinors χ , χ must obey the relation |χ |2 ± |χ |2 = c e±A where c ± ± 1 2 1 2 are constants. We can, without loss of generality solve these conditions in terms of unit norm spinors χi0 and an angle ζ as 1 A χ1 = √ e 2 2 1 + sin ζ χ10 , 1 A χ2 = √ e 2 2 1 − sin ζ χ20 , c+ = 1, c− = e2A sin ζ (5.1) To make a Mink3 solution AdS4 requires us to fix the dependence of e2A on the AdS radius, but since e2A sin ζ is constant, we must either set ζ = c− = 0 or fix ζ such that it also depends on the AdS radius. The latter contradicts the assumption of an SO(2, 3) isometry, so we conclude that AdS4 requires c− = 0; i.e. equal 7d spinor norms. 8 Strictly speaking section 3.1.1 contains AdS × S 5 which can be expressed as a non-compact AdS solution. We will 4 5 disregard such higher dimensional AdS solutions. 216 N.T. Macpherson et al. / Nuclear Physics B 933 (2018) 185–233 more, the background is only SO(2, 3) invariant if the internal metric is independent of r, which fixes ρ and the xi to scale as ρ ∼ r 1/2 and xi ∼ r. Keeping this in mind, we parametrise x1 = r q(μ) sin θ cos φ , x2 = r q(μ) sin θ sin φ , x3 = r q(μ) cos θ , ρ = r 1/2 h(μ) , where q(μ) and h(μ) are undetermined functions of some coordinate μ, and the (θ, φ) directions parametrise a 2-sphere such that the R3 spanned by xi is written in polar coordinates with radius r q(μ). Now we have to ensure that the metric is diagonal with respect to the r-direction, i.e. set gr μ = 0, and that it shows the 1/r 2 behaviour for grr , which amounts to imposing grr = e2à /r 2 . These two conditions lead to the following expressions for à and  in terms of q(μ), h(μ) and independent of (θ, φ):   q  (μ) 4à e = q(μ) q(μ) − h(μ)  , 2h (μ) q(μ) q  (μ) e = −2e−à . h(μ) h (μ) These expressions imply, once inserted in the first eq. of (4.15e), the following ODE for the q and h functions:   q  (μ) h (μ)2 + h(μ) h (μ) = h(μ) h (μ) q  (μ) ,   which can be solved in closed form as h = h q(μ) and also implies the Bianchi identities of the fluxes. As h is a function of q, rather than μ, we can use diffeomorphism invariance to fix q such that h is simple, without loss of generality we choose 2L3 μ cos2 , k 2 where L and k are constants. This leads to μ h(μ) = −2L3/2 sin . 2 The resulting metric is of the form  2L μ  2 μ μ ds 2 = cos ds (AdS4 ) + L2 dμ2 + 4 sin2 ds 2 (S 3 ) + cos2 ds 2 (S 2 ) k 2 2 2 (5.3) q(μ) = with fluxes k F2 = − Vol(S 2 ) , 2 F4 = 3 Vol(AdS4 ) . L (5.4) This is the IIA reduction of AdS4 × S 7 /Zk with length scale L and k D6-branes, as in eq. (2.8) of [52]. The fact that (θ, φ) are isometry directions of this solution means that there is an additional SU (2)S 2 symmetry due to the round S 2 factor in the metric and fluxes. The spinors of this solution are then charged under SU (2)S 2 and just one of the SU (2)’s of S 3 (see the ν dependence N.T. Macpherson et al. / Nuclear Physics B 933 (2018) 185–233 217 of (4.22)), SU (2)+ say. Since S 2 and S 3 appear as a product the spinors are actually charged under SU (2)+ × SU (2)S 2 which realises an enhanced SO(4) R-symmetry as required by the N = 4 super-conformal algebra in 3d – SU (2)− , under which the spinors are not charged, is a flavour symmetry. 6. Type II with a single Killing spinor In the previous two sections, we have worked out the supersymmetry conditions making use of the pure spinor equations (2.4), which are valid only in case |χ1 |2 = |χ2 |2 . Note that this is a necessary condition for the existence of D-branes which do not break background supersymmetry. The supersymmetry condition for a Dp -brane is given by (p) 1 = 2 . Since (p) is unitary, squaring this equation leads to the conclusion that left- and right-handside must have equal norm.9 We will examine the simplest non-equal norm case, namely the one where 2 = 0 . (6.1) We could either make use of the generalised geometrical reformulation of supersymmetry which incorporates |χ1 |2 − |χ2 |2 = 0 as deduced in appendix D, or use the actual Killing spinor equations. Considering the simplicity of this case, we will use the latter. Much of the work has however already been done: the conditions for seven-dimensional pure NSNS solutions have been deduced in [29,56,57] up to some ansätze. We will merely show that the ansätze made in [29,56,57] (no warp factor, no external NSNS flux) are in fact enforced by supersymmetry, and then proceed to plug in the decomposition resulting from M7 = S 3 × M4 . This leads to a pair of explicit pure NS backgrounds: the NS5-brane and the U-dual to the IIB conical backgrounds of section 3.2.1. We will also analyse the seven-dimensional RR-sector, which is new, but the conclusion is that all RR-fluxes vanish. 6.1. Seven-dimensional decomposition Our starting point are the democratic supersymmetry equations, which read as follows for 2 = 0:     1 1 /  1 = ∇M − H / M 1 = 0 , λF/ M 1 = 0 . (6.2) ∂/φ − H 2 4 As can be seen, the NSNS and RR sectors decouple. We impose a similar 3 + 7 decomposition as before: the metric and RR flux is given by (2.1), while the Killing spinor 1 is given by (2.2) and 2 = 0. We generalize the NSNS 3-form flux by allowing a term h e3A Vol3 . Using the convention γμνρ = μνρ , plugging the above decompositions into (6.2) leads to the following 7d equations:     1 1 1 m mnp (7) np + ih χ = ∇m − Hmnp γ ∂m φγ − Hmnp γ χ =0 (6.3a) 12 2 8   1 A e h − ieA ∂m Aγ m χ = 0 (6.3b) 4 9 The argument is slightly more complicated due to the fact that Spin(1, 9) spinors do not admit a non-trivial norm, and hence one should decompose to Spin(9) first. See [53] for details. Also note that strictly speaking, the norms need only be equivalent on the brane. 218 N.T. Macpherson et al. / Nuclear Physics B 933 (2018) 185–233 λf χ = λf γm χ = 0 . (6.3c) From (6.3b) it follows that ∂m A = h = 0 . (6.4) As can be seen, the NSNS and RR sector split and can thus be analysed independently. The existence of a globally defined nowhere-vanishing Spin(7) Majorana spinor χ reduces the structure group of M7 to G2 . More concretely, the following bilinears can be defined: ϕmnp = −iχ † γmnp χ , (7 ϕ)mnpq = χ † γmnpq χ , (6.5) where we have normalized χ . The other bilinears, i.e., the 1-, 2-, 5- and 6-form vanish. As has been deduced in [29,56,57], (6.3a) can be rewritten in terms of the G2 -structure as dϕ ∧ ϕ = d(e−2 7 ϕ) = d(e−2 ϕ) − e−2 7 H = 0 . (6.6) We will analyse the solution to the NSNS sector by requiring a further splitting of M7 = S 3 × M4 . On the other hand, we will show that the RR-fluxes vanish for any M7 . 6.2. NSNS sector Considering the case M7 = S 3 × M4 , we further decompose the spinor as   χ = ξ ⊗ sin(α/2)η + cos(α/2)γ̂ η + m.c., η† η = 1, η† γ̂ η = 0 , (6.7) with m.c. the Majorana conjugate. This leads to a further reduction of the structure group. Since S 3 is parallelisable, it has trivial structure group, leading to a Spin(4) structure group on M4 . Generically, the structure group need not reduce on M4 .10 In the case where either η+ or η− is nowhere vanishing, the structure group reduces to SU (2), in case both are nowhere-vanishing, the structure group is trivial. As everywhere else, our analysis is purely local and we will work with a local trivial structure, parametrising possible vanishing of either chiral spinor by the angle α. First, as in [58], we make use of an auxiliary SU (3)-structure (J, ) to express the G2 -structure as ϕ = −v2 ∧ J − Im, 1 7 ϕ = J ∧ J + Re ∧ v2 . 2 (6.8) Next, we decompose the SU (3)-structure in terms of the vielbeine as 1 K1 K2 K3 J = − (K1 ∧ w1 + K2 ∧ w2 + K3 ∧ v1 ),  = eiα ( + iw1 ) ∧ ( + iw2 ) ∧ ( + iv1 ) 2 2 2 2 (6.9) Inserting this into (6.6), one finds 10 Note that on any M with a spin structure, an SU (2)-structure can be found [54] [55]. However, this is not necessarily 7 the structure group defined by the spinors we are making use of, and so even when splitting M7 = M3 × M4 with M3 parallelisable, M4 need not admit a globally well-defined SU (2)-structure; consider for example M7 = S 3 × S 4 . N.T. Macpherson et al. / Nuclear Physics B 933 (2018) 185–233 219 d(e3C−2 cos αv2 ) = d(e2C−2 ui ) + νeC−2 (2ui ∧ v2 + sin αij k uj ∧ uk ) = 0 , (6.10a) 1 d(e2C−2 (sin αui ∧ v2 + ij k uj ∧ uk )) − νij k eC−2 cos αuj ∧ uk ∧ v2 = 0 , 2 (6.10b) dα ∧ v1 ∧ w1 ∧ w2 = 0 , u = (v1 , w1 , w2 ), (6.10c) e−2 7 H = −d(e−2 cos αv1 ∧ w1 ∧ w2 ) − νVol(S 3 ) ∧ d(e3C−2 sin α). (6.10d) When α = π2 , by taking linear combinations, exterior derivatives and wedge products with the vielbein of the equations in (6.6), one can derive   ij k d(eC− cos α) − νe− v2 ∧ uj ∧ uk  + e5C+12 cos4 α d(e−3C+2 sec α tan α) ∧ v2 ] ∧ ui = 0, where, since ui form a basis of independent 1-forms, the terms in square parentheses must vanish. This is sufficient to conclude that dα = 0 , C = C(ρ) ,  = (ρ) , ρ = eC− . (6.11) It is then not hard to establish that dui ∧ uj = 0 , i = j, (6.12) in a similar fashion. This means we can locally parametrise v2 = ν sec αe dρ , dui = ci ij k uj ∧ uk , dci = 0 . (6.13) Plugging this back into (6.6) we find that ci = e−C ν tan α, (6.14) so either dC = 0 or c = α = 0. Case 1: When α = 0, ui are the vielbeine of T 3 so we can simply take ui = dxi . (6.15) All that is left to do is calculate H and impose its Bianchi identity. We find H = ν∂ρ (e2 )ρ 3 Vol(S 3 ) . (6.16) Closure of the flux then implies that e2 is harmonic, leading to   c e2 = gs2 1 + 2 , ρ gs , c ∈ R , (6.17) which is consistent with the definition of ρ. Finally, we note that the metric is given by ds 2 = ds 2 (R1,5 ) + e2 dρ 2 + ρ 2 ds 2 (S 3 ) . (6.18) This is an NS5-brane, dual to the D5-brane solution in section 3.1.2 [41]. Case 2: When 0 < α < π2 , ui span the vielbeine of another S 3 so we take ui = eC̃ ν̃ K̃i , d K̃i + ij k d K̃j ∧ d K̃k , d C̃ = 0 , 2 2 (6.19) 220 N.T. Macpherson et al. / Nuclear Physics B 933 (2018) 185–233 where consistency requires that ν̃eC cos α + eC̃ ν sin α = 0 , (6.20) and find H = −2e3C ∂ρ (e− ) cos2 αVol(S 3 ) − 2e3C̃ ∂ρ (e− ) cos α sin αVol(S̃ 3 ) . (6.21)   By definition of ρ, the Bianchi identity is satisfied. After redefining r = exp e−C cos αρ , it follows that the metric is given by ds 2 = ds 2 (R1,2 ) + dr 2 + e2C ds 2 (S 3 ) + cot2 αe2C ds 2 (S̃ 3 ) . (6.22) Note that in IIB, this solution can be obtained from the solution of section 3.2.1 by means of the following S-duality transformation (up to redefining some constants):  → − , ds 2 → e− ds 2 , F3 → −H . (6.23) 6.3. RR-sector The NSNS sector has been analysed by imposing a further decomposition Spin(7) → Spin(3) × Spin(4) on the spinor. On the other hand, we will analyse the RR-sector in full generality. Let us consider the RR-flux constraints equations (6.3c), repeated here for convenience: λf χ = λf γm χ = 0 . (6.24) For type IIA, we have that λf = f0 − f 2 − i f 3 − i f 1 , (6.25) where we have defined f3 = 7 f4 , f1 = 7 f6 . For type IIB, one finds λf = f 1 − f 3 − i f 2 − if0 , (6.26) with f2 = 7 f5 , f0 = 7 f7 , hence up to some field redefinitions, the supersymmetry constraints are identical. The fluxes, a priori irreducible representations of SO(7), decompose into representations of G2 as follows: 7 → 7, 21 → 7 + 14, 35 → 1 + 7 + 27. Concretely, we parametrise p f2|mn = ϕmnp f2 + f2|mn q q f3|mnp = f3 ϕmnp + ψmnpq f3 + f3|q[m ϕnp] , (6.27) pq where the 14 satisfies f2|mn = 12 ψmnpq f2 and the 27 corresponds to a symmetric traceless 2-tensor. Furthermore, we have introduced the notation ψ = 7 ϕ for convenience. Making use of the G2 -structure identities (A.5), (A.6), we find f 1 χ = f1|m γ m χ f 1 γm χ =f1|m χ − iϕmnp f1n γ p χ f 2 χ = 3if2|m γ m χ f 2 γm χ =3if2|m χ − ϕmnp f2n γ p χ − 2f2|mn γ n χ f 3 χ = 7if3 χ − 4f3|m γ m χ f 3 γm χ = − if3 γm χ + 4f3|m χ + 6if3|mn γ n χ . (6.28) Inserting the above into (6.24) and comparing representation by representation, it follows that all RR fluxes vanish. N.T. Macpherson et al. / Nuclear Physics B 933 (2018) 185–233 221 Acknowledgements We would like to thank Alessandro Tomasiello for collaboration in the early stages of this project. D.P. would like to thank Noppadol Mekareeya for discussion. J.M. is grateful to the Physics Department of Milano-Bicocca U., the ITP at KU Leuven and SISSA for the warm hospitality. Throughout this project N.M. has been variously funded by INFN; the European Research Council under the European Union’s Seventh Framework Program (FP/2007–2013); ERC Grant Agreement n. 307286 (XD-STRING); and the Italian Ministry of Education, Universities and Research under the Prin project “Non-Perturbative Aspects of Gauge Theories and Strings” (2015MP2CX4). J.M. is supported by the FPI grant BES-2013-064815 of the Spanish MINECO, and the travel grant EEBB-I-17-12390 of the same institution. He also acknowledges partial support through the Spanish and Regional Government Research Grants FPA2015-63667-P and FC-15-GRUPIN-14-108. D.P. is funded by ERC Grant Agreement n. 307286 (XD-STRING). Appendix A. Conventions & identities We decompose ten-dimensional gamma matrices as μ = σ3 ⊗ eA γμ ⊗ I , m = σ1 ⊗ I ⊗ γm , (A.1) and seven-dimensional gamma matrices as γα(7) = eC σα ⊗ γ̂ , γa(7) = I ⊗ γa , B7 = σ2 ⊗ B4 , B4 B4∗ = −I . (A.2) A.1. M7 We consider gamma matrices satisfying γmnpqrst = imnpqrst , 1 γ(7−n) = (−1) 2 n(n−1) i 7 γ(n) . (A.3) A nowhere-vanishing Spin(7) spinor defines a G2 -structure on M7 by means of the bilinear ϕmnp = −iχ † γmnp χ . (A.4) Defining ψ = 7 φ = χ † γmnpq χ , the G2 -structure satisfies the following identities [59]: ψmnrs ψpq rs = −2ψmnpq + 4δmp δnq − 4δmq δnp , ψmnrs ϕp rs = −4ϕmnp , nrs n = 6δm , ϕmrs ϕ (A.5) as well as γmn χ = iϕmnp γ p χ , γmnp χ = iϕmnp χ − ψmnpq γ q χ , γmnpq χ = −4iϕ[mnp γq] χ + ψmnpq χ . (A.6) 222 N.T. Macpherson et al. / Nuclear Physics B 933 (2018) 185–233 A.2. S 3 We consider Pauli-matrices playing the role of gamma-matrices. They satisfy σαβγ = iαβγ , 1 σ(3−n) = (−1) 2 n(n−1) 3 σ(n) , (A.7) with α, β, γ = 1, 2, 3 indices on S 3 . The non-vanishing spinor bilinears of S 3 are given by ξ c† σ α ξ =  1 α 1 K1 + iK2α , ξ † σ α ξ = K3α . 2 2 (A.8) The real 1-forms Ki , i = 1, 2, 3, define a trivial structure on S 3 (i.e., a vielbein, up to normalisation). Note that S 3 is parallelisable and hence the trivial structure is globally well-defined. We will always normalise the volume form as K1 ∧ K2 ∧ K3 = −8Vol(S 3 ) , (A.9) regardless of which specific vielbein is used. Appendix B. The bispinors of M4 We consider gamma matrices satisfying 1 γabcd = abcd , γ(4−n) = (−1) 2 n(n+1) 4 γ(n) γ̂ , (B.1) with γ̂ = γ1234 the chirality matrix. Given a globally well-defined nowhere vanishing chiral spinor η+ , one can construct the bilinears † Jab = iη+ γab η+ , c† ωab = iη+ γab η+ (B.2) which furnish an SU (2)-structure. Given two globally well-defined nowhere vanishing chiral spinors of opposite chirality, the structure group reduces to a trivial structure [60]. Generically, supersymmetry requires a nowhere vanishing spinor η, which can admit a chiral locus. This ensures that, although the structure group of M4 cannot be globally reduced, it is possible to reduce the structure group of the generalised cotangent bundle T M4 ⊕ T ∗ M4 to SU (2) × SU (2), completely analogously to the well-known situation of SU (3)-structures [30]. Since the supersymmetry constraints are local, we will always work with the vielbeine determining the local trivial structure. Using the conventions of [61] with η = (η+ , η− ), we set † v = v1 + iv2 = η− γa η+ dx a , c† w = w1 + iw2 = η− γa η+ dx a . (B.3) Although some care must be taken on the chiral locus, where the above 1-forms all vanish, it turns out that no solutions exist on the chiral locus, as discussed in sections 3 and 4. We can expand the locally defined 4d components of the Killing spinors η1,2 in terms of η, satisfying η† η = 1 , η† γ̂ η = 0 (B.4) as α α η + sin γ̂ η , η2 = aη + bγ̂ η + cηc + d γ̂ ηc 2 2 where a, b, c and d are subject to η1 = cos (B.5) |a|2 + |b|2 + |c|2 + |d|2 = 1 . (B.6) N.T. Macpherson et al. / Nuclear Physics B 933 (2018) 185–233 223 We can then calculate the 4d bispinors appearing in (3.1a)–(3.1f) and (4.1a)–(4.1f), where to do so we find it useful to parametrise a = a1 + ia2 , b = b1 + ib2 , c = c1 + ic2 , d = d1 + id2 , (B.7) for ai , bi , ci and di real. However we first note that in both IIA and IIB we must solve the 0-form constraints (ψγ̂2 )0 = (Imψγ̂1 )0 = 0 , which reduce to α α α α α α b2 cos + a2 sin = d2 cos + c2 sin = d1 cos + c1 sin = 0 . 2 2 2 2 2 2 We can solve these in general by fixing α α α a2 = λ1 cos( ) b2 = − λ1 sin( ) c1 = λ2 cos( ) , 2 2 2 α α α c2 = −λ3 cos( ) d1 = − λ3 sin( ) d2 = − λ1 sin( ) , 2 2 2 which turns (B.6) into a12 + b12 + λ21 + λ22 + λ23 = 1. (B.8) (B.9) (B.10) (B.11) In terms of this parametrisation the 4d bispinors are given by 1 ψ+ = a1 − i λ1 − i b1 v1 ∧ v2 − (λ2 − iλ3 )v1 ∧ (w1 − i w2 ) + (i a1 + λ1 )w1 ∧ w2 + b1 v 1 ∧ v 2 ∧ w 1 ∧ w 2 , 1 = (a1 − i λ1 )v1 − i b1 v2 − (λ2 − i λ3 )(w1 − i w2 ) + (i a1 + λ1 )v1 ∧ w1 ∧ w2 ψ− + b1 v 2 ∧ w 1 ∧ w 2 , 2 = −(λ2 + iλ3 ) − (a1 + i λ1 )v1 ∧ (w1 − i w2 ) − i b1 v2 ∧ (w1 − i w2 ) ψ+ − i(λ2 + i λ3 )w1 ∧ w2 , 2 = −(λ2 + i λ3 )v1 − (a1 + i λ1 )(w1 − i w2 ) − i b1 v1 ∧ v2 ∧ (w1 − i w2 ) ψ− + (λ3 − i λ2 )v1 ∧ w1 ∧ w2 , (B.12) ψγ̂1 + = b1 − (i a1 + λ1 )v1 ∧ v2 − i (λ2 − i λ3 )v2 ∧ (w1 − i w2 ) + i b1 w1 ∧ w2 + (a1 − i λ1 )v1 ∧ v2 ∧ w1 ∧ w2 , ψγ̂1 − = −b1 v1 + (i a1 + λ1 )v2 + (λ3 + i λ2 )v1 ∧ v2 ∧ (w1 − i w2 ) − i b1 v1 ∧ w1 ∧ w2 − (a1 − i λ1 )v2 ∧ w1 ∧ w2 , ψγ̂2 + = (i λ2 − λ3 )v1 ∧ v2 − b1 v1 ∧ (w1 − i w2 ) − i(a1 + i λ1 )v2 ∧ (w1 − i w2 ) − (λ2 + i λ3 )v1 ∧ v2 ∧ w1 ∧ w2 , ψγ̂2 − = (λ3 − i λ2 )v2 + b1 (w1 − i w2 ) + i(a1 + iλ1 )v1 ∧ v2 ∧ (w1 − i w2 ) + (λ2 + i λ3 )v2 ∧ w1 ∧ w2 . 224 N.T. Macpherson et al. / Nuclear Physics B 933 (2018) 185–233 Appendix C. The SU (2) doublets of S 3 There exist two independent spinors on S 3 that obey the Killing spinor relations i ∇a ξ± = ± γa ξ± , 2 (C.1) each of which preserves two supercharges. Additionally the global isometry group of S 3 can be decomposed as SO(4) = SU (2)+ × SU (2)− , so S 3 supports two sets of SU (2) Killing vectors i , i = 1, 2, 3, that are dual to one forms that obey K± 1 dKi± ± ij k Kj± ∧ Kk± , 2 (C.2) i.e. the right-/left-invariant forms of SU (2). It is possible to use the spinors on S 3 to construct SU (2)± doublets. Consider the following vector with spinor entries  α ξ± α ξ± = . (C.3) ξ±c These transform under the action of the spinorial Lie derivative as11 i β LK ± ξ±a = ± (σi )a b ξ± , LK ± ξ∓a = 0, (C.5) i i 2 for σi the Pauli matrices, which means that ξ±a transforms as a doublet under local SU (2)± transformations and a singlet under SU (2)∓ . Appendix D. Supersymmetry conditions for three-dimensional external spacetimes In [31], supersymmetry conditions for 3+7 dimensional compactifications are given in terms of bispinors. The repackaging of the supersymmetry conditions was done under the following conditions: • The external space is Minkowski. • The spinors have equivalent length. • The NSNS flux H does not have an external component. In this section, we will look at relaxing the latter two conditions to obtain more general solutions. Our starting point will be the ten-dimensional bispinor description of the supersymmetry constraints, as described in [37]: 11 The spinorial Lie derivative along a Killing vector K is defined as 1 LK  = K μ ∇μ  + (dK)μν γ μν . 8 (C.4) The easiest way to see that this leads to the claimed transformation property, is to parametrise the vielbein on S 3 as e1 = 12 dθ, e2 = 12 sin θdφ, e3 = 12 (dψ + cos θdφ) and take the flat space gamma-matrices to be the Pauli matrices i i i 0 , ξ = e− 2 ψσ3 ξ 0 for ξ 0 constant 2d spinors. The SU (2) forms are σi . Then (C.1) is solved by ξ+ = e 2 θ σ1 e 2 φσ3 ξ+ − ± − ± i i i then precisely Ki+ = −iTr(σi dgg −1 ), Ki− = −iTr(σi g −1 dg) for g = e 2 φσ3 e 2 θ σ2 e 2 ψσ3 . The result is then not hard to show. N.T. Macpherson et al. / Nuclear Physics B 933 (2018) 185–233 225   dH e−  + K̃ ∧ F + ιK F = 0 (D.1a) dK̃ = ιK H (D.1b)  LK g10 = 0     1 =0, e+1 ·  · e+2 , MN ±dH e−  · e+2 + e d † e−2 e+2  − F 2       1 e+1 ·  · e+2 , dH e−  · e+1 − e d † e−2 e+1  − F MN = 0 , 2 (D.1c) (D.1d) (D.1e) with  = 1 ⊗ ¯2 , KM = 1 (¯1 M 1 + ¯2 M 2 ) , 32 K̃M = 1 (¯1 M 1 − ¯2 M 2 ) . 32 (D.2) The final two equations are known as the pairing equation; we refer to [37] for more details, and will follow along the lines of section 4.2 of that reference in the following. We consider the case where the Killing spinors are given by (2.2). Due to the properties of Spin(1, 2), we can define 1 1 (D.3) ζ̄ γμ ζ = vμ , ζ̄ γμν ζ = (3 v)μν , 2 2 with the other bilinears vanishing. Since we are considering flat space, ζ are covariantly constant, hence dv = 0. Making use of the spinor decomposition, it follows that 1 1 8e−A  = v ∧ ∓ − 3 v ∧ ± , K = eA (|χ1 |2 + |χ2 |2 )v , K̃ = eA (|χ1 |2 − |χ2 |2 )v, 8 8 (D.4) where we have defined + + i− = 8e−A χ1 ⊗ χ2† and K, K̃ should be read as 1-forms in ten dimensions. Using the flux decomposition F = f + e3A Vol3 ∧ 7 λ(f ) , H = H3 + e3A hVol3 . (D.5) We will first solve (D.1c): since by construction, v, K are Killing vectors, we must have |χ1 |2 + |χ2 |2 = c+ eA . (D.6) Next let us consider (D.1b), which leads to c+ e3A h = 0 , |χ1 |2 − |χ2 |2 = c− e−A . (D.7) Next, let us consider (D.1a). We find that dH3 e3A− ± = c+ e3A 7 λ(f ) dH3 e2A− ∓ = c− f . (D.8) In addition, the fact that c− = 0 does not change the argument of [37], so the pairing equations remain unchanged, leading to (f, ∓ ) = 0 . (D.9) 226 N.T. Macpherson et al. / Nuclear Physics B 933 (2018) 185–233 Appendix E. M-theory The focus of this paper are backgrounds in type IIA and type IIB. In this appendix, we will discuss M-theory backgrounds on R1,2 × S 3 × M5 . Given equivalent internal spinor norms, our (massive) IIA classification is complete (up to finding solutions to PDE). Therefore, a significant number of backgrounds one would obtain from a similar analysis of M-theory are those which one can obtain from uplifting our massless IIA backgrounds. Novel solutions from a complete M-theory analysis would be backgrounds satisfying one of the two conditions: either M5 does not admit an S 1 factor to be integrated out to perform the dimensional reduction to IIA, or the internal component of the Killing spinor on M8 = S 3 × M5 is such that after the reduction, the resulting seven-dimensional internal components of the IIA spinors are not of equal norm. Such a full M-theory classification is beyond the scope of this paper. Instead, we aim to make contact with the literature of M-theory on R1,2 × M8 , which is much studied (see for example [62–67]). We will derive the decomposed supersymmetry conditions, and give several simple classes of solutions. For N = 1 solutions to the supersymmetry constraints on R1,2 × M8 , the Killing spinor  decomposes as  = ξ ⊗ (χ+ + χ− ) , (E.1) where ξ is a Majorana spinor of Spin(1, 2) and χ± are chiral Majorana spinors of Spin(8). Generically, χ± can have zeroes, and the structure group of M8 is SO(8), although a Spin(7)-structure can be defined on the auxiliary space M8 × S 1 [66]. In the case where one of the two does not vanish, the structure group reduces to Spin(7) [65]. If both chiral spinors have no zeroes, both of them define a Spin(7)-structure: the intersection of the two leads to a reduction of the structure group to G2 .12 The reduction of the structure group leads to the existence of globally defined invariant tensors. Instead, we will work locally, and consider patches where either one or both are non-zero. E.1. Spin(7) holonomy Let us first examine the case with  = ζ ⊗ χ+ . (E.2) Following the conventions of [65], the general solution to the M-theory supersymmetry constraints with these ansätze is that ds 2 = e2 ds 2 (R1,2 ) + e− ds 2 (M8 ) , G = Vol3 ∧ d(e3 ) + F , (E.3) where ds 2 (M8 ) a metric of Spin(7) holonomy. The four-form F lies in the 27 of Spin(7), i.e., it satisfies F pqr m npqr = 0 , (E.4) with mnpq = χ̄ γmnpq χ the invariant four-form defining the Spin(7)-structure. In addition, the Bianchi identity and equation of motion for F require that F is harmonic and satisfies 12 Note that this is only the case for two Spin(7)-structures defined in terms of opposite chirality spinors. Given two same chirality globally well-defined nowhere vanishing spinors, the structure group instead reduces to Spin(6) ≃ SU (4). N.T. Macpherson et al. / Nuclear Physics B 933 (2018) 185–233 227 1 d(8 d(e−3 )) + F ∧ F = 0 (E.5) 2 away from M2-brane sources. Let us now impose M8 = S 3 × M5 . The internal metric and Killing spinor decompose as   1 2 2C 2 3 2 ⊗ (ξ ⊗ η + ξ c ⊗ ηc ) , ds (M8 ) = e ds (S ) + ds (M5 ) , χ+ = (E.6) 0 and the gamma matrices decompose as γα(8) = σ1 ⊗ σα ⊗ I , γa(8) = σ2 ⊗ I ⊗ γa , B (8) = σ3 ⊗ σ2 ⊗ B5 , (E.7) with the charge conjugation matrix satisfying B5 ∗ = −B5 , α an index on S 3 and a an index on M5 . The pseudoreal Spin(5) spinor η of unit norm gives rise to an SU (2)-structure on M5 , via [68] 1 1 η ⊗ η† = (1 + V ) ∧ e−iJ , η ⊗ ηc† = (1 + V ) ∧ ω . (E.8) 4 4 The (local) SU (2)-structure consists of a real one-form V , real two-form J and a complex twoform ω such that J ∧ J = 12 ω ∧ ω∗ and ιV J = ιV ω = J ∧ ω = ω ∧ ω = 0. By making use of this decomposition, the Spin(7) four-form decomposes in terms of the I × SU (2)-structure as 1 eC̃ =− J ∧J + V ∧ (K1 ∧ Reω + K2 ∧ Imω + K3 ∧ J ) 2 2 (E.9) e2C̃ (E.10) (dK1 ∧ Reω + dK2 ∧ Imω + dK3 ∧ J ) − νe3C̃ V ∧ Vol(S 3 ) . 4 Using the decomposed , we examine the supersymmetry conditions. First, we consider the flux component F , which we decompose as −ν F = e3C Vol(S 3 ) ∧ F1 + 5 F̃1 , (E.11) with F1 , F̃1 one-forms on M5 . Inserting this and (E.9) into (E.4), it follows from (m, n) = (a, b) ∗ = F a J = 0. Hence that F1 ∼ F̃1 , F1a Va = 0 and from (m, n) = (α, a) that F1a ωab = F1a ωab 1 ab F1 = F̃1 = 0, hence F = 0. The equation of motion for the flux (E.5) thus reduces to the following constraint on the warp factors: d(e3C 5 d(e−3 )) = 0 . (E.12) Next, the requirement that M8 is of Spin(7) holonomy is equivalent to the closure of , which is equivalent to d(e3C̃ V ) = d(e2C̃ J ) + 2νeC̃ V ∧ J = d(e2C̃ ω) + 2νeC̃ V ∧ ω = d(J ∧ J ) = 0 . (E.13) In general, this means that locally V = e−3C dτ , and we will write ds 2 (M8 ) = e2C ds 2 (S 3 ) + e−6C dτ 2 + ds42 (E.14) Let us give some simple classes of examples for which the above conditions are solved. • In the case that C = C(τ ), the metric ds42 is conformally Calabi–Yau. Let J = e2(W −C) J˜, ω = e2(W −C) ω̃. Then provided that we define W (τ ) to satisfy 228 N.T. Macpherson et al. / Nuclear Physics B 933 (2018) 185–233 W  + νe−4C = 0 , (E.15) we find that d J˜ = d ω̃ = 0. • By taking C = − 14 log(4τ ), we find that dJ = dω = 0, hence ds42 is a Calabi–Yau metric.13 Introducing ρ = (4τ )1/4 , the metric reduces to ds 2 (M8 ) = dρ 2 + ρ 2 e2C ds 2 (S 3 ) + ds42 , (E.16) and thus M8 = R4 × Y2 , where Y2 is a Calabi–Yau two-fold. • Next, let us examine Sasaki–Einstein structures, as well as a class of generalizations. It can be shown that any five-dimensional Sasaki–Einstein can be defined by means of a set of real forms (Ṽ , ωj ), j = 1, 2, 3 with Ṽ a one-form and ωj two-forms. These satisfy [15] dω1 = d(ω2 + iω3 ) + 3i Ṽ ∧ (ω2 + iω3 ) = 0 . (E.17) A more general class of spaces are the so-called hypo manifolds [69], satisfying dω1 = d(Ṽ ∧ ω2 ) = d(Ṽ ∧ ω3 ) = 0, which themselves are a subclass of balanced manifolds [70], satisfying d(ω1 ∧ ω1 ) = d(Ṽ ∧ ω2 ) = d(Ṽ ∧ ω3 ) = 0 . (E.18) By setting J = ω1 , ec V = Ṽ , Reω = ω2 , Imω = ω3 , it follows that any solution to the supersymmetry constraints is a balanced metric. On the other hand, any solution to the supersymmetry constraints which is hypo automatically is such that ds42 is Calabi–Yau. This leads to the conclusion that the spinors do not define a Sasaki–Einstein on M5 , as the base space of a Sasaki–Einstein manifold is not Ricci-flat. Another way to see this is to note that Sasaki–Einstein metrics can be written as a fibration over a Kähler–Einstein base, but it is clear that since the supersymmetry constraints are invariant under permutations of (J, Reω, Imω), ds42 cannot be non-Calabi–Yau Kähler. E.2. G2 -structure Next, let us examine the case where both internal chiral Killing spinors are (locally) nonvanishing. Again following [65], the Killing spinor is given by  = e− θ ⊗ (χ+ + χ− ) , leading to the solution   2 2 2 1,2 2 ds = e ds (R ) + ds (M8 ) , G = e3 (Vol3 ∧ f + F ). (E.19) (E.20) This time, the metric is not of special holonomy. Instead, it allows a G2 -structure with non-trivial torsion. The norms of χ± can be parametrised as |χ± |2 = 1 ± sin ζ , (E.21) with sin ζ a function of M8 such that the norms of χ± are non-vanishing. The bilinears of χ± defining the G2 -structure are given by 13 We have redefined τ to absorb the sign ν. N.T. Macpherson et al. / Nuclear Physics B 933 (2018) 185–233 Km = 1 1 χ+† γm(8) χ− , ϕmnp = χ † γ (8) χ− . cos ζ cos ζ + mnp 229 (E.22) In terms of these, the constraints d(e3 cos ζ K) = 0 , (E.23a) K ∧ d(e6 ιK 8 ϕ) = 0 , (E.23b) d(e12 cos ζ ϕ ∧ ιK 8 ϕ) = 0 , (E.23c) cos ζ d(ϕ) ∧ ϕ + 4 8 dζ − 2 cos ζ 8 f = 0 , (E.23d) d(e3 sin ζ ) − e3 f = d(e6 cos ζ ϕ) + e6 8 F − e6 sin ζ F = 0, (E.23e) are locally equivalent to the supersymmetry conditions. Next, we split M8 = S 3 × M5 , leading to the following decomposition of the metric and flux: ds 2 (M8 ) = e2C ds 2 (S 3 ) + ds 2 (M5 ) , F = e3C F1 ∧ Vol(S 3 ) + F4 . The spinors decompose as   1 χ+ = 1 + sin ζ ⊗ (ξ ⊗ η1 + ξ c ⊗ η1c ) , χ− = 0 (E.24)   0 1 − sin ζ ⊗ (ξ ⊗ η2 + ξ c ⊗ η2c ) 1 (E.25) and the gamma matrices again decompose as (E.7). The Spin(5) spinors can be expanded in a common basis as b η1 = η, η2 = a0 η + aηc + wη, |a0 |2 + |a|2 + |b|2 = 1 (E.26) 2 where b can be made real by rotating the 1-form w and η is unit norm. We assume w = w1 + iw2 is locally non-vanishing. As a result, a second locally non-vanishing 1-form can be defined as 1 (E.27) u = ιw∗ ω , 2 with ω defined as in (E.8). We thus see that the local SU (2)-structure defined on M5 by η reduces further to a trivial structure, with the local vielbein defined by (V , w1, w2 , u1 , u2 ). We now express the G2 -structure (E.22) in terms of the trivial structure of S 3 × M5 . The first bilinear we calculate is eC (Ima0 K1 + Rea0 K2 − ImaK3 ) − bu2 − ReaV , 2 and the only way to make this compatible with (E.23a) is to set K= (E.28) a0 = Ima = 0 (E.29) so that the spinors η1,2 are nowhere parallel. We are now free to parametrise b = cos α, Rea = sin α and rotate to a frame where (E.30) 230 N.T. Macpherson et al. / Nuclear Physics B 933 (2018) 185–233 K = V. (E.31) Having done this the other bilinears take the form ϕ = − cos αe1 ∧ e2 ∧ e3 − e3C sin αVol(S 3 ) − ν − e2C cos αei ∧ dKi , 4 1 eC Ki ∧ (u1 ∧ ei + ij k sin α ej ∧ ek ), 2 2 ιK 8 ϕ = u1 ∧ (sin αe1 ∧ e2 ∧ e3 − e3C cos α ∧ Vol(S 3 )) +ν e2C 1 eC (sin α u1 ∧ ei + ij k ej ∧ ek ) ∧ dKi − cos αij k u1 ∧ ej ∧ ek ∧ Ki , 4 2 4 (E.32) where we have defined e = (w1 , w2 , − u2 ), (E.33) for ease of presentation. Remark that ϕ ∧ ιK 8 ϕ = 7ν Vol7 , (E.34) where Vol7 is the volume form of the manifold spanned by the warped left-invariant forms of S 3 and the vielbein, with orientation  C  eC eC e K1 , K2 , K3 , u1 , e1 , e2 , e3 . 2 2 2 Inserting these definitions for ϕ and ιK 8 ϕ into (E.23a)–(E.23e) lead to the 5d conditions d(e3 cos ζ V ) = d(e6+3C cos αu1 ) ∧ V = d( e−6−3C u1 ) ∧ e 1 ∧ e 2 ∧ e 3 = 0 , cos3 α cos2 ζ (E.35a) d(e6+2C cos ζ cos αei ) + ν e6+C cos ζ (2u1 ∧ ei + sin αij k ej ∧ ek ) = 0 , (E.35b)   1 d(e6+2C (sin αu1 ∧ ei + ij k ej ∧ ek )) + νe6+C cos αij k u1 ∧ ej ∧ ek ∧ V = 0 , 2 (E.35c) d(e−2C cos αij k ej ) ∧ ek ∧ u1 ∧ V = ij k u1 ∧ du1 ∧ ej ∧ ek = 0 , (E.35d) − 2 cos ζ dα ∧ e1 ∧ e2 ∧ e3 + 2 5 dζ − cos ζ 5 f = d(e3 sin ζ ) − e3 f = 0 , (E.35e) d(e6+3C cos ζ sin α) + e6+3C (− 5 F4 + sin ζ F1 ) = 0 , (E.35f) d(e6 cos ζ cos αe1 ∧ e2 ∧ e3 ) + e6 (5 F1 − sin ζ F4 ) = 0 , (E.35g) where we have used that cos ζ = 0 and one can show that cos α = 0 is inconsistent with supersymmetry. 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