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In the conformal frame for $g={\Omega }^2\eta$, identify the spinor fibres using the chosen vierbein. If ${ϵ}_0$ is a flat constant spinor field, the spin connection of a conformally flat metric gives ${\nabla }_aϵ={\hat{\gamma }}_a{\hat{\gamma }}^c({\partial }_c\log \Omega )ϵ/2$. Contracting with curved ${\gamma }^a={\Omega }^{-1}{\hat{\gamma }}^a$ yields ${\gamma }^a{\nabla }_aϵ=n{\Omega }^{-1}{\hat{\gamma }}^c({\partial }_c\log \Omega )ϵ/2$. Thus ${\nabla }_aϵ={\gamma }_a{\gamma }^b{\nabla }_bϵ/n$: the rescaled parallel spinor is a twistor spinor.

An orthonormal coframe in spacetime can be rotated by a position-dependent Lorentz matrix without changing the spacetime metric tensor. Its spin connection transforms as ${\omega }^{\prime }=\Lambda \omega {\Lambda }^{-1}-d\Lambda {\Lambda }^{-1}$. A spinor field transforms through a chosen spin lift ${\psi }^{\prime }=S(\Lambda )\psi$, not through the vector matrix $\Lambda$. With $S^{-1}{\gamma }^{A}S={\Lambda }^A{}_B{\gamma }^B$, an infinitesimal parameter gives $S=1+\frac{1}{4}{\alpha }_{AB}{\gamma }^{AB}+O({\alpha }^2)$, where ${\gamma }^{AB}=\frac{1}{2}[{\gamma }^A,{\gamma }^B]$. Consistent global transition lifts require a spin structure; a $2\pi$ spatial rotation lifts to minus the identity on spinors.

A parallel spinor field is annihilated by the spinor covariant derivative. Its value at one point determines it on a connected domain by parallel transport. The spinor curvature identity implies $R_{\mu \nu AB}{\gamma }^{AB}ϵ=0$, but this local condition alone need not settle global existence: the spin structure and holonomy also matter. Lorentzian parallel spinors need not force the whole Ricci tensor to vanish, unlike the positive-definite case.

The positive energy theorem concerns the total gravitational energy of isolated systems, rather than a pointwise gravitational energy density. Its standard asymptotically flat form isHere are the hypotheses used in a spinorial proof. Take smooth, complete, three-dimensional initial data in general relativity $(\Sigma ,h,k)$ with a spin structure, no inner boundary, and finitely many asymptotically Euclidean ends. Assume the usual differentiable falloff $h_{ij}-{\delta }_{ij}=O(r^{-1})$, $\partial h=O(r^{-2})$, $k_{ij}=O(r^{-2})$, with sufficiently decaying further derivatives and integrable matter densities. The Hamiltonian constraint and momentum constraint must hold, and the matter must satisfy the dominant energy condition. These hypotheses exclude a naked singularity or an unaccounted boundary flux from the argument. An oriented three-manifold admits a spin structure; in higher dimensions the spinorial proof needs this as an additional assumption. Work in signature $(-+++)$ and units $G=c=1$.

Let $n$ be the future unit normal, choose $k_{ij}=-g({\nabla }_{i}n,e_j)$, and define $\mu =T_{ab}{n}^a{n}^b$ and $J_i=-T_{ab}{n}^a{e}_i^b$. With this convention the Hamiltonian constraint and momentum constraint readThe dominant energy condition implies $\mu \ge |J{|}_h$. The asymptotic charges are the ADM energy and ADM momentum,These are boundary charges; a sign condition on $\mu$ alone does not visibly determine their sign.

Choose gamma matrices obeying $\{{\gamma }^a,{\gamma }^b\}=2g^{ab}$, with spatial matrices Hermitian and ${\gamma }^{\hat{0}}$ anti-Hermitian. The relevant derivative is the Sen spinor connection, not just the intrinsic spin connection:The symmetric extrinsic curvature makes $\mathcal{D}={\gamma }^i{D}_i^{(3)}-\frac{1}{2}(\mathrm{tr}k){\gamma }^{\hat{0}}$. This is an elliptic differential operator with the ordinary spatial Dirac operator as principal part. Solve the Witten spinor equation $\mathcal{D}ϵ=0$, with $ϵ\to {ϵ}_{\infty }$ constant at the end under consideration and zero asymptotic constants at any other ends.

The central algebraic ingredient is the Witten-Nester identity. Define $\overline{ϵ}=-{ϵ}^{†}{\gamma }^{\hat{0}}$, the future-directed Dirac current $K^a=\overline{ϵ}{\gamma }^aϵ$, and the Nester two-formIn an orthonormal frame $K^0=|ϵ{|}^2$. Since $-{\gamma }^{\hat{0}}{\gamma }^i{q}_i$ is Hermitian with eigenvalues $\pm 1$ for any unit spatial vector $q$, $|\mathbf{K}|\le K^0$. Thus the Dirac current is future causal. Differentiate the Nester two-form by the product rule. The antisymmetric second-derivative terms reduce using the spinor curvature identityThe Clifford algebra and the first Bianchi identity turn the curvature contraction into the Einstein tensor. The spatial derivative terms give the difference of two squares. Projecting onto $\Sigma$ therefore givesThe constraints are exactly the normal components of the Einstein field equations needed here. Moreover,For a solution of the Witten spinor equation, both terms left in this divergence are nonnegative.

There is an essential analytic step: an asymptotically constant spinor field solving the Witten spinor equation must actually exist. Extend the chosen constant smoothly from infinity to a field ${ϵ}_0$. Then $\mathcal{D}{ϵ}_0=O(r^{-2})$ is square-integrable. For compactly supported fields $\eta$, the same Witten-Nester identity has zero boundary flux and givesThe asymptotically Euclidean Hardy inequality in Euclidean space, local elliptic estimates and this identity supply the coercive estimate in the weighted first-order Sobolev space of decaying corrections. One way to see why a compact-region error cannot spoil coercivity is to assume the estimate fails and take a normalized sequence whose Dirac operator tends to zero. Local compactness and the estimate on the ends produce a decaying homogeneous solution. Integrating the identity with cutoffs shows that such a solution is parallel for the Sen spinor connection. A parallel field tending to zero at infinity is zero, contradicting its normalization. The end estimates exclude escape of the normalized sequence to infinity.

More constructively, on this completed correction space solvefor every compactly supported $\chi$, using the Lax-Milgram theorem. Set $ϵ={ϵ}_0+\eta$. Since $\mathcal{D}$ is formally skew-adjoint in the positive spatial spinor inner product, the residual $\xi =\mathcal{D}ϵ$ is a square-integrable weak solution of $\mathcal{D}\xi =0$. Elliptic regularity makes it smooth. Cutoff integration of the identity excludes a nonzero square-integrable homogeneous solution: it would be parallel, whereas a nonzero parallel field has nonzero limiting norm on an asymptotically Euclidean end and cannot be square-integrable. Hence $\xi =0$. Weighted elliptic estimates then give the required asymptotic decay of $\eta$. This supplies elliptic existence of a Witten spinor, rather than assuming it from the sign of a formal integral.

Finally integrate the Witten-Nester identity over $\Sigma$. The ADM boundary term of the Nester two-form isTo identify this boundary term, expand the asymptotic spin connection to first order in $h-\delta$. Its flux is half of $({\partial }_j{h}_{ij}-{\partial }_i{h}_{jj})s^i{K}_{\infty }^0$; the extrinsic curvature contribution is $-(k_{ij}-h_{ij}\mathrm{tr}k)s^j{K}_{\infty }^i$. Their coefficients give precisely the two charges displayed above. The derivative of the decaying spinor correction has zero leading integrated flux by antisymmetry, and the higher-order terms vanish with the falloff. ThusChoose a normalized asymptotic spinor field that is an eigenvector with eigenvalue $+1$ of $-{\gamma }^{\hat{0}}{\gamma }^i{q}_i$. Then $K_{\infty }^0=1$ and ${\mathbf{K}}_{\infty }=q$. Choosing $q=\mathbf{P}/|\mathbf{P}|$ proves $E\ge |\mathbf{P}|$; if $\mathbf{P}=0$, any such choice gives $E\ge 0$. This is positivity in every asymptotic frame, not only the zero-momentum case.

The zero-energy rigidity also follows from the vanishing integrals. If $E=0$, then $\mathbf{P}=0$ and the construction for a basis of asymptotic spinors produces a basis of parallel fields. Their spinor curvature identity forces the spatially pulled-back spacetime curvature to vanish; the nonnegative matter terms vanish too. The resulting Gauss–Codazzi equations are those of a hypersurface in Minkowski spacetime, and its vacuum development is flat. Completeness and the ordinary Euclidean asymptotic end rule out a nontrivial flat quotient. The initial hypersurface need not have $k=0$: a curved spacelike slice of Minkowski spacetime also has zero ADM energy. The proof works by converting an asymptotic energy charge into a nonnegative bulk integral, with the spinor equation and its existence supplying the crucial bridge.

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.

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.

Lorentz symmetry conserves angular momentum and boost charges. For that scalar example, $T^{\mu \nu }$ is symmetric, so the Lorentz transformation currents can be writtenTheir divergence is $T^{\rho \sigma }-T^{\sigma \rho }=0$. Spatial pairs give angular momentum, and time-space pairs give the conserved boost charges. For a spinor field or a vector field, the intrinsic field variation supplies an additional spin current; equivalently the Belinfante-Rosenfeld stress-energy tensor combines spin and orbital pieces into a symmetric stress tensor. The nonunitarity of the finite component Spinor representation of the Lorentz group does not obstruct conservation of these physical spacetime charges.

Global phase symmetry conserves particle charge. For a complex scalar field with density ${\partial }_{\mu }{\Phi }^{\ast }{\partial }^{\mu }\Phi -V({\Phi }^{\ast }\Phi )$, the constant transformation $\Phi \mapsto e^{-i\alpha }\Phi$ leaves the density invariant. Using $Q_{\Phi }=-i\Phi$ and $Q_{{\Phi }^{\ast }}=i{\Phi }^{\ast }$ gives the Noether currentFor a Dirac field, the corresponding variations $Q_{\psi }=-i\psi$, $Q_{\bar{\psi }}=i\bar{\psi }$ yield instead $j^{\mu }=\bar{\psi }{\gamma }^{\mu }\psi$, the Dirac current. Its divergence vanishes by the Dirac equation and adjoint Dirac equation. The Yukawa interaction of part 2 also preserves this phase symmetry, because each bilinear has one $\psi$ and one $\bar{\psi }$. Thus it preserves fermion number and forbids annihilation of two fermions into a single scalar. These currents express charge or particle-minus-antiparticle-number conservation, rather than conservation of total particle count in arbitrary interacting processes.

A global symmetry in field theory uses a spacetime-independent parameter and ordinarily relates physically distinct configurations. A gauge symmetry instead has arbitrary local parameters and represents redundancy in the description. For example, the Maxwell field is unchanged physically by $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\alpha (x)$: its electromagnetic field tensor is unchanged. For charged matter use $D_{\mu }={\partial }_{\mu }+iq{A}_{\mu }$ together with $\psi \mapsto e^{-iq\alpha (x)}\psi$. Differentiating shows $D_{\mu }^{\prime }{\psi }^{\prime }=e^{-iq\alpha }{D}_{\mu }\psi$, so the coupled Dirac action is locally gauge invariant. A local phase applied to free matter alone produces extra derivatives of $\alpha$ and is not an invariance; the compensating gauge field is necessary.

Local arbitrariness gives identities among field equations, rather than an independent physical charge for every function $\alpha$. In the pure Maxwell field, action invariance under a compactly supported $\alpha$ impliesThus ${\partial }_{\nu }{E}^{\nu }=0$ holds off shell, directly because $F^{\mu \nu }$ is antisymmetric. This exemplifies the Noether second theorem. Gauge fixing and the Gupta-Bleuler null-state quotient remove redundant photon components; they do not remove a physical global charge. Gauge transformations that are nontrivial at a boundary can carry boundary charges, so the redundancy statement concerns transformations satisfying the chosen boundary conditions. Finally, discrete symmetries such as the sign reversal of a real scalar or parity symmetry in quantum field theory can be important without furnishing a continuous-parameter Noether current.

Choose a local orthonormal coframe $e^A$ with $g={\eta }_{AB}{e}^A{e}^B$. The Levi-Civita connection is represented by the connection 1-forms satisfyingThe gamma matrices satisfy the Clifford algebra relations $\{{\gamma }^A,{\gamma }^B\}=2{\eta }^{AB}I$. The matrices $\frac{1}{4}[{\gamma }^A,{\gamma }^B]$ represent the infinitesimal orthogonal transformations on spinor fields; lifting the frame connection to this representation defines the spinor covariant derivativeBoth frame indices are summed, so each antisymmetric pair occurs twice. Under a local Lorentz transformation with spin lift $S$, the spin connection transforms as ${\Omega }_{\mu }\mapsto S{\Omega }_{\mu }{S}^{-1}-({\partial }_{\mu }S)S^{-1}$ and $ϵ\mapsto Sϵ$; consequently ${\nabla }_{\mu }ϵ\mapsto S{\nabla }_{\mu }ϵ$. This establishes covariance. Global construction requires a compatible spin structure (and the appropriate orientation and time orientation); the formulas themselves work on a local orthonormal frame patch.

The supplied gamma matrices have squares $-I,I,I$ and anticommute pairwise, so $\eta =\mathrm{diag}(-1,1,1)$. Work on a connected regular patch with $r>0$ and $V\ne 0$. Since the metric depends only on $V^2$, we may use its positive smooth square root $V>0$. TakeTheir exterior derivatives are $d{e}^0=-V^{\prime }{e}^0\wedge e^1$, $d{e}^1=0$, and $d{e}^2=(V/r)e^1\wedge e^2$. Solving Cartan's first structure equation and imposing Lorentzian connection-form antisymmetry givesIn particular ${\omega }_{01}=-V{V}^{\prime }dt$ and ${\omega }_{12}=-Vd\theta$. The parallel spinor equations are thereforeDifferentiate the angular equation with respect to $r$ and use ${\partial }_rϵ=0$. This gives $V^{\prime }{\gamma }^1{\gamma }^2ϵ=0$. Since $({\gamma }^1{\gamma }^2{)}^2=-I$, this matrix is invertible. A parallel spinor that vanishes at any point vanishes everywhere on the connected domain, by uniqueness of parallel transport. Thus a nonzero parallel spinor implies $V^{\prime }=0$ throughout the patch. There are no nonzero parallel spinors unless $V$ is constant. The zero spinor field is always a solution, so “nonzero” is essential in interpreting the printed assertion. A smooth original $V$ of either sign is constant on such a patch exactly when $|V|$ is constant; $V=0$ is excluded because the given metric is not defined there.

Write the constant as $c=|V|>0$, and introduce $T=ct$, $\rho =r/c$, and $\varphi =c\theta$. The metric becomesA circle at small $\rho$ has circumference $cP\rho$ if $\theta$ has period $P$, whereas its proper radius is $\rho$. Smoothness requires circumference divided by radius to tend to $2\pi$, not a multiple of $2\pi$. Conversely, when $cP=2\pi$, the coordinates $X=\rho \cos \varphi$, $Y=\rho \sin \varphi$ exhibit a smooth flat spatial disk. HenceFor a prescribed period $2\pi$, this requires $c=1$; otherwise there is a conical singularity at the apex.

For constant $V=c$, the time and radial equations make the components independent of $t,r$. The angular equation has all solutionswhere ${ϵ}_0$ is an arbitrary constant two-component spinor field and ${\gamma }^1{\gamma }^2=-{\gamma }^0$. There are two independent real solutions for real spinor fields, or two independent complex solutions if complex spinors are used. At the smooth angular period, these polar-frame components change sign. This is exactly the polar-frame parallel spinor transformation under the spin lift of a full rotation: the rotating polar orthonormal frame does not extend through the origin. The spin structure extending over the disk uses this antiperiodic polar trivialization; in a Cartesian orthonormal frame the same parallel spinors have constant components and extend smoothly. Requiring periodic components in the rotating frame would select the other spin structure on the punctured disk, which does not extend over its center.

Use coordinate indices $\mu ,\nu$ and Lorentz-frame indices $A,B$, and take ${\eta }_{AB}=\mathrm{diag}(-1,1,1,1)$. Locally choose four independent one-forms $e^A=e^A{}_{\mu }d{x}^{\mu }$ satisfyingOne construction begins with a smooth timelike vector field in a sufficiently small neighborhood, normalizes it, and applies the Gram-Schmidt process to its spacelike orthogonal complement. The dual orthonormal coframe in spacetime gives the vierbein; equivalently its matrix factorizes the metric as $g=e^T\eta e$. The inverse matrix gives the frame $e_A=e_A{}^{\mu }{\partial }_{\mu }$ with $e_A{}^{\mu }{e}^B{}_{\mu }={\delta }_A^B$. This is a local construction: a global frame is not guaranteed by a metric, and a globally defined spinor field requires a compatible spin structure.

The metric does not determine a unique vierbein. A local Lorentz transformation changes the coframe components byThe coordinate index is unchanged in this frame transformation. Restrict to the proper orthochronous Lorentz group for its connected spin lift; orientation-reversing transformations require further structure and are not part of this local spin-group discussion.

Let constant gamma matrices obey the Clifford algebra $\{{\gamma }^A,{\gamma }^B\}=2{\eta }^{AB}$, and write ${\gamma }^{AB}=\frac{1}{2}[{\gamma }^A,{\gamma }^B]$. A spinor field transforms by the spin lift of the Lorentz transformation, not by its vector matrix:For $\Lambda =1+\alpha +O({\alpha }^2)$, with ${\alpha }_{AB}=-{\alpha }_{BA}$, the Clifford algebra identityshows explicitly thatIndeed the commutator of ${\gamma }^C$ with this infinitesimal spin matrix is ${\alpha }^C{}_B{\gamma }^B$. A finite transformation connected to the identity is obtained by exponentiating these generators along a path. Its lift has two possible signs, so spinors are not ordinary Lorentz vectors: a $2\pi$ spatial rotation acts by $-1$, and $4\pi$ by $+1$. Consistent transition lifts define the spin structure; the metric alone does not select it.

For the metric connection, take the Levi-Civita connection and determine the spin connection from the tetrad postulateIt is equivalently determined by $d{e}^A+{\omega }^A{}_B\wedge e^B=0$ and ${\omega }_{\mu AB}=-{\omega }_{\mu BA}$. Explicitly, in terms of the inverse frame,Thus both the coordinate connection and the Lorentz connection 1-form are determined by the metric and chosen frame; no independent torsion has been supplied.

An ordinary derivative fails covariance because ${\partial }_{\mu }(S\psi )=S{\partial }_{\mu }\psi +({\partial }_{\mu }S)\psi$. Seek ${\nabla }_{\mu }\psi =({\partial }_{\mu }+{\Omega }_{\mu })\psi$. Covariance requiresThe coframe's connection obeys ${\omega }^{\prime }=\Lambda \omega {\Lambda }^{-1}-d\Lambda {\Lambda }^{-1}$. Applying the spin representation of the same Lorentz algebra, whose infinitesimal generator was just determined, suppliesThis proves the transformation law rather than choosing the coefficient arbitrarily. It is the spinor covariant derivative induced by the spin connection. With curved matrices ${\gamma }^{\nu }(x)=e_A{}^{\nu }{\gamma }^A$, the same coefficient and tetrad postulate giveso differentiation is compatible with Clifford multiplication and $\{{\gamma }^{\mu }(x),{\gamma }^{\nu }(x)\}=2g^{\mu \nu }(x)$.

Finally, expand the commutator of these derivatives:The spin representation preserves the Lorentz algebra bracket. Consequently $d\Omega +\Omega \wedge \Omega =\frac{1}{4}{R}_{AB}{\gamma }^{AB}$, where $R^A{}_B=d{\omega }^A{}_B+{\omega }^A{}_C\wedge {\omega }^C{}_B$. In components the spinor curvature identity isFor the full second derivative, ${\nabla }_{\mu }({\nabla }_{\nu }\psi )$ also includes $-{\Gamma }_{\mu \nu }^{\rho }{\nabla }_{\rho }\psi$, because its first derivative has a covector index. This term cancels on antisymmetrization for the torsion-free connection. With unit-weight brackets, $X_{[\mu \nu ]}=(X_{\mu \nu }-X_{\nu \mu })/2$, the requested expression is thereforeThis convention distinguishes the antisymmetrized second derivative from the full commutator by its factor of two. In a noncoordinate orthonormal frame, an operator commutator must additionally subtract the derivative along the frame-vector commutator; using the full tensorial second derivative incorporates that correction.

The quantized fermions obey a Clifford algebra, $\{{\hat{\psi }}^a,{\hat{\psi }}^b\}=g^{ab}$. When $M$ is a spin manifold, they act on the spinor bundle andthe Hilbert space of square-integrable spinor fields. The supercharge becomes the Dirac operator, the Hamiltonian is one half of its square, and because $M$ is even dimensional, $(-1{)}^F$ is the spinor chirality operator.

A spin structure is a consistent choice of lifting oriented orthonormal frame changes to their spin double cover, permitting globally defined spinor fields. For a spatial circle in a worldsheet, the two choices give periodic and antiperiodic fermions, hence the Ramond sector and Neveu–Schwarz sector. Local field equations do not choose the global spin structure.