SUPERGRAVITY P. van NIEUWENHUIZEN To Joel Scherk 0370 ...

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340 P. van Nieuwenhuizen, SupergravityWe can now define the dynamics of gauge supersymmetry. It consists of the following equations ofmotionRAn = AgAn. (33)They follow from the following actionI=Jdmxcro\/_g[R+(n_m+2)A]. (34)11 = m — n.The symbol \‘—g is the superdeterminant (see appendix) and we used thatThere isgnAg”also the familiar Bianchi identity(_)A(RHA —~g’MR)~A=0. (35)Clearly (for A non-zero) flat space (gn11 Smn) is no solution.We turn now to the question whether gauge supersymmetry contains global supersymmetry. Thiscan be analyzed by looking at the vacuum expectation value of the field equations. Consider thatVA = (01 VA~4I0)is given byV,.m(O) = 5~, V,.”~= 0, VO,m(O) = _~(6Fm)0,, V0,”’~°~ = ~ (36)This is the most general form for V1%” which is invariant under constant transformations with~A(O)= (~i’F~0’, E”). (37)The functions (CF~)0~must be symmetric in (ai, /3j) in order that the metric in g(O)1%fl satisfy theEinstein equations (33). (A tree level statement.) Hence= y”F0(s) + iy”-y 5F 0(a). (38)Any global tensor QAH (i.e., a constant tensor invariant under ~~~0))is of the form= ki’q,.,,, Q,..a, = kl(0F,.)aA,= k2,11C~+ k1(OF~)~1(ëF,.)~1.Squaring Vm(O) according to g = V~V one finds for g~°~ the same result as in Q,,,. but with k1 = 1 andk2,0 = kSq where k is the k in i~. In the tree approximation g~hobeysprovidedRAn(g~)= Ag~)~ (40)—2k 2F,.F” = A, —k2 Tr(F~F,.)= An,.,.. (41)(39)
P. van Nieuwenhuizen, Supergravitv 341These equations follow from the Bose—Bose and Fermi—Fermi part of (40); the Bose—Fermi content isredundant.If one remains at the tree level, then this implies that N = 2 (trace the second equation). However, inthe presence of quantum loop correction, calculated with the globally supersymmetric gauge fixing term,the corrections to (0JgAn~0)are obtained by differentiating the effective potential with respect to g~°h.The result must again be of the form of Q,.,. but also R 1%~must be of this form (with calculablecoefficients). The only difference is that in (37) one must replace A by A — K’ and A — A’ respectively. In thiscase it may turn out that tracing of (37) 2 selects particular N values or perhaps none. In any case, (37)+ 1 unknowns (namely, the elements of F(s) and F(a), and A).represents Hence, if there N(N are + 1) solutions, + 1 equations therefor must N be extra symmetries.To investigate the residual symmetries of the vacuum, one solves the Killing equations(0) — — ~A+fl+AH42The general solution is (see Phys. Rev. 10 (1978) 2759):= a” +A”,.x~+EF”0= ~(Ao-r~o~’ + “‘ — M”’,~10~’ (43)where the matrix M must satisfy (CM) + (MTC) = 0, and [M, F”] = 0. Lorentz covariance impliesM”’,3J = (M0)~5~ + (M1)’1(i75)”,9 so that the two N x N matrices M0 and M1 are antisymmetric in i andj.The important point is now that corresponding to the global symmetries of the vacuum metric given byM, the full dynamical theory has the corresponding local symmetries. In particular, the component fieldsare found inside the metric tensors as followsg,.~(z)= g,.,.(x) + Otfr,.,.(x) += cui,.,ai(x)+ (~M’)a,47’,.+ (gym) e”(x)+.... (44)The fields qS~are Yang—Mills fields and I denotes the generators of the internal group M. Also, uponexplicit evaluation, it is found that the action in (34) has as unbroken symmetries only the spacetime andlocal supersymmetries, plus the Yang—Mills symmetries associated with M. All other symmetriescontained in ~ (z) are broken spontaneously, hence there are no fields with vanishing masses with spinsexceeding 2. One might call the symmetry group which is given by the Killing vectors in (39) the groupGsp(N/4)R.,.~since it contains the contracted symplectic group Sp(4) (hence the spacetime symmetries)plus the group 0 of Yang—Mills invariances plus its grading.Let us repeat the analysis performed above on the metric now on the super-vielbein. Requiring that= 0 with V/’~°~ and e” ~ given before one finds that= A~”(x), E~”’(Z)~o~ = 0Eajtm (z)~°~ = 0, Eaj”(Z)~°~= ~(A. O•)baSli + p,j.bi (45)In other words, of the whole original tangent symmetry group Osp(3, 1/N) only the diagonal part0(3, 1) x G remains unbroken.
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