# 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

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

Therefore,

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

Consequently,

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

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

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

and

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

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

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

and

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

and

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

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.