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Choose the gauge covariant derivative $D_{\mu }={\partial }_{\mu }+ig{A}_{\mu }$ and matter transformation ${\varphi }^{\prime }=h\varphi$. Covariance $D_{\mu }^{\prime }{\varphi }^{\prime }=h{D}_{\mu }\varphi$ forces $A_{\mu }^{\prime }=h{A}_{\mu }{h}^{-1}+(i/g)({\partial }_{\mu }h)h^{-1}$. For $h=1-ig\omega +O({\omega }^2)$, expansion gives the displayed adjoint covariant derivative variation. With an orthonormal Hermitian Lie-algebra basis, $\delta A_{\mu }^a={\partial }_{\mu }{\omega }^a-g{f}^{abc}{A}_{\mu }^b{\omega }^c$. Thus Lorenz-gauge variation is $\delta (\partial \cdot A)=\partial \cdot \mathcal{D}\omega$. Choosing the Faddeev-Popov operator as $\mathcal{M}=-\partial \cdot \mathcal{D}$, with its field-independent sign absorbed in normalization, gives ghost action $-\bar{c}\partial \cdot \mathcal{D}c$; consistent propagator and vertex signs follow from this choice.

This gauge-invariant higher-derivative density has equationThe extra commutator comes from varying the connection inside the adjoint covariant derivative: $\delta (D_{\mu }{F}_{\nu \rho })=D_{\mu }\delta F_{\nu \rho }+[\delta A_{\mu },F_{\nu \rho }]$. Two uses of gauge-covariant integration by parts give the first term, and Killing-form invariance gives the second. In an abelian limit it reduces to $\square {\partial }_{\mu }{F}^{\mu \nu }=0$. Keeping the covariant derivative order is essential because their commutator is curvature.

In three space dimensions, rescale $A_i(x)$ to $\lambda A_i(\lambda x)$ and $\Phi (x)$ to $\Phi (\lambda x)$. The gauge curvature then scales with two derivatives, and the adjoint covariant derivative of the Higgs field with one. Magnetic, Higgs-gradient and potential energies give the displayed formula. Stationarity requires $V_B=V_D+3V_U$, which permits nonzero energies. In the Bogomolny-Prasad-Sommerfield monopole limit $V_U=0$, the Bogomolny equations give $V_B=V_D$.

For the trace inner product and adjoint covariant derivative, compatibility gives $\Delta |\Phi {|}^2=2|D\Phi {|}^2+2⟨\Phi ,D_i{D}_i\Phi ⟩$. The gauge-theory Bianchi identity and $B_i=\pm D_i\Phi$ imply $D_i{D}_i\Phi =0$. Therefore $\Delta |\Phi {|}^2=2|D\Phi {|}^2$, equal to the energy density $|B{|}^2+|D\Phi {|}^2$ on a Bogomolny-Prasad-Sommerfield monopole. If the energy density is instead defined with an overall factor $1/2$, this Laplacian is twice that density.

An adjoint scalar has adjoint covariant derivative $D_{\mu }\varphi ={\partial }_{\mu }\varphi +[A_{\mu },\varphi ]$. A symmetric invariant bilinear form on a Lie algebra contracts the gauge field strength and scalar kinetic terms to give gauge invariance. On a compact real form the positive internal metric is $-\kappa$, because the Killing form itself is negative definite; complex scalar components require the associated Hermitian pairing. The Killing form is degenerate on Abelian factors, where another invariant metric is needed for a nondegenerate kinetic term.

Absorb the gauge coupling into the connection $A_{\mu }=A_{\mu }^a{T}_a$ and use $D_{\mu }={\partial }_{\mu }+A_{\mu }$. The anti-Hermitian generators make $U(x)=\exp [{\omega }^a(x)T_a]$ unitary for real parameters. With a matter field transforming as ${\psi }^{\prime }=U\psi$, require $D_{\mu }^{\prime }{\psi }^{\prime }=U{D}_{\mu }\psi$. Applying both sides to an arbitrary field givesExpanding $U=1+\omega +O({\omega }^2)$ and $U^{-1}=1-\omega +O({\omega }^2)$ yields the infinitesimal gauge transformation with an anti-Hermitian connectionThe adjoint covariant derivative is $D_{\mu }\omega ={\partial }_{\mu }\omega +[A_{\mu },\omega ]$. In components,The trace normalization makes the structure constants totally antisymmetric: $f_{abc}=-\mathrm{Tr}([T_a,T_b]T_c)$ and cyclicity of the trace supply antisymmetry in the remaining indices. Replacing $U=e^{\omega }$ by $e^{-\omega }$ reverses the infinitesimal parameter convention and gives $\delta A=D\omega$; mixing the two conventions would change the ghost operator incorrectly. The trace normalization also gives the invariant quadratic form $-\mathrm{Tr}(A_{\mu }{A}_{\nu })=A_{\mu }^a{A}_{\nu }^a$.

A gauge-invariant functional integral counts all representatives along each gauge orbit. The gauge condition $\mathcal{F}[A]=0$ selects a representative, and its delta functional enforces that selection. Assume a local patch where each orbit intersects the gauge slice once and the gauge-condition derivative has no residual zero modes. DefineChanging the orbit integration variable from $\omega$ to $\mathcal{F}[A^{\omega }]$ givesInsert this Faddeev-Popov gauge-orbit identity into the unfixed path integral. The action and measure are invariant under a change $A\mapsto A^{\omega }$, so integration over $\omega$ factors out as the gauge-group volume. Dividing by that redundant volume leaves the fixed integral and its Faddeev-Popov determinant. In a perturbative patch with fixed determinant sign or phase, the absolute value is replaced by the oriented determinant, up to a field-independent normalization:Here $D_{\mu }^{cb}={\delta }^{cb}{\partial }_{\mu }+f_{dbc}{A}_{\mu }^d$. This is a functional determinant of the gauge-condition Jacobian, not an arbitrary extra weight. The local invertibility assumption matters: residual gauge transformations or multiple orbit intersections, described by the Gribov ambiguity, prevent treating this derivation as a global unique gauge slice without further qualifications.

Both extra factors can be incorporated into an enlarged action. An auxiliary real field $b^a$ Fourier-represents the delta functional, while a Grassmann Gaussian integral over independent Faddeev-Popov ghost fields $c,\bar{c}$ represents the determinant:Thus the original fixed integral can be written with action $S+\int b\mathcal{F}+\int \bar{c}Mc$ and ordinary integration over the extra fields, with no explicit delta functional or determinant. The anticommuting fields give a determinant rather than the inverse determinant obtained from commuting Gaussian fields.

For the usual Gaussian family of gauges, add $\xi b^a{b}^a/2$ and integrate out $b$. Completing the square producesThis is Gaussian gauge fixing with an auxiliary field; equivalently one averages the condition $\mathcal{F}=f$ with its Gaussian weight. Gauge-invariant observables are unchanged apart from normalization. For the Lorenz gauge ${\mathcal{F}}_a={\partial }^{\mu }{A}_{\mu }^a$, the chosen transformation convention gives $M=-{\partial }^{\mu }{D}_{\mu }$. The ghost action is $-\int \bar{c}{\partial }^{\mu }{D}_{\mu }c$, or $\int ({\partial }^{\mu }\bar{c})D_{\mu }c$ after integration by parts. The dependence of $D$ on $A$ accounts for the ghost interactions.

Use the Minkowski metric convention $\eta =\mathrm{diag}(1,-1,-1,-1)$. On the Adjoint representation, the adjoint covariant derivative is $D_{\mu }X={\partial }_{\mu }X+[A_{\mu },X]$. Set $a_{\mu }=\delta A_{\mu }$ and $\chi =\delta \Phi$. ThenThe trace obeys $\mathrm{Tr}(X[Y,Z])=\mathrm{Tr}([Z,X]Y)$, and covariant integration by parts has the ordinary boundary term because the trace of a commutator is zero. Thus compactly supported variations giveThe trace pairing on $\mathfrak{su}(2)$ is nondegenerate, so the Euler-Lagrange field equations areIn particular the sign of the scalar mass term follows from the plus sign in the supplied trace Lagrangian density; it should not be guessed from a different metric convention. This is the adjoint Yang-Mills-Higgs trace variation.

For the static Bogomolny-Prasad-Sommerfield monopole, take the usual purely magnetic extension ${\partial }_0{A}_i={\partial }_0\Phi =0$ and temporal gauge $A_0=0$. At $m=0$, the equations reduce towith the $\nu =0$ equation identically zero. We now verify both spatial equations from $B_i=D_i\Phi$ without assuming them. The gauge-theory Bianchi identity gives $D_i{B}_i=0$, hence immediately $D_i{D}_i\Phi =0$. Also $F_{ji}={\epsilon }_{jik}{B}_k$, soHere ${\epsilon }_{jik}{\epsilon }_{jk\ell }=-2{\delta }_{i\ell }$. This proves that the Bogomolny equations imply all static zero-mass Yang-Mills equations and the scalar equation. The specification of $A_0$ matters: the spatial Bogomolny equations by themselves do not constrain an arbitrary extra electric potential, so the implication is for their standard purely magnetic static extension.

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.

Covariance of the fundamental gauge covariant derivative requires $D_{\mu }^{f\prime }(g\chi )=g{D}_{\mu }^f\chi$ for every column $\chi$. Expanding the left side and cancelling $g{\partial }_{\mu }\chi$ gives $({\partial }_{\mu }g)+A_{\mu }^{\prime }g=g{A}_{\mu }$. Therefore the necessary and sufficient Yang-Mills gauge transformation isIn operator notation $D_{\mu }^{f\prime }=g{D}_{\mu }^f{g}^{-1}$. Expanding its commutator on an arbitrary column shows $[D_{\mu }^f,D_{\nu }^f]\chi =F_{\mu \nu }\chi$: the second derivatives cancel, as do the terms containing a single derivative of $\chi$. Conjugating this identity proves the gauge field strength transformationThus the gauge field strength transforms homogeneously even though the potential has an inhomogeneous derivative term.

For the adjoint scalar field, write $B_{\mu }=({\partial }_{\mu }g)g^{-1}$ and ${\Phi }^{\prime }=g\Phi g^{-1}$. Differentiating the inverse matrix givesThe extra terms cancel, proving $D_{\mu }^{\prime }{\Phi }^{\prime }=g(D_{\mu }\Phi )g^{-1}$ for the adjoint covariant derivative. Its curvature is obtained by another direct expansion:where the Jacobi identity combines the last two terms.

For the equations of motion take compactly supported variations, and assume $e\ne 0$, as required by the Lagrangian coefficient. Invariance of the Killing form impliesIndeed, the two connection terms sum to $\kappa ([A_{\mu },X],Y)+\kappa (X,[A_{\mu },Y])=0$. This is the gauge-covariant integration by parts identity. The needed variations areAntisymmetry of the gauge field strength combines its two variation terms. Applying gauge-covariant integration by parts to the action $S=\int \mathcal{L}{d}^{4}x$ givesHere $\kappa ([\delta A_{\nu },\Phi ],D^{\nu }\Phi )=\kappa (\delta A_{\nu },[\Phi ,D^{\nu }\Phi ])$ fixes the sign of the scalar current. Since the Killing form is nondegenerate on a semisimple Lie algebra, independent variations give the coupled Yang-Mills equations and scalar equationEquivalently, $D_{\mu }{F}^{\mu \nu }=e^2[D^{\nu }\Phi ,\Phi ]$. These signs use the stated Minkowski metric, with upper spatial derivatives $D^i=-D_i$.

It remains to verify every equation for the static Bogomolny equations. The hypotheses imply $F_{0i}=0$ and $D_0\Phi =0$, so the temporal gauge equation is automatically zero. The gauge-theory Bianchi identityfollows, for example, from the Jacobi identity for the fundamental gauge covariant derivatives, using their commutator $F_{ij}$. Define $B_i=\frac{1}{2}{\epsilon }_{ijk}{F}_{jk}$. Contracting the gauge-theory Bianchi identity with ${\epsilon }_{ijk}$ gives $D_i{B}_i=0$. The Bogomolny equations say $B_i=e{D}_i\Phi$, so $e\ne 0$ implies $D_i{D}_i\Phi =0$. In the static configuration $D_{\mu }{D}^{\mu }\Phi =-D_i{D}_i\Phi$, verifying the scalar equation.

For the spatial gauge equation, use $F^{ij}=F_{ij}$ and $D^j\Phi =-D_j\Phi$. The Bogomolny equations, the curvature commutator proved above, and the contraction of two Levi-Civita symbols givesince ${\epsilon }_{ijk}{\epsilon }_{ik\ell }=-2{\delta }_{j\ell }$. This is exactly $D_i{F}^{ij}+e^2[\Phi ,D^j\Phi ]=0$. Thus the static first-order Bogomolny system satisfies all four gauge equations and the scalar equation, with no extra ansatz or boundary assumption needed for this local implication. This is the mechanism by which Bogomolny monopole equations imply Yang-Mills-Higgs equations.

Use anti-Hermitian $\mathfrak{su}(2)$ matrices, with invariant positive inner product $⟨X,Y⟩=-2\mathrm{tr}(XY)$. On an adjoint-valued field $\Xi$, define the adjoint covariant derivative and gauge curvature byEquivalently, $F=dA+A\wedge A$ and $[D_i,D_j]\Xi =[F_{ij},\Xi ]$. Put $B_i=\frac{1}{2}{ϵ}_{ijk}{F}_{jk}$, with ${ϵ}_{123}=1$. The monopole Bogomolny equations are $B_i=D_i\Phi$.

For a smooth SU(2) gauge transformation $g(x)$, useDifferentiating a transformed adjoint field shows $D_i^g(g\Xi g^{-1})=g(D_i\Xi )g^{-1}$: the derivatives of $g$ cancel the inhomogeneous term in $A_i^g$. Applying the commutator of these derivatives gives $F_{ij}^g=g{F}_{ij}{g}^{-1}$, and therefore $B_i^g=g{B}_i{g}^{-1}$. Both sides of $B_i=D_i\Phi$ transform identically, establishing gauge invariance of the solution condition. Individual matrix components are covariant, rather than invariant numbers.

The monopole boundary condition is a nonzero asymptotic Higgs norm,together with finite three-dimensional Yang-Mills-Higgs energy. Standard monopole asymptotics have $D\Phi =O(r^{-2})$ and $F=O(r^{-2})$. The normalized Higgs field at infinity defines a map $S_{\infty }^2\to SU(2)/U(1)\simeq S^2$, whose topological degree is the magnetic charge. A nonzero charge gives a genuine non-Abelian magnetic monopole; the nonzero Higgs boundary value alone also allows the zero-charge vacuum. Thus the relevant nontrivial solutions are Bogomolny-Prasad-Sommerfield monopoles in the zero-Higgs-potential limit.

Now form the four-dimensional connection with $\tau$-independent components. Its gauge curvature hasThe self-duality sign requires an orientation. Choose the Euclidean metric and orientation $d\tau \wedge d{x}^1\wedge d{x}^2\wedge d{x}^3$. For cyclic $(i,j,k)$, the Hodge star operator satisfiesUsing $F_{jk}={ϵ}_{jki}{B}_i$ and $d{x}^i\wedge d\tau =-d\tau \wedge d{x}^i$, the curvature isEach term has negative star eigenvalue, soThis proves the Anti-self-dual Yang-Mills equations in the stated convention. The orientation of a monopole lift matters: with $d{x}^1\wedge d{x}^2\wedge d{x}^3\wedge d\tau$ instead, the same connection is self-dual. With that opposite orientation, the anti-self-dual lift would use $-\Phi d\tau$.

To fix the action normalization, take the Euclidean Yang-Mills actionBoth sums are over ordered index pairs. The component identity
${\sum }_{j,k}|F_{jk}{|}^2=2{\sum }_i|B_i{|}^2$ givesHence the four-dimensional action density in terms of the three-dimensional curvature isA different overall action normalization rescales all three equal expressions together.

A Yang-Mills instanton is a smooth finite-Euclidean-action self-dual or anti-self-dual connection in pure Yang-Mills theory, with the usual asymptotic pure-gauge behavior. Physically the term generally refers to a nontrivial field; its asymptotic gauge winding labels an instanton number. The Bianchi identity $D\mathcal{F}=0$, together with $\ast \mathcal{F}=-\mathcal{F}$, implies $D\ast \mathcal{F}=0$, so the first-order equations already imply the second-order Yang-Mills equations. Finite action is a separate requirement.

Here ${\mathcal{L}}_4$ is independent of $\tau$. If the monopole has nonzero curvature, the density is positive on an open set in ${\mathbb{R}}^3$, and integrating its infinite $\tau$-extent givesFor a finite-energy monopole this can also be written as its positive action per unit length times ${\int }_{\mathbb{R}}d\tau$. Thus a nontrivial monopole lift is not a Yang-Mills instanton on ${\mathbb{R}}^4$. The flat solution $A_i=0$, $\Phi ={\Phi }_0$ constant is the zero-action exception: it is a vacuum, or a trivial charge-zero instanton if that terminology includes flat connections. It is not a counterexample to the nontrivial-monopole assertion.

More generally, finite Yang-Mills action forbids nontrivial translation symmetry, even when the translation preserves a connection only up to gauge. Such a symmetry would preserve its positive gauge-invariant action density. A nonflat field has a ball on which this density is bounded below by some positive constant. Repeated translations by a nonzero symmetry vector produce infinitely many disjoint copies of a sufficiently small such ball, all with the same positive action contribution, contradicting finiteness. Continuous translation invariance is a special case. Therefore a nonflat ASDYM instanton is localized in all four Euclidean directions and has no nonzero translational symmetry. Translating an instanton still produces another instanton with a shifted center; these position parameters belong to the instanton moduli space, rather than being invariances of one localized solution.

Choose the convention ${\psi }^{\prime }=g\psi$ for a field in the defining representation and $D_{\mu }={\partial }_{\mu }+A_{\mu }$. Requiring $D_{\mu }^{\prime }{\psi }^{\prime }=g{D}_{\mu }\psi$ gives the Yang-Mills gauge transformationThe gauge field strength is the curvature $[D_{\mu },D_{\nu }]=F_{\mu \nu }$, soFor an adjoint field $Y$, the adjoint covariant derivative is $D_{\mu }Y={\partial }_{\mu }Y+[A_{\mu },Y]$, and direct substitution gives $D_{\mu }^{\prime }{Y}^{\prime }=g(D_{\mu }Y)g^{-1}$. Consequently $D_{\mu }{F}_{\nu \rho }$ transforms by conjugation too.

The Killing-form invariance under a Lie-group adjoint action gives $\kappa (gY{g}^{-1},gZ{g}^{-1})=\kappa (Y,Z)$. Its proof is that the adjoint matrices are conjugated by the same invertible map, leaving their trace product unchanged. With $s=\kappa (F_{\mu \nu },F^{\mu \nu })$, both $s^p$ and $\kappa (D_{\mu }{F}_{\nu \rho },D^{\mu }{F}^{\nu \rho })$ are therefore gauge invariant.

For their field equations, take a compactly supported variation $a_{\mu }=\delta A_{\mu }$. Curvature varies byThe invariant form permits gauge-covariant integration by parts, because ${\partial }_{\mu }\kappa (Y,Z)=\kappa (D_{\mu }Y,Z)+\kappa (Y,D_{\mu }Z)$. Antisymmetry of $F$ givesIntegrating, the power of the Yang-Mills invariant gives the equationFor $p=1$ this recovers the Yang-Mills equations. For larger $p$, keep the displayed product form, including at $s=0$.

For the derivative-squared action, write $B_{\mu \nu \rho }=D_{\mu }{F}_{\nu \rho }$. Varying the derivative as well as the curvature givesIts first variation isIntegrate the first term once, substitute $\delta F_{\nu \rho }=D_{\nu }{a}_{\rho }-D_{\rho }{a}_{\nu }$, and integrate again. It becomes $4\int \kappa (a_{\nu },D_{\mu }{D}^2{F}^{\mu \nu })d^{4}x$, where $D^2=D_{\lambda }{D}^{\lambda }$. Invariance converts the second term into $2\int \kappa (a_{\nu },[F_{\rho \sigma },D^{\nu }{F}^{\rho \sigma }])d^{4}x$. Nondegeneracy of the Killing form and arbitrary variations yield the derivative-squared Yang-Mills action equationThe derivative order matters because covariant derivatives generally do not commute. In an abelian limit the commutator term vanishes and the equation reduces to $\square {\partial }_{\mu }{F}^{\mu \nu }=0$, a useful check on the fourth-order equation for the potential.

For the adjoint covariant derivative $D_i={\partial }_i+[A_i,\cdot ]$, the gauge curvature isThe matrix-valued wedge product gives one commutator in each $d{x}^i\wedge d{x}^j$ coefficient; $[D_i,D_j]\Phi =[F_{ij},\Phi ]$ follows directly. The given trace defines a real positive inner product on the anti-Hermitian algebra $\mathfrak{su}(2)$.

The additional terms in $⟨D_j\Phi ,\Psi ⟩+⟨\Phi ,D_j\Psi ⟩$, compared with ordinary differentiation, areThus trace cyclicity proves the compatibility identityWith compactly supported variations, this gives covariant integration by parts without boundary contributions.

Use $F_{ij}={ϵ}_{ijk}{B}_k$, so $|B{|}^2=\frac{1}{2}{\sum }_{i,j}|F_{ij}{|}^2$. Write the independent variations as $\delta A_i=a_i$ and $\delta \Phi =\chi$. The curvature and Higgs derivative variations areThe variation of the Yang-Mills theory magnetic energy isFor the Higgs field derivative energy, trace cyclicity also gives $⟨X,[a,\Phi ]⟩=⟨[\Phi ,X],a⟩$. HenceArbitrariness of the variations proves the Euler-Lagrange field equationsThe gauge equation can equivalently be written $(D\times B{)}_j+[\Phi ,D_j\Phi ]=0$, where $(D\times B{)}_j={ϵ}_{jik}{D}_i{B}_k$. This equivalent form also checks the sign of the commutator source.