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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.

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.

In flat three-dimensional spacetime with polar spatial orthonormal coframe $(dT,d\rho ,\rho d\varphi )$, the spin connection has ${\omega }_{12}=-d\varphi$. Thus every parallel spinor has the displayed components. Since $({\gamma }^1{\gamma }^2{)}^2=-1$, these components change sign under $\varphi \mapsto \varphi +2\pi$. This is the lift of a full rotation to the spin structure extending over the disk; the polar frame itself does not extend over the origin. In a Cartesian frame the same parallel spinors have constant components.

The vacuum Witten-Nester identity makes a zero-energy Witten spinor parallel under the Sen spinor connection on a complete asymptotically flat slice. In an orthogonal static slicing, extrinsic curvature vanishes, so this spinor is intrinsically parallel. Contracting its spin-curvature integrability condition gives zero spatial Ricci tensor; in three dimensions this implies flatness. The static vacuum equation $D_i{D}_{j}N=N{R}_{ij}(h)$ then gives a parallel lapse gradient, which asymptotic normalization forces to vanish. The time-independent spinor consequently extends to a spacetime parallel spinor, with Minkowski spacetime the complete regular asymptotically flat example.

A twistor spinor has vanishing gamma-trace-free part of its spinor covariant derivative. In $n$ dimensions the displayed equation uses gamma matrices satisfying $\{{\gamma }_a,{\gamma }_b\}=2g_{ab}$. It is weaker than being a parallel spinor. In four dimensions it implies $[{\gamma }_a{\nabla }_b+{\gamma }_b{\nabla }_a-g_{ab}{\gamma }^c{\nabla }_c/2]ϵ=0$.