= Solution
\b[The harmonic equation, including critical points.] All derivatives in this argument use the three-dimensional <Levi-Civita connection> of $\gamma$. Write $U_i=\nabla_iU$, $|\nabla U|^2=\gamma^{ij}U_iU_j$ and $\Delta_\gamma U=\gamma^{ij}\nabla_i\nabla_jU$. The curvature relation for the <static vacuum conformal spatial metric> gives $R=2|\nabla U|^2$. Substituting it into the <contracted Bianchi identity> yields
$$
\nabla^iR_{ij}
=2(\Delta_\gamma U)U_j+2U^i\nabla_iU_j
=\frac12\nabla_jR
=2U^i\nabla_jU_i.
$$
The Hessian of a scalar is symmetric, so the last terms cancel:
$$
(\Delta_\gamma U)U_j=0.
$$
Where $\nabla U\ne0$, 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 gives
$$
\boxed{\Delta_\gamma U=0\quad\text{everywhere}.}
$$
This avoids incorrectly dividing by a gradient at its zeros.
\b[Integration by parts and rigidity.] The usual whole-space energy identity is
$$
0=\int U\Delta_\gamma U\,dV_\gamma
=-\int|\nabla U|^2dV_\gamma+
\lim_{R\to\infty}\int_{S_R}U\,n^i\nabla_iU\,dS_\gamma.
$$
With 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 $\varepsilon>0$, the region $\Omega_\varepsilon=\{U>\varepsilon\}$ has compact closure. Completeness and the absence of an inner boundary ensure no missing boundary pieces; standard <asymptotic flatness> and $U\to0$ 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 gives
$$
0=-\int_{\Omega_\varepsilon}|\nabla U|^2dV_\gamma
-\varepsilon\int_{\partial\Omega_\varepsilon}|\nabla U|\,dS_\gamma.
$$
Both terms are nonpositive, so both vanish. A nonempty component with $U>\varepsilon$ cannot have zero gradient throughout and boundary value $\varepsilon$. Thus $\Omega_\varepsilon$ is empty. Choose arbitrarily small regular values and apply the same argument to $-U$; it follows that
$$
\boxed{U=0,\qquad R_{ij}(\gamma)=0.}
$$
This is <vanishing harmonic function by level-set integration>. It also makes explicit why an inner boundary would invalidate the conclusion.
\b[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 $R_{ijpq}=0$: \b[$\gamma$ 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 becomes
$$
\boxed{ds^2=-dt^2+dx^2+dy^2+dz^2.}
$$
Therefore <horizonless static vacuum rigidity> gives \b[<Minkowski spacetime> as the sole solution under these hypotheses.]
\b[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,
$$
U=\frac12\log(1-2M/r),\qquad
\gamma=dr^2+r(r-2M)d\Omega^2,
$$
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.
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