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The harmonic equation, including critical points. All derivatives in this argument use the three-dimensional Levi-Civita connection of $\gamma$. Write $U_i={\nabla }_{i}U$, $|\nabla U{|}^2={\gamma }^{ij}{U}_i{U}_j$ and ${\Delta }_{\gamma }U={\gamma }^{ij}{\nabla }_i{\nabla }_{j}U$. The curvature relation for the static vacuum conformal spatial metric gives $R=2|\nabla U{|}^2$. Substituting it into the contracted Bianchi identity yieldsThe Hessian of a scalar is symmetric, so the last terms cancel:Where $\nabla U\ne 0$, this implies ${\Delta }_{\gamma }U=0$. On the interior of the critical set $\{\nabla U=0\}$, $U$ is locally constant and its Laplacian is also zero. Every other critical point is a limit of noncritical points, so continuity of the Laplacian givesThis avoids incorrectly dividing by a gradient at its zeros.

Integration by parts and rigidity. The usual whole-space energy identity isWith standard static asymptotic falloff $U=O(R^{-1})$ and $\nabla U=O(R^{-2})$, the last term is $O(R^{-1})$ and vanishes. Positivity of the spatial Riemannian metric then makes $U$ constant, and its limiting value fixes that constant to zero.

There is also a direct integration by parts proof requiring only the stated vanishing of $U$ at infinity, rather than an assumed flux decay rate. For a regular value $\epsilon >0$, the region ${\Omega }_{\epsilon }=\{U>\epsilon \}$ has compact closure. Completeness and the absence of an inner boundary ensure no missing boundary pieces; standard asymptotic flatness and $U\to 0$ on all ends keep this positive level set away from infinity. Its outward unit normal is $n=-\nabla U/|\nabla U|$. Multiplying the harmonic equation by $U$ and integrating givesBoth terms are nonpositive, so both vanish. A nonempty component with $U>\epsilon$ cannot have zero gradient throughout and boundary value $\epsilon$. Thus ${\Omega }_{\epsilon }$ is empty. Choose arbitrarily small regular values and apply the same argument to $-U$; it follows thatThis is vanishing harmonic function by level-set integration. It also makes explicit why an inner boundary would invalidate the conclusion.

From zero Ricci curvature to Minkowski spacetime. In dimension three the Schouten tensor is $S_{ij}=R_{ij}-R{\gamma }_{ij}/4$, so it vanishes here. The three-dimensional curvature reconstruction then gives $R_{ijpq}=0$: $\gamma$ is a flat Riemannian manifold. This use of three-dimensional curvature from the Ricci tensor is essential; zero Ricci tensor would not by itself imply flatness in four dimensions.

A complete connected flat spatial manifold has Euclidean universal cover. Standard asymptotic flatness, with an ordinary Euclidean end, excludes a nontrivial free Euclidean quotient: a nontrivial fixed-point-free Euclidean isometry contains a translational or screw component, and its cyclic quotient has at most quadratic volume growth, incompatible with a three-dimensional Euclidean end. Extra identifications cannot restore cubic growth. Thus the complete spatial metric is globally Euclidean, not just locally flat. With $U=0$ and the standard global time coordinate the four-metric becomesTherefore horizonless static vacuum rigidity gives Minkowski spacetime as the sole solution under these hypotheses.

Allowing horizons. A black hole exterior can be static and asymptotically flat without being Minkowski. The Schwarzschild spacetime is the basic example. Its lapse function vanishes at the event horizon, so $U$ is unbounded below there; the horizon introduces an inner end or boundary in the spatial reduction, and the previous energy argument no longer has its hypotheses. In fact for Schwarzschild,so this conformal quotient terminates at $r=2M$ and is not the complete nonsingular quotient used above.

With the usual regularity, connected-horizon and global hypotheses of the static vacuum black-hole uniqueness theorem, the nontrivial black-hole exterior is Schwarzschild. Rotating Kerr black holes are stationary but not static and therefore are not alternatives in this question. This statement concerns a regular domain outside the horizon; the Schwarzschild interior contains a curvature singularity. If “globally static” and “nonsingular everywhere” are retained literally, Schwarzschild does not meet them. Allowing horizons means relaxing the horizonless global hypotheses to the corresponding static exterior problem.

Use the standard Euclidean convention $e^{-S_E}$ and restore the curvature normalization explicitly. Let $\phi$ be the conventional dimensionless dilaton, whose worldsheet term isFor constant ${\phi }_0$, the Gauss-Bonnet theorem gives $S_{\phi }={\phi }_0\chi (\Sigma )$. A closed surface of genus $g$ has Euler characteristic $\chi =2-2g$, so its contribution is weighted byThusThis is the dilaton normalization and Euler-characteristic weighting of string perturbation theory. If the unit curvature coefficient in the printed expression is retained literally in this Euclidean convention, then $\phi =4\pi \Phi$ and $g_s=e^{4\pi {\Phi }_0}$. If $\Phi$ denotes the conventional dilaton, the usual omitted $1/(4\pi )$ is understood and $g_s=e^{{\Phi }_0}$. One must specify that normalization before exponentiating the field. Lorentzian overall signs are fixed consistently by Wick rotation.

For the sigma-model beta function, usewith vanishing two-form and dilaton. Perform the background field expansion of a string sigma model covariantly by setting $X={\exp }_{\bar{X}}\eta$. In Riemann normal coordinates, the quadratic fluctuation action isThe covariant derivative contains the pulled-back target connection. This expansion accounts for the connection vertices as well as the normal-coordinate metric expansion; keeping only a single metric tadpole would not give a covariant answer.

Integrating the Gaussian fluctuations gives $\frac{1}{2}\mathrm{Tr}\log \Delta$, whereContracting the curvature vertex with the short-distance propagator traces $R^a{}_{cad}$ to the Ricci tensor $R_{cd}$. In dimensional regularization at $d=2-ϵ$, the logarithmic integral is $\int d^{d}p/(2\pi {)}^d(p^2+{\mu }^2{)}^{-1}=1/(2\pi ϵ)+O(1)$. With the stated curvature convention the divergent effective action isIt is cancelled by the metric counterterm $\delta g_{ab}={\alpha }^{\prime }{R}_{ab}/ϵ$. Equivalently, write $g^{\mathrm{bare}}={\mu }^{-ϵ}(g+{\alpha }^{\prime }R/ϵ+\cdots )$. Differentiating the bare coupling at fixed value gives the one-loop metric beta function of a string sigma modelThis loop is a loop of two-dimensional fluctuation fields: its expansion is in $\alpha \prime$, not a change of worldsheet genus or an extra power of $g_s$. The sign here defines the renormalization scale to increase toward the ultraviolet. For a round target sphere, positive Ricci curvature makes its squared radius increase with that scale, consistent with the usual asymptotic freedom of the inverse-radius sigma-model coupling. With the printed kinetic coefficient $1/2$ taken literally, $2\pi {\alpha }^{\prime }=1$ and the leading coefficient is $R_{ab}/(2\pi )$. Requiring the metric beta function to vanish gives $R_{ab}=0$ at this order. Full bosonic-string Weyl consistency also requires the other anomaly coefficients, including the critical-dimension condition when the dilaton vanishes.

For T-duality, choose coordinates $(y,x^i)$ adapted to the isometry and use ${\alpha }^{\prime }=1$ units for the local Buscher rules. Assume $g_{yy}\ne 0$, with a spacelike isometry for the usual unitary circle duality. In light-cone worldsheet coordinates take the convention that $E=g+B$ multiplies ${\partial }_{+}{X}^a{\partial }_{-}{X}^b$. Gauge translations of $y$, replace ${\partial }_{\pm }y$ by $A_{\pm }$, gauge fix $y=0$, and impose flatness with a multiplier $\tilde{y}$. The relevant first-order Lagrangian isIntegrating out $\tilde{y}$ gives a locally pure-gauge connection and hence the original model. Instead integrate the last term by parts and solve the algebraic equations for $A_{\pm }$:Substitution givesReading its symmetric and antisymmetric parts yields the Buscher dual of a metric with vanishing two-form:The spectator metric is the Schur complement of $g_{yy}$. Reversing the orientation of $\tilde{y}$ reverses the displayed mixed two-form signs; our multiplier and $E=g+B$ conventions fix them.

The metric rules arise already from the classical algebraic elimination. For quantum equivalence the regulated Gaussian measure also produces the Buscher dilaton shiftFor a compact isometry the multiplier periodicity and flat-connection sectors implement the exchange of momentum and winding. On a circle, restoring dimensions gives $R^{\prime }={\alpha }^{\prime }/R$. Thus the calculation derives the local dual background, while the measure and global sectors explain how it becomes a string-theory duality.

Use $A_{\beta }={\partial }_{\mu }{h}^{\mu }{}_{\beta }$ and $\square ={\eta }^{\mu \nu }{\partial }_{\mu }{\partial }_{\nu }$. The linearized Ricci tensor and scalar obeyThe background Ricci tensor and Ricci scalar vanish, so varying the metric in the scalar-curvature term contributes nothing at first order beyond this trace. The Einstein tensor perturbation is consequentlyAll contractions use the Minkowski metric. Direct differentiation gives ${\partial }^{\alpha }\delta G_{\alpha \beta }=0$, the linearized contracted Bianchi identity, which checks the relative signs. No time derivatives or gauge-dependent terms have been discarded.

Substitute $T^a{}_{bc}={\ell }^a{g}_{bc}$ into the preceding identity, using metric compatibility. Set $M_{bd}={\ell }_{b;d}+{\ell }_b{\ell }_d$ and $S=g^{bd}{M}_{bd}={\nabla }_a{\ell }^a+{\ell }_a{\ell }^a$. Contract the first and third curvature indices, in the stated curvature convention, to obtainIf the two Ricci tensors agree, then $M_{bd}=S{g}_{bd}$. Taking the trace in dimension $n>1$ gives $S=nS$, so $S=0$ and $M=0$. Conversely $M=0$ makes the Ricci difference vanish. Thus $\boxed{{\tilde{R}}_{bd}=R_{bd}⟺{\ell }_{b;d}+{\ell }_b{\ell }_d=0}$.

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.

All free indices below are spatial, and $K={\gamma }^{\mu \nu }{K}_{\mu \nu }$. Contract the vacuum Ricci tensor using $g^{\mu \nu }={\gamma }^{\mu \nu }-n^{\mu }{n}^{\nu }$:Reversing both antisymmetric curvature pairs identifies the last term as $-E_{\alpha \beta }$. The spatial Gauss–Codazzi equations for a spatial hypersurface then giveFor the magnetic part of the Weyl tensor, moving the normal to the first slot of the metric volume tensor introduces a minus sign:The remaining curvature indices are spatial, so the normal projection in the Gauss–Codazzi equations for a spatial hypersurface givesThe interchange $\mu \leftrightarrow \nu$ makes the two spatial covariant derivative terms equal. Thus the constants, with the paper's orientation and extrinsic-curvature convention, areIn particular, the sign of the magnetic expression must include the minus from moving $n$ past the first volume-tensor index. These formulas reconstruct the spatial electric part of the Weyl tensor and magnetic part of the Weyl tensor from the induced metric and the extrinsic curvature of a spatial hypersurface.

Fix the Riemann curvature tensor conventionFor the limiting affine connection, every product of two nonzero connection coefficients is zero, because its only nonzero coefficients have spatial upper index and both lower indices $t$. Its only possibly nonzero curvature components, apart from antisymmetry in the last two indices, areThe Ricci tensor is thereforeAll other components vanish, so the limiting connection is a connection with vanishing Ricci tensor exactly when $\varphi$ satisfies the Laplace equation on spatial Euclidean space. This is Ricci-flatness, not necessarily vanishing curvature: $\varphi =x^2-y^2$ is a harmonic function but has nonzero Hessian matrix and hence a nonflat limiting connection. Reversing the curvature sign convention reverses the displayed Ricci sign but leaves the equivalence unchanged.

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.

Use the intended positive-mass family: write $b=r_{+}>0$ and $a=r_{-}=2P^2/b\ge 0$, so $R^2=r(r-a)$, $V=1-b/r$ and $ab=2P^2$. Lowering the index of the Killing vector field gives $k^{♭}=-Vdt$, henceFor the stipulated orientation and Lorentzian signature, $\sqrt{-g}=R^2\sin \theta$ and the Hodge star operator obeysThe minus sign follows from $g^{tt}{g}^{rr}=-1$. Thus the Komar mass of the magnetic dilaton black hole isIts asymptotic mass is $M=r_{+}/2$.

For any Killing vector field, its antisymmetric derivative satisfies ${\nabla }_a(d{k}^{♭}{)}^{ab}=-2R^b{}_c{k}^c$. Applying Stokes theorem between two spheres relates the difference of their Komar mass to this Ricci source. It vanishes in vacuum but need not vanish for the nonzero Ricci tensor of Einstein-Maxwell-dilaton theory. Here, explicitly, $d{M}_{\mathrm{Komar}}/dr=P^2/r^2$. The potential $A=P\cos \theta d\phi$ is only a local gauge potential near the polar axes; the mass computation uses its globally defined field strength and the metric, so does not depend on this gauge-patch issue.

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.

For the torsion-free Levi-Civita connection, cyclically summing the displayed coordinate expression and using ${\Gamma }^a{}_{bc}={\Gamma }^a{}_{cb}$ gives the algebraic first Bianchi identityApply the Jacobi identity to three covariant-derivative commutators, or differentiate the coordinate formula in normal coordinates. The third-derivative terms and the derivatives of quadratic connection terms cancel cyclically, giving the differential second Bianchi identityContracting the first and third curvature indices and then contracting once more yieldsor the contracted Bianchi identity

For the metric variation $h_{ab}=\delta g_{ab}$,while the two double-divergence terms in the supplied metric variation of scalar curvature become a boundary term after integration by parts for tensor fields. The Einstein-Hilbert action therefore contributes $G_{ab}+\Lambda g_{ab}$. Varying the Proca action with respect to the metric gives the Proca stress-energy tensor, and the full Einstein-Proca theory equation is

Since $\delta F_{ab}=2{\nabla }_{[a}\delta A_{b]}$, variation with respect to $A_b$ and one integration by parts give the covariant Proca equationTaking its covariant divergence givesbecause the resulting contractions pair the symmetric Ricci tensor with the antisymmetric electromagnetic field tensor. Since $m\ne 0$, the Lorenz constraint in Proca theory follows:Moreover, $F=dA$ and $d^2=0$, so the Electromagnetic Bianchi identity is

Finally, use that identity to differentiate the matter tensor. The result groups naturally asBoth terms vanish by the Proca equation and its Lorenz constraint in Proca theory. Thus ${\nabla }^a{T}_{ab}=0$, exactly as required by the contracted Bianchi identity applied to the Einstein field equations. The vector equation is therefore consistent with the gravitational equation.

Lower the free index of $K$. The Killing equation makes the last two indices of ${\nabla }_{\nu }{\nabla }_{\mu }{K}_{\alpha }$ antisymmetric after the appropriate derivative is moved. Combining the three resulting Ricci identities cyclically givesApplying the covector form of the Ricci identity and then the first Bianchi identity reduces this toRaising $\alpha$ proves the second covariant derivative of a Killing vector identityContracting $\mu$ and $\nu$ and using the definition and symmetries of the Ricci tensor gives

The Killing equation gives ${\nabla }_{\mu }{K}^{\mu }=0$ and ${\nabla }_{[\mu }{K}_{\lambda ]}={\nabla }_{\mu }{K}_{\lambda }$. Commuting derivatives once and contracting the Ricci identity givesTherefore, using the contracted Bianchi identity and symmetry of $R_{\mu \beta }$,Since ${\nabla }_{[\mu }{K}_{\lambda ]}$ is antisymmetric, swapping the two contracted index names shows that the left side equals its derivative-antisymmetrized form:Now set $A_{\mu \lambda }={\nabla }_{[\mu }{K}_{\lambda ]}$. The right side is the contracted commutator $[{\nabla }^{\mu },{\nabla }^{\lambda }]A_{\mu \lambda }$. By the curvature commutator on a covariant tensor, it is a sum of contractions of the symmetric Ricci tensor with the antisymmetric tensor $A$, and hence vanishes. Thusso the Killing vector preserves scalar curvature.

Contracting the Ricci tensor gives the Ricci scalarHence the time-time component of the Einstein tensor is $G_{00}=R_{00}-\frac{1}{2}{g}_{00}R=3H^2$. A comoving perfect fluid in general relativity has $T_{00}=\rho$, so the time-time Einstein field equations give the flat Friedmann equation

For the Friedmann-Lemaitre-Robertson-Walker metric in Cartesian spatial coordinates, the only time-dependent components are $g_{ij}=a^2{\delta }_{ij}$. The Christoffel symbol formula therefore givesSubstituting these into the definition of the Ricci tensor givesThus, with no summation over $i$, the requested components are ${\Gamma }^0{}_{ii}=a\dot{a}$, ${\Gamma }^i{}_{0i}={\Gamma }^i{}_{i0}=\dot{a}/a$, and $R_{ii}=a\ddot{a}+2{\dot{a}}^2$.

Contracting the Riemann curvature tensor gives the nonzero components of the Ricci tensor:The Ricci scalar is the contraction with the inverse metric. ThereforeIt is independent of both coordinates, so the hyperbolic plane has constant negative scalar curvature.

Varying the Ricci tensor and contracting givesSubstitution of part (i), followed by metric compatibility, reduces the divergence toThus the metric variation of scalar curvature isand consequently

Substituting the forms from part (b) into Cartan's second structure equation gives the independent mixed-index curvature 2-formsEvery form is proportional to its corresponding basis 2-form, so the curvature operator is diagonal. The diagonal Ricci tensor components are contractions of these sectional curvature coefficients. In each case the coefficients cancel in the pattern $2-1-1=0$, with the Lorentzian sign included when the time direction is contracted. HenceThis is consistent with Schwarzschild spacetime being a vacuum solution away from $r=0$.

Define the Ricci tensor by contracting the first and third Riemann indices:Contract the first Bianchi identity in its upper index and third lower index. The middle term vanishes by antisymmetry in the first pair, while antisymmetry in the final pair turns the last term into $-R_{\gamma \alpha }$. Thusso the Ricci tensor is symmetric.

For the stated constant-curvature form,The contracted Bianchi identity isUsing $\nabla g=0$ givesorIn four dimensions,More generally, $K$ is constant for every $n>2$. In dimension $n=2$, the contracted Bianchi identity gives no such conclusion, and the Gaussian curvature may vary from point to point.