352 P. <strong>van</strong> Nieuwenhuizen, SupergravityKA,~ is not allowed (it would violate Maxwell gauge invariance)~‘= -~R(e,~— e~[~F’~’clr~ + ~ EFai 3,,ô + ~$Y8] + CKEL~ôe m&fr~.K (12)K1D~(W)E+ D(F~”~. — 86~f’~”~) fta~y&= EiF[,~4~1. (C, D, E are constants.)Remarkably enough, the ~ terms are obtained by putting in the gravitino action ~(w+ ~)as connectionand also taking ~(F+ F) in the Noether-type coupling. I do not know why this “minimal” solution yieldsall cl’ terms. (In N = 2, d = 4 supergravity, and also in the d = 4 Maxwell—Einstein system one finds thesame features.) Roughly speaking, the covariantization terms in F have twice as many ~‘s as the termswith F, so one needs half of them to obtain a supercovariant field equation, but unlike the N = 2 model,the four cl”s in the F terms do not appear symmetrically, so that this argument is not complete.The spin connection w solves ôI/öw = 0 and is given by(Z)~mn= A~m~ — ~ (13)In d = 4, a. = c~,but here they differ. This illustrates once again that by going to more complicatedtheories one finds out which equalities are a coincidence and which are not.The gauge algebra is the usual one except that the Lorentz parameter istm’2 = ~W + 2(T~ — 24em’2e1F~)Pa,~ (14)Awhile two supersymmetry transformations on A~ lead to a gauge transformation 5A~,,.=3~A~+ cyclic terms, where= —~E 2r,~v — 1 ~ (15)Terms such as ~always come from converging P in {Q, Q} = P + more, into~ I.!.”\ ~ /.~ALA0gen.coord.~ ) — °gauge~~ ~A—The gauge algebra reads[5~( ~), ôQ(E2)] = ögen.coord.(~’) + 50(—r~)+ ÔL(Atm’2 in (14)) + 6M~welI(A,~in (15)) (17)and this suggests strongly that the N = 1, d = 11 model can be obtained by gauging Osp(1/32) sinceSp(32) is spanned by Jtm, F~’2,F”2’2’~.(These are the matrices satisfying the definition of symplecticmatrices CM + MTC = 0 with antisymmetric C. We recall that CPA ~ One would like to
P. <strong>van</strong> Nieuwenhuizen, Supergravity 353consider e~and w~”2as the gauge fields of Ftm, and Fm’2. However, A,~ cannot be considered as thegauge field of fm’2~st(it does not have enough indices and it is totally antisymmetric), and this is aproblem for future research [559].6.3. The N = 8 model in 4 dimensions by dimensional reductionIn this subsection we discuss the important and pioneering work of Cremmer and Julia [96,99]. Aswe saw, in ordinary dimensional reduction all fields depend only on x1, . . . , x~instead of on all elevencoordinates. One then proceeds to compute the various quantities appearing in the action: inversevielbein and connections. One writes the four-dimensional part of the elfbeins as ö~’em1.where ô is thedeterminant of the 7 X 7 part of the elfbein, and by choosing y appropriately, the Einstein action of thereduced theory has the usual form.In this section we discuss how the massless N = 8 model in d = 4 dimensions is obtained by ordinary(i.e. not accompanied by spontaneous symmetry breaking) dimensional reduction. This analysis will alsolead us to deduce that the d = 4 theory has as internal symmetry an E(7) global x SU(8) local. However,the SU(8) local has not 63 gauge fields, so that this SU(8) cannot serve as a grand unification ofSU(3)x SU(2) x U(l). Rather, the connections are nonlinear functions of other fields, in the same wayas the spin connection w(e) gauges SO(3, 1) but is not an independent gauge field.In d = 11 dimensions, one has general coordinate transformationsOVA =~“öHVA+(öA~”)Vll, (A,H=1,11). (1)Splitting A = (A, a) with A, ~u,v = 1, 4 and a, f3, y = 5, 11 one has under ordinary dimensional reduction1,... , x’~)restricted to ~ . , x~)a global symmetry group generated bywith fields ~(x= (~= 0, ~“ = M”~x~). (2)Indeed, OVA again depends only on x’,. . . , x4 as follows fromOV~0,OVa =(~a~)V13M~aV1~. (3)One must in fact require that oar = M~ = 0 in order that the reduced Hilbert action is still0Ainvariant (peR)under since the thisaction specialis class a scalar of general density. coordinate Usually one transformations. discards this total <strong>To</strong> see derivative this, note since thatcoordinates O(eR) =tend toinfinity, but now 7 coordinates have become little circles and one must require that 9,~’2= 0 = M’~a.Thus, in d = 4 one expects a global group SL(7, R).The Lorentz group 0(10, 1) in d = 11 acts on tetrads asOeM = ~M~N (4)Splitting the flat index M into (m, a) with m, n,~ = 1, 4 and a, b, c,= 5, 11, we haveoem = wmen + meaôea = wae’2 + (J)aeb (5)Thus we can fix the parameters w’~,.= _~~a by putting e’2~= 0. In this case the vielbein has the formtn aM.I~L ~Len ‘~0 e,~ (6)
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