262 P. <strong>van</strong> Nieuwenhuizen. Supergraviti’(ii) a massive (2, ~,~,1) ghost multiplet with M 2 =(iii) two massive (~,0~,0~)multiplets, the first containing 5, F and one mode of y~R. the othercontaining t9 A, the scalar in R + R2 and the second mode in y R. Both have M2 = ~y/a.If one would not have known these auxiliary fields, one would have predicted the existence of auxiliaryfields which lead to the encircled states(2, ~),(2, ~,~,CD), ~,0~,®)‘ U’ (I (ö~)). (5)Of course, this does not uniquely determine which field representation should yield these states, but theresults are so simple that one can hardly miss AM, S and F. It also explains that a chiral gauge field isneeded in N = 1 conformal supergravity.One can turn these arguments around and deduce some knowledge about auxiliary fields from therequirement that all states should fill up multiplets [168]. In the N = 2 model, one of the linearizedinvariants contains R2 + R . yJ’~y•R + 2[8M(t,,,. — 2~h’2(3MA~— 8~AM))]2where the auxiliary field t,,~rotates into the gravitino field equation! [169](A similar thing happens in the N = 8 model in 11dimensions [559].)Adding the Poincaré action, the states constitute one massless (2, ~,~,1) multiplet andthe massive (1, (~)4, 05) multiplet. Adding the invariant which starts with R~,., a massive(2, (~)4, (1)6, (~)4, 0~)multiplet is produced. A curious feature is that an entire (1, (~)4, (0)~)multipletdevelops, generated by the auxiliary fields. All N = 2 auxiliary fields occur in these multiplets (seesection 6).2.12. On-shell counterterms and non-linear invariancesOne approach to renormalizability has been to investigate which supersymmetric counterterms arenon-zero on-shell, based on the assumption that counterterms must be supersymmetric. This assumptionmay not be correct as the following counter examples show. In Yang—Mills theory, the Lorentz invariantW,. (W,. n FM,.) is not a Yang—Mills invariant, nor does it <strong>van</strong>ish on-shell, but its variation <strong>van</strong>isheson-shell. If it would be produced in one-loop corrections as a counterterm, it would clearly be incorrectto first write down all possible off-shell invariants and then to study which of them <strong>van</strong>ish on-shell.As a second example [165], consider the globally supersymmetric Wess—Zumino model~ 2?~’= MA)2 )~2A/A+~F 2+~G2 (1)= —F(A2 + B2)+ 2GAB + ~(A + iy5B)A.Both ~ and ~2’g are separately invariant under t5(A + iB) = i(1 + y5)A, 8(F + iG) = iJ(1 + ‘y5)A, and= /(A — iy5B) + (F + iy5G)e. The one-loop divergences £2? are proportional to only ~2?~”; this is apeculiarity of this model which has been explained in several ways, but which fact we will take forgranted. It is now clear that if one eliminates F and G from £9~’~ + ,2?g, from £2? and from thetransformation rules by substituting the equations of motion obtained from .2?~”+ ,~2’g,then ~ + ,2’g isstill invariant but £2’ is no longer invariant. In fact one finds£2? = 2-(aMA) 2 ~(8MB)2— — ~IA + ~g2(A2+ B2)2 (2)6 £2’ = ig(A — iysB)21A with l.A = IA — 2g(A + iy5B)A. (3)
P. <strong>van</strong> Nieuwenhuizen, Supergravity 263One can also add masses and consider non-renormalizable models such as the Lang—Wess model. Inall these cases one finds the same phenomenon. Thus there are three possibilities for counterterms:(i) £2’ is off-shell invariant;(ii) £2? =0 on-shell but not off-shell invariant;(iii) 8(&2’) = 0 but &.2’~0 on-shell, nor off-shell invariant.It is easy to construct many examples of case (iii). Take any invariant action, thus 61 = ôçb’I,61,g + (&/i”,g)I,j 1 =0.Differentiating so that this identity with respect to some parameter g in the action, one finds81’.g <strong>van</strong>ishes on-shell but I~gitself is not invariant in general.Thus to prove the finiteness of the S-matrix it is incorrect to demonstrate that all invariant counterterms<strong>van</strong>ish on-shell. It is correct to show that on-shell no £2? exists which is on-shell invariant. In that case,one must use the on-shell transformation rules and, for example, to show that no £2’ exists withoutpurely gravitational sector, requires very detailed analysis. Details of such analysis have not yet beenpublished.From a more general point of view [165],the generating functional in the background field methodZ(~)=fdx exp i[I(cb +x) — I(cb) — I(~).~x] (4)explains these examples. If the original action is invariant under 4’ -+ g(4’) then in general thebackground functional is invariant under background symmetry transformations 4’-~g(çb) and x -~g(4’ + x) — g(çb), provided g(4’) is linear in 4’. (In that case I.~xis again an invariant.) In our examples,however, g(4) was after elimination of auxiliary fields no longer linear in fields.In order to connect these considerations to supergravity, one must find arguments which show thatcase (iii) counterterms do not occur there. In most of the analysis of on-shell invariants, auxiliary fieldswere not considered since theywere unknown at the time. Now, however, one can discuss the influenceof nonclosure of the algebra on invariance of counterterms. Since the commutator of two symmetrytransformations is again a symmetry of the original action, in supergravity with open algebra (withoutauxiliary fields) there are infinitely many field-dependent transformations under which any trueinvariant of the theory should be invariant. In most cases this leaves the original action as the only trueinvariant, and dangerous counterterms should then be of type (iii) with respect to these symmetries.It should be noted that in supergravity the transformation laws with auxiliary fields are still non-linear.However, the one- and two-loop finiteness remains true, since it was proved on-shell using helicityarguments [512,283] or Noether methods [123].2.13. Canonical decomposition of the actionThe Hamiltonian approach to supergravity, using path-integrals, first demonstrated four-ghostcouplings in supergravity (Fradkin and Vasiliev [226]).As an introduction we discuss the “p, q”decomposition of the action. This action analysis is the counterpart of the analysis of the field equations,and, of course, we will again find that there are two physical modes. However, unlike in the fieldequations, one does not need to choose a gauge for the actions.We consider the free linear spin 3/2 action first, and make a spacetime split [126]11= ~E”fIIiys( 7j494 — 749j)lIJk + E”~4y5%84rk. (1)
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