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With vanishing extrinsic curvature, the momentum constraint is identically satisfied and the vacuum Hamiltonian constraint reduces to $\mathcal{R}=0$. To evaluate that Ricci scalar, write ${\psi }_i={\partial }_i\psi$, raise these intermediate indices with ${\delta }^{ij}$, and use ${\gamma }^{ij}={\psi }^{-4}{\delta }^{ij}$. The Levi-Civita connection isSubstitution into the stated curvature convention givesTracing with ${\gamma }^{ij}$ cancels the gradient-square terms:Here the Laplacian and norm on the right are those of the flat Euclidean metric. Thus the time-symmetric conformally flat vacuum initial data constraints becomeThe equivalence uses $\psi \ne 0$. This identity is also the three-dimensional specialization of scalar curvature under conformal rescaling; locally the conformal exponent is $2\log |\psi |$, so either fixed nonzero sign of $\psi$ gives the same metric.

On the punctured region $r>0$, choose a positive constant $a$ and setFor a radial function in three dimensions, $\Delta \psi =r^{-2}{\partial }_r(r^2{\partial }_r\psi )$. Since $r^2{\partial }_r\psi =-a$, this harmonic function has $\Delta \psi =0$ at every $r>0$, is nonconstant and positive, and approaches one at infinity. It defines the requested time-symmetric conformally flat vacuum initial data on that punctured or exterior region. One often writes $a=M/2$, giving the time-symmetric spatial metric of the Schwarzschild metric in isotropic coordinates.

A puncture or inner boundary is necessary if global regularity was intended. A smooth harmonic function on all of ${\mathbb{R}}^3$ that tends to one at infinity must equal one everywhere: applying the maximum principle for harmonic functions to larger and larger balls bounds $|\psi -1|$ by its arbitrarily small boundary values. The nonconstant example is therefore not a smooth solution through $r=0$, and no such globally regular example exists under those stronger assumptions.