 Solution

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.