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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.

Use metric signature $+---$ and the parity matrix $P^{\mu }{}_{\nu }=\mathrm{diag}(1,-1,-1,-1)$. If a Dirac field transforms as ${\psi }^{\prime }(x)={\eta }_P{S}_P\psi (x_P)$, the chain rule changes the spatial derivatives' signs. Covariance of its Dirac equation therefore requiresThe Clifford algebra gives precisely these identities for $S_P={\gamma }^0$. In an irreducible Dirac representation any other solution differs by a scalar phase, because its ratio with ${\gamma }^0$ commutes with all gamma matrices. With $|{\eta }_P|=1$, the Dirac mass bilinear and kinetic term consequently obeyThe factor ${\gamma }^0$ compensates the sign reversal of spatial derivatives. A transformation containing only the argument change would not preserve the massive Dirac equation. This establishes parity symmetry of the free or vector-coupled Dirac theory; it does not make a chiral weak interaction parity symmetric.

For charge conjugation, transpose the adjoint Dirac equation to obtain $(i{\gamma }^{\mu T}{\partial }_{\mu }+m){\bar{\psi }}^T=0$. Multiplying by $C$ gives the original Dirac equation for ${\psi }^c=C{\bar{\psi }}^T$ ifOne may choose $C$ unitary; in four dimensions a conventional choice in a standard Dirac or chiral basis is $C=i{\gamma }^2{\gamma }^0$, with $C^T=-C$. Its overall phase and ${\eta }_C$ do not affect the following bilinear transformation. Reordering the Grassmann fermion fields gives, for a color matrix $X$ commuting with the gamma matrices,Thus a vector color current changes sign and transposes its color generator under charge conjugation.

Write the Hermitian color connection as ${\mathcal{A}}_{\mu }=A_{\mu }^a{T}^a$. The interaction coming from $D_{\mu }={\partial }_{\mu }+ig{\mathcal{A}}_{\mu }$ is $-g\bar{\psi }{\gamma }^{\mu }{\mathcal{A}}_{\mu }\psi$. Its charge-conjugation symmetry requires ${\mathcal{A}}_{\mu }^C=-{\mathcal{A}}_{\mu }^T$. Parity acts on the Lorentz index as on a vector connection, so combining the two givesIn components, ${\mathcal{A}}_0\mapsto -{\mathcal{A}}_0^T(x_P)$ and ${\mathcal{A}}_i\mapsto +{\mathcal{A}}_i^T(x_P)$. The transpose is a color-space transpose, independent of the spinor matrix $C$. Intrinsic fermion phases cancel between a field and its adjoint.

Define the matrix gauge field strength unambiguously byThe derivative terms transform with the expected parity factors and a minus transpose. For the commutator, the crucial identity is $[X^T,Y^T]=-[X,Y{]}^T$, so the nonlinear term has the same transformation as the derivative terms. This is the CP transformation of non-Abelian field strength:For example, ${\mathcal{F}}_{0i}\mapsto +{\mathcal{F}}_{0i}^T(x_P)$ while ${\mathcal{F}}_{ij}\mapsto -{\mathcal{F}}_{ij}^T(x_P)$. With $\mathrm{tr}(T^a{T}^b)=\frac{1}{2}{\delta }^{ab}$, the color sum in the theta operator is twice a trace of two field strengths. The two charge-conjugation minus signs cancel, and transposition reverses the trace factors without changing their trace. The four parity matrices contracted with the Levi-Civita symbol contribute $\det P=-1$. HenceThe theta operator is CP odd; a generic fixed nonzero coefficient violates CP. Setting the coefficient to zero removes this term. Although the density is locally a total derivative, nontrivial gauge topology means that it need not be irrelevant to the quantum theory. If a properly normalized theta angle is identified periodically, invariance at special values equivalent to their negatives is a separate global question; it does not change the CP odd transformation of the local operator.