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A central charge in four-dimensional extended supersymmetry is a Lorentz scalar even generator appearing in $\{Q_{\alpha }^A,Q_{\beta }^B\}={ϵ}_{\alpha \beta }{Z}^{AB}$. It commutes with the displayed translation, Lorentz algebra, and supercharge generators. An added R-symmetry automorphism can act on the central charges.

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.

The Clifford algebra gives $S^{\mu \nu }=\frac{1}{2}{\gamma }^{\mu }{\gamma }^{\nu }-\frac{1}{2}{\eta }^{\mu \nu }I$. Reordering one more gamma matrix gives $[S^{\mu \nu },{\gamma }^{\rho }]={\eta }^{\nu \rho }{\gamma }^{\mu }-{\eta }^{\mu \rho }{\gamma }^{\nu }$. The commutator derivation identity then givesso these matrices represent the Lorentz algebra. This convention has no extra factor of $i$; the spatial generators are $-\frac{i}{2}{ϵ}^{jkl}\mathrm{diag}({\sigma }^l,{\sigma }^l)$.

In a real-generator convention, $M^{\rho \sigma }=\frac{1}{4}[{\gamma }^{\rho },{\gamma }^{\sigma }]$ obeys $[M^{\rho \sigma },{\gamma }^{\mu }]=g^{\sigma \mu }{\gamma }^{\rho }-g^{\rho \mu }{\gamma }^{\sigma }$. Applying this identity to the commutator of two gamma matrices proves the Lorentz algebra relations. With the matching vector generators and real antisymmetric parameters, $S=\exp (\frac{1}{2}{\Omega }_{\rho \sigma }{M}^{\rho \sigma })$ obeys $S^{-1}{\gamma }^{\mu }S={\Lambda }^{\mu }{}_{\nu }{\gamma }^{\nu }$. A convention using Hermitian Lorentz generators instead must consistently insert factors of $i$.

Take ${\partial }_{\pm }=({\partial }_{\tau }\pm {\partial }_{\sigma })/2$ on the strip $0\le \sigma \le \pi$. Varying the Grassmann-odd worldsheet Majorana fermions gives ${\partial }_{-}{\psi }_1=0$, ${\partial }_{+}{\psi }_2=0$ and, up to the orientation of the strip boundary, the surface termAt each endpoint it vanishes if ${\psi }_1=\eta {\psi }_2$ with $\eta =\pm 1$, with the same relation on variations. A common sign can be absorbed by redefining one chiral field, so set ${\eta }_0=+1$. Since ${\psi }_1=f(\tau +\sigma )$ and ${\psi }_2=f(\tau -\sigma )$, the second endpoint imposes $f(u+2\pi )={\eta }_{\pi }f(u)$. Equal endpoint signs give the periodic Ramond sector, and opposite endpoint signs give the antiperiodic Neveu–Schwarz sector:These gluing choices preserve the Neumann-endpoint worldsheet supersymmetry conditions and give integer or half-integer worldsheet Majorana fermion modes, respectively.

First fix a normalization issue in the original PDF. With its stated canonical string oscillator algebra, the standard superconformal current mode iswithout the printed factor $1/2$. To check the Ramond supercurrent zero-mode square directly, pair positive and negative integer modes:The zero-mode Clifford relations give $({\alpha }_0\cdot {\psi }_0{)}^2={\alpha }_0^2/2$. Different $Q_m$ anticommute, and so do $Q_m$ with the zero-mode term. The two same-sign terms in $Q_m^2$ vanish by antisymmetry of the fermions and symmetry of the bosonic product. The cross terms giveThe second term comes from commuting ${\alpha }_m$ through ${\alpha }_{-m}$, so the frequency factor $m$ is essential. ThereforeHere $L_0^R$ uses vacuum-annihilating Ramond normal ordering; its string intercept is zero. A plane-to-cylinder central shift must be included consistently if a different zero-mode convention is used.

For the mode $F_r=G_r/2$ actually printed, the algebra instead gives $\{F_0,F_0\}=L_0^R/2$. Already the oscillator-free term gives $\{F_0,F_0\}={\alpha }_0^2/4$, whereas $2L_0^R={\alpha }_0^2$. Thus the requested identity is not valid with both that prefactor and the supplied string oscillator algebra. Removing the prefactor gives the identity just derived. Renaming $L_0^R/4$ as $L_0$ would fix this isolated equality but would spoil the usual Virasoro normalization and the subsequent mass formula. We use the standard normalized modes $G_r$ in the remaining physical constraints; their vanishing is unaffected by a common nonzero rescaling.

The RNS string physical-state conditions in the two sectors areThe last condition is the fermionic zero-mode constraint; its square implies the Ramond mass-shell condition. Physical null string states are quotiented. Matter central charge is $3d/2$, while the reparameterization worldsheet ghost fields and superconformal ghosts contribute $-26+11=-15$. Thus $d=10$ is the critical dimension of the RNS superstring, with $a_{NS}=1/2$ and $a_R=0$.

To solve the constraints, choose $X^{\pm }=(X^0\pm X^9)/\sqrt{2}$, so ${\eta }_{+-}=-1$, and work on a patch with ${\alpha }_0^{+}=\sqrt{2{\alpha }^{\prime }}{p}^{+}\ne 0$. In light-cone gauge in string theory fix all nonzero ${\alpha }_n^{+}$ and all ${\psi }_r^{+}$ to zero. For the Ramond sector, the printed restriction to nonzero fermion modes is not by itself complete: one must also reduce the longitudinal fermionic zero modes using the zero-mode supersymmetry constraint. The usual reduced light-cone description sets the remaining ${\psi }_0^{+}$ to zero as well and retains the transverse Clifford algebra. Covariantly this reduction is the Dirac equation on the ground-state particle polarization; it is not an extra unconstrained canonical string oscillator.

With this reduction the string oscillator generators have the formRepeated $i$ runs over the eight transverse directions. The coefficient in $L_n^{\perp }$ comes from the fermion worldsheet stress tensor $-\psi \partial \psi /2$; for $n=0$, positive and negative $r$ combine to give the occupation energy $r{\psi }_{-r}{\psi }_r$. Thus the light-cone super-Virasoro constraints are explicitly solved byHere $a$ is $1/2$ in the NS sector and zero in the R sector. The zero-mode equation says $2p^{+}{p}^{-}=p_i^2+(N-a)/{\alpha }^{\prime }$. All longitudinal excitations are consequently determined by transverse bilinears. The critical dimension and string intercepts are necessary for the resulting quantum theory to preserve the target Lorentz algebra.

The frequency-weighted Neveu–Schwarz level operator isThe PDF's displayed number operator omits the fermion frequency weights and the $r=1/2$ term. With the stated anticommutators that cannot be a level operator: it would assign no level to the first fermionic excitation. The bosonic term already includes its frequency through $[{\alpha }_n,{\alpha }_{-n}]=n$, whereas the fermionic term must include its frequency explicitly.

One independent way to obtain the string intercept is the transverse zero-point energy. Each periodic boson contributes $\frac{1}{2}{\sum }_{n\ge 1}n=-1/24$. Each antiperiodic real fermion contributes $-\frac{1}{2}{\sum }_{r\ge 1/2}r=-1/48$, with the minus sign due to fermionic normal ordering. For example a common exponential cutoff givesSubtracting the local divergent terms leaves $8(-1/24-1/48)=-1/2$, so the NS string intercept is $a=1/2$. In the Ramond sector the integer fermions instead contribute $+1/24$ per component and cancel the bosonic contribution, giving $a_R=0$.

The unprojected NS sector oscillator vacuum has $N=0$ andIt is a spacetime scalar tachyon. The first excitation ${\psi }_{-1/2}^i|0;p⟩$ has $N=1/2$ and is massless, with eight vector particle polarizations. Covariantly its particle polarization obeys $p\cdot ϵ=0$ modulo the longitudinal null particle polarization, leaving the same eight degrees of freedom.

The R sector oscillator vacuum is also massless, but it is degenerate because $\{{\psi }_0^{\mu },{\psi }_0^{\nu }\}={\eta }^{\mu \nu }$. The operators ${\Gamma }^{\mu }=\sqrt{2}{\psi }_0^{\mu }$ give the target Clifford algebra, and $G_0|u;p⟩=0$ becomes $p_{\mu }{\Gamma }^{\mu }u=0$. At nonzero null momentum this Dirac equation leaves sixteen unprojected on-shell spinor states. Equivalently the eight transverse zero modes give the sixteen-dimensional spin representation $8_s\oplus 8_c$ of $SO(8)$. Thus the Ramond ground state describes spacetime fermions, not a single scalar oscillator vacuum.

To define the GSO projection explicitly, let $f_{NS}$ count excited NS fermions and give its vacuum even parity. ThenIt retains odd fermion number, removes the tachyon, and retains the eight massless vector particle polarizations. In the Ramond sector let ${\Gamma }_{\ast }$ be transverse chirality and $f_R$ count nonzero-mode fermions. Choose one consistent chirality sign $s=\pm 1$ and useEach operator squares to itself. At the ground level $P_R$ retains one eight-dimensional spinor chirality, so the retained spectrum has eight massless bosonic and eight massless fermionic states. The parity convention can instead assign the NS vacuum odd parity and write a plus projector; the retained states are identical.

Equality continues at every mass level. Let $q$ count ${\alpha }^{\prime }{M}^2$, suppressing common momentum and endpoint-label factors, and set $D(q)={\prod }_{n\ge 1}(1-q^n{)}^8$. The projected string oscillator generating functions areThe Ramond factor eight follows because either chirality projector selects eight ground states for every nonzero-oscillator parity. The Jacobi abstruse identity makes the bracket equal to $16q^{1/2}{\prod }_{n\ge 1}(1+q^n{)}^8$, soThis proves level-by-level equality of bosonic and fermionic multiplicities. The appropriate spacetime supercharge from an RNS spin field relates the two sectors; the projection ensures that it acts within the retained state space. These sector and chirality conventions agree with the primary superstring lecture notes. Equal total counts alone are not the definition of the projection; they follow from the stated parity selection and the string oscillator identity.

For the whole paper use $\eta =\mathrm{diag}(1,-1,-1,-1)$, natural units and the usual Dirac adjoint $\bar{\psi }={\psi }^{†}{\gamma }^0$. The Lorentz algebra generators below have the real-generator convention of the PDF, with no extra factor of $i$.

Construct the Lorentz-spinor generators from a Clifford algebra asThey are antisymmetric in $\rho ,\sigma$. Reordering a third gamma matrix with the Clifford algebra yieldsUse the commutator derivation identity on the two factors in $M^{\tau \nu }=\frac{1}{4}[{\gamma }^{\tau },{\gamma }^{\nu }]$:This derives every Lorentz algebra commutation relation, so the matrices $M^{\rho \sigma }$ form its spinor representation. Their finite exponentials describe the connected Spinor representation of the Lorentz group, more precisely the lift to its double cover.

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.

Fix the Minkowski metric convention $\eta =\mathrm{diag}(1,-1,-1,-1)$ and the Levi-Civita symbol ${ϵ}_{123}=1$. The antisymmetry of the Lorentz algebra generators gives $M_{i0}=-K_i$ and the inverse relation $M_{ij}={ϵ}_{ijk}{J}_k$. Inserting one temporal index in each generator immediately yieldsTwo Lorentz boosts therefore generate a rotation through their commutator; the minus sign distinguishes this algebra from the rotation algebra in four-dimensional Euclidean space.

For a spatial generator and a boost, the same Lorentz algebra givesContracting with ${ϵ}_{iab}/2$ givesThus the three Lorentz boosts transform as a spatial vector under rotations.

Finally, the all-spatial bracket becomesUse $M_{ab}={ϵ}_{abr}{J}_r$ in the double contraction with ${ϵ}_{iab}{ϵ}_{jcd}/4$. The epsilon contraction identity reduces it to $[J_i,J_j]=i{ϵ}_{ijk}{J}_k$. One can check the sign directly: $J_1=M_{23}$ and $J_2=M_{31}$ give $[J_1,J_2]=i{M}_{12}=i{J}_3$; cyclic permutations give the other nonzero brackets. The requested coefficients areThese are rotation and boost commutators with a fixed metric signature. The PDF does not explicitly specify the signature. Keeping its generator convention but choosing $\eta =\mathrm{diag}(-1,1,1,1)$ reverses every displayed algebra coefficient: the pairs become $(1,0),(0,-1),(-1,0)$. More generally, if $s={\eta }_{00}=\pm 1$, the three nonzero coefficients are $A_1=-s$, $B_2=s$ and $A_3=s$. Specifying the Minkowski metric is therefore necessary to make the numerical signs unambiguous.