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For a left left-acting BRST differential with $sA=Dc$, $s\bar{c}=b$, $sb=0$, choose $\Psi ={\bar{c}}^a(\partial \cdot A^a+\xi b^a/2)$. The graded Leibniz rule gives $s\Psi =b\partial \cdot A+\xi b^2/2-\bar{c}\partial \cdot Dc$. Integrating the auxiliary Gaussian yields $-(\partial \cdot A{)}^2/(2\xi )-\bar{c}\partial \cdot Dc$. At $\xi =1$ this is Feynman gauge. Opposite antighost or exact-action sign conventions require a corresponding change of fermion.

// Target: relativistic-quantum-field.bigb

In conformal gauge, let the odd left-acting BRST differential act by $sX=c^{+}{\partial }_{+}X+c^{-}{\partial }_{-}X$ and $s{c}^{\pm }=c^{\pm }{\partial }_{\pm }{c}^{\pm }$. The graded Leibniz rule gives the displayed expression for $s^{2}X$. It vanishes using the ghost equations of motion ${\partial }_{-}{c}^{+}={\partial }_{+}{c}^{-}=0$, but it is not zero for arbitrary off-shell fields. For independent Grassmann variables $\xi ,\rho$, take $c^{+}=\xi$, $c^{-}=\rho {\sigma }^{+}$ and $X={\sigma }^{-}$; then $s^{2}X=-\xi \rho \ne 0$. Each ghost itself satisfies $s^2{c}^{\pm }=0$ algebraically. With the eliminated-antighost transformation $s{b}_{\pm \pm }=i{\theta }_{\pm \pm }/{\alpha }^{\prime }$ and $s{\theta }_{\pm \pm }=0$ on the solution space, the antighosts are nilpotent too. This on-shell closure is compatible with an off-shell total-derivative variation of the gauge-fixed action. A full off-shell BRST quantization uses the complete gauge transformation and, where required, auxiliary fields rather than silently dropping mixed ghost derivatives.

The graded Leibniz rule terms of two connections cancel, making their difference $C^{\infty }$-linear in the section. It is therefore an element of $A^1(\mathrm{End}E)$. Conversely adding any such form to a connection gives another connection, making the space of connections an affine space.

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 $\xi$.

Write $f\circ g={\sum }_{i=0}^{p-1}(-1{)}^{i(q-1)}f{\circ }_{i}g$ for insertion of a degree-$q$ cochain into a degree-$p$ one. With the unsigned Hochschild cup product and the left graded Leibniz rule, take $[f,g]=(-1{)}^{(p-1)(q-1)}f\circ g-g\circ f$. It descends to Hochschild cohomology and gives a Gerstenhaber algebra. The alternative insertion bracket $f\circ g-(-1{)}^{(p-1)(q-1)}g\circ f$ differs by the displayed degree sign and obeys the corresponding right rule. Degree-one brackets are commutators of derivations in either convention.

The parity grading records whether a homogeneous element has even or odd Grassmann degree. Homogeneous elements obey $XY=(-1{)}^{|X||Y|}YX$. An odd differential uses the graded Leibniz rule; this is the sign bookkeeping behind left-acting BRST differential calculations.

On an oriented boundaryless Riemannian manifold, apply Stokes theorem to the compactly supported form $f\ast \alpha$. The graded Leibniz rule gives $\int df\wedge \ast \alpha =-\int fd\ast \alpha$. The Hodge star operator identifies these top forms with the scalar integrands displayed above. This proves that the codifferential on one-forms is $\delta =-\ast d\ast$, the formal adjoint of the exterior derivative. Compact support of $f$ suffices.

On a closed oriented Riemannian manifold, let $\alpha$ have degree $r-1$ and $\beta$ degree $r$. The graded Leibniz rule and Stokes theorem give $\int d\alpha \wedge \ast \beta =(-1{)}^r\int \alpha \wedge d\ast \beta$. The codifferential convention $\delta \beta =(-1{)}^{n(r+1)+1}\ast d\ast \beta$, combined with the square of the Hodge star, gives $\ast \delta \beta =(-1{)}^{r}d\ast \beta$. This proves that the codifferential is the formal adjoint of the exterior derivative. A boundary requires the corresponding boundary term or specified boundary conditions.

Using the graded Leibniz rule, $s^{2}A=D(sc)+g(Dc)\times c=0$ because $D(c\times c)=2(Dc)\times c$. The graded Jacobi identity gives $s^{2}c=g^2(c\times c)\times c/2=0$. BRST nilpotence on the antighost and Nakanishi-Lautrup field is immediate. No field equations are required.

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.

On an oriented Riemannian $n$-manifold, the metric volume form is the unique top form taking value one on every positively oriented orthonormal tangent basis. In a positive coordinate chart it isFor another positive coordinate chart $y$, write $A=\partial x/\partial y$. Then $g_y=A^{\mathsf{T}}{g}_{x}A$, so $\sqrt{\det g_y}=\det A\sqrt{\det g_x}$, since $\det A>0$. The wedge of coordinate differentials acquires the same determinant factor. The two local expressions agree, proving coordinate independence and smooth global existence. Positivity of the chosen charts is what fixes the sign of the volume form.

The metric induces the inner product on exterior powers of a cotangent space. Define the Hodge star operator $\ast :{\Lambda }^p{T}^{\ast }M\to {\Lambda }^{n-p}{T}^{\ast }M$ byfor forms of the same degree, and apply it fiberwise to smooth forms. In an oriented orthonormal coframe, if $I$ is an increasing $p$-index and $I^c$ its ordered complement, $\ast e^I=ϵ(I,I^c)e^{I^c}$. Exchanging the two blocks has sign $(-1{)}^{p(n-p)}$, so

Use the real $L^2$ inner product $(\alpha ,\beta )={\int }_M\alpha \wedge \ast \beta$. For a $p$-form $\alpha$ and a $(p+1)$-form $\beta$, the graded Leibniz rule and the given boundaryless Stokes identity giveThe codifferential acting on $\beta$ has sign $(-1{)}^{n(p+2)+1}$. Applying the star once more and using its square on the $(n-p)$-form $d\ast \beta$ givesThe exponent reduction uses parity, since $p^2\equiv p(\mathrm{mod}2)$. Combining the two displayed identities proves $(d\alpha ,\beta )=(\alpha ,\delta \beta )$. Symmetry of the real inner product supplies preciselyThus Hodge integration by parts in arbitrary degree identifies $\delta$ as the formal adjoint of $d$. The displayed bilinear identity extends to complex differential forms by complex linearity. For positive norm identities on complex forms, use the Hermitian inner product with conjugation instead.

With the nonnegative convention, the Hodge Laplacian isIt is the form-valued extension of the Laplace-Beltrami operator, whose action on functions is $-\mathrm{div}\mathrm{grad}$. A harmonic differential form is a smooth form in $\ker \Delta$. The adjoint identity givesIf $\Delta \alpha =0$, both norms vanish, and smoothness impliesThe closed compact hypotheses are essential for the absence of boundary terms and this global energy conclusion.

The Hodge decomposition theorem states that on a closed oriented Riemannian manifoldis an $L^2$-orthogonal direct sum, with finite-dimensional harmonic space ${\mathcal{H}}^p$. Every smooth form has unique harmonic, exact and coexact components, although primitives of the latter two need not be unique. Equivalently every de Rham cohomology class has a unique harmonic representative. An exact harmonic form belongs to both the harmonic and exact summands; their orthogonality makes its squared norm zero. Therefore $\boxed{\alpha \text{ exact and harmonic}⟹\alpha =0}$. The same conclusion follows directly from $\alpha =d\eta$ and $\Vert \alpha {\Vert }^2=(\delta \alpha ,\eta )=0$.

The exterior derivative is the real-linear degree-one operator $d:{\Omega }^k\to {\Omega }^{k+1}$ characterized by its usual derivative on functions, the graded Leibniz ruleand $d^2=0$. In local coordinates,These rules determine it uniquely, and the coordinate formula shows its existence and invariance under coordinate changes. It also commutes with pullback of a differential form.

On the unit ball let $V=x^i{\partial }_i$ be the radial vector field, and let $F_t(x)=tx$. For $k\ge 1$ define the radial homotopy operator byHere the vectors are identified using the ambient Euclidean coordinates. Equivalently $h_k\omega ={\int }_0^1{t}^{-1}{F}_t^{\ast }({\iota }_V\omega )dt$, where ${\iota }_V$ is the interior product. The apparent factor $t^{-1}$ is harmless: the coordinate coefficients of its integrand contain $t^{k-1}$, so the integral is smooth at $t=0$ and depends smoothly on $x$. This is a linear map to ${\Omega }^{k-1}$; set $h_0=0$.

For $t>0$, differentiation of the coordinate pullback and Cartan's magic formula giveFor completeness, Cartan’s identity follows by checking it on functions and coordinate one-forms: both sides give $V(f)$ on $f$ and $d(V(x^i))$ on $d{x}^i$. Both are degree-zero derivations of the exterior algebra, so the checks imply the identity on every local form. Integrating from zero to one provesFor positive $k$, the pullback by the constant map is zero, so this is the required identity. If $d\omega =0$ and $k\ge 1$, it follows that $\omega =d(h_k\omega )$. A chart about any point contains such a ball, proving the Poincaré lemma. In degree zero the formula is instead $h_{1}df=f-f(0)$: a constant function cannot be the image of the uncorrected homotopy identity. This degree-zero correction to radial homotopy specifies the necessary positive-degree convention in the printed request.

To treat the sphere, cover $S^2$ by the complements of its north and south poles. Each is diffeomorphic to ${\mathbb{R}}^2$, where the same star-shaped homotopy applies. A closed differential one-form $\alpha$ therefore has local primitives $f_N$ and $f_S$. On their overlap $d(f_N-f_S)=0$. This overlap is connected, so the difference is a constant $c$. Replacing $f_N$ by $f_N-c$ makes the primitives agree, and they glue smoothly to a global function. ThusThis proves directly that closed one-forms on the two-sphere are exact.

For the Hodge decomposition theorem, assume a compact oriented Riemannian manifold without boundary. The Hodge star operator is characterized by $\alpha \wedge \ast \beta =⟨\alpha ,\beta ⟩{\mathrm{vol}}_g$ and obeys $\ast \ast \alpha =(-1{)}^{k(n-k)}\alpha$. The formal $L^2$ adjoint of $d$ on $k$-forms is $\delta =(-1{)}^{n(k+1)+1}\ast d\ast$. Define the Hodge Laplacian $\Delta =d\delta +\delta d$ and the space ${\mathcal{H}}^k=\ker \Delta$ of harmonic differential forms. Integration by parts, with no boundary term, givesHence a form is harmonic exactly when it is both closed and coclosed. The Hodge decomposition theorem states that ${\mathcal{H}}^k$ is finite-dimensional and that there is the orthogonal decompositionIf a closed form has decomposition $\alpha =h+d\beta +\delta \gamma$, then $⟨\alpha ,\delta \gamma ⟩=⟨d\alpha ,\gamma ⟩=0$. Orthogonality consequently forces $\delta \gamma =0$, so its de Rham cohomology class is represented by $h$. An exact harmonic form $h=d\beta$ has $\Vert h{\Vert }^2=⟨\delta h,\beta ⟩=0$, proving uniqueness. Thus

Every top-degree form is $\omega =f{\mathrm{vol}}_g$, and $d\omega =0$ for dimensional reasons. Its Hodge star is $f$, so the formula for $\delta$ shows that $\delta \omega =0$ exactly when $df=0$. On a connected manifold this says that $f$ is constant. Therefore the harmonic top-degree forms on a closed oriented manifold form the spaceAny such smooth manifold admits a Riemannian metric; the assertion is valid for every choice of that metric.

Keep the metric and hence the Hodge star fixed. Write $\alpha =\delta A$, so $\delta F=d\alpha$, and use the Bianchi identity for an Abelian p-form $dF=d^{2}A=0$. The bilinear pairing $U\wedge \star V$ on real four-forms is symmetric even for a fixed indefinite metric. Consequently the variation of the kinetic term is $d\alpha \wedge \star F$.

The two differentiated four-form factors in the cubic term have the same sign, since moving a four-form through another four-form or through a three-form contributes an even exponent. ThusThe graded Leibniz rule givesUsing the Generalized Stokes theorem, the complete first variation is thereforeFor compactly supported variations, a closed $M$, or boundary conditions removing this last term, the Euler-Lagrange equation and the Bianchi identity for an Abelian p-form areThe coefficient three follows from the normalization actually present in the action; inserting a differently normalized supergravity coupling would change it. This is the eleven-dimensional three-form action variation.

Under the gauge transformation $A\mapsto A+d\Lambda$, the four-form $F$ is unchanged because $d^2\Lambda =0$. The kinetic density is unchanged, and the cubic density changes byIt is an exact eleven-form, so the action changes only by a boundary integral. The Euler-Lagrange equations depend only on $F$ and are invariant even when this boundary integral is nonzero. Invariance of the action itself additionally requires suitable boundary or support conditions.

Choose a basis $T_a$ of the Lie algebra, with $[T_b,T_c]=c^a{}_{bc}{T}_a$. On the Matrix Lie group setLeft multiplication by a constant matrix preserves $\lambda$, while right multiplication preserves $\rho$. Differentiating $g^{-1}g=I$ gives $d(g^{-1})=-g^{-1}(dg)g^{-1}$. Apply the graded Leibniz rule to obtainBecause the wedge product antisymmetrizes the matrix products, these Maurer-Cartan equations becomeFollow the vector-field naming used in the question: $L_a(g)=T_{a}g$ is right-invariant and generates left multiplication, whereas $R_a(g)=g{T}_a$ is left-invariant and generates right multiplication. They are dual to ${\rho }^a$ and ${\lambda }^a$, respectively. For any one-form $\eta$, $d\eta (X,Y)=X(\eta (Y))-Y(\eta (X))-\eta ([X,Y])$. Evaluating the two Maurer-Cartan equations on the corresponding invariant fields yieldsFor the last identity, the left and right multiplication flows commute. The opposite sign for the right-invariant vector fields is therefore intrinsic; the letters attached to the two families are only a convention.

A connection on a vector bundle $E\to M$ is a linear map $D:\Gamma (E)\to {\Omega }^1(M;E)$ with $D(fs)=df\otimes s+fDs$. Write a local frame as a row $e=(e_1,\dots ,e_r)$, and a section as $s=eu$ with a coefficient column $u$. If the connection one-form is the matrix $A={\sum }_i{A}_{i}d{x}^i$, thenThis fixes the matrix convention for all subsequent signs.

For an $E$-valued differential form of degree $k$, represented locally by a column $\alpha$ of ordinary forms, its covariant exterior derivative isThe defining graded Leibniz rule with a scalar form $\omega$ of degree $j$ isIt follows by moving the one-form $A$ past $\omega$, which introduces $(-1{)}^j$. It uniquely extends the connection on sections.

The induced endomorphism bundle connection is characterized on an endomorphism section $T$ by $(D^{\mathrm{End}}T)s=D(Ts)-T(Ds)$. For a matrix-valued form $\beta$ of degree $k$, the extension isIndeed it is exactly the formula that makesThe endomorphism-valued exterior product uses matrix composition as well as the ordinary exterior product, so factors cannot in general be interchanged.

Define the curvature form of a connection by $D^2$ acting on sections, or equivalently byIt is tensorial in both vector arguments and in $s$: the derivatives of multiplying functions cancel by the connection rule. For example $D^2(fs)=D(dfs+fDs)=f{D}^{2}s$, since the two $df\wedge Ds$ terms cancel and $d^{2}f=0$. Thus curvature is a two-form with values in the endomorphism bundle. Locally, applying the graded rule givessoFor a change of frame of a vector-bundle connection $e^{\prime }=eg$, the coefficients satisfy $u^{\prime }=g^{-1}u$, and differentiation givesSquaring yields $F^{\prime }=g^{-1}Fg$, precisely the transformation rule of an endomorphism-valued two-form. This independently proves that the local expressions define one global curvature form of a connection.

For the Bianchi identity, use the endomorphism formula with degree two:Every term cancels with its indicated sign. Thereforewhere $d_A$ here means the induced derivative on the endomorphism bundle.

For a line bundle, the endomorphism bundle is canonically the trivial scalar bundle: every fibre endomorphism is a scalar times the identity. Scalar multiplication commutes, so $A\wedge A=0$ and the endomorphism covariant derivative on scalar-valued forms reduces to $d$. The Bianchi identity consequently says $dF=0$.

If $D^{\prime }$ is another connection, their difference is $C^{\infty }$-linear in sections because the Leibniz terms cancel. Hence $D^{\prime }-D=\eta$ is a global endomorphism-valued one-form; for a line bundle it is an ordinary global scalar one-form. Locally $A^{\prime }=A+\eta$, soThus the two closed curvatures differ by an exact form, proving the connection-independent curvature class of a line bundle:No global trivialization of the line bundle itself was assumed; it is its endomorphism bundle that has the canonical scalar identification.

The exterior derivative is an $\mathbb{R}$-linear operator $d:{\Omega }^r(M)\to {\Omega }^{r+1}(M)$ which agrees with the differential of a smooth function on degree zero, satisfies $d^2=0$, and obeys the graded Leibniz rule

For the uniqueness of the exterior derivative from its axioms, first note that these axioms force locality. If $\alpha$ vanishes near $p$, choose a smooth bump function $\chi$ equal to one near $p$ and supported where $\alpha =0$. Then $0=d(\chi \alpha )=d\chi \wedge \alpha +\chi d\alpha$ gives $d\alpha (p)=0$. Thus $d\alpha (p)$ depends only on the form near $p$.

In a coordinate chart write $\alpha ={\sum }_I{a}_{I}d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_r}$. Since $d(d{x}^i)=d^2{x}^i=0$, the axioms forceThis also proves existence. Define $d$ by that formula in each chart. The product rule for partial derivatives gives the graded Leibniz rule, and symmetry of mixed partial derivatives against antisymmetry of the wedge product of differential forms gives $d^2=0$. On overlaps it still agrees with the differential of every function, by the chain rule, so the preceding uniqueness argument in the second coordinate system makes the formulas agree. They therefore glue to a global exterior derivative.

For a differential one-form $\omega ={\sum }_i{a}_{i}d{x}^i$, the exterior derivative of a one-form evaluated on vector fields follows directly:In $X(\omega (Y))-Y(\omega (X))$, the additional terms are ${\sum }_i{a}_i[X(Y^i)-Y(X^i)]=\omega ([X,Y])$, where the bracket is the Lie bracket of vector fields. Hence

The de Rham cohomology iswith zero denominator in degree zero. The Poincaré lemma states that a closed differential form of positive degree on a star-shaped open subset of Euclidean space is an exact differential form; consequently every closed positive-degree form is locally exact on a smooth manifold.

For the suspension isomorphism for top de Rham cohomology of spheres, put $U=S^n\setminus \{\text{south pole}\}$ and $V=S^n\setminus \{\text{north pole}\}$. Both are diffeomorphic to ${\mathbb{R}}^n$, while $W=U\cap V$ deformation retracts onto the equatorial $S^{n-1}$. The Mayer--Vietoris sequence for de Rham cohomology comes from the short exact sequence of cochain complexesThe final map is surjective: a partition of unity ${\chi }_U+{\chi }_V=1$ lets a form $\eta$ on $W$ be lifted by extending ${\chi }_V\eta$ to $U$ and $-{\chi }_U\eta$ to $V$ by zero. The Poincaré lemma makes $H^{n-1}(U),H^{n-1}(V),H^n(U),H^n(V)$ vanish when $n>1$. Exactness givesIf $j:S^{n-1}\hookrightarrow W$ is the equatorial inclusion, homotopy invariance of de Rham cohomology makes $j^{\ast }$ an isomorphism. Thereforeis injective, indeed an isomorphism. Concretely, choose local primitives $d{\alpha }_U=\omega {|}_U$, $d{\alpha }_V=\omega {|}_V$, and send $[\omega ]$ to $[j^{\ast }({\alpha }_U-{\alpha }_V)]$. Their difference is closed; changing either primitive changes it by an exact form, so this agrees with the cohomological construction.

Use the standard boundaryless-manifold convention, so $M$ is a closed manifold. Its orientation and Riemannian metric determine a unique positive Riemannian volume form ${\mu }_g$ taking value one on every positively oriented orthonormal frame. In an oriented atlas it isFor the coordinate invariance of the Riemannian volume form, on an overlap put $J=\partial y/\partial x$. Then $g_x=J^T{g}_{y}J$ and $d{y}^1\wedge \cdots \wedge d{y}^n=(\det J)d{x}^1\wedge \cdots \wedge d{x}^n$. The oriented transition has $\det J>0$, so the square root of the metric determinant and the coordinate volume transform together. The local forms therefore agree and define ${\mu }_g$ globally.

For real $r$-forms the Hodge star operator is defined byThe pointwise wedge pairing is nondegenerate, so this uniquely defines a smooth map $\ast :{\Omega }^r(M)\to {\Omega }^{n-r}(M)$. On an oriented orthonormal coframe, it sends a basis wedge to the signed complementary wedge. Applying it twice gives

For Hodge star eigenvalues on graded differential forms, extend $\ast$ complex-linearly to the direct sum of all form degrees. This interpretation matters: on a fixed degree it is an endomorphism only in middle degree. In positive odd dimension, $r(n-r)$ is even for every $r$, so ${\ast }^2=I$ and only $\pm 1$ can occur. Both occur, with eigenforms $1+{\mu }_g$ and $1-{\mu }_g$.

In positive even dimension, ${\ast }^2$ is $+I$ on even-degree forms and $-I$ on odd-degree forms. Hence ${\ast }^4=I$ and the only possible eigenvalues are $\pm 1,\pm i$. The first two occur as above. For the latter, choose a smooth local orthonormal frame and multiply its first dual differential one-form by a nonzero smooth bump function supported in the chart. This gives a form $\eta$ for which $\eta$ and $\ast \eta$ are independent. For $\lambda =\pm i$,Thus all four occur. Consequently, for $n\ge 1$,On degree $m$ in dimension $2m$, the Hodge star has eigenvalues $\pm 1$ for even $m$ and $\pm i$ for odd $m$; signed complementary middle-degree wedges show both occur for $m\ge 1$. In zero dimensions it is multiplication by the signed unit volume at each point, so only the signs actually present occur.

For Hodge integration by parts in arbitrary degree, let $\alpha \in {\Omega }^{r-1}(M)$ and $\beta \in {\Omega }^r(M)$. The graded Leibniz rule and Stokes theorem on the closed manifold giveUsing the given codifferential and the square of the Hodge star,Thereforewhich proves that $\delta$ is the formal adjoint of $d$. The complex version inserts conjugation in the usual Hermitian inner product.

The Hodge decomposition theorem gives an $L^2$-orthogonal decompositionwhere ${\mathcal{H}}^k(M)=\ker \Delta$ is finite dimensional and consists of the harmonic differential forms. Each de Rham cohomology class has a unique harmonic representative. Equivalently, with harmonic projection $P_{\mathcal{H}}$ there is a smooth Green operator of the Hodge Laplacian $G$ satisfying $\Delta G=G\Delta =I-P_{\mathcal{H}}$.

The solvability condition for the Hodge Poisson equation isNecessity follows from $⟨\Delta \alpha ,h⟩=⟨\alpha ,\Delta h⟩=0$. Conversely the orthogonality condition gives $P_{\mathcal{H}}\beta =0$, so ${\alpha }_0=G\beta$ solves the Poisson equation for differential forms.

The nonnegativity of the Hodge Laplacian identityidentifies its kernel with closed and coclosed forms. Thus the affine space of solutions of the Hodge Poisson equation isFor completeness, if a closed form is decomposed as $\omega =h+d\xi +\delta \zeta$, then $d\delta \zeta =0$, and $\Vert \delta \zeta {\Vert }^2=⟨\zeta ,d\delta \zeta ⟩=0$. Hence its cohomology class is represented by $h$. An exact harmonic form $h=d\xi$ must vanish, since $\Vert h{\Vert }^2=⟨\delta h,\xi ⟩=0$. This proves the canonical isomorphism ${\mathcal{H}}^k(M)\cong H_{\mathrm{dR}}^k(M)$, so the solution set is an affine space whose translation vector space is isomorphic to the requested de Rham cohomology.

The gauge-invariant Yang-Mills action integrates over physically equivalent configurations along each gauge orbit. At the quadratic level its gauge-field kernel has longitudinal zero directions: in momentum space $K_{\mu \nu }(k)=-k^2{\eta }_{\mu \nu }+k_{\mu }{k}_{\nu }$ satisfies $K_{\mu \nu }{k}^{\nu }=0$. It therefore has no inverse from which to construct a gauge-boson propagator. Gauge fixing both removes the redundant orbit integration and supplies an invertible quadratic operator away from the ordinary physical propagator poles.

For the covariant condition ${\partial }^{\mu }{A}_{\mu }=0$, the Faddeev-Popov determinant is that of ${\partial }^{\mu }{D}_{\mu }$. Its representation by odd Faddeev-Popov ghost fields and antighost fields produces the displayed ghost term. In a nonabelian theory $D_{\mu }$ depends on the gauge field, so the determinant is not an irrelevant constant: ghost interactions must be retained. The even Nakanishi-Lautrup field $b$ writes gauge fixing in a form with off-shell BRS symmetry. For $\xi \ne 0$, its algebraic equation is $b=-\partial \cdot A/\xi$, giving $-(\partial \cdot A{)}^2/(2\xi )$ after elimination. At $\xi =0$ it imposes the gauge condition as a Lagrange multiplier.

For the nilpotence proof, write the adjoint bracket as $[X,Y{]}^a=f^{abc}{X}^b{Y}^c$, with coefficients in a Grassmann algebra. If $|X|,|Y|$ are their parities, this is a graded Lie bracket:Thus two odd fields have a symmetric bracket, in particular $[c,c]$ need not vanish. The transformations become $s{A}_{\mu }=D_{\mu }c$, $sc=-[c,c]/2$, $s\bar{c}=-b$, $sb=0$, and $s$ obeys the left graded Leibniz rule.