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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 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.

The Noether gauging procedure starts by allowing a constant symmetry parameter to depend on position. For an internal symmetry written as $\delta {\varphi }^i={ϵ}^A{T}_A{\varphi }^i$, integration by parts identifies the Noether current:up to a boundary term. Introduce a gauge field $A_{\mu }^A$ and add $-g{A}_{\mu }^A{j}_A^{\mu }$, with the inhomogeneous gauge transformation $\delta A_{\mu }^A=g^{-1}{\partial }_{\mu }{ϵ}^A+\cdots$. This cancels the displayed variation. The current coupling has its own variation, so one iterates: add the needed higher-order interactions and corrections to the gauge transformations, and require closure of the local symmetry algebra. For an ordinary gauge theory, this reorganizes derivatives into gauge covariant derivatives and fixes the nonlinear gauge field strength. Gauging is therefore more than replacing a constant parameter by a function.

For rigid supersymmetry, the corresponding Noether current is the spinor-valued supercurrent $S^{\mu }$. A local Majorana spinor parameter gives a variation proportional to $({\partial }_{\mu }\overline{ϵ})S^{\mu }$. A gravitino ${\Psi }_{\mu }$, with an inhomogeneous transformation $\delta {\Psi }_{\mu }\sim {\kappa }^{-1}{\partial }_{\mu }ϵ$, cancels it through a coupling proportional to $\kappa {\bar{\Psi }}_{\mu }{S}^{\mu }$. The commutator of two rigid supersymmetry transformations is a translation. With local parameters, the translation parameter becomes position dependent, so the completion must also gauge translations: it introduces the vierbein, diffeomorphism invariance of general relativity, and local Lorentz transformations. The gravitino and graviton thus belong to the same supergravity multiplet.

For pure minimal four-dimensional supergravity, start with the free massless Fierz-Pauli action and the free Rarita-Schwinger field kinetic term. The Noether gauging procedure replaces the former by the Einstein-Hilbert action and couples the latter to the vierbein and spin connection. In a canonical normalization the leading transformations can be writtenAn overall sign or phase changes with the curvature and Majorana spinor conventions, but the relative normalization must be used consistently. To see why these terms fit together, first work to quadratic order in fermions, so the spin connection is torsion free. Varying the Rarita-Schwinger field kinetic term under $\delta {\Psi }_{\mu }={\nabla }_{\mu }ϵ/\kappa$ produces an antisymmetric double spinor covariant derivative. The spinor curvature identity and the Einstein tensor contraction of spin curvature give, in the matching curvature convention,The resulting Einstein tensor term cancels the variation of the Einstein-Hilbert action under $\delta e_{\mu }{}^a$. This is the essential first nontrivial cancellation in the construction.

At higher fermion order, the variations of the vierbein and spin connection produce further terms. Treating the spin connection as independent makes its Euler-Lagrange field equation algebraic. Its solution is a gravitino-induced torsion connection. With ${\Psi }_a=e_a{}^{\mu }{\Psi }_{\mu }$ and ${\gamma }^{ab}=[{\gamma }^a,{\gamma }^b]/2$, a conventional canonical normalization givesEquivalently, with $T^a=d{e}^a+{\tilde{\omega }}^a{}_b\wedge e^b$, the torsion tensor obeysThe Majorana Grassmann bilinear interchange makes the right-hand side antisymmetric in $\mu ,\nu$. Eliminating the spin connection generates the four-fermion completion. In the 1.5-order formalism one substitutes its solution but varies the other fields holding the connection fixed: the omitted chain-rule term is $(\delta S/\delta \omega )\delta \tilde{\omega }=0$. The remaining cubic fermion variations cancel by the Clifford algebra and Fierz rearrangements. This produces the displayed supergravity action in its compact connection-dependent form. Calling this formulation on shell means that no independent auxiliary fields have been retained and that closure of the supersymmetry algebra uses the propagating Euler-Lagrange field equations; it does not mean that action invariance is obtained only by setting every variation to zero on a solution.

In the antisymmetrized kinetic term, ${\Psi }_{\sigma }$ is a spinor-valued one-form and the derivative is conventionally writtenThis is the Lorentz-covariant exterior-derivative convention for that term. If one instead uses the full affine covariant derivative of a vector-spinor, a coordinate-index connection is also present and the antisymmetrization must include its torsion tensor correction. The crucial physical distinction from pure metric general relativity is the spin connection contains gravitino-induced contorsion, rather than being the torsion-free Levi-Civita connection. It reduces to the ordinary spinor covariant derivative when the gravitino vanishes.

The chirality matrix is covariantly constant under the metric-preserving spin connection and anticommutes with every ${\gamma }^{\mu }$. HenceThus multiplication by ${\gamma }_5$ reverses the sign of the mass parameter. At $m=0$ it maps solutions into solutions of the same Dirac equation.

Use the spinor curvature identityHere the rough Laplacian includes the affine connection on the index of the first derivative. Clifford expansion givesIn the last contraction, the four-gamma part vanishes by the first Bianchi identity, the two-gamma parts vanish by symmetry of the Ricci tensor, and the scalar contractions give $-2R$. Therefore the Lichnerowicz spinor-square formula isMultiplying the first-order equation by $-/\nabla +m$ now provesThe explicit curvature coefficient is fixed by the minimal spinor coupling; it is not an arbitrary scalar-field curvature coupling.

For the general metric-preserving connection, define $T^{\rho }{}_{\mu \nu }={\Gamma }_{\mu \nu }^{\rho }-{\Gamma }_{\nu \mu }^{\rho }$ and let $R^T$ be its curvature. Distinguish the spin connection acting on the spinor from the full second derivative acting also on its derivative index. Their antisymmetric relation isThus the torsion correction to the Dirac square gives the exact squared field equationGeneral torsion removes the curvature symmetries used in the preceding simplification. The curvature contraction may now contain nonscalar Clifford terms, in addition to the displayed first-order torsion term; simply replacing $R$ by a torsionful scalar in the torsion-free answer is not valid.

Equivalently, write the connection as Levi-Civita plus contorsion tensor $C_{\mu ab}$. Then $/{\nabla }_T=/{\nabla }_0+Q$, where $Q={\gamma }^{\mu }{C}_{\mu ab}{\gamma }^{ab}/4$, and the square contains $\{/{\nabla }_0,Q\}+Q^2$. This exhibits the extra couplings through derivatives of contorsion, first-order spinor derivatives and quadratic contorsion. Metric compatibility still preserves ${\gamma }_5$, so the mass-sign result remains valid. Spin therefore couples directly to curvature through its connection and the universal $R/4$ term, and to torsion through additional spin-dependent terms. The scalar reduction in the torsion-free square does not mean the full spinor transport is insensitive to the remaining Riemann curvature.

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.

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.

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.