The BRST variation of the gauge fixing condition

Question

Following Polchinski volume I, p 126 onwards, The BRST variation of fields $\phi^{i}$ is given by $$\delta_{B} \phi_{i} = - i \epsilon c^{\alpha} {\delta}_{\alpha}\phi_{i} \; .\tag{4.2.6a}$$ My question is as follows; why is the variation of a function of the fields, ${F}^A(\phi)$, which in this contexts defines the gauge fixing condition, $$\delta_{B} F(\phi) = - i \epsilon c^{\alpha} {\delta}_{\alpha}F^{A}(\phi) \; ?\tag{1}$$

If one uses standard variational techniques (in this case the chain rule) one should arrive at

\begin{align*} \delta_{B} F(\phi) &= \delta_{B}\phi_{i} \frac{\delta}{\delta \phi_{i}} F^{A}(\phi) \\ &=-i \epsilon {c}^{\alpha}\delta_{\alpha} \phi_{i} \frac{\delta}{\delta \phi_{i}} F^{A}(\phi)\tag{2}. \end{align*} In previous posts here, and here the users then equate $\phi_{i} \frac{\delta}{\delta \phi_{i}} F^{A}(\phi)$ with ${F}^{A}(\phi)$ why can we do this? Naively if ${F}^{A}(\phi)$ was, say, a polynomial in the fields, $\phi^{n}$ with $n > 1$ then one would expect to arrive at the variation

$$\delta_{B} F(\phi) = - i \color{red}{n} \epsilon c^{\alpha} {\delta}_{\alpha}F^{A}(\phi) \; .\tag{3}$$

What am I doing wrong that has sent me on such a divergence?

Answer 1

If $F^A(\phi)=0$ is a gauge fixing condition, the path integral measure will include a factor of $\delta \left( F^A(\phi) \right)$, where $\delta()$ is a delta functional. Usually this delta is then reexpressed as a path integral so that it can be exponentiated, leading to a "gauge-fixing" term in the action, which is equation (4.2.4) in page 126 of Polchinski.

The presence of this delta functional means that we can treat $F^A(\phi)$ as being linear in $\phi$, since this is only nonzero in an infinitesimal region around $\phi=0$ where we can ignore all but the linear term of the expansion of $F^A(\phi)$ in powers of $\phi$. The case of gauge fixing the metric in the bosonic string is even simpler, since $F^A$ is linear from the start: $F \left( g \right) = g_{a b} - \hat{g}_{a b}$, where $\hat{g}$ is the fiducial metric of choice $(\hat{g}_{ab} = e^{2\omega(\sigma)} \delta_{ab}$ for conformal gauge, for instance).

If $F^A(\phi)$ is linear in $\phi$, the BRST variation goes right through it: \begin{equation} \delta_B F^A(\phi) = F^A( \delta_B \phi ) = F^A( -i \epsilon c^\alpha \delta_\alpha \phi_i ) = -i \epsilon c^\alpha \delta_\alpha F^A(\phi). \end{equation}

Answer 2

1. It seems the underlying issue is the meaning of $\delta_{\alpha}$. In Ref. 1 $\delta_{\alpha}$ is a Grassmann-even (non-infinitesimal) linear derivation/vector field (which satisfies Leibniz rule) in the Lie algebra of gauge transformations. [The Leibniz rule is how the order of the polynomial $F$ is correctly accounted for in OP's eq. (1), which seems to be OP's main question.] The subindex $\alpha$ is a Lie algebra index of the gauge Lie algebra.

2. In contrast, the BRST transformation $\delta_B$ in Ref. 1. is an infinitesimal Grassmann-even linear derivation/vector field. It contains an infinitesimal Grassmann-odd parameter $\epsilon$.

References:

1. J. Polchinski, String Theory Vol. 1, 1998; Section 4.2.