298 P. <strong>van</strong> Nieuwenhuizen, SupergravityI = J d~x[aR,,~, m”(M)R,,j”(M) 0,,,~+ $R,,,,(Q)y5(—C’)R,,,,~(S)+ yR,,,~(A)R~(D) + ~R,,.,(P)R,,.,.(K)]&”’” (4)(note that R,~(S)starts with ~ + ~ For internal symmetries such as electromagnetism, theaction is unique and non-affine, namely the Maxwell action. Hence, since the internal symmetriesconsist here of only the U(1) chiral invariance, we add its actionI = J d~x[8R,,~(A)R,,~(A)g”~g”~’Vg] (5)Under local M, D, A gauge transformations (given by oh”,. (D,,,e)”’ with e equal to Amn, AD, AA) theaction is invariant if one uses the rule of transformation of curvatures. Under local K-gauges, the actionis invariant if the P curvature <strong>van</strong>ishes and if the 0-curvature is chiral-dualR,.,, m(P) = 0, R,.~(Q)+ ~ ,.~“=R,,,~(Q)y = 50. (6)Local S-invariance follows then iffl=2iy=6=—8a, ~=0. (7)As we will discuss shortly, local 0-invariance follows if one imposes one more constraintR,,,,(0)o” = 0. (8)Thus all constraints are derived from a dynamical model by requiring various gauge invariances. Inprinciple, the transformation rules of fields are modified when one solves these constraints. However, asone easily verifies, all constraints are M, K, D, A, S invariant, and thus for these symmetries all fieldskeep their transformation rules 3h/~= (DE)A and the action remains M, K, D, A, S invariant.One can solve the constraints above. The P-curvature yields the torsion equation,my4’~)+(e”btm — em,,b”). (9)(0,, = _w mn(e)_.i(4’ — * yn4’m + ~pThus there is the same torsion as in ordinary supergravity, induced by gravitinos, plus torsion inducedby the dilation field.The solution of the 0-constraints eliminates the conformal supersymmetry gauge field= (y”(S~v+(7~~~v), 5,.,, = ~ /J,
P. <strong>van</strong> Nieuwenhuizen, Supergravity 299transformation rules for the dependent fields arise after solving the constraints (see subsection 3.2)= ~Rmn(0)y,. 0 (11)= y” (y5R,,,,,(A) + ~1~,,,,(A)) 0. (12)These results and a similar result for O’f”,. will be derived in subsection 4.We now discuss the 0-invariance of the action. If all fields would transform as given by the structureconstants alone, one would have81 = 8a Jd4x [1~~(0) y5y0,e0R,,,,~(K)+2iR~~(S)(ysR,.,,(A)+~R,.,,(A))eo— ë,.yy . 4’)R,,,,(A)R~~(A) + 2~0y”4’0R,.,,(A)R’”’(A)]. (13)The extra terms O’w,,”’ and 8’4i,. give an extra 8’I, whose general form was derived on page 281.However, since Qi”~ is constant and proportional to “~ for A = M, 5, the Bianchi identity holdsso that the D,,R,,. terms <strong>van</strong>ishes, and also the 3,,Q terms <strong>van</strong>ish. One is left withS’I = 8a J d 4x[28~w,.m~1~~”~(K) ~ + 41~”(S)y,~,y~5’4,, + 4ü~,.(y 5R”~(A) + ~1~”’(A))o’q 5,,]. (14)The R(Q) R(K) terms in 81 cancel against the R(K) term in 8’I if and only if 8lw,.mn is as given above.For this to be the case, one needs both 0-constraints; with the dual-chiral constraint alone nocancellation would occur. Furthermore, substituting O’~,,.all other terms in 81+ 8’I cancel at once.We now show that the dilation field cancels from the action. The remaining independent fields areem,,, 4’~,,,A,,, b,. and ftm,. while ffll,. is nonpropagating and is expressed in terms of other fields bysolving its own field equation (just as ~,.mn in ordinary supergravity). Let us consider how these fieldstransform under K-gauges. Only b,. transforms, and since 8Kb,. = 2~K,,. one can gauge away b,.altogether. This one might have expected since there are as many K-gauge degrees of freedom asb-components. However, since the action was K-invariant to begin with, it must therefore beindependent of b,,.The action is thus unique and determines in turn the constraints [523].In tact, as we shall see, thesolution of the ftm,, -field equation is at the same time the solution of an extra constraint [535].At thelinearized level, these results were obtained using superspace methods [191].Let us close this subsection by showing that there are equal numbers of boson and fermion states.Since the graviton and gravitino have higher derivative actions, one has two spin 2 states (El2 counts astwo El) and three spin ~states (Jill counts as three )T~.Indeed, one finds one massless (2,~)multipletand one massive (2, ~,~,1) multiplet [191,526].4.3. Constraints and gauge algebraFor ordinary supergravity, one can formulate the theory in second order formalism either byimposing a constraint on curvatures (R14,,tm (F) = 0) or by solving a field equation belonging to the gauge
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