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

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

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252 P. <strong>van</strong> Nieuwenhuizen, SupergravityIf the algebra closes, B =0. In the second and third terms of B” one can bring ~“change in sign. Moreover, since qIk()Yk =77~k(yi it follows thatto the right withoutB” = B”(—~~’. (16)We now write r = C’~A1,770 = C’3j12 and ~“verify a very useful identityCM3 in order to make contact with (10). One may~ + two cyclic terms = C’3A 3C”C”(—)”(3A ,A2). (17)In other words, using ghost fields the cyclicity is automatic. With i~u—)” = —~j~and= 7l~$,kR,,C”AC’3C” (18)one finds upon contracting B” with two antighosts and dropping the common factor 3A1A2, thefollowing result (after substantial relabeling of indices)C*AFA,IC*TFT.,Bhi = C*)FA jC*TEI[T1~öfö$a — 77y$,kRa R’y.kfl~a+R’7kn~’,,(—)”]C’~’AC’3C”. (19)This result looks indeed very much like (10) if one compares (11) with the expression between squarebrackets in (19). In fact, these expressions are equal. <strong>To</strong> see this note that i + I is even since B has thesymmetry in (16) but the two antighosts in (19) have symmetry (—) 1X~±1)~Since always either i or j inB” is fermionic, see (15), ii = 1. Thus the terms with B” in (19), coming from the triple commutator ofgauge transformations, are indeed the same terms as what is left of 61 (quantum).If one could argue that B” = 0, then one would have completed the proof of BRST invariance.However, B is nonzero and we proceed to solve it from (13). On-shell R ~AAA...=0 and defining R A suchthat R’A 0 on-shell (parts <strong>van</strong>ishing on-shell are proportional to I.,,, and can be incorporated into thenon-closure function q), it follows that AA = I(cl),A”. Hence (13) becomes I(cl),(RIAAA 1 + B” +cyci.) = 0 which tells us that the second factor is a gauge invariance of the classical action. ThusR&AAAE.. + B” + cycl. = R’AX” + I(cl),1M”with I(cl),,I(cl).,M” =‘O and this yields finallyB” = (RIAXIA + IC] MJII) + (— )“~(j ~ 1) (20)where M” has the same symmetry in (ji) and (ii), namely that of n”. Thus M” is completely symmetric.We will now show that in supergravity the function M <strong>van</strong>ishes while X is only nonzero when Arefers to local Lorentz rotations. From (15) it follows that B” does not contain a~por aae terms. Indeed,i~ and fl~o,kand R’71, do not contain any derivatives, while R”7 does not contain a derivative of fieldsand ~ contains only a derivative of a field (of the tetrad) if r refers to local Lorentz rotations and a, /3to local supersymmetry (in this case the commutator is proportional to üj~m12(e, 4’) as we saw). Thus Mi”<strong>van</strong>ishes. The Jacobi identity thus becomes[R’AA”a’3y + IiR’AX~’p7(— )‘~ 1]C”AC’3C°( — )~= 0 (21)

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