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The Brauer morphism retains precisely the fixed group-basis elements under conjugation by $D$:It is unital and surjective because $k{C}_G(D)\subseteq (kG{)}^D$. For multiplicativity, fix $x\in C_G(D)$. Its coefficient in a product is ${\sum }_{gh=x}{a}_g{b}_h$. Conjugation by $D$ permutes these pairs and the summand is constant on each orbit. Nonsingleton orbits have p-divisible size and contribute zero. Singleton pairs have both entries in $C_G(D)$, so they give exactly the coefficient in ${\mathrm{Br}}_D(a){\mathrm{Br}}_D(b)$. This proves the ring-homomorphism assertion.

Orbit sums form a basis of $(kG{)}^D$. A nonfixed orbit sum of $g$ is ${\mathrm{Tr}}_{C_D(g)}^D(g)$, a proper-subgroup relative trace. Conversely any proper-subgroup trace has zero coefficient on a fixed basis element, since that coefficient is multiplied by $[D:Q]=0$ in $k$. ThusThere is no assertion that this whole kernel is nilpotent.

Brauer first main theorem bijects the modular blocks of $kG$ with defect $D$ and those of $k{N}_G(D)$ with defect $D$, matching their nonzero Brauer images. Here is a proof with its auxiliary ingredients.

Write $N=N_G(D)$, $C=C_G(D)$ and defineThese are ideals in the indicated commutative algebras, since an invariant multiplier passes through a relative trace. The Brauer morphism and relative trace identityfollows by letting $D$ act on $G/D$. Fixed cosets have representatives in $N$; projection onto $C$ gives equal contributions on each other orbit, whose p-divisible size makes the contribution zero. Thus ${\mathrm{Br}}_D$ surjects $I_D(G)$ onto $J_D$.

To prove the trace criterion for defect groups, choose a p-subgroup $E$ of least order with a given block idempotent $b$ in $I_E(G)$. Such a choice exists because a Sylow $P$ gives $b={\mathrm{Tr}}_P^G(b/[G:P])$. Write $b={\mathrm{Tr}}_E^G(a)$ and replace $a$ by $ba$. If ${\mathrm{Br}}_E(b)=0$, the kernel formula and trace transitivity give $b\in {\sum }_{Q<E}{I}_Q(G)$. But $bZ(kG)$ is an Artinian local ring, and a finite sum of proper ideals cannot contain its identity. Multiplication by $b$ therefore forces $b\in I_Q(G)$ for some smaller $Q$, a contradiction. Hence ${\mathrm{Br}}_E(b)\ne 0$.

If $F$ is not conjugate into $E$, its action on $G/E$ has no fixed coset and the same orbit argument gives ${\mathrm{Br}}_F(I_E(G))=0$. Every subgroup with nonzero Brauer image of $b$ is therefore conjugate into $E$. Its maximal such subgroups are exactly the conjugates of $E$, proving the definition's conjugacy assertion andAmong block idempotents in this ideal, the Brauer image at $D$ survives precisely when their defect is $D$.

We also use primitive idempotents under a surjection of commutative Artinian ideals. Decompose the ambient algebras into local factors. The component of an ideal in a local factor is either the whole factor, possessing its identity, or a proper ideal in its nilpotent radical, possessing no nonzero idempotent. A nonzero quotient of a local algebra is local. A surjection of these ideals consequently bijects its surviving primitive idempotents with those of the target. Entire factors may be killed; the whole kernel need not be nilpotent.

Apply this result to $I_D(G)\to J_D$. Its surviving primitive idempotents are exactly the modular blocks of $G$ with defect $D$. Apply it also to $I_D(N)\to J_D$, where the target is identical since $N_N(D)=N$ and $C_N(D)=C$. Its surviving primitive idempotents are exactly the modular blocks of $N$ with defect $D$. Matching images in $J_D$ proves Brauer first main theorem, in both directions and with uniqueness.

For the last construction, the intended hypothesis is that $D$ is a defect group of the source modular block $b$. Put $L=N_H(D)$ and use the theorem inside $H$ to obtain the corresponding modular block $c$ of $kL$. The centralizer condition gives $C_H(D)=C_G(D)$, so ${\mathrm{Br}}_D$ maps $Z(kG)$ into $Z(kL)$. Let ${\omega }_c:Z(kL)\to F_c$ be the central character of a block, retaining its residue field $F_c$ rather than assuming it equals $k$. The unital compositesends exactly one global block idempotent to one, because their orthogonal idempotent images sum to one in a field. Define $b^G$ to be this unique modular block. This is block induction, independent of the chosen $H$-conjugate of $D$. It needs no containment of $N_G(D)$ in $H$, and its defect may be larger than $D$. With that containment it agrees with the first-main-theorem correspondence.

The printed PDF does not require $D$ to be a defect group of $b$. Its literal conditions do not always define the usual induced modular block. For example, over an algebraically closed field of characteristic three, take $G=A_5$, $H=A_4$ as a point stabilizer, and $D=⟨(123)⟩$. Then $D{C}_G(D)=D\le H$. The ordinary degree-three irreducible of $A_4$ gives a defect-zero modular block $b$ on reduction. The two distinct ordinary degree-three irreducibles of $A_5$ give distinct defect-zero modular blocks $B_{+}$ and $B_{-}$. Both restrict to that same character of $A_4$, with values $3,-1,0$ on its identity, double transpositions and three-cycles.

The standard defect-zero lifting theorem says that these reductions are simple projective modules and their block algebras are full matrix algebras. Thus both module restrictions $B_{+}{\downarrow }_{H\times H}$ and $B_{-}{\downarrow }_{H\times H}$ contain the same modular block bimodule $b$; indeed each module restriction is isomorphic to it here. The actual defect of $b$ is 1, whose centralizer is $G$, not a subgroup of $H$. The unrelated subgroup $D=C_3$ cannot select between the two modular blocks. This is block induction requires the block's defect centralizer; the qualified construction above answers the intended request.

Take finite-dimensional modules over $k$ and finite $R$-free lattices over the complete valuation ring, the usual integral modular representation theory convention. If a completion has several local factors, use the factor supporting the modular block. A block of a group algebra is an indecomposable two-sided ideal $B=RGb$, where $b$ is a primitive central idempotent. The identity decomposes as the sum of orthogonal block idempotents, and $M={⨁}_{b}bM$ for every module. Hence an indecomposable module belongs to $B$ exactly when $bM=M$.

A defect group of a block in characteristic $p$ is a maximal p-subgroup $D$ with ${\mathrm{Br}}_D(b)\ne 0$. Equivalently, it is minimal with $b\in {\mathrm{Tr}}_D^G((kG{)}^D)$; the trace criterion for defect groups and conjugacy are proved below. For a complete discrete valuation ring, use the corresponding reduced modular block; modular reduction of block idempotents preserves its defect groups. The equivalent bimodule description says that the module vertices of $B$, with $(x,y)$ acting by $a\mapsto xa{y}^{-1}$, are conjugates of $\Delta D=\{(d,d):d\in D\}$. This basic characterization follows from the relative-projectivity trace criterion for the bimodule action.

Fix a Sylow subgroup $P$ containing $D$. The regular group algebra is the permutation module $R[G]$ for $G\times G$, and multiplication by $b$ is its block projection. Restriction to a subgroup containing a chosen module vertex retains a summand with that module vertex: a module source at that module vertex is a summand of the module restriction, and the Mackey restriction formula and minimality prevent all the restricted summands from having smaller module vertices. Thus $B{\downarrow }_{P\times P}$ has an indecomposable summand with module vertex $\Delta D$.

For a finite p-group $Q$, the transitive permutation module $k[Q/L]$ is indecomposable: the group algebra of a p-group in characteristic p is local, and this module has a one-dimensional top, incompatible with a sum of two nonzero modules. Its module vertex is $L$, because it is induced from the trivial $L$-module and its Brauer quotient of a module at $L$ has the nonzero fixed coset $L$. The fixed-basis rule for this quotient follows by removing nonsingleton orbit sums, which are proper-subgroup traces. It excludes a smaller module vertex. Reduction gives the same statements for permutation lattices. Therefore Krull-Schmidt decomposition identifies the chosen block summand with an entire transitive $P\times P$ orbit in $G$.

Choose a point $c$ in that orbit whose stabilizer is exactly $\Delta D$. Being fixed by every $(d,d)$ gives $c\in C_G(D)$. Its full stabilizer isEquality with $\Delta D$ gives $P\cap c^{-1}Pc=D$. Conjugating, and using that $c$ centralizes $D$, provesThis is defect groups are centralizer-conjugate Sylow intersections.

If $Q◃G$ is a p-subgroup, then $QP$ is a p-subgroup for every Sylow $P$, so $QP=P$. It lies in both Sylow factors above. Thus every normal p-subgroup of $G$ is contained in $D$.

Finally put $N=N_G(D)$, choose $S\in {\mathrm{Syl}}_p(N)$, and extend it to a Sylow $P$ of $G$. Then $D\le S$ and $P\cap N=S$. Apply the proved intersection result to this $P$. Its element $c$ belongs to $C_G(D)\le N$, so $cP{c}^{-1}\cap N=cS{c}^{-1}$ is Sylow in $N$ too. Hence $D=S\cap cS{c}^{-1}$. The p-core $O_p(N)$ is contained in both factors, whereas normality of $D$ in $N$ gives the opposite inclusion. Therefore $D=O_p(N_G(D))$, so $D$ is a p-radical subgroup. The centralizer condition on the conjugating element is important for this deduction.

Use Euclidean worldsheet conventions with ${ϵ}^{12}=1$. Including a target metric, a Kalb–Ramond field $B_{\mu \nu }$ and the dilaton $\Phi$ givesThe imaginary coefficient of the antisymmetric term comes from Euclidean continuation; the Lorentzian action has the corresponding real coupling. For constant ${\Phi }_0$, the Gauss-Bonnet theorem gives $S_{\Phi }={\Phi }_0\chi (\Sigma )$, with Euler characteristic $\chi =2-2g$ on a connected closed oriented surface. Thus dilaton Euler-characteristic weighting givesAdding one handle multiplies a contribution by $g_s^2$, identifying $g_s$ as the string coupling. With canonically normalized external closed-string states, $n$ vertices add $g_s^n$, giving the string genus expansion $g_s^{2g-2+n}$.

Before gauge fixing, the Polyakov path integral for inserted string vertex operators is schematicallyThe quotient removes descriptions related by worldsheet diffeomorphisms and Weyl transformations. Gauge fixing has a nontrivial Jacobian, the Faddeev-Popov determinant. Anticommuting vector ghosts $c^a$ and symmetric trace-free antighosts $b_{ab}$ exponentiate this determinant as a local worldsheet ghost action. These worldsheet ghost fields implement the gauge quotient; they are not additional spacetime particles. Their zero modes must be treated separately rather than included in an invertible determinant.

Three kinds of geometry enter the scattering calculation. The worldsheet moduli describe genuinely different conformal shapes that remain after local gauge fixing. Each such shape must be integrated over. The mapping class group consists of orientation-preserving diffeomorphisms modulo those continuously deformable to the identity; it identifies different markings of the same conformal surface, so only one representative of each class is counted. The worldsheet conformal Killing group is the continuous residual conformal symmetry of a chosen surface. It moves insertion positions without changing the geometry and must be divided out, usually by fixing enough vertex positions.

For a torus, write $z\sim z+1\sim z+\tau$, with $\mathrm{Im}\tau >0$. Its shape modulus is $\tau$, its mapping class group acts through the modular group, and its connected worldsheet conformal Killing group acts by translations. Thus the shape integral runs over the standard fundamental domain of the modular group, while one insertion position can be fixed by translation. On the sphere, three positions can instead be fixed by Möbius transformations; for genus at least two the continuous conformal Killing group is trivial.

Now use the half-normalized worldsheet diffeomorphism ghost operator in the question. A vector field $v^a$ can preserve the representative metric after a compensating Weyl transformation precisely when $P_{1}v=0$. The trace of the metric variation sets $\delta \omega =\frac{1}{2}{\nabla }_a{v}^a$, leaving the Conformal Killing equationThe kernel of this operator gives the infinitesimal worldsheet conformal Killing group.

For a surviving infinitesimal worldsheet modulus, choose its metric variation $t_{ab}$ orthogonal to all gauge variations. Orthogonality to Weyl transformations requires $h^{ab}{t}_{ab}=0$. On the closed surface, the formal adjoint of the worldsheet conformal Killing operator follows by integration by parts:Orthogonality to every diffeomorphism variation is therefore the adjoint-kernel condition, givingIn a local conformal coordinate these imply ${\partial }_{\bar{z}}{t}_{zz}=0$ and its conjugate: the moduli variations are represented by holomorphic quadratic differentials. Multiplying the definition of $P_1$ by two changes its adjoint by two but changes neither kernel.

For four tachyon vertex operators on the torus, there is one complex shape modulus, four complex insertion positions and one complex translation to remove. The worldsheet integration count after conformal gauge fixing is thereforeFix the fourth insertion at $z_4=0$. A schematic gauge-fixed expression isHere the antighost pair supplies the modulus measure, and the ghost pair at the fixed vertex absorbs the translation zero modes. The three unfixed positions and $\tau$ are the four complex integrations. The punctured Riemann surface moduli dimension gives the same result, $3g-3+n=4$. Target-spacetime zero-mode integrals producing momentum conservation are separate from this requested count of worldsheet integrations.

The string sigma model in a spacetime metric is the Polyakov actionTake conformal gauge $h_{ab}=\mathrm{diag}(-1,1)$, and choose spatial period $2\pi$ for the closed string. Substitution of the plane-wave metric givesThe stipulated light-cone gauge in string theory $X^{+}=\tau$ sets ${\dot{X}}^{+}=1$ and $X^{\prime +}=0$. Its longitudinal term $-T\int {\dot{X}}^{-}$ is a total time derivative. The transverse string sigma model in an isotropic plane wave is thereforeThe resulting Euler-Lagrange equations are the massive wave equationsThese conventions fix the mass term directly from $X^{+}=\tau$. More generally $X^{+}=\kappa \tau$ would replace $m$ by $|m\kappa |$, with $p^{+}=2\pi T\kappa$ for this spatial period.

Use the real orthonormal Fourier basisFor the plane-wave string mode frequencies, let ${\omega }_0=|m|$ and ${\omega }_{n,c}={\omega }_{n,s}=\sqrt{n^2+m^2}$. The complete real mode expansion for nonzero frequencies isEvery term obeys the wave equation and is periodic. Orthonormality and completeness let its coefficients match arbitrary appropriate initial position and velocity data. For $m=0$, replace the zero-mode bracket by $q_0^I(0)+\tau p_0^I/T$; the zero mode becomes a free particle rather than an oscillator.

The canonical momentum density is ${\Pi }_I=T{\dot{X}}^I$. The transverse Hamiltonian isWrite $X^I={\sum }_a{q}_a^I{f}_a$ and $p_a^I=T{\dot{q}}_a^I$. Since $\int f_a{f}_b={\delta }_{ab}$ and $\int f_a^{\prime }{f}_b^{\prime }=n_a^2{\delta }_{ab}$, substitution gives the plane-wave string mode HamiltonianThis is a collection of independent harmonic oscillators. The last equality follows by substituting the sine-and-cosine time dependence; each oscillator energy is constant.

For canonical quantization of each nonzero-frequency mode, putThen $H_{\perp }={\sum }_{I,a}{\omega }_a(a_a^{I†}{a}_a^I+1/2)$. Opposite travelling-wave oscillator combinations give the equivalent formFor $m\ne 0$, the formal zero-point energy is $E_0=\frac{24}{2}[|m|+2{\sum }_{n\ge 1}\sqrt{n^2+m^2}]$. At $m=0$, replace the zero-mode occupation term by ${\sum }_I(p_0^I{)}^2/(2T)$. The zero-point expression is divergent and needs a regulator or a specified normal ordering convention; no mass-independent intercept has been assumed.

The remaining Virasoro constraints determine the longitudinal coordinate:Periodicity of $X^{-}$ imposes ${\int }_0^{2\pi }{\Pi }_I{X}^{\prime I}d\sigma =0$, the closed-string level matching condition. Labelling the travelling oscillators by their opposite worldsheet momenta gives ${\sum }_{n\ge 1,I}n(N_n^I-{\tilde{N}}_n^I)=0$. The light-cone energy $p^{-}=-p_{+}$ equals $T\int ({\dot{X}}^{-}+m^2{X}^I{X}^I)d\sigma =H_{\perp }$, consistent with the positive mass term in the Hamiltonian.

This is the classical reduction in the specified background. As a full quantum background with constant dilaton and no Kalb–Ramond field, this metric has $R_{++}=24m^2$ and fails the lowest-order sigma-model beta function equation when $m\ne 0$; additional background fields would be required for quantum Weyl invariance.

Use a mostly-plus target metric tensor and the standard open-string mode expansion convention ${\alpha }_0^{\mu }=\sqrt{2{\alpha }^{\prime }}{p}^{\mu }$. The integer string oscillator level is $N={\sum }_{n\ge 1}{\alpha }_{-n}\cdot {\alpha }_n$, whose eigenvalues are nonnegative integers. Then $L_0=N+{\alpha }^{\prime }{p}^2$, so the string mass-shell condition givesThe ground state is a tachyon, and the level-one state is massless. The physical string state conditions in old covariant string quantization areOnly the positive-mode Virasoro constraints annihilate a ket; their negative-mode partners annihilate the corresponding bra. Imposing both signs on every ket would contradict the nonzero Virasoro central extension.

The central charge comes from normal ordering the quadratic oscillator expressions. Classically their Poisson brackets have no central term. Quantum mechanically, moving annihilation operators through creation operators produces double contractions. From the oscillator commutator one obtains $[L_m,{\alpha }_n^{\mu }]=-n{\alpha }_{m+n}^{\mu }$, which gives the noncentral part $(m-n)L_{m+n}$.

The coefficient is fixed by the free-boson Virasoro central term. On a formal zero-momentum vacuum, for $m>0$,The two contraction pairings giveHere ${\eta }_{\mu \nu }{\eta }^{\mu \nu }=D$; the time coordinate contributes one, just as each spatial coordinate does. Comparing with the Virasoro algebra provesThis counts covariant target coordinates. It is distinct from the $D-2$ physical transverse polarizations. Including the central charge of reparameterization ghosts gives $c_{\mathrm{total}}=D-26$; cancellation of the worldsheet Weyl anomaly gives the usual critical dimension of the bosonic string $D=26$.

A general level-one state isThe zero-mode constraint requires $k^2=0$. Since $[L_1,{\alpha }_{-1}^{\mu }]={\alpha }_0^{\mu }$, the remaining nontrivial positive-mode constraint is $k\cdot ϵ=0$; all $L_m$ with $m\ge 2$ act trivially at this level. Thus the polarization vector is transverse to a nonzero null momentum.

There is also a null string state $L_{-1}|0;k⟩=\sqrt{2{\alpha }^{\prime }}k\cdot {\alpha }_{-1}|0;k⟩$. It is physical when $k^2=0$ and orthogonal to every physical polarization. The null-state quotient of a string therefore identifiesChoose $k=(E,E,0,\dots ,0)$. Transversality sets ${ϵ}^1={ϵ}^0$, and the null-state identification removes this common component. The remaining $D-2$ spatial components have positive norm. Hence the physical level-one states form a massless vector with $D-2$ polarizations, or 24 at the critical dimension. The removed component is gauge redundancy. States violating the constraints, including independent timelike negative-norm polarizations, are unphysical; physical null states are removed by the quotient rather than counted as additional particles.

Put $s=q^2>0$ and neglect electron and final-quark masses compared with $\sqrt{s}$. First fix a final color charge; its multiplicity will be restored for the inclusive rate. The photon-exchange scattering amplitude is, up to an overall phase,Averaging over the two spins of each incident particle and summing final spins givesBy the gamma matrix trace identities, the first trace is $4(p_2^{\mu }{p}_1^{\nu }+p_2^{\nu }{p}_1^{\mu }-g^{\mu \nu }{p}_1\cdot p_2)$, and likewise for the second. Contracting the two tensors cancels the terms proportional to $(p_1\cdot p_2)(k_1\cdot k_2)$, givingIn the centre-of-mass frame all four energies are $\sqrt{s}/2$. The first pair of scalar products equals $s(1-\cos \theta )/4$ each, and the second pair equals $s(1+\cos \theta )/4$ each. Hence $\overline{|\mathcal{M}{|}^2}=e^4{Q}_q^2(1+{\cos }^2\theta )$.

For clarity, the printed flux hint has a factor-of-four inconsistency: when $\mathcal{F}=4\sqrt{(p_1\cdot p_2{)}^2-m_1^2{m}_2^2}=2s$, the prefactor multiplying the spin-averaged amplitude is $1/\mathcal{F}$, not $1/(4\mathcal{F})$. Alternatively, $1/(4F_0)$ is correct if $F_0=\sqrt{(p_1\cdot p_2{)}^2-m_1^2{m}_2^2}=s/2$. Use this massless two-body scattering flux normalization, keeping the spin average separate.

To obtain the two-body Lorentz-invariant phase space directly, integrate the momentum-conserving spatial delta function, leaving ${\mathbf{k}}_2=-{\mathbf{k}}_1$. The remaining radial integral is $\int dk\delta (\sqrt{s}-2k)=1/2$, so $d{\Phi }_2/d\Omega =1/(32{\pi }^2)$. The relativistic scattering cross-section is consequentlySince $\int d\Omega (1+{\cos }^2\theta )=2\pi {\int }_{-1}^1(1+x^2)dx=16\pi /3$, the integrated electron-positron annihilation into a quark pair result, for this fixed color, isThe first two printed formulas are thus color-stripped. Distinct final colors are orthogonal states and must be summed, producing $N_c=3$. At leading order each accessible quark flavor contributes independently. Away from thresholds and narrow resonances, quark-hadron duality relates the inclusive parton rate to the observed inclusive electron-positron annihilation into hadrons, givingThis formula concerns the photon channel, assumes $\sqrt{s}$ is large compared with the relevant quark masses and the strong-interaction scale, and does not include Z boson exchange.

Beyond leading order, an exactly exclusive pair of colored quarks is not an observable asymptotic state because of confinement and hadronization. Moreover its virtual corrections have infrared divergences; real soft or collinear gluon emission cannot be excluded if a finite rate is wanted. The inclusive sum includes both the corrected two-parton channel and channels such as $q\bar{q}g$, with their degenerate soft and collinear limits. Real and virtual infrared divergences then cancel for an infrared-safe observable. A specified jet observable can also be meaningful if it defines an infrared-safe observable; a bare two-quark final state is not.

A convenient inclusive calculation uses the optical theorem. For the electromagnetic hadronic current $j_{\mu }={\sum }_f{Q}_f{\bar{q}}_f{\gamma }_{\mu }{q}_f$, defineThe hadronic electromagnetic-current spectral density is obtained from its discontinuity. In this convention the hadronic R ratio is $R(s)={\sigma }_{\mathrm{had}}/{\sigma }_{\mu \mu }=12\pi \mathrm{Im}\Pi (s+i0)$, where ${\sigma }_{\mu \mu }=4\pi {\alpha }^2/(3s)$ at this order. Cutting the forward amplitude sums all allowed hadronic intermediate states, so perturbative Quantum chromodynamics applied to this inclusive quantity implements the required sum of real and virtual contributions.

Finally, write $a={\alpha }_s({\mu }^2)$, $C=1.99-0.11n_f$, $L=\ln (s/{\mu }^2)$, and ${\beta }_0=11-2n_f/3$. The given truncated correction isDifferentiate with respect to $t=\ln {\mu }^2$, using the stated one-loop beta function $da/dt=-{\beta }_0{a}^2/(4\pi )$ and $dL/dt=-1$. The derivative of $a/\pi$ cancels the explicit logarithmic derivative at order $a^2$. Keeping the residual derivative of the truncated expression givesThus the principle of minimal sensitivity, applied to this expression with nonzero coupling and ${\beta }_0\ne 0$, selects the one-loop stationary scale for the hadronic annihilation correctionThis is a prescription based on the residual order-$a^3$ scale dependence of an order-$a^2$ approximation. At strictly order $a^2$, the derivative already vanishes for every renormalization scale; the prescription does not assert exact scale dependence of the observable. It is useful only while the selected running coupling remains perturbative. If $a=0$ or ${\beta }_0=0$, the stated derivative imposes no unique scale.

Start with the Dirac equation $(i{\gamma }^{\mu }{\partial }_{\mu }-m)\psi =0$. The equation for the Dirac adjoint is $i({\partial }_{\mu }\bar{\psi }){\gamma }^{\mu }+m\bar{\psi }=0$. Transpose it and multiply by the charge-conjugation matrix C to obtainFor this to be the original Dirac equation multiplied by minus one, the necessary matrix identity isUsing $C^{†}=C^{-1}$, the charge conjugation of the Dirac adjoint follows directly:The final step uses $({\gamma }^0{)}^T{C}^{-1}=-C^{-1}{\gamma }^0$ and $({\gamma }^0{)}^2=1$.

Let $j_V^{\mu }=\bar{\psi }{\gamma }^{\mu }\psi$, $j_A^{\mu }=\bar{\psi }{\gamma }^{\mu }{\gamma }^5\psi$, and $j_L=j_V-j_A$. For anticommuting fields, or normal-ordered operator bilinears, the minus sign in the transformed Dirac adjoint cancels the minus sign from interchanging the two fields. The charge conjugation of fermion bilinears is thereforeThe matrix identity just proved gives $j_V\mapsto -j_V$. Also $C({\gamma }^5{)}^T{C}^{-1}={\gamma }^5$, obtained by transposing its four gamma matrices and reversing their order. Henceso $j_A\mapsto j_A$ and $j_L\mapsto -j_R$, where $j_R=j_V+j_A$. Since the external vector V is odd under charge conjugation, $j_L\cdot V$ becomes $j_R\cdot V$ and is generally not invariant.

For parity symmetry in quantum field theory, write $P^{\mu }{}_{\nu }=\mathrm{diag}(1,-1,-1,-1)$. The given spinor transformation gives $j_V\mapsto P{j}_V$, $j_A\mapsto -P{j}_A$, and $V\mapsto PV$, with arguments evaluated at the reflected point. The contracted interaction again becomes $j_R\cdot V$. Each of parity and charge conjugation interchanges the two chiral contractions; applying both restores $j_L\cdot V$. This particular neutral current interaction is CP invariant, although neither P nor C is separately a symmetry. This conclusion uses the stipulated C-odd vector transformation; it is not a claim that all interactions of the Standard Model preserve CP symmetry.

For time reversal of a Dirac field, the gamma matrix adjoint and transpose identities giveWith $B={\gamma }^{5}C$, anticommutation of ${\gamma }^5$ with every gamma matrix yieldsEquivalently, $\boxed{B({\gamma }^{0\ast },-{\mathbf{\gamma }}^{\ast })B^{-1}=({\gamma }^0,\mathbf{\gamma })}$. Since time reversal is an antiunitary operator, the factor i in ${\gamma }^5=i{\gamma }^0{\gamma }^1{\gamma }^2{\gamma }^3$ must also be conjugated. ThusUsing the stated transformations of the field and its Dirac adjoint, the transformed bilinear consequently isThe external vector has the same component signs. Their contraction is unchanged, proving time reversal of a real neutral chiral vector interaction. A real coupling is understood; an antiunitary operator also conjugates coefficients, and relative complex weak phases need not satisfy this invariance.

A charged weak box contribution to kaon mixing is shown below. The incoming $d\bar{s}$ and outgoing $s\bar{d}$ connect through two charged W bosons and two internal up-type quarks. The black arrows are fermion flow, so the lower line's arrows oppose the momentum of the corresponding antiquark. The blue arrows specify the direction of positive charge carried by the charged propagators. Each vertex is a charged weak current vertex, with the appropriate CKM matrix element; the internal flavors are summed over $u,c,t$.

Use precisely the printed exchange-positive CP convention for neutral kaon mixing. In the flavor basis, CP symmetry acts by $S=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$. Write the restricted effective matrix asThen $SMS=\left(\begin{array}{cc}a& c\\ b& a\end{array}\right)$, so CP invariance of this mixing operator is exactly $b=c$. Therefore CP noninvariance in this neutral-kaon mixing block means $M_{12}\ne M_{21}$. Violation elsewhere in a larger Hamiltonian would not, by itself, prove an inequality for this particular block.

For neutral-kaon mass matrix diagonalization, choose correlated roots $r^2=b$, $s^2=c$, and eigenvalues ${\lambda }_{\pm }=a\pm rs$. The vectors $(r,s{)}^T$ and $(r,-s{)}^T$ are the corresponding eigenvectors, as direct multiplication verifies. Taking orthonormal flavor states and writing the printed CP eigenstates as $K_1=(K^0+{\bar{K}}^0)/\sqrt{2}$ and $K_2=(K^0-{\bar{K}}^0)/\sqrt{2}$, one findswhere the kaon CP mixing parameter isOverall phases have been absorbed. The kaon mixing square-root branch convention must be chosen continuously so that $r=s$ in the CP-conserving limit for the eigenstate labeled plus. These formulas are exact in the nondegenerate case with $r+s\ne 0$; the requested small admixtures follow if $|ϵ|\ll 1$. Smallness is an additional weak-CP-violation assumption, not a consequence of $b\ne c$ alone. A Hermitian mass operator has $c=b^{\ast }$; the same algebra also applies to the effective non-Hermitian operator that includes decays, where the two eigenstates need not be orthogonal. At a fully degenerate $b=c=0$, a CP eigenbasis can simply be chosen.

The derivative convention uses hypercharge Y with $Q=T_3+Y$. For each lepton family $l=e,\mu ,\tau$, the left-handed fields form an SU(2) doublet $L_l=({\nu }_l,l{)}_L^T$, with $Y=-1/2$. Their weak isospin values are $T_3=+1/2,-1/2$ and their electric charges are $0,-1$. The right-handed charged lepton $l_R$ is an SU(2) singlet, with $T_3=0$, $Y=-1$ and charge $-1$. The minimal model used here has no right-handed neutrino. A sterile right-handed neutrino, if added as a singlet with $Y=0$, would have no charged weak current interaction. Antiparticles carry the conjugate representations and opposite electric charges.

Expanding ${\bar{L}}_{l}i{\gamma }^{\mu }{D}_{\mu }{L}_l$ supplies $-g{\bar{L}}_l{\gamma }^{\mu }{A}_{\mu }^a{\sigma }_a{L}_l/2$. The off-diagonal entries of $A^a{\sigma }_a$ are $\sqrt{2}{W}^{+}$ and $\sqrt{2}{W}^{-}$. With the chiral projector $P_L=(1-{\gamma }^5)/2$, one has ${\bar{\nu }}_L{\gamma }^{\mu }{l}_L=\bar{\nu }{\gamma }^{\mu }{P}_{L}l$. Thus the two W boson interactions areFor quarks, separate rotations diagonalize the up-type and down-type mass terms. If their left-handed rotations are $U_{uL}$ and $U_{dL}$, the charged interaction contains the CKM matrix $V=U_{uL}^{†}{U}_{dL}$:There is an implicit sum over color charge. This quark mixing permits transitions between different families, rather than just the paired fields in a single doublet.