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

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362 P. van Nieuwenhuizen. Supergravi~’yAn often used matrix representation of the gamma matrices is/0 lcTk\0 )‘ ~11 0’—i)’\lUk/0 —10For two-component spinor formalism one uses another representation, see appendix E. A Majoranarepresentation is, for example,10 —10B. Majorana spinors10 iU2\ 11 072 0 )‘ “~ ‘\0 —1/0 1U3‘~‘~~iU3 0~A four-component spinor A”(a = 1,4) transforms under Lorentz transforms actions as(A’)” = = ~[y,~’ ‘y,,] (1)4 are purely imaginary (in our conventions s, i’ = 1, 4 and {y,,, y~}= 28,.,,). Thewhere Dirac conjugate w’~are real spinor and w’ A4y30 (A ~ (y 0 transforms thus as(~)~ = A,~(exp— (2)because o,.,. are antihermitian (y,. is always taken to be hermitian). The Majorana conjugate spinor A T isdefined byAT 0 = A 0C 00 (3)and is required to transform in the same way as the Dirac conjugate spinor. Note that A and AT are ingeneral complex. We will always write the index of A as a superscript and that of A as a subscript. FromoX0 = A0(C )“ (exp ~(4)it follows that the matrix C01,must satisfy (note the position of the indices)C00(o,,~~~(C’)”~ =(5)The most general solution of this equation isC= a(1+ y5)Co+/3(1— y5)Co(6)where C0 is a special solution, usually called “the charge conjugation matrix”, and discussed below, anda and /3 are arbitrary constants. In two-component formalism, C0 is diagonal and equal to ( ,.~,,, ~4B),and since dotted and undotted spinors transform independently, while cr,,,. is also diagonal in dotted—undotted space, the appearance of two arbitrary constants is clear.A Majorana spinor is per definition a spinor whose Dirac conjugate is proportional to its Majoranaconjugate (note again the position of indices)
P. van Nieuwenhuizen, Supergravity 363A,. = A~C 1,0b, b arbitrary. (7)Hence AT= AtC*yb* = ATCY4C*7(b*b) from which it follows that CY4C*Y must be a positivenumber times the unit matrix. This is indeed the case in four dimensions (see below), but, for example,in five dimensions this is not the case and there one needs an internal index in order to define Majoranaspinors. We will always scale C such thatTC. (8)A=AIn order that the relation A = ATC is maintained in time, one has for a spinor whose field equation isthe Dirac equation (y”ö,. + M)A = 0C C-1—— T 9X(C-’7~TCä,.+M)OJ )‘L — ‘/~~ ()For massless spinors, one finds the weaker conditionC~y,.Cl=a~y,.T, a2+1. (10)(One easily proves that a2 = +1 from (5).) In four dimensions one finds solutions with a = +1 and —1, but,for example, in five dimensions a = +1. However, although in five dimensions a matrix C exists, one needsmore than one spinor in order to define a Majorana spinor, as we have seen above.Since the dotted and undotted components of A are mixed in the Dirac equation, one now gets thestronger condition Cy,.C-1 = ±y,,Tfor C, and it is clear that if a C exists, it is unique up to a constant.This matrix C is the charge conjugation matrix. That a C exists follows from the fact that the matrices(—y4,,”) (as well as +y4,,T) satisfy the same group multiplication table as y,,. Hence, since according toSchur’s lemma there is only one inequivalent four-dimensional representation of the finite group with 32elements spanned by y-matrices, y,. and ~,.T are indeed equivalent, as are y,. and +y,.T.For spin 3/2 fields one can essentially repeat their steps, and the same results are obtained.In a general representation of {y,,, y,.} = 20,,,. with, however, hermitian y,,, the charge conjugationmatrix C is antisymmetricCr=_C. (11)However, only in special representations is it true that (C- 1)as = — C,.1, (as should already be anticipatedfrom the position of the indices). To see this, note that using Cy,.C-’ = ay,,T twice, one findsTCy,.C-lCr= y,,. (12)(Cl)Hence C-1C~commutes with y,,. Thus C-ICr= kI, or CT= kC so that (cl)T C= k2C and k = ±1.Thus (Cy,)T= ak(Cy,.), (Q,.)T k(Cu) (Cy5y,.)T —akCy5y,. and (Cy5)T= kCy5. Clearly a = +1and a = —1 are both allowed (C = y~y~and C = y~in a Majorana representation). Since the 16 Diracmatrices are linearly independent and there can be at most 6 antisymmetric matrices, k = —1 and C isantisymmetric.Taking the hermitian conjugate of Cy,.C-’ = ay,,T one finds that CrC= 11 and clearly 1>0. Thus.
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