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Complex two-by-two matrices of determinant one form a complex three-dimensional Lie group, or a real six-dimensional Lie group. Their congruence action on Hermitian two-by-two matrices gives the Lorentz spinor double cover with kernel $\{I,-I\}$. The two Weyl spinor representations are $A$ and $(A^{†}{)}^{-1}$. Treating the group as real is essential for the second, antiholomorphic representation.

Use the Lorentz spinor double cover $SL(2,\mathbb{C})\to S{O}^{+}(1,3)$. For a lift with distinct eigenvalues $\lambda ,{\lambda }^{-1}$, diagonalize and choose logarithms $z,-z$; this gives a traceless matrix logarithm. For a repeated eigenvalue, choose the sign of the lift so it is $I+N$, where $N^2=0$, and use $\log (I+N)=N$. The two lifts have the same Lorentz image. Exponentiating the differential of the covering map therefore expresses every proper orthochronous transformation as the exponential of a real element of the Lie algebra of the Lorentz group. This argument does not assert that every element of $SL(2,\mathbb{C})$ itself has a traceless logarithm.

The complexified Lorentz algebra splits into two commuting ${\mathfrak{sl}}_2$ factors. Extending two homogeneous polynomial representations of SU2 therefore gives every finite-dimensional irreducible complex group representation of the Lorentz spinor double cover. The kernel sign permits descent precisely when $j_L+j_R$ is an integer. These field group representations are generally nonunitary for a positive-definite inner product; infinite-dimensional unitary Lorentz group representations belong to a different construction.

Fix $m>0$, $k=(m,0)$, and a rotationless Lorentz boost $L(p)$ taking $k$ to $p$ on the positive-energy mass shell. The little group on the Lorentz spinor double cover is SU(2). Its spin-$s$ irreducible representation induces a unitary representation on $L^2(d^{3}p/(2p^0))\otimes {\mathbb{C}}^{2s+1}$. Here $W(\Lambda ,p)=L(\Lambda p{)}^{-1}\Lambda L(p)$ is the Wigner rotation, lifted to SU(2) when necessary, and the translation convention is $U(I,a)|p⟩=e^{ip\cdot a}|p⟩$. The little-group cocycle proves the group law, and Wigner's classification gives irreducibility. Every nonnegative integer or half-integer spin is allowed on the double cover; descent to the ordinary proper orthochronous Poincaré group requires integer spin, because its $2\pi$ rotation must act as the identity.

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.

A two-component Weyl spinor is a complex spinor transforming in one of the two inequivalent group representations $(1/2,0)$ and $(0,1/2)$ of the Lorentz spinor double cover $\mathrm{SL}(2,\mathbb{C})$. An undotted left-handed field ${\chi }_{\alpha }$ transforms by a matrix $M\in \mathrm{SL}(2,\mathbb{C})$; its complex conjugate ${\bar{\chi }}_{\dot{\alpha }}$ transforms by $M^{\ast }$. Dotted and undotted indices distinguish the two chirality representations. Raising a dotted index uses the invariant epsilon tensor and converts the complex conjugate transformation into the equivalent inverse-adjoint form. Rotations act by the same SU(2) generators on the two chiralities, while boosts have opposite generators, so the two complex representations cannot be identified by a complex-linear Lorentz invariant map.

The antisymmetric invariant tensors ${ϵ}_{\alpha \beta }$ and ${ϵ}_{\dot{\alpha }\dot{\beta }}$ raise and lower spinor indices; $M^TϵM=ϵ$ proves invariance of the contracted bilinear $\chi \psi ={\chi }^{\alpha }{\psi }_{\alpha }$. The components of a fermionic Weyl spinor are odd Grassmann variables, so this bilinear is symmetric under exchange of the two spinors: the antisymmetric epsilon sign cancels the Grassmann sign. In particular, choosing ${ϵ}_{12}=1$ and ${ϵ}^{12}=-1$ gives $\chi \chi =2{\chi }_1{\chi }_2$, generally nonzero. This Weyl spinor bilinear exchange identity permits a Majorana mass term for a single suitable Weyl spinor. With mostly-minus metric signature and ${\bar{\sigma }}^{\mu }=(I,-\mathbf{\sigma })$, the free massless kinetic term is $i{\chi }^{†}{\bar{\sigma }}^{\mu }{\partial }_{\mu }\chi$, and its field equation of motion is $\boxed{{\bar{\sigma }}^{\mu }{\partial }_{\mu }\chi =0}$.

Use mostly-minus metric signature, $\eta =\mathrm{diag}(1,-1,-1,-1)$. The Weyl sigma matrices are ${\sigma }^{\mu }=(I,\mathbf{\sigma })$ and ${\bar{\sigma }}^{\mu }=(I,-\mathbf{\sigma })$, with the three Pauli matricesThe Pauli matrix multiplication law ${\sigma }_i{\sigma }_j={\delta }_{ij}I+i{ϵ}_{ijk}{\sigma }_k$ immediately gives ${\sigma }^{\mu }{\bar{\sigma }}^{\nu }+{\sigma }^{\nu }{\bar{\sigma }}^{\mu }=2{\eta }^{\mu \nu }I$: the $00$ case is $2I$, the mixed cases cancel, and the spatial cases give $-2{\delta }_{ij}I$. The spinor indices of ${\sigma }_{\alpha \dot{\beta }}^{\mu }$ express the correspondence between a Lorentz four-vector and a Hermitian $2\times 2$ matrix. For $X=x^{0}I+x^i{\sigma }_i$, $\det X=(x^0{)}^2-|\mathbf{x}{|}^2$; $X\mapsto MX{M}^{†}$ preserves this determinant for $M\in \mathrm{SL}(2,\mathbb{C})$, yielding the Lorentz spinor double cover.

In four-dimensional N=1 supersymmetry, the odd supercharges $Q_{\alpha }$ and ${\bar{Q}}_{\dot{\alpha }}=Q_{\alpha }^{†}$ transform as conjugate Weyl spinors. Their Super-Poincaré algebra isThe ordinary translation and Lorentz generators satisfy the Poincare algebra, and the Lorentz generators act on the supercharges by the corresponding group representations on Weyl spinors. Tracing the mixed anticommutator gives $P_0=\frac{1}{4}{\sum }_{\alpha }\{Q_{\alpha },Q_{\alpha }^{†}\}$, proving energy positivity in global supersymmetry. For a massive rest-frame state, $P_{\mu }=(M,0,0,0)$, the rescaled charges $Q_{\alpha }/\sqrt{2M}$ obey the canonical anticommutation relations. Acting with their adjoints builds a supermultiplet containing states of alternating fermion parity.

Use units $c=\hslash =1$ and the Minkowski metric $g=\mathrm{diag}(1,-1,-1,-1)$. Composing two affine Poincaré group transformations, with the rightmost applied first, givesThe translations form a normal subgroup, so this is the semidirect product of the Lorentz group with ${\mathbb{R}}^{1,3}$. The transformations $(\mathrm{diag}(1,R),0)$, $R\in SO(3)$, fix time and the spatial origin. Their products and inverses correspond to $R{R}^{\prime }$ and $R^{-1}$, hence they form a subgroup isomorphic to SO(3).

A rotationless Lorentz boost changes the velocity of a frame without adding a spatial rotation. For $|\mathbf{v}|<1$ a convenient active convention iswith $B(0)=I$. It satisfies $B^{T}gB=g$, $\det B=1$, and $B(m,0)=(m\gamma ,m\gamma \mathbf{v})$. Collinear boosts form a one-parameter subgroup with additive rapidity, but general boosts do not. For nonzero boosts along $x$ and $y$, $(B_x{B}_y{)}_{02}={\gamma }_x{\gamma }_y{v}_y$ while $(B_x{B}_y{)}_{20}={\gamma }_y{v}_y$: their product is not symmetric, whereas every rotationless boost above is symmetric. The product contains a Wigner rotation. Infinitesimally, the Poincare algebra gives $[K_i,K_j]=-i{ϵ}_{ijk}{J}_k$ for Hermitian generators, so the boost generators do not close amongst themselves.

There is a group-versus-cover qualification in the spin range: an honest unitary representation of the proper orthochronous Poincaré group permits integer spin only. Half-integer spin requires its double cover $SL(2,\mathbb{C})⋉{\mathbb{R}}^{1,3}$, or a projective unitary representation of the ordinary group. The construction below gives the requested range on that double cover and descends for integer spin.

Fix $m>0$ and the rest momentum $k=(m,0)$. Its little group consists of rotations: if a proper orthochronous Lorentz transformation fixes $k$, its time column is $(1,0)$, the metric condition forces the corresponding time row, and the remaining block belongs to SO(3). On the Lorentz spinor double cover, this little group is SU(2). Choose its irreducible spin-$s$ representation, with rest-frame states $|k,s,\sigma ⟩$, $\sigma =-s,-s+1,\dots ,s$, and matrices $D^{(s)}$, the Wigner D-matrices.

For each $p$ on the forward mass shell, $p^2=m^2$, $p^0>0$, choose the standard rotationless Lorentz boost $L(p)=B(\mathbf{p}/p^0)$ and defineThe labels are canonical spin projections, defined by this choice of boosts. For any proper orthochronous $\Lambda$, factorThe last factor fixes $k$ and hence is a Wigner rotation. On the double cover, use its lift to SU(2). Taking the translation convention $U[I,a]|p,s,\sigma ⟩=e^{ip\cdot a}|p,s,\sigma ⟩$ yields the massive induced representation of the Poincare double cover:The argument of the translation phase is $\Lambda p$, since $U[\Lambda ,a]=U[I,a]U[\Lambda ,0]$. The Wigner rotation identityproves the Lorentz part of the representation law; the phases multiply according to the affine product already derived.

Normalize the generalized momentum states by $⟨p,s,\sigma |q,s,\tau ⟩=2p^0(2\pi {)}^3{\delta }^3(\mathbf{p}-\mathbf{q}){\delta }_{\sigma \tau }$. The invariant mass shell measure $d^{3}p/(2p^0)$ and unitarity of $D^{(s)}$ make the induced representation unitary on $L^2(d^{3}p/(2p^0))\otimes {\mathbb{C}}^{2s+1}$. This covariant normalization accounts for the absence of an energy square-root factor in the displayed transformation law. To see irreducibility, translations force an operator in the commutant to act within momentum fibres; the little-group action and Schur lemma make its action scalar on each irreducible spin fibre. Lorentz transitivity on the forward mass shell makes this scalar constant. The commutant is therefore scalar, so the unitary representation is irreducible. Conversely, the simultaneous translation spectrum and its little-group representation give the massive positive-energy sector of Wigner's classification.

Thus every $m>0$ and $s\in \{0,\frac{1}{2},1,\dots \}$ gives such an irreducible representation on the double cover. Its central element over a $2\pi$ spatial rotation acts as $(-1{)}^{2s}$. Since this rotation is the identity in the ordinary Poincaré group,

For the parity operator, put $\Pi =\mathrm{diag}(1,-1,-1,-1)$ and $\bar{p}=\Pi p=(p^0,-\mathbf{p})$. The extension obeys $PU[\Lambda ,a]P^{-1}=U[\Pi \Lambda \Pi ,\Pi a]$. At rest, $P$ commutes with the rotations; Schur lemma therefore gives $P|k,s,\sigma ⟩={\eta }_P|k,s,\sigma ⟩$ with a common phase $|{\eta }_P|=1$. Since $\Pi L(p)\Pi =L(\bar{p})$, the parity action on canonical massive spin states isFor the usual reflection convention $P^2=I$, ${\eta }_P=\pm 1$ is the intrinsic parity. Canonical spin is an axial quantity and its label $\sigma$ is unchanged; a helicity label instead changes sign because momentum is reversed. Either intrinsic parity extends a given massive spin representation.

First use the actual PDF boost bracket $[K_i,K_j]=-i{ϵ}_{ijk}{J}_k$; replacing the final $J_k$ by $K_k$, as the TeX transcription does, would be a different and incorrect algebra. Expanding the three brackets yieldsThusThe two sets have SU(2)-type brackets. Their compact real form exponentiates to $SU(2)\times SU(2)$, whereas their full complexification exponentiates to $SL(2,\mathbb{C})\times SL(2,\mathbb{C})$. This does not turn the physical real Lorentz group into a compact product. With Hermitian $J,K$, $A^{†}=B$, and the two factors are tied by the Lorentz reality condition; the connected physical spin cover is $SL(2,\mathbb{C})$.

For the right-handed spinor representation, the supplied sigma formula gives ${\bar{\sigma }}^{ij}={ϵ}_{ijk}{\sigma }_k/2$ and ${\bar{\sigma }}^{0i}=i{\sigma }_i/2$. ThereforeIt is the $(0,1/2)$ representation in this definition of $A,B$. These are finite-dimensional representation matrices; the boosts are not Hermitian for a positive-definite spinor inner product, as expected for a noncompact group. They are distinct from the operator-component commutator coefficients discussed in Question 1.

For the group map, encode a real Minkowski vector by the Hermitian matrixThe trace identity $\frac{1}{2}\mathrm{Tr}({\bar{\sigma }}^{\mu }{\sigma }_{\nu })={\delta }^{\mu }{}_{\nu }$ recovers its contravariant components. For $N\in SL(2,\mathbb{C})$, $X^{\prime }=NX{N}^{†}$ is Hermitian and has the same determinant. Expanding $X^{\prime }$ in that basis givesThe coefficients are real because the trace of a product of Hermitian matrices is real. Equality of determinants for all $x$ proves ${\Lambda }^T\eta \Lambda =\eta$. The action for $N_1{N}_2$ is the composition of the two actions, proving the homomorphism law.

Since $SL(2,\mathbb{C})$ is connected and the identity maps to the identity, its image has determinant $+1$ and preserves time orientation. More directly, positive-definite matrices representing future timelike vectors remain positive definite under $X\mapsto NX{N}^{†}$. The kernel is $\{I,-I\}$: if every Hermitian $X$ is fixed, $X=I$ first makes $N$ unitary, and then commutation with all Hermitian matrices makes it a scalar with determinant one.

Conversely, $SU(2)$ gives every spatial rotation, and positive Hermitian determinant-one matrices $\exp (\mathbf{b}\cdot \mathbf{\sigma }/2)$ give every pure boost. Rotation-boost decomposition yields every proper orthochronous Lorentz transformation. Thus the formula gives the Lorentz spinor double coverIt is a map into $SO(1,3)$ as asked, with precisely its identity component as image; it does not produce disconnected time-reversing transformations.

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.

Write $u=(a,b{)}^T$. The Pauli matrices giveHenceTherefore $\boxed{U_{\mu }{U}^{\mu }=0}$, with $U^0>0$ for $u\ne 0$: the bilinear is a future null Lorentz four-vector, and $u=0$ gives the zero vector.

For $S=e^{\rho {\sigma }_z/2}=\mathrm{diag}(e^{\rho /2},e^{-\rho /2})$, the transformed components are $a^{\prime }=e^{\rho /2}a$ and $b^{\prime }=e^{-\rho /2}b$. Thus $a^{\prime \ast }{b}^{\prime }=a^{\ast }b$, while the two squared magnitudes acquire factors $e^{\rho }$ and $e^{-\rho }$. Substitution givesThe matrix preserves the Minkowski metric and is an active boost along $+z$ of rapidity $\rho$, with velocity $c\tanh \rho$. Relative to the passive standard-boost convention of part i, its velocity parameter has the opposite sign.

For $S=e^{i\theta {\sigma }_z}$, $a^{\prime }=e^{i\theta }a$ and $b^{\prime }=e^{-i\theta }b$, so $a^{\prime \ast }{b}^{\prime }=e^{-2i\theta }{a}^{\ast }b$. ConsequentlyThis is an active right-handed spatial rotation through $-2\theta$ about $z$. The minus sign follows from the specified positive $i\theta {\sigma }_z$ exponent; the familiar positive-angle spin rotation uses $e^{-i\phi {\sigma }_z/2}$. The doubled angle exhibits the Lorentz spinor double cover: $u$ and $-u$ give the same four-vector.