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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.

Put $r^2=y_1^2+y_2^2$ and write the Riemannian metric as a conformal rescaling of a Riemannian metric:The base metric is flat, so its Ricci scalar is zero and its Laplacian is $\Delta ={\partial }_1^2+{\partial }_2^2$. The formula for scalar curvature under conformal rescaling simplifies in dimension two toDifferentiate explicitly:Since $e^{-2\omega }=1+r^2$, the required Ricci scalar is