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A smooth gauge field of finite positive Euclidean Yang-Mills action cannot be invariant under a nonzero translation, even up to gauge transformation. The positive gauge-invariant action density would be periodic along that translation. A ball on which the density is bounded below by a positive constant has infinitely many disjoint translated copies, making the action infinite. Continuous translation invariance is a special case. Consequently a nonflat translation-invariant field satisfying the Anti-self-dual Yang-Mills equations on ${\mathbb{R}}^3\times \mathbb{R}$ is not a finite-action Yang-Mills instanton. Flat connections are the zero-action exception. Translations still act on instanton moduli, sending a localized instanton to a distinct centered configuration.

The scale $\rho >0$ in a BPST instanton changes the width of its action density while leaving total action unchanged. This reflects conformal invariance of four-dimensional classical Yang-Mills action. As $\rho \to 0$, gauge curvature concentrates at the center and the smooth family approaches a bubbling boundary, not another smooth zero-size instanton.

The Adjoint representation of a Lie algebra acts on the algebra itself: ${\mathrm{ad}}_X(Y)=[X,Y]$. In the basis $T_b$, its matrices areFor any algebra element $X$, the Jacobi identity givesTherefore the $D\times D$ matrices satisfy the same Lie bracket relations:This construction does not require that the algebra be simple or that the adjoint action be faithful.

For the field calculation, write $A_{\mu }=A_{\mu }^a{T}_a$ and $\lambda ={\lambda }^a{T}_a$. In the derivative-plus-connection convention the Yang-Mills field strength isThe sign of the parameter here is the negative of the matter-field parameter in the usual gauge transformation in the derivative-plus-connection convention. Work with the stated $\lambda$ throughout. Expanding the variation givesThe second derivatives cancel by commutativity of partial derivatives, and all remaining derivatives of $\lambda$ cancel by antisymmetry of the Lie bracket. The Jacobi identity makes the last two terms $[[A_{\mu },A_{\nu }],\lambda ]$. HenceUnlike the connection, the curvature has no inhomogeneous derivative term: it transforms in the Adjoint representation.

The quadratic form ${\kappa }_{ab}=\mathrm{tr}(T_a^{\mathrm{ad}}{T}_b^{\mathrm{ad}})$ is the Killing form. Its symmetry follows from cyclicity of the matrix trace. Using the just-proved adjoint commutators,ThusEquivalently, each adjoint generator is skew with respect to this invariant bilinear form. This invariance holds even when the Killing form is degenerate; no inverse form is needed here.

The spacetime metric used to raise Lorentz indices is unaffected by the internal gauge transformation. Since $\delta F=-{\mathrm{ad}}_{\lambda }F$, invariance givesTherefore the Killing-form contraction of the field strengths is gauge invariant, as required for the internal contraction in a Yang-Mills action.

Use an SU(2) connection one-form $A=A_{\mu }d{x}^{\mu }$ with Skew-Hermitian generators $T_a=-i{\sigma }_a/2$, so $\mathrm{tr}(T_a{T}_b)=-{\delta }_{ab}/2$. Absorb the gauge coupling into $A$, and write its gauge curvature as $F=dA+A\wedge A$. The Yang-Mills gauge transformation convention is $A^h=h^{-1}Ah+h^{-1}dh$, giving $F^h=h^{-1}Fh$.

In four dimensions the Second Chern form and Second Chern number of the associated fundamental rank-two bundle areHere $M$ is a closed oriented four-manifold. This sign follows by expanding the total Chern class $\det (1+iF/(2\pi ))$ and using $\mathrm{tr}F=0$. The first Chern class vanishes for SU(2); the second is the degree-four characteristic class relevant here. With the same curvature convention the commonly positive self-dual instanton number isBoth sign conventions occur; fixing them at the outset avoids identifying opposite-oriented integers.

The form is gauge-invariant by conjugation and cyclicity of the matrix trace. The gauge-theory Bianchi identity $D_{A}F=0$ makes it closed. To see that its integral is independent of the connection, vary $A$:On a closed $M$ the integral of this exact form vanishes. This is Chern-Weil connection transgression, showing that $k_C$ depends on the bundle's topology rather than on a particular gauge field.

Locally the same four-form is an exterior derivative. The Chern-Simons 3-formsatisfiesFor verification, differentiating gives $\mathrm{tr}(dA\wedge dA+2dA\wedge A\wedge A)$; expansion of $F\wedge F$ gives these terms and $\mathrm{tr}{A}^4$. Graded cyclicity moves the first one-form through the other three, changing the sign, so $\mathrm{tr}{A}^4=0$. This local exactness does not make the Second Chern number zero on every closed manifold: on a nontrivial bundle the gauge potentials and ${\omega }_3$ are only patchwise defined.

For the usual finite-action sector on ${\mathbb{R}}^4$, impose pure-gauge behavior $A\to h^{-1}dh$ at infinity. Compactification gives a bundle over $S^4$, with transition map $h:S^3\to SU(2)$. The Maurer-Cartan equation gives $d(h^{-1}dh)=-(h^{-1}dh{)}^2$, henceWith the outward boundary orientation, Stokes theorem therefore givesThese are opposite choices of the winding number, and either becomes the topological degree of $h$ after specifying the orientation of the target SU(2) sphere. Thus integrality also follows from ${\pi }_3(SU(2))=\mathbb{Z}$. Smooth deformations preserving the asymptotic sector cannot change it.

The positive Euclidean Yang-Mills action isDefine $\Vert X{\Vert }^2=-\int \mathrm{tr}(X\wedge \ast X)$. Since the Hodge star has square one on two-forms in Euclidean dimension four,Choosing $s=\mathrm{sgn}Q$ gives the Yang-Mills instanton Bogomolny boundThe self-dual Yang-Mills equations and their anti-self-dual version imply the Yang-Mills equations $D_A\ast F=0$ by the Bianchi identity. Their finite-action solutions are Yang-Mills instantons.

For a concrete self-dual example, the BPST instanton with center $a$ and size $\rho >0$ haswhere ${\eta }_{\mu \nu }^a$ are self-dual 't Hooft symbols. Their contraction is ${\sum }_{a,\mu ,\nu }({\eta }_{\mu \nu }^a{)}^2=12$. Thus $F_{\mu \nu }^a{F}_{\mu \nu }^a=192{\rho }^4/(|x-a{|}^2+{\rho }^2{)}^4$, andThis gives $S=8{\pi }^2/g^2$ and $Q=1$. Reversing duality gives $Q=-1$. The free center, size and framed gauge orientation illustrate the instanton moduli space.

On a three-dimensional slice $\Sigma$, there is no four-form Chern integral intrinsic to the slice. Its transgression is the Chern-Simons number of a gauge field, in the instanton-charge conventionIt is a real number for a general connection, not necessarily an integer, and is not strictly gauge invariant. Set $\theta =h^{-1}dh$ and $\bar{\theta }=dh{h}^{-1}$. Substituting $A^h=h^{-1}Ah+\theta$, using $d\theta =-{\theta }^2$ and graded trace cyclicity, gives the gauge change of the Chern-Simons three-formThe mixed terms collect into the displayed exact form; the last term is also fixed by the pure-gauge case $A=0$. Therefore on a closed $\Sigma$, or with boundary conditions making the exact-form integral vanish,Thus $N_{\mathrm{CS}}\mathrm{mod}\mathbb{Z}$, or $\exp (2\pi i{N}_{\mathrm{CS}})$, is gauge invariant. Gauge maps homotopic to the identity have $w=0$; maps of nonzero winding are large transformations even when they equal the identity at infinity.

For a Yang-Mills vacuum on ${\mathbb{R}}^3$, zero magnetic energy requires $F_{ij}=0$. On simply connected space this is a pure gauge $A=h^{-1}dh$, and $N_{\mathrm{CS}}=w(h)$ is an integer after fixing the asymptotic trivialization. Quotienting only by transformations homotopic to the identity leaves integer-labelled vacuum representatives. Allowing all winding transformations identifies them classically; quantum states can instead transform by a phase, producing the theta vacuum. A three-dimensional pure-gauge representative carries a winding number but is not a localized positive-energy monopole. In fact no static finite-energy lump in three-dimensional pure Yang-Mills theory exists with the usual decay conditions: $A_i^{(s)}(x)=s{A}_i(sx)$ gives $E(s)=sE(1)$, and stationarity forces $E=0$.

Finally, on an oriented spacetime slab $[t_{-},t_{+}]\times \Sigma$, with no side-boundary contribution, the transgression identity givesA Yang-Mills instanton therefore interpolates between vacuum representatives whose Chern-Simons numbers differ by its integer charge. In a four-dimensional quantum theory a Yang-Mills theta term weights such sectors by $e^{i\theta Q}$, making $\theta$ periodic modulo $2\pi$. In an intrinsically three-dimensional theory an action $2\pi \ell N_{\mathrm{CS}}$ has a gauge-invariant phase $e^{iS}$ only for integer level $\ell$ under all winding gauge maps. These statements distinguish the gauge-invariant integer four-dimensional Chern charge from the three-dimensional connection-dependent quantity defined modulo integers.

An instanton is a localized finite-action classical solution of a Euclidean field theory. Its role differs from that of a static spatial classical field-theory soliton: an instanton is localized in Euclidean spacetime and contributes to quantum transition amplitudes, rather than persisting as a stationary real-time particle. Wick rotation changes the oscillatory weight $e^{iS/\hslash }$ into $e^{-S_E/\hslash }$ when the continuation and boundary conditions are valid. The Euclidean path integral is consequently dominated semiclassically by stationary Euclidean configurations. Perturbation theory around one vacuum misses other saddles whose weights have the nonanalytic form $e^{-S_I/\hslash }$.

A useful first example is quantum tunnelling between two degenerate minima in quantum mechanics. For coordinate $q$ with mass $m$, the Euclidean action iswith the vacuum energy subtracted from $V$. Its Euler-Lagrange equation is $m\ddot{q}=V^{\prime }(q)$, equivalent to motion in the inverted potential $-V$. A finite-action trajectory joining zero-energy minima has conserved Euclidean energy $m{\dot{q}}^2/2-V=0$. A forward instanton therefore satisfies $\dot{q}=\sqrt{2V/m}$, while an anti-instanton has the opposite orientation. Completing the square gives the action boundwhich these zero-energy paths saturate. This is the Euclidean counterpart of the first-order kink argument, with Euclidean time replacing the spatial coordinate.

For the quartic double-well instanton, chooseThe first-order equation in the interval $-a<q<a$ is $\dot{q}=\omega (a^2-q^2)/(2a)$. Separation givesThe center ${\tau }_0$ is arbitrary. Its derivative ${\dot{q}}_I$ is an instanton translation zero mode: differentiating the Euclidean field equation shows that it lies in the kernel of the quadratic fluctuation operator $-m{\partial }_{\tau }^2+V^{\prime \prime }(q_I)$. A naive Gaussian determinant therefore has a zero eigenvalue. Replace its integration by the collective-coordinate integral over ${\tau }_0$, and take the determinant only over nonzero modes. The resulting instanton fluctuation prefactor combines that Jacobian with the determinant ratio against the vacuum. The action controls the exponential suppression; the prefactor is essential for a quantitative rate or splitting.

When $S_I/\hslash \gg 1$, crossings are rare and their typical separations greatly exceed their widths. In the dilute instanton gas, ordered center integrals for $n$ well-separated crossings over Euclidean duration $\mathcal{T}$ produce ${\mathcal{T}}^n/n!$. If a single-crossing rate is $\kappa =K{e}^{-S_I/\hslash }$, a symmetric two-well system has even-crossing and odd-crossing sums proportional to $\cosh (\kappa \mathcal{T})$ and $\sinh (\kappa \mathcal{T})$. The spectral interpretation gives levels $E_{\mathrm{well}}\mp \hslash \kappa$ andThis is a real tunnelling level splitting. A Euclidean bounce instead describes false vacuum decay: it leaves and returns to a metastable vacuum, and a characteristic negative fluctuation mode supplies the imaginary part interpreted as decay. A bounce must not be confused with a stable double-well instanton merely because both use Euclidean saddles.

The central gauge-theory example is a Yang-Mills instanton on Euclidean four-space. State conventions carefully: take anti-Hermitian generators $T_a=-i{\sigma }_a/2$, with $\mathrm{Tr}(T_a{T}_b)=-{\delta }_{ab}/2$, connection $A=A_{\mu }d{x}^{\mu }$, and gauge curvature $F=dA+A\wedge A$. With coupling $g$, the positive Yang-Mills action isThe Hodge star operator squares to one on Euclidean two-forms. Under the usual finite-action boundary conditions, the connection approaches a pure gauge at infinity, defining a map $S_{\infty }^3\to SU(2)\simeq S^3$. Its winding degree is the integer instanton numberThis is unchanged by smooth deformations preserving the boundary sector. Geometrically this is, up to the stated sign convention, the second Chern number of the compactified bundle. Different sign conventions for trace, curvature or orientation must be changed together.

Use the positive norm $\Vert F{\Vert }^2=-\int \mathrm{Tr}(F\wedge \ast F)$. Decomposition into self-dual and anti-self-dual curvature givesThe Yang-Mills instanton Bogomolny bound is saturated by $F=\ast F$ for positive $k$, or $F=-\ast F$ for negative $k$. These first-order equations imply the full Yang-Mills equations: the Bianchi identity $D_{A}F=0$ becomes $D_A\ast F=0$ when $\ast F=\pm F$. Thus the topological minimum is a genuine classical saddle, not just an arbitrary field with integer charge. Self-duality alone is a local differential condition; finite action is an additional requirement.

An explicit unit-charge example is the BPST instanton. Let $y=x-x_0$, $\rho >0$, and choose the self-dual symbols ${\eta }_{ij}^a={ϵ}_{aij}$, ${\eta }_{i4}^a={\delta }_{ai}$ and ${\eta }_{4i}^a=-{\delta }_{ai}$ for $i,j=1,2,3$, using orientation $d{x}^1\wedge d{x}^2\wedge d{x}^3\wedge d{x}^4$. In regular gauge,The curvature is self-dual. Since ${\sum }_{a,\mu ,\nu }({\eta }_{\mu \nu }^a{)}^2=12$, its squared component curvature is $192{\rho }^4/(y^2+{\rho }^2{)}^4$. Radial integration yieldsso $k=1$. Replacing self-dual symbols by anti-self-dual ones produces the opposite charge with the same positive action.

The center has four coordinates and $\rho$ is an instanton size modulus. Classical four-dimensional Yang-Mills action is scale invariant, so varying $\rho$ changes the action-density width but not the total action. If gauge transformations must tend to the identity at infinity, three global gauge orientations also remain, giving eight parameters for the one-instanton framed instanton moduli space. If those global orientations are quotiented out, the unframed one-instanton space has five parameters. A zero-size limit is a singular concentration of curvature, not another smooth instanton. Higher-charge solutions likewise possess instanton moduli spaces; their collective-coordinate integrals replace the corresponding zero-mode determinants.

In a Euclidean-time description, topologically distinct gauge vacua are separated by a change of Chern-Simons number; an instanton interpolates between them, with the change equal to $k$ in compatible orientation conventions. Theta-weighted instanton sectors give $Z(\vartheta )={\sum }_k{e}^{ik\vartheta }{Z}_k$, connecting topology to the vacuum angle. With fermions, the gauge-field topology also controls fermionic zero modes and chiral selection rules through the chiral anomaly. These phenomena explain why instantons encode physics beyond small fluctuations around a single vacuum.

Instantons link Euclidean classical solutions, topological sectors and exponentially suppressed quantum effects. Their interpretation requires the boundary conditions, normalization, fluctuation spectrum and measure over collective coordinates. A dilute approximation is controlled only when its individual-action suppression and separation assumptions hold; classical scale moduli and their quantum corrections can make size integrals delicate rather than automatically finite.

The Adjoint representation of a Lie algebra acts on the Lie algebra itself by ${\mathrm{ad}}_X(Y)=[X,Y]$. The Jacobi identity givesThus it is a Lie algebra representation. In the basis $T_a$ with $[T_a,T_b]=c^c{}_{ab}{T}_c$, its matrices areThe column index labels the basis vector being acted on; this convention fixes the matrix signs.

Work in finite dimension over characteristic zero, as in the usual real or complex gauge algebras. A simple Lie algebra is nonabelian and has no nonzero proper Lie algebra ideal. A semisimple Lie algebra has no nonzero solvable ideal; equivalently it is a direct sum of simple ideals. The nondegenerate Killing form is $\kappa (X,Y)=\mathrm{tr}({\mathrm{ad}}_X{\mathrm{ad}}_Y)$. Its invariance follows from the representation identity and trace cyclicity:

For the reduction, take a nonzero proper ideal $\mathfrak{a}$, if one exists, and define its Killing-orthogonal complement ${\mathfrak{a}}^{\perp }$. Invariance makes it an ideal: for $y\in {\mathfrak{a}}^{\perp }$, $x\in L$, $z\in \mathfrak{a}$, $\kappa ([x,y],z)=-\kappa (y,[x,z])=0$. Moreover $[\mathfrak{a},{\mathfrak{a}}^{\perp }]=0$, because for $u\in \mathfrak{a}$, $v\in {\mathfrak{a}}^{\perp }$, and any $x\in L$,and the global Killing form is nondegenerate. Their intersection is consequently an abelian ideal: its elements lie in both commuting ideals. Semisimplicity forces that intersection to be zero. The dimension formula for a nondegenerate bilinear form then givesThe summands commute and are semisimple, since a solvable ideal in either summand would also be an ideal of $L$. Repeating on smaller summands terminates in simple ideals. This proves the orthogonal ideal splitting for a nondegenerate Killing form and describes the required reduction; abelian one-dimensional factors do not occur in a semisimple algebra.

If $\kappa (T_a,T_b)=\ell {\delta }_{ab}$ with $\ell \ne 0$, invariance givesso $c_{abc}=-c_{acb}$ after lowering the output index with $\delta$. The bracket already gives $c_{abc}=-c_{bac}$. These two adjacent transpositions generate all permutations, proving that the structure constants of a Lie algebra are completely antisymmetric. This is the antisymmetry of Killing-lowered structure constants; the scalar $\ell$ cancels in the chosen normalization.

For the gauge calculation write $\Lambda ={\lambda }_a{t}_a$, $A_{\mu }=A_{\mu a}{t}_a$, and $D_{\mu }={\partial }_{\mu }+A_{\mu }$. The representation generators here obey $[t_a,t_b]=c_{abc}{t}_c$; factors of $i$ and the gauge coupling are absorbed in this connection convention. Expanding the varied derivative givesFor this to equal $\Lambda D_{\mu }\varphi$, choose the gauge transformation in the derivative-plus-connection conventionThen all extra terms cancel, proving covariance of the gauge covariant derivative for arbitrary space-dependent parameters.

Apply both derivatives to a test field. The second derivatives commute and the terms involving one derivative of the field cancel, leaving the Yang-Mills field strengthThus its components areThe already obtained variation can be written $\delta D_{\mu }=[\Lambda ,D_{\mu }]$. The operator Jacobi identity then yieldsorFinally, with fixed space-time metric,The contracted product is symmetric in $a,c$, while $c_{abc}$ is antisymmetric in those indices. Hence the Yang-Mills action density is gauge invariant, with the sign convention derived rather than assumed.

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.

Because $D_{\mu }$ is even, $D_{\mu }[c,c]=[D_{\mu }c,c]+[c,D_{\mu }c]=2[D_{\mu }c,c]$. Varying both the field and the connection in $D_{\mu }c$ now givesFor the ghost,The last equality is the graded Jacobi identity applied to three copies of the odd field $c$. In components the remaining expression is $\frac{1}{2}{f}^{abc}{f}^{bde}{c}^d{c}^e{c}^c$; its coefficient is antisymmetrized over $c,d,e$, and the ordinary Jacobi identity for the structure constants makes that antisymmetrization zero. Finally $s^2\bar{c}=-sb=0$ and $s^{2}b=0$. HenceThis is off-shell nilpotence of the Yang-Mills BRST quartet.

To verify invariance of the complete Lagrangian, $s{F}_{\mu \nu }=[F_{\mu \nu },c]$, obtained from $[D_{\mu },D_{\nu }]c=[F_{\mu \nu },c]$. Thus $s(F_{\mu \nu }^a{F}^{\mu \nu a})=0$ because the product of the two curvature components is symmetric in their color labels while $f^{abc}$ is antisymmetric. The remaining terms vary asThe last term vanishes by the already proved $s^{2}A=0$, and the first and third terms cancel pointwise. ThereforeEquivalently, the covariant gauge-fixing density as a BRST variation is $s\Psi$ with $\Psi =-{\bar{c}}^a(\partial \cdot A^a+\xi b^a/2)$. The graded sign from moving $s$ past $\bar{c}$ produces the positive ghost term, and $s^2\Psi =0$ proves its invariance. No integration-by-parts qualification is needed for the density as written.

Let $Q=Q_{\mathrm{BRS}}$ be the BRS charge. The physical state space is its BRST cohomology, usually at ghost number zero:Nilpotence ensures $\mathrm{im}Q\subseteq \ker Q$. A state $Q|\chi ⟩$ is orthogonal to every closed state because $⟨\psi |Q|\chi ⟩=⟨Q\psi |\chi ⟩=0$, and its norm is $⟨\chi |Q^2|\chi ⟩=0$. Thus adding an exact state changes neither physical inner products nor amplitudes. Self-adjointness here is relative to the stated indefinite Hermitian form; a nonzero self-adjoint nilpotent operator could not exist on a positive-definite state space. Positivity must be proved on the quotient, not assumed on the gauge/ghost space.

Fix a color label and a nonzero momentum $k$, and take the usual nontrivial asymptotic charge, $\alpha \ne 0$. Nilpotence on a vector state gives $Q|c⟩=0$, because at least one component of $k$ is nonzero. Nilpotence on the antighost state then givesThe two coefficients are complex conjugates as required by self-adjointness: using the given ghost pairing,Thus the antighost-to-vector coefficient is ${\alpha }^{\ast }$, as the printed bar notation indicates. These conditions are the massless shell condition and the consistency of the charge with the inner products.

A general one-particle vector representative is $|\epsilon ⟩={\epsilon }^{\mu }|A_{\mu }⟩$. It is closed precisely when $k_{\mu }{\epsilon }^{\mu }=0$. The vector state $k^{\mu }|A_{\mu }⟩=Q|\bar{c}⟩/\bar{\alpha }$ is exact, so representatives obeyThe ghost state is also exact: choose $v^{\mu }$ with $\alpha v\cdot k=1$, then $Q(v^{\mu }|A_{\mu }⟩)=|c⟩$. The antighost state is not closed because its image is a nonzero longitudinal vector. More explicitly, closure of a general sum of vector, ghost and antighost states forces the antighost coefficient to vanish and the vector polarization to be orthogonal to $k$; its ghost component disappears in the quotient. Thus the full one-particle cohomology already reduces toWith the mostly-plus Minkowski metric, choose a Lorentz frame with $k^{\mu }=(\omega ,0,0,\omega )$, $\omega >0$. Closure gives ${\epsilon }^3={\epsilon }^0$, and the norm isSubtracting a multiple of $k$ sets ${\epsilon }^0={\epsilon }^3=0$. Consequently positive one-particle BRST cohomology consists of two transverse positive-norm polarizations per color, represented by $|A_1⟩,|A_2⟩$. A nonzero class has strictly positive norm; the zero-norm longitudinal direction was precisely the exact direction removed.

The qualifications $k\ne 0$ and $\alpha \ne 0$ are essential to this one-particle argument. The zero-momentum sector requires separate treatment, and the trivial-charge case cannot yield the asserted removal of negative-norm vector states from the stated relations alone. Abstract nilpotence and self-adjointness by themselves do not guarantee positive cohomology.

Take a local gauge transformation $g(x)$ acting on the adjoint field by ${\Phi }^{\prime }=g\Phi g^{-1}$. The connection convention $D_{\mu }={\partial }_{\mu }+[A_{\mu },\cdot ]$ is compatible withIndeed, differentiating ${\Phi }^{\prime }$ gives the extra commutator $[({\partial }_{\mu }g)g^{-1},{\Phi }^{\prime }]$, exactly cancelled by the inhomogeneous term in the transformed gauge potential.

For a direct proof of the Yang-Mills gauge transformation of curvature, put ${\Omega }_{\mu }=({\partial }_{\mu }g)g^{-1}$ and $B_{\mu }=g{A}_{\mu }{g}^{-1}$. Differentiation givesSubstituting $A_{\mu }^{\prime }=B_{\mu }-{\Omega }_{\mu }$ into the Yang-Mills field strength cancels all terms involving $\Omega$ and leavesThis direct calculation also applies when the adjoint action has a kernel.

Expand $D_{\mu }{F}_{\nu \rho }+D_{\nu }{F}_{\rho \mu }+D_{\rho }{F}_{\mu \nu }$. The mixed second derivatives of $A$ cancel in pairs. The first-derivative terms from differentiating $[A_{\nu },A_{\rho }]$ cancel the terms from $[A_{\mu },{\partial }_{\nu }{A}_{\rho }-{\partial }_{\rho }{A}_{\nu }]$. The remaining sum isby the Jacobi identity. Thus the gauge-theory Bianchi identity is

For the variation of the Yang-Mills action, write $a_{\mu }=\delta A_{\mu }$. Differentiating the curvature gives $\delta F_{\mu \nu }=D_{\mu }{a}_{\nu }-D_{\nu }{a}_{\mu }$. Invariance of the Killing form impliesThis is invariant integration by parts for a gauge covariant derivative. Using symmetry of the Killing form, antisymmetry of $F$, and compactly supported variations,Since the Killing form of a semisimple Lie algebra is nondegenerate and the variations are arbitrary, the Yang-Mills equations are

For the constant Yang-Mills theta term, symmetry under exchanging the two antisymmetric index pairs givesup to the boundary variation of the Yang-Mills theta term. Contracting the gauge-theory Bianchi identity with ${ϵ}^{\mu \nu \rho \sigma }$ makes the last integrand vanish. Therefore a constant theta term does not alter the bulk Yang-Mills equations: $D_{\mu }{F}^{\mu \nu }=0$. Its variation is a boundary term, so this conclusion uses compact support or boundary conditions that remove that term.

Finally apply $D^{\rho }$ to the gauge-theory Bianchi identity, obtainingThe adjoint covariant derivative obeys $[D_{\alpha },D_{\beta }]X=[F_{\alpha \beta },X]$. Commute $D^{\rho }$ past the other derivative. The differentiated divergences vanish by the Yang-Mills equations, leavingIn the last line the second commutator needs its sign tracked carefully: directly,Thus the simplified sum gives the covariant wave equation for the Yang-Mills field strengthAll spacetime derivatives and index manipulations here use the fixed flat metric.