= 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 means
$$
\boxed{\nabla_\mu\epsilon=0.}
$$
It is an overdetermined equation and expresses an unbroken <supersymmetry> of the background. In the cosmological deformation the corresponding equation is $\mathcal D_\mu\epsilon=0$; 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 current>
$$
K^\mu=\bar\epsilon\gamma^\mu\epsilon
$$
obeys $\nabla_\nu K_\mu=0$. In particular $\nabla_{(\nu}K_{\mu)}=0$, so \b[$K$ is a <Killing vector>], and in fact is parallel. For the real cosmological deformation, $\nabla_\mu\epsilon=-(m/2)\gamma_\mu\epsilon$ and $\nabla_\mu\bar\epsilon=(m/2)\bar\epsilon\gamma_\mu$ instead give $\nabla_\mu K_\nu=(m/2)\bar\epsilon[\gamma_\mu,\gamma_\nu]\epsilon$, whose symmetric part again vanishes.
The causal property uses an ordinary commuting spinor. Take an orthonormal frame with $(\gamma^0)^\dagger=-\gamma^0$, $(\gamma^i)^\dagger=\gamma^i$, and $\bar\epsilon=\epsilon^\dagger\gamma^0$. Then $K^0=-\epsilon^\dagger\epsilon$. Reverse its time orientation and write $V=-K$, so
$$
V^0=\epsilon^\dagger\epsilon,\qquad V^i=\epsilon^\dagger\alpha^i\epsilon,\qquad\alpha^i=-\gamma^0\gamma^i.
$$
The 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 $|u_iV^i|\le V^0$. Choose $u$ along the spatial part of $V$ to obtain
$$
\boxed{K_\mu K^\mu=V_\mu V^\mu=-(V^0)^2+|\mathbf V|^2\le0.}
$$
A 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 $\epsilon_\infty$ at spatial infinity. Define the real <Nester two-form>
$$
B^{\mu\nu}=\bar\epsilon\gamma^{\mu\nu\rho}\nabla_\rho\epsilon-\overline{\nabla_\rho\epsilon}\gamma^{\mu\nu\rho}\epsilon.
$$
Antisymmetry removes second derivatives except through the <Riemann curvature tensor>. Differentiation and the three-gamma <spinor curvature identity> used above give
$$
\nabla_\nu B^{\mu\nu}=2(\nabla_\nu\bar\epsilon)\gamma^{\mu\nu\rho}\nabla_\rho\epsilon+G^{\mu\nu}K_\nu.
$$
The 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 $\chi_i=\nabla_i\epsilon$. Since $\gamma^0\gamma^{0ij}=-\gamma^{ij}$,
$$
2(\nabla_i\bar\epsilon)\gamma^{0ij}\nabla_j\epsilon=2\left(\sum_i|\chi_i|^2-\left|\sum_i\gamma^i\chi_i\right|^2\right).
$$
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 equation>
$$
\gamma^i\nabla_i\epsilon=0,\qquad\epsilon\longrightarrow\epsilon_\infty.
$$
The 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 $G_{\mu\nu}=\kappa^2T_{\mu\nu}$ now give
$$
Q[\epsilon_\infty]=\frac1{\kappa^2}\int_{S_\infty}B^{0i}dS_i
=\frac2{\kappa^2}\int_\Sigma|\nabla_i\epsilon|^2dV+\int_\Sigma T_{\mu\nu}n^\mu V^\nu dV\ge0.
$$
The <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>:
$$
Q[\epsilon_\infty]=-P_\mu V_\infty^\mu
=\epsilon_\infty^\dagger\bigl(E\,1-P_i\alpha^i\bigr)\epsilon_\infty.
$$
For example its energy term uses $B^{0i}=\tfrac12(\partial_jh_{ij}-\partial_ih_{jj})|\epsilon_\infty|^2$ 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 proves
$$
\boxed{E\ge|\mathbf P|\ge0.}
$$
<Killing 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.
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