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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.

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.