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Take a complete smooth asymptotically flat spacelike initial-data hypersurface $\Sigma$, carrying a spin structure, without inner boundary. Let $n$ be its future unit normal, $h$ its induced metric, and use the extrinsic curvature convention $k_{ij}=-g({\nabla }_{i}n,e_j)$. Assume the Einstein constraints, the dominant energy condition, and the standard decay sufficient for finite ADM charges. In one asymptotically Cartesian end it is enough here to work with $h_{ij}={\delta }_{ij}+q_{ij}$, $q=O(r^{-1})$, $\partial q=O(r^{-2})$, $k=O(r^{-2})$, and the corresponding differentiated decay and integrability assumptions. The desired conclusion is positivity of the whole ADM four-momentum, not merely of the energy in one selected frame.

Introduce a commuting auxiliary spinor $ϵ$; this is not an anticommuting gravitino field. With $\overline{ϵ}=-{ϵ}^{†}{\gamma }^{\hat{0}}$ its causal Dirac spinor current is $K^{\mu }=\overline{ϵ}{\gamma }^{\mu }ϵ$, future causal. The Nester two-form isIt is real in these conventions and antisymmetric in $\mu ,\nu$. Differentiating, the two terms quadratic in first derivatives become equal after exchanging $\nu ,\rho$. The remaining terms are curvature commutators. The Einstein tensor contraction of spin curvature gives ${\gamma }^{\mu \nu \rho }{\nabla }_{\nu }{\nabla }_{\rho }=G^{\mu }{}_{\lambda }{\gamma }^{\lambda }/2$; the adjoint term supplies the other half. ThusThis is the central spinor-curvature identity, derived by differentiation rather than posited as a positivity formula.

Project onto $\Sigma$ and put $b^i=n_{\mu }{B}^{\mu i}$. In an adapted orthonormal frame $n_{\mu }=(-1,0,0,0)$, Clifford algebra givesIndeed the expansion of $|{\gamma }^i{\nabla }_iϵ{|}^2$ is ${\sum }_i|{\nabla }_iϵ{|}^2+({\nabla }_iϵ{)}^{†}{\gamma }^{ij}{\nabla }_jϵ$. With a Gaussian normal extension of the slice, the symmetric second fundamental form has zero contraction with the antisymmetric $B$, so the projected divergence is the intrinsic spatial divergence. This proves the Witten-Nester identityHere $D_i$ on $b$ denotes the spatial metric derivative. On spinors, ${\nabla }_i$ is the Sen spinor connection, the restriction of the spacetime connection, not just the intrinsic three-dimensional one:The extrinsic-curvature term is essential when the data have momentum.

Choose $ϵ$ to solve the Witten spinor equation, $\mathcal{D}ϵ=0$, approaching a prescribed constant ${ϵ}_{\infty }$ in the chosen end and zero constants in any other ends. The Einstein constraints identify $G_{\mu \nu }{n}^{\mu }{K}^{\nu }=8\pi G{T}_{\mu \nu }{n}^{\mu }{K}^{\nu }$. The dominant energy condition makes this nonnegative because both $n$ and $K$ are future causal. Consequently the right-hand side of the identity becomes a sum of nonnegative terms.

Existence of that spinor is an analytic part of the proof, not something established by the positivity identity alone. Extend ${ϵ}_{\infty }$ smoothly using a cutoff and solve $\mathcal{D}u=-\mathcal{D}{ϵ}_0$ for a decaying correction. The spatial principal symbol ${\gamma }^i{\xi }_i$ squares to $|\xi {|}_h^2$, so the equation is elliptic. For a compactly supported spinor, the integrated identity givesTogether with the weighted asymptotic estimates this gives the coercivity needed for elliptic existence of a Witten spinor. A decaying homogeneous solution would be parallel by the same identity and hence zero, excluding the kernel; the appropriate weighted elliptic solvability gives the correction. Under the stated smooth decay assumptions it has $ϵ-{ϵ}_{\infty }=O(r^{-1})$ and first derivative $O(r^{-2})$. The rigorous analytic step is supplied by the spinor existence proof. This also explains why completeness, asymptotic data and the absence of uncontrolled inner boundary terms are hypotheses of the argument.

It remains to identify the boundary integral with the physical ADM charge. In the asymptotic symmetric coframe $e_i{}^a={\delta }_i{}^a+q_{ia}/2+O(r^{-2})$, the leading spatial spin connection isInsert this and the Sen correction into $b^i$. The real scalar part of ${\gamma }^{ij}{\gamma }^{kl}$ is ${\delta }^{il}{\delta }^{jk}-{\delta }^{ik}{\delta }^{jl}$; its two-gamma part has imaginary expectation and drops out. It follows that the intrinsic connection contributes $\frac{1}{2}({\partial }_j{q}_{ij}-{\partial }_i{q}_{jj})K_{\infty }^{\hat{0}}$. For the other term, symmetry of $k$ removes the three-spatial-gamma product, leaving $-(k_{ij}-k{h}_{ij})K_{\infty }^{\hat{j}}$. Thus the ADM boundary term of the Nester two-form has the expansionThe constant-spinor derivative contribution integrates to zero on a large coordinate sphere. Its coefficient is antisymmetric in $i,j$; the corresponding tangential vector on the sphere is divergence free, so integration by parts removes it. Equivalently, the flux of ${\gamma }^{ij}{\partial }_{j}u$ has identically zero divergence, with its asymptotic leading term having zero flux. The quadratic correction terms are $O(r^{-3})$ and their surface integrals vanish in the limit.

Define the ADM energy and ADM momentum with the same conventions:The boundary calculation then gives, including every normalization factor,This is the energy associated with the asymptotic spinor's causal translation. In the zero-momentum rest frame it is the ADM mass times $K_{\infty }^{\hat{0}}$; more generally it proves the full energy-momentum bound.

Finally apply the divergence theorem to the Witten identity. There is no inner boundary, and the spinors assigned zero at other ends give no charge there. ThereforeAsymptotic constant spinors can produce every future null direction. Normalize $K_{\infty }^{\hat{0}}=1$ and choose its spatial direction along $\mathbf{P}$; the inequality givesThis establishes the positive energy theorem and nonnegativity of the invariant mass when it is defined. For zero ADM energy, repeat the identity for a basis of asymptotic spinors: all become parallel and all matter contractions vanish. Their curvature integrability forces the spacetime curvature to vanish on the initial surface; the standard rigidity conclusion is that the data embed in Minkowski spacetime. Merely having a parallel null spinor in a local wave geometry would not by itself prove this global rigidity: completeness and asymptotic flatness remain essential.

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.