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Factor the odd parameter on the left, $\delta =ϵs$. The resulting left-acting BRST differential obeys the graded Leibniz ruleFor the odd Grassmann field $c$, the bracket in the transformation is a graded commutator: $[c,c{]}_{\mathrm{gr}}=2c^2$, not the identically zero ordinary commutator of a matrix with itself. Thus $sc=i{c}^2$, while $s{A}_{\mu }=D_{\mu }c$, $s\bar{c}=h$ and $sh=0$.

On the ghost,On the gauge field, variation of the connection and the adjoint covariant derivative givesHere $D_{\mu }$ is even and therefore obeys the ordinary product rule. Also $s^2\bar{c}=sh=0$ and $s^{2}h=0$, without using any field equation; this is off-shell nilpotence supplied by the Nakanishi-Lautrup field.

Applying the graded Leibniz rule twice cancels the two cross terms:The square is consequently an even graded derivation. Since it vanishes on every generator, it vanishes inductively on every polynomial in the fields. Hence $s^2\mathcal{O}=0$ for every such operator. This genuine result is stronger than the automatic vanishing obtained by merely setting ${ϵ}^2=0$; two independent transformation parameters also give a vanishing commutator.

Put the anticommuting parameter on the left and write $\delta \Phi =ϵs\Phi$. Then $s$ is a left-acting BRST differential: it is odd and satisfies the graded product rule $s(UV)=(sU)V+(-1{)}^{|U|}U(sV)$. The Faddeev-Popov ghost field $c$ and antighost field $b$ are odd, whereas $A_{\mu }$ and the Nakanishi-Lautrup field $h$ are even. Use $s{A}_{\mu }=D_{\mu }c$, $sc=-[c,c]/2$, $sb=h$, $sh=0$.

A total derivative in the Lagrangian variation must be included in the Noether current. For constant $ϵ$, the ghost covariant derivative has $s(D_{\mu }c)=0$, by the Jacobi identity and the odd statistics of $c$. The Yang-Mills term is invariant, and the remaining variation isThe term proportional to $\xi$ does not vary because $sh=0$.

To obtain the signs without an ambiguity about fermionic canonical momenta, now allow $ϵ=ϵ(x)$. In particular,The coefficients of ${\partial }_{\mu }ϵ$ in the Yang-Mills, ghost, and multiplier terms are respectivelyThe positive sign of the last ghost expression results from moving ${\partial }_{\mu }ϵ$ past the odd ${\partial }^{\mu }{b}^a$. Subtracting the total-derivative term $K^{\mu }$ therefore gives the Yang-Mills BRST Noether currentIndeed, the full localized variation is $\delta \mathcal{L}=({\partial }_{\mu }ϵ)j^{\mu }+{\partial }_{\mu }(ϵK^{\mu })$. The Noether theorem then gives ${\partial }_{\mu }{j}^{\mu }=0$ on the field equations. With spatial boundary terms vanishing, the corresponding BRST charge in four spacetime dimensions isIt is odd and has ghost number one. Overall generator phases depend on the convention relating this Noether charge to quantum commutators; one may use $s\mathcal{O}=i[Q_{\mathrm{BRST}},\mathcal{O}\}$. The displayed current fixes the classical Noether normalization for the left-parameter convention.