Ngô Quốc Anh

January 14, 2013

Hyperbolicity of the 3+1 system of the Einstein equations under the harmonic slicing

Filed under: Uncategorized — Tags: — Ngô Quốc Anh @ 2:20

Let us first recall the evolution equation of K in this note

\displaystyle \frac{\partial }{{\partial t}}{K_{\alpha \beta }} = - {\nabla _\alpha }{\nabla _\beta }N + N({\text{Ric}_{\alpha \beta }}-\gamma _\alpha ^\mu \gamma _\beta ^\nu {\overline {\text{Ric}} _{\mu \nu }} + K{K_{\alpha \beta }} - 2{K_{\alpha \mu }}K_\beta ^\mu ) + {\mathcal L_{\vec \beta}}{K_{\alpha \beta }},

which can be rewritten as the following

\displaystyle {\overline {{\text{Ric}}} _{\alpha \beta }} = - \frac{1}{N}{\nabla _\alpha }{\nabla _\beta }N + {\text{Ri}}{{\text{c}}_{\alpha \beta }} + K{K_{\alpha \beta }} - 2{K_{\alpha \mu }}K_\beta ^\mu - \frac{1}{N}(\frac{\partial }{{\partial t}} - {\mathcal{L}_{\vec \beta }}){K_{\alpha \beta }}

for all \alpha ,\beta > 0 since in our setting \gamma_i^j=\delta_i^j for all i,j>0. The quantities \text{Ric}_{0\beta} and \text{Ric}_{00} come from the Codazzi and Ricci equations. To be precise, we have, by the Codazzi equation,

\displaystyle {\overline {{\text{Ric}}} _{0\beta }} = {\nabla _h}K_\beta ^h - {\partial _\beta }K

and, by the Ricci equation,

\displaystyle {\overline {{\text{Ric}}} _{00}} = \frac{1}{N}(\frac{\partial }{{\partial t}} - {\mathcal{L}_{\vec \beta }})K + \frac{1}{N}{\nabla ^i}{\nabla _i}N + {K_{ij}}{K^{ij}}.

It is clear that the infinitesimal variation of the Ricci curvature \delta R_{mu\nu} corresponding to an infinitesimal \delta g variation of the space metric is

\displaystyle\delta {R_{\mu \nu }} = \frac{1}{2}({\nabla ^h}{\nabla _{(i}}\delta {g_{j)h}} - {\nabla _h}{\nabla ^h}\delta {g_{ij}} - {\nabla _j}{\partial _i}({g^{hk}}\delta {g_{hk}})),

where the notation (i j) is nothing but ij+ji. This formula can be applied to \frac{\partial}{\partial t} and \mathcal L_{\vec \beta} to get

\displaystyle (\frac{\partial }{{\partial t}} - {\mathcal{L}_{\vec \beta }}){R_{\mu \nu }} = \frac{1}{2} \Big({\nabla ^h}{\nabla _{(i}}(\frac{\partial }{{\partial t}} - {\mathcal{L}_{\vec \beta }}){g_{j)h}} - {\nabla _h}{\nabla ^h}(\frac{\partial }{{\partial t}} - {\mathcal{L}_{\vec \beta }}){g_{ij}} - {\nabla _j}{\partial _i}({g^{hk}}(\frac{\partial }{{\partial t}} - {\mathcal{L}_{\vec \beta }}){g_{hk}}) \Big).

Thanks to the evolution equation for the metric g, i.e. 2NK_{ij}=-(\frac{\partial }{{\partial t}} - {\mathcal{L}_{\vec \beta }})g_{ij}, we get

\displaystyle (\frac{\partial }{{\partial t}} - {\mathcal{L}_{\vec \beta }}){R_{\mu \nu }} = - {\nabla ^h}{\nabla _{(i}}N{K_{j)h}} + {\nabla _h}{\nabla ^h}(N{K_{ij}}) + {\nabla _j}{\partial _i}(\underbrace {{g^{hk}}(N{K_{hk}})}_{NK}.

We now exchange the order of covariant derivatives of the first term on the right hand side of the preceding equation as follows

\displaystyle - {\nabla ^h}{\nabla _{(i}}N{K_{j)h}} = - {\nabla _{(i}}{\nabla ^h}N{K_{j)h}} - 2NR_{ijm}^hK_h^m - N{R_{m(i}}K_{j)}^m.

This can be proved as follows: First we have

\displaystyle {\nabla _h}{\nabla _i}{\beta _{jk}} - {\nabla _i}{\nabla _h}{\beta _{jk}} = - R_{hij}^p{\beta _{pk}} - R_{hik}^p{\beta _{jp}},

we obtain

\displaystyle\begin{array} {lcl}{\nabla ^h}{\nabla _i}{\beta _{jk}} - {\nabla _i}{\nabla ^h}{\beta _{jk}} &=& {g^{hl}}({\nabla _l}{\nabla _i}{\beta _{jk}} - {\nabla _i}{\nabla _l}{\beta _{jk}}) \hfill \\ &=& - {g^{hl}}R_{lik}^p{\beta _{jp}} - {g^{hl}}R_{lij}^p{\beta _{pk}}. \hfill \\ \end{array}


\displaystyle {\nabla ^h}{\nabla _i}(N{K_{jk}}) = {\nabla _i}{\nabla ^h}(N{K_{jk}}) - N{g^{hl}}R_{lik}^p{K_{jp}} - N{g^{hl}}R_{lij}^p{K_{pk}}.


\displaystyle\begin{array} {lcl}{\nabla ^h}{\nabla _i}(N{K_{jh}}) &=& {\nabla _i}{\nabla ^h}(N{K_{jh}}) - N{g^{hl}}R_{lih}^p{K_{jp}} - N{g^{hl}}R_{lij}^p{K_{ph}} \hfill \\ &=& \displaystyle {\nabla _i}{\nabla ^h}(N{K_{jh}}) - N\underbrace {R_i^p}_{{R_{im}}{g^{mp}}}{K_{jp}} - NR_{lij}^pK_p^l \hfill \\ &=& {\nabla _i}{\nabla ^h}(N{K_{jh}}) - N{R_{im}}K_j^m - NR_{mij}^pK_p^m.\end{array}

Exchanging i and j and adding together we get the estimate.

This helps us to write

\displaystyle\begin{array}{lcl} {\Omega _{ij}} &=& \displaystyle (\frac{\partial }{{\partial t}} - {\mathcal{L}_{\vec \beta }}){R_{ij}} + {\nabla _{(i}}({N^2}{R_{0j)}}) \hfill \\&=& \displaystyle - (\frac{\partial }{{\partial t}} - {\mathcal{L}_{\vec \beta }})({N^{ - 1}}\big(\dfrac{\partial }{{\partial t}} - {\mathcal{L}_{\vec \beta }}){K_{ij}} \big) + (\frac{\partial }{{\partial t}} - {\mathcal{L}_{\vec \beta }} )(K{K_{ij}}- 2{K_{im}}K_j^m) \hfill \\&& \displaystyle - (\frac{\partial }{{\partial t}} - {\mathcal{L}_{\vec \beta }})({N^{ - 1}}{\nabla _j}{\partial _i}N) - N{\nabla _i}{\partial _j}K - {\nabla _{(i}}({K_{j)h}}{\partial ^h}N) \hfill \\&& \displaystyle - 2NR_{ijm}^hK_h^m - N{R_{m(i}}K_{j)}^m + {\nabla _h}{\nabla ^h}(N{K_{ij}}) + K{\nabla _j}{\partial _i}N. \end{array}

We shall eliminate at the same time the following two terms {\nabla _i}{\partial _j}K and (\frac{\partial }{{\partial t}} - {\mathcal{L}_{\vec \beta }})({N^{ - 1}}{\nabla _j}{\partial _i}N). Indeed, since

\displaystyle\frac{\partial }{{\partial t}}{\nabla _j}{\partial _i}N = {\nabla _j}{\partial _i}\frac{\partial }{{\partial t}}N - \frac{1}{2}{g^{kl}}({\nabla _{(i}}\frac{\partial }{{\partial t}}{g_{j)l}} - {\nabla _l}\frac{\partial }{{\partial t}}{g_{ij}}){\partial _k}N


\displaystyle {\mathcal{L}_{\vec \beta }}{\nabla _j}{\partial _i}N = {\nabla _j}{\partial _i}{\mathcal{L}_{\vec \beta }}N - \frac{1}{2}{g^{kl}}({\nabla _{(i}}{\mathcal{L}_{\vec \beta }}{g_{j)l}} - {\nabla _l}{\mathcal{L}_{\vec \beta }}{g_{ij}}){\partial _k}N,

we find that

\displaystyle\begin{array}{lcl} \displaystyle (\frac{\partial }{{\partial t}} - {\mathcal{L}_{\vec \beta }}){\nabla _j}{\partial _i}N &=&{\nabla _j}{\partial _i}(\dfrac{\partial }{{\partial t}} - {\mathcal{L}_{\vec \beta }})N - \frac{1}{2}({\nabla _{(i}}\underbrace {(\frac{\partial }{{\partial t}} - {\mathcal{L}_{\vec \beta }}){g_{j)l}}}_{ - 2N{K_{j)l}}} - {\nabla _l}\underbrace {(\frac{\partial }{{\partial t}} - {\mathcal{L}_{\vec \beta }}){g_{ij}}}_{ - 2N{K_{ij}}}){\partial ^l}N \hfill \\ &=&\displaystyle {\nabla _j}{\partial _i}(\frac{\partial }{{\partial t}} - {\mathcal{L}_{\vec \beta }})N + ({\nabla _{(i}}(N{K_{j)l}} - {\nabla _l}(N{K_{ij}})){\partial ^l}N.\end{array}

Therefore, the third order terms in N and the second order terms in K in the equation for K can be written in terms of

\displaystyle {C_{ij}} = \frac{1}{N}{\nabla _j}{\partial _i} \Big((\frac{\partial }{{\partial t}} - {\mathcal{L}_{\vec \beta }})N + {N^2}K \Big).

Consequently, the condition C_{ij}=0 should hold which yields the following condition

\displaystyle (\frac{\partial }{{\partial t}} - {\mathcal{L}_{\vec \beta }})N + K{N^2} = 0.

Following the previous note, we can easily check that the lapse function N has to verify the condition

\displaystyle N = \alpha \sqrt {\det \gamma }

with \alpha satisfying

\displaystyle (\frac{\partial }{{\partial t}} - {\mathcal{L}_{\vec \beta }})\alpha = 0.

Once we can remove all high order derivatives, we see that the Einstein equations R_{\alpha\beta}=\rho_{\alpha\beta} imply the following wave equation

\displaystyle N\square {K_{ij}} = {Q_{ij}} + {\Theta _{ij}},

where we set

\displaystyle\square {K_{ij}} = - {N^{ - 2}}{(\frac{\partial }{{\partial t}} - {\mathcal{L}_{\vec \beta }})^2}{K_{ij}} + {\nabla ^h}{\nabla _h}{K_{ij}}


\displaystyle {\Theta _{ij}} = (\frac{\partial }{{\partial t}} - {\mathcal{L}_{\vec \beta }}){\rho _{ij}} + {\nabla _{(i}}({N^2}{\rho _{0j)}})


\displaystyle\begin{array}{lcl} {Q_{ij}} &=&\displaystyle - {K_{ij}}(\frac{\partial }{{\partial t}} - {\mathcal{L}_{\vec \beta }})K + 2{g^{hm}}{K_{m(i}}(\frac{\partial }{{\partial t}} - {\mathcal{L}_{\vec \beta }}){K_{j)h}} + 4N{g^{hl}}{g^{mk}}{K_{lk}}{K_{im}}{K_{jh}} \hfill \\ &&\displaystyle+ (2{\nabla _{(i}}{K_{j(l}}){\partial ^l}N - 2K{N^{ - 1}}{\partial _i}N{\partial _j}N - 2{\partial _{(i}}N{\partial _{j)}}K \hfill \\&&\displaystyle - 3{\partial _h}N{\nabla ^h}{K_{ij}} - {K_{ij}}{\nabla ^h}{\nabla _h}N - {N^{ - 1}}{K_{ij}}{\partial ^h}N{\partial _h}N \hfill \\ &&\displaystyle+ {N^{ - 1}}{K_{h(i}}{\partial _{j)}}N{\partial ^h}N + ({\nabla _{(i}}{\partial ^h}N){K_{j)h}} + 2NR_{ijm}^hK_h^m \hfill \\ &&\displaystyle+ N{R_{m(i}}K_{j)}^m - 2K{\nabla _j}{\partial _i}N.\end{array}

In the final step, we show that any solution of the above wave equation verifies the Einstein equation. Indeed, we first write the stress-energy tensor T_{\alpha\beta} as the following

\displaystyle T_{\alpha\beta}=\rho_{\alpha\beta} - \frac{1}{2} g_{\alpha\beta} \rho.

Then we transform the Einstein equation into the following from \Sigma_{\alpha\beta}=0 with

\displaystyle \Sigma_{\alpha\beta} = (R_{\alpha\beta} - \rho_{\alpha\beta}) - \frac{1}{2} g_{\alpha\beta} ( R-\rho).

Clearly, solving the equation \Sigma_{\alpha\beta} =\Theta_{\alpha\beta} is equivalent to solving

\displaystyle {\nabla _0}{\Sigma _{ij}} + {\nabla _{(i}}({N^2}\Sigma _{j)}^0) - {g_{ij}}{\nabla _h}({N^2}{\Sigma ^{0h}}) + {f_{ij}} = 0,

where f_{ij} is linear and homogeneous in \Sigma_{ij}. By using the Bianchi identities, we finally obtain a quasi third order system for \Sigma_{ij} with principal operator the hyperbolic operator (\frac{\partial }{{\partial t}} - {\mathcal{L}_{\vec \beta }})\square. The vanishing for t=0 of \Sigma_{ij} results from the constraint equations. Moreover, one can see that all derivatives of order \leq 2 of \Sigma_{ij} also vanish. It follows from the uniqueness theorem for hyperbolic systems that \Sigma_{ij}=0 all the time.

See also:

Leave a Comment »

No comments yet.

RSS feed for comments on this post. TrackBack URI

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Create a free website or blog at

%d bloggers like this: