York Decomposition of Symmetric Tensors

The York decomposition separates a symmetric tensor into trace, transverse-traceless, and longitudinal parts relative to a chosen spatial metric. Applied to extrinsic curvature, it isolates the mean curvature and turns the momentum constraint into a vector elliptic equation. The split is geometric but not automatic: boundary conditions, asymptotic decay, topology, and conformal-Killing fields determine existence and uniqueness.

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

Helpful background. The Einstein Equation supplies the momentum constraint; Formulations of General Relativity orients its role in the initial-value problem; and Presymplectic Potentials and Currents prepares the distinction between constrained data and reduced phase space.

Trace and longitudinal operators

On a three-dimensional Riemannian manifold $(\Sigma,\gamma_{ij})$, split a symmetric tensor $S_{ij}$ into trace and trace-free parts:

$$S_{ij}=A_{ij}+\frac13\gamma_{ij}S, \qquad \gamma^{ij}A_{ij}=0.$$

For a vector field $W^i$, define the conformal Killing or longitudinal operator

$$(\mathbb LW)_{ij} =D_iW_j+D_jW_i -\frac23\gamma_{ij}D_kW^k.$$

It is symmetric and trace free. Its divergence is the vector Laplacian

$$(\Delta_{\mathbb L}W)^i \equiv D_j(\mathbb LW)^{ij} =D_jD^jW^i +\frac13D^iD_jW^j +{}^{(3)}R^i{}_jW^j.$$

The curvature term follows from commuting spatial covariant derivatives. The principal symbol is

$$\sigma(\Delta_{\mathbb L})(k)^i{}_j =\lvert k\rvert^2\delta^i{}_j+\frac13k^ik_j,$$

up to the conventional overall sign of an elliptic operator. Its eigenvalues are positive for nonzero $k_i$, so the operator is elliptic. Ellipticity does not remove its global kernel.

The York split

Seek

$$A^{ij}=A_{\mathrm{TT}}^{ij}+(\mathbb LW)^{ij},$$

where

$$\gamma_{ij}A_{\mathrm{TT}}^{ij}=0, \qquad D_jA_{\mathrm{TT}}^{ij}=0.$$

Taking a divergence gives

$$\Delta_{\mathbb L}W^i=D_jA^{ij}.$$

After solving this vector elliptic equation with declared boundary or asymptotic data, define

$$A_{\mathrm{TT}}^{ij} =A^{ij}-(\mathbb LW)^{ij}.$$

The result is transverse by construction. On a closed manifold, or when the boundary term vanishes, the two pieces are $L^2$ orthogonal:

$$\int_\Sigma A_{\mathrm{TT}}^{ij}(\mathbb LW)_{ij}\,dV_\gamma =-2\int_\Sigma W_iD_jA_{\mathrm{TT}}^{ij}\,dV_\gamma =0.$$

This orthogonality explains the name “decomposition,” but it relies on the domain of the operator. With a boundary,

$$2\int_{\partial\Sigma} n_jA_{\mathrm{TT}}^{ij}W_i\,dA$$

must be retained and matched to the chosen boundary conditions.

Gourgoulhon derives the conformal longitudinal operator and transverse split in Gourgoulhon 2007, §6.4, pp. 91–95, PDF. Bartnik and Isenberg place it in the conformal constraint method in Bartnik and Isenberg 2004, §4.1, pp. 14–19, PDF.

Application to extrinsic curvature

With $K_{ij}=-\gamma_i{}^\mu\gamma_j{}^\nu\nabla_\mu n_\nu$ and $j_i=-\gamma_i{}^\mu n^\nu T_{\mu\nu}$, write

$$K_{ij}=A_{ij}+\frac13\gamma_{ij}K.$$

The momentum constraint becomes

$$D_jA^{ij}-\frac23D^iK=8\pi G j^i.$$

Choose a trace-free seed tensor $M^{ij}$ and set

$$A^{ij}=M^{ij}+(\mathbb LW)^{ij}.$$

Then

$$\Delta_{\mathbb L}W^i =\frac23D^iK+8\pi G j^i-D_jM^{ij}.$$

In constant-mean-curvature vacuum data, $D_iK=0$ and $j^i=0$. If the seed is already transverse, $W^i=0$ is a solution. Away from CMC, the mean-curvature gradient couples the vector equation to the conformal factor through the physical covariant derivative and conformal weights; solving the momentum constraint in isolation can then be misleading.

The TT part is often called freely specifiable data. That phrase is local to a particular conformal construction. The Hamiltonian constraint still restricts the physical metric and the norm of $A_{ij}$, and boundary or regularity conditions restrict which TT tensors are admissible.

Kernels and solvability

A conformal Killing field $V^i$ satisfies

$$(\mathbb LV)_{ij}=0.$$

Therefore $W^i$ is not unique: $W^i+V^i$ generates the same longitudinal tensor. On a closed manifold, the vector equation is solvable only when its source $F^i$ is orthogonal to every conformal Killing field:

$$\int_\Sigma V_iF^i\,dV_\gamma=0.$$

This is the Fredholm compatibility condition. One can fix uniqueness by requiring $W^i$ to be orthogonal to the kernel, but that does not create solvability if the source violates the condition.

On an asymptotically flat end, decay conditions can remove translations, rotations, or growing modes from the allowed kernel. On a bounded region, Dirichlet, Neumann-like, or mixed conditions select a different operator domain. Inner black-hole boundaries may couple $W^i$ to expansion or quasiequilibrium conditions. There is no boundary-independent York split.

What TT does and does not mean

A transverse-traceless tensor contains two components per point after a local Fourier-space count, which suggests gravitational radiation. But several qualifications matter:

• transverse is defined with respect to the selected spatial metric;
• a TT piece of $K_{ij}$ is initial data, not a gauge-invariant waveform at null infinity;
• global TT tensors depend on topology and boundary conditions;
• conformal rescaling changes the physical interpretation and normalization;
• Coulombic or stationary data can have nontrivial trace-free structure.

Thus York’s split organizes the constraints; it does not by itself identify the radiative observables of a spacetime development.

Independent checks

On flat $\mathbb R^3$, take a plane-wave vector

$$W^i=A^ie^{ik_jx^j}.$$

Then, ignoring an overall Fourier sign,

$$\Delta_{\mathbb L}W^i \longleftrightarrow -\left(\lvert k\rvert^2\delta^i{}_j+\frac13k^ik_j\right)A^j.$$

For $A\cdot k=0$, the magnitude is $\lvert k\rvert^2$; for $A^i$ parallel to $k^i$, it is $4\lvert k\rvert^2/3$. Both are nonzero when $k\ne0$, confirming ellipticity. At $k=0$, constant translations lie in the kernel—exactly the global ambiguity expected without decay or boundary conditions.

Common pitfalls

Elliptic does not mean invertible. Conformal Killing fields form a kernel, and the source must satisfy an adjoint compatibility condition.

TT does not automatically mean observable radiation. It is a slice-dependent geometric decomposition used to solve constraints.

Boundary terms cannot be dropped by notation. Orthogonality and uniqueness depend on the operator domain and the data prescribed on every end or boundary.

Exercise: uniqueness modulo conformal Killing fields

Suppose $W_1^i$ and $W_2^i$ solve the same York vector equation on a closed manifold. Show that their difference is a conformal Killing field.

Solution

Let $V^i=W_1^i-W_2^i$. Then

$$\Delta_{\mathbb L}V^i=0.$$

Multiply by $V_i$, integrate, and use integration by parts:

$$0 =\int_\Sigma V_i\Delta_{\mathbb L}V^i\,dV_\gamma =-\frac12\int_\Sigma (\mathbb LV)_{ij}(\mathbb LV)^{ij}\,dV_\gamma.$$

The nonnegative integrand must vanish, so $(\mathbb LV)_{ij}=0$. Hence $V^i$ is a conformal Killing field. Imposing $L^2$ orthogonality to that kernel selects a unique representative when the source satisfies the compatibility conditions.

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.