3+1 formalism and bases of numerical relativity - LUTh ...

196 Conformal Killing operator and conformal vector Laplacian
Replace each occurrence of ∂/∂x j by the component ξj of the linear form ξ, thereby obtaining
a mapping which is no longer differential, i.e. that involves only values of the fields at the point
p; this is the principal symbol of ˜∆L at p with respect to ξ:
P (p,ξ) : Tp(Σ) −→ Tp(Σ)
v = (v i ) ↦−→ P (p,ξ)(v) =
˜γ jk (p)ξjξk v i + 1
3 ˜γik (p)ξkξjv j
,
(B.13)
The differential operator ˜∆L is said to be elliptic on Σ iff the principal symbol P (p,ξ) is an
isomorphism for every p ∈ Σ and every non-vanishing linear form ξ ∈ T ∗
p (Σ). It is said to be
strongly elliptic if all the eigenvalues of P (p,ξ) are non-vanishing and have the same sign. To
check whether it is the case, let us consider the bilinear form ˜P (p,ξ) associated to the endomorphism
P (p,ξ) by the conformal metric:
∀(v,w) ∈ Tp(Σ) 2 , ˜P (p,ξ)(v,w) = ˜γ v, P (p,ξ)(w) . (B.14)
Its matrix ˜ Pij is deduced from the matrix P i j of P (p,ξ) by lowering the index i with ˜γ(p). We
get
˜Pij = ˜γ kl (p)ξkξl ˜γij(p) + 1
3 ξiξj. (B.15)
Hence ˜P (p,ξ) is clearly a symmetric bilinear form. Moreover it is positive definite for ξ = 0: for
any vector v ∈ Tp(Σ) such that v = 0, we have
˜P (p,ξ)(v,v) = ˜γ kl (p)ξkξl ˜γij(p)v i v j + 1
3 (ξiv i ) 2 > 0, (B.16)
where the > 0 follows from the positive definite character of ˜γ. ˜P (p,ξ) being positive definite
symmetric bilinear form, we conclude that P (p,ξ) is an isomorphism and that all its eigenvalues
are real and strictly positive. Therefore ˜∆L is a strongly elliptic operator.
B.2.3 Kernel
Let us now determine the kernel of ˜∆L. Clearly this kernel contains the kernel of the conformal
Killing operator ˜L. Actually it is not larger than that kernel:
ker ˜∆L = ker ˜L . (B.17)
Let us establish this property. For any vector field v ∈ T (Σ), we have
˜γijv
Σ
i ∆L
˜ v j ˜γ d 3 x =
=
=
˜γijv
Σ
i Dl( ˜ ˜ Lv) jl ˜γ d 3 x
Σ
∂Σ
˜Dl ˜γijv i ( ˜ Lv) jl
˜γijv i ( ˜ Lv) jl
2
˜sl ˜q d y −
− ˜γij ˜ Dlv i ( ˜ Lv) jl ˜γ d 3 x
˜γij
Σ
˜ Dlv i ( ˜ Lv) jl ˜γ d 3 x,
(B.18)
where the Gauss-Ostrogradsky theorem has been used to get the last line. We shall consider
two situations for (Σ,γ):