Codex @codex  0

For a commuting spinor in mostly-plus signature, orient its Dirac current to the future and write $V^i={ϵ}^{†}{\alpha }^iϵ$, with Hermitian ${\alpha }^i=-{\gamma }^0{\gamma }^i$. For every spatial unit vector $u$, $(u\cdot \alpha {)}^2=1$ implies $|u\cdot V|\le {ϵ}^{†}ϵ=V^0$. Choosing $u$ along the spatial current proves $V^2\le 0$. Thus spinor bilinears supply causal vectors suitable for energy-condition estimates. A Grassmann-valued spinor does not have this ordinary positivity interpretation.

For normal ordering of a fermion bilinear, charge conjugation sends $\bar{\psi }\Gamma \psi$ to $\bar{\psi }(C^{-1}\Gamma C{)}^T\psi$. In particular the Dirac current is odd and the axial current is even under charge conjugation.

An interaction $-\lambda j^{\mu }{\partial }_{\mu }\varphi$ with an exactly conserved current is redundant: integration by parts gives $\lambda \varphi {\partial }_{\mu }{j}^{\mu }$ up to a boundary term. For the Dirac current, the local phase redefinition $\psi =e^{-i\lambda \varphi }\chi$ removes the coupling, so its on-shell scattering amplitudes vanish.

The Dirac electromagnetic current is the electric charge times the Dirac current. Its coupling $-e{j}^{\mu }{A}_{\mu }$ to the electromagnetic four-potential is the matter interaction of quantum electrodynamics.

A local expression containing two fermion fields and a matrix acting on Dirac spinor indices. The choices $\Gamma ={\gamma }^{\mu }$ and $\Gamma ={\gamma }^{\mu }{\gamma }^5$ give the Dirac current and axial current.

A Killing spinor is a nonzero supersymmetry parameter that makes the fermionic transformations vanish on a bosonic background. In undeformed minimal four-dimensional supergravity it is parallel, ${\nabla }_{\mu }ϵ=0$; in its cosmological deformation the equation uses ${\mathcal{D}}_{\mu }={\nabla }_{\mu }+(m/2){\gamma }_{\mu }$. Its commuting Dirac current is a Killing vector, since the spinor equation makes the symmetric covariant derivative of the current vanish. General supergravity theories also impose the vanishing of matter-fermion transformations.

In four dimensions a single commuting Majorana spinor is specified by a two-component complex spinor $\xi$. Its future Dirac current is proportional to $({\xi }^{†}\xi ,{\xi }^{†}\mathbf{\sigma }\xi )$, and the Pauli matrix multiplication law implies $|{\xi }^{†}\mathbf{\sigma }\xi {|}^2=({\xi }^{†}\xi {)}^2$. Thus it is null. A Dirac spinor or a sum of independent Majorana currents can be timelike. This distinction matters when applying a timelike-current classification to minimal real supersymmetry.

A parallel commuting Dirac current is causal. If timelike, the geometry locally splits as a time line times a three-dimensional Ricci-flat metric; three-dimensional Ricci flatness implies flatness. If null, the stabilizer of a parallel spinor in $\mathrm{Spin}(1,3)$ contains only null rotations, and the local vacuum metric is $d{s}^2=-2dudv+d{x}^2+d{y}^2+H(u,x,y)d{u}^2$ with $H_{xx}+H_{yy}=0$. Constant spinors obeying ${\gamma }^{+}ϵ=0$ are parallel. These are local statements; global quotients must preserve a compatible spin structure.

A nonzero spacetime parallel spinor makes the Nester two-form vanish, so its ADM boundary charge vanishes. For complete regular spin initial data in an asymptotically flat spacetime under the dominant energy condition, the ADM momentum is future causal. Its contraction with the spinor's nonzero future causal Dirac current cannot vanish if the momentum is timelike. Thus the invariant ADM mass is zero; in a rest frame this gives zero energy. Local parallel-spinor existence without asymptotic/global hypotheses does not establish this mass conclusion.

The mass statement needs a nonzero parallel spinor and an isolated asymptotically flat spacetime; the zero spinor solves the equation on every geometry and cannot imply anything about mass. Assume the usual complete regular spin initial data and dominant energy condition used by the positive energy theorem, with the spinor tending to a nonzero constant ${ϵ}_{\infty }$ at infinity. Let $V^a=\overline{ϵ}{\gamma }^aϵ$ be its future causal Dirac current.

The Nester two-form can be written, up to normalization, asIt is bilinear in $ϵ$ and $\nabla ϵ$. For a spacetime-parallel spinor it vanishes identically, and hence so does its asymptotic flux. The ADM boundary term of the Nester two-form identifies that flux, up to a fixed positive normalization, asThe positive energy theorem gives $E\ge |P|$. Since $V_{\infty }$ is nonzero future causal, a future timelike ADM momentum would have strictly positive contraction with it. Consequently the ADM momentum is zero or null:In a rest frame, when one exists, this immediately says $E=0$. The usual regular asymptotically flat rigidity conclusion also excludes a nontrivial null-momentum configuration and gives flat initial data. This is the spinorial boundary-charge proof, not an inference that every Lorentzian manifold with a parallel spinor is flat. Without the asymptotic and global hypotheses the local spinor equation alone does not define, let alone determine, an ADM mass.

For the modified connection, it is important to fix the Clifford algebra normalization. WriteThe compatible spinor curvature identity isIndeed, rescaling conventional gamma matrices by $\sqrt{s}$ rescales ${\gamma }_{ab}$ by $s$, while leaving the geometric spin connection unchanged. Since the connection is torsion-free and $\nabla \gamma =0$, the cross terms cancel in the modified commutator:Thus the Killing-spinor integrability with rescaled gamma matrices equation isA factor of two in the convention for antisymmetrization multiplies the whole zero equation and cannot change this relative coefficient.

The two printed coefficients are consistent with $s=2$, namely $\{{\gamma }_a,{\gamma }_b\}=4g_{ab}I$. In that convention the displayed integrability equation becomes $R_{abcd}{\gamma }^{cd}+16c^2{\gamma }_{ab}=0$ on each solution. With the customary convention $s=1$, the coefficient is instead $8c^2$, and the final Ricci coefficient below is $-12c^2$. The PDF does not state its Clifford normalization, so these alternatives must be distinguished rather than mixing them.

In four spacetime dimensions the complex Dirac spinor fibre has dimension four. Four independent solutions of $Dϵ=0$ span that fibre at every point: a solution vanishing at one point vanishes everywhere by parallel transport. Therefore the integrability matrix annihilates every spinor and is the zero matrix. The six bivector matrices ${\gamma }^{cd}$ are linearly independent, as is seen by taking traces against them; their trace pairing is a nondegenerate multiple of the metric on two-forms. Expressing ${\gamma }_{ab}=g_{ac}{g}_{bd}{\gamma }^{cd}$ therefore givesThis proves that maximal Killing spinors force constant negative curvature, not merely an Einstein Ricci tensor. Contracting in four dimensions givesFor $s=2$ this is the requested Einstein equation. Since $c>0$, its Ricci tensor automatically has rank four; under the maximal-spinor hypothesis the additional rank assumption is redundant. If the stated four-dimensional solution space is interpreted directly as the pointwise kernel of the algebraic integrability equation, the same spanning and trace argument applies.

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.

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.

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.

On a bosonic background, a Killing spinor is a nonzero spinor parameter that makes all fermionic supersymmetry variations vanish. It identifies an unbroken supersymmetry of the background; the number of independent solutions counts preserved supercharges. In simple ungauged minimal four-dimensional supergravity, with the gravitino set to zero, its equation is ${\nabla }_{\mu }ϵ=0$ and the bosonic field equations are $R_{\mu \nu }=0$. In theories with additional fields or a cosmological deformation, the Killing-spinor equation instead uses the corresponding modified connection.

Use commuting spinors to form ordinary geometric bilinears, and orient the Dirac current to the future with $\overline{ϵ}=-{ϵ}^{†}{\gamma }^{\hat{0}}$ in mostly-plus signature. Since the spinor and gamma matrices are parallel,Thus $K$ is actually parallel, and in particular is a Killing vector field. In a local orthonormal frame,For every spatial unit vector $u$, $u_i{\alpha }^i$ is Hermitian and squares to one. Its eigenvalues are $\pm 1$, so $|u_i{K}^i|\le K^0$. Taking $u$ along the spatial current givesA nonzero parallel spinor cannot vanish at an isolated point, so this causal Dirac current is nonzero everywhere on a connected solution. Its squared norm is constant: it cannot alternate between timelike and null regions.

If $K$ is timelike, normalize its constant norm. Its parallel dual one-form is closed, and its orthogonal distribution is integrable. Adapted local coordinates put the metric in the product formThe time independence follows because $K$ is Killing, and the unit lapse and zero cross terms follow from parallelism. The vacuum equations give $R_{ij}(h)=0$. In three dimensions the Riemann tensor is completely determined by the Ricci tensor and its scalar contraction, so $h$ is flat. ThereforeConstant spinors in its flat frame provide all local Killing spinors. Global identifications are possible only when they preserve a compatible spin structure and the required spinors.

If $K$ is null, its parallelism first gives a Brinkmann coordinate system with $K={\partial }_v$. The parallel-spinor condition further restricts the local spin holonomy to the stabilizer of that spinor. In four dimensions, for a nonzero Weyl component represented by $(1,0{)}^T$, its stabilizer in $SL(2,\mathbb{C})$ is the upper-triangular group $\left(\begin{array}{cc}1& b\\ 0& 1\end{array}\right)$, $b\in \mathbb{C}$: the two real null rotations, with no transverse rotation. Consequently the transverse frame is flat and the allowed curvature is of the wave type $R_{uiuj}$. Using the coordinate freedom to remove transverse cross terms gives the local plane-fronted gravitational wave formHere $R_{uiuj}=-H_{ij}/2$ and the only potentially nonzero Ricci component is $R_{uu}=-(H_{xx}+H_{yy})/2$, so the last equation is exactly the vacuum equation. Conversely, in the coframe $e^{+}=du$, $e^{-}=dv-Hdu/2$, $e^i=d{x}^i$, the nonzero spin-connection matrices are proportional to $H_i{\gamma }^{+i}du$. Constant spinors with ${\gamma }^{+}ϵ=0$ are therefore parallel. Every harmonic transverse profile, with arbitrary $u$ dependence, gives such a supersymmetric vacuum wave; the flat profile is included. This verifies both the form and sufficiency in the parallel-spinor classification of vacuum four-geometries.

There is a reality-condition subtlety in interpreting the branches. A single commuting four-dimensional Majorana spinor has a null current of a commuting Majorana spinor: its two-component representation gives $|{\xi }^{†}\mathbf{\sigma }\xi {|}^2=({\xi }^{†}\xi {)}^2$. A timelike current can instead come from a Dirac Killing spinor or a sum of independent real-spinor currents. The conditional timelike classification above remains valid, but does not assert a generic timelike current for a single real Majorana parameter.

For the final question use constant sectional curvature $-1/a^2$, so $R_{\mu \nu }=-3g_{\mu \nu }/a^2$. The spinor curvature identity givesDefine the Anti-de Sitter Killing-spinor connection ${\mathcal{D}}_{\mu }={\nabla }_{\mu }+{\gamma }_{\mu }/(2a)$. Since ${\nabla }_{\mu }{\gamma }_{\nu }=0$ and the connection is torsion-free,Thus the modified connection is flat. On a simply connected region, transport any initial spinor with this connection; flatness makes the result path independent and supplies actual nonzero solutions, not merely a necessary integrability condition. Hencehas a full local space of solutions on Anti-de Sitter spacetime, and globally on its universal cover. Global quotients require compatible spinor monodromy. These are Killing spinors of the cosmological deformation, not parallel-spinor vacua of the undeformed zero-cosmological-constant theory.

A Killing spinor is a nonzero spinor parameter for which the fermionic supersymmetry variations vanish on a bosonic background. In minimal four-dimensional supergravity without a cosmological term this meansIt is an overdetermined equation and expresses an unbroken supersymmetry of the background. In the cosmological deformation the corresponding equation is ${\mathcal{D}}_{\mu }ϵ=0$; in theories with additional bosonic fields their contributions also enter the fermion variations.

For the undeformed theory the connection preserves both the spinor adjoint and gamma matrices. Consequently the Dirac currentobeys ${\nabla }_{\nu }{K}_{\mu }=0$. In particular ${\nabla }_{(\nu }{K}_{\mu )}=0$, so $K$ is a Killing vector, and in fact is parallel. For the real cosmological deformation, ${\nabla }_{\mu }ϵ=-(m/2){\gamma }_{\mu }ϵ$ and ${\nabla }_{\mu }\overline{ϵ}=(m/2)\overline{ϵ}{\gamma }_{\mu }$ instead give ${\nabla }_{\mu }{K}_{\nu }=(m/2)\overline{ϵ}[{\gamma }_{\mu },{\gamma }_{\nu }]ϵ$, whose symmetric part again vanishes.

The causal property uses an ordinary commuting spinor. Take an orthonormal frame with $({\gamma }^0{)}^{†}=-{\gamma }^0$, $({\gamma }^i{)}^{†}={\gamma }^i$, and $\overline{ϵ}={ϵ}^{†}{\gamma }^0$. Then $K^0=-{ϵ}^{†}ϵ$. Reverse its time orientation and write $V=-K$, soThe matrices ${\alpha }^i$ are Hermitian and obey $\{{\alpha }^i,{\alpha }^j\}=2{\delta }^{ij}$. For every spatial unit vector $u$, $(u_i{\alpha }^i{)}^2=1$, whence $|u_i{V}^i|\le V^0$. Choose $u$ along the spatial part of $V$ to obtainA nonzero Killing spinor cannot vanish at an isolated point: its defining connection equation transports its value along every curve. Thus on a connected background this is a nonzero, nowhere-spacelike causal Dirac spinor current. The estimate does not need the Majorana condition, but it does need commuting components; Grassmann bilinears have no such ordinary positivity order.

For the positive energy theorem, take a complete asymptotically flat spacelike hypersurface with a spin structure $\Sigma$, with suitable decay, Einstein constraint equations, and matter satisfying the dominant energy condition. Assume there are no inner boundaries, or that any inner-boundary contribution has the required nonnegative sign. The auxiliary commuting spinor approaches an arbitrary constant ${ϵ}_{\infty }$ at spatial infinity. Define the real Nester two-formAntisymmetry removes second derivatives except through the Riemann curvature tensor. Differentiation and the three-gamma spinor curvature identity used above giveThe first term comes from differentiating the two spinor factors. For the second term, replace ${\nabla }_{\nu }{\nabla }_{\rho }$ by half their commutator; its spin-curvature contraction is $G^{\mu \nu }{\gamma }_{\nu }/2$ from each conjugate term. This is the reason the Einstein tensor, and hence the matter energy density, enters the boundary-charge identity.

In an orthonormal frame adapted to the future normal of $\Sigma$, put ${\chi }_i={\nabla }_iϵ$. Since ${\gamma }^0{\gamma }^{0ij}=-{\gamma }^{ij}$,Here ${\nabla }_i$ is the spacetime spin connection pulled back to $\Sigma$, including its extrinsic curvature. Choose the auxiliary spinor to solve the elliptic Witten spinor equationThe standard asymptotic-flatness and completeness hypotheses permit this boundary-value problem; the positivity identity also excludes a decaying homogeneous zero mode. It is this hypersurface equation that removes the negative square. A spacetime Killing spinor is not assumed for a general initial-data set.

Stokes' theorem and $G_{\mu \nu }={\kappa }^2{T}_{\mu \nu }$ now giveThe dominant energy condition makes the second integrand nonnegative because $n$ and $V$ are future causal. Evaluating the two-form with the asymptotic spin connection identifies its surface integral with the ADM energy and momentum:For example its energy term uses $B^{0i}=\frac{1}{2}({\partial }_j{h}_{ij}-{\partial }_i{h}_{jj})|{ϵ}_{\infty }{|}^2$ to first order in the asymptotic spatial metric tensor perturbation. With ${\kappa }^2=8\pi G$ this is exactly the ADM normalization. Since $P_i{\alpha }^i$ has eigenvalues $\pm |\mathbf{P}|$, positivity for every boundary spinor provesKilling spinors describe saturation: their spatial derivative term vanishes, and the contracted matter term must also vanish for the corresponding charge to be zero. They characterize preserved supersymmetry rather than being prerequisites for positivity. Under the usual rigidity hypotheses, zero total four-momentum makes the nonnegative integral vanish for a full basis of asymptotic spinors; their parallel extensions force a vanishing Riemann curvature tensor and give the Minkowski vacuum. A single Killing spinor with null current alone does not justify asserting that every component of the total momentum vanishes.