= Solution
Remove the common Grassmann parameter and write the <BRST transformation> as the odd derivation $s$:
$$
sA_\mu^a=(D_\mu c)^a,
\qquad
sc^a=-\frac g2\epsilon^{abc}c^bc^c,
\qquad
s\bar c^a=B^a,
\qquad
sB^a=0.
$$
The last two equations immediately give $s^2\bar c^a=s^2B^a=0$. Applying $s$ to the ghost and using the graded product rule gives
$$
s^2c^a=\frac{g^2}{4}\epsilon^{abc}
\left(\epsilon^{bde}c^dc^ec^c-\epsilon^{cde}c^bc^dc^e\right)=0.
$$
The cancellation is the <Jacobi identity> for the $SU(2)$ structure constants together with anticommutation of the ghost fields. For the gauge field, the component calculation is
$$
s^2A_\mu^a=(D_\mu sc)^a
+g\epsilon^{abc}(D_\mu c)^bc^c.
$$
The graded product rule gives
$$
(D_\mu sc)^a=-\frac g2\epsilon^{abc}
\left[(D_\mu c)^bc^c+c^b(D_\mu c)^c\right]
=-g\epsilon^{abc}(D_\mu c)^bc^c,
$$
so the two terms cancel. Thus $s^2$ vanishes on every elementary field.
For two independent Grassmann parameters, $\delta_1=\eta_1s$ and $\delta_2=\eta_2s$ satisfy $\delta_1\delta_2O=\eta_1\eta_2s^2O=0$. Since $s$ obeys the graded Leibniz rule, induction extends $s^2O=0$ from the generators $A,c,\bar c,B$ to every polynomial $O(A,c,\bar c,B)$. Hence the BRST transformations are nilpotent.
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