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HETEROTIC INSTANTONS AND SOLITONS

arXiv:hep-th/9110070v1 29 Oct 1991CERN-TH 6286/91DFTT 42/91HETEROTIC INSTANTONS AND SOLITONSIN ANOMALY-FREE SUPERGRAVITYI. Pesando1 2Dipartimento di Fisica Teorica dell’Universit`a di TorinoInstituto Nazionale di Fisica Nucleare, Sezione di Torinovia P.Giuria 1, I-10125 Turin, ItalyA.

K. Tollst´en3CERN, CH-1211 Geneva 23, SwitzerlandAbstractWe extend the classical heterotic instanton solutions to all orders in α′using the equations of anomaly-free supergravity, and discuss the relationbetween these equations and the string theory β-functions.CERN-TH 6286/91DFTT 42/91October 19911Work supported in part by Ministero dell’Universit`a e della Ricerca Scientifica e Tecnologica2email I PESANDO@TO.INFN.IT3email TOLLSTEN@CERNVM

1IntroductionDuring the past few years the study of classical (low energy) equations for 10-dimensional superstrings has yielded a number of interesting solutions, see forinstance [1, 2, 3, 4, 5, 6, 7]. The standard method of finding these is to solvethe equations of motion of ordinary supersymmetric Einstein-Yang-Mills theory(augmented by the Lorentz Chern-Simons term in the definition of the antisym-metric tensor), which equal the string β-functions to the lowest order.

To showthat a solution obtained this way is also a solution to string theory, one then triesto construct the corresponding superconformal sigma-model.However, from the space-time point of view it might appear somewhat un-satisfactory to look for supersymmetric solutions using a set of manifestly non-supersymmetric equations. A supersymmetrization of the coupled Einstein-Yang-Mills theory including the Lorentz Chern-Simons term would also naturally ex-tend the equations to higher orders in α′, which is interesting in itself.

Such asupersymmetrization has actually already been performed, both in the normalcase (with a 3-form H) [8, 9, 10] using an important observation by Bonora,Pasti and Tonin [11], and in the dual case [12]. The equations of this model,which we call the Anomaly-Free Supergravity (AFS), give α′ corrections to thestandard equations, leading to implicit equations for the physical fields.

In thegeneral case one must then expand to all orders in α′. AFS is hence a classicallyconsistent, all orders in α′ (on-shell) supersymmetric theory which incorporatesthe Green-Schwarz condition for anomaly cancellations [13].Fortunately, for the purpose of generalizing the classical solutions of [2, 3], itwill turn out to be sufficient to use the original implicit equations.

This is whatis done in this paper. We find a solution to the AFS equations in a closed formcontaining only O(α′0) and O(α′1) terms.

This solution, which is really a familyof solutions containing the ones by Callan, Harvey, and Strominger referred toabove, consists of a non-linear differential equation for the field appearing in themetric, which, in general, might have to be solved as an expansion in α′. Alsothe relation between the dilaton and the metric contains derivatives.It should perhaps be pointed out here that although AFS agrees with theeffective string theory to lowest order and contains terms to all orders, it isextremely unlikely that it will turn out to be equivalent to the massless effectivestring theory.

It does not incorporate the ζ(3) terms in an obvious fashion, and ithas been shown that AFS can at least in principle be extended to a non-minimalversion containing extra representations which can accommodate such terms [14].We believe, however, that the minimal AFS provides a better approximation tostring theory than the one normally used, and, as is argued at the end of thispaper, it might even provide a necessary condition for a solution to be a solutionof string theory.1

2The instanton solutionWe will here study the generalization to AFS of the heterotic five-brane solutionby Strominger [2]. We follow his calculation closely, only making a slightly moregeneral ansatz.

As in the lowest order case, the solution also turns out to incorpo-rate the wormhole solution [3, 4] related to the solution with a five-brane source `ala Duffand Lu [7]. Since AFS corrections are rather complicated, we find it mostconvenient to work directly in the variables of [10], instead of performing the fieldredefinitions [15, 2] to obtain the σ-model variables and a flat five-brane metric.We use mainly the conventions of [9, 10].

For instance, gMN = (+, −, ..., −),{γM, γN} = 2gMN and a p-form is defined as ω(p) = ωM1...Mp dzM1 ∧... ∧dzM2.However, to make comparison easier, we use the index conventions of [2], thatis, M, N, P, ... are 10-dimensional space-time indices, and A, B, C... are the cor-responding tangent space indices.We now want to find a maximally symmetric, supersymmetric solution, thatis, we want a solution to the AFS equations with all spinorial fields equal to zero,and with a (non-zero) Majorana-Weyl spinor ǫ satisfying [9, 10]δψM = DMǫ + 136γMγA1A2A3ǫ T A1A2A3 = 0,(1a)δλ = −2i γAǫ ∂Aφ + i γA1A2A3ǫ ZA1A2A3 = 0,(1b)δχ = −14γA1A2ǫ F A1A2 = 0. (1c)To the lowest order, T A1A2A3 and ZA1A2A3 in the gravitino and dilatino transfor-mation equations are both proportional to HA1A2A3.

This is no longer the casein AFS. Here instead, we have the torsion 4TA1A2A3 = −3 e−43φHA1A2A3 −2γ1 e−43φWA1A2A3(2)where γ1 ∼α′ andWA1A2A3 = 12TA1A2A3 + 3 TB1B2[A1RB1B2A2A3] + 3 TB1[A1A2RB1A3]+4 TB1B2[A1T B2B3A2TB1A3]B3 −( 227 + h1)TA1A2A3T 2(3)with h1 a free parameter reabsorbable in a redefinition of the dilaton:e43 φ′= e43φ −2h1γ1 T 2(4)4Here, as in the following, we drop all fermionic terms.2

It is this torsion which turns up in covariant derivatives and the curvature tensor;D = D(Ω), R = R(Ω), and Ω= ω + T, and which also occurs in the gravitinotransformation law, while in the gaugino transformation law we haveZA1A2A3 = 16 TA1A2A3 + 6γ1 e−43φW (5)A1A2A3(5)withW (5)A1A2A3 = 136TA1A2A3 −(19 + 32h1) TB1B2[A1RB1B2A2A3] +(29 +3h1) TB1[A1A2RB1A3]+ 136DB1TB2[A1A2TB1B2A3]+ (14 + 3h1)T B1B2[A1DA2TA3]B1B2−( 554 + 53h1) 15! ǫA1A2A3B1B2B3C1C2C3C4T B1B2B3DC1T C2C3C4−( 518 + 356 h1) 15!

ǫA1A2A3B1B2B3C1C2C3C4T B1B2B3T DC1C2T C3C4D+(59 + 5h1) TB1B2[A1T B2B3A2TB1A3]B3 + ( 518 + 52h1) TB1[A1A2TA3]B2B3T B1B2B3−( 1108 + 736h1) TA1A2A3T 2. (6)Obviously the field redefinitions in for instance [15] would lead to a rather longcalculation which is not needed for the present purpose.

If we further demandthat the solution fulfilD[MHNP Q] = −4 Tr(F[MNFP Q]) −γ1 Tr(R[MNRP Q]),(7)with the traces defined just as the sum over the group indices, we will also auto-matically satisfy the bosonic equations of motion [10]. In order to find a five-branesolution we split up space-time intozM −→(ya, xµ);a = 0, 1, ..5;µ = 6, ..9(8)and assume a metric of the formgMN =e2A−e2A...−e2A−e2B...−e2B.

(9)3

Here A = A(r) and B = B(r) are arbitrary scalar fields which, as well as all thefields in the following, depend only on r = (δµνxµxν)12. Our strategy will now beto solve (1a)-(1c) with ZA1A2A3 and TA1A2A3 regarded as independent fields andonly put the solution into (5) afterwards.We start by studying the dilatino equation (1b).

Just like in [2] it can besolved by defining chiral spinors 51√g6ǫa1...a6γa1...a6ǫ± = ±6! ǫ±1√g4ǫµ1...µ4γµ1...µ6ǫ± = ±4!

ǫ±(10)with g6 = −det(gab) = e12A, g4 = det(gµν) = e8B, and by puttingZµ1µ2µ3 ∼ǫνµ1µ2µ3 ∂νφ(r) eC(r),(11)and the rest of the components to zero. We then immediately findZ±µ1µ2µ3 = ∓13ǫνµ1µ2µ3 ∂νφ(r) e−4B.

(12)Note that the factor e−4B =1√g4 is exactly what is needed to make Z±µ1µ2µ3 atensor in x-space. Proceeding to the gravitino equation, we make a similar, butindependent ansatz:Tµ1µ2µ3 = ǫνµ1µ2µ3 ∂νD(r) eE(r).

(13)The M = a component of (1c) is then0 = ∂aǫ± + 12γaγµǫ± ∂µA ∓16γaγµǫ± ∂µD eE+4B(14)which can be solved by making ǫ independent of ya, D± = ±3A±(+constant),and E = −4B. For M = µ we get0 = ∂µǫ± ∓16ǫ± ∂µD + 12γµνǫ± ∂ν(B ± 23D).

(15)A solution isǫ± = eA/2η±(16)with η± constant chiral and antichiral spinors,B = −2A + constant,(17)5The ǫ...’s are here defined as tensor densities;1√g ǫ... are the proper tensors in respectivespace.4

and henceTµ1µ2µ3 = ±ǫνµ1µ2µ3 ∂νA e8A. (18)The constant in (17) can be absorbed in a constant rescaling of the coordinatesand is dropped below.

The gaugino equation (1c) is now directly solved by the(anti)instanton configurationFµν = ±12ǫρσµνFρσe8A. (19)In the γ1 = 0 case we would have had Zµνρ ∼Tµνρ and hence directly A ∼φ.For γ1 ̸= 0 we have to insert our solution into (5).

After a straightforward, butcumbersome calculation, we find that onlyW (5)µ1µ2µ3 = ±ǫνµ1µ2µ3 e8A [ 112∂ρ∂ρ∂νA + 12∂ρ∂ρA∂νA−(12 + 9h1)∂ρA∂ρ∂νA −(32 + 27h1)∂ρA∂ρA∂νA](20)is different from zero. Using (20) and contracting with ǫµ1µ2µ3µe8A we get∂µφ = −32∂µA −18γ1e−43 φ[ 112∂ρ∂ρ∂µA + 12∂ρ∂ρA∂µA−(12 + 9h1)∂ρA∂ρ∂µA −(32 + 27h1)∂ρA∂ρA∂µA](21)which can directly be integrated toe−43φ+2A = k −2γ1e2A(∂µ∂µA −3(1 + 18h1)∂µA∂µA)(22)with k constant.

For γ1 = 0 we have to choose k > 0, and it can be put equal toone by once more rescaling the coordinates. However, in the AFS case, there arealso solutions for k ≤0 as we shall see below, so we choose to keep k as a freeparameter.

Finally, our solution has to satisfy (7). Again we find that only the[µνρσ] component is different from zero, and it yields∂µ{∂µAe−43φ+2A + γ1e−4A[∂ν∂ν∂µA + 2∂ν∂νA∂µA−6(1 + 18h1)∂νA∂νA∂µA]} = −43e−4ATr(FµνF µν)(23)We insert (22), and use∂µ = −e4A∂µF µν = e8AFµν(24)and obtain5

∇2[k e−6A + 6γ1 ∇2A] = −8 Tr F 2(25)Here ∇2 =1r3∂∂rr3 ∂∂r is the Laplacian in four-dimensional Euclidean space.Comparing the expression we use to those of other authors, see for instance[16, 17] and also [15, 2], we find agreement for α′ =κ22g2 and γ1 = −2α′ and, inparticular,Tr(FµνF µν) =18 · 30α′ Tr(FµνF µν)Strominger(26)Hence, our solutions are exactly the same (as they should be ) for γ1 = 0 (φ = −12φS ). The general solution given our ansatz is then any A(r) and φ(r)satisfyingk e−6A + 6γ1 ∇2A = k e−6A0 + k′r2 + 8α′r2 + 2ρ2(r2 + ρ2)2(27a)e43φ+2A = k −2γ1 e2A(∇2A −3(1 + 18h1) (∇A)2)(27b)together with (9), (12), (17), and (18).

As a consistency check, as well of [10] asof the calculations above, the equations of motions were also studied, and foundto be linear combinations of (derivatives of) the equations (27).3Analysis of the solution and discussion.In equations (27) we have three, so far arbitrary, integration constants. However,the solution is not physical, and might not even exist for all values of k, k′, andA0.

For instance, the constants have to be chosen so that A and φ are real. Wewill here first restrict ourselves to a discussion of a few cases already mentioned inthe literature.

All these solutions can, of course, be extended to multi-instantonor multi-wormhole solutions in the standard fashion. Afterwards, we will giveexamples of other exact, but mainly unphysical, solutions.

The instanton solution[2] (gauge solution in [3, 4]) has k′ = 0. If we assume that A can be written as apower series in 1r at infinity, we findA = A0 + A1r2 + O 1r4,r →∞.

(28)The γ1 part gives then no contribution at infinity and the calculation of mass,axion charge, and the Bogomoln’yi bound still give the same result as in the paperby Strominger. Furthermore, since we still have the same continuous symmetries,the zero-modes should remain unchanged, and a solution of this form giving areal φ is then indeed an extension of [2] to all orders in α′.If we instead let ρ →0 we obtain the generalization of the neutral solution[3], related to [7].

Some care must be taken in this case since it is not obvious6

that taking the limit ρ →0 in (27) and then solving for A and φ will give thesame result as the other way round.Perhaps the most important example, however, is that of the symmetric solu-tion of [3, 4], which is argued to have no higher order corrections. In order to beable to define a Lorentz connection which equals the Yang Mills potential we usethe “original” version of the instanton potential, see e.g.

[18]. In analogy withthe authors quoted above we thus putAµ = ¯Σµν ∂ν log1 + ρ2r2= −6 ¯Σµν ∂νA(29)with ρ2 = nα′e6A0.

That isA = A0 −16 log1 + ρ2r2. (30)Inserting this into (27a) we getk e−6A01 + ρ2r2−8α′ρ4r2(r2 + ρ2)2 = k e−6A0 + k′r2 + 8α′ r2 + 2ρ2(r2 + ρ2)2,(31)which is satisfied if we choosek′ = (nk −8)α′.

(32)Since A also satisfies∇2A = 6 (∇A)2,(33)we can eliminate the correction term in (27b) if we choose the parameter of AFSh1 = 118. (34)We obtainφ = −32A + constant(35)The symmetric solution is hence a solution also to AFS for the choice of h1 in(34).

Since a particular choice of h1 just corresponds to a field redefinition (4),this value of h1 must give the same choice of φ as in the references above.So far, the symmetric solution is the only one we have given explicitly, onlyassuming that there exist well-behaved solutions of (27) of the neutral and gaugetype too, albeit not in a closed form. The symmetric solution is, however, notthe only example of a simple solution of (27), although the others we have founddo not, in general, have an immediate physical interpretation.Both for the7

symmetric solution and for these new ones, we have cancellations between tr F 2and tr R2 so they do not have a proper limit as γ1 →0. They also have k ≤0.In (27a) we have already implicitly assumed that A has a well-defined value,A0, as r →∞, so that the metric is Minkowski at infinity, and that k ̸= 0.

Wenow relax these constraints and put k e−6A0 = k′′. For k′′ = 0 we find the solutionA = A0 + 13 log1 + r2ρ2+ 13 logrρ,(36)with k < 0 and eφ < 0 if we assume (34).Putting k = 0 we can also addnon-logarithmic terms to A, and we find another solutionA = A0 + A2r2 −16 log1 + r2ρ2−k′24α′ logrρ−k′′96α′r2,(37)which has two free integration constants, A0 and A2, and is hence the generalsolution for k = 0.

In order to remove the essential singularities, we must chooseA2 and k′′ as zero, and certain values of k′ might then yield interesting solutions.The effective Lagrangian of string theory should also contain higher orderterms multiplied by the transcendental coefficient ζ(3) [19]. It is very hard toimagine how these could occur within the framework of AFS, although it hasbeen suggested that they might depend on the boundary conditions chosen whensolving (3) to construct an effective Lagrangian [9].

Another, perhaps more likely,source is to note [14] that AFS as used here is a minimal supersymmetric extensionof the anomaly-free Einstein-Yang-Mills theory, and that it is possible to extendit, by relaxing the constraints used in solving the Bianchi identities. One can thenaccommodate precisely the superfield needed [20] for the ζ(4) R4-terms.

Theymight then act as counterterms, the precise value of the coefficients being decidedfrom cancellation of divergences, as suggested in [21].To correspond to the string β-functions, all equations of AFS should then beaugmented by ζ(3) terms which, for simple non-transcendental solutions such asthe ones given above, have to be satisfied separately. Since it is argued, using theσ-model approach, that the symmetric solution is really a solution to the string[3, 4], we can assume that the ζ(3) equations are indeed satisfied separately in thiscase.

Hence, a necessary condition for all “normal” classical solutions to stringtheory should be that they satisfy the AFS equations. It would be interesting tostudy other solutions like the black five-brane one [6], and also to search for newexotic compactifications, in this framework.

It would of course also be interestingto derive the full, non-minimal AFS, and introduce the right ζ(3) coefficients, butjudging from the derivation of the equations of motion for AFS [10] this mightrequire an unrealistic amount of computing power.8

Acknowledgement We thank Dieter L¨ust for useful discussions.References[1] A. Dabholkar, G. Gibbons, J. A. Harvey and F. R. Ruiz, Nucl.

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