Stabilising the supersymmetric Standard Model on the Z_6' orientifold
aa r X i v : . [ h e p - t h ] A ug Stabilising the supersymmetric Standard Modelon the Z ′ orientifold December 6, 2018
David Bailin & Alex Love Department of Physics & AstronomyUniversity of SussexBrighton BN1 9QH, U.K.
Abstract
Four stacks of intersecting supersymmetric fractional D6-branes on the Z ′ orientifold have previ-ously been used to construct consistent models having the spectrum of the supersymmetric StandardModel, including a single pair of Higgs doublets, plus three right-chiral neutrino singlets. How-ever, various moduli, K¨ahler moduli and complex-structure moduli, twisted and untwisted, remainunfixed. Further, some of the Yukawa couplings needed to generated quark and lepton masses areforbidden by a residual global symmetry of the model. In this paper we study the stabilisation ofmoduli using background fluxes, and show that the moduli may be stabilised within the K¨ahler cone.In principle, missing Yukawa couplings may be restored, albeit with a coupling that is suppressedby non-perturbative effects, by the use Euclidean D2-branes that are pointlike in spacetime, i.e. E2-instantons. However, for the models under investigation, we show that this is not possible. [email protected] Introduction
The attraction of using intersecting D6-branes in a bottom-up approach to constructing the StandardModel is by now well known [1], and indeed models having just the spectrum of the Standard Modelhave been constructed [2, 3]. The four stacks of D6-branes wrap 3-cycles of an orientifold T / Ω , wherethe six extra spatial dimensions are assumed to be compactified on a 6-torus T and Ω is the world-sheet parity operator; the use of an orientifold is essential to avoid the appearance of additional vector-like matter. However, non-supersymmetric intersecting-brane models lead to flavour-changing neutral-current (FCNC) processes induced by stringy instantons that can only be suppressed to levels consistentwith current bounds by choosing a high string scale, of order TeV, which in turn leads to fine tuningproblems [4]. It is therefore natural, and in any case of interest in its own right, to construct intersecting-brane models of the supersymmetric Standard Model. A supersymmetric theory is not obliged to havea low string scale, so the instanton-induced FCNC processes may be reduced to rates well below theexperimental bounds by choosing a sufficiently high string scale without inducing fine-tuning problems.To construct a supersymmetric theory [5, 6, 7, 8, 9], instead of T , one starts with an orbifold T /P ,where P is a point group which acts as an automorphism of the lattice defining T ; this has the addedadvantage of fixing (some of) the complex structure moduli. An orientifold is then constructed as beforeby quotienting the orbifold with the action of the world-sheet parity operator Ω . In previous papers[10, 11, 12] we have studied orientifolds with the point group P = Z ′ , and derived models having thespectrum of the supersymmetric Standard Model plus three right-chiral neutrinos. The 6-torus factorisesinto three 2-tori T = T × T × T , with T k ( k = 1 , , parametrised by the complex coordinate z k .Then the generator θ of the point group P = Z ′ acts on z k as θz k = e πiv k z k (1)where ( v , v , v ) = 16 (1 , , − (2)This requires that T and T are SU (3) root lattices, so that θ is an automorphism, and this in turnfixes the complex structure moduli U , for T , to be U = U = e iπ/ ≡ α . (Note that the G and SU (3) root lattices are the same, contrary to our previous assertions.) Since θ acts on T as a reflection, θz = − z , its lattice is arbitrary. The embedding R of Ω acts antilinearly on all z k and we may choosethe phases so that R z k = ¯ z k ( k = 1 , , (3)This too must be an automorphism of the lattice, and this requires the fundamental domain of each torus T k to be in one of two orientations, denoted A and B , relative to the Re z k axis. In the A orientation of T the basis vector e = R is real, whereas in the B orientation e = R e − iπ/ ; for both orientationsthe second basis vector e = αe . Similarly for the basis vectors e and e of T . For T , the basisvector e = R is real in both orientations, but the real part of the complex structure U ≡ e /e satisfies Re U = 0 in the A orientation, and Re U = 1 / in the B orientation. Thus e = iR Im U in A , and e = R (1 / i Im U ) in B .The models having the spectrum of the supersymmetric Standard Model, with which we are con-cerned in this paper, arise only in the AAA and
BAA orientations. They include four stacks of (super-symmetric) fractional D6-branes, each wrapping the three large spatial dimensions and a 3-cycle of thegeneral form κ = 12 (cid:16) Π bulk κ + Π ex κ (cid:17) (4)where Π bulk κ = X p =1 , , , A κp ρ p (5)is an untwisted point-group invariant bulk 3-cycle, and Π ex κ = X j =1 , , , ( α j ǫ j + ˜ α j ˜ ǫ j ) (6)2s an exceptional cycle. The four basis bulk 3-cycles ρ p ( p = 1 , , , and their bulk coefficients A κp are defined in [10], the latter being expressed in terms of the wrapping numbers ( n κk , m κk ) of the basis1-cycles π k − , π k of T k ( k = 1 , , . θ acts as a Z reflection in T and T and therefore has sixteenfixed points at f i,j = 12 ( σ e + σ e ) ⊗
12 ( σ e + σ e ) (7)where σ , , , = 0 , , and we use Honecker’s [6, 13] notation in which i, j = 1 , , , correspondto the pairs ( σ , σ ) or ( σ , σ )1 ∼ (0 , , ∼ (1 , , ∼ (0 , , ∼ (1 , (8)The eight exceptional 3-cycles ǫ j , ˜ ǫ j ( j = 1 , , , and their coefficients α κj , ˜ α κj are defined (in [10]) interms of collapsed 2-cycles at f i,j times a 1-cycle in T , the coefficients being determined by the wrap-ping numbers ( n κ , m κ ) of the basis 1-cycles π , π of T . Supersymmetry requires that the bulk part ofthe fractional brane passes through the fixed points associated with the exceptional piece. If, for exam-ple, ( n κ , m κ ) = (1 ,
0) mod 2 , then, depending on the choice of Wilson lines, only the exceptional cycleswith α , , ˜ α , or α , , ˜ α , non-zero are allowed; similarly, for the (0 ,
1) mod 2 case, only α , , ˜ α , or α , , ˜ α , may be non-zero, and for the (1 ,
1) mod 2 case, only α , , ˜ α , or α , , ˜ α , may be non-zero.Orientifold invariance requires that we also include D6-branes wrapping the orientifold image κ ′ ≡ R κ of each 3-cycle κ , and the action of R on the basis 3-cycles ρ p , ǫ j , ˜ ǫ j is also given in [10]. The preciseform of the 3-cycles associated with the four stacks is given in [11, 12] and need not concern us for thepresent. D6-branes carry Ramond-Ramond (RR) charge and are coupled electrically to the 7-form RRgauge potential C . So too is the O6-plane, a topological defect associated with the orientifold actionwhich has − units of RR charge.The massive version of the effective supergravity describing compactified type IIA string theory inthe presence of background fluxes has action [14, 15] S IIA = 12 κ Z d x √− g (cid:18) e − φ [ R + 4( ∂φ ) − | H | ] − [ | F | + | F | + m ] (cid:19) − κ Z (cid:16) B ∧ dC ∧ dC + 2 B ∧ dC ∧ F bg + C ∧ H bg ∧ dC − m B ∧ B ∧ B ∧ dC + m B ∧ B ∧ B ∧ B ∧ B (cid:19) − µ X κ N κ Z M × κ d ξ e − φ √− g + √ µ X κ N κ Z M × κ C (9)where κ = (2 π ) α ′ is the 10-dimensional Newtonian gravitational constant and µ = (2 π ) − α ′− / is the unit of D6-brane RR charge. The sum over κ is understood to include all D6-brane stacks, theirorientifold images κ ′ , and the O6 -brane π O6 with charge − µ , and N κ is the number of D6-branes inthe stack wrapping the 3-cycle κ . The field strengths associated with the Kalb-Ramond field B and theRR fields C , are H = dB + H bg (10) F = dC + m B (11) F = dC + F bg − C ∧ H − m B ∧ B (12)where H bg and F bg are background fluxes, and the mass m is the background value of F . The pres-ence of the fluxes generally deforms the original metric. The direct product of the four-dimensionalMinkowski space and the compactified (Calabi-Yau) space is replaced by a warped product [16, 17]which, as we shall see, introduces a potential for (some of) the moduli. dC is the Hodge dual of F , thefield strength associated with the 7-form RR gauge field C . One effect of the m term is that a piece ofthe F ∧ ∗ F term in (9) couples H bg to C F ∧ ∗ F ⊃ m H bg ∧ C (13)3o that this term also contributes to the C tadpole equation. The requirement that there are no RR C tadpoles is therefore generalised [18] to µ X κ N κ ( κ + κ ′ ) − O6 ! + 14 κ Π m H bg = 0 (14)where Π m H bg is the 3-cycle of which m H bg is the Poincar´e dual. In the models presented in [11, 12]tadpole cancellation requires that Π m H bg , and hence m H bg , is non-zero.In general, we must also address the question of whether the total K-theory charge [19] is zero.The presence of K-theory charge may be exhibited by the introduction of a “probe” Sp (2) ≃ SU (2) brane π probe . For a consistent theory we require that there are an even number of chiral fermions in thefundamental representation of Sp (2) . Thus the additional constraint [20, 8] is that X κ N κ κ ∩ π probe = 0 mod 2 (15)where the sum is over all D6-branes, but not including their orientifold images, and π probe is any 3-cyclethat is its own orientifold image π probe = π probe ′ (16)although this may be too strong a constraint. It follows that [20, 8] π probe = 12 (cid:16) Π bulkprobe + Π exprobe (cid:17) (17)where, on the AAA lattice, Π bulkprobe = A ρ + A ( ρ + 2 ρ ) (18) Π exprobe = X j =1 , , , ˜ α j (2 ǫ j + ˜ ǫ j ) (19)The two independent (supersymmetric) possibilities are A p = (1 , , ,
0) ˜ α j = t (1 , t , ,
0) or t (0 , , , t ) (20) or A p = (0 , , ,
2) ˜ α j = t (1 , , t ,
0) or t (0 , , , t ) (21)with t , t = ± . In our models, in particular in the model deriving from the fourth entry in Table 1of reference [12], the contributions to the left-hand side of (15) from the stacks b and c are necessarily even , the former because N b = 2 , and the latter because it is zero. For the remaining stacks, we findthat a ∩ π probe = − d ∩ π probe for both cases (20) and (21) above. Thus the K-theory constraint (15) is satisfied. The same is true of the other models on the AAA lattice, as well as for the
BAA cases too.All of the models that we have considered have the attractive feature that they have the spectrum ofthe supersymmetric Standard Model, including a single pair of Higgs doublets, plus three right-chiralneutrino singlets. In the presence of these suitably chosen background fields m and H bg the models are consistent string theory vacua. Nevertheless, despite the attraction of having “realistic” spectra, they aredeficient. First, there are many unfixed moduli, K¨ahler moduli, complex structure moduli, axions and thedilaton, all of which have unobserved massless quanta unless they are stabilised. We shall see later thatthe non-zero background flux H bg required by tadpole cancellation stabilises one linear combinationof the (axion) moduli. Tadpole cancellation generally ensures the absence of anomalous U (1) gaugesymmetries in the models; the associated gauge boson acquires a string-scale mass via the generalisedGreen-Schwarz mechanism, and the U (1) survives only as a global symmetry. However, some of thesurviving global symmetries forbid the Yukawa couplings required to generate mass terms for some ofthe quarks and leptons. This is the second deficiency of these models. Further, as noted previously in[11], there is a surviving unwanted U (1) B − L gauge symmetry, associated with baryon number B minuslepton number L . In addition, in all of the models that we constructed, the U (1) stack associated withthe fractional 3-cycle c has the property that c = c ′ , where c ′ is the orientifold image of c . This means4hat the U (1) c gauge symmetry is enhanced to SP (2) = SU (2) , so that the models actually have assurviving gauge symmetry group SU (3) colour × SU (2) L × SU (2) R × U (1) B − L . The weak hyperchargeis given by Y = ( B − L ) + T R , and the matter is in the following representations ( n , n L , n R ) B − L of SU (3) colour × SU (2) L × SU (2) R × U (1) B − L : Q L = ( , , ) (22) q cL = (¯ , , ) − (23) L = ( , , ) − (24) ℓ cL , ν cL = ( , , ) (25) H u,d = ( , , ) (26)In addition, the models we have constructed cannot yield gauge coupling constant unification. A stack κ gives rise to a gauge group factor with coupling constant g κ given [21, 22] by α κ ≡ πg κ = m Vol( κ )(2 π ) g string K κ (27)where Vol( κ ) is the volume of the 3-cycle κ and K κ = 1 for a U ( N κ ) stack. The consistency of ourtreatment with the supergravity approximation requires that the contribution of the bulk part of the frac-tional 3-cycle Vol(Π bulk κ ) to Vol( κ ) is large compared to the contribution from the exceptional part Vol(Π ex κ ) , so we need only consider the former in evaluation g κ . As derived in [10], for a supersym-metric stack, the quantity Z κ = e e e [ A κ − A κ + U ( A κ − A κ ) + e iπ/ ( A κ + A κ U )] > (28)is real and positive. Here A κp ( p = 1 , , , are the bulk wrapping numbers, e k − ( k = 1 , , are thebasis vectors of T k , and U is the complex structure of T ; the complex structure of T , is fixed by the Z ′ orbifold symmetry to be U , = e iπ/ . Then Vol( κ ) p T / Z ′ ) = Z κ | e e e | p | Im U | (29)The solutions for the AAA lattice given in Table 1 of [12], in which U = − i/ √ , all have Z a = 2 | e e e | and Z b = | e e e | (30)Using equation (27) above, it follows that at the string scale m string the coupling strengths for the SU (3) colour and SU (2) L groups satisfy α α = 12 (31)which is clearly inconsistent with the “observed” unification α = α at the scale m X ≃ × GeV. We reach the same conclusion for the solutions on the
BAA lattice given in Table 6 of [12], inwhich U = − i √ . Thus, running from the string scale to the TeV scale with the three-generationsupersymmetric Standard Model spectrum, none of our solutions can reproduce the measured valuesof the non-abelian coupling strengths of the SU (3) colour and SU (2) L gauge groups. In fact the onlysupersymmetric models obtained in [10] yielding three chiral generations Q L of quark doublets via ( a ∩ b, a ∩ b ′ ) = (2 , or (1 , , having no chiral matter in symmetric representations of the gauge groups,and not too much in antisymmetric representations, that also produce non-abelian coupling constantunification, are the two solutions on the BAB lattice given in Table 15 of that paper. We showed in [12]that neither model can have just the required Standard Model spectrum, but it is of interest to see whatcan be achieved if we relax this constraint and allow additional vector-like matter but not extra chiralexotics. This requires at least two U (1) stacks (both of which must be d -type in the terminology ofthat paper). The best we can do yields two additional vector-like Higgs doublets H u + H d ) and fouradditional vector-like charged lepton singlets ℓ cL + ¯ ℓ cL ) , and in any case the weak hypercharge U (1) Y α Y = 3 α / as required by the “observed” standard-model unification. Wehave not pursued this any further. The one-loop gauge threshold corrections to (27) have been computedby Gmeiner and Honecker [23]. However, for the models under consideration, these are very small andthe above conclusion is unaffected. Another possibility that in principle might yield a realistic model isto start with an SU (3) colour stack a and an SU (2) L stack b satisfying ( a ∩ b, a ∩ b ′ ) = (3 , or (0 , , andto require gauge coupling constant unification α = α . Following the work of Gmeiner and Honecker[9], we know at the outset that there are no such models that yield the standard-model spectrum and satisfy tadpole cancellation without the introduction of non-zero background flux H bg . However, sincewe have entertained the presence of such flux, it is of interest to know how far one can get with suchmodels. We have searched for solutions satisfying both of these criteria, but have found none.Finally, the presence of a non-zero flux H bg means that there may also arise a Freed-Witten anomaly[24]. In the presence of D6-branes the localised Bianchi identity associated with the stack κ imposes theconstraint [25] H bg ∧ [ κ ] = 0 (32)where [ κ ] is the 3-form that is the Poincar´e dual of κ . Since H bg is odd under the orientifold action R ,only the R -even part of [ κ ] , deriving from the R -odd part of κ , can contribute to the anomaly. We havestudied this in Appendix A. Our conclusion in all cases is that there is a non-zero anomaly deriving fromthe SU (3) stack a and also from one of the U (1) stacks.The deficiencies detailed above mean that our models can only be considered as semi-realistic. Nev-ertheless, it is of interest to see the extent to which the first two deficiencies can be remedied in modelswith a realistic spectrum. In this paper we study the fixing of moduli using background fluxes, the sta-bility of these solutions and their consistency with the supergravity approximation in which they arederived. We also investigate the utility of non-perturbative effects, so-called E2-instantons, to stabiliseaxion moduli and to repair the missing Yukawa couplings. In this and the following section we parallel the the treatment given by DeWolfe et al. [14] of the Z × Z orientifold. It has been shown by Grimm and Louis [26] that the effective four-dimensional theory de-riving from type IIA supergravity compactified on a Calabi-Yau 3-fold is an N = 2 supergravity theory.The moduli space is the product of two factors, one containing the vector multiplets (which includethe K¨ahler moduli), and the other the hypermutiplets (which include the complex structure moduli anddilaton). The metric on each space is derived from a K¨ahler potential, K K and K cs respectively. Theorientifold projection R to an N = 1 supergravity reduces the size of each moduli space.Consider first the K¨ahler moduli. The complexified K¨ahler form J c = B + iJ (33)is odd under the action of R and can therefore be expanded in terms of the R -odd (1 , -forms. In ourcase, on the Z ′ orbifold, we have three untwisted, invariant (1 , -forms w k ( k = 1 , , defined by w k ≡ dz k ∧ d ¯ z k (no summation) (34)There are also eight θ -twisted sector invariant harmonic (1 , -forms e (1 ,j ) , ˆ w j , ( j = 1 , , , , definedas follows. Associated with each of the 16 fixed points f i,j , defined in (7), is a localised (1 , -form e ( i,j ) ≡ ω k, ¯ ℓ dz k ∧ d ¯ z ℓ ( k, ℓ = 1 , (35)After blowing up the fixed point using the Eguchi-Hanson EH metric [27], ω k, ¯ ℓ has the form ω k ¯ ℓ = a ( u ) δ k ¯ ℓ + b ( u )( z k − Z k )(¯ z ℓ − ¯ Z ℓ ) (36)when the fixed point f i,j is at ( z , z ) = ( Z , Z ) ∈ T × T . The functions a ( u ) and b ( u ) are given by a ( u ) = u − ( λ + u ) − / λ (37) b ( u ) = a ′ ( u ) (38)6ith λ the blow-up parameter and u ≡ | z − Z | + | z − Z | (39)Under the action of the point group generator θ these (1 , forms transform as e (1 ,j ) → e (1 ,j ) (40) e (4 ,j ) → e (6 ,j ) → e (5 ,j ) → e (4 ,j ) (41)Thus the eight invariant θ -twisted (1 , forms are e (1 ,j ) and ˆ w j ≡ e (4 ,j ) + e (5 ,j ) + e (6 ,j ) ( j = 1 , , , (42)We denote the blow-up parameter associated with e (1 ,j ) by λ j . Point-group invariance (41) requires that e (4 ,j ) , e (5 ,j ) and e (6 ,j ) all have the same blow-up parameter, which we denote by ˆ λ j . All of the invariant θ -twisted (1 , forms given above are odd under the action of R , so in general we may expand thecomplexified K¨ahler form as J c = X k =1 , , t k iw k + X j =1 , , , ( T j ie (1 ,j ) + ˆ T j i ˆ w j ) (43)where t k = b k + iv k (44) T j = B j + iV j (45) ˆ T j = ˆ B j + i ˆ V j (46) b k , B j , ˆ B j are associated with the Kalb-Ramond field B , and the K¨ahler moduli v k , V j , ˆ V j with theK¨ahler form J . The K¨ahler potential K K for the K¨ahler moduli is given by K K = − log Z Z T / Z ′ J ∧ J ∧ J ! (47) = − log
323 Vol v v v − π Vol X j v ( λ j V j + 3ˆ λ j ˆ V j ) (48)where Vol , are the coordinate volumes of T and T respectively. Thus Vol = Y k =1 , , Vol k (49)where Vol k = R k − Im U k (50)As previously noted, the SU (3) lattice used for T , has U = α = U , so that Im U = √ / U .For the models found in [11, 12], Im U = − / √ on the AAA lattice and −√ on the BAA lattice. Itis convenient to absorb the coordinate volumes into the moduli, so we make the redefinitions t k Vol k → t k (51) T j πλ j → T j (52) ˆ T j π ˆ λ j → ˆ T j (53)and then K K = − log v v v − X j v ( V j + 3 ˆ V j ) (54)7ote that, unlike in the Z × Z case discussed in [14], the twisted moduli V j and ˆ V j are inextricablycoupled to the untwisted modulus v .The complex structure moduli are obtained by expanding the holomorphic (3 , -form Ω in terms ofthe basis 3-forms. There are four Z ′ -invariant untwisted 3-forms, defined as in [12] by σ ≡ dz ∧ dz ∧ dz (55) σ ≡ dz ∧ dz ∧ d ¯ z (56) σ ≡ d ¯ z ∧ d ¯ z ∧ dz = σ (57) σ ≡ d ¯ z ∧ d ¯ z ∧ d ¯ z = σ (58)Hence R ( σ ± σ ) = ± ( σ ± σ ) (59) R ( σ ± σ ) = ± ( σ ± σ ) (60)The invariant θ -twisted 3-forms ω j , ˜ ω j ( j = 1 , , , are also as defined in [12] as ω j ≡ [ α ( e (4 ,j ) − e (5 ,j ) ) + ( e (5 ,j ) − e (6 ,j ) )] ∧ dz (61) ˜ ω j ≡ [( e (4 ,j ) − e (5 ,j ) ) + α ( e (5 ,j ) − e (6 ,j ) )] ∧ d ¯ z (62)Then R ( ω j ∓ α ˜ ω j ) = ± ( ω j ∓ α ˜ ω j ) on AAA (63) R (˜ ω j ∓ αω j ) = ± (˜ ω j ∓ αω j ) on BAA (64)As above, it is convenient to factorise out coordinate volumes, so that the K¨ahler potential K cs forthe complex structure moduli is independent of them. Then on the AAA lattice we may expand theholomorphic 3-form as
Ω = 1 √ Vol [ Z ( σ + σ ) − g ( σ − σ ) + Z ( σ + σ ) − g ( σ − σ )]+ X j π ˆ λ j √ Vol [ Y j α ( ω j − α ˜ ω j ) − f j α ( ω j + α ˜ ω j )] (65)On the BAA lattice ω j and ˜ ω j are interchanged. In both cases Z , and Y j are associated with the R -evenforms, and g , , f j with the R -odd ones. It is easy to show that the complex conjugates of the twisted3-forms are given by ¯ ω j = α ˜ ω j (66) ¯˜ ω j = α ω j (67)The orientifold constraint requires that R Ω = ¯Ω (68)which gives Z , , g , , Y j , f j are real (69)The required K¨ahler potential is K cs = − log i Z T / Z ′ Ω ∧ ¯Ω ! (70) = − log −
163 ( Z g − Z g ) + 48 X j Y j f j (71)8he R projection projects out half of the moduli of the N = 2 theory, including half of the uni-versal hypermultiplet; the dilaton and one axion survive. The surviving moduli are all contained in thecomplexified 3-form Ω c ≡ C + 2 i Re( C Ω) (72)where C is the RR 3-form gauge potential, and C is the “compensator” that incorporates the dilatondependence C ≡ e − D + K cs / (73)with the four-dimensional dilaton D defined by e D ≡ √ e φ + K K / (74)Since C is even under the action of R we may expand it as C = 1 √ Vol [ x ( σ + σ ) + x ( σ + σ )] + X j π ˆ λ j √ Vol X j α ( ω j − α ˜ ω j ) (75)on the AAA lattice; as before, in the
BAA case we interchange ω j and ˜ ω j . Expanding Ω c as in (65), onthe AAA lattice Ω c = 1 √ Vol [ N ( σ + σ ) − T ( σ − σ ) + N ( σ + σ ) − T ( σ − σ )]+ X j πλ j √ Vol [ M j α ( ω j − α ˜ ω j ) − S j α ( ω j + α ˜ ω j )] (76)with the usual interchange for the BAA case. Then the surviving moduli are the expansion of Ω c in H , i.e. the R -even states with moduli N k = x k + 2 iCZ k ( k = 0 , (77) M j = X j + 2 iCY j ( j = 1 , , , (78)in both cases.The potential V arising after dimensionally reducing the massive type IIA supergravity is V = e K X i,j = { t k ,T j , ˆ T j ,N k ,M j } K ij D i W D j W − | W | + m e K Q Im W Q (79)where the K¨ahler potential K = K K + K Q with K Q = − (cid:18) Z Re( C Ω) ∧ ∗ Re( C Ω) (cid:19) (80)It follows from (65) that on the AAA lattice
Re( C Ω) = 1 √ Vol [ CZ ( σ + σ ) + CZ ( σ + σ )] + X j π ˆ λ j √ Vol α ( ω j − α ˜ ω j ) (81)Also, since Ω is the holomorphic (3 , -form, ∗ Ω = − i Ω and ∗ Re( C Ω) = Re ∗ ( C Ω) = 1 √ Vol [ iCg ( σ − σ ) + iCg ( σ + σ )] + i X j π ˆ λ j √ Vol g j α ( ω j + α ˜ ω j ) (82)so that e − K Q / = C e − K cs = e − D (83)9here K cs is given in (71), and the last equality follows from the definition (73). The same result followson the BAA lattice. Like the K¨ahler moduli t k , the complex structure moduli N , , M j enter the K¨ahlerpotential only via their imaginary parts. The superpotential [28, 29, 30] is W = W Q + W K where W Q ( N k , M j ) ≡ Z Ω c ∧ H bg (84) W K ( t k , T j , ˆ T j ) ≡ e + Z J c ∧ F bg − Z J c ∧ J c ∧ F bg − m Z J c ∧ J c ∧ J c (85)and e ≡ Z F bg (86)We note that W Q depends only on the NS-NS flux H bg and W K only on the RR fluxes F bgn ( n =0 , , , . H is odd under the action of R , so that, analogously to (75), we may expand its backgroundvalue as H bg = i √ Vol [ p ( σ − σ ) + p ( σ − σ )] + X j iπ ˆ λ j √ Vol P j α ( ω j + α ˜ ω j ) (87)on the AAA lattice, with ω ↔ ˜ ω on BAA . As shown in [12], flux quantisation requires that the coeffi-cients are quantised. On the
AAA lattice ( p , p ) = − π α ′ √ Vol √ R R R ( n + 3 n , n − n ) with n , n ∈ Z (88) P j = − π α ′ √ Vol R ˆ n j with ˆ n j ∈ Z (89)where n , and ˆ n j respectively are associated with the flux of H bg through the 3-cycles ρ , and ǫ j ;note that p , , P j are independent of the coordinate scales R , , . For the solution discussed in § | n ˆ n j | = 12 for j = 4 , . Thus | n | = 1 , , , , , .For j = 1 , we get | n ˆ n | = 12 | − t c | = 0 , , which is always consistent with these values of n .Cancellation of the untwisted part proportional to ρ + 2 ρ requires that the corresponding values of n satisfy | n | = 1296 , , , , , , and of n satisfy | n | = (1 + t c )(144 , , , , , .( t c = ± is one of the Wilson lines associated with the stack c .)Alternatively, on the BAA lattice ( p , p ) = π α ′ √ Vol R R R ( n + 3 n , − n + 3 n ) with n , n ∈ Z (90) P j = 2 π α ′ R √ ˜ n j with ˜ n j ∈ Z (91)where n , and ˜ n j respectively are associated with the flux of H bg through the 3-cycles ρ , and ˜ ǫ j . In thiscase tadpole cancellation of the exceptional parts requires that | n ˜ n j | = 12 , so that | n | = 1 , , , , , .Then, n = 0 and the corresponding values of n satisfy | n | = 432 , , , , , .The form (87) for H bg gives W Q ( N k , M j ) = −
83 ( N p − N p ) + 24 X j M j P j (92)The background fluxes F bg and F bg that appear in W K have similar expansions. Since F is odd underthe action of R and F even F bg = X k =1 , , k f k iw k + X j =1 , , , F j πλ j ie (1 ,j ) + ˆ F j π ˆ λ j i ˆ w j ! (93)10 bg = 1Vol X k =1 , , Vol k e k ˜ w k + X j =1 , , , E j Vol πλ j w ∧ e (1 ,j ) + ˆ E j Vol π ˆ λ j w ∧ ˆ w j ! + X j =1 , , , G j π λ j e (1 ,j ) ∧ e (1 ,j ) + ˆ G j π ˆ λ j ˆ w j ∧ ˆ w j ! (94)where ˜ w k = dz i ∧ d ¯ z i ∧ dz j ∧ d ¯ z j where ( i, j, k ) = cyclic (1 , , (95)The constant term e in W K , defined in (86), arises from the Hodge dual F bg of F polarised in the non -compact directions. All of these fluxes, including F bg , are quantised, the general constraint beingthat for any closed ( p + 2) -cycle Σ p +2 µ p Z Σ p +2 F p +2 = 2 πn with n ∈ Z (96)with µ p = (2 π ) − p α ′− ( p +1) / the electric charge of a D p -brane. For the present, we set F bg = 0 , andthen W K ( t k , T j , ˆ T j ) = e − X k =1 t k e k + 4 X j ( T j E j + 3 ˆ T j ˆ E j ) + 4 t X j ( G j + 3 ˆ G j ) − m t t t − X j t ( T j + 3 ˆ T j ) (97)The advantage of this formalism is that we may immediately identify supersymmetric vacua by theirvanishing F -terms: F i = D i W ≡ ∂ i W + W ∂ i K = 0 (98)for every chiral superfield i . For the complex-structure moduli, taking i = N k , M j , we get p k + 2 ie D W ( Cg k ) = 0 ( k = 0 , (99) P j + 2 ie D W ( Cf j ) = 0 ( j = 1 , , , (100)As in [14], the imaginary parts of these equations are degenerate. Using (84) and (77) ... (78), they givethe single constraint Re W = 0 (101)which fixes only one linear combination of the axions x , x and X j
83 ( x p − x p ) − X j X j P j = Re W K (102)This degeneracy derives from the fact that the coeffcients p k , P j that determine H bg are real , and there-fore have insufficient degrees of freedom to stabilise both the complex structure moduli and their axionicpartners. As we shall discuss later, in §
5, E2-instantons can lift the remaining degeneracy. The real partsgive e − K cs / p k g k = e − K cs / P j f j = Q (103)where Q ≡ Im W e D (104)Then (103) determines the moduli g k , f j up to an overall scale fixed by Q . Finally, using (83), (104)gives e − φ = √ e K K / Im WQ (105)11hich fixes the dilaton once the other moduli are all fixed [14]. It follows from (99), (100), (71), (83)and (92) that Im W Q + 2 iW = 0 (106)Thus, using (101), when the complex structure moduli satisfy their field equations W K + Im W Q = 0 (107)and the vacuum value of the superpotential is determined entirely by the K¨ahler moduli W = − i Im W K ( t k , T j , ˆ T j ) (108)Vanishing F-terms for the K¨ahler moduli in (98) give e k + m t t t t k − iW e K K v v v v k = 32 δ k X j h m ( T j + 3 ˆ T j )+ 2( G j + 3 ˆ G j ) − iW e K K ( V j + 3 ˆ V j ) i (109) E j + m t T j = 4 iW e K K v V j (110) ˆ E j + m t ˆ T j = 4 iW e K K v ˆ V j (111)Using (101), the imaginary parts of these equations require that Im( ∂ i W K ) = 0 for i = t k , T j , ˆ T j (112)The simplest solution of these is b k = 0 = B j = ˆ B j (113)and then the above equations reduce to e k = 32 δ k X j [2( G j + 3 ˆ G j ) − X ( V j + 3 ˆ V j )] + X v v v v k (114) E j = Xv V j (115) ˆ E j = Xv ˆ V j (116)where X ≡ m + 4 iW e K K = m + 4Im W K e K K (117)using (108). They couple the untwisted volume modulus v to the twisted volume moduli V j , ˆ V j . Solvingfor all moduli in terms of v and X gives v = e Xv (118) v = e Xv (119) V j = E j Xv (120) ˆ V j = ˆ E j Xv (121)Substituting these into the v equation gives ˜ e Xv = e e − X j ( E j + 3 ˆ E j ) ≡ F ( e k , E j , ˆ E j ) (122)where ˜ e ≡ e − X j ( G j + 3 ˆ G j ) (123)12hen (54) gives e K K = 3 X e v (124)and the definition (117) yields W K e v = 1 − m X (125)Substituting (118) ...(121) and (113) into (97), it then follows that when the K¨ahler moduli satisfy theirfield equations, X = 35 m (126)so that | v | = s F ( e k , E j , ˆ E j )3˜ e m (127)Thus the background fluxes e k , E j , ˆ E j , G j , ˆ G j and m fix v and X , and hence, via equations (118)...(121), the remaining K¨ahler moduli.The effective supergravity theory is a justifiable approximation [14] so long as the volumes v k , V j , ˆ V j are large enough that the O( α ′ ) corrections are negligible and the string coupling g s is small enough toneglect corrections. Further, to remain within the K¨ahler cone we require that the untwisted volumes arelarge compared with the blow-up volumes, i.e. v k ≫ V j , ˆ V j ≫ . Since (the non-zero value of) m is fixed by the RR tadpole cancellation condition (14), and we have set F bg = 0 , the question then iswhether there are choices of the background 4-form flux F bg for which these constraints are obeyed.It follows from equations (118) ... (121) that v , /V j = e , /E j , so that the K¨ahler cone constraintsrequire that e , e ≫ E j , and similarly for ˆ E j . Hence F ∼ e e . Then the constraints v k ≫ requirethat e e ≫ ˜ e m , e ˜ e ≫ e m and e ˜ e ≫ e m , and these imply that e , ˜ e , e ≫ m . For theblow-up volumes, similarly, the constraints v k ≫ V j , ˆ V j ≫ require that e , e , e e / ˜ e ≫ E j , ˆ E j ≫ p e e / ˜ e m . All of these are easily arranged. In general, besides the supersymmetric vacua identified in the previous section, we expect there to beadditional vacua that are non-supersymmetric. To identify these we should find the effective potentialin the four-dimensional Einstein frame, in which the four-dimensional Einstein-Hilbert action has thestandard normalisation. However, the axion fields x k and X j , defined in (75), enter the ten-dimensionalaction (9) only via the C ∧ H bg ∧ dC term in the Chern-Simons piece. This term is only non-zero if dC is “polarised” in the four-dimensional spacetime directions, i.e. dC = f d x ≡ F ; it has no physicaldegrees of freedom and can be treated as a Lagrange multiplier. The part of the action involving F hasthe form S = − κ Z ( F ∧ ∗ F + 2 F ∧ X ) (128)where X = F bg + B ∧ F bg + C ∧ H bg − m B ∧ B ∧ B (129)The equation of motion for F gives ∗ F + X = 0 (130)Then subsituting back gives S = − κ Z X ∧ ∗ X (131)which is stationary when X = 0 . The equation that stabilises the axion follows from Z X = 0 = Z (cid:16) F bg + B ∧ F bg + C ∧ H bg − m B ∧ B ∧ B (cid:17) (132)13sing (43), (75), (87) and (94) this gives
83 ( x p − x p ) − X j X j P j = e − X j b j e j + 4 X j ( B j E j + 3 ˆ B j ˆ E j ) + 4 b X j ( G j + 3 ˆ G j ) − m b b b + 2 m b X j ( B j + 3 ˆ B j ) (133)This fixes the same linear combination of the axions x , x and X j as in (102), and indeed, using (97),the value agrees with that found in the supersymmetric treatment when the Kalb-Ramond fields b k , B j and ˆ B j have the values given in (113).The remaining moduli are stabilised by minimising the effective potential V in the Einstein framewith metric g Eµν . We pass to this frame by redefining the four-dimensional metric g µν = e φ Vol( M ) g Eµν (134)where
Vol( M ) is the volume of the compact space M = T / Z ′ Vol( M ) ≡ Z T / Z ′ d y √ g (135)with g the determinant of the 6-dimensional metric. Invariance of the 6-dimensional K¨ahler metricunder the action of the point group and the orientifold projection R requires that ds = γ dz d ¯ z + γ dz d ¯ z + γ dz d ¯ z (136)where the γ i ( i = 1 , , are real and positive. In the θ -twisted sector there are 16 Z fixed points f i,j ∈ T × T with i, j = 1 , , , , defined in (7) and (8). These fixed points are blown up using theEguchi-Hanson EH metric ds = g k, ¯ ℓ dz k d ¯ z ℓ (137)where k, ℓ = 1 , and g k, ¯ ℓ = Γ[ A ( u ) δ k ¯ ℓ + B ( u )( z k − Z k )(¯ z ℓ − ¯ Z ℓ )] (138)when f i,j is at ( z , z ) = ( Z , Z ) ∈ T × T . The functions A ( u ) and B ( u ) are given by A ( u ) = u − ( λ + u ) / λ (139) B ( u ) = A ′ ( u ) (140)with λ the blow-up parameter and u as defined in (39). In general, both the twisted modulus Γ andthe blow-up parameter λ depend on the fixed point f i,j with which they are associated. However, thetransformation property (41) of the twisted 2-forms, or rather the analogous property of the twisted 2-cycles, shows that ˆΓ j and ˆ λ j , associated with f ,j , f ,j and f ,j , are independent of the T fixed point i = 4 , , ; the corresponding parameters for f ,j are denoted by Γ j and λ j . In the untwisted sector thereare then three real moduli and Vol( M ) = 16 Y k =1 , , Vol( T k ) = 16 γ γ γ Vol (141)where Vol is defined in (49) and (50). The (4-dimensional) volume of the blow-up is Vol( f i,j ) = Γ π λ (142)taking ≤ u . λ . The local analysis that we carry out here is valid provided that the volume of theblow-up modes is small compared with the untwisted volume Vol( T )Vol( T ) of the 4-torus containingthem, i.e. provided that Γ π λ ≪ Vol( T )Vol( T ) . Blowing up f i,j in this manner removes a volume14 ol( f i,j ) from the untwisted volume Vol( T )Vol( T ) . With g Eµν as given in (134), the effective potential V is defined by S = 1 κ Z d x q − det( g E )( − V ) (143)Taking F bg = 0 , as in (97), there are four contributions to VV = V H + V F + V m + V BI (144)deriving respectively from the | H | , | F | , m and the Born-Infeld terms in (9). With H bg given by(87), we find V H = h e φ Vol ( M ) (145)where h = 23 ( p + p ) + 6 X j P j (146)on both lattices. As noted previously, h is fixed by the integers given in equations (88) ... (91), indepen-dently of the coordinate scales R , , . Similarly, with F bg given by (94), we find V F = e φ Vol ( M ) X k =1 , , e k Vol( T k ) + 16 Vol( M )Vol( T ) X j ( E j + 3 ˆ E j )++ 16 Vol( M )Vol( T ) X j " G j Vol( f (1 ,j ) ) + 3 ˆ G j Vol( f (4 ,j ) ) (147)where Vol( T k ) = γ k Vol k for k = 1 , , (148)with Vol k defined in (50). Likewise V m = m e φ M ) = µ m e φ Vol( M ) (149)with µ = 1 / .As in [14], the only terms relevant to the stabilisation of the twisted moduli are V F and V m , sincethe former dominates as Vol( f i,j ) → and the latter as Vol( M ) → ∞ . In equation (149) we may write Vol( M ) = Vol ( M ) −
16 Vol( T ) X j [Vol( f ,j ) + 3Vol( f ,j )] (150)where Vol ( M ) = Vol( T )Vol( T )Vol( T ) / is the volume with no blow up. Then, minimising thepotential gives Vol( f ,j ) = | G j |√ | m | (151) Vol( f ,j ) = | ˆ G j |√ | m | (152)and we are justified in using this local treatment provided that the F bg fluxes are chosen so that | G j , ˆ G j | ≪ √ | m | Vol( T )Vol( T ) (153)With these values for the blow-up volume V F = V F + V F (154)15here V F = e φ Vol ( M ) X k =1 , , e k Vol( T k ) + 16 Vol( M )Vol( T ) X j ( E j + 3 ˆ E j ) (155) V F = √ e φ | m | Vol( T )6Vol ( M ) X j ( | G j | + 3 | ˆ G j | ) (156)The Born-Infeld term gives V BI = µ κ e φ Vol ( M ) X κ N κ Z κ d ξ p det( g ) (157)and using the (bulk part of the) tadpole cancellation condition given in (14), we can rewrite this as V BI = − e φ Vol ( M ) Z Π m Hbg d ξ p det( g ) (158)where Π m H bg is the 3-cycle of which the field m H bg is the Poincar´e dual; H bg is given in (87). Forthe two cases of interest, as shown in [12], Π m H bg = √ Vol R R R m [( p − p ) ρ − ( p + p )( ρ + 2 ρ )] − i √ Vol (1 − α ) R X j P j (2 ǫ j + ˜ ǫ j ) (159) = − √ Vol √ R R R m [( p + p ) ρ + ( p − p )( ρ + 2 ρ )] − i √ Vol (1 − α ) R X j P j ( ǫ j + 2˜ ǫ j ) (160)for AAA and
BAA respectively. To calculate the integral in (158), we use the result [31] quoted in [14],since Π m H bg is a special Lagrangian 3-cycle. The holomorphic 3-form Ω , defined in (65), is normalisedby demanding that i Z M Ω ∧ ¯Ω = 1 = 163 ( Z g − Z g ) + 48 X j Y j f j (161) ≡ G ( Z , Z , Y j ) (162)Then, according to the calibration formula Z Π m Hbg d ξ p det( g ) = p M ) Z Π m Hbg (Ω + ¯Ω) (163)So V BI = − b | m | e φ Vol / ( M ) (164)where b = 2 √ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)
23 ( p Z − p Z ) + 6 X j Y j P j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (165)for both lattices.The various contributions to V are homogeneous in Vol( T k ) . Hence at the stationary point X k Vol( T k ) ∂V∂ Vol( T k ) = 6 V H + 7 V F + 5 V F + 3 V m + 92 V BI (166) ∂V∂φ = 2 V H + 4 V F + 4 V F + 4 V m + 3 V BI (167)16liminating V F gives V H = 8 V F + 16 V m + 3 V BI (168)Also, we require that ∂V /∂ Vol( T k ) = 0 , which gives | e Vol( T ) | = | e Vol( T ) | ≡ y == | e Vol( T ) | − T )Vol( T ) X j ( E j + 3 ˆ E j ) + 9 √ | m | Vol( M ) T )Vol( T ) X j ( | G j | + 3 | ˆ G j | ) (169)(with y > ). It follows that | e Vol( T ) | = y (1 + ǫ ) − ηy Vol( M ) (170)where ǫ ≡ | e e | X j ( E j + 3 ˆ E j ) (171) η ≡ √ | m e e | X j ( | G j | + 3 | ˆ G j | ) (172)The requirement (153) that justifies the local treatment gives | G j , ˆ G j | ≪ √ | m | y | e e | (173)so that η ≪ m y ) (174)We may also write Vol( M ) in terms of y : Vol( M ) = 16 Y k Vol( T k ) = y (1 + ǫ ) / (36 | e e e | + ηy ) / (175)so that | e Vol( T ) | = y (1 + ǫ )1 + ηy | e e e | (176)Defining x ≡ e φ p Vol( M ) (177)it follows from (168) that h = (cid:18) µm + 32 η y (cid:19) x − b | m | x (178)which fixes x as a function of y . Hence | m | x = 3 b µ + η m y ) s µ + η m y ) h b (179)and at the stationary point, we may eliminate the dilaton and express the potential in terms of y alone: V = A Vol( M ) + B Vol( M ) (180)17here Vol( M ) is given by (175) and A ≡ hx + µm x − b | m | x (181) B ≡ x y (1 + ǫ ) (cid:18) ηy | e e e | + ηy ) (cid:19) (182)with x given by (178). Since B > , it is easy to see that the potential V → + ∞ as y → . Similarly, V → as y → ∞ . The limit is approached from above or below depending upon the sign of A in thisregion. If A < , then there is certainly an anti-de-Sitter minimum at a finite value of y ; otherwise, noconclusion can be reached without a more detailed consideration of the parameters. It follows from (181)and (178) that A ∼ x ( h − µm x ) as y → ∞ (183)In the same limit, (179) gives | m | x ≃ b µ r µ b ! (184)Then A < if and only if b > µh (185)To proceed further, we need to know the dependence of the moduli g , , f j that appear in (161) on Z , , Y j . For simplicity, we consider only the bulk contributions g , and assume that these derive froma homegeneous quadratic prepotential G , defined in (162), of the form G ( Z , Z ) = αZ + 2 βZ Z + γZ (186)with α, β and γ (real) constants (not functions of Z /Z ). Then the moduli g , are given by g = − ∂ G ∂Z = − αZ + βZ ) (187) g = ∂ G ∂Z = 2( βZ + γZ ) (188)The question we address is whether G may be chosen so that (185) is always satisfied. Keeping only thebulk contributions, the minimum value of b = 329 ( Z p − Z p ) (189)subject to the constraint (161) that G ( Z , Z ) = 3 / is b = 3 γp + 2 βp p + αp αγ − β (190)Evidently, we may ensure that (185) is satisfied by choosing α, β, γ sufficiently small. On the AAA lattice, p p = 3 + 2 t c = 5 , (191)so that the minimum value of b and h are b = 3 p ( αγ − β ) [15 γ + 6 β + α + 4 t c (3 γ + β )] (192) = 3 p ( αγ − β ) (25 γ + 10 β + α, γ + 2 β + α ) (193) h = 83 p (7 + 6 t c ) (194) = 83 p (13 , (195)18or t c = +1 , − respectively. On the BAA lattice, since p = p in this case, b and h have the samevalues as in the t c = − case for the AAA lattice.The untwisted part of the 4-form flux F bg given in equation (94). It is specified by the quanti-ties e k Vol k / Vol ( k = 1 , , . Using (169), the ratios of the metric moduli γ i /γ j = e j Vol j /e i Vol i are specified for a given value of F bg . The minimisation of V fixes y / | e e e | , and F bg also speci-fies the combination | e e e | / Vol . Thus, the stabilisation fixes the overall scale of the metric moduli ( γ γ γ ) = ( y / | e e e | ) ( | e e e | / Vol ) in terms of the specified background fluxes. Similarly, the(untwisted) background flux H bg , defined in equation (87), is specified by p , / √ Vol . Thus equation(179) fixes x/ √ Vol in terms of the background fluxes m and H bg . With x defined in (177), it followsthat x/ √ Vol ≃ e φ √ γ γ γ , and since the moduli γ , , have already been fixed, this result stabilisesthe dilaton φ in terms of the background fluxes. The argument may be extended to include the twistedmoduli. Since we have taken F bg = 0 , the | F | and | F | terms in the the action S IIA , given in (9), are at leastquadratic in the fields B , there being no Z ′ -invariant 1-form fields C . The Chern-Simons terms havealready been accounted for in the minimisation of X . Thus the whole action S IIA is at least quadraticin the moduli fields b k , B j , ˆ B j defined in (44) ... (46) and (51) ... (53), and we may consistently setall of their expectation values to be zero, as in (113) in the supersymmetric case. However, there arefluctuations b k ( x ) , B j ( x ) , ˆ B j ( x ) around this solution, and the B ∧ B ∧ ∗ F bg contribution to | F | canmake the solution unstable if the mass matrix for the fluctuations has a negative eigenvalue.After eliminating the Lagrange multiplier F ≡ dC , the effective action deriving from this field isgiven in (131) with X in (132). With the B -moduli set to zero, the stabilised linear combination of theaxions given in (102) reduces to
83 ( x p − x p ) − X j X j P j = e (196)The B ∧ F bg + C ∧ H bg piece in X is linear in the fluctuation fields and the above stabilised combinationof axion fields. Hence the action (131) mixes them and we need to consider the quadratic terms, includingkinetic terms, for both sets of fields simultaneously. The unstabilised (orthogonal) axion fields are, ofcourse, massless.The kinetic terms for the B field fluctuations derive from the contribution − κ Z d x √− g e − φ | H | ⊃ − κ Z d x √− ge − φ dB ∧ ∗ dB (197) = − κ Z d x √− g E L K ( B ) (198)where the kinetic Lagrangian density is L K ( B ) = 12 X k =1 , , ( ∂ µ ˜ b k )( ∂ µ ˜ b k ) + 12 X j =1 , , , [( ∂ µ ˜ B j )( ∂ µ ˜ B j ) + ( ∂ µ ˜ˆ B j )( ∂ µ ˜ˆ B j )] (199)with ∂ µ ˜ b k = g µνE ∂ ν ˜ b k etc. , and the fields ˜ b k , ˜ B j , ˜ˆ B j defined so that they are canonically normalised: ˜ b k ≡ b k Vol( T k ) (200) ˜ B j ≡ s T )Vol( M ) B j (201) ˜ˆ B j ≡ s T )Vol( M ) ˆ B j (202)19uadratic terms in these fields arise from − κ Z d x √− g [ | F | + | F | ] ⊃ − κ Z [ m B ∧ ∗ B − m B ∧ B ∧ ∗ F bg ]= − κ Z d x √− g E e φ Vol ( M ) X k (cid:20) m Vol( M )˜ b k ˜ b k + 43 m ˜ b ˜ b ˜ b e k Vol( T k )˜ b k (cid:21) + X j (cid:20) m Vol( M ) − m e Vol( T ) (cid:21) ( ˜ B j + ˜ˆ B j ) + 16 m s M )Vol( T ) ˜ b X j ( ˜ B j E j + √ B j ˆ E j ) − m Vol( T )˜ b ˜ b X j ( G j + 3 ˆ G j ) − m Vol( M ) √ X j (cid:20) ˜ B j s j + ˆ˜ B j ˆ s j (cid:21) (203)where s j , ˆ s j are respectively the signs of G j /m , ˆ G j /m , and the last term follows when we substitutethe stabilised values (151) and (152) of the blow-up volumes.The kinetic terms for the C fluctuations arise from − κ Z d x √− g | F | ⊃ − κ Z dC ∧ ∗ dC (204) = − κ Z d x √− g E e φ Vol( M )
43 ( ∂ ( x − x )) + 43 ( ∂ ( x + x )) + 6 X j ( ∂X j ) (205) = − κ Z d x √− g E ( ∂ ˜ x ) + ( ∂ ˜ x ) + X j ( ∂ ˜ X j ) (206)and the canonically normalised fields ˜ x , and ˜ X j are given by ˜ x ≡ s M ) 2 e φ ( x − x ) (207) ˜ x ≡ s M ) 2 e φ ( x + x ) (208) ˜ X j ≡ s M ) 2 e φ X j (209)Quadratic terms in b k and x , arise from (131) S = − κ Z X ∧ ∗ X ⊃ − κ Z ( B ∧ F bg + C ∧ H bg ) ∧ ∗ ( B ∧ F bg + C ∧ H bg ) (210)As noted previously, the only coupled combination of axion fields corresponds to the stabilised axion,whose normalised field ˜ a is given in terms of the rescaled fields ˜ x , , ˜ X j by p x − p x − X j P j X j ∝ ( p + p )˜ x + ( p − p )˜ x − √ X j P j ˜ X j ≡ N ˜ a (211)where N = √ p + p + 36 X j P j / (212)We shall consider only the untwisted contibutions. Then the quadratic terms deriving from (210) are S = − κ Z d x √− g E e φ ( M ) X k Vol( T k )˜ b k e k + e − φ q M )( p + p )˜ a ! (213)20tability requires that the eigenvalues of the mass matrix are all positive. However the uncoupled axionis massless, so the best we can hope for is that the remaining four mass eigenstates are non-tachyonic.The mass matrix deriving from (213) may be written in the form m = 2 e φ m Vol( M ) α γ αγ ( s + s s α ) α ( s + s s αγ ) s αβγαγ ( s + s s α ) 1 + α αγ ( s + s s α ) s αβα ( s + s s αγ ) αγ ( s + s s α ) 1 + α γ s αβγs αβγ s αβ s αβγ β (214)where α ≡ | e e e || m | y (215) β ≡ p + p ) / √ m x (216) γ ≡ (cid:18) ηy | e e e | (cid:19) / (1 + ǫ ) − / (217)and s , , = ± are the signs of e , , . The general expressions for the eigenvalues are too large to betractable, but positive-definiteness is ensured provided that the following quantities are all positive: tr( m ) = X i ( m ) i = β + 3 + a (1 + 2 γ ) (218) det( m ) = Y i ( m ) i = β (1 − a + 2 a γ − a γ ) ≡ d ( a ) (219) X i,j ( m ) i ( m ) j = 3 β + 6 a γ + 2 a γ + 3 + a ≡ d ( a ) (220) X i,j,k ( m ) i ( m ) j ( m ) k = 3 β + 1 + 4 a γ + 4 a γ − a − a β γ − a β ≡ d ( a ) (221)where a ≡ s s s α (222)When γ > / it is obvious that for large, positive values of a ≫ all of these are positive. Thequestion is whether there are other values, in particular negative values, for which we also have positive-definiteness, and what can be said when γ ≤ / . By inspection it is clear that the trace is automaticallypositive. For general (non-zero) values of β and γ , det ( m ) = d ( a ) > provided that a ≡ γ (cid:16) − p γ (cid:17) < a < a (223) or a < a (224)where a = 14 γ (cid:16) p γ (cid:17) , a = 1 for γ > (225) a = 1 , a = 14 γ (cid:16) p γ (cid:17) for γ < (226)Note that a is always negative, and a , positive. In the special case that γ = 0 , the function d ( a ) = β (1 − a ) is positive only in the range − < a < .We also require that d ( a ) is positive. Evidently this is always the case for a > , so we need onlyconsider whether negative values of a lead to stronger constraints than those already derived. Accordingto (223), the most negative value that we need to consider is a = a , which satisfies d ( a ) = 0 . For thisvalue of a it follows that d ( a ) = 3 β + 4 a (1 + 2 γ ) > (227)21urther, d ( a ) has only a single real (negative) root, so the positivity of P i,j ( m ) i ( m ) j gives no furtherconstraints.Finally, we require also that d ( a ) is positive. It is convenient to write d ( a ) = N ( a ) − β D ( a ) (228)where N ( a ) ≡ (4 γ − a + 4 γ a + 1 (229) = ( a + 1)[(4 γ − a + a − a + 1] (230) D ( a ) ≡ (2 γ + 1) a − (231)The special case in which γ = 0 is easy to analyse. In this case N ( a ) = 1 − a is positive only in therange − < a < in which d ( a ) is also positive. Since D ( a ) = a − is negative throughout this range,it is only for values of a in this range that we have positive definiteness. The case in which γ = 1 / isalso easy to analyse. Positivity of d ( a ) requires that either − √ a < a < a = 1 or a > √ .The function N ( a ) = 1 + a is positive only when a > − . Thus N ( a ) is positive in both of theseranges, while D ( a ) = 3( a − / is negative in the region a < a < a , but positive in a > a . Itfollows that m is positive definite for any value of β when a is in the range a < a < a , but only forvalues of β < N ( a ) /D ( a ) in the range a > a .The full analysis of the conditions in which d ( a ) and d ( a ) are both positive for general values of γ is given in Appendix B. The conclusions are as follows: For values of a in the range a < a < a , themass matrix m is positive definite for all values of γ and all values of β . If . . γ < / , thereis in addition a finite region a < a < a in which m is positive definite but only for values of β thatare bounded above by N ( a ) /D ( a ) . Finally, if γ > / , there is an infinite region a > a in which m is positive definite, again for values of β that are bounded above by N ( a ) /D ( a ) . Here a , , , defined inequations (223), (225) and (226), specify the regions in which d ( a ) > , and a is the root of the cubicfactor in equation (230).Although we have been discussing the conditions under which the (untwisted) mass eigenstates arenon-tachyonic, in principle this is too strong a requirement in the anti-de Sitter space of our vacuumsolutions. Tachyonic mass eigenstates are stable provided that they satisfy the Breitenlohner-Freedmanbound [32, 33] m i ≥ m BF ≡ − | V min | (232)where −| V min | is the value of the potential at the anti-de Sitter minimum. The massless uncoupled axionobviously satisfies the bound, so it will not generate instability. However, determining which values of a lead to other mass eigenstates that satisfy this weaker constraint is something that can only be done when V min has actually been calculated, and this in turn requires a detailed consideration of the parameters, asalready noted. The expectation or, more accurately, the hope is that when the anti-de Sitter minimum islifted to Minkowski, in the manner of KKLT [34], then the tachyonic states will be lifted too. However,as Conlon has noted [35], it is not clear that all tachyons will be lifted by this mechanism. The upliftingis generally rather poorly controlled, and it is at least plausible that there may remain tachyons in theMinkowski space. We have so far fixed only one linear combination of the axion fields. As noted previously, we may usenon-perturbative effects to stabilise the remaining axions. The non-perturbative effects under discussionare D p -branes that wrap non-trivial cycles in the compactification space M , and that are pointlike in M . Their world-volume is ( p + 1) -dimensional and spacelike, so they are Euclidean D p -branes, calledE p -branes or E p -instantons for short. In type IIA string theory, p is even and p +1 ≤ . Hence p = 0 , , .Since there are no non-trivial 1- and 5-cycles on the orientifold T / Z ′ with which we are concerned,only E2-branes are relevant. The instanton action S inst is given by [36] S inst = 2 π (cid:18) g s Z Ξ Re Ω − i Z Ξ C (cid:19) (233)22here Ξ is the 3-cycle wrapped by the E2-brane, and Ω is the holomorphic 3-form. Evidently an E2-instanton is coupled to the axion fields in C and can lift (some of) the degeneracy of the axions thatare not stabilised by the background flux m H bg . Quite generally, we may expand Ξ in terms of theuntwisted 3-cycles ρ p ( p = 1 , , , and the exceptional 3-cycles ǫ j , ˜ ǫ j ( j = 1 , , , , so that Ξ = 12 X p Z p ρ p + 12 X j ( z j ǫ j + ˜ z j ˜ ǫ j ) (234)where Z p , z j , ˜ z j are integers. Supersymmetry constrains these coefficients. On the AAA lattice, it re-quires that Z − Z + Z = 0 (235) Z − Z + Z > (236)As displayed in equations (189) ... (194) of reference [12], there are three types of supersymmetric3-cycle: Z p = (1 , , ,
0) mod 2 ( n k , m k ) = (1 ,
0; 1 ,
0; 1 ,
0) mod 2 (237) = (1 , θ, , θ ) mod 2 ( n k , m k ) = (1 , θ,
1; 1 ,
1) mod 2 (238) = (0 , , ,
0) mod 2 ( n k , m k ) = (0 ,
1; 1 ,
1; 0 ,
1) mod 2 (239)(with θ = 0 , ) called respectively c -, d θ - and e -type. They are associated with exceptional parts having c : ( z , ; ˜ z , ) or ( z , ; ˜ z , ) = (0 ,
0; 1 ,
1) mod 2 (240) d θ : ( z , ; ˜ z , ) or ( z , ; ˜ z , ) = ( θ, θ ; 1 ,
1) mod 2 (241) e : ( z , ; ˜ z , ) or ( z , ; ˜ z , ) = (0 ,
0; 1 ,
1) mod 2 (242)With the general form of the instanton’s 3-cycle Ξ given in equation (234), and with C on the AAA lattice given in equation (75), we find Im S inst = − π / [(2 Z − Z )( x + x ) + Z ( x − x )] + 3 / √ X j X j z j (243) ∝ Z − Z ) r / ˜ x + Z r − / ˜ x ] + 1 √ X j ˜ X j z j (244)using the canonically normalised fields defined in equations (207), (208) and (209). In general, this isquite different from the combination a given in equation (211) that is stabilised by the background flux.Evidently a separate instanton is required for each of the unstabilised axions.Similarly, on the BAA lattice, supersymmetry requires that Z − Z − Z = 0 (245) Z − Z + 2 Z > (246)and, as displayed in equations (275) ... (280) of reference [12], again there are three types of supersym-metric 3-cycle: Z p = (0 , , ,
0) mod 2 ( n k , m k ) = (1 ,
0; 1 ,
1; 1 ,
0) mod 2 (247) = ( θ, , θ,
1) mod 2 ( n k , m k ) = (1 ,
1; 1 , θ ; 1 ,
1) mod 2 (248) = (0 , , ,
1) mod 2 ( n k , m k ) = (0 ,
1; 0 ,
1; 0 ,
1) mod 2 (249)(with θ = 0 , ) called respectively c -, d θ - and e -type. They are associated with exceptional parts having c : ( z , ; ˜ z , ) or ( z , ; ˜ z , ) = (1 ,
1; 0 ,
0) mod 2 (250) d θ : ( z , ; ˜ z , ) or ( z , ; ˜ z , ) = (1 , θ, θ ) mod 2 (251) e : ( z , ; ˜ z , ) or ( z , ; ˜ z , ) = (1 ,
1; 0 ,
0) mod 2 (252)23n this case, we find that Im S inst = − πR / [ Z ( x + x ) − ( Z − Z ) Z ( x − x )] − / √ X j X j ˜ z j (253) ∝ Z r / ˜ x − ( Z − Z ) r − / ˜ x ] − √ X j ˜ X j ˜ z j (254)Again, this is generally quite different from the combination a given in equation (211).As noted in the Introduction, the surviving global U (1) symmetries in our models forbid some ofthe Yukawa couplings that are needed to give non-zero masses to the quarks and leptons via the Higgsmechanism. Consider, for example, the model described by the fourth solution in Table 1 of reference[12]. The weak hypercharge Y is a linear combination of the U (1) charges Q a,c,d associated respectivelywith the SU (3) c stack a , and the U (1) stacks c, d . Y = 16 Q a + y c Q c + 12 Q d (255)where y c = ± . Using equations (63) and (66) of that paper, the intersection numbers of a with the SU (2) L stack b and its orientifold image b ′ are given by ( a ∩ b, a ∩ b ′ ) = (1 ,
2) if ( − τ a + τ b = 1 (256) = (2 ,
1) if ( − τ a + τ b = − (257)thereby generating the required total of Q L quark doublets (with Y = ). Similarly, using equation(247), the U (1) stack d and its orientifold image d ′ have intersection numbers ( a ∩ d, a ∩ d ′ ) = (0 , (258)so that there are no quark-singlet states q cL at these intersections. Choosing y c = − in (255), the Higgsdoublet H u with Y = arises at the intersection of b with the U (1) stack c , while H d with Y = − arises at the intersection with its orientifold image c ′ : ( b ∩ c, b ∩ c ′ ) = (1 , (259)The quark singlets arise at the intersections of a with c and c ′ ( c ∩ a, c ′ ∩ a ) = (3 , (260)the former giving u cL and the latter d cL . First, consider the case described by (257). u -quark massterms arising from the two Q L states at a ∩ b require the coupling of the states at a ∩ b , b ∩ c and c ∩ a ,which is allowed by the conservation of Q a , Q b and Q c . However, the u -quark mass term arising fromthe Q L state at a ∩ b ′ requires the coupling of the states at a ∩ b ′ , b ∩ c and c ∩ a , which is allowed by theconservation of Q a and Q c , but not by Q b , since the product has ∆ Q b = 2 . Similarly, only two d -quarkmass terms are allowed by conservation of Q b . The alternative choice described by (256) allows onlyone quark mass term for both u - and d -type quarks.We also have ( d ′ ∩ b, d ′ ∩ b ′ ) = (1 ,
2) if ( − τ b + τ d χ = − − τ a + τ b (261) = (2 ,
1) if ( − τ b + τ d χ = 1 = ( − τ a + τ b (262)which generate the required total of L lepton doublets (with Y = − ), while the lepton singlets arisefrom ( c ′ ∩ d ′ , c ∩ d ′ ) = (3 , (263)the former giving the ℓ cL charged lepton singlets, and the latter the ν cL the neutrino singlet states. Forthe case (257) under consideration, equation (261) gives the location of the lepton doublets. The charged24epton mass term arising from the lepton doublet at d ′ ∩ b require the coupling of the states at d ′ ∩ b , b ∩ c ′ and c ′ ∩ d ′ , which is allowed by the conservation of Q b , Q c and Q d . However, the charged lepton massterms arising from the two lepton doublets at d ′ ∩ b ′ require couplings that again have ∆ Q b = 2 . Similarly,only one neutrino mass term is allowed by conservation of Q b . The alternative choice described by (262)allows two lepton mass term for both charged leptons and neutrinos. Thus, at the perturbative level, afterelectroweak symmetry breaking, we either have two massive quark generations and one massive leptongeneration, or vice versa . In the model discussed in reference [11], the same correlation is obtained.In both cases the missing couplings can only be provided by non-perturbative instanton effects. Thesegenerate terms in the superpotential W of the form W ≃ Y i Φ i e − S inst (264)that violate the global U (1) symmetries that survive after the Green-Schwarz mechanism breaks anyanomalous U (1) gauge symmetry [36]; here Φ i are the (generally charged) matter superfields and S inst is the action of the non-perturbative instanton. Such a term is allowed if and only if the gauge transforma-tion of the matter field product Q i Φ i under an anomalous U (1) gauge transformation is cancelled by thetransformation of the exponential factor induced by the shift of Im S inst under the U (1) transformation[37]. Under a U (1) κ gauge transformation, associated with the stack κ , parametrised by Λ κ , in whichthe 1-form vector potential A κ is shifted by δA κ = d Λ κ (265)the imaginary part Im S inst of the instanton action (233) is shifted by δ (Im S E2 ) = Λ κ Q κ ( E ) (266)where Q κ ( E ) is the U (1) κ charge of the instanton, given by Q κ ( E ) = − Ξ ∩ N κ ( κ − κ ′ ) (267)To repair the missing Yukawa couplings we require that Q b ( E ) = − (268) Q a ( E ) = 0 = Q c ( E ) = Q d ( E ) (269)The general form of Ξ is given in equation (234) Then, using our solution for the SU (2) L stack on the AAA lattice given in Table 1 and equation (66) of reference [12], it follows from (268) above that ( − τ b +1 [ z + ( − τ b z ] = 1 (270)so that z or z , but not both, are odd. We also require, as in (269), that the instanton has zero charge withrespect to the other U (1) charges. For Q c this is guaranteed, since c = c ′ . Further, since a − a ′ = d ′ − d in our solution, Q a ( E
2) = 0 ensures that Q d ( E
2) = 0 . Thus, there is just one further constraint, whichyields Z − Z − Z + ( − τ a [ z + ( − τ a z ] = 0 (271)It follows from (270) that Ξ is of d -type, as defined in equation (241), and it is easy to find solutionswith all of the desired properties. For example Ξ = 12 ( ρ − ρ − ρ + ρ ) + 12 ( − τ a +1 (cid:2) ǫ + ( − τ a ǫ − ˜ ǫ − ( − τ a ˜ ǫ (cid:3) (272)with τ a = τ b mod 2 (273)The above solution gives (Ξ ∩ b, Ξ ∩ b ′ ) = ( − , − (274)25hus the required total instanton charge (268) derives from one (massless) particle at the intersection of Ξ with b , and two at the intersections of Ξ with b ′ . To repair the missing u -quark Yukawa, for example,we need a 5-point coupling in which both b and b ′ intersect the fractional 3-cycle Ξ of the instanton: ( a ∩ b ′ )( b ′ ∩ Ξ)(Ξ ∩ b )( b ∩ c )( c ∩ a ) (275)Since Ξ ∩ b = − , we should interpret it as one intersection with Q b = +1 , rather than -1 intersectionswith Q b = − . However, since b ′ ∩ Ξ = 2 is positive, the coupling (275) does not then conserve Q b , andwe conclude that we can not repair the Yukawa with this E2-instanton. Further, equation (268) requiresthat Ξ ∩ b − Ξ ∩ b ′ = 1 which can only be satisfied with non-zero Ξ ∩ b and Ξ ∩ b ′ when they havethe same sign, as in the above solution. Consequently Ξ ∩ b and b ′ ∩ Ξ cannot have the same sign inany of the solutions, and they therefore contribute zero to the total Q b charge in (275). Hence we cannotrepair the Yukawa with any of the single E2-instanton solutions of the constraints. The same conclusionfollows for the model discussed in reference [11], as well as for the models on the BAA lattice given inTable 6 of reference [12].
All of the models that we have considered have the attractive feature that they have the spectrum ofthe supersymmetric Standard Model, including a single pair of Higgs doublets, plus three right-chiralneutrino singlets. In the presence of the previously derived non-zero background field strength m H bg they are also free of RR tadpoles, and therefore constitute consistent string-theory models. We showed in § one of the axion moduli. Further, we found that it is easy tochoose a non-zero background field strength F bg that stabilises the K¨ahler and complex-structure moduliassociated with the supersymmetric mininima at values within the K¨ahler cone in which the supergravityapproximation is valid. In § non -supersymmetric stationary points ofthe effective potential, and in § all of the axion moduli can only be achieved by the use of non-perturbative instantoneffects, and these were discussed in §
5. In principle, such effects might also restore the missing quark andlepton Yukawa couplings to the Higgs doublets that are needed to generate masses when the electroweaksymmetry is spontaneously broken. However, we also showed that this does not happen for the particularmodels of interest here.
Acknowledgements
It is a pleasure to thank Joe Conlon and Andrei Micu for very helpful conversations, and also GabrieleHonecker and Timm Wrase for informative correspondence.
Appendix A: The Freed-Witten anomaly
We need to assess whether any of our stacks κ = a, b, c, d gives a non zero Freed-Witten anomaly ∆ κ ≡ H bg ∧ [ κ ] (276)where the background flux H bg in general has the form given in (87), and [ κ ] is the 3-form that is thePoincar´e dual of the fractional 3-cycle κ . The general form for the bulk part is [Π bulk κ ] = X p A κp η p (277)where η p ( p = 1 , , , are the 3-forms that are the Poincar´e duals of the bulk 3-cycles ρ p , given in eqns(329) ... (332) of reference [12]. As noted previously, we need only consider the R -even part of [ κ ] . On26he AAA lattice, [Π bulk κ ] + [Π bulk κ ] ′ = A κ ( η + 2 η ) + (2 A κ − A κ ) η (278) = 6 R R R [3 A κ ( σ − σ − σ + σ ) − (2 A κ − A κ )( σ + σ + σ + σ )] (279)where the terms on the right-hand side are defined in eqns (55) ... (58). Then H bg ∧ [Π bulk κ ] = − iR R R √ Vol [3 A κ ( p + p ) + (2 A κ − A κ )( p − p )] w ∧ w ∧ w (280)where the (1,1)-forms w k ( k = 1 , , are defined in eqn (34). Similarly, the general form for theexceptional part is [Π ex κ ] = X j ( α κj χ j + ˜ α κj ˜ χ j ) (281)where χ j , ˜ χ j ( j = 1 , , , are the Poincar´e duals of the exceptional 3-cycles ǫ j , ˜ ǫ j ; these are definedin eqns (369) and (370) of reference [12]. Then, on the AAA lattice, [Π ex κ ] + [Π ex κ ] ′ = X j (2 ˜ α κj − α κj ) ˜ χ j (282) = X j R π ˆ λ j α Vol (2 ˜ α κj − α κj )( ω j − α ˜ ω j ) (283)where ω j and ˜ ω j are defined in eqns (61) and (62). It follows that H bg ∧ [Π ex κ ] = X i,j iR π ˆ λ j Vol / (2 ˜ α κj − α κj ) P j e ( i,j ) ∧ e ( i,j ) ∧ w (284)Here e ( i,j ) are the localised (1,1)-forms defined in (35).Consider first the solution on the AAA lattice given in § κ = a, b, c, d respectively we have A κ ( p + p ) + (2 A κ − A κ )( p − p ) = (6 p , , , − p ) (285)so that cancellation of (the bulk part of) the Freed-Witten anomaly ∆ bulk κ for the κ = a and d stacksrequires that p = 0 . It follows from (88) that this in turn requires that n = 3 n and it is evident fromthe discussion following eqn (88) that this is not satisfied by any of our solutions. Thus ∆ bulk κ = 0 forthe stacks κ = a and d . Correspondingly, for the exceptional parts we find α κj − α κj = − t a (1 , , , t a ) (286) = 2 t b (1 , , t b , (287) = (0 , , , (288) = 3 t a (1 , , , t a ) (289)The localisation of the (1,1)-forms e ( i,j ) means that the cancellation of (the exceptional part of) theFreed-Witten anomaly ∆ ex κ for κ = a, b and d requires that P = 0 = P = P , and hence that ˆ n = 0 = ˆ n = ˆ n . We have already noted that tadpole cancellation requires that | n ˆ n j | = 12 for j = 4 , so the last of these can not be satisfied; with the choice t c = 1 , though, we can satisfy the firstof these. Thus, ∆ ex κ = 0 at least for the stacks κ = a and d . A very similar analysis, with the sameconclusions, applies to the other solutions derived from Table 1. The solutions on the BAA lattice arediscussed in § ∆ bulk κ = 0 for the stacks κ = a and d . 27 ppendix B: Positive definiteness of the axionic fluctua-tions It is clear from its definition that D ( a ) ≡ (2 γ + 1) a − is negative for a − < a < a + , where a ± ≡ ± r γ + 1 (290)and positive elsewhere. For future information, it is easy to verify that a ≡ γ (cid:16) − p γ (cid:17) > a − ∀ γ (291) γ (cid:16) p γ (cid:17) > a + for γ < (292) γ (cid:16) p γ (cid:17) < a + for γ > (293)The analysis of N ( a ) = (4 γ − a + 4 γ a + 1 , defined in equation (229), is more complicated. Forall values of γ it has a root at a = − , a saddle point at a = 0 , and one other stationary point at a = a D ≡ γ − γ (294)When < γ < / this stationary point is at a positive value of a and is a maximum. In this case, N ( a ) is positive for a < a < a , and negative elsewhere; here − a < a < and a > a D > is the(positive) root α ( γ ) of the cubic factor in equation (230); in the special case γ = 0 , for example, a = 1 .It is easy to verify that N ( a − ) ≤ (actually for all values of γ , with equality only when γ = 1 ); thus a − < a < a . Further, N ( a + ) is negative for < γ . . , vanishes when γ ≃ . , andis positive for all other values of γ ; it follows that a < a + for < γ . . , but a + < a for . . γ < / . Alternatively, when γ > / , a D is negative and N ( a D ) is a minimum. In thiscase, N ( a ) is negative for a < a < a , and positive elsewhere, and now both a and a are negativeroots of N ( a ) , with a < a D < a ; for γ < the position of the stationary point satisfies a D < − ,whereas for γ > we find a D > − . Thus when γ < , − a < a and a is the root of thecubic in equation (230), whereas for γ > , − a < a and a is the root of the cubic. Obviously, a + > a for values of γ in this range. These considerations lead us to consider three ranges of valuesfor γ , with the signs of the functions given in the associated Tables. • < γ . . Region a N ( a ) D ( a ) d ( a ) I a < a − - + -II a − < a < − - -III − < a < a + - +IV a < a < a + - -V a + < a - + -Table 1: Signs of the functions N ( a ) , D ( a ) , d ( a ) when < γ < . We need to identify the regions in which both d ( a ) and d ( a ) are positive. With a defined inequation (223), we note that D ( a ) is negative for any value of γ . Similarly, N ( a ) is positive(actually for any value of γ ). It follows that a is in region III of Table 1; this is consistent withthe observation above that a < a in this case. From equation (226) we see that a = 1 for valuesof γ in this range. Since D (1) = 2( γ − < , in this case, and N (1) = 8 γ > , it follows that28 is also in region III of Table 1. Finally, using the value of a given in (226), we find that D ( a ) is positive, (actually for any γ < ; it vanishes when γ = 1 , and is negative for all other values.)For values of γ . . , N ( a ) is negative; (it vanishes when γ ≃ . , and is positive forall other values.) It follows that a is in region V of Table 1 in which d ( a ) is negative. Thus theonly region in which both d ( a ) and d ( a ) are positive is a < a < a = 1 , and this is the case forall values of β ; note that this range does not require the solution of the cubic. • . < γ < The properties of the functions given above show that in this case a is in region III of Table 2, asis a , and if γ . . then a is in region V; otherwise it is region IV. Thus, if γ . . ,the only region in which both d ( a ) and d ( a ) are positive is again a < a < a = 1 , and as beforethis is the case for all values of β . However, in the case γ & . , N ( a ) and D ( a ) are bothpositive in the region a < a < a , so that both d ( a ) and d ( a ) are positive here too, provided that β < N ( a ) /D ( a ) . The determination of a requires the solution of the cubic, which we discussbelow. Region a N ( a ) D ( a ) d ( a ) I a < a − - + -II a − < a < − - -III − < a < a + + - +IV a + < a < a + +V a < a - + -Table 2: Signs of the functions N ( a ) , D ( a ) , d ( a ) when . < γ < / • γ > In this case we conclude that a is in region IV of Table 3, as is a , and a is in region V. Thusagain both d ( a ) and d ( a ) are positive in the region a < a < a , and this is the case for allvalues of β . As noted previously, there is a further region a > a in which positive-definitenesssis assured provided that β < N ( a ) /D ( a ) . Since N ( a ) grows with a more rapidly than D ( a ) , theupper bound on β grows with a .Region a N ( a ) D ( a ) d ( a ) I a < a + +II a < a < a − - + -III a − < a < a - -IV a < a < a + + - +V a + < a + +Table 3: Signs of the functions N ( a ) , D ( a ) , d ( a ) when γ > / . For γ < a = − , whereas for γ > a = − .To solve the cubic we write N ( a ) , defined in equation (230) in the form N ( a ) = ( a + 1)(4 γ − C ( a ) (295)where C ( a ) has the form C ( a ) = a + c a + c a + c (296)with c = 14 γ − − c = c (297)29he roots of C ( a ) are found by first changing variables from a to y , where a = y − c (298)to cast it in the canonical form y + py = q (299)with p ≡ c − c = 2(1 − γ )3(4 γ − (300) q ≡
127 (9 c c − c − c ) = 4(45 γ − γ − γ − (301)Equation (299) is solved by making Vieta’s substitution y = w − p w (302)so that w = q ± r q p (303) = 2(45 γ − γ − γ − ± γ − r γ − γ + 127 (304)Thus we can solve for any value of γ = 1 / . Consider first the plus sign in equation (304). For ≤ γ < / , the root α ( γ ) of the cubic C ( a ) = 0 increases monotonically from α (0) = 1 with α ( γ ) → + ∞ as γ → / from below. Thus in this case a > . For γ > / , the root α ( γ ) isnegative and monotonically increasing as γ increases, with α ( γ ) → −∞ as γ → / from above,and α ( γ ) → as γ → ∞ . For the solution corresponding to the minus sign in (304) one has to bemore careful because w (1 /
6) = 0 = p (1 / which makes the evaluation of y undefined at that value of γ ; one has to do the ranges ≤ γ < / and / < γ < / separately, even though there’s nothingspecial about the cubic for this value of γ . However, the conclusion is that both signs give the samevalue of the root α ( γ ) for any given value of γ .The following two examples illustrate the procedure. • γ = 0 . Since γ > . , we expect positive-definiteness for some values of a in region IV of Table 2,besides those in − . a < a < a = 1 in region III. For this value of γ equations (226),(304), (302) and (298) yield a = 3 .
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