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Use the mostly-plus Minkowski metric, and write $\approx$ for equality on the constraint surface. Assume that the mechanical constraints are locally independent. They are first-class constraints whenThus their Poisson brackets vanish on the constraint surface, and their Hamiltonian flows preserve that surface. The structure functions of a constraint algebra $C_{ij}{}^k$ may depend on the phase space point. The finite real span of the constraints is a Lie algebra if it closes with constant structure coefficients, in a suitable choice of generators. The Jacobi identity then gives the usual conditions on the structure constants of a Lie algebra. With general structure functions the finite real span need not close, even though the Poisson bracket of all smooth functions is itself a Lie bracket.

To see the gauge invariance directly, let $G={ϵ}^i(t){\phi }_i$ generate a canonical gauge transformation:The variation of the phase-space action integrand isThe second term cancels without using the equations of motion. Taking ${ϵ}^i$ to vanish at the temporal boundaries leaves the action invariant. Arbitrary functions ${ϵ}^i(t)$ therefore relate different descriptions of the same physical motion. This reasoning also works with structure functions; constant structure coefficients are only needed for the finite-dimensional Lie algebra claim.

For a closed string, choose $0\le \sigma <2\pi$ and periodic fields. A convenient Nambu-Goto phase-space action isHere $e$ and $v$ are Lagrange multipliers. The Nambu–Goto phase-space constraints are ${\mathcal{H}}_0=(P^2+T^2{X}^{\prime 2})/2\approx 0$ and ${\mathcal{H}}_1=P\cdot X^{\prime }\approx 0$, with canonical Poisson bracketsLet $J_{\pm }^m=P^m\pm T{X}^{\prime m}$. Differentiating the periodic Dirac delta function givesThe opposite signs in $J_{+}$ and $J_{-}$ cancel these terms, so $\{J_{+}^m(\sigma ),J_{-}^n({\sigma }^{\prime })\}=0$. Replace the original constraints by the equivalent chiral densitiesTheir mixed Poisson brackets vanish. Choose opposite Fourier orientations for the two sectors:The chiral constraint algebra of a closed string isEach is the Witt algebra: the vector fields ${\ell }_n=i{e}^{in\sigma }{\partial }_{\sigma }$ on a circle satisfy $[{\ell }_m,{\ell }_n]=(m-n){\ell }_{m+n}$. Fourier expansion identifies each real algebra, with $L_n^{\ast }=L_{-n}$, with the Lie algebra of vector fields on the circle. The two commuting copies give ${\mathrm{Diff}}_1\oplus {\mathrm{Diff}}_1$, not a quantum central extension.

For an open string, allowed boundary conditions must remove the endpoint term in the variation of the action, consistently with the allowed endpoint variations. The spatial boundary term isIt expresses the open-string endpoint momentum flux. In the temporal gauge for a string $X^0=t$, take a boundary-adapted parametrization with $v=0$ at the ends. Fixing $X^i(t,a)=0$ gives $\delta X^i(t,a)=0$, a Dirichlet boundary condition. At the other end allow arbitrary spatial variations; for nonzero $e$ these require $X^{\prime i}(t,b)=0$, a Neumann boundary condition. Also $X^{\prime 0}=0$ in this temporal gauge for a string, so $X^{\prime m}(t,b)=0$. The constraint at this free-end string boundary condition reduces to $P^2=0$. Hamilton's equation $\dot{X}=eP+v{X}^{\prime }$ consequently gives ${\dot{X}}^2=0$ there. Since ${\dot{X}}^0=1$, the free endpoint has spatial speed one. This is the null motion of a free string endpoint.

A straight rotating string with one fixed endpoint supplies the required solution in at least two spatial dimensions. Set $e=1/T$, $v=0$, andTake $P_m=T{\dot{X}}_m$. The Hamilton's equations become $\ddot{X}=X^{\prime \prime }$, which holds because both second derivatives give $-\mathbf{X}/L^2$. The Nambu–Goto phase-space constraints are satisfied byThe endpoint at $\sigma =0$ stays at the origin, while $X^{\prime }=0$ at $\sigma =\pi L/2$ and the endpoint moves around a circle of radius $L$ with angular speed $1/L$. At each time the whole string lies on a straight radial segment. Its spatial proper length isThe velocity is everywhere perpendicular to the segment, so this also equals the sum of local rest-frame lengths. The induced worldsheet metric becomes degenerate at the null free endpoint, as expected for the limiting free-end solution.

Use gamma matrices satisfying the Clifford algebra relation $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2{\eta }^{\mu \nu }$, with ${\gamma }^{0†}={\gamma }^0$ and ${\gamma }^{i†}=-{\gamma }^i$. The Dirac adjoint is $\bar{\Psi }={\Psi }^{†}{\gamma }^0$. Take the Dirac actionIt differs from the manifestly Hermitian form with $(i/2)\bar{\Psi }{\gamma }^{\mu }{\overleftrightarrow{\partial }}_{\mu }\Psi$ only by a boundary term. Treat $\Psi$ and $\bar{\Psi }$ as independent variables when applying the principle of stationary action. Varying $\bar{\Psi }$ givesVarying $\Psi$ and integrating by parts gives the adjoint Dirac equation, $i({\partial }_{\mu }\bar{\Psi }){\gamma }^{\mu }+m\bar{\Psi }=0$. Multiplying the Dirac equation by $i{\gamma }^{\nu }{\partial }_{\nu }+m$ also gives $(\square +m^2)\Psi =0$, so its dispersion relation is Lorentz invariant.

For a Lorentz transformation $x^{\prime }=\Lambda x$, the field transforms in the Spinor representation of the Lorentz group:Since ${\partial }_{\mu }^{\prime }=({\Lambda }^{-1}{)}^{\nu }{}_{\mu }{\partial }_{\nu }$, this identity gives the Lorentz covariance of the Dirac operator:Thus every solution is carried to another solution. The field is a spinor rather than a four-vector; the transformation of the gamma matrices supplies the necessary covariance. The Dirac action is Lorentz invariant because its integrand is a scalar and $d^{4}x$ is invariant.

For electric charge $q$, use the gauge covariant derivative $D_{\mu }={\partial }_{\mu }+iq{A}_{\mu }$. The minimal electromagnetic coupling of a Dirac field isThe local gauge transformations in this convention areDirect substitution gives $D_{\mu }^{\prime }{\Psi }^{\prime }=e^{-iq\alpha }{D}_{\mu }\Psi$, which proves gauge covariance of the charged Dirac equation. The mass and kinetic terms are invariant, and $F_{\mu \nu }^{\prime }=F_{\mu \nu }$, so the full action has gauge invariance. The conserved current is $j^{\mu }=q\bar{\Psi }{\gamma }^{\mu }\Psi$. It transforms as a four-vector under Lorentz transformations, and varying $A_{\mu }$ gives ${\partial }_{\nu }{F}^{\nu \mu }=j^{\mu }$.

To make the infinitesimal spinor transformation explicit, write ${\Lambda }^{\mu }{}_{\nu }={\delta }^{\mu }{}_{\nu }+{\omega }^{\mu }{}_{\nu }$ with ${\omega }_{\mu \nu }=-{\omega }_{\nu \mu }$. DefineThe Lorentz generators from gamma-matrix commutators have precisely the needed algebra: $[{\sigma }^{\mu \nu },{\gamma }^{\rho }]=2i({\eta }^{\nu \rho }{\gamma }^{\mu }-{\eta }^{\mu \rho }{\gamma }^{\nu })$, which verifies $S^{-1}{\gamma }^{\rho }S={\gamma }^{\rho }+{\omega }^{\rho }{}_{\nu }{\gamma }^{\nu }$ to first order. The infinitesimal transformation of a Dirac field is thereforeAt the same coordinate argument, the orbital change must also be included:Both formulas describe the same transformation; their arguments differ.

Finally, the gamma adjoint identities imply ${\sigma }^{\mu \nu †}={\gamma }^0{\sigma }^{\mu \nu }{\gamma }^0$ and hence the pseudo-unitarity of the spinor Lorentz representation, $S^{†}{\gamma }^{0}S={\gamma }^0$. It follows thatThus the Dirac scalar bilinear $\bar{\Psi }\Psi$ is a Lorentz scalar. Infinitesimally the spin terms in $\delta (\bar{\Psi }\Psi )$ cancel; at fixed coordinates only the ordinary scalar orbital transformation remains.

Use the weak hypercharge normalization $Q=T_3+Y$. Write $U(x)\in SU(2{)}_L$ and the hypercharge transformation as $e^{i\beta (x)Y}$. A left-handed doublet transforms as ${\psi }_L\mapsto e^{i\beta Y_L}U{\psi }_L$, while a right-handed singlet transforms as ${\psi }_R\mapsto e^{i\beta Y_R}{\psi }_R$. The Higgs doublet transforms as $\varphi \mapsto e^{i\beta /2}U\varphi$.

The representations per fermion generation are:

There is no right-handed neutrino in the minimal Standard Model. If such a sterile singlet were added, its hypercharge would be zero. The two hypercharge columns give equivalent normalizations, not different physical assignments. This is the electroweak representation and hypercharge table.

Yukawa interactions use $\varphi$ for down-type quarks and charged leptons, but $\tilde{\varphi }=i{\sigma }^2{\varphi }^{\ast }$ for up-type quarks. The conjugate doublet transforms as $e^{-i\beta /2}U\tilde{\varphi }$, using the pseudoreality of the $SU(2)$ fundamental representation. For example, the hypercharge sums in the down, charged-lepton, and up terms areThe doublet indices contract between ${\bar{\psi }}_L$ and the scalar, and colour indices contract in the quark terms. This verifies gauge invariance.

The indices $i,j=1,2,3$ label fermion generations, specifying which left-handed family and right-handed family are coupled. They are not weak-isospin indices. Each fermion type has its own complex Yukawa matrix. After symmetry breaking these matrices determine masses; the mismatch of the two quark left-handed mass rotations produces the CKM matrix. Left fields are weak doublets, right fields are weak singlets, and the scalar is a hypercharge-$1/2$ doublet in the stated convention.

A Non-Abelian gauge theory promotes an internal group representation symmetry to a spacetime-dependent transformation and introduces a gauge field so that differentiation transforms covariantly. The symmetry is a gauge redundancy: physically meaningful quantities must be invariant under local changes of field representative. We work in four-dimensional Minkowski spacetime with signature $(+---)$, and use skew-Hermitian matrix generators throughout.

For the compact Simple Lie group $G$, its real Lie algebra has a negative-definite Killing form. Hence $B=-\kappa$ is a positive invariant bilinear form on a Lie algebra. Choose a $B$-orthonormal basis $T_a$, $a=1,\dots ,\dim G$, with $[T_a,T_b]=f_{ab}{}^c{T}_c$. Invariance makes $f_{abc}=B([T_a,T_b],T_c)$ fully antisymmetric. This construction uses the Adjoint representation and works independently of any chosen scalar representation.

Let ${\mathcal{A}}_{\mu }={\mathcal{A}}_{\mu }^a{T}_a$ be a Lie-algebra-valued gauge potential, with the coupling absorbed into this connection. If $R$ is the given finite-dimensional irreducible representation of $G$, its Lie algebra representation $\rho =dR$ definesFor a finite local transformation $U(x)\in G$, require ${\Phi }^{\prime }=R(U)\Phi$ and $D_{\mu }^{\prime }{\Phi }^{\prime }=R(U)D_{\mu }\Phi$. Differentiating the first equation forces the gauge-field transformation lawThus the gauge potential transforms inhomogeneously; an ordinary derivative alone would leave an uncancelled derivative of $U$.

The commutator of gauge covariant derivatives determines the gauge field strength:The Lie algebra representation preserves the bracket, so this formula holds for any $R$, including a trivial scalar representation. The gauge field strength itself is defined in the gauge algebra. The transformation of the connection impliesIt is therefore covariant, rather than invariant as a Lie-algebra-valued quantity.

For infinitesimal $U=1+ϵ+O({ϵ}^2)$, the same equations give the infinitesimal gauge transformation with an anti-Hermitian connection:Substitution into $\delta (D_{\mu }\Phi )$ cancels the terms containing ${\partial }_{\mu }ϵ$ and gives $\delta (D_{\mu }\Phi )=\rho (ϵ)D_{\mu }\Phi$, the gauge covariance of a scalar covariant derivative.

The Yang-Mills theory has Lagrangian densityThis is the Killing-form Yang-Mills Lagrangian. Its sign uses a positive internal form $B$ and the stated spacetime signature. In a $B$-orthonormal basis it is $-{\mathcal{F}}_{\mu \nu }^a{\mathcal{F}}^{a\mu \nu }/(4g_{\mathrm{YM}}^2)$. Gauge invariance follows because $B$ is invariant under the adjoint group action. Infinitesimally, the variation is proportional toAfter rescaling ${\mathcal{A}}_{\mu }=g_{\mathrm{YM}}{A}_{\mu }$, the gauge coupling appears in $D_{\mu }={\partial }_{\mu }+g_{\mathrm{YM}}\rho (A_{\mu })$ and in the non-Abelian part of the canonically normalized field strength. Squaring that field strength gives cubic and quartic gauge-field self-interactions. An ordinary local quadratic mass term in the gauge potential is not invariant because its transformation contains ${\partial }_{\mu }ϵ$.

For the scalar field kinetic term, compactness supplies a positive invariant Hermitian form $h$ on its representation space. One can obtain it by averaging any positive form with normalized Haar measure, the unitarization of a compact-group representation. Then $R(U)$ is unitary for $h$, and $\rho (ϵ)$ is anti-Hermitian. The Yang-Mills theory coupled to an arbitrary scalar representation hasHere $V$ is any real gauge-invariant scalar potential, so $V(R(U)\Phi )=V(\Phi )$. For example,is available for every unitary representation, with $\lambda \ge 0$ giving a bounded quartic contribution. Other invariant interactions may exist for particular representations; the potential is not required to depend only on $h(\Phi ,\Phi )$.

The scalar kinetic term is gauge-invariant because both covariant derivatives transform by the same unitary $R(U)$. Infinitesimally its variation vanishes byThe potential is invariant by its defining condition. For a real irreducible representation, use the averaged positive symmetric form instead and write the kinetic term as $\frac{1}{2}h(D_{\mu }\Phi ,D^{\mu }\Phi )$; all transformation and invariance arguments remain valid. Expansion of the covariant kinetic term produces both the linear gauge-scalar interaction and the quadratic term needed by local gauge invariance.

The invariant inner products, gauge covariant derivatives and gauge field strength together construct the full theory for every compact Simple Lie group and every supplied finite-dimensional irreducible representation of the scalar fields. Both finite and infinitesimal transformation laws have been included, and every sign follows the same anti-Hermitian convention.

Use Lie algebra generators with $[T_b,T_c]=f^a{}_{bc}{T}_a$, and define the adjoint gauge covariant derivative by $(D_{\mu }X{)}^a={\partial }_{\mu }{X}^a+f^a{}_{bc}{A}_{\mu }^b{X}^c$. Varying the gauge field strength gives $\delta F_{\mu \nu }=D_{\mu }\delta A_{\nu }-D_{\nu }\delta A_{\mu }$. Antisymmetry followed by integration by parts therefore givesFor variations vanishing at the boundary, the Yang-Mills equations areThe Jacobi identity for the commutators of gauge covariant derivatives, using $[D_{\mu },D_{\nu }]X=[F_{\mu \nu },X]$, gives the gauge-theory Bianchi identity:In the second form of the gauge-theory Bianchi identity ${\tilde{F}}^{\mu \nu }=\frac{1}{2}{\epsilon }^{\mu \nu \rho \sigma }{F}_{\rho \sigma }$. The identity is a geometric consequence of the definition of curvature, not a second dynamical field equation.

For the usual non-Abelian Yang-Mills theory, the quantum running coupling becomes strong at low energies. Dimensional transmutation generates a scale absent from the classically scale-invariant equations. The expected confining dynamics and massive colour-singlet spectrum involve confinement and a mass gap, rather than freely propagating weakly coupled coloured waves. These nonperturbative effects are not captured by simply solving the classical equations; this is not a claim of a mathematical proof of the Yang-Mills mass gap.

Write the gauge condition as ${\chi }^a[A]=f^a[A]$ to distinguish it from the structure constants. Under an infinitesimal gauge transformation, ${\delta }_{\omega }{A}_{\mu }=D_{\mu }\omega$. Define the Faddeev-Popov operator by ${\mathcal{M}}^{ab}=\delta {\chi }^a[A+D\omega ]/\delta {\omega }^b{|}_{\omega =0}$. In a Euclidean convention, the gauge-fixing and ghost additions can be chosen asA nonlocal kernel would require a double integral; a local functional $\chi$ instead makes $\mathcal{M}$ a local differential operator. Choosing a local gauge functional thus retains a local gauge-fixed action and the ordinary local interaction structure of perturbative quantum field theory.

The fields $c^a$ and ${\overline{c}}^a$ are independent Grassmann-valued fields in the Adjoint representation, Lorentz scalars with ghost numbers $+1$ and $-1$. They are the Faddeev-Popov ghost fields: their functional integral produces $\det \mathcal{M}$, the Faddeev-Popov determinant correcting the volume element along a gauge orbit. They are internal fields, not physical asymptotic particles. Their statistics supply a minus sign for every closed ghost loop. No auxiliary field is needed in the displayed gauge-fixing representation.

In axial gauge, ${\chi }^a=n^{\mu }{A}_{\mu }^a$, soWith the Fourier transform convention $e^{ik\cdot x}$, the ghost Feynman rules in the original normalization of $A$ areMomentum conservation accompanies the vertex; each ghost loop has the extra minus sign. With a canonically normalized field $A=g{A}_{\mathrm{can}}$, the vertex is instead $-g{f}^a{}_{bc}{n}_{\mu }$. The gauge fixing term is quadratic and introduces no further interaction vertex.

The Euclidean quadratic kernel of $A$ is $g^{-2}[k^2{\delta }_{\mu \nu }-k_{\mu }{k}_{\nu }+{\xi }^{-1}{n}_{\mu }{n}_{\nu }]$. For $n\cdot k\ne 0$, its inverse isThis follows by multiplication with the kernel: the $k^2\delta -kk$ piece gives $I-n{k}^T/(n\cdot k)$, and the ${\xi }^{-1}nn$ piece supplies the missing $n{k}^T/(n\cdot k)$. The strict axial-gauge propagator is thereforeIt obeys $n^{\mu }{D}_{\mu \nu }=0$. Every ghost attachment to an internal gauge propagator therefore vanishes in the strict limit; equivalently, $\mathcal{M}=n\cdot \partial$ on the gauge slice. Thus the interacting Faddeev-Popov ghosts decouple. Residual gauge transformations and the poles at $n\cdot k=0$ require compatible boundary conditions and an axial-gauge pole prescription.

After Wick rotation, the corresponding Minkowski propagator has ${\delta }_{\mu \nu }$ replaced by ${\eta }_{\mu \nu }$ and the usual overall factor $-i$, with a compatible causal prescription. The fixed $n$ makes the gauge-fixed propagator nonmanifestly Lorentz covariant. Nevertheless, gauge invariance makes physical observables independent of this gauge-choice vector, so their perturbative predictions are Lorentz invariant. Already at tree level, contraction with conserved external currents removes all terms containing $k_{\mu }$ or $k_{\nu }$, leaving $g^{2}j\cdot j^{\prime }/k^2$. The quantum Ward identities give the analogous cancellation in complete physical amplitudes; arbitrary gauge-dependent Green functions need not be independent of $n$.

Use the real Lie-algebra convention of the question, with a gauge covariant derivative $D_{\mu }={\partial }_{\mu }+A_{\mu }$. The gauge field strength is its curvature:Write $a_{\mu }={\delta }_X{A}_{\mu }/ϵ=-{\partial }_{\mu }X+[X,A_{\mu }]$. To first order in $ϵ$,Substitute $a_{\mu }$. The mixed second derivatives of $X$ cancel. The terms containing first derivatives of $X$ cancel in pairs. The remaining terms areThe Jacobi identity combines the last two into $[X,[A_{\mu },A_{\nu }]]$. ConsequentlyThe transformation is homogeneous even though the connection transformation contains an inhomogeneous derivative term.

For a finite-dimensional Lie algebra, define its Adjoint representation by ${\mathrm{ad}}_X(Y)=[X,Y]$ and its Killing form byIt is a symmetric bilinear form by cyclicity of the trace. The Jacobi identity gives ${\mathrm{ad}}_{[Z,X]}=[{\mathrm{ad}}_Z,{\mathrm{ad}}_X]$. Set $A={\mathrm{ad}}_Z$, $B={\mathrm{ad}}_X$, $C={\mathrm{ad}}_Y$. ThenThis proves the invariant bilinear form on a Lie algebra property, without assuming simplicity or nondegeneracy.

For definiteness use the Minkowski metric $\eta =\mathrm{diag}(1,-1,-1,-1)$ and take a real compact semisimple Lie algebra as the gauge algebra. Its positive internal metric is $B=-\kappa$. A Killing-form Yang-Mills Lagrangian with the coupling absorbed into the connection isThe spacetime metric is unchanged by internal gauge transformations. HenceThus gauge invariance follows directly from invariance of the Killing form. Other overall conventions are possible, but the energy sign must be checked rather than inferred from a prefactor in isolation.

For physical kinetic terms the internal form must be real, nondegenerate and positive definite after choosing the overall sign. If ${\mathcal{E}}_i=F_{0i}$ and ${\mathcal{B}}_i=\frac{1}{2}{ϵ}_{ijk}{F}_{jk}$, the above convention has Lagrangian $[B({\mathcal{E}}_i,{\mathcal{E}}_i)-B({\mathcal{B}}_i,{\mathcal{B}}_i)]/(2g_{\mathrm{YM}}^2)$ and physical energy densityThe canonical Hamiltonian density also contains the nondynamical multiplier and a spatial divergence. With ${\Pi }_i={\mathcal{E}}_i/g_{\mathrm{YM}}^2$ and $D_i{\Pi }_i={\partial }_i{\Pi }_i+[A_i,{\Pi }_i]$, integration by parts using the invariant bilinear form on a Lie algebra givesThe Gauss law constraint in gauge theory sets $D_i{\Pi }_i=0$. If the boundary flux vanishes, or the appropriate boundary contribution is included, the integrated physical Hamiltonian is the positive energy displayed above.

An indefinite internal form would give gauge-field polarizations with opposite kinetic signs. A degenerate form would fail to supply a kinetic term for some directions. The compactness criterion from the Killing form says that negative-definiteness of the Killing form of a real finite-dimensional algebra is equivalent to compact semisimplicity; thus the pure Killing-form construction selects compact semisimple real forms. One cannot use a complex-bilinear Killing form on arbitrary complex field components as if it were a positive Hermitian metric.

This does not prohibit Abelian gauge theories. A compact Abelian factor has zero Killing form, so it needs a separately chosen positive invariant bilinear form on a Lie algebra, rather than the Killing form. More generally an algebra with a positive invariant metric on a Lie algebra is compact reductive, namely a direct sum of a compact semisimple algebra and an Abelian center. A noncompact group can also share the same compact Lie algebra through global covering choices in an Abelian factor; positivity is a statement about the algebra and internal metric, not by itself a classification of global gauge-group topology. Quantum matter anomalies and global restrictions would require additional input; no matter content is specified here.

For the finite matrix transformation, let $g(x)\in SU(N)$ and regard $g$ as a multiplication operator. The identity ${\partial }_{\mu }{g}^{-1}=-g^{-1}({\partial }_{\mu }g)g^{-1}$ givesTaking commutators of these differential operators cancels the adjacent multiplication operators $g^{-1}g$, yieldingThis argument keeps the derivatives acting on test fields and avoids treating $D_{\mu }$ as just a matrix.

There is a normalization issue in the printed last paragraph. The standard Killing form defined through the adjoint trace on $\mathfrak{su}(N)$ isnot simply $\mathrm{tr}(XY)$. For example, with $N=2$ and $X=Y=-i{\sigma }_3/2$, the defining trace is $-1/2$, whereas the adjoint trace is $-2$. The printed trace formula can be used as a rescaled invariant form, with the constant absorbed into the gauge coupling; it has exactly the invariance needed here. For either normalization,by cyclicity. This proves finite Yang-Mills gauge transformation invariance, with the healthy sign chosen for anti-Hermitian gauge fields. The normalization discrepancy is not a failure of gauge invariance.

Finally let $g(x)=e^{ϵX(x)}=I+ϵX(x)+O({ϵ}^2)$. Since $X\in \mathfrak{su}(N)$ is traceless and skew-Hermitian, this exponential lies in $SU(N)$. Expanding the finite formula givesThus the stated infinitesimal transformation is the derivative of the finite transformation at the identity. It describes transformations in the identity component; arbitrary global or large gauge transformations need not be generated by one globally defined infinitesimal parameter.

Use anti-Hermitian representation matrices for the unitary representation, so $R(X{)}^{†}=-R(X)$. Absorb the gauge coupling into the connection, consistently with the transformation law in the PDF. The appropriate gauge covariant derivative isThis has a plus sign because the inhomogeneous part of the connection transformation is $-ϵ{\partial }_{\mu }X$. Its variation isThus gauge covariance of a scalar covariant derivative givesThe cancellation uses that $R$ is a Lie algebra representation, including its commutator identity.

The gauge field strength isIt is the curvature appearing in $[D_{\mu },D_{\nu }]\varphi =R(F_{\mu \nu })\varphi$. The index positions in the PDF are related by the constant Minkowski metric; lowering the index in its transformation law gives the convention used here.

Take signature $(+,-,-,-)$, let $B$ be a real symmetric gauge-invariant internal form, and let $⟨\cdot ,\cdot {⟩}_V$ be the positive Hermitian form preserved by the unitary matter representation. One consistent Lorentz transformation scalar Lagrangian density isHere $g_{\mathrm{YM}}>0$, $m^2\ge 0$, and $\lambda \ge 0$ give a simple stable potential. The coupling is in the kinetic normalization because it was absorbed into $A_{\mu }$; it is not inserted again into $D_{\mu }$. Taking a zero or another gauge-invariant potential is also possible. A positive nondegenerate $B$ gives the usual physical gauge kinetic energy, as for a compact gauge algebra. More generally the invariance calculation below applies to any symmetric invariant bilinear form on a Lie algebra, including the Killing form.

Under the infinitesimal gauge-field transformation law,Invariance and symmetry of $B$ give the cancellation. For matter, both $\varphi$ and $D_{\mu }\varphi$ transform by $R(X)$, so anti-Hermiticity givesfor either pairing used in the Lagrangian. Its potential is a function of the invariant scalar norm. Therefore $\boxed{{\delta }_X\mathcal{L}=0}$, establishing gauge invariance of this Yang-Mills theory coupled to a scalar. Contraction of spacetime indices with the Minkowski metric gives Lorentz invariance.

For the Adjoint representation, define the Lie algebra structure constants byInternal superscripts here are basis labels, as in the paper; spacetime indices alone are raised with the Minkowski metric. The adjoint covariant derivative and curvature have componentsThese formulas involve no extra factor of $i$, because the generators use the Lie-algebra convention rather than Hermitian physics generators.