BRST Quantization of the Bosonic String: Nilpotence of BRST transformation (Polchinski)

Currently I am studying string theory and I encountered a bunch of interrelated problems in the context of BRST quantization which I can't solve for myself although I tried hard for some days.

My question concerns the BRST transformation of the bosonic string in eq. (4.3.1) of Polchinski's book. In the paragraph following these transformations, Polchinski says that "the reader can check nilpotence up to the equations of motion". I tried but wasn't able to complete the proof - here my calculations:

\begin{align} \delta_\mathrm{B}\delta_\mathrm{B}'X^\mu&=\delta_\mathrm{B}\left[\mathrm{i}\varepsilon'\left(c\partial X^\mu+\bar{c}\bar{\partial}X^\mu\right)\right]\\[2pt] &=\mathrm{i}\varepsilon'\left[(\delta_\mathrm{B}c)\partial X^\mu+c(\delta_\mathrm{B}\partial X^\mu)+(\delta_\mathrm{B}\bar{c})\partial X^\mu+\bar{c}(\delta_\mathrm{B}\bar{\partial} X^\mu)\right]\\[2pt] &=\mathrm{i}\varepsilon'\left[(\mathrm{i}\varepsilon c\partial c)\partial X^\mu+c(\mathrm{i}\varepsilon(c\partial\partial X^\mu+\bar{c}\underbrace{\bar{\partial}\partial X^\mu}_{=0}))+(\mathrm{i}\varepsilon\bar{c}\bar{\partial}\bar{c})\bar{\partial}X^\mu+\bar{c}(\mathrm{i}\varepsilon(c\underbrace{\partial\bar{\partial}X^\mu}_{=0}+\bar{c}\bar{\partial}\bar{\partial}X^\mu))\right]\\[2pt] &=-\varepsilon'\left[\varepsilon c\partial c\partial X^\mu+c\varepsilon c\partial\partial X^\mu+\varepsilon\bar{c}\bar{\partial}\bar{c}\bar{\partial}X^\mu+\bar{c}\varepsilon\bar{c}\bar{\partial}\bar{\partial}X^\mu\right]\\[2pt] &=-\varepsilon'\varepsilon \left[c\partial c\partial X^\mu-\underbrace{c c}_{=0}\partial\partial X^\mu+\bar{c}\bar{\partial}\bar{c}\bar{\partial}X^\mu-\underbrace{\bar{c}\bar{c}}_{=0}\bar{\partial}\bar{\partial}X^\mu\right]\\[2pt] &=-\varepsilon'\varepsilon \left[c\partial c\partial X^\mu+\bar{c}\bar{\partial}\bar{c}\bar{\partial}X^\mu\right]\\[2pt] &=? \end{align}

Here I used the equation of motion for the field $X^\mu$ and exploited $c^2=0=\bar{c}^2$. At this point, however, I do not see why the remaining terms should vanish.

Hint: The infinitesimal BRST transformation $$\delta_{\mathrm{B}}\left(F[X,c,\ldots]\right)~:=~F[X+\delta_{\mathrm{B}}X,c+\delta_{\mathrm{B}}c,\ldots]-F[X,c,\ldots]\tag{A}$$ acts "under" the spacetime derivatives marked in red, i.e. OP's second line should read \begin{align} \delta_{\mathrm{B}}\delta_\mathrm{B}'X^\mu &\stackrel{(4.3.1a)}=\delta_\mathrm{B}\left[\mathrm{i}\varepsilon'\left(c\color{red}{\partial} X^\mu+\bar{c}\color{red}{\bar{\partial}}X^\mu\right)\right]\\[2pt] &~~~\stackrel{(A)}=~~~\mathrm{i}\varepsilon'\left[(c+\delta_{\mathrm{B}}c)\color{red}{\partial} (X^\mu+\delta_{\mathrm{B}}X^\mu)+(\bar{c}+\delta_{\mathrm{B}}\bar{c})\color{red}{\bar{\partial}}(X^\mu+\delta_{\mathrm{B}}X^\mu)-c\color{red}{\partial} X^\mu-\bar{c}\color{red}{\bar{\partial}}X^\mu\right]\\[2pt] &~~~=~~~\mathrm{i}\varepsilon'\left[(\delta_\mathrm{B}c)\color{red}{\partial} X^\mu +c\color{red}{\partial}(\delta_\mathrm{B}X^\mu) +(\delta_\mathrm{B}\bar{c})\color{red}{\bar{\partial}} X^\mu +\bar{c}\color{red}{\bar{\partial}}(\delta_\mathrm{B}X^\mu)\right]\\[2pt] &~~~=~~~\ldots\tag{B} \end{align}
• 1. What is the explanation that $\delta_\mathrm{B}$ acts under the spacetime derivatives? Considering the general BRST transformation (4.2.6a), the field $\phi_i$ can be $\textbf{any}$ non-ghost field. 2. In (4.2.22a) however, this very general transformation behaviour collapses to the transformation of $X^\mu$ and $e$ (without any derivatives!). Why is this the case? May 22, 2017 at 5:46
• 1. I updated the answer. 2. There is a derivative wrt. $\tau$ in eqs. (4.2.22a) & (4.2.22a) if you look in Polchinski's book. It's denoted with a dot. May 22, 2017 at 8:50