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

P. van Nieuwenhuizen, SupergravOy 397SU 8, while if S0~owere a subgroup of SU8 then the 8 of SU8 would have to split up in ten l’s of SO~~which is impossible). So supergravity rules out SU6, SU7, SU8, E6 and SO10, but allows SU5 as the grandunified group.Let us finally discuss how to determine which subsets of SU5 spin 1/2 multiplets have vanishing axialanomaly. The axial anomaly of a triangle graph (this is enough to consider) due to fermions in an St]5representation R which are minimally coupled to the SU5 gauge fields, is given byAabc(R) = tr[{Ta(R), Tb(R)}TC(R)] = dabcA(R)where dabc = tr[{Aa, A6}A~]and Ta(R) and Aa are the SU5 generators in the R and adjoint representation,respectively. To compute A(R) for the various spin 1/2 multiplets in216 = (10,3) + (5, ~)+ (45, 1) + (1,3) + (5, 1) + (1,6) + (24,3) + (5,8)504= (45, 3)+ ~40,3)+ ~24,1)+ (15, 1)+ (10, 1)+ (10,8)+ (10,6)+ (10,3)+ (5,3)+ (5,3)one uses that A(R1 + R2) = A(R1)+A(R2), while A(R1 ®R2) = A(R1) x dim R2 + dim R1 x A(R2).Since all representations of compact Lie algebras can be taken unitary, A(R) = —A(R). With theserules one can ex ress all A(R) in terms of A(S) and A(10). For example, A(10®5)= A(10)+A(40),where 9 ®~=~+9~and= 40 = — trace (5 x 10— 10 = 40 components).The reader easily computes the other A(R), and relates them to elements of FABC or PA. As iswell known, for SU5 one has A(5) = A(10). To prove this by elementary means, note that 5 splits upunder SU3 into 3 + 1+ 1, while 10 (= 5 X 5 antisymmetrized) splits up under St]3 into 3 + 3 + 3 + 1.Consider now in Aai,c(R) the indices a, b, c to be of the subgroup SU3, and make a similaritytransformation such that in Ta(R) the St]3 matrices appear in the left upper corner. The dabc are, ofcourse, invariant under this change of basis, but dabc (SU5) = dabc (SU3) for these a, b, c. Consequently,A(5, SU5) = A(3, SU3) and A(10, SU5) = A(3, SU3).With these results one finds several subsets which are free from axial anomalies (one of which isgiven below). However, decomposing each set under St]3 (color) X U1 (em) and requiring vector-likecouplings to these groups, there is a unique subset which has more chiral fermions than all others(45 + 45)+ 4(24) + 9(10 + 10) + 3(5 + 5) + 3(5 + 10) + 9(1).All fermions in this set can acquire masses (between 1015 and 10’~GeV), except the three chiral (5+ 10)multiplets. This is the result mentioned above.There are other approaches to phenomenology. One of them (due to E. Witten) starts from thed = 11 dimensional model of section 6.2, and assumes that the 11 —4 = 7 dimensional internal space is acoset space G/H. If one takes G = SU3 x SU2 X U1, the minimal H which still allows a nontrivial actionof each of the factors SU3, St]2 and U1 of G on G/H is H= SU2 x U1 x U1. Remarkably, this space isexactly 7-dimensional. If one decomposes the elfmetric 2K’~a (y), g~in where the the wayrectangular we have discussed matrix K’~a(y) in section are 6.3, theand Killing replaces vectors theof7 the vector 7-dimensional fields A5.a (x, y-spacey) by A5. (a (x~ = 1, 12 and a = 1, 7), and if one substitutes this ansatz