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On two-by-two Hermitian matrices, the determinant of $X=tI+x\cdot \sigma$ is $t^2-|x{|}^2$. The action $X\mapsto AX{A}^{†}$ preserves it for $A\in SL(2,\mathbb{C})$. A kernel element is unitary and commutes with every Hermitian matrix, hence is $\pm I$. The derivative is injective and both real dimensions are six, giving the full proper orthochronous Lorentz group as the connected image.

For $M=\mathrm{diag}(-1,1,1,1)$,Every such matrix has the block formso the Lie algebra and the Lorentz group have dimension six. At $A\in O(1,3)$, the tangent space is $A\mathfrak{o}(1,3)$.

The RNS string adds worldsheet Majorana fermions ${\psi }^{\mu }$ to the coordinate fields $X^{\mu }$. The new fields are spinors on the worldsheet but vectors under the target Lorentz group; this distinction matters when interpreting their oscillator states. In a flat target, a convenient Lorentzian convention for the free gauge-fixed action iswhere ${\rho }^a$ are two-dimensional gamma matrices. The bosonic and fermionic kinetic terms are related by worldsheet supersymmetry. Before fixing local worldsheet supersymmetry, one introduces a worldsheet gravitino along with the worldsheet metric. After superconformal gauge fixing, the residual matter constraints are the Virasoro constraints and the constraints generated by the superconformal current $G\propto \psi \cdot \partial X$.

Each real worldsheet Majorana fermion contributes $1/2$ to the chiral central charge, while each coordinate contributes one. The reparameterization bc system contributes $-26$, and fixing local worldsheet supersymmetry also produces commuting superconformal ghosts of weights $(3/2,-1/2)$, with central charge $11$. ThusIn light-cone gauge in string theory there are eight physical transverse coordinate fields and eight transverse worldsheet Majorana fermions.

On the doubled spatial interval of an open string, the two allowed fermionic periodicities areThey arise from the relative signs of the left-right fermion gluing conditions at the two ends. Thus the Ramond sector has integer-mode oscillators $d_n$, including $d_0$, and the Neveu–Schwarz sector has half-integer oscillators $b_r$, $r\in \mathbb{Z}+1/2$, without zero modes. In either sector,A closed string has independent left- and right-moving periodicities, giving four sector choices rather than two.

The Ramond zero-mode Clifford algebra isHence its ground states carry a Spinor representation of the Lorentz group. The zero-mode supercurrent constraint reduces on a ground state to the Dirac equation $p_{\mu }{\Gamma }^{\mu }u=0$. In the Neveu–Schwarz sector the oscillator vacuum instead carries a scalar representation. Consequently the open string NS sector describes spacetime bosons, while the R sector describes spacetime fermions. Exciting a target-vector oscillator does not change this statistics assignment.

The string intercepts follow from the transverse zero-point energy. Per boson the integer-mode contribution is $\frac{1}{2}\zeta (-1)=-1/24$. Per fermion it is the negative of the corresponding half-sum. Since $\zeta (-1,1/2)=1/24$, the eight transverse pairs giveThus $a_{\mathrm{NS}}=1/2$, $a_{\mathrm{R}}=0$, and the open string mass formula is ${\alpha }^{\prime }{M}^2=N-a$. This distinguishes the lowest states explicitly:

The supersymmetric GSO projection keeps odd oscillator fermion number in the NS sector. Defining $F_{\mathrm{osc}}$ so that the NS vacuum has $F_{\mathrm{osc}}=0$, its projector isIt removes the scalar tachyon and keeps the massless vector. In the R sector, fermion parity contains the zero-mode chirality operator, so one may writeHere $F_{\mathrm{osc}}$ counts nonzero-mode fermionic excitations. On the ground states, this keeps one Majorana-Weyl spinor chirality: 16 real covariant components, or eight on-shell polarizations after the Dirac equation. The projected massless open-string states are an eight-polarization gauge vector and an eight-polarization gaugino. The compatible projected tower exhibits spacetime supersymmetry; unprojected worldsheet supersymmetry by itself did not eliminate the tachyon or equalize these massless counts.

For a closed string, both chiral zero-mode constraints must hold:This shifted closed-string level matching is particularly important in the mixed sectors. The lowest states and their spacetime nature are as follows.

For type IIA superstring theory, the two R sector ground-state chiralities are opposite. The Ramond–Ramond potentials can be represented by $C_1,C_3$, with transverse counts $8+56=64$. The two mixed-sector gravitinos have opposite chiralities, and the ten-dimensional theory is nonchiral. For type IIB superstring theory, the two R sector chiralities agree; the Ramond–Ramond potentials are $C_0,C_2,C_4$, with self-dual five-form field strength, giving $1+28+35=64$. Its two gravitinos have the same chirality. Thus both projected type II theories have 128 massless bosonic and 128 massless fermionic polarizations and no tachyon, while their chiralities and form spectra differ.

If unoriented open strings are included in the complete theory, the supersymmetric example is type I string theory. Worldsheet orientation reversal projects the closed type IIB superstring theory spectrum and identifies the two mixed sectors; the surviving closed fields include the graviton, dilaton, a Ramond–Ramond potential $C_2$, and one gravitino and dilatino. The open states form vector and gaugino multiplets with Chan-Paton factors; tadpole and anomaly cancellation select $SO(32)$. This is an additional orientation projection, not part of the GSO projection itself. The heterotic string instead pairs one supersymmetric chiral side with a bosonic chiral side carrying an internal gauge sector; it does not have a Ramond choice on both sides.

Use the four-dimensional real vector space of Hermitian matricesFor $A\in SL(2,\mathbb{C})$, $X\mapsto AX{A}^{†}$ preserves determinant, hence defines a Lorentz transformation. Positive definite $X$ stays positive definite, so future timelike vectors remain future timelike. Connectedness also fixes the orientation, giving an image in $S{O}_0(3,1)$.

If the action is trivial, $X=I$ makes $A$ unitary. The remaining condition says it commutes with every Hermitian matrix, forcing $A=\lambda I$ with ${\lambda }^2=1$. Infinitesimally $aX+X{a}^{†}=0$ first makes $a$ anti-Hermitian and then scalar; tracelessness forces $a=0$. Both real Lie-algebra dimensions are six, so the image is open. Polar decomposition shows $SL(2,\mathbb{C})$ is connected, with connected unitary factor $SU(2)$ and a contractible positive Hermitian determinant-one factor. The image is therefore the full proper orthochronous Lorentz group.This is the Hermitian-matrix double cover of SO0(3,1). The full determinant-one Lorentz group also contains transformations reversing time orientation, so the identity-component qualification cannot be omitted.

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 $SO(3,1)$.

Combining the real transformations givesLet $v=\tanh \psi$, so $\gamma =\cosh \psi$ and $\gamma v=\sinh \psi$. The usual rotation about the positive $x$-axis has the $yz$ block $\left(\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right)$. Taking $\boxed{n=(1,0,0),\theta =-i\psi }$ gives exactly the displayed transformation, because $\cos (-i\psi )=\cosh \psi$ and $\sin (-i\psi )=-i\sinh \psi$. This is a complex rotation, not a real spatial rotation.

Complex rotations preserve the bilinear dot product, without conjugation. Henceis invariant, giving the two independent real Lorentz invariantsTheir completeness refers to scalar invariants of the full Lorentz group. For a generic non-null field, a boost along $E\times B$ makes the fields parallel. Indeed with $s=|E\times B|$ and $A=E^2+B^2$, the required speed solves $s(1+v^2)-Av=0$; the smaller root is $v=[A-\sqrt{(E^2-B^2{)}^2+4(E\cdot B{)}^2}]/(2s)<1$. For $s=0$ the fields are already parallel. A real rotation then gives $E=en$, $B=bn$, leaving just $e^2-b^2$ and $eb$ as independent scalar parameters. Thus there are precisely two independent continuous scalar invariants on generic orbits; null fields have special orbits and the invariants do not distinguish zero from a nonzero null field.

If the printed phrase is read as invariance under only this single-axis boost, it admits extra fixed components $F_x$, so the assertion of precisely two must be understood in its Lorentz-scalar sense.

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.

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.

Complexifying the Clifford algebra removes the signature dependence of its representation dimension. Pair the $2k$ gamma matrices into operators $a_i,a_i^{†}$ with $\{a_i,a_j^{†}\}={\delta }_{ij}$ and $\{a_i,a_j\}=0$. The $k$ commuting occupation operators have eigenvalues zero or one. Starting from a Clifford vacuum, applying every subset of the $k$ fermionic creation operators produces $2^k$ independent states. Conversely, these operators construct matrices on a $2^k$-dimensional space, and their products span its full matrix algebra. Thus an irreducible complex Clifford algebra module in $D=2k$ hasA general Clifford module is a direct sum of these; the statement concerns the minimal spinor.

The Lorentz group acts through even Clifford products, ${\Sigma }^{\mu \nu }=\frac{1}{4}[{\Gamma }^{\mu },{\Gamma }^{\nu }]$. More precisely this is a representation of its double cover, the Spin group. The normalized product ${\Gamma }_{\ast }$ of all $2k$ matrices squares to one and anticommutes with each ${\Gamma }^{\mu }$, while commuting with every ${\Sigma }^{\mu \nu }$. Its eigenspaces are the two Weyl spinor representations, so the Dirac spinor is reducible under the connected spin group:Each summand is irreducible. The Clifford module itself remains irreducible because an individual ${\Gamma }^{\mu }$ interchanges the summands. In $D=2k+1$ one can use the old ${\Gamma }_{\ast }$ as the additional matrix, with an appropriate phase for its signature. No independent chirality matrix remains: the product of all $2k+1$ matrices is central and fixed to a scalar on an irreducible module. There are two choices of this central sign for the complex Clifford algebra, but each restricts to the same irreducible $2^k$-dimensional spinor representation of the connected spin group. An odd-dimensional spinor has no Weyl splitting. Majorana spinor reality conditions depend additionally on dimension and signature; they must be imposed separately from this complex dimension count.

For the massless field counts, use the rotational part $SO(D-2)$ of the little group and real physical polarizations. A massless p-form gauge field has only transverse, antisymmetric components after its gauge redundancy and equations of motion are imposed, giving $\left(\genfrac{}{}{0ex}{}{D-2}{p}\right)$ polarizations. A graviton has a symmetric traceless transverse tensor, giving $(D-2)(D-1)/2-1=D(D-3)/2$. A massless fermion obeys a Dirac equation which halves its off-shell spinor components. In ten dimensions a Majorana-Weyl spinor has sixteen real components off shell and eight physical components; in nine dimensions a Majorana spinor has sixteen off shell and eight physical components; in eleven dimensions a Majorana spinor has thirty-two off shell and sixteen physical components. A gravitino is a transverse vector-spinor with its gamma trace removed, leaving $(D-3)$ times the corresponding physical spinor dimension. Thus the ten-dimensional Majorana-Weyl gravitino has $7\times 8=56$ polarizations, the nine-dimensional Majorana gravitino has $6\times 8=48$, and the eleven-dimensional Majorana gravitino has $8\times 16=128$.

The common Neveu-Schwarz sector of type IIA supergravity and type IIB supergravity consists of a metric tensor, the Kalb–Ramond field $B_2$, and the dilaton $\varphi$. Their counts are respectively $35$, $\left(\genfrac{}{}{0ex}{}{8}{2}\right)=28$, and $1$. The Ramond-Ramond sectors and fermions distinguish the theories. In type IIA supergravity the independent bosonic potentials and counts areThere are two Majorana-Weyl gravitini of opposite chirality, giving $56+56=112$, and two Majorana-Weyl dilatini of opposite chirality, giving $8+8=16$. Their total is $128$. With gravitini labelled ${\psi }^{+}$ and ${\psi }^{-}$, the corresponding dilatini have signs $-$ and $+$ respectively. Type IIA is nonchiral.

In type IIB supergravity, the independent bosonic potentials and counts areThe five-form field strength associated with $C_4$ is a self-dual differential form, so only half the seventy transverse four-form components are independent. In Lorentzian ten dimensions the Hodge star operator on five-forms squares to one, making a real self-duality condition possible. The two Majorana-Weyl gravitini have the same chirality and contribute $112$ states. Both Majorana-Weyl dilatini have the opposite chirality to those gravitini and contribute $16$ states. Type IIB is chiral, with $128+128$ bosonic and fermionic polarizations.

For dimensional reduction, retain the massless zero modes on a flat circle, with no flux, gauging or fermion twist. A metric splits into a lower-dimensional metric, one vector and one scalar. An ordinary $p$-form splits into a $p$-form and a $(p-1)$-form, according as it has no compact index or one. This preserves counts by Pascal's identity, $\left(\genfrac{}{}{0ex}{}{D-2}{p}\right)=\left(\genfrac{}{}{0ex}{}{D-3}{p}\right)+\left(\genfrac{}{}{0ex}{}{D-3}{p-1}\right)$.

Reducing type IIA supergravity to nine dimensions gives the following decompositions; each number is a count of physical states:Thus the bosonic maximal nine-dimensional supergravity multiplet contains one graviton, three massless vectors, two massless two-forms, one massless three-form and three real scalars, withEach ten-dimensional Majorana-Weyl gravitino becomes one nine-dimensional Majorana gravitino and one Majorana spin-one-half field, giving $56=48+8$. Each ten-dimensional dilatino gives one more Majorana spin-one-half field. The fermionic multiplet therefore has two gravitini and four spin-one-half fields, with $2(48)+4(8)=128$. The fields belong to one maximal gravity supermultiplet, not separate interacting matter multiplets.

Reducing type IIB supergravity gives $g_{\mu \nu },A_{\mu }^{\mathrm{metric}},\rho$ from the metric; one two-form and one vector each from $B_2$ and $C_2$; and the two scalars $\varphi ,C_0$. The four-form potential yields a nine-dimensional four-form and three-form, but the ten-dimensional self-dual differential form condition relates their field strengths. Retaining either one gives $\left(\genfrac{}{}{0ex}{}{7}{3}\right)=\left(\genfrac{}{}{0ex}{}{7}{4}\right)=35$ independent polarizations, not seventy. The same bosonic count, three scalars, three vectors, two two-forms and one three-form, follows. The fermionic reduction likewise gives two gravitini and four spin-one-half fields. These are the same maximal nine-dimensional supergravity spectrum, as expected from T-duality. The spectrum is nonchiral in nine dimensions, since the odd-dimensional spinor representation has no independent Weyl chirality; the ten-dimensional distinction is lost on restriction to nine-dimensional Lorentz symmetry.

Eleven-dimensional supergravity has a metric with $11(8)/2=44$ states, a three-form with $\left(\genfrac{}{}{0ex}{}{9}{3}\right)=84$, and a Majorana gravitino with $128$. For direct reduction on a flat two-torus, write the internal indices as $i=1,2$. The metric gives $g_{\mu \nu }$, two vectors $g_{\mu i}$ and three symmetric components $g_{ij}$; the three-form gives $A_{\mu \nu \rho }$, two two-forms $A_{\mu \nu i}$ and one vector $A_{\mu 12}$. HenceThe eleven-dimensional Majorana spinor restricts to two nine-dimensional Majorana spinors. The vector-spinor consequently yields two gravitini and four spin-one-half fields, giving $128=2(48)+4(8)$. This is again the same $128+128$ maximal nine-dimensional multiplet.

For reduction of massless type IIA supergravity to four dimensions, use a flat six-torus and retain all zero modes, without flux or projections. Internal indices $i=1,\dots ,6$ giveThe multiplicities $21,15,20$ come respectively from a symmetric pair of six internal indices, $\left(\genfrac{}{}{0ex}{}{6}{2}\right)$, and $\left(\genfrac{}{}{0ex}{}{6}{3}\right)$. Before dualizing, this gives one graviton, $28$ vectors, $63$ scalars, seven two-forms and one three-form. The seven two-forms provide seven scalar polarizations, whereas the three-form has no local propagating polarization. Thus the final bosonic spectrum is one graviton, $28$ vectors and $70$ scalars, with $2+56+70=128$ states.