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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.

Work over $\mathbb{R}$ or $\mathbb{C}$, in finite dimension and characteristic zero. The Killing form of a Lie algebra is the trace form of its Adjoint representation:It is bilinear and symmetric by the cyclic property of the trace. It is an invariant bilinear form on a Lie algebra, since the Jacobi identity gives ${\mathrm{ad}}_{[x,y]}=[{\mathrm{ad}}_x,{\mathrm{ad}}_y]$ and henceEquivalently, $\kappa ([z,x],y)+\kappa (x,[z,y])=0$. A Lie algebra automorphism preserves the form, because it conjugates the adjoint matrices. Its radical of a bilinear form ${\mathfrak{r}}_{\kappa }=\{x:\kappa (x,\mathfrak{g})=0\}$ is an ideal of a Lie algebra by invariance. The center of a Lie algebra lies in this radical, so the form vanishes for an abelian algebra. On a direct sum of ideals the summands are orthogonal and the form restricts to their own Killing forms.

A short argument proves the requested implication without assuming the Cartan criterion for semisimplicity. Let $\mathfrak{a}$ be an abelian ideal of a Lie algebra, $x\in \mathfrak{a}$, and $y\in \mathfrak{g}$. The map ${\mathrm{ad}}_x$ takes $\mathfrak{g}$ into $\mathfrak{a}$ and vanishes on $\mathfrak{a}$, while ${\mathrm{ad}}_y$ preserves $\mathfrak{a}$. Therefore ${\mathrm{ad}}_x{\mathrm{ad}}_y$ has image in $\mathfrak{a}$ and is zero on $\mathfrak{a}$. In a basis extending a basis of $\mathfrak{a}$, both diagonal blocks are zero, so its trace vanishes. This proves that abelian ideals lie in the radical of the Killing form.

If the solvable radical $\mathfrak{r}$ were nonzero, its derived series of a Lie algebra would have a last nonzero term $\mathfrak{a}$. The Jacobi identity makes every derived term an ideal of $\mathfrak{g}$, and the last one is abelian. Thus $0\ne \mathfrak{a}\subseteq {\mathfrak{r}}_{\kappa }$, contradicting nondegeneracy. We obtainThe converse holds as well in characteristic zero; together these implications are the Cartan criterion for semisimplicity.

For a complex simple Lie algebra, that criterion gives nondegeneracy on $\mathfrak{g}$. Let $\mathfrak{h}$ be a Cartan subalgebra. Use the standard root-space decompositionFor $h,h^{\prime }\in \mathfrak{h}$ and a root vector $e_{\alpha }$, choose $t\in \mathfrak{h}$ with $\alpha (t)\ne 0$. Invariance givesThus $\mathfrak{h}$ is orthogonal to every nonzero root space. If $h\in \mathfrak{h}$ is also orthogonal to $\mathfrak{h}$, it is orthogonal to all of $\mathfrak{g}$ and hence is zero. This establishes nondegeneracy of the Killing form on a Cartan subalgebra:The same invariance calculation shows that $\kappa ({\mathfrak{g}}_{\alpha },{\mathfrak{g}}_{\beta })=0$ unless $\alpha +\beta =0$. Opposite root spaces are therefore paired nondegenerately. Tracing the adjoint action on the root-space decomposition yieldssince the root spaces of a complex semisimple algebra are one-dimensional and its adjoint action on $\mathfrak{h}$ is zero.

The relevant Euclidean subspace of a Cartan subalgebra iswhere the $h_{{\alpha }_i}$ are the simple coroots, normalized by ${\alpha }_i(h_{{\alpha }_i})=2$. The standard Euclidean property is that $\kappa$ is real and positive definite on this space. The displayed trace formula explains it: $\kappa (h,h)$ is a sum of real squares, and the roots span ${\mathfrak{h}}^{\ast }$, so all squares vanish only for $h=0$. Via the nondegenerate form, the roots can consequently be regarded as vectors in a real Euclidean normed vector space. The induced dual inner product makes root reflections orthogonal and allows root lengths and angles to be encoded by Cartan integers and the Dynkin diagram.

This positivity is on a specified real subspace; the complex Killing form is a bilinear form, not a positive Hermitian inner product. On a compact real form the Killing form is instead negative definite. For example the Killing form of the special linear Lie algebra gives $\kappa (X,Y)=4\mathrm{tr}(XY)$ for ${\mathfrak{sl}}_2$, so $\kappa (L_0,L_0)=8$ but $\kappa (i{L}_0,i{L}_0)=-8$.

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.

To make the Killing-form sign explicit, write $B=s\kappa$ and $⟨\varphi ,\psi {⟩}_V=s{\kappa }^{ab}({\varphi }^a{)}^{\ast }{\psi }^b$. The literal contraction requested in the paper corresponds to $s=1$. For a compact real form of a complex semisimple gauge algebra, its actual positive inner product has $s=-1$, since the Killing form is negative definite. With a complex adjoint scalar, the fully expanded Killing-form Lagrangian for an adjoint scalar isAll repeated internal labels are summed; $s^2=1$ explains the absence of $s$ from the quartic term. Invariance is equivalently the component identity $f^{ca}{}_d{\kappa }^{db}+f^{cb}{}_d{\kappa }^{ad}=0$. For a real adjoint scalar, omit complex conjugation and insert $1/2$ in the scalar kinetic and quadratic mass terms; the same invariance proof applies.

The PDF's identification of the Killing form with an inner product thus requires a sign convention. It also requires nondegeneracy if it is to give ordinary kinetic terms: on an Abelian gauge factor the Killing form vanishes, so an additional invariant metric is needed there. These qualifications do not alter the covariance identities or the formal gauge invariance of the displayed Killing-form contractions.

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

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

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

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

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

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

A bilinear form $B$ on a Lie algebra is invariant whenequivalently $B([x,y],z)+B(y,[x,z])=0$.

Choose a basis $(x_i)$ of $\mathfrak{g}$ and its $B$-dual basis $(x^i)$. The Casimir element iswhere $U(\mathfrak{g})$ is the universal enveloping algebra. The tensor ${\sum }_i{x}_i\otimes x^i$ corresponds under $B:\mathfrak{g}\cong {\mathfrak{g}}^{\ast }$ to the identity endomorphism, so invariance of $B$ makes it fixed by the diagonal adjoint action. Applying multiplication $U(\mathfrak{g})\otimes U(\mathfrak{g})\to U(\mathfrak{g})$ givesfor every $x\in \mathfrak{g}$. Thus $\Omega$ lies in the center of an associative algebra of $U(\mathfrak{g})$.

A positive invariant metric is a positive-definite real symmetric invariant bilinear form on a Lie algebra. Each adjoint map is skew-adjoint. Invariance gives $[\mathfrak{g},\mathfrak{g}{]}^{\perp }=\mathfrak{z}(\mathfrak{g})$, so the algebra splits orthogonally into its Abelian center and its derived algebra. The latter has zero center, and $\kappa (X,X)=\mathrm{tr}({\mathrm{ad}}_X^2)=-\Vert {\mathrm{ad}}_X{\Vert }^2$ is strictly negative for nonzero $X$. Thus it is compact semisimple; the whole algebra is compact reductive. Conversely, a compact semisimple summand has metric $-\kappa$ and the center can carry any positive metric. The Killing form alone vanishes on the center.

The usual uniqueness theorem assumes a complex simple Lie algebra, or a real form with simple complexification. For a complex simple algebra regarded as real, both $\mathrm{Re}{\kappa }_{\mathbb{C}}$ and $\mathrm{Im}{\kappa }_{\mathbb{C}}$ define a real symmetric invariant bilinear form on a Lie algebra, and they are linearly independent. The real Killing form is $2\mathrm{Re}{\kappa }_{\mathbb{C}}$. For ${\mathfrak{sl}}_2(\mathbb{C})$ and $H=\mathrm{diag}(1,-1)$, ${\kappa }_{\mathbb{C}}(H,H)=8$ and ${\kappa }_{\mathbb{C}}(H,iH)=8i$, proving the independence. This real algebra is simple: its real complexification is two complex simple ideals exchanged by conjugation, and a real ideal would complexify to a conjugation-stable sum of these ideals, hence either zero or the whole algebra.

For a finite-dimensional complex representation of a solvable Lie algebra, the Lie theorem gives a basis in which all representing matrices are upper triangular. Commutators are strictly upper triangular, and their products with upper triangular matrices have zero trace. Thus the derived algebra lies in the radical of a bilinear form. In particular, a nonabelian solvable Lie algebra can carry an invariant bilinear form on a Lie algebra that is nondegenerate without that form coming from a finite-dimensional representation trace.

// Target: semisimple-lie-algebra.bigb