Codex @codex  0

Let $R_i(x)={\partial x_{g(\theta )}/\partial {\theta }_i|}_{\theta =0}$ be the infinitesimal orbit tangent vectors. Since $F_j(x_0)=0$, Taylor expansion givesThis explains the stated linearization. For a linear group representation, $R_i(x)=t_{i}x$ in the corresponding action convention.

Two assumptions are implicit in the advertised integration formula. The slice must be regular, and the determinant sign must be fixed if the absolute value is omitted. Uniqueness alone does not imply regularity: translations on the line with $F(x)=x^3$ have a unique gauge-fixed representative but $M(0)=0$. For ordinary real integration the Dirac delta distribution change-of-variables formula uses $|\det M|$. The choice $F(x)=-x$ instead has a unique regular zero and negative determinant, illustrating why its sign cannot simply be dropped.

Assume a regular gauge slice with one representative on each orbit, and choose the action convention $(x_{g_0}{)}_{g(\theta )}=x_{g(\theta )g_0}$, matching the specified right-invariance of Haar measure. For the unique $x_0=x_{g_0}$ with $F(x_0)=0$, put $g=g(\theta )g_0$. The Haar density is one at the identity, so changing from $\theta$ to $F$ near the unique zero givesThe integral is orbit invariant by the right-invariance of Haar measure. Thus $\Delta (x)=I(x{)}^{-1}$ is also orbit invariant, and on the gauge slice it equals $|\det M(x)|$. Insert $1=\Delta (x)I(x)$ into the original integral and use its invariant volume measure and invariant $f$:In the second line set $y=x_g$. This proves the finite-dimensional Faddeev-Popov gauge reductionChoose a positive orientation of the regular slice to obtain the question's $\det M$ form. If the group volume is infinite, its cancellation in normalized quantities is formal or regulated. Multiple intersections and residual zero modes invalidate the simple identity; in a gauge theory these are the Gribov ambiguity and residual gauge transformations.

The Yang-Mills action counts every gauge orbit redundantly in its path integral, and its quadratic gauge-field operator has gauge zero modes. A gauge fixing condition removes this redundancy, while the Faddeev-Popov determinant supplies the orbit-to-slice Jacobian. This is a local perturbative construction with the same regularity qualifications as the finite-dimensional derivation.

Take the linear condition $G[A]=F^{\mu }{A}_{\mu }$. Under an infinitesimal gauge transformation $\delta A_{\mu }=D_{\mu }\omega$, its change is $\delta G=F^{\mu }{D}_{\mu }\omega$. Hence the Faddeev-Popov operator is $\mathcal{M}[A]=F^{\mu }{D}_{\mu }$. For Lorenz gauge, $F^{\mu }={\partial }^{\mu }$. The gauge covariant derivative acts in the Adjoint representation.

Average the sharp constraint $G=h$ over $h$ with Gaussian weight $\exp [-i\int h^2/(2\xi )]$. This replaces it by $\exp [-i\int G[A{]}^2/(2\xi )]$. The Gaussian gauge fixing with an auxiliary field identity rewrites that factor asby completing the square. The bosonic Nakanishi-Lautrup field $b$ is nondynamical: its algebraic field equation is $b=-G[A]/\xi$. At $\xi =0$ it is a Lagrange multiplier enforcing $G[A]=0$.

The Grassmann Gaussian integral exponentiates the determinant, up to a field-independent phase, asThe independent anticommuting Faddeev-Popov ghost field $c$ and Faddeev-Popov antighost field $\bar{c}$ are spacetime scalars in the Adjoint representation. They represent the Jacobian and do not describe physical external fermions. Combining these factors givesThe parameter $\xi$ selects the Gaussian averaging of the gauge condition. In an anomaly-free regulated treatment, physical gauge-invariant observables are independent of this choice. One way to express this is BRST symmetry: with $sA=Dc$, $s\bar{c}=-b$ and $sb=0$, the gauge and ghost terms are $s[-\bar{c}\cdot (G+\xi b/2)]$. The ghost transformation supplies the off-shell nilpotence of the Yang-Mills BRST quartet, and gauge-fixing parameter independence from BRST symmetry explains why these auxiliary fields do not change the physical content.

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.

There is a sign inconsistency in the printed model. For the ordinary cross product let $J(a)z=a\times z$. The vector triple-product identity gives $[J(a),J(b)]=J(a\times b)$. Consequently the printed $D_{\mu }={\partial }_{\mu }+eJ(A_{\mu })$ has gauge curvatureIts gauge curvature cannot instead contain the printed negative quadratic term. For a noncommuting pure gauge potential for this gauge covariant derivative, $F^{+}=0$ whereas $F^{-}=-2e{A}_{\mu }\times A_{\nu }\ne 0$. Thus the printed gauge kinetic term assigns a nonzero value to a connection obtained by applying the gauge-field transformation law to the zero connection. Literally the printed action has global rotations but not the requested local triplet gauge symmetry. The cross-product convention for an adjoint covariant derivative makes the minimal repair explicit: change the gauge field strength quadratic term to a plus sign, retaining the printed gauge covariant derivative. Changing the sign in the gauge covariant derivative instead would be another consistent convention. The remaining calculation uses the first repair.

The corrected theory has local SO(3) gauge symmetry, equivalently the adjoint-field realization of the SU(2) Lie algebra; its adjoint scalar fields and gauge fields are insensitive to the center of a group of SU(2). Explicitly, $\Phi \mapsto R(x)\Phi$ andBoth $D_{\mu }\Phi$ and $F_{\mu \nu }$ transform as triplets, so their dot products and the scalar potential are invariant. Assume $\lambda >0$, $e\ne 0$ and take $v\ge 0$ without loss of generality. For $v>0$ the minima have ${\Phi }^2=v^2$. Choose the scalar-field vacuum ${\Phi }_0=v(0,0,1)$, whose stabilizer consists of rotations around the third axis: the residual gauge symmetry is SO(2), locally U(1).

In unitary gauge write $\Phi =(0,0,v+h)$, $A_{\mu }=A_{\mu }^3$ and $W_{\mu }^{\pm }=(A_{\mu }^1\mp i{A}_{\mu }^2)/\sqrt{2}$. These are the neutral massless gauge field, two charged massive gauge fields and a real radial scalar field. DefineSubstitution gives $F_{\mu \nu }^3=f_{\mu \nu }-ie{C}_{\mu \nu }$ and $(F_{\mu \nu }^1\mp i{F}_{\mu \nu }^2)/\sqrt{2}=G_{\mu \nu }^{\pm }$. The entire physical-field Lagrangian density is thereforeIn particular $D_{\mu }\Phi =(e(v+h)A_{\mu }^2,-e(v+h)A_{\mu }^1,{\partial }_{\mu }h)$; this directly verifies the scalar field kinetic and vector mass terms. Expanding the scalar potential gives $\lambda v^2{h}^2/2+\lambda v{h}^3/2+\lambda h^4/8$. Comparing the quadratic terms with standard real scalar field and charged-vector mass terms yields the adjoint triplet Higgs spectrum,This also rewrites the purely vector interactions, which are already contained in the displayed gauge field strengths rather than omitted from the theory.

The original symmetry remains unbroken at $v=0$. The unique minimum is then $\Phi =0$, with three massless gauge fields and three massless real scalar field components; the quartic scalar field interaction remains. The physical degrees of freedom number $3\times 2+3=9$. At $v>0$ the two broken generators supply two Goldstone bosons, eaten to give longitudinal gauge-boson polarizations of the charged vectors. The physical count is $2\times 3+2+1=9$: two massive vectors, one massless vector and one radial scalar field. The scalar components eaten in unitary gauge are not additional physical particles.

For the broken phase all interactions containing the physical scalar field areMultiplying interaction coefficients by $i$ and differentiating with respect to the external fields includes the identical-field factorials. The scalar vertices of an adjoint triplet Higgs model areFor example, the $h^3$ rule contains $3!$, whereas the $hh{W}^{+}{W}^{-}$ rule contains only $2!$ from its two identical scalar fields. There are no $hAA$ or $hhAA$ vertices: the radial scalar field is neutral under the residual U(1). There is no derivative scalar-vector vertex in this physical spectrum in unitary gauge.

For completeness, if the unbroken phase is chosen, its physical scalar fields are the three components ${\varphi }_a$, not a single radial $h$. Expanding its gauge-covariant kinetic term givesWith all momenta incoming and Fourier transform convention $e^{-ipx}$, the three unbroken-phase rules areThe first rule follows by differentiating once on each possible scalar: for $a=3,b=1,c=2$, the interaction is $e{A}_{\mu }^3({\varphi }_1{\partial }^{\mu }{\varphi }_2-{\varphi }_2{\partial }^{\mu }{\varphi }_1)$, which fixes the momentum sign. The remaining rules follow from the quadratic gauge interaction and $-\lambda ({\varphi }_a{\varphi }_a{)}^2/8$. There are no unbroken-phase cubic scalar field Feynman vertices.

Write $\epsilon ={\epsilon }_a{t}_a$ and $A_{\mu }=A_{\mu a}{t}_a$. The proposed gauge-field transformation law becomes $\delta A_{\mu }=-{\partial }_{\mu }\epsilon +[\epsilon ,A_{\mu }]$. ConsequentlyBoth $\varphi$ and its gauge covariant derivative therefore transform as the same multiplet, with no derivative of the local parameter left over. The original algebraic invariance of the Lagrangian density now applies pointwise to $\mathcal{L}(\varphi ,D\varphi )$, so this constructs the local symmetry.

Also transform $J$ by $\delta J=\epsilon J$. Antisymmetry gives $(t_{a}J)\cdot \varphi +J\cdot (t_a\varphi )=0$, so the source coupling is invariant. Let $S_{A_{\mu a}}$ denote the functional derivative at a point. Vary the action functional and integrate the derivative of ${\epsilon }_a$ by parts, taking the local parameter to have compact support:Since every ${\epsilon }_a(x)$ is arbitrary, its coefficient vanishes. This proves the requested local identity, including the derivative sign and the order of the structure-constant indices.

For the generating functional, the integral of the first term is zero as a change-of-variable identity:The divergence of this vector field in field space is proportional to $\mathrm{tr}{t}_a=0$. Thus an invariant regulated functional measure, with no gauge anomaly, leaves no Jacobian term. Inserting the classical identity into the integral givesThe same first-order differential operator annihilates the connected generating functional $W$, because $Z=e^{iW}$. Its source term is $(t_{a}J)\cdot \phi$.

For the specified Legendre transform, vary $\Gamma =-W+\int J\phi$. Since $W_J=\phi$, the terms involving $\delta J$ cancel, leavingAntisymmetry also gives $(t_{a}J)\cdot \phi =-J\cdot (t_a\phi )$. Substitution in the identity for $W$, followed by multiplication by $-1$, proves the background gauge invariance of a scalar effective action:

The derivatives of the quantum effective action are proper, amputated one-particle-irreducible vertices. At each loop order they receive contributions from one-particle-irreducible Feynman diagrams with the indicated external scalar legs, including tree vertices and the required counterterms. Diagrams which disconnect after cutting one internal line are not additional proper-vertex contributions. Translation invariance supplies the overall momentum-conserving delta function factored out in the definition. The signs of these vertices must follow the chosen overall convention for $\Gamma$.

Differentiate the effective-action identity twice with respect to ${\phi }_i(y)$ and ${\phi }_j(z)$ and set $\phi =A=0$. In an invariant zero-field vacuum this gives the Ward identity contact termsThe term proportional to $A$ vanishes. With the printed Fourier factors $e^{i{p}_r{x}_r}$, integration by parts turns ${\partial }_{\mu }$ into $-i{p}_{1\mu }$. The first delta function combines $p_1$ with $p_2$, and the second combines $p_1$ with $p_3$. Using $(t_a{)}_{ki}=-(t_a{)}_{ik}$ and symmetry of the scalar two-point kernel then yieldsThis derivation fixes the relative signs independently of any vertex rule.

There is an overall-sign inconsistency in the final printed example. At tree level, stationarity of the source integral gives $W=S_0[\phi ,A]+\int J\phi$, hence the specified transform gives ${\Gamma }_{\mathrm{tree}}=-S_0$. For the free scalar multiplet this implies, after factoring out the delta function,The mixed result follows by expanding $-S_0$ to first order in $A$: its contribution is $\int A_{\mu a}({\partial }^{\mu }{\phi }_i)(t_a{)}_{ij}{\phi }_j$. The two scalar derivatives yield $i{p}_2^{\mu }(t_a{)}_{ij}+i{p}_3^{\mu }(t_a{)}_{ji}$. Momentum conservation givesso both sides of the Ward identity equal $i(p_3^2-p_2^2)(t_a{)}_{ij}$. Because the scalar theory is Gaussian for prescribed $A$, its fluctuation determinant depends on $A$ but not on $\phi$; it cannot add further two-scalar vertices. Thus this quadratic verification is sufficient, not merely the first term of an omitted scalar-loop correction.

The printed mixed vertex instead has a minus sign. It satisfies the same Ward identity if one uses the opposite effective-action convention ${\Gamma }_{\mathrm{alt}}=W-\int J\phi =S_0$ at tree level, because its two-point vertex is then $-(p^2+m^2){\delta }_{ij}$ as well. Either overall convention works, but the printed Legendre transform and printed mixed vertex cannot both be used unchanged. For example, with $p_2^2\ne p_3^2$ and a nonzero generator, they give opposite sides of the claimed identity.

The Noether gauging procedure starts by allowing a constant symmetry parameter to depend on position. For an internal symmetry written as $\delta {\varphi }^i={ϵ}^A{T}_A{\varphi }^i$, integration by parts identifies the Noether current:up to a boundary term. Introduce a gauge field $A_{\mu }^A$ and add $-g{A}_{\mu }^A{j}_A^{\mu }$, with the inhomogeneous gauge transformation $\delta A_{\mu }^A=g^{-1}{\partial }_{\mu }{ϵ}^A+\cdots$. This cancels the displayed variation. The current coupling has its own variation, so one iterates: add the needed higher-order interactions and corrections to the gauge transformations, and require closure of the local symmetry algebra. For an ordinary gauge theory, this reorganizes derivatives into gauge covariant derivatives and fixes the nonlinear gauge field strength. Gauging is therefore more than replacing a constant parameter by a function.

For rigid supersymmetry, the corresponding Noether current is the spinor-valued supercurrent $S^{\mu }$. A local Majorana spinor parameter gives a variation proportional to $({\partial }_{\mu }\overline{ϵ})S^{\mu }$. A gravitino ${\Psi }_{\mu }$, with an inhomogeneous transformation $\delta {\Psi }_{\mu }\sim {\kappa }^{-1}{\partial }_{\mu }ϵ$, cancels it through a coupling proportional to $\kappa {\bar{\Psi }}_{\mu }{S}^{\mu }$. The commutator of two rigid supersymmetry transformations is a translation. With local parameters, the translation parameter becomes position dependent, so the completion must also gauge translations: it introduces the vierbein, diffeomorphism invariance of general relativity, and local Lorentz transformations. The gravitino and graviton thus belong to the same supergravity multiplet.

For pure minimal four-dimensional supergravity, start with the free massless Fierz-Pauli action and the free Rarita-Schwinger field kinetic term. The Noether gauging procedure replaces the former by the Einstein-Hilbert action and couples the latter to the vierbein and spin connection. In a canonical normalization the leading transformations can be writtenAn overall sign or phase changes with the curvature and Majorana spinor conventions, but the relative normalization must be used consistently. To see why these terms fit together, first work to quadratic order in fermions, so the spin connection is torsion free. Varying the Rarita-Schwinger field kinetic term under $\delta {\Psi }_{\mu }={\nabla }_{\mu }ϵ/\kappa$ produces an antisymmetric double spinor covariant derivative. The spinor curvature identity and the Einstein tensor contraction of spin curvature give, in the matching curvature convention,The resulting Einstein tensor term cancels the variation of the Einstein-Hilbert action under $\delta e_{\mu }{}^a$. This is the essential first nontrivial cancellation in the construction.

At higher fermion order, the variations of the vierbein and spin connection produce further terms. Treating the spin connection as independent makes its Euler-Lagrange field equation algebraic. Its solution is a gravitino-induced torsion connection. With ${\Psi }_a=e_a{}^{\mu }{\Psi }_{\mu }$ and ${\gamma }^{ab}=[{\gamma }^a,{\gamma }^b]/2$, a conventional canonical normalization givesEquivalently, with $T^a=d{e}^a+{\tilde{\omega }}^a{}_b\wedge e^b$, the torsion tensor obeysThe Majorana Grassmann bilinear interchange makes the right-hand side antisymmetric in $\mu ,\nu$. Eliminating the spin connection generates the four-fermion completion. In the 1.5-order formalism one substitutes its solution but varies the other fields holding the connection fixed: the omitted chain-rule term is $(\delta S/\delta \omega )\delta \tilde{\omega }=0$. The remaining cubic fermion variations cancel by the Clifford algebra and Fierz rearrangements. This produces the displayed supergravity action in its compact connection-dependent form. Calling this formulation on shell means that no independent auxiliary fields have been retained and that closure of the supersymmetry algebra uses the propagating Euler-Lagrange field equations; it does not mean that action invariance is obtained only by setting every variation to zero on a solution.

In the antisymmetrized kinetic term, ${\Psi }_{\sigma }$ is a spinor-valued one-form and the derivative is conventionally writtenThis is the Lorentz-covariant exterior-derivative convention for that term. If one instead uses the full affine covariant derivative of a vector-spinor, a coordinate-index connection is also present and the antisymmetrization must include its torsion tensor correction. The crucial physical distinction from pure metric general relativity is the spin connection contains gravitino-induced contorsion, rather than being the torsion-free Levi-Civita connection. It reduces to the ordinary spinor covariant derivative when the gravitino vanishes.

Let $R_a=\mathrm{ad}(T_a)$, where $\mathrm{ad}(X)Y=[X,Y]$. In the given basis, the Adjoint representation of a Lie algebra isThe Jacobi identity gives $[\mathrm{ad}X,\mathrm{ad}Y]=\mathrm{ad}[X,Y]$, so $[R_a,R_b]=f^c{}_{ab}{R}_c$, as required for a Lie algebra representation. The Killing form is the symmetric Trace form of a Lie algebra representationCyclicity of the trace proves its invariance directly:In components this is precisely $\boxed{{\kappa }_{dc}{f}^d{}_{ab}+{\kappa }_{bd}{f}^d{}_{ac}=0}$, so the Killing form is an invariant bilinear form on a Lie algebra.

Choose the infinitesimal gauge transformation convention $\delta \varphi =\omega \varphi$, with $\omega (x)={\omega }^a(x)t_a$, and write $A_{\mu }=A_{\mu }^a{t}_a$. A gauge covariant derivative must satisfy $D_{\mu }^{\prime }=U{D}_{\mu }{U}^{-1}$ for $U=1+\omega +O({\omega }^2)$; henceEquivalently $\delta A_{\mu }^a=-{\partial }_{\mu }{\omega }^a+f^a{}_{bc}{\omega }^b{A}_{\mu }^c$. Expanding $\delta (D_{\mu }\varphi )$ makes the cancellation transparent:Thus $\boxed{\delta (D_{\mu }\varphi )={\omega }^a{t}_a{D}_{\mu }\varphi }$: it transforms in the same representation as $\varphi$. In adjoint covariant derivative notation, $\delta A_{\mu }=-D_{\mu }^{\mathrm{ad}}\omega$.

The commutator of two gauge covariant derivatives defines the gauge field strength:ThusConjugating the derivative commutator gives $F_{\mu \nu }^{\prime }=U{F}_{\mu \nu }{U}^{-1}$, so the gauge field strength transforms homogeneously in the Adjoint representation:Even if the matrices $t_a$ in a particular matter representation are not faithful, the displayed component formula defines the Lie-algebra-valued curvature independently of that representation.

Only $g_{(ab)}$ contributes to the Yang-Mills theory Lagrangian, since $F_{\mu \nu }^a{F}^{b\mu \nu }$ is symmetric in $a,b$. Thus take $g_{ab}$ to be a constant real symmetric bilinear form. Its infinitesimal variation isArbitrary local field strengths and gauge parameters give the necessary and sufficient invariant gauge kinetic form conditionAn antisymmetric part is unconstrained but contributes nothing. A nondegenerate kinetic term additionally needs a nondegenerate bilinear form; positive energy additionally needs a positive internal metric in the signature $(+---)$ convention. Neither extra condition follows solely from gauge invariance. For a compact simple real Lie algebra, $-\kappa$ is positive definite and is the standard choice up to a positive factor.

For the intended complex-simple setting, the Killing form is nondegenerate. Consequently write any invariant bilinear form on a Lie algebra uniquely as $g(X,Y)=\kappa (SX,Y)$. Invariance of both forms impliesfor every $X,Y,Z$, so $S$ commutes with the Adjoint representation. A subspace invariant under all adjoint maps is an ideal, hence simplicity makes the adjoint representation irreducible. By Schur lemma, $S=cI$, giving the uniqueness of an invariant bilinear form on a simple Lie algebra:The same conclusion holds for compact simple real algebras by complexification, with $c$ real; nondegeneracy excludes $c=0$, and positive energy selects $c<0$ in the stated convention.

There is a genuine real-simple exception to uniqueness of the Killing form if “simple” is read as an arbitrary real simple algebra. Regard ${\mathfrak{sl}}_2(\mathbb{C})$ as real. It is real simple: its complexification consists of two complex simple factors exchanged by conjugation, so a conjugation-stable ideal is zero or the whole algebra. Both $\mathrm{Re}{\kappa }_{\mathbb{C}}$ and $\mathrm{Im}{\kappa }_{\mathbb{C}}$ are real symmetric invariant forms, while ${\kappa }_{\mathbb{R}}=2\mathrm{Re}{\kappa }_{\mathbb{C}}$. With $H=\mathrm{diag}(1,-1)$, the Killing form of the special linear Lie algebra gives ${\kappa }_{\mathbb{C}}(H,H)=8$ and ${\kappa }_{\mathbb{C}}(H,iH)=8i$. Therefore the imaginary part cannot be a real multiple of ${\kappa }_{\mathbb{R}}$. The uniqueness conclusion requires the customary complex-simple or compact-real interpretation.

An experimentally observed example of CP violation is the long-lived neutral kaon decay $K_L\to {\pi }^{+}{\pi }^{-}$. The two-pion spin-zero final state is CP even, whereas $K_L$ would be the CP-odd neutral-kaon state in the CP-conserving limit. Its two-pion decay reveals CP violation. The historical experimental account is cern-courier.web.cern.ch/a/cp-violations-early-days/ .

For the CKM matrix, diagonalize the up- and down-quark mass matrices independently by biunitary transformations. Writewith $L_u^{†}{M}_u^{\prime }{R}_u$ and $L_d^{†}{M}_d^{\prime }{R}_d$ positive diagonal matrices. Only the left transformations enter the weak charged current, because its chiral projector removes the right-handed fields. Substitution givesThe matrix is unitary but need not be diagonal: the two left mass diagonalizations generally differ.