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Since the chirality matrix commutes with spinor Lorentz generators, $j_5^{\mu }=\bar{\psi }{\gamma }^{\mu }{\gamma }^5\psi$ transforms as a vector under proper Lorentz transformations and as a pseudovector under parity. Contracting two copies gives the Lorentz scalar $j_{5\mu }{j}_5^{\mu }$; the extra parity signs cancel. Lorentz invariance of this interaction does not require current conservation.

Use natural units and the Minkowski metric $g=\mathrm{diag}(1,-1,-1,-1)$. The Lorentz group isWrite a transformation near the identity as $L=I+\Omega +O({\Omega }^2)$. The defining relation gives ${\Omega }^{T}g+g\Omega =0$, so ${\omega }_{\mu \nu }=g_{\mu \rho }{\Omega }^{\rho }{}_{\nu }$ is antisymmetric. ChooseThere are $\left(\genfrac{}{}{0ex}{}{4}{2}\right)=\boxed{6}$ independent real parameters: three spatial rotations and three Lorentz boosts. Differentiating $(e^{t\Omega }{)}^{T}g{e}^{t\Omega }$ with respect to $t$ shows it is constant and equals $g$. Thus exponentiating these infinitesimal generators gives Lorentz transformations connected to the identity. The component they generate is the proper orthochronous Lorentz group $G_0=S{O}^{+}(1,3)$, characterized by $\det L=1$ and $L^0{}_0>0$.

The local exponential calculation alone establishes products of exponentials. The single-exponential assertion also holds here. Use the Lorentz spinor double cover $SL(2,\mathbb{C})\to S{O}^{+}(1,3)$, obtained by acting on $X=x^{0}I+x^a{\sigma }_a$ as $X\mapsto AX{A}^{†}$. Its determinant is the Minkowski norm. A lift $A$ with distinct eigenvalues $\lambda ,{\lambda }^{-1}$ is diagonalizable and has a traceless logarithm with eigenvalues $z,-z$, where $e^z=\lambda$. With repeated eigenvalues, choose between $A$ and $-A$ so that the lift is $I+N$ with $N^2=0$; its traceless logarithm is $N$. Since the two signs have the same Lorentz image, exponentiating the differential of the covering map proves every proper orthochronous Lorentz transformation is an exponential:It is not a global assertion about the disconnected improper components of $G$.

For Dirac spinors, choose gamma matrices satisfying the Clifford algebra $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2g^{\mu \nu }I$ and setThe Clifford relations giveThey also give the stipulated Lorentz commutator with every $M$ replaced by $\Sigma$: commute $\Sigma$ through the two gamma factors defining the other generator and collect the four terms. Consequently this is a representation of the Lie algebra of the Lorentz group, and exponentiation yieldsThe Dirac adjoint $\bar{\psi }={\psi }^{†}{\gamma }^0$ transforms as ${\bar{\psi }}^{\prime }=\bar{\psi }{S}^{-1}$, since $S^{†}{\gamma }^{0}S={\gamma }^0$.

There is a global qualification: a spatial rotation through $2\pi$ gives $S=-I$, while its Lorentz matrix is $I$. Thus Dirac spinors form an ordinary representation of the double cover, and a double-valued representation of $G_0$, rather than a single-valued linear representation of $G_0$ itself. Fermion bilinears are unaffected by this sign.

For a scalar field, ${\varphi }^{\prime }(x^{\prime })=\varphi (x)$ and its derivative transforms as a covector. A Lagrangian density must have all Lorentz indices contracted into scalars using invariant tensors; arbitrary scalar potentials are allowed. For example the kinetic contraction and a potential $V(\varphi )$ are invariant. Lorentz invariance by itself does not restrict the potential to quartic order: that restriction would be an additional four-dimensional power-counting requirement.

Define ${\gamma }^5=i{\gamma }^0{\gamma }^1{\gamma }^2{\gamma }^3$. It commutes with all ${\Sigma }^{\rho \sigma }$, so the axial current $j_5^{\mu }=\bar{\psi }{\gamma }^{\mu }{\gamma }^5\psi$ transforms as a vector under $G_0$. Hence all three proposed contractions are invariant under $G_0$. If invariance under parity or the full improper Lorentz group is also required, the distinction is:The axial current is a pseudovector: under an improper Lorentz transformation it has the extra determinant sign. The mixed derivative–axial contraction is therefore a pseudoscalar, while the square of the axial current has two cancelling signs. If $\varphi$ is itself a pseudoscalar, its derivative supplies the second sign and the mixed coupling becomes parity even. This is the axial derivative coupling and parity distinction; parity invariance must not be silently assumed from proper Lorentz invariance.

In four dimensions, $[\varphi ]=1$, $[\psi ]=3/2$ and $[{\partial }_{\mu }]=1$. The three operators have mass dimensions $4,5,6$, respectively. Thus the latter two require dimensionful couplings and are not power-counting renormalizable interactions, but this does not invalidate their Lorentz invariance or their use in an effective field theory.

Continue with metric signature $+---$. The Pauli matrices obey $\{{\sigma }^i,{\sigma }^j\}=2{\delta }^{ij}{I}_2$. Block multiplication in the chiral gamma-matrix representation givesandHence $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2g^{\mu \nu }{I}_4$, the required Clifford algebra. Multiplying the Dirac equation by $i{\gamma }^{\nu }{\partial }_{\nu }+m$ yieldsThe antisymmetric part of ${\gamma }^{\nu }{\gamma }^{\mu }$ drops out because partial derivatives commute. Therefore every component of the Dirac spinor satisfies the Klein-Gordon equation. The converse is not true: four arbitrary scalar solutions need not obey the first-order Dirac equation.

The Lorentz group consists of real invertible matrices $\Lambda$ preserving the Minkowski metric: ${\Lambda }^{T}g\Lambda =g$. Writing $\Lambda =I+\omega +O({\omega }^2)$ gives ${\omega }_{\mu \nu }=-{\omega }_{\nu \mu }$, so there are six continuous parameters: three spatial rotation angles and three boost rapidities. The proper orthochronous Lorentz group is the connected component with determinant $+1$ that preserves time orientation. Parity and time reversal are additional discrete operations and are not encoded by the six real parameters in an exponential near the identity.

For the real-generator convention used here, vector generators can be writtenTake ${\Omega }_{\rho \sigma }=-{\Omega }_{\sigma \rho }$ and use the same parameters in the spinor representation. The Lorentz-spinor generators from a Clifford algebra areTo verify the Lorentz algebra, first commute a generator with one gamma matrix:For example, this follows by moving ${\gamma }^{\tau }$ through each product using the Clifford anticommutator. Applying the commutator derivation rule to $\frac{1}{4}[{\gamma }^{\tau },{\gamma }^{\nu }]$ then givesNo extra factor of $i$ belongs in these generators with this commutator convention. Hermitian-generator conventions shift the factors of $i$ into the brackets and the exponential instead.

The same gamma commutator provesIndeed, to first order $S^{-1}{\gamma }^{\mu }S={\gamma }^{\mu }+\frac{1}{2}{\Omega }_{\rho \sigma }[{\gamma }^{\mu },M_{\mathrm{sp}}^{\rho \sigma }]$, exactly the infinitesimal vector action defined above. Exponentiating proves the finite relation. The spinor field transforms as ${\psi }^{\prime }(x^{\prime })=S(\Lambda )\psi (x)$ with $x^{\prime }=\Lambda x$; combining this relation with ${\partial }_{\mu }^{\prime }=({\Lambda }^{-1}{)}^{\nu }{}_{\mu }{\partial }_{\nu }$ preserves the Dirac equation. A $2\pi$ rotation gives $S=-I_4$, so the spinor matrices give the double-cover spin representation, rather than a single-valued ordinary representation of the Lorentz group itself.

For a boost in direction $i$, $M_{\mathrm{sp}}^{0i}=\frac{1}{2}{\gamma }^0{\gamma }^i$ is Hermitian, with eigenvalues $\pm \frac{1}{2}$. A real rapidity $\eta$ therefore produces eigenvalues $e^{\pm \eta /2}$ in $S$, whose moduli are not one. The finite-component spinor representation is not unitary for boosts, although the rotation matrices are unitary. The Lorentz group is noncompact; this does not forbid the unitary infinite-dimensional action on the physical Hilbert space of states.

Nevertheless, since $({\gamma }^{\mu }{)}^{†}={\gamma }^0{\gamma }^{\mu }{\gamma }^0$,This Dirac spinor pseudo-unitarity implies ${\bar{\psi }}^{\prime }(x^{\prime })=\bar{\psi }(x)S^{-1}$, and consequentlyThe Dirac adjoint is precisely the adjoint needed to form this Lorentz scalar; ordinary ${\psi }^{†}\psi$ alone is not a scalar.

Let ${\gamma }^5=i{\gamma }^0{\gamma }^1{\gamma }^2{\gamma }^3=\mathrm{diag}(-I_2,I_2)$. It anticommutes with every gamma matrix and commutes with every $M_{\mathrm{sp}}^{\rho \sigma }$, so $S^{-1}{\gamma }^{5}S={\gamma }^5$. Thus the axial current transforms as a four-vector under proper Lorentz transformations:Under parity it is an axial vector, acquiring the additional pseudovector sign. Contracting two axial currents cancels that sign and uses the invariant metric. Therefore $j_{5\mu }{j}_5^{\mu }$ is compatible with Lorentz invariance, including parity invariance of this contraction. Lorentz invariance of the interaction does not require the axial current to be conserved.

The strong-interaction matrix element between two spin-zero pseudoscalar mesons has only $p^{\mu }$ and $k^{\mu }$ available. The product of the two intrinsic parities is positive. An axial current would require a pseudovector constructed from these momenta, but an expression involving the Levi-Civita symbol needs three independent four-vectors and therefore vanishes. This is a consequence of parity conservation in the hadronic matrix element, not of parity conservation in the weak interaction. The vector current can have the two independent structures $p+k$ and $p-k$. Its coefficients are Lorentz scalars; with $p^2=m_K^2$ and $k^2=m_{\pi }^2$ fixed, their only varying invariant is $s=q^2$. HenceThese are the pseudoscalar-to-pseudoscalar form factors. With relativistically normalized states they are dimensionless. The vanishing axial matrix element and this decomposition explain the two equalities separately.

Write $q_1$ and $q_2$ for the outgoing electron and antineutrino momenta. From the Fermi interaction, an invariant scattering amplitude, up to an irrelevant overall sign or phase, isThe CKM matrix element $V_{us}$ multiplies the quark current in the convention of this interaction. Let ${\ell }_{\mu }=\bar{u}(q_1){\gamma }_{\mu }(1-{\gamma }^5)v(q_2)$. For massless leptons, the massless Dirac equation and chirality matrix anticommutation giveIn the second term move $/q_2$ through the chiral projector before applying $/q_{2}v=0$. Since $p+k=2p-q$, this transverse massless leptonic current givesThe disappearance of $f_{-}$ uses the massless approximation; for a massive charged lepton its contraction is proportional to the lepton mass.

Use the fermion spin sum and the supplied gamma matrix trace identities. The symmetric part of the leptonic tensor isThe Levi-Civita symbol term is antisymmetric and drops out when contracted with $p^{\mu }{p}^{\nu }$. ThusThere is no initial-spin average because the kaon is spinless. For the integrated massless leptonic tensor, keep every factor of $2\pi$ explicit and define the unnormalized two-lepton Lorentz-invariant phase spaceThe leptons are massless, so $q_i^0=|{\mathbf{q}}_i|$. Its trace is $g^{\mu \nu }{I}_{\mu \nu }=\pi s$. ThereforeThe three Lorentz-invariant phase-space measures and their momentum delta function contribute $1/[8(2\pi {)}^5]$, in addition to $1/(2m_K)$ in the decay rate. Combining them with the spin sum givesThis massless semileptonic pseudoscalar decay rate uses a two-lepton integral over future-timelike $q$ and the pion integral is restricted to the physically allowed region. The null endpoint follows by continuity. The coefficient has mass dimension $-5$, so the complete expression has mass dimension one, as a decay rate must in natural units.

In the kaon centre-of-momentum frame, put $E=k^0$ and $\kappa =|\mathbf{k}|$. ThenThe dimensionally consistent Källén function isThe pion-only mass term must have fourth power: the second power printed in the PDF is dimensionally inconsistent. This repair also follows directly from squaring $E=(m_K^2+m_{\pi }^2-s)/(2m_K)$. Angular integration and the change of variable givewhere the negative sign reverses the endpoints. Combining this with the bracket $\lambda /4$ yieldsThus $\boxed{B=G_F^2|V_{us}{|}^2/(192{\pi }^3{m}_K^3),a=0,b=(m_K-m_{\pi }{)}^2}$. The lower limit is the minimum invariant mass of two massless leptons; at the upper limit the pion is at rest. The coefficient has mass dimension $-7$, while $ds{\lambda }^{3/2}$ has dimension eight. The Källén function also shows why the differential decay rate vanishes at zero pion momentum.

Use the printed gamma matrix conventions throughout, in particular ${\gamma }^5=-i{\gamma }^0{\gamma }^1{\gamma }^2{\gamma }^3$. The transpose identity for ${\gamma }^0$ gives $C{\gamma }^0=-{\gamma }^{0}C$, and its Hermitian conjugate gives the corresponding identity for $C^{†}$. Thus the Dirac adjoint of the charge-conjugated field isReordering the anticommuting components of the fermion bilinear supplies a minus sign:Invariance for arbitrary fields therefore requires $C^{†}C=I$, so the charge-conjugation matrix is a unitary matrix. This is the required normalization constraint; it is not a bosonic reordering calculation.

For a spinor matrix $\Gamma$, the same calculation and the given $C^{-1}=C^T$ giveWith $V^{\mu }=\bar{\psi }{\gamma }^{\mu }\psi$ and $A^{\mu }=\bar{\psi }{\gamma }^{\mu }{\gamma }^5\psi$, the supplied transpose identities implyConsequently the vector current is C-odd and the axial current is C-even: $V^{\mu }\mapsto -V^{\mu }$ and $A^{\mu }\mapsto A^{\mu }$.

For parity, put $\Lambda =\mathrm{diag}(1,-1,-1,-1)$. The gamma matrix identities ${\gamma }^0{\gamma }^{\mu }{\gamma }^0={\Lambda }^{\mu }{}_{\nu }{\gamma }^{\nu }$ and ${\gamma }^0{\gamma }^5{\gamma }^0=-{\gamma }^5$ giveThus $V^0$ is P-even and $\mathbf{V}$ P-odd, whereas $A^0$ is P-odd and $\mathbf{A}$ P-even. The axial current transforms as a pseudovector.

Both the D meson and kaon are spin-zero pseudoscalars. The strong interaction states obey parity symmetry in quantum field theory. Consequently an axial current matrix element between them would have to be a pseudovector formed from only $p$ and $k$, which is impossible: an totally antisymmetric tensor would require more linearly independent vectors. ThusLorentz covariance then leaves two linearly independent vectors for the vector current matrix element, giving the pseudoscalar-to-pseudoscalar form factorsThe vanishing axial current here follows from strong interaction parity symmetry in quantum field theory, not parity symmetry in quantum field theory of the weak interaction.

To obtain the requested decay formula, use naive factorization of a nonleptonic meson decay: approximate the four-quark matrix element by the product of the ${\bar{D}}^0\to K^{+}$ current matrix element and the vacuum-to-pion matrix element. This is an additional hadronic approximation; tree-level weak vertices alone do not establish it, and nonfactorizable Quantum chromodynamics effects can change the result. The vacuum-to-pion vector current matrix element vanishes by parity symmetry in quantum field theory, while the specified pion decay constant normalization givesThe $\sqrt{2}$ cancels the $1/\sqrt{2}$ in the four-fermion interaction. Up to an irrelevant overall phase, the scattering amplitude is thereforeFor $m_{\pi }=0$, $q^2=0$, so only $f_{+}(0)$ survives. Integrating the two-body decay phase space in the D meson rest frame givesHenceThis coefficient uses exactly the stated pion decay constant convention and the stated massless-pion approximation.

Use the Papkovich–Neuber representation in the formwhere $\mathbf{\Phi }$ and $\chi$ are harmonic functions. A torque is a axial vector, so rotational covariance and decay select the harmonic vector fieldHere $\mathbf{x}\cdot \mathbf{\Phi }=0$ and $\nabla \cdot \mathbf{\Phi }=0$, soThis rotlet equals the rotating sphere in Stokes flowwhen $\mathbf{\Omega }=\mathbf{G}/(8\pi \mu a^3)$. It satisfies the no-slip boundary condition on $r=a$, decays at infinity, and its Newtonian fluid stress tensor transmits the applied couple $\mathbf{G}$.

The body is invariant under the reflections and half-turns that preserve its $z$-axis. A polar vector force $F{\mathbf{e}}_z$ can therefore produce only the polar velocity $U_z{\mathbf{e}}_z$; every rotation is excluded because angular velocity is a pseudovector. Hence only $U_z$ is nonzero when $\mathbf{F}=(0,0,F)$ and $\mathbf{G}=0$.

For $\mathbf{F}=(F,0,0)$, the surviving symmetry-allowed components arewhile $U_y,U_z,{\Omega }_x,{\Omega }_z$ vanish. The coupling between translation in $x$ and rotation about $y$ is permitted because the two horizontal rods lie at opposite vertical offsets.