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BRST cohomology identifies physical states and observables with BRST-closed objects modulo BRST-exact ones. A change of gauge-fixing fermion changes the action by a BRST-exact term and therefore leaves BRST-cohomology classes unchanged when the measure has no BRST anomaly.

For the left graded Leibniz rule and $s\bar{c}=-b$, $sb=0$, the displayed gauge-fixing fermion gives $b\partial \cdot A+\xi b^2/2+\bar{c}\partial \cdot Dc$. The ghost signs follow from odd parity. BRST nilpotence implies invariance of this density without using field equations. Eliminating the Nakanishi-Lautrup field gives $-(\partial \cdot A{)}^2/(2\xi )$ for nonzero $\xi$.

For an admissible family of gauge-fixing fermions, variation inserts a BRST-exact operator into normalized correlations. If the measured operators are BRST-closed, the graded BRST Ward identity makes that insertion vanish. This extends gauge-parameter independence to changes of the gauge functional itself, provided the action and measure preserve BRST symmetry and the change produces no boundary contribution. It is the perturbative statement; global gauge-orbit problems require additional analysis.

A gauge-fixing fermion is a Grassmann-odd functional whose BRST transformation $s\Psi$ supplies the gauge-fixing and ghost terms. Nilpotence gives $s(s\Psi )=0$ immediately.

Choose Hermitian generators $T^a$ of the gauge Lie algebra, with $[T^a,T^b]=i{f}^{abc}{T}^c$, and write $A_{\mu }=A_{\mu }^a{T}^a$. For the usual compact gauge algebra choose $\mathrm{tr}(T^a{T}^b)={\delta }^{ab}/2$. More generally use a nondegenerate invariant symmetric bilinear form on the algebra in place of this trace contraction; a completely arbitrary component metric would not give gauge invariance.

A matter vector is useful to fix conventions: with ${\psi }^{\prime }=U(x)\psi$ and $D_{\mu }={\partial }_{\mu }-ig{A}_{\mu }$, require $D_{\mu }^{\prime }{\psi }^{\prime }=U{D}_{\mu }\psi$. Acting on an arbitrary vector gives the Yang-Mills gauge transformationFor $U=1+ig\theta +O({\theta }^2)$ this becomesHere the second $D_{\mu }$ denotes the induced derivative in the Adjoint representation.

Define the curvature by $[D_{\mu },D_{\nu }]=-ig{F}_{\mu \nu }$. Expanding the commutator derivesSince commutators of the transformed derivatives conjugate, $F_{\mu \nu }^{\prime }=U{F}_{\mu \nu }{U}^{-1}$. Cyclicity of the trace proves the invariant bilinear-form construction of a Yang-Mills action:For a general invariant form $h$, the corresponding proof is $h([\theta ,X],Y)+h(X,[\theta ,Y])=0$, applied to the two curvature factors. No matter field is required in this pure-gauge Lagrangian.

Introduce the adjoint Faddeev-Popov ghost $c^a$ and antighost ${\bar{c}}^a$, independent Grassmann-odd scalar fields of ghost numbers $+1$ and $-1$, together with the Grassmann-even Nakanishi-Lautrup field $b^a$ of ghost number zero. The bar labels the independent antighost, not an additional propagating complex-conjugate matter field. Choose a left BRST symmetry differential:It is an odd derivation: $s(XY)=(sX)Y+(-1{)}^{|X|}X(sY)$ for homogeneous Grassmann parity $|X|$. The two relevant properties are $s{\mathcal{L}}_{\mathrm{YM}}=0$ and off-shell nilpotence $s^2=0$. The first is gauge invariance with parameter replaced by the ghost, and the second follows from the Lie algebra's graded Jacobi identities. ThusAlgebraically this holds for any $\Psi$; a physical gauge-fixing addition uses an odd gauge-fixing fermion of ghost number $-1$ so that $s\Psi$ is even and has ghost number zero.

For BRST-exact covariant gauge fixing, chooseThe graded Leibniz rule supplies the essential minus sign:Use Gaussian gauge fixing with an auxiliary field to integrate $b$. Pointwise,Translation of the regulated oscillatory Gaussian leaves only a field-independent normalization. The remaining Lagrangian is thereforeThis is covariant Feynman gauge, with gauge parameter one. Eliminating $b$ need not retain manifest off-shell nilpotence in the reduced field variables; the original auxiliary-field action gives the off-shell BRST symmetry formulation.

Finally, the ungauge-fixed quadratic momentum kernel isGauge directions are null vectors, so there is no inverse defining a vector propagator, and the functional integral also counts an infinite gauge orbit volume. The added gauge-fixing term cancels the longitudinal part, leaving $Q_{\mu \nu }^{ab}=-{\delta }^{ab}{p}^2{\eta }_{\mu \nu }$. With the vacuum prescription, the propagators areThe ghost action represents the Faddeev-Popov determinant and supplies ghost-vector vertices; closed ghost loops have the fermionic minus sign. Expanding the cubic and quartic Yang-Mills terms and this ghost interaction now defines consistent perturbative Feynman rules. Physical mass-shell poles remain, whereas the gauge degeneracy has been removed. Possible global gauge-fixing obstructions and boundary zero modes are separate from the perturbative inverse around the trivial vacuum.

For adjoint-valued quantities write $(X\times Y{)}_a={\epsilon }_{abc}{X}_b{Y}_c$. Substituting the infinitesimal transformation into the gauge field strength and using the Jacobi identity givesA useful way to organize the calculation is $[D_{\mu },D_{\nu }]X=g{F}_{\mu \nu }\times X$: the gauge field strength transforms covariantly because the gauge covariant derivative does. Thus $\delta (F_a^{\mu \nu }{F}_{\mu \nu a})=2g{F}_a^{\mu \nu }{\epsilon }_{abc}{F}_{\mu \nu b}{\lambda }_c=0$ by antisymmetry. This proves gauge invariance of the Yang-Mills action.

The Yang-Mills equations are $D_{\mu }{F}^{\mu \nu }=0$. They determine physical evolution only modulo gauge redundancy: gauge transformations with arbitrary spacetime-dependent parameters take solutions to equivalent solutions. In Hamiltonian mechanics, $A_0$ has no independent quadratic time-derivative term, and the corresponding equation is a Gauss law constraint in gauge theory. A gauge parameter can vanish with its necessary derivatives on the initial slice yet change the later potential energy, so the potential energy itself is not uniquely fixed by initial physical data.

In the uncorrected path integral, integration along gauge orbits overcounts equivalent fields. At the perturbative level the quadratic operator of the kinetic term has gauge zero modes and no inverse, so it does not supply a quantum field theory propagator. Gauge fixing together with the Faddeev-Popov determinant removes this obstruction in the local perturbative construction.

For the odd BRST transformation, products obey the graded Leibniz rule. The adjoint product with Grassmann variables satisfiesIn particular, $c\times c$ is not zero, and $c\times Dc=Dc\times c$. The supplied transformations givesince the even covariant derivative obeys $D(c\times c)=2(Dc)\times c$. For the ghost,The final equality is the graded Jacobi identity; in components it is the Jacobi identity for ${\epsilon }_{abc}$ contracted with the totally antisymmetric product of three ghosts. Also $s^2\bar{c}=-sb=0$ and $s^{2}b=0$. The square of an odd derivation is an even derivation, because its two mixed product terms cancel. Hence $s^2=0$ on every polynomial in the fields, off shell, without imposing field equations.

Let $\Psi =\int d^{d}x[{\partial }^{\mu }{\bar{c}}_a{A}_{\mu a}+\xi {\bar{c}}_a{b}_a/2]$ be the gauge-fixing fermion. Since $sF=gF\times c$, $s{\mathcal{L}}_{\mathrm{YM}}=0$. The BRST-exact operator in ${\mathcal{L}}_q={\mathcal{L}}_{\mathrm{YM}}-s\psi$ has zero variation by BRST nilpotence, soTo see the resulting kinetic terms and signs, expand the exact term with $s\bar{c}=-b$:After integration by parts,The auxiliary Nakanishi-Lautrup field $b$ can be completed into a square and integrated out, giving the gauge-fixing term $-(\partial \cdot A_a{)}^2/(2\xi )$ for $\xi \ne 0$. At $\xi =0$, retain $b$ as the multiplier enforcing the gauge condition rather than divide by $\xi$. The Grassmann Gaussian integral gives the determinant of the gauge-condition operator $-\partial \cdot D$. Field-independent normalizations cancel in the normalized integral; insertions involving $b$ can be handled before integration or through its source terms.

At $g=0$ the gauge-fixed quadratic operator is invertible with the usual prescription, and the ghosts have a quadratic kinetic term. The remaining terms are cubic gauge interactions of order $g$, quartic gauge interactions of order $g^2$, and a ghost-gauge interaction of order $g$. Expanding them and using Wick contractions gives a perturbative expansion for any defined polynomial insertion $X$ of the fields. The exponent is the gauge-fixed action $S_q$, as printed in the PDF; the converted TeX incorrectly substitutes $S_A$.

At $\xi =1$, the free gauge action, up to a boundary term, is $\frac{1}{2}\int A_{\mu a}\square A_a^{\mu }$. Its Fourier kernel is $-p^2{\eta }_{\mu \nu }{\delta }_{ab}$. With the mostly-plus convention used above, the Feynman-gauge adjoint propagator isThis is consistent with the scalar-line convention in Question 2.

Finally define the odd functional $B=\frac{1}{2}\int d^{d}x{\bar{c}}_a{b}_a$. Then ${\partial }_{\xi }{S}_q=-sB$. Differentiating the normalized expectation, including its denominator, givesfor an insertion with no explicit $\xi$ dependence. A physical gauge-invariant insertion $X[A]$ is BRST-closed, since its infinitesimal gauge variation vanishes also when the parameter is replaced by $c$. The assumed BRST Ward identity gives $⟨sB⟩=0$ and $⟨s(XB)⟩=0$. For an even $X$ with $sX=0$, the latter says $⟨XsB⟩=0$. ThereforeThe same conclusion holds for any BRST-closed insertion. Mere color-singlet invariance of an arbitrary ghost-dependent expression need not imply $sX=0$; the physical-observable qualification is essential. The proof uses the invariant-measure assumptions encoded in the stated Ward identity.

Fix the current sign convention by defining the localized variation as ${\delta }_{ϵ}S=-\int d^{d}x({\partial }_{\mu }{ϵ}^a)j_a^{\mu }$. For a first-derivative Lagrangian invariant without a boundary term under constant parameters, this means $j_a^{\mu }=-\partial \mathcal{L}/\partial ({\partial }_{\mu }\varphi )t_a\varphi$; include the usual improvement term if the constant variation is a total derivative. On solutions, arbitrary compactly supported parameters imply ${\partial }_{\mu }{j}_a^{\mu }=0$. The Noether charge is $Q_a=\int d^{d-1}x{j}_a^0$ and is conserved if the spatial flux vanishes. This sign convention matches the printed Ward identity; reversing the current also reverses the corresponding generator convention.

Assume an invariant regulated functional measure, invariant vacuum boundary conditions and no quantum anomaly. Changing variables in the normalized path integral gives $0=⟨{\delta }_{ϵ}X⟩+i⟨X{\delta }_{ϵ}S⟩$. Integration by parts then yieldsFor a product of scalar fields, the local Ward identity contact terms areAway from the insertions this is current conservation. Time-ordering or the distributional functional identity supplies the contact terms.

For the gauge theory write $[X,Y{]}^a=f^{abc}{X}^b{Y}^c$ and use the left-acting BRST differential $s$, with ${\delta }_{ϵ}=ϵs$. It obeys the graded Leibniz rule andAssume the structure constants satisfy the Jacobi identity and the dot product is invariant; antisymmetry alone would not be enough. Since $c$ and $D_{\mu }c$ are odd, their Lie brackets are symmetric in these two arguments. ConsequentlyAlso $s{F}_{\mu \nu }=D_{\mu }{D}_{\nu }c-D_{\nu }{D}_{\mu }c=[F_{\mu \nu },c]$. The graded Jacobi identity gives $s[c,c]=0$, so $s^{2}c=0$; the other three fields have zero second variation immediately. For independent odd parameters, ${\delta }_{{ϵ}^{\prime }}{\delta }_{ϵ}=-{ϵ}^{\prime }ϵs^2=0$. These are off-shell BRST nilpotence identities because the Nakanishi-Lautrup field $b$ is retained.

The Yang-Mills theory variation is proportional to $F^{\mu \nu }\cdot [F_{\mu \nu },c]=0$. The gauge-fixing variation is $({\partial }^{\mu }b)\cdot D_{\mu }c$, while the ghost variation is its negative; the $b^2$ term does not vary. Equivalently these terms are $s\Psi$ for the gauge-fixing fermion $\Psi =({\partial }^{\mu }\bar{c})\cdot A_{\mu }+\xi \bar{c}\cdot b/2$. BRST nilpotence makes this expression invariant. Ghost-number scaling also leaves every term invariant: the ghost and antighost factors carry opposite weights.

Here are the two explicit currents in the same sign convention as the Ward identity. Localizing the even ghost parameter gives coefficient $({\partial }_{\mu }\theta )[\bar{c}\cdot D^{\mu }c-({\partial }^{\mu }\bar{c})\cdot c]$. Localizing the odd parameter, keeping it on the left, gives coefficientThe last sign comes from moving the odd parameter through the odd antighost derivative. Since the current was defined as minus this coefficient,These are the ghost-number Noether current and the BRST current in derivative-b gauge fixing. If the opposite Noether sign is used, both displayed currents acquire an overall minus sign. Integrating the gauge-fixing term by parts changes the Noether representative by the associated boundary improvement; mixing the two Lagrangian conventions without that improvement gives incorrect signs.

With no BRST anomaly, the conserved odd BRST charge has $Q_B^2=0$. The BRST cohomology identifies closed states $Q_B|\psi ⟩=0$ modulo exact states $Q_B|\chi ⟩$. Nilpotence puts every exact state in the closed space. In the usual indefinite gauge-fixed state space, a Hermitian BRST charge makes exact states orthogonal to closed states; the standard no-ghost/positivity assumptions then give a physical inner product on the quotient. The physical sector is its ghost-number-zero component,The ghost number assigns $+1$ to $c$, $-1$ to $\bar{c}$ and zero to gauge and auxiliary fields. Gauge-invariant observables and the chosen vacuum have ghost number zero; unphysical ghost excitations are removed in BRST pairs. Thus physical representatives are expected to satisfy $Q_G|\psi ⟩=0$. This zero-grading selection is part of the physical-state prescription, not a consequence of nilpotence alone.

Use the Minkowski metric, path-integral weight $e^{iS}$ and the Abelian gauge theory transformation $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\omega$. The gauge functional $F[A]=\partial \cdot A+A^2$ varies asThus the Faddeev-Popov operator is ${\mathcal{M}}_A=\square +2A\cdot \partial$. It depends on the gauge field despite the gauge group being Abelian: the ghosts interact because this gauge condition is nonlinear.

Choose the gauge-fixing fermion $\Psi =\int d^{4}x\bar{c}(F[A]+\xi h/2)$. The gauge-fixed action $S=S_{\mathrm{Maxwell}}+s\Psi$ isHere the tensor $F_{\mu \nu }={\partial }_{\mu }{A}_{\nu }-{\partial }_{\nu }{A}_{\mu }$ is distinct from the scalar gauge functional $F[A]$. With $\xi =0$, integrating over $h$ imposes the exact printed constraint. For nonzero $\xi$, eliminating $h=-F[A]/\xi$ instead givesThis version displays the additional gauge-dependent cubic and quartic gauge-field vertices as well as the ghost interaction; the strict condition is its $\xi \to 0$ limit.

For the Fourier transform convention $c(x)=\int d^{4}p(2\pi {)}^{-4}{e}^{-ipx}c(p)$, the quadratic ghost kernel is $p^2$. With the ordering $⟨c(p)\bar{c}(q)⟩$,The term $-2\bar{c}{A}^{\mu }{\partial }_{\mu }c$ has Fourier coefficient $2i{p}_{\mu }$, where $p$ is the incoming ghost momentum. Multiplication by $i$ in the Feynman rule givesThese signs refer to the displayed action, Fourier convention and ghost ordering. Reversing the ghost/antighost convention changes corresponding signs consistently. A closed ghost loop has the additional minus sign from Grassmann variables.

A gauge-invariant observable is BRST-closed: replacing its infinitesimal gauge parameter by the ghost gives $s\mathcal{O}=0$. Change the gauge functional continuously, or interpolate between two admissible choices, by changing the gauge-fixing fermion to ${\Psi }_t$. The action changes by the BRST-exact operator ${\partial }_{t}S=s({\partial }_t{\Psi }_t)$.

For a normalized correlator of $\mathcal{O}={\prod }_i{\mathcal{O}}_i$ with each insertion BRST-closed, differentiation of the functional integral givesThe BRST Ward identity says $⟨sX⟩=0$ for an invariant measure and action, with appropriate boundary conditions. Since $s\mathcal{O}=0$, the graded Leibniz rule makes the first insertion an exact variation of $\mathcal{O}{\partial }_t{\Psi }_t$ up to its harmless parity sign, and both terms vanish. ThusPhysical gauge-invariant correlation functions are independent of this gauge condition, even though individual gauge-field and ghost Feynman diagrams change. This argument assumes an admissible perturbative gauge fixing, a BRST symmetry-preserving regulator/measure and no uncanceled boundary contribution. A global failure of those assumptions is not settled by the formal local calculation.

Write $F[A]={\partial }_{\mu }{A}_{\mu }+i{A}_{\mu }{A}_{\mu }$ and choose the U(1) gauge symmetry convention $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\alpha$. Its infinitesimal variation isA formal BRST symmetry construction uses $s{A}_{\mu }={\partial }_{\mu }c$, $sc=0$, $s\bar{c}=ih$, $sh=0$ and the gauge-fixing fermion $\Psi =\int \bar{c}(F-i\xi h/2)d^{4}x$. Acting with the odd differential givesFor a real admissible gauge functional, integration over $h$ gives $F^2/(2\xi )$ in the Euclidean action, and $\xi \to 0$ imposes the gauge condition. Overall constant phases in the Faddeev-Popov determinant can be absorbed in the measure convention. Here nonlinear Abelian gauge fixing makes ${\mathcal{O}}_A$ depend on $A$. The action for the Faddeev-Popov ghost fields contains the interaction $-2i\bar{c}{A}_{\mu }{\partial }_{\mu }c$, so the Faddeev-Popov ghost fields cannot be discarded as in a linear Abelian gauge, even though the $U(1)$ Adjoint representation is trivial.

There is a genuine reality obstruction for a complex Euclidean gauge condition in the source. For a Hermitian $U(1)$ gauge field, each Euclidean component $A_{\mu }$ is real. The real and imaginary parts of $F=0$ separately require $\partial \cdot A=0$ and ${\sum }_{\mu }{A}_{\mu }^2=0$, hence $A=0$. For example $A_2=B{x}_1$, with other components zero and $B\ne 0$, has curvature $F_{12}=B$; no gauge transformation can put it on this slice because curvature is gauge invariant. Thus the stated condition is not a gauge fixing of general real Euclidean configurations. The boxed action is the intended formal complex-gauge construction. A legitimate complexified contour prescription would be additional data, not an ordinary real delta-functional enforcing this printed condition.

The Grassmann field $c$ is odd and the adjoint scalar field $\varphi$ is even. The two printed signs are consistent with a right-acting BRST differential, whose graded Leibniz rule isThe bracket between two odd fields is graded, so $[c,c]=2c^2$. Applying this rule to the ordinary Lie bracket $[c,\varphi ]=c\varphi -\varphi c$ givesThe second bracket here is graded because both its entries are odd. The graded Jacobi identity gives $[c,[c,\varphi ]]=\frac{1}{2}[[c,c],\varphi ]$. ThereforeThe side of the odd derivation is essential. With the usual left graded Leibniz rule and the same two printed signs, the result would be $-[[c,c],\varphi ]$, which is generally nonzero. A left-acting convention must reverse one of those signs. All subsequent BRST symmetry formulas here use the right-acting convention.

Choose an anti-Hermitian basis $T^a$ with $[T^b,T^c]=f^{abc}{T}^a$, an invariant positive invariant bilinear form on a Lie algebra, and the adjoint covariant derivative $D_{\mu }X={\partial }_{\mu }X+[A_{\mu },X]$. Couplings are absorbed into this convention; restoring $D_{\mu }={\partial }_{\mu }+g[A_{\mu },\cdot ]$ multiplies each ghost interaction vertex below by $g$. The associated BRST charge acts as $Q{A}_{\mu }=-D_{\mu }c$, $Q\bar{c}=ih$, and $Qh=0$.

Write $F^a=\partial \cdot A^a+(n\cdot \partial {\varphi }^a)/\sqrt{2}$. For the gauge-fixing fermion $\Psi =\int {\bar{c}}^a(F^a-i{h}^a/2)$, the right graded Leibniz rule givesThe Gaussian functional integral over the Nakanishi-Lautrup field produces the positive gauge fixing term $F^a{F}^a/2$. The ghost operator is $-\partial \cdot D-(n\cdot \partial ){\mathrm{ad}}_{\varphi }/\sqrt{2}$.

Now use the canonical free kinetic terms $F_{\mu \nu }^a{F}_{\mu \nu }^a/4+(\partial {\varphi }^a{)}^2/2$. At nonzero momentum in Euclidean space $p$, put $\eta =n\cdot p$. The quadratic kernel for $(A_{\mu }^a,{\varphi }^a)$, per color, isThe transverse gauge-field kernel plus the gauge-fixing longitudinal term has become $p^2{\delta }_{\mu \nu }$. The scalar quantum field theory propagator is the inverse Schur complement, not merely the inverse of the scalar diagonal entry:Hence the free adjoint-scalar propagator in scalar-dependent gauge fixing isFor completeness the mixed quantum field theory propagator is $⟨A_{\mu }^a{\varphi }^b{⟩}_0=-{\delta }^{ab}\eta p_{\mu }/(\sqrt{2}{p}^4)$; ignoring this mixing would give an incorrect scalar answer.

The scalar propagator is independent of the gauge vector $n$, not of the momentum component parallel to $n$. For a unit $n$, $p^2=p_{\perp }^2+p_{\parallel }^2$, and the free scalar quantum field theory propagator still depends on $p_{\parallel }$. Thus the literal momentum-independence clause in the PDF is false for the standard minimally coupled massless scalar action; the cancellation above establishes the natural gauge-vector-independence statement. The usual massless zero mode in field theory at $p=0$ needs a separate infrared prescription.

Finally, integration by parts gives the ghost action in an unambiguous convention:Use $e^{ip\cdot x}$ for the Fourier transform of every field, with all momenta incoming. Let the antighost carry color $a$ and momentum $q$, the boson color $b$ and momentum $k$, and the ghost color $c$ and momentum $p$, so $q+k+p=0$. Expansion of $e^{-S_{\mathrm{gh}}}$ gives the ghost vertices in scalar-dependent gauge fixingThe free Faddeev-Popov ghost field propagator is ${\delta }^{ab}/p^2$ and every closed ghost loop contributes a minus sign. There are no further ghost interaction vertices in this gauge. Factors of $i$ and an overall ghost-vertex sign depend on the Fourier transform and ghost-ordering conventions; the displayed ghost action fixes both here.

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.

The gauge-invariant Yang-Mills and matter part ${\mathcal{L}}_0$ is BRST invariant because a BRST variation is a gauge transformation with ghost-valued parameter. For the gauge-fixing fermionthe graded Leibniz rule givesup to the common sign convention used to define the ghost term. This is precisely the gauge-fixing, auxiliary-field, and ghost sector of the displayed Lagrangian. Therefore

Write the gauge-fixing and ghost action as the BRST transformation of the gauge-fixing fermion:with harmless rescalings of $B$ accommodating the convention $s{\bar{c}}^a=\xi B^a$ used in the question. The gauge-invariant action obeys $sS[A]=0$, while nilpotence givesThus the entire gauge-fixed action is BRST invariant. Eliminating the auxiliary field $B^a$ by its algebraic field equation reproduces the axial gauge-fixing term and the ghost action found in part (b).

Changing $n^{\mu }$ to $u^{\mu }$ changes the gauge-fixing fermion byThe corresponding change of the action is the BRST-exact term $s({\Psi }_u-{\Psi }_n)$. For a BRST-closed observable $\mathcal{O}$, invariance of the functional measure implies that the variation of its expectation value is the expectation of a total BRST variation and vanishes:Therefore physical states and observables, which are classes in BRST cohomology, do not depend on the fixed axial-gauge vector, provided there is no BRST anomaly or boundary contribution.