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For a perturbation $h_{xx}=2a(t-z)$, $h_{yy}=2b(t-z)$ with all other components zero, the linearized Vacuum Einstein equations give $(a+b{)}^{\prime \prime }=0$. Put $s=a+b$ and choose ${\xi }_x=sx/2$, ${\xi }_y=sy/2$, ${\xi }_t=-s^{\prime }(x^2+y^2)/4$, ${\xi }_z=s^{\prime }(x^2+y^2)/4$. Under the linearized coordinate gauge transformation, the mixed time/transverse and longitudinal/transverse components cancel, while the other time/longitudinal changes are proportional to $s^{\prime \prime }=0$. The remaining components are $h_{xx}=a-b$, $h_{yy}=-(a-b)$, $h_{xy}=0$, a transverse-traceless gauge perturbation of plus polarization. This explicitly removes a nonzero affine trace without assuming it vanishes by boundary conditions. If $(a-b{)}^{\prime \prime }=0$ too, all linearized curvature vanishes.

For $g=-d{t}^2+A(u{)}^{2}d{x}^2+B(u{)}^{2}d{y}^2+d{z}^2$, $u=t-z$, with nowhere-zero smooth $A,B$, take $e^0=dt$, $e^1=Adx$, $e^2=Bdy$, $e^3=dz$, $\ell =du$. In signature $(-,+,+,+)$, Cartan's first structure equation and metric compatibility give the lower-index connection 1-forms ${\omega }_{01}={\omega }_{13}=-(A^{\prime }/A)e^1$ and ${\omega }_{02}={\omega }_{23}=-(B^{\prime }/B)e^2$. Cartan's second structure equation gives ${\Theta }_{01}={\Theta }_{13}=-(A^{\prime \prime }/A)\ell \wedge e^1$ and ${\Theta }_{02}={\Theta }_{23}=-(B^{\prime \prime }/B)\ell \wedge e^2$. All other independent curvature 2-forms vanish: in particular the $12$ component cancels between the time and longitudinal terms. Contracting $R^a{}_{bad}$ yields $\mathrm{Ric}=-(A^{\prime \prime }/A+B^{\prime \prime }/B)du\otimes du$. Thus the Vacuum Einstein equations are $A^{\prime \prime }/A+B^{\prime \prime }/B=0$, while nonzero $A^{\prime \prime }/A=-B^{\prime \prime }/B$ gives opposite transverse tidal curvatures and plus polarization relative to these axes.

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.

Work on a regular patch with a smooth nonzero conformal factor. Its sign is constant there, so replace it by $|\Omega |$ if necessary and assume $\Omega >0$. Put $u=\log \Omega$, and let all partial derivatives, index raising on $u_i,{\Omega }_i$, and the operators $\Delta ,|\nabla \cdot {|}^2$ in the following formulas use the flat Cartesian metric $\delta$. The orthonormal coframe $E^i=\Omega d{x}^i$ hasThe proposed connection 1-forms reduce toLowering the first frame index with $\delta$ makes these antisymmetric, which establishes metric compatibility. Moreover,Thus Cartan's first structure equation has zero torsion. Uniqueness of the metric-compatible torsion-free connection identifies these forms as the Levi-Civita connection; this also explains why merely guessing a skew connection would not have sufficed.

To calculate the Riemann curvature tensor, use Cartan's second structure equation. Direct differentiation and multiplication giveIntroduce the Schouten tensor in Cartesian coordinate components,The preceding calculation groups into the curvature 2-formsKeep the convention $R_{ijkl}=g({\partial }_i,R({\partial }_k,{\partial }_l){\partial }_j)$, with $R(X,Y)=[{\nabla }_X,{\nabla }_Y]-{\nabla }_{[X,Y]}$, as in the first solution. Converting frame indices to coordinate indices gives the curvature of a conformally Euclidean metric:These are coordinate components. Components in the dual orthonormal frame are ${\Omega }^{-4}{R}_{ijkl}$, hence the same expression with the overall factor ${\Omega }^{-2}$. This distinction is important when contracting the Riemann curvature tensor.

Contract with $g^{ik}={\Omega }^{-2}{\delta }^{ik}$. In four dimensions the Ricci tensor becomes $R_{jl}=2S_{jl}+{\delta }_{jl}{\sum }_k{S}_{kk}$, and thereforeTaking the remaining metric trace yields the Ricci scalarThe cancellation of the gradient-squared terms in the last expression is special to dimension four.

For the Tolman wormhole, compute on $r>0$:Substitution into the Ricci tensor and Ricci scalar givesFor example, the coordinate radial eigenvalue is $-12\lambda /(r^2+\lambda {)}^2$, while each of the three tangential eigenvalues is $4\lambda /(r^2+\lambda {)}^2$. Their sum vanishes, but none vanishes when $\lambda \ne 0$. The Vacuum Einstein equations with zero cosmological constant require $R_{ij}-\frac{1}{2}R{g}_{ij}=0$ and hence $R_{ij}=0$. The metric is scalar-flat but not a vacuum solution for any nonzero $\lambda$; only $\lambda =0$ gives flat vacuum space. A nonzero cosmological constant does not rescue it: its trace would force $R=4\Lambda$, so $R=0$ would first force $\Lambda =0$.

The smooth wormhole interpretation uses $\lambda >0$; $r=0$ is another asymptotically flat end, not an included origin, as inversion $\rho =\lambda /r$ shows. For $\lambda <0$, the conformal factor vanishes at $r=\sqrt{-\lambda }$, where the metric is degenerate. The local non-vacuum calculation remains valid on each regular region away from that sphere, but such a metric is not a smooth wormhole across it.

Using the affine parameter and the normalization requested for the geodesic Lagrangian,The coordinate $v$ is cyclic. Its Euler-Lagrange equation gives $\ddot{u}=0$. The transverse coordinate equations give $\ddot{x}+H_x{\dot{u}}^2=0$ and $\ddot{y}+H_y{\dot{u}}^2=0$. For the remaining coordinate,Its Euler-Lagrange equation, after using $\ddot{u}=0$, isCompare with ${\ddot{x}}^a+{\Gamma }_{bc}^a{\dot{x}}^b{\dot{x}}^c=0$. With indices $(1,2,3,4)=(u,v,x,y)$, the complete list of nonzero Christoffel symbols, including symmetry of the lower indices, isAll contracted symbols ${\Gamma }_{ad}^d$ vanish. For $R_{11}$, the first derivative term in the supplied Ricci tensor formula is ${\partial }_v{H}_u+{\partial }_x{H}_x+{\partial }_y{H}_y=H_{xx}+H_{yy}$, since $H$ is independent of $v$. The second derivative term vanishes because the trace symbols vanish. The third term vanishes for the same reason, and every possible product in the last term requires a zero symbol with upper index $1$ or a lower $2$, so it also vanishes. ThusThe Vacuum Einstein equations with zero cosmological constant require $R_{ab}=0$, and therefore $\boxed{H_{xx}+H_{yy}=0}$. The wave profile of this plane-fronted gravitational wave is harmonic in the transverse plane for each fixed $u$.

For the Vacuum Einstein equations with zero cosmological constant, $R_{\alpha \beta }=0$, and its tensor contraction gives $R=0$. Every Ricci tensor and scalar curvature term in the Weyl tensor consequently disappears:Thus the Weyl tensor contains all vacuum curvature. Here vacuum means Ricci-flat; a nonzero cosmological constant would retain a constant-curvature contribution to the Riemann curvature tensor.

Let $E_a$ denote the frame dual to the coframe. For the convention just stated, the Ricci tensor is obtained from the curvature 2-forms through the Ricci one-forms ${\mathcal{R}}_b={\sum }_a{\iota }_{E_a}{\Theta }^a{}_b=R_{bd}{e}^d$. In particular ${\iota }_{E_1}(\ell \wedge e^1)=-\ell$ and likewise for index two. It follows thatThusThe scalar curvature vanishes even before imposing vacuum because $du$ is null, but the full tensor vanishes exactly whenThis is the Vacuum Einstein equations in this family. The nonzero-coframe hypothesis makes division by $A$ and $B$ legitimate. A nonzero $P=-Q$ can leave the Riemann curvature tensor nonzero even in vacuum.

At each point, apply the indefinite-signature version of the Gram-Schmidt process to any local coframe, choosing one timelike and three spacelike one-forms. This produces an orthonormal coframe in spacetime $\{{\theta }^a\}$ satisfyingIt is unique only up to a local Lorentz transformation. The torsion-free connection 1-forms are then determined by Cartan's first structure equation and metric compatibility,Cartan's second structure equation gives the curvature 2-formsand contraction gives the Ricci tensor $R_{bd}=R^a{}_{bad}$.

For the displayed metric, takeThen $d{\theta }^0=0$ and $d{\theta }^i=-{\theta }^0\wedge {\theta }^i$. The nonzero connection forms are thereforewith ${\omega }^i{}_j=0$. A second application of the structure equations givesEquivalently,so this is a constant sectional curvature spacetime with $K=1$. In four dimensions,The Vacuum Einstein equations with a cosmological constant are $G_{ab}+\Lambda g_{ab}=0$, and henceThe metric is the contracting flat slicing of de Sitter spacetime with curvature radius one.

Using the stated Christoffel symbols in the definition of the Riemann curvature tensor, the two nontrivial contractions areHenceand the vacuum Einstein field equations reduce toThis is the diagonal Rosen coordinates for a plane gravitational wave equation.

For the plus-polarized wave, setComparison of the transverse metric components gives, to linear order,orThereforeThe Vacuum Einstein equations are consequently satisfied at linear order. A real gravitational wave is obtained by taking the real part of the complex plane-wave notation.

A timelike Killing vector with hypersurface orthogonality admits local coordinates with $k={\partial }_t$ and $d{s}^2=N(\mathbf{x}{)}^{2}d{t}^2-h_{ij}(\mathbf{x})d{x}^{i}d{x}^j$. The only connection coefficients with time indices that can be nonzero are ${\Gamma }_{0i}^0={\partial }_i\log N$ and ${\Gamma }_{00}^i=N{h}^{ij}{\partial }_{j}N$. Substitution into the Ricci tensor gives $R_{0i}=0$. Thus $R_{ab}{k}^b$ has only its time component and is proportional to $k_a$. Equivalently $k_d{R}^d{}_{[a}{k}_{b]}=0$. This local statement does not require the Vacuum Einstein equations and is independent of the overall curvature-sign convention.

Rescaling the spatial metric by the squared static lapse function gives a useful three-dimensional quotient metric $\gamma$. In Vacuum Einstein equations, its Ricci tensor obeys $R_{ij}(\gamma )=2{\partial }_{i}U{\partial }_{j}U$, while $U$ is harmonic. The contracted Bianchi identity recovers the harmonic equation without dividing at critical points.

For $\lambda >0$, this four-dimensional Euclidean metric has two asymptotically flat ends. Inversion $\rho =\lambda /r$ puts the metric near the $r=0$ end into the same form with $\rho \to \infty$. On $r>0$, its Ricci scalar vanishes because $\Delta (r^{-2})=0$, but its Ricci tensor is $R_{ij}=4\lambda ({\delta }_{ij}-4x_i{x}_j/r^2)/(r^2+\lambda {)}^2$. It therefore does not satisfy the Vacuum Einstein equations unless $\lambda =0$. For negative $\lambda$ the conformal factor vanishes at a finite radius, so that metric is not a smooth wormhole across that sphere.

A vacuum spacetime has vanishing matter stress-energy tensor. In $D>2$ dimensions, tracing the Vacuum Einstein equations gives $R=2D\Lambda /(D-2)$ and hence $R_{ab}=2\Lambda g_{ab}/(D-2)$; in four dimensions this reduces to $R_{ab}=\Lambda g_{ab}$. With zero cosmological constant it is Ricci flat, although its full Riemann curvature tensor can be nonzero: a vacuum Kerr black hole and vacuum gravitational waves illustrate that distinction.