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For an ordinary scalar field, its derivative is a polar covector, while the axial current is a pseudovector. Their contraction is invariant under the proper orthochronous Lorentz group but changes sign under parity. If the field is a pseudoscalar, the extra transformation sign makes this coupling parity even. In four spacetime dimensions its mass dimension is five, so Lorentz invariance alone does not imply power-counting renormalizability.

Identify Minkowski vectors with Hermitian matrices $X=x^{0}I+x^a{\sigma }_a$. The action $X\mapsto SX{S}^{†}$ preserves $\det X=(x^0{)}^2-|\mathbf{x}{|}^2$. It gives a surjective homomorphism from $SL(2,\mathbb{C})$ to the proper orthochronous Lorentz group with kernel $\{I,-I\}$. Polar decomposition retracts the covering group onto SU(2), making it simply connected. This cover permits half-integer spin group representations that do not descend to the Lorentz group itself.

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.

Identify a Minkowski vector $(t,x,y,z)$ with the Hermitian matrix representation of Minkowski four-vectorsFor $A\in SL(2,\mathbb{C})$, the real-linear action $X\mapsto AX{A}^{†}$ preserves Hermiticity and the determinant, so it preserves the Lorentz quadratic form. It also preserves the cone of positive-definite Hermitian matrices, hence the future timelike cone. The group $SL(2,\mathbb{C})$ is connected, by polar decomposition of an invertible complex matrix into $SU(2)$ and positive determinant-one Hermitian matrices. Thus its image lies in the proper orthochronous Lorentz group.

If $AX{A}^{†}=X$ for every Hermitian $X$, taking $X=I$ makes $A$ unitary; the remaining equations then say it commutes with every Hermitian matrix. Their real span is all Hermitian matrices and their complex span is all complex matrices, so $A$ is scalar. The determinant condition leaves precisely $A=\pm I$. A discrete kernel gives an injective differential, and the real dimensions of both groups are six. The image is therefore open and equals the connected Lorentz group:The quotient does not include the disconnected time-reversing component of $SO(3,1)$.

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 signature $\eta =\mathrm{diag}(1,-1,-1,-1)$, the numerical matrices ${\sigma }_{\mu }=(I,{\sigma }_i)$ and ${\bar{\sigma }}_{\mu }=(I,-{\sigma }_i)$ given in the question, and raise genuine tensor indices with $\eta$. A real Minkowski spacetime vector corresponds to the Hermitian matrixFor SL(2,C) matrices, $X^{\prime }=AX{A}^{†}$ is Hermitian and has the same determinant. It therefore induces a real linear Lorentz transformation; composition of the congruence actions agrees with matrix multiplication. The kernel consists of $\pm I$: preservation of $X=I$ makes a kernel element unitary, and preservation of every Hermitian $X$ makes it scalar.

Polar decomposition of an invertible complex matrix connects every determinant-one complex matrix to its unitary factor, so the group is connected. The action gives the Lorentz spinor double cover of the proper orthochronous Lorentz group. It covers rotations via $A_R=e^{-i\vartheta \mathbf{n}\cdot \mathbf{\sigma }/2}$ and boosts via the positive Hermitian matrices below. Every proper orthochronous transformation is a boost followed by a rotation, since one can first match its image of the future unit time vector and then use its rotation stabilizer. The congruence construction does not cover spatial parity or time reversal. In particular parity has determinant $-1$ as a four-vector transformation; all transformations continuously produced from SL(2,C) have determinant $+1$.

Set $c_{\theta }=\cosh (\theta /2)$, $s_{\theta }=\sinh (\theta /2)$ and $N=\mathbf{n}\cdot \mathbf{\sigma }$. The Pauli matrix multiplication law gives $N^2=I$. The proposed boost matrix is $A_B=c_{\theta }I+s_{\theta }N$, with eigenvalues $e^{\pm \theta /2}$, determinant one and $A_B^{†}=A_B$. Split $\mathbf{x}=x_{\parallel }\mathbf{n}+{\mathbf{x}}_{\perp }$. The perpendicular Pauli part anticommutes with $N$, so multiplication givesThis is an active Lorentz boost. The image of a rest worldline has velocity $\mathbf{v}=\tanh \theta \mathbf{n}$, fixing the velocity sign convention.

For two nonzero boosts, write $A^{\prime }=c^{\prime }I+s^{\prime }{\mathbf{n}}^{\prime }\cdot \mathbf{\sigma }$ and $A=cI+s\mathbf{n}\cdot \mathbf{\sigma }$. Their product isThe last term is anti-Hermitian. Thus $A^{\prime }A$ is not Hermitian unless the axes are collinear; a pure boost has Hermitian lifts $\pm A_B$, so it cannot give the same Lorentz transformation. The noncollinear boost obstruction from Pauli products is thereforeThat rotation is a Wigner rotation. A zero-rapidity factor is the trivial exception, regardless of the arbitrary axis assigned to it.

Define $J_i=\frac{1}{2}{ϵ}_{ijk}{M}_{jk}$ and $K_i=M_{0i}$. The printed Lorentz algebra brackets giveSet $L_i=(J_i+i{K}_i)/2$, $R_i=(J_i-i{K}_i)/2$. ThenThis is the chiral decomposition of the complex Lorentz algebra. The two copies are the complexifications of the SU(2) algebras conventionally labelled left and right. They are not two independent compact real subalgebras of the real Lorentz algebra: $L,R$ are complex linear combinations of the real generators. The distinction is needed for noncompact boosts.

To verify the infinitesimal two-component action, putIts trace vanishes by antisymmetry, so $\det A=1+O({\omega }^2)$. The Pauli matrix multiplication law impliesSince $a^{†}=\frac{1}{4}{\omega }^{\mu \nu }{\bar{\sigma }}_{\nu }{\sigma }_{\mu }$, antisymmetrizing this identity yieldsConsequently $AX{A}^{†}=X+{\sigma }_{\mu }{\omega }^{\mu }{}_{\nu }{x}^{\nu }+O({\omega }^2)$, exactly the claimed infinitesimal coordinate transformation with ${\omega }^{\mu }{}_{\nu }={\omega }^{\mu \rho }{\eta }_{\rho \nu }$.

The two Weyl spinor representations transform asBoth maps preserve group multiplication. For $\Lambda =e^{\omega }$, choose its spin lift and write the finite matrices asFor rotations these coincide as the SU(2) doublet; for boosts their generator signs are opposite. Any complex-linear intertwiner would commute with all rotations and hence be scalar by the Schur lemma, but a nonzero scalar cannot intertwine the opposite boost matrices. They are therefore inequivalent. Infinitesimally their $K_i$ matrices are $-i{\sigma }_i/2$ and $+i{\sigma }_i/2$, while their $J_i$ matrices are both ${\sigma }_i/2$. Thus they have chiral labels $(1/2,0)$ and $(0,1/2)$. The finite matrices are representations of the spin cover; choosing $A$ or $-A$ matters for spinors even though their four-vector transformations coincide.

Using the supplied block gamma matrices, the generator of the Dirac spinor isThe antisymmetry of $\omega$ removes the symmetric Clifford part. Hence

There is a conjugation-order error in the last displayed gamma identity in the PDF. The Clifford algebra gives, with ${\gamma }^{\rho }={\eta }^{\rho \sigma }{\gamma }_{\sigma }$,Exponentiating this linear commutator action proves the inverse Lorentz action on gamma matricesThe second identity is the order required for the requested bilinears when $\psi \mapsto S\psi$. An explicit countercheck to the printed order is a positive boost along the third axis: ${\omega }^{03}=-\theta$ giveswhereas the printed right side has a plus sign. This cannot be repaired by dropping index raising; the same raising convention is needed in the preceding coordinate transformation.

The adjoint relation supplied in the question gives ${\Omega }^{†}{\gamma }^0=-{\gamma }^0\Omega$, hence the Dirac spinor pseudo-unitarity identity $S^{†}{\gamma }^{0}S={\gamma }^0$. The Dirac adjoint therefore transforms as $\bar{\psi }\mapsto \bar{\psi }{S}^{-1}$. It now follows that the Dirac scalar bilinear isthe vector isand the antisymmetric second-rank tensor isThese are respectively a Lorentz scalar, Lorentz four-vector and Lorentz tensor. The source's inconsistent gamma identity is replaced by its correct inverse/order pair; all three transformation laws then follow with the stated spinor transformation.

The proper orthochronous Lorentz group is the connected Lorentz group in the question. Its double cover is $SL(2,\mathbb{C})$, viewed as a real Lie group. Identify a spacetime vector with the Hermitian matrixThe action $X\mapsto SX{S}^{†}$ for $S\in SL(2,\mathbb{C})$ preserves this determinant and hence the Minkowski metric. The group is connected, so its image is proper and orthochronous. Its kernel consists of $\pm I$: a matrix in the kernel first preserves $I$, hence is unitary, and then commutes with every Hermitian matrix, so is scalar; determinant one forces the two signs. Matrices in $SU(2)$ generate spatial rotations, and positive Hermitian determinant-one matrices generate boosts. Rotations and boosts generate the connected Lorentz group, so the action is onto. Polar decomposition also gives $SL(2,\mathbb{C})\cong SU(2)\times {\mathbb{R}}^3$ as a manifold, proving that it is simply connected. This establishes the Lorentz spinor double cover.

In an anti-Hermitian rotation-generator convention, the Lorentz algebra brackets areThe negative sign in the last bracket distinguishes boosts from Euclidean four-dimensional rotations. After complexification, setA direct bracket calculation gives $[A_i,A_j]={ϵ}_{ijk}{A}_k$, $[B_i,B_j]={ϵ}_{ijk}{B}_k$ and $[A_i,B_j]=0$. Thus the chiral decomposition of the complex Lorentz algebra isComplexification matters: the real Lorentz algebra is not the compact real algebra $\mathfrak{su}(2)\oplus \mathfrak{su}(2)$.

Each spin-$j$ homogeneous polynomial representation of SU2 extends from $SU(2)$ to $SL(2,\mathbb{C})$ as $D_j(S)={\mathrm{Sym}}^{2j}S$. Its complex-conjugate extension uses $\overline{S}$. The finite-dimensional irreducible complex group representations of the covering group are thereforeThe two separate complexified Lie-algebra factors act irreducibly on the two spin spaces, so their tensor product is irreducible. Conversely an invariant complex subspace for the real group is invariant under its complexified Lie algebra; the highest-weight classification for the two ${\mathfrak{sl}}_2$ factors gives exactly these tensor products. This constructs all finite-dimensional complex Lorentz representations.

Again $-I\in SL(2,\mathbb{C})$ acts by $(-1{)}^{2j_L+2j_R}$. Therefore the irreducible representations of the connected Lorentz group itself, in this finite-dimensional complex category, areIf the sum is a half-integer, the group representation is a group representation of the spin cover, or a projective group representation of the Lorentz group, and is not an ordinary single-valued group representation of the group named in the question.

The scalar $(0,0)$ and four-vector $(1/2,1/2)$ descend. The left and right Weyl spinors, $(1/2,0)$ and $(0,1/2)$, do not. Their direct sum is a Dirac spinor, reducible under the connected group; parity exchanges its two chiral summands. The group representations $(1,0)$ and $(0,1)$ describe the two complex chiral parts of an antisymmetric tensor. On the rotation subgroup, self-duality of $SU(2)$ irreducibles identifies the conjugate spin space with the usual spin space, so the Clebsch-Gordan decomposition for SU2 gives the same spin range $|j_L-j_R|,\dots ,j_L+j_R$ as in part (i).

These are group representations used for fields, and they need not be unitary for a positive-definite inner product. Indeed no nontrivial finite-dimensional group representation of this group is unitary: if it were, its differential would embed the simple real Lorentz algebra into an algebra of skew-Hermitian matrices. The trace form $-\mathrm{tr}(XY)$ would give an invariant positive-definite form $b$ on that algebra. Invariance and the boost brackets would then force $-b(J_3,J_3)=b([K_1,K_2],J_3)=b(K_1,[K_2,J_3])=b(K_1,K_1)$, impossible for positive-definite $b$. Equivalently nontrivial boosts in these polynomial group representations have real exponential rather than phase eigenvalues.

The qualification about dimension is necessary because a noncompact group also has infinite-dimensional unitary group representations. They too can be built using $SU(2)$ spin spaces, but not by a single finite pair. For example, the rotation content of induced Lorentz representations is obtained from normalized induced representations of the upper triangular subgroup of $SL(2,\mathbb{C})$, with $n\in \mathbb{Z}$, $\rho \in \mathbb{R}$ and unitary characters $(a/|a|{)}^n|a{|}^{2i\rho }$ on its diagonal $(a,a^{-1})$, trivial on its unipotent part. In the compact picture this uses functions on $SU(2)$ satisfyingExpanding functions into $SU(2)$ matrix coefficients selects a single right-torus weight from each spin space, giving the rotation contentNormalized induction supplies the boost action, coupling these infinitely many rotation spaces. The central sign is $(-1{)}^n$, so the even-$n$ family descends to the connected Lorentz group and has integer rotation spins. This explains both the finite-dimensional field construction and why a classification of unitary group representations cannot simply be identified with the finite two-spin labels.

For the stated matrix exponential parametrization in the connected proper orthochronous Lorentz group, use the corresponding Spinor representation of the Lorentz group:The spacetime argument is inverse-transformed because this is an active transformation of the field at a fixed coordinate $x$. Equivalently, with $x^{\prime }=\Lambda x$, ${\psi }^{\prime }(x^{\prime })=D(\Lambda )\psi (x)$. The identity $[S^{\mu \nu },{\gamma }^{\rho }]={\eta }^{\nu \rho }{\gamma }^{\mu }-{\eta }^{\mu \rho }{\gamma }^{\nu }$ integrates to $D^{-1}{\gamma }^{\mu }D={\Lambda }^{\mu }{}_{\nu }{\gamma }^{\nu }$, which ensures covariance of the Dirac equation.

Strictly, $D$ depends on a lift to the Spin group, not just on the final Lorentz transformation: two lifts differ by sign. The specified generator exponential chooses a lift along its continuous path from the identity. This qualification is essential for the full-rotation comparison in part (g).

With $\eta =\mathrm{diag}(1,-1,-1)$, the defining condition for the $(1+2)$-dimensional Lorentz group isTaking determinants gives $(\det \Lambda {)}^2=1$, hence $\det \Lambda =\pm 1$. The norm of the zeroth column givesso $|{\Lambda }^0{}_0|\ge 1$. The Proper orthochronous Lorentz group is the component with $\det \Lambda =1$ and ${\Lambda }^0{}_0\ge 1$.