2.2 York–Lichnerowicz conformal decompositions

For general initial-data configurations, the most widely used class of constraint decompositions are the York–Lichnerowicz conformal decompositions. At their heart are a conformal decomposition of the metric and certain components of the extrinsic curvature, together with a transverse-traceless decomposition of the extrinsic curvature.

First, the metric is decomposed into a conformal factor ψ multiplying an auxiliary 3-metric [69107108]:

γ ≡ ψ4γ&tidle; . (16 ) ij ij
The auxiliary 3-metric &tidle;γij is often called the conformal or background 3-metric, and it carries five degrees of freedom. Its natural definition is given by
−1∕3 &tidle;γij = γ γij, (17 )
leaving &tidle;γ = 1, but we are free to choose any normalization for &tidle;γ. Using (16View Equation), we can rewrite the Hamiltonian constraint (14View Equation) as
&tidle; 2 1 &tidle; 1 5 2 1 5 ij 5 ∇ ψ − 8ψ R − 8ψ K + 8ψ KijK = − 2πG ψ ρ, (18 )
where &tidle;∇2 ≡ &tidle;∇i∇&tidle; i is the scalar Laplace operator, and ∇&tidle; i and R&tidle; are the covariant derivative and Ricci scalar associated with &tidle;γij. Equation (18View Equation) is a quasilinear elliptic equation for the conformal factor ψ, and we see that the Hamiltonian constraint naturally constrains the 3-metric.

The conformal decomposition of the Hamiltonian constraint was proposed by Lichnerowicz. But, the key to the full decomposition is the treatment of the extrinsic curvature introduced by York [109110]. This begins by splitting the extrinsic curvature into its trace and tracefree parts,

Kij ≡ Aij + 13γijK. (19 )
The decomposition proceeds by using the fact that we can covariantly split any symmetric tracefree tensor as follows:
𝒮ij ≡ (𝕃X )ij + T ij. (20 )
Here, T ij is a symmetric, transverse-traceless tensor (i.e., ∇jT ij = 0 and T i= 0 i) and
ij i j j i 2 ij ℓ (𝕃X ) ≡ ∇ X + ∇ X − 3γ ∇ ℓX . (21 )
After separating out the transverse-traceless portion of 𝒮ij, what remains, (𝕃X )ij, is referred to as its “longitudinal” part. We now want to apply this transverse-traceless decomposition to the tracefree part of the extrinsic curvature Aij. However, the conformal decomposition of the metric leaves us with at least two ways to proceed.

The goal of the decomposition is to produce a coupled set of elliptic equations to be solved with some prescribed boundary conditions. We have already reduced the Hamiltonian constraint to an elliptic equation being solved on a background space in terms of differential operators that are compatible with the conformal 3-metric. In the end, we want to reduce the momentum constraints to a set of elliptic equations based on differential operators that are compatible with the same conformal 3-metric. But, the longitudinal operator (21View Equation) can be defined with respect to any metric space. In particular, it is natural to consider decompositions with respect to both the physical and conformal 3-metrics.

2.2.1 Conformal transverse-traceless decomposition

Let us first consider decomposing Aij with respect to the conformal 3-metric [111116Jump To The Next Citation Point]. As we will see, when certain assumptions are made, this decomposition has the advantage of producing a simpler set of elliptic equations that must be solved. The first step is to define the conformal tracefree extrinsic curvature ij A&tidle; by

Aij ≡ ψ −10 &tidle;Aij or A ≡ ψ−2A&tidle; . (22 ) ij ij
Next, the transverse-traceless decomposition is applied to the conformal extrinsic curvature,
A&tidle;ij ≡ (𝕃&tidle;X )ij + &tidle;Qij. (23 )
Note that the longitudinal operator &tidle;𝕃 and the symmetric, transverse-tracefree tensor Q&tidle;ij are both defined with respect to covariant derivatives compatible with &tidle;γij.

Applying equations (16View Equation), (19View Equation), (21View Equation), (22View Equation), and (23View Equation) to the momentum constraints (15View Equation), we find that they simplify to

i 2 6 i 10 i &tidle;Δ ð•ƒX = 3ψ ∇&tidle; K + 8πG ψ j , (24 )
whereUpdateJump To The Next Update Information
Δ&tidle; Xi ≡ &tidle;∇ (𝕃&tidle;X )ij = ∇&tidle;2Xi + 1&tidle;∇i(∇&tidle; Xj ) + R&tidle;i Xj, (25 ) 𝕃 j 3 j j
and we have used the fact that
¯∇j 𝒮ij = ψ− 10∇&tidle;j (ψ10𝒮ij) (26 )
for any symmetric tracefree tensor ij 𝒮.

In deriving equation (24View Equation), we have also used the fact that Q&tidle;ij is transverse (i.e. &tidle;∇j &tidle;Qij = 0). However, in general, we will not know if a given symmetric tracefree tensor, say M&tidle;ij, is transverse. By using (20View Equation) we can obtain its transverse-traceless part &tidle;Qij via

&tidle;Qij ≡ M&tidle; ij − (&tidle;𝕃Y )ij, (27 )
and using the fact that if &tidle;Qij is transverse, we find
&tidle;∇ &tidle;Qij ≡ 0 = &tidle;∇ &tidle;M ij − Δ&tidle; Y i. (28 ) j j 𝕃
Thus, Eqs. (27View Equation) and (28View Equation) give us a general way of constructing the required symmetric transverse-traceless tensor from a general symmetric traceless tensor.

Using the linearity of &tidle;𝕃, we can rewrite (23View Equation) as

&tidle;ij &tidle; ij &tidle; ij A = (𝕃V ) + M , (29 )
where
V i ≡ Xi − Y i. (30 )
Similarly, using the linearity of &tidle; Δ ð•ƒ, we can rewrite (24View Equation) as
&tidle; i 2 6&tidle; i &tidle; &tidle;ij 10 i Δ ð•ƒV = 3ψ ∇ K − ∇j M + 8πG ψ j . (31 )
By solving directly for V i, we can combine the steps of decomposing M&tidle; ij with that of solving the momentum constraints.

After applying (19View Equation) and (22View Equation) to the Hamiltonian constraint (18View Equation), we obtain the following full decomposition, which I will list together here for convenience:

γij = ψ4 &tidle;γij, ij −10 ij 1 −4 ij K = ψ A&tidle; + 3ψ &tidle;γ K, A&tidle;ij = (&tidle;𝕃V )ij + &tidle;M ij, (32 ) i 2 6 i ij 10 i &tidle;Δ ð•ƒV − 3ψ &tidle;∇ K = − ∇&tidle;j M&tidle; + 8 πG ψ j , &tidle;∇2ψ − 1ψ &tidle;R − -1ψ5K2 + 1ψ−7A&tidle; A&tidle;ij = − 2πG ψ5 ρ. 8 12 8 ij

In the decomposition given by (32View Equation), we are free to specify a symmetric tensor &tidle;γ ij as the conformal 3-metric, a symmetric tracefree tensor &tidle; ij M, and a scalar function K. Then, with given matter energy and momentum densities, ρ and i j, and appropriate boundary conditions, the coupled set of constraint equations for ψ and V i are solved. Finally, given the solutions, we can construct the physical initial data, γij and Kij.

The decomposition outlined above has the interesting property that if we choose K to be constant and if the momentum density vanishes4, then the momentum constraint equations fully decouple from the Hamiltonian constraint. As we will see later, this simplification has proven to be useful.

2.2.2 Physical transverse-traceless decomposition

Alternatively, we can decompose Aij with respect to the physical 3-metric [828384]. We decompose the extrinsic curvature as

Aij ≡ (¯ð•ƒW )ij + Qij. (33 )
In this case, the longitudinal operator ¯ð•ƒ and the symmetric transverse-tracefree tensor ij Q are both defined with respect to covariant derivatives compatible with γij.

Applying equations (16View Equation), (19View Equation), (21View Equation), (33View Equation), and (26View Equation) to the momentum constraint (15View Equation), we find that it simplifies to

&tidle; i &tidle; ij &tidle; 2&tidle; i 4 i Δ𝕃W + 6(𝕃W ) ∇j ln ψ = 3∇ K + 8πG ψ j, (34 )
where we have used the fact that
¯ ij −4 &tidle; ij (𝕃W ) = ψ (𝕃W ) . (35 )

As in the previous section, we will obtain the symmetric transverse-traceless tensor Qij from a general symmetric tracefree tensor &tidle;ij M by using (20View Equation). In this case, we take

ij − 10 ij ij Q ≡ ψ M&tidle; − (¯ð•ƒZ ) , (36 )
and use the fact that Qij is transverse, to obtain
&tidle; i &tidle; ij &tidle; −6&tidle; &tidle;ij Δ ð•ƒZ + 6(𝕃Z ) ∇j ln ψ = ψ ∇j M . (37 )
Again, we can define
Vi ≡ W i − Zi, (38 )
and use the linearity of &tidle;𝕃 and &tidle;Δ ð•ƒ to combine the process of obtaining the transverse-traceless part of M ij and solving the momentum constraints. We obtain the following full decomposition, which I will list together here for convenience5:
4 γij = ψ &tidle;γij(, ) Kij = ψ− 4 A&tidle;ij + 1γ&tidle;ijK , 3 &tidle;Aij = (&tidle;𝕃V )ij + ψ −6M&tidle;ij, (39 ) i ij 2 i −6 ij 4 i Δ&tidle;𝕃V + 6 (𝕃&tidle;V ) &tidle;∇j lnψ = 3∇&tidle; K − ψ ∇&tidle;j M&tidle; + 8πG ψ j , ∇&tidle;2 ψ − 1ψR&tidle; − 1-ψ5K2 + 1ψ5 &tidle;A &tidle;Aij = − 2πG ψ5ρ. 8 12 8 ij

In the decomposition given by (39View Equation), we are again free to specify a symmetric tensor &tidle;γij as the conformal 3-metric, a symmetric tracefree tensor &tidle;M ij, and a scalar function K. Then, with given matter energy and momentum densities, ρ and i j, and appropriate boundary conditions, the coupled set of constraint equations for ψ and i V are solved. Finally, given the solutions, we can construct the physical initial data, γij and Kij.

Notice that, while very similar to the decomposition from Section 2.2.1, the sets of equations are distinctly different. In general, if we make the same choices for the freely specifiable data in both decompositions (i.e., we choose &tidle;γij, &tidle; ij M, and K the same), we will produce two different sets of initial data. Both will be equally valid solutions of the constraint equations, but they will have distinct physical properties.

There is at least one exception to this. Assume we have a valid set of initial data γij and Kij, which satisfies the constraint equations (14View Equation) and (15View Equation). For any everywhere-positive function Ψ, we define our freely specifiable data as follows:

−4 &tidle;γij ≡ Ψ γij, M&tidle; ij ≡ Ψ10 (Kij − 1γijK ) , (40 ) i 3 K ≡ K i.
Then the solution to both sets of equations, assuming we use correct boundary conditions, will be ψ = Ψ and i V = 0, which yields the original data as the solution for each decomposition.
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