Ngô Quốc Anh

December 10, 2012

Construction of spacetimes via solutions of the vacuum Einstein constraint equations and the propagation of the gauge condition

Filed under: Riemannian geometry — Tags: — Ngô Quốc Anh @ 23:10

A couple of days ago, we showed that the Einstein equations are essentially hyperbolic under the harmonic gauge. To be precise, the solvability of the equations $\displaystyle {\overline {{\text{Ric}}}} - \frac{1}{2}\overline g \text{Scal}_{\overline g} +\Lambda \overline g = \kappa\overline T,$

is equivalent to solving the following hyperbolic system $\displaystyle - \frac{1}{2}{\overline g ^{km}}{\overline g _{ij,km}} = \Psi_{ij}((\overline g_{pq})_{0\leqslant p,q \leqslant n},(\overline g_{pq,r})_{0\leqslant p,q,r \leqslant n}),$

provided $\displaystyle {\lambda ^\alpha } = 0$ for all $\alpha = \overline {0,n}$. Later on, when we consider the Cauchy problem for the Einstein equations on some appropriate framework, the first and second fundamental forms need to verify some constraint equations, [here and here].

By tracing the above equation, we obtain $\displaystyle \text{Scal}_{\overline g} + \frac{2n}{2 - n}\Lambda = \frac{2\kappa}{2 - n}\text{trace}_{\overline g}(\overline T),$

which helps us to write $\displaystyle\overline {{\text{Ric}}} = \kappa \left( {\overline T - \frac{\overline g}{{n - 2}}{\text{trace}}_{\overline g}(\overline T )} \right) + \frac{2}{{n - 2}}\Lambda \overline g.$

In the vacuum case, the above equation is nothing but $\displaystyle {\overline {{\text{Ric}}}}=\frac{2}{{n - 2}}\Lambda \overline g.$

The spacetime $V$ we choose will be the product $M \times [0,+\infty)$. What we really need is to construct a metric $\overline g$ on $V$ such that the Einstein equations fulfill. In fact, thanks to the hyperbolic system showed above, we aim to find a suitable initial condition such that

• The gauge condition can propagate in time;
• The hyperbolic system is solvable for small time $t$.

While the latter is standard (we shall touch this issue later), the former needs to study and this is the main point of this notes.

In this entry, given a solution $(g,K)$ of the constraint equations on a manifold $(M,g)$ of the dimension $n$, we shall construct an appropriate spacetime $(V,\overline g)$ of the dimension $n+1$. Recall that $\displaystyle\begin{array}{lcl} {\overline {{\text{Ric}}} _{ij}} + \frac{1}{2}({{\overline g}_{\alpha i}}\lambda _{,j}^\alpha + {{\overline g}_{\alpha j}}\lambda _{,i}^\alpha ) &=&- \frac{1}{2}{{\overline g}^{km}}{{\overline g}_{ij,km}} + {\text{lower order terms}} \hfill \\&=& \kappa \left( {{{\overline T}_{ij}} - \frac{1}{{n - 2}}{\text{trac}}{{\text{e}}_{\overline g}}(\overline T){{\overline g}_{ij}}} \right) + \frac{2}{{n - 2}}\Lambda {{\overline g}_{ij}} + \frac{1}{2}({{\overline g}_{\alpha i}}\lambda _{,j}^\alpha + {{\overline g}_{\alpha j}}\lambda _{,i}^\alpha ). \end{array}$

Initially, we set $\displaystyle\begin{gathered} {\overline g _{ij}} \;\,= {g_{ij}},1 \leqslant i,j \leqslant n, \hfill \\ {\overline g _{00}} \,\,= - 1, \hfill \\ {\overline g _{0j}} \,\,= 0,1 \leqslant j \leqslant n, \hfill \\ {\overline g _{ij,0}} = - 2{K_{ij}},1 \leqslant i,j \leqslant n. \hfill \\ \end{gathered}$

As a consequence of the choice above, we also have $\overline g _{ij,k}$ for non-zero $i,j,k$. To fully construct the initial condition, we still need to find $\overline g_{0j,o}$.

First, in view of the gauge condition, $\displaystyle {\lambda ^\alpha } = {\square _{\overline g }}{x^\alpha } =0$, we find that $\displaystyle {\lambda ^\alpha } = \frac{1}{{\sqrt {\det \overline g } }}{(\sqrt {\det \overline g } \,{\overline g ^{ij}}x_{,i}^\alpha )_{,j}} = \overline g _{,j}^{\alpha j} + \frac{1}{2}{\overline g ^{\alpha j}}{\overline g ^{pq}}{\overline g _{pq,j}} = 0.$

Therefore, at $t=0$, we get that $\displaystyle\begin{gathered} {\lambda ^0} = \overline g _{,0}^{00} + \frac{1}{2}{\overline g ^{00}}{\overline g ^{pq}}{\overline g _{pq,0}}, \hfill \\ {\lambda ^\alpha } = \overline g _{,0}^{\alpha 0} + \underbrace {\sum\limits_{j = 1}^n {\left( {\overline g _{,j}^{\alpha j} + \frac{1}{2}{{\overline g }^{\alpha j}}{{\overline g }^{pq}}{{\overline g }_{pq,j}}} \right)} }_{\text{already determined}},1 \leqslant \alpha \leqslant n. \hfill \\ \end{gathered}$

Hence, in order to guarantee that $\lambda^\alpha \equiv 0$ at $t=0$ for all $\alpha=\overline{0,n}$, we first select $\overline g _{,0}^{\alpha 0}$ such that $\lambda^\alpha =0$ at $t=0$. Once this task is done, we can determine $\overline g _{,0}^{00}$ at $t=0$ such that $\lambda^0=0$ at $t=0$.

In other words, the initial data that preserves the gauge condition can be found in this way. Within a small time, the hyperbolic system mentioned above always admits a solution $\overline g$. That is to say that the reduced Einstein equations have solution, i.e., $\displaystyle \mathop{\overline {\text{Ric}}}\limits_{(h)}{}^{ij}=\kappa \left( {{{\overline T}^{ij}} - \frac{1}{{n - 2}}{\text{trace}}_{\overline g}(\overline T){{\overline g}^{ij}}} \right)+\frac{2}{{n - 2}}\Lambda \overline g^{ij},$

where the $(0,2)$-tensor $\mathop{\overline {\text{Ric}}}\limits_{(h)}$ is given by $\displaystyle\mathop{\overline {\text{Ric}}}\limits_{(h)}{}_{ij}={\overline {{\text{Ric}}} _{ij}} + \frac{1}{2}({\overline g _{\alpha i}}\lambda _{,j}^\alpha + {\overline g _{\alpha j}}\lambda _{,i}^\alpha ).$

However, it is not necessary to have that $\overline g$ solves the Einstein equations unless the gauge condition remains valid within the small time. We shall prove this affirmatively.

We now consider how could the gauge condition propagate in time so long as the metric $\overline g$ solves the reduced equations. As we shall see, all $\lambda^\alpha$ satisfy a homogeneous linear wave equation which is a consequence of the Bianchi identities with vanishing initial time derivatives which come from the constraint equations.

First, since the metric $\overline g$ solves $\displaystyle - \frac{1}{2}{{\overline g}^{km}}{{\overline g}_{ij,km}} + {\text{lower order terms}} = \kappa \left( {{{\overline T}_{ij}} - \frac{1}{{n - 2}}{\text{trace}}_{\overline g}(\overline T){{\overline g}_{ij}}} \right)$

and thanks to the Einstein equation for the $\text{Ric}$ curvature, we immediately have $\displaystyle {\overline {{\text{Ric}}} _{ij}} + \frac{1}{2}({{\overline g}_{\alpha i}}\lambda _{,j}^\alpha + {{\overline g}_{\alpha j}}\lambda _{,i}^\alpha ) = \frac{2}{{n - 2}}\Lambda {{\overline g}_{ij}}.$

Let us write $\displaystyle {L_{\alpha \beta }} = \frac{1}{2}{\overline g _{\beta \mu }}{\partial _\alpha }{\lambda ^\mu } + \frac{1}{2}{\overline g _{\alpha \mu }}{\partial _\beta }{\lambda ^\mu }-\frac{2}{{n - 2}}\Lambda {{\overline g}_{\alpha \beta}}.$

Using this, we have $\displaystyle {\overline {{\text{Ric}}} ^{\alpha \beta }} - \frac{1}{2}{\overline g ^{\alpha \beta }}\text{Scal}_{\overline g} = - ({L^{\alpha \beta }} - \frac{1}{2}{\overline g ^{\alpha \beta }}L),$

where $L$ is the trace of $L^{\alpha\beta}$. Since the Einstein tensor is divergence free, we obtain $\displaystyle {\nabla _\lambda }({L^{\lambda \mu }} - \frac{1}{2}{g^{\lambda \mu }}L) = 0.$

Keep in mind that $\nabla _\lambda \overline g=0$. A simple calculation shows that $\displaystyle\begin{gathered} 0 = {\nabla _\lambda }({\overline g ^{\alpha \lambda }}{\overline g ^{\beta \mu }}{L_{\alpha \beta }} - \frac{1}{2}{\overline g ^{\lambda \mu }}{\overline g ^{\alpha \beta }}{L_{\alpha \beta }}) \hfill \\ \,\,\,\,= \frac{1}{2}{\nabla _\lambda } \Big({\overline g ^{\alpha \lambda }}{\overline g ^{\beta \mu }}({\overline g _{\beta \mu }}{\partial _\alpha }{\lambda ^\mu } + {\overline g _{\alpha \mu }}{\partial _\beta }{\lambda ^\mu }) - \frac{1}{2}{\overline g ^{\lambda \mu }}{\overline g ^{\alpha \beta }}({\overline g _{\beta \mu }}{\partial _\alpha }{\lambda ^\mu } + {\overline g _{\alpha \mu }}{\partial _\beta }{\lambda ^\mu }) \Big) \hfill \\ \,\,\,\,= \frac{1}{2}{\overline g ^{\alpha \lambda }}\underbrace {{{\overline g }^{\beta \mu }}{{\overline g }_{\beta \mu }}}_1{\nabla _\lambda }({\partial _\alpha }{\lambda ^\mu }) + \frac{1}{2}{\overline g ^{\beta \mu }}\underbrace {{{\overline g }^{\alpha \lambda }}{g_{\alpha \mu }}}_{\delta _\mu ^\lambda }{\nabla _\lambda }({\partial _\beta }{\lambda ^\mu }) \hfill \\ \qquad- \frac{1}{4}{\overline g ^{\lambda \mu }}\underbrace {{{\overline g }^{\alpha \beta }}{{\overline g }_{\beta \mu }}}_{\delta _\mu ^\alpha }{\nabla _\lambda }({\partial _\alpha }{\lambda ^\mu }) - \frac{1}{4}{\overline g ^{\lambda \mu }}\underbrace {{{\overline g }^{\alpha \beta }}{{\overline g }_{\alpha \mu }}}_{\delta _\mu ^\beta }{\nabla _\lambda }({\partial _\beta }{\lambda ^\mu }) \hfill \\ \,\,\,\,= \frac{1}{2}{\overline g ^{\alpha \lambda }}{\nabla _\lambda }({\partial _\alpha }{\lambda ^\mu }) + \frac{1}{2}{\overline g ^{\beta \mu }}{\nabla _\lambda }({\partial _\beta }{\lambda ^\lambda }) - \frac{1}{2}{\overline g ^{\lambda \mu }}{\nabla _\lambda }({\partial _\beta}{\lambda ^\beta}). \hfill \\ \end{gathered}$

Thanks to $\displaystyle {\square _{\overline g }} = {\overline g ^{\lambda \alpha}}{\nabla _\lambda }({\partial _\alpha }(\cdot))$

we have proved that $\displaystyle\frac{1}{2}{\square _{\bar g}}{\lambda ^\mu } + B_{jq}^p(({{\bar g}_{rs}}),({{\bar g}_{rs,l}}))\lambda _{,p}^q = 0.$

The above long calculation also shows that $\displaystyle\begin{array}{lcl} {\overline {{\text{Ric}}} _{\alpha \beta }} - \frac{1}{2}{\overline g _{\alpha \beta }}{\text{Scal}}_{\overline g} + \Lambda {{\overline g}_{\alpha \beta }} &=&\displaystyle - {L_{\alpha \beta }} + \frac{1}{2}{\overline g _{\alpha \beta }}L + \Lambda {{\overline g}_{\alpha \beta }} \hfill \\&=&\displaystyle - \frac{1}{2}{\overline g _{\beta \mu }}{\partial _\alpha }{\lambda ^\mu } - \frac{1}{2}{\overline g _{\alpha \mu }}{\partial _\beta }{\lambda ^\mu } + \frac{2}{{n - 2}}\Lambda {{\overline g}_{\alpha \beta }} \hfill \\&&\displaystyle + \frac{1}{2}{\overline g _{\alpha \beta }}{\overline g ^{pq}}\left( {\frac{1}{2}{{\overline g }_{q\mu }}{\partial _p}{\lambda ^\mu } + \frac{1}{2}{{\overline g }_{p\mu }}{\partial _q}{\lambda ^\mu } - \frac{2}{{n - 2}}\Lambda {{\overline g}_{pq}}} \right) + \Lambda {{\overline g}_{\alpha \beta }} \hfill \\&=&\displaystyle - \frac{1}{2}{\overline g _{\beta \mu }}{\partial _\alpha }{\lambda ^\mu } - \frac{1}{2}{\overline g _{\alpha \mu }}{\partial _\beta }{\lambda ^\mu } + \frac{2}{{n - 2}}\Lambda {{\overline g}_{\alpha \beta }} \hfill \\&&\displaystyle + \frac{1}{4}{\overline g _{\alpha \beta }}\underbrace {{{\overline g }^{pq}}{{\overline g }_{q\mu }}}_{\delta _\mu ^p}{\partial _p}{\lambda ^\mu } + \frac{1}{4}{\overline g _{\alpha \beta }}\underbrace {{{\overline g }^{pq}}{{\overline g }_{p\mu }}}_{\delta _\mu ^q}{\partial _q}{\lambda ^\mu } - \frac{1}{{n - 2}}\Lambda {\overline g _{\alpha \beta }}\underbrace {{{\overline g }^{pq}}{{\overline g}_{pq}}}_n + \Lambda {{\overline g}_{\alpha \beta }} \hfill \\&=&\displaystyle - \frac{1}{2}{\overline g _{\beta \mu }}{\partial _\alpha }{\lambda ^\mu } - \frac{1}{2}{\overline g _{\alpha \mu }}{\partial _\beta }{\lambda ^\mu } + \frac{1}{2}{\overline g _{\alpha \beta }}{\partial _\mu }{\lambda ^\mu }. \end{array}$

In view of the vacuum setting and the constraint equations [here and here], there holds $\displaystyle (\text{Eins}+\Lambda \overline g)_{\alpha 0} = 0$

at $t=0$. Hence, at $t=0$, we get $\displaystyle - \frac{1}{2}{\overline g _{0\mu }}{\partial _\alpha }{\lambda ^\mu } - \frac{1}{2}{\overline g _{\alpha \mu }}{\partial _0}{\lambda ^\mu } + \frac{1}{2}{\overline g _{\alpha 0}}{\partial _\mu }{\lambda ^\mu } = 0.$

By using $\mu=\alpha=0$, i.e. the Hamiltonian constraint, we obtain $\partial_0\lambda^0=0$. We still need to prove that $\partial_0\lambda^\mu=0$ for all $\mu=\overline{1,n}$. Indeed, thanks to $\partial_\alpha\lambda^\mu=0$ for any $\alpha, \mu>0$, we know that $\displaystyle 0 = - \frac{1}{2}{\overline g _{0\mu }}\underbrace {{\partial _\alpha }{\lambda ^\mu }}_0 - \frac{1}{2}{\overline g _{\alpha \mu }}{\partial _0}{\lambda ^\mu } + \frac{1}{2}{\overline g _{\alpha 0}}\underbrace {{\partial _\mu }{\lambda ^\mu }}_ 0 = - \frac{1}{2}\sum\limits_{\mu = 1}^n {{{\overline g }_{\alpha \mu }}{\partial _0}{\lambda ^\mu }} .$

Since this is true for all $\alpha=\overline{1,n}$ and the matrix $(g_{\alpha \mu})$ is invertible, we have that all $\partial_0\lambda^\alpha$ vanish at $t=0$ for all $\alpha>0$. Since $\lambda^\alpha$ satisfy a homogeneous linear hyperbolic system with vanishing initial data, $\lambda^\alpha$ vanish identically by the uniqueness for solutions of the Cauchy problem for the hyperbolic evolutions.