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Nilpotency of BRST operator in gravity

Physics Asked by Pratik Chatterjee on January 18, 2021

I am going through the BRST quantisation in Perturbative quantum gravity and looked at the papers of Nishijima and Ojima. I am confused about the closure of the BRST operator; I.e $s^2=0$, particularly in the case of the ghost field.

The BRST transform of the ghost field is:
$$sc^{mu} = c^{lambda}nabla_{lambda}c^{mu}$$

Therefore, $$s^2c^{mu}= sc^{lambda}nabla_{lambda}c^{mu}-c^{lambda}nabla_{lambda}sc^{mu}= c^{nu}nabla_{nu}c^{lambda}nabla_{lambda}c^{mu}-c^{lambda}nabla_{lambda}c^{nu}nabla_{nu}c^{mu}-c^{lambda}c^{nu}nabla_{lambda}nabla_{nu}c^{mu}=- c^{lambda}c^{nu}nabla_{lambda}nabla_{nu}c^{mu} $$

So I find $s^2 c^{mu}$ is not vanishing here. What mistake am I doing here?

Here is the link of the paper by Nishijima:
https://doi.org/10.1143/PTP.60.272

The particular equation numbers are: 2.25 and 2.26, in page number 275.

One Answer

The anticommutativity of the $c$'s means expression is a commutator $$ c^lambda c^nu nabla_lambdanabla_nu c^mu=frac 12 (c^lambda c^nu-c^nu c^lambda)nabla_lambdanabla_nu c^mu = frac 12 c^lambda c^nu [nabla_lambda, nabla_nu] c^mu = frac 12 c^lambda c^nu {R^mu}_{alphalambda nu}c^alpha =frac 12 c^alpha c^lambda c^nu {R^mu}_{alphalambda nu} $$ which vanishes (for torsion free) by the first Bianchi identity.

Answered by mike stone on January 18, 2021

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