String Theory and M-Theory

4.7 SCFT and BRST 143
These contribute ĉ = −10, and so the superconformal anomaly cancels for
D = 10.
As in the case of the bosonic string theory, the quantum action has a
global fermionic symmetry, namely BRST symmetry. In this case the transformations
that leave the Lagrangian invariant up to a total derivative are
δX µ = η(c∂X µ − i
2 γψµ ), (4.144)
δψ µ = η(c∂ψ µ − 1
2 ψµ ∂c + 2iγ∂X µ ), (4.145)
δc = η(c∂c − γ 2 ), (4.146)
δb = ηTB, (4.147)
δγ = η(c∂γ − 1
γ∂c), (4.148)
2
δβ = ηTF. (4.149)
These transformations are generated by the BRST charge
QB = 1
(cT
2πi
matter
B +γT matter
F +bc∂c− 1 3
cγ∂β −
2 2 cβ∂γ −bγ2 )dz. (4.150)
The transformations of b and β, in particular, correspond to the basic equations
and
{QB, b(z)} = TB(z) (4.151)
[QB, β(z)] = TF(z). (4.152)
As in the case of the bosonic string, the BRST charge is nilpotent, Q2 B = 0,
in the critical dimension D = 10. The proof is a straightforward analog
of the one given for the bosonic string theory and is left as a homework
problem. One first uses Jacobi identities to prove that [{QB, Gr}, βs] and
[{QB, Gr}, bm] vanish if ĉ = 0. This implies that {QB, Gr} cannot depend
on the γ or c ghosts. Since it has positive ghost number, this implies that it
vanishes. It follows (using the superconformal algebra and Jacobi identities)
that [QB, Ln] must also vanish. Hence QB is superconformally invariant for
ĉ = 0. In this case [Q2 B , bn] = [QB, Ln] = 0 and [Q2 B , βr] = {QB, Gr} = 0,
which implies that Q2 B cannot depend on the c or γ ghosts. Since it also has
positive ghost number, it vanishes. Thus nilpotency follows from ĉ = 0.
As a result of nilpotency, it is again possible to describe the physical states