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In the complex-conjugate convention for charge conjugation of a Dirac field, a unitary matrix $B$ obeys the displayed gamma matrix identity. For $M^{\rho \sigma }=\frac{1}{4}[{\gamma }^{\rho },{\gamma }^{\sigma }]$, two minus signs cancel to give $M^{\rho \sigma }B=B{M}^{\rho \sigma \ast }$. Real Lorentz transformation parameters therefore give $SB=B{S}^{\ast }$, so $B{\psi }^{\ast }$ has the same Spinor representation of the Lorentz group transformation law as $\psi$. Complex conjugating the free Dirac equation and multiplying by $B$ proves that the conjugate also solves it for real mass. This $B$ is the matrix multiplying ${\psi }^{\ast }$; the convention multiplying ${\bar{\psi }}^T$ instead uses $B({\gamma }^0{)}^T$.

A chiral spinor superfield carries a Spinor representation of the Lorentz group index and satisfies ${\bar{D}}_{\dot{\alpha }}{\Psi }_{\beta }=0$. The chiral field-strength superfield is a fermionic example: its lowest component is the Weyl spinor gaugino. Its spinor index distinguishes it from a scalar chiral superfield, even though both obey the same chirality constraint.

The generators in the Spinor representation of the Lorentz group satisfy $M^{†}{\gamma }^0+{\gamma }^{0}M=0$, so the finite transformation preserves the indefinite form $S^{†}{\gamma }^{0}S={\gamma }^0$. This makes $\bar{\psi }\psi$ a Lorentz scalar through the Dirac adjoint. Boost generators are Hermitian rather than anti-Hermitian: a boost has eigenvalues $e^{\pm \eta /2}$ and is not unitary in the ordinary positive-definite component norm. Rotations are unitary. This component representation is distinct from the unitary action on the physical state space.

Use natural units and the Minkowski metric $g=\mathrm{diag}(1,-1,-1,-1)$. Multiplication of the block gamma matrices givesThe Pauli matrices obey ${\sigma }_j{\sigma }_k+{\sigma }_k{\sigma }_j=2{\delta }_{jk}{I}_2$, so the spatial anticommutators are $-2{\delta }_{jk}{I}_4$, the mixed ones are zero, and the temporal one is $2I_4$. ThusBecause partial derivatives commute, only the symmetric part of ${\gamma }^{\mu }{\gamma }^{\nu }$ survives in their contraction. Multiplying the Dirac equation on the left by its opposite-mass factor therefore gives the flat-space Dirac factorizationHence every component satisfies $\boxed{(\square +m^2)\psi =0}$, the Klein-Gordon equation. This implication does not remove the first-order spinor constraints imposed by the Dirac equation.

For Lorentz covariance, transform coordinates by $x^{\prime }=Lx$ and the Dirac spinor by ${\psi }^{\prime }(x^{\prime })=S(L)\psi (x)$. The chain rule gives ${\partial }_{\mu }^{\prime }=(L^{-1}{)}^{\nu }{}_{\mu }{\partial }_{\nu }$. ConsequentlyThe cancellation uses $L^{\mu }{}_{\rho }(L^{-1}{)}^{\nu }{}_{\mu }={\delta }^{\nu }{}_{\rho }$. Thus the Lorentz transformation maps every solution to another, establishing the Lorentz covariance of the Dirac operator.

The infinitesimal identity follows directly from the Clifford algebra. Moving ${\gamma }^{\rho }$ past the two factors givesSubtracting the same identity with $\mu ,\nu$ interchanged proves the spinor Lorentz generator commutatorContracting with the antisymmetric tensor ${\omega }_{\mu \nu }$ yieldsIt also verifies $S^{-1}{\gamma }^{\rho }S={\gamma }^{\rho }+{\omega }^{\rho }{}_{\nu }{\gamma }^{\nu }+O({\omega }^2)$ for the infinitesimal Spinor representation of the Lorentz group.

For a positive-energy four-momentum, put $\psi =u(p)e^{-ip\cdot x}$, $E=\sqrt{{\mathbf{p}}^2+m^2}>0$. The Dirac equation becomes $(/p-m)u=0$, where the Feynman slash notation uses $/p={\gamma }^{\mu }{p}_{\mu }=E{\gamma }^0-\mathbf{\gamma }\cdot \mathbf{p}$. Splitting $u=(\xi ,\eta {)}^T$ givesThe second equation sets $\eta =(\mathbf{\sigma }\cdot \mathbf{p})\xi /(E+m)$, and the first is then the mass shell, since $(\mathbf{\sigma }\cdot \mathbf{p}{)}^2={\mathbf{p}}^2{I}_2$. Choosing two orthonormal two-spinors ${\chi }_s$ gives the Dirac plane waves in the standard representation:The negative-frequency solutions are independentlywith orthonormal ${\eta }_s$. They are the antiparticle modes after field quantization. The normalization gives $u_r^{†}{u}_s=v_r^{†}{v}_s=2E{\delta }_{rs}$ and ${\overline{u}}_r{u}_s=2m{\delta }_{rs}$, ${\overline{v}}_r{v}_s=-2m{\delta }_{rs}$, using the Dirac adjoint $\overline{u}=u^{†}{\gamma }^0$.

For a spatial rotation, ${\sigma }^{ij}={ϵ}_{ijk}\mathrm{diag}({\sigma }_k,{\sigma }_k)$. Hence the intrinsic rotation generators areFor a massive particle in its rest frame, the two independent spinors therefore carry the two-dimensional spin one-half representation, with $J_3$ eigenvalues $\pm 1/2$ and Casimir $s(s+1)=3/4$. A $2\pi$ rotation acts as $-I$, the spinor sign under a full spatial rotation. Thus the particle has spin $\boxed{s=1/2}$; four spinor components encode particle and antiparticle sectors, not four spin states of one particle. In the massless limit the corresponding physical labels are helicities $\pm 1/2$ rather than a rest-frame spin basis.

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 signature $(-,+,+,+)$ and $c=1$. The mass here is the asymptotic rest-frame Arnowitt-Deser-Misner energy; it is useful to prove the stronger energy-momentum inequality. Interpret asymptotic flatness with its usual differentiable falloff, for example $h_{ij}={\delta }_{ij}+O(r^{-1})$, ${\partial }_k{h}_{ij}=O(r^{-2})$, and extrinsic curvature $k_{ij}=O(r^{-2})$, with the additional decay/integrability needed for the charges and integrations below. The complete nonsingular slice has no inner boundary. An oriented three-dimensional slice admits a spin structure. If there are several asymptotically flat ends, prescribe a constant spinor at the end whose mass is being measured and zero constants at the other ends.

Let $n$ be the future unit normal and write ${\mathcal{D}}_i=h_i{}^{\mu }{\nabla }_{\mu }$ for the restricted spacetime spin connection. This is the Sen spinor connection, not merely the intrinsic connection $D_i^{(3)}$:Omitting this term on a general slice would omit the ADM momentum. Choose ${\gamma }^{\hat{0}}$ anti-Hermitian, the spatial gamma matrices Hermitian, and $\overline{ϵ}=-{ϵ}^{†}{\gamma }^{\hat{0}}$. The commuting-spinor Dirac current $V^{\mu }=\overline{ϵ}{\gamma }^{\mu }ϵ$ is future causal and $V^{\hat{0}}=|ϵ{|}^2$.

The Nester two-form supplies the needed integration identity. DefineHere multiple gamma indices denote antisymmetrized products. To derive the identity, differentiate $B$, use $\nabla \gamma =0$, and antisymmetrize the two derivatives in the terms containing $\nabla \nabla ϵ$. Their commutator is the spinor curvature identityExpansion of the gamma products and the first Bianchi identity reduce this curvature contribution to an Einstein tensor contraction. It vanishes for the Vacuum Einstein equations. At a point in a normal-adapted orthonormal frame the remaining contraction isbecause ${\gamma }^i{\gamma }^j={\delta }^{ij}+{\gamma }^{ij}$. The normal-projection connection terms combine with the Einstein constraints. Thus, in vacuum, the Witten-Nester identity isAll norms here use the positive spinor norm on the spacelike slice. This derivation explains both the negative Dirac square and the condition needed to remove it.

Solve the Witten spinor equation $\mathcal{D}ϵ=0$ with $ϵ\to {ϵ}_{\infty }$. The existence step is part of the argument. Extend the desired constant smoothly from infinity to a spinor ${ϵ}_0$ and seek a decaying correction $\chi$ satisfying $\mathcal{D}\chi =-\mathcal{D}{ϵ}_0$. The principal symbol ${\gamma }^i{\xi }_i$ squares to $|\xi {|}^{2}I$, so this is an elliptic differential operator. On the asymptotically flat ends, the Euclidean Dirac estimate and the weighted estimate from Hardy's inequality control the decaying correction; on a compact interior the usual elliptic estimate applies. Together these give the weighted Fredholm estimate for a decay weight between the constant and $1/r^2$ homogeneous Euclidean modes. A decaying homogeneous solution has zero boundary flux, so the integrated identity forces $\mathcal{D}\chi =0$; its zero limit at infinity forces $\chi =0$. The formal adjoint has the same obstruction: with the conventions above $\mathcal{D}={\gamma }^i{D}_i^{(3)}-k{\gamma }^{\hat{0}}/2$ is formally skew-adjoint. Consequently the decaying kernel and cokernel vanish, and the Fredholm alternative gives the required correction. Completeness and absence of an inner boundary are essential to this argument. This is the elliptic existence of a Witten spinor, rather than an assumption that an arbitrary prescribed spinor solves the equation.

For completeness, identify the asymptotic flux. Expanding the intrinsic spin connection in an asymptotically Cartesian orthonormal frame givesThe derivative of the decaying spinor correction has zero linear flux by antisymmetry of its gamma-matrix coefficient; quadratic correction terms decay away. Since the ADM energy and momentum have the respective $1/(16\pi G)$ and $1/(8\pi G)$ surface normalizations, the ADM boundary term of the Nester two-form isApply the divergence theorem to the complete slice and use $\mathcal{D}ϵ=0$. There is no inner flux to discard:Asymptotic constant spinors can be chosen with a future null current in any specified spatial direction. Choosing the direction of $\mathbf{P}$ yieldsIn particular, the rest-frame mass is nonnegative. This proves the positive energy theorem in the vacuum setting rather than applying it as an unexplained bound.

If this mass is zero, then $E=0$ in the asymptotic rest frame and the inequality gives $\mathbf{P}=0$. The same identity with any nonzero ${ϵ}_{\infty }$ has zero right-hand side. Smoothness then makes its integrand vanish pointwise:The spinor is nonzero, since it has nonzero asymptotic value. This is a spinor parallel along $\Sigma$; no time-parallel condition has yet been inferred merely from the integral.

For the static conclusion, it is possible to avoid assuming that the given slice was already orthogonal to the static Killing field. Because $E=\mathbf{P}=0$, repeat the zero-energy construction for four linearly independent asymptotic Dirac spinors. Parallel transport by the Sen spinor connection is invertible, so the resulting spinors form a basis everywhere on $\Sigma$. Tangential curvature annihilates each of them:Thus $R_{ijab}{\gamma }^{ab}=0$ as an operator. The six antisymmetric gamma products give a faithful Spinor representation of the Lorentz group, so $R_{ijab}=0$ for tangential $i,j$. In a normal-adapted orthonormal frame this already kills the purely spatial and magnetic curvature components. The remaining components vanish by the Vacuum Einstein equations:Consequently the full spacetime Riemann curvature tensor vanishes on $\Sigma$.

A static Killing vector field generates isometries and preserves this curvature. Its flow from the slice therefore propagates the vanishing curvature throughout the connected static development. Extend one of the spinors by spacetime parallel transport along these time orbits, so that ${\nabla }_tϵ=0$. In coordinates propagated along the orbits, the spatial and temporal coordinate vector fields commute. Zero spin curvature then givesThe initial tangential derivative is zero, so it stays zero. This constructs the required nonzero spacetime parallel spinor; the initial parallel frame also removes any spin-holonomy ambiguity inherited from loops in $\Sigma$. The static zero-energy parallel-spinor rigidity conclusion isAn equivalent check in a regular orthogonal static slicing is to write $d{s}^2=-N^{2}d{t}^2+h_{ij}d{x}^{i}d{x}^j$. The slice has zero extrinsic curvature. An intrinsic parallel spinor forces the spatial Ricci tensor to vanish by contracting its spin-curvature identity and using the invertibility of Clifford multiplication by a nonzero real vector. By three-dimensional curvature from the Ricci tensor, $h$ is flat. The static vacuum equation $D_i{D}_{j}N=N{R}_{ij}(h)$ then makes the lapse gradient parallel, and asymptotic normalization forces $N=1$. The temporal spin connection vanishes as well.

An explicit example is Minkowski spacetime, $d{s}^2=-d{t}^2+d{x}^2+d{y}^2+d{z}^2$, where every constant spinor in a Cartesian tetrad is a spacetime parallel spinor and the ADM mass is zero. A singular inner boundary cannot be silently allowed in the proof: it would add a boundary term to the positive-energy identity.

Start with the Clifford algebra $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2{\eta }^{\mu \nu }$. For the dimension count, complexify and rescale the timelike matrix so that the generators obey the Euclidean complex relation $\{{\gamma }_a,{\gamma }_b\}=2{\delta }_{ab}$. This changes neither the complex representation dimension nor its decomposition under the complexified even algebra. In $D=2k$, pair the generators intoDirect substitution gives $\{a_i,a_j\}=\{b_i,b_j\}=0$ and $\{a_i,b_j\}={\delta }_{ij}$. The commuting operators $n_j=b_j{a}_j$ satisfy $n_j^2=n_j$, so their occupation eigenvalues are zero or one. An irreducible module contains an all-zero occupation state $|0⟩$, obtained by lowering the occupied entries of a simultaneous eigenvector. The statesare nonzero and independent by their distinct joint occupation eigenvalues. They span an invariant module; irreducibility makes this the full module. Conversely this Fock construction supplies a representation on exactly these $2^k$ states. Products of occupation projectors and raising/lowering operators give its matrix units, so the generated algebra is the full matrix algebra. Therefore the minimal complex Dirac spinor has

This module is irreducible under the full Clifford algebra, but the Spinor representation of the Lorentz group is generated by the even bilinears ${\Sigma }^{\mu \nu }=\frac{1}{4}[{\gamma }^{\mu },{\gamma }^{\nu }]$. The chirality matrix in the Euclideanized basis isIts two eigenspaces therefore remain invariant under Lorentz transformations. Each has half the Fock states, since any gamma matrix bijects them. The even algebra connects every pair of occupation states of the same parity and has no further invariant subspaces within either one. HenceThese are the irreducible Weyl spinors. Globally they are ordinary representations of $\mathrm{Spin}(D-1,1)$, the double cover; calling them representations of $SO(D-1,1)$ uses the conventional projective spinor terminology.

For $D=2k+1$, adjoin ${\gamma }_{2k+1}=\pm {\gamma }_{\ast }$. It anticommutes with the old matrices and gives a Clifford module of the same dimension $2^k$. The new even generators $[{\gamma }_a,{\gamma }_{2k+1}]$ interchange the two old chirality eigenspaces, so the odd-dimensional Lorentz spinor is irreducible of complex dimension $2^k$, with no Weyl splitting. The product of all odd-dimensional gamma matrices is central and acts as a scalar; it cannot be a new anticommuting chirality operator. Its two signs distinguish the two complex odd-Clifford modules, whose restrictions to Spin group are equivalent. Majorana spinor reality conditions, when the signature permits them, are additional conditions and should not be confused with complex chirality.

For the reduction, take the massless zero modes on a flat seven-torus with no flux or projections. The eleven-dimensional fields are the metric $G_{MN}$, a three-form $A_{MNP}$ and a Majorana gravitino ${\Psi }_M$. Split $M=(\mu ,i)$ with $\mu =0,\dots ,3$ and $i=1,\dots ,7$. The metric supplies $G_{\mu \nu }$, seven vectors $G_{\mu i}$ and 28 scalar fields $G_{ij}=G_{ji}$. The three-form supplies a four-dimensional three-form $A_{\mu \nu \rho }$, seven two-forms $A_{\mu \nu i}$, $\left(\genfrac{}{}{0ex}{}{7}{2}\right)=21$ vectors $A_{\mu ij}$ and $\left(\genfrac{}{}{0ex}{}{7}{3}\right)=35$ scalars $A_{ijk}$.

A massless two-form in four dimensions has one physical particle polarization and can be dualized to a scalar by $H_3=\ast d\sigma$; the seven two-forms thus supply seven more scalars. The four-dimensional three-form has no local particle polarization, since its four-form field strength is locally a constant times the volume form. The total bosonic content is consequentlyThe eleven-dimensional Majorana spinor has 32 real components, decomposing into eight four-dimensional Majorana spinors. Accordingly ${\Psi }_{\mu }$ supplies eight gravitini, while ${\Psi }_i$ supplies $7\times 8=56$ spin-$1/2$ fields after diagonalizing the gravitino/matter kinetic terms. The resulting theory has 32 real supercharges, hence is four-dimensional N=8 supergravity, withThe on-shell counts check the reduction: the metric has $11(11-3)/2=44$ polarizations and the three-form has $\left(\genfrac{}{}{0ex}{}{9}{3}\right)=84$, totaling 128. In four dimensions $2+28\cdot 2+70=128$ bosonic polarizations match $8\cdot 2+56\cdot 2=128$ fermionic ones. Higher Kaluza–Klein levels, fluxes or projections would be additional choices beyond this massless toroidal reduction.

A vierbein is a local orthonormal coframe, consisting of four linearly independent covectors $e^{\mu }{}_a$ such thatHere $a,b$ label spacetime coordinates, while $\mu ,\nu$ label a Lorentz frame with $\eta =\mathrm{diag}(-1,1,1,1)$. Its inverse $e_{\mu }{}^a$ obeys $e_{\mu }{}^a{e}^{\mu }{}_b={\delta }_b^a$ and $e_{\mu }{}^a{e}^{\nu }{}_a={\delta }_{\mu }^{\nu }$. The vierbein converts tensor indices between the coordinate and orthonormal descriptions.

The metric is unchanged under an arbitrary local Lorentz transformation $e^{\mu }{}_a\mapsto {\Lambda }^{\mu }{}_{\nu }(x)e^{\nu }{}_a$, since ${\Lambda }^T\eta \Lambda =\eta$. This is local frame redundancy, with corresponding transformations of fields and connections; it is distinct from coordinate diffeomorphism covariance. A generic curved spacetime does not have global Lorentz isometries. Rather, local tangent frames have Lorentz freedom, and the field equations can be written covariantly under their changes. On flat spacetime, global Lorentz transformations are also spacetime isometries.

A spinor field is a section of the spinor bundle associated to a spin structure. Local components transform in a Spinor representation of the Lorentz group, more precisely a representation of its spin double cover. For the proper orthochronous group in four dimensions, that cover is ${\mathrm{Spin}}^{+}(1,3)\simeq SL(2,\mathbb{C})$. Thus a spinor field transforms as $ϵ\mapsto S(\Lambda )ϵ$ under frame changes, rather than as a coordinate-index tensor. The double-valued lift of a frame rotation requires a consistent spin structure globally; local spinor calculations alone do not choose that global structure. One can use a Dirac spinor for the gamma-matrix calculation below.

Write ${\omega }_a{}^{\mu }{}_{\nu }$ for the Lorentz-frame spin connection. Expanding the tetrad postulate givesMultiply by the inverse vierbein to solve for the connection:With the torsion-free Levi-Civita connection, metric compatibility makes ${\omega }_{a\mu \nu }={\eta }_{\mu \rho }{\omega }_a{}^{\rho }{}_{\nu }$ antisymmetric in $\mu ,\nu$. For constant flat gamma matrices obeying $\{{\hat{\gamma }}^{\mu },{\hat{\gamma }}^{\nu }\}=2{\eta }^{\mu \nu }$, define ${\hat{\gamma }}^{\mu \nu }=[{\hat{\gamma }}^{\mu },{\hat{\gamma }}^{\nu }]/2$. The spinor covariant derivative isUnder a local Lorentz transformation, the matrix connection acquires the inhomogeneous term needed to make ${\nabla }_aϵ$ transform as a spinor. The derivative of the vierbein in the boxed formula is therefore essential.

For the conformally flat metric, choose $e^{\mu }{}_a=\Omega {\delta }^{\mu }{}_a$ and $e_{\mu }{}^a={\Omega }^{-1}{\delta }_{\mu }^a$. Let ${\ell }_a={\partial }_a\log \Omega$ and raise these coordinate indices temporarily with the flat metric $\eta$. Directly evaluating the Christoffel symbols givesSince ${\partial }_a{e}^{\mu }{}_b=\Omega {\ell }_a{\delta }_b^{\mu }$, the connection formula yieldsThis is the spin connection of a conformally flat metric in the specified conformal frame. It is frame dependent, unlike an invariant curvature tensor.

Set $L={\hat{\gamma }}^c{\ell }_c$. Antisymmetry of $\omega$ makes its action on spinorsFor $ϵ={\Omega }^{1/2}{ϵ}_0$ with ${\partial }_a{ϵ}_0=0$, we have ${\partial }_aϵ={\ell }_aϵ/2$. The Clifford algebra gives $\{{\hat{\gamma }}_a,L\}=2{\ell }_a$, soThe curved gamma matrices are ${\gamma }_a=\Omega {\hat{\gamma }}_a$ and ${\gamma }^a={\Omega }^{-1}{\hat{\gamma }}^a$. ConsequentlyThus the rescaled field is a twistor spinor. Multiplying this last equation by ${\gamma }_b$, symmetrizing in $a,b$ and using $\{{\gamma }_a,{\gamma }_b\}=2g_{ab}$ provesThis is the conformal rescaling of a parallel spinor to a twistor spinor. It does not claim that the rescaled spinor remains parallel in the conformal metric; the nonzero connection and the derivative of ${\Omega }^{1/2}$ combine precisely into its gamma-trace part.

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.

For $x^{\prime }=\Lambda x$, choose the corresponding spin liftThe spinor field transforms asThe infinitesimal Clifford algebra identity from part (b) exponentiates to $S^{-1}{\gamma }^{\mu }S={\Lambda }^{\mu }{}_{\nu }{\gamma }^{\nu }$, establishing Lorentz covariance of the Dirac operator. At fixed coordinates the infinitesimal variation also includes the orbital term $-{\Omega }^{\mu }{}_{\nu }{x}^{\nu }{\partial }_{\mu }\psi$.

A spatial rotation generator $M^{ij}$ is anti-Hermitian, but a Lorentz boost generator is Hermitian. For example,Its eigenvalues are $\pm \frac{1}{2}$. A nonzero rapidity $\zeta$ gives boost eigenvalues $e^{\pm \zeta /2}$, which are not all on the unit circle. This proves that the finite-dimensional component Spinor representation of the Lorentz group cannot be unitary for a positive-definite component inner product, even after a change of basis. It does not contradict a unitary action on the physical space of quantum states.

Consequently ${\psi }^{\prime †}(x^{\prime }){\psi }^{\prime }(x^{\prime })={\psi }^{†}{S}^{†}S\psi$ need not equal ${\psi }^{†}\psi$: it is the time component of the Dirac current, not a Lorentz scalar. The appropriate invariant form is indefinite. From $({\gamma }^{\mu }{)}^{†}={\gamma }^0{\gamma }^{\mu }{\gamma }^0$ one obtainsThis Dirac spinor pseudo-unitarity gives ${\bar{\psi }}^{\prime }(x^{\prime })=\bar{\psi }(x)S^{-1}$ and hence the Dirac scalar bilinear

A symmetry of a classical field theory maps solutions to solutions; for the Noether theorem we require a continuous variational symmetry of the action, allowing its density to change by a total divergence. This is stronger than simply observing an accidental symmetry of one solution. Consider first-derivative fields ${\varphi }_a$ and defineThe Euler-Lagrange field equations are $E_a=0$. For a fixed-coordinate variation ${\delta }_0{\varphi }_a=\epsilon Q_a$, the product rule gives the off-shell identityThis identity is the proof's essential integration-by-parts step.

For a general spacetime symmetry write $x^{\prime \mu }=x^{\mu }+\epsilon {\xi }^{\mu }$ and ${\varphi }_a^{\prime }(x^{\prime })={\varphi }_a(x)+\epsilon {\Delta }_a$. At a fixed point $Q_a={\Delta }_a-{\xi }^{\nu }{\partial }_{\nu }{\varphi }_a$, while the volume-element variation contributes ${\partial }_{\mu }(\mathcal{L}{\xi }^{\mu })$. Suppose action invariance takes the formCombining with the preceding identity provesThis is the Noether current for a spacetime symmetry. It also covers internal symmetries, for which $\xi =0$. A constant parameter corresponds to one Noether current; several independent parameters give several currents. Integrating the continuity equation givesThus the Noether charge is conserved when the fields have suitable falloff or no flux through the spatial boundary. These are classical identities; quantum symmetries may additionally require absence of an quantum anomaly.

Translations conserve energy and momentum. If the density has no explicit coordinate dependence, take the active fixed-coordinate variation $Q_a={\partial }_{\nu }{\varphi }_a$ for translation in direction $\nu$. Then ${\delta }_0\mathcal{L}/\epsilon ={\partial }_{\nu }\mathcal{L}$, and the internal-form proof with $K^{\mu }={\delta }^{\mu }{}_{\nu }\mathcal{L}$ gives the canonical energy-momentum tensorIts four spatial integrals are the conserved energy and momentum. For a real scalar field with density $\frac{1}{2}(\partial \varphi {)}^2-V(\varphi )$ this becomes $T^{\mu \nu }={\partial }^{\mu }\varphi {\partial }^{\nu }\varphi -{\eta }^{\mu \nu }\mathcal{L}$. Direct differentiation gives ${\partial }_{\mu }{T}^{\mu \nu }=(\square \varphi +V^{\prime }(\varphi )){\partial }^{\nu }\varphi$, which vanishes by the field equation.