 Solution

The BRST operator $Q_B$ is Grassmann odd and represents the gauge symmetry on the gauge-fixed state space. Requiring two successive BRST transformations to vanish meansThis nilpotence makes physical states a BRST cohomology. If $[Q_B,S_0]=0$ and $\Psi$ is the gauge-fixing fermion, the graded Jacobi identity givesso $S=S_0+\{Q_B,\Psi \}$ is BRST invariant.

A holomorphic field of conformal weight $h$ has the Laurent expansionUnder the state–operator correspondence, $\varphi (z)|0⟩$ must be regular at the origin. Terms with $n>-h$ have negative powers, soFor the anticommuting bc system,Separating creation and annihilation modes and summing the geometric series for $|z|>|w|$ gives the bc ghost operator-product expansion

The BRST current built from the matter and ghost stress tensors has an operator-product expansion with $c$ whose residue givesFor a matter Virasoro primary operator $\Phi (z,\bar{z})$ of weights $(1,1)$, the two standard string vertex operators areThe first is the local, unintegrated vertex. Using $Q_{B}c=c\partial c$, $Q_B\bar{c}=\bar{c}\bar{\partial }\bar{c}$, and the weight-$(1,1)$ transformation of $\Phi$, the terms cancel pairwise and $Q_{B}U=0$. For the integrated vertex,so $\{Q_B,V\}=0$ on a closed worldsheet because the variation is a total derivative. Both vertices therefore represent the same BRST cohomology class.