Conformal Decomposition of the Einstein Constraints

The conformal method converts the four Einstein constraint equations into a coupled elliptic system for a positive scalar and a vector potential. Its characteristic exponents are not decorative choices: in three spatial dimensions they make the momentum constraint transform cleanly and put the Hamiltonian constraint into Lichnerowicz form. The method constructs vacuum data here; matter can be included only after its conformal weights and physical energy conditions have been specified.

Required background. Hypersurface Normals and Tangential Projectors supplies the spatial geometry and extrinsic curvature, while PDEs and Green Functions for Gravitation supplies elliptic operators, boundary data, and kernel conditions.

Helpful background. The Einstein Equation supplies the parent field equation; Formulations of General Relativity locates the construction within the initial-value problem; and Presymplectic Potentials and Currents prepares the later distinction between constrained data and physical phase space.

Conformal free data and weights

Let $(\Sigma,\bar\gamma_{ij})$ be a three-dimensional Riemannian manifold. The semi-decoupling version of the York method starts with

• a conformal metric $\bar\gamma_{ij}$;
• a symmetric transverse-traceless tensor $\bar A_{\mathrm{TT}}^{ij}$;
• a mean-curvature function $K$.

The unknowns are a positive function $\psi$ and a vector field $W^i$. Define extrinsic curvature by $K_{ij}=-\gamma_i{}^\mu\gamma_j{}^\nu\nabla_\mu n_\nu$ and write

$$\gamma_{ij}=\psi^4\bar\gamma_{ij}, \qquad K_{ij}=A_{ij}+\frac13\gamma_{ij}K,$$

and assign the trace-free contravariant tensor the weight

$$A^{ij}=\psi^{-10}\bar A^{ij}, \qquad \bar A^{ij}=\bar A_{\mathrm{TT}}^{ij}+(\bar{\mathbb L}W)^{ij}.$$

Equivalently, $A_{ij}=\psi^{-2}\bar A_{ij}$. The powers $4,-10,-2$ belong to three spatial dimensions. In $n$ dimensions the metric power becomes $4/(n-2)$, so copying these exponents into another dimension is incorrect.

The divergence identity that selects the weight $-10$ is

$$D_jA^{ij}=\psi^{-10}\bar D_j\bar A^{ij}$$

for trace-free $\bar A^{ij}$. Thus no derivatives of $\psi$ contaminate this divergence. Bartnik and Isenberg state the free and determined fields and the corresponding weights in Bartnik and Isenberg 2004, §4.1, eqs. (47)–(51), pp. 12–14, PDF.

Vector and Lichnerowicz equations

In vacuum, with cosmological constant set to zero for this construction, the constraints are

$${}^{(3)}R+\frac23K^2-A_{ij}A^{ij}=0, \qquad D_jA^{ij}-\frac23D^iK=0.$$

The scalar curvature transforms as

$${}^{(3)}R =\psi^{-4}\bar R-8\psi^{-5}\bar\Delta\psi,$$

whereas $A_{ij}A^{ij}=\psi^{-12}\bar A_{ij}\bar A^{ij}$. Substitution yields the Lichnerowicz equation

$$\bar\Delta\psi -\frac18\bar R\psi +\frac18\bar A_{ij}\bar A^{ij}\psi^{-7} -\frac1{12}K^2\psi^5=0.$$

The momentum constraint becomes

$$\bar\Delta_{\mathbb L}W^i =\frac23\psi^6\bar D^iK, \qquad \bar\Delta_{\mathbb L}W^i \equiv\bar D_j(\bar{\mathbb L}W)^{ij}.$$

These are coupled elliptic equations because the vector source contains $\psi^6\bar D^iK$ and the scalar equation contains $W^i$ through $\bar A^{ij}$. When $K$ is constant, the vector equation loses its scalar dependence. If the conformal metric has no relevant conformal-Killing kernel, one first solves for $W^i$ and then solves the scalar equation. This is the constant-mean-curvature simplification, not a generic decoupling theorem.

Gourgoulhon derives the conformal transformations of curvature and the constraint equations in Gourgoulhon 2007, §§6.3–6.5, pp. 87–98, PDF.

Direct exponent check

Every nonlinear power follows from a short scaling calculation. Multiplying the Hamiltonian constraint by $\psi^5$ gives

$$-8\bar\Delta\psi +\bar R\psi -\bar A_{ij}\bar A^{ij}\psi^{-7} +\frac23K^2\psi^5=0.$$

The powers are therefore fixed as follows:

• curvature contributes $\psi$ after the common factor is removed;
• the trace-free norm contributes $\psi^{5-12}=\psi^{-7}$;
• the mean-curvature term contributes $\psi^5$;
• raising the index on $D_iK$ contributes $\psi^{-4}$, which combines with the divergence weight $\psi^{-10}$ to give the vector source $\psi^6\bar D^iK$.

This check also fixes which metric raises each barred or unbarred index. Mixing the two metrics silently changes the powers.

Solvability is additional information

Choosing free data does not guarantee a physical solution. One must require $\psi>0$, supply boundary or asymptotic conditions, satisfy compatibility with conformal Killing fields, and determine whether the semilinear scalar equation admits a solution. Yamabe class, vanishing or nonvanishing TT data, and the behavior of $K$ all matter. Far from constant mean curvature, existence, uniqueness, and nonuniqueness require results beyond this construction.

Matter introduces another choice rather than a universal extra term. For example, assigning a conformal scaling to the physical energy density changes the power of $\psi$ in the scalar equation. A useful matter model must preserve the matter constraints and state whether its barred fields are fixed sources or determined variables.

Common pitfalls

Free data are not unconstrained physical observables. TT and mean-curvature seeds still have global regularity, boundary, kernel, and solvability restrictions.

CMC is a structural simplification. When $\bar D_iK\ne0$, the scalar and vector equations are genuinely coupled.

A positive solution is essential. A sign-changing solution would make the physical metric degenerate and is not admissible initial data.

Exercise: time-symmetric conformally flat data

Let $\bar\gamma_{ij}=\delta_{ij}$, $K=0$, and $\bar A^{ij}=0$ on a vacuum region. Derive the equation for $\psi$ and verify the one-puncture solution.

Solution

All curvature and extrinsic-curvature terms vanish, so the Lichnerowicz equation reduces to

$$\delta^{ij}\partial_i\partial_j\psi=0.$$

On $\mathbb R^3\setminus\{0\}$,

$$\psi=1+\frac{GM}{2r}$$

is harmonic because $\Delta(1/r)=0$ away from the puncture. It approaches one at spatial infinity and produces the isotropic spatial metric of a time-symmetric Schwarzschild slice. The physical manifold omits $r=0$; for $M\gt0$ this limit is an additional asymptotic end, as explained in Black-Hole Initial Data. If one instead extends $1/r$ as a distribution on auxiliary Euclidean $\mathbb R^3$, then $\Delta(1/r)=-4\pi\delta^{(3)}(x)$. That auxiliary distribution is not a physical point source on the punctured vacuum slice.

References

• Bartnik, Robert, and Jim Isenberg. “The Constraint Equations.” In The Einstein Equations and the Large Scale Behavior of Gravitational Fields, 1–38. Basel: Birkhäuser, 2004. arXiv:gr-qc/0405092.
• Gourgoulhon, Éric. 3+1 Formalism and Bases of Numerical Relativity. Lecture notes, arXiv:gr-qc/0703035v1 (2007). arXiv.