# Ngô Quốc Anh

## December 8, 2012

### Derivation of the Einstein constraint equations: The momentum constraint

Filed under: Riemannian geometry — Tags: — Ngô Quốc Anh @ 3:28

This entry is a continuation of the previous entry where we showed in detail the derivation of the Hamiltonian constrain in general relativity. Today, we derive the so-called momentum constraint equation.

First, let us recall the Codazzi equation which is given by the following identity

$\displaystyle {R^ \bot }(X,Y)Z = ({\nabla _X}\mathrm{I\!I})(Y,Z) - ({\nabla _Y}\mathrm{I\!I})(X,Z).$

Sometimes, we call it the Codazzi–Mainardi equation or the Ricci identity, which expresses the curvature of the normal bundle in terms of the second fundamental form. Using this, we obtain

$\displaystyle\sum\limits_{i = 1}^n {{R^ \bot }({e_i},Y){e_i}} = \sum\limits_{i = 1}^n {({\nabla _{{e_i}}}\mathrm{I\!I})(Y,{e_i})} - \sum\limits_{i = 1}^n {({\nabla _Y}\mathrm{I\!I})({e_i},{e_i})}$

for any tangent vector $Y$. Since $R(e_i,Y)e_i$, for all $i \geqslant 1$, belongs to the tangent space, there hold

$\displaystyle\left\langle {R({e_i},Y){e_i},{e_0}} \right\rangle = 0 \quad \forall i = \overline{1,n}$

and thus we can write

$\displaystyle\left\langle {{R^ \bot }({e_i},Y){e_i},{e_0}} \right\rangle = \left\langle {R({e_i},Y){e_i},{e_0}} \right\rangle\quad \forall i = \overline{1,n}.$

Moreover, by the antisymmetric property of the Riemann curvature tensor, i.e., $R_{ijkl}=-R_{ijlk}$, we know that

$\displaystyle\left\langle {R({e_0},{e_i}){e_0},{e_0}} \right\rangle = 0\quad \forall i = \overline{1,n}$

which immediately implies

$\displaystyle\left\langle {R({e_0},Y){e_0},{e_0}} \right\rangle = 0.$

Hence, we get that

$\displaystyle\text{Ric}_{\overline g }(Y,{e_0}) = \sum\limits_{i = 1}^n {\left\langle {R({e_i},Y){e_i},{e_0}} \right\rangle } .$

Therefore,

$\displaystyle \text{Ric}_{\overline g }(Y,{e_0}) = \sum\limits_{i = 1}^n {\left\langle {{R^ \bot }({e_i},Y){e_i},{e_0}} \right\rangle } = \sum\limits_{i = 1}^n {\left\langle {({\nabla _{{e_i}}}\mathrm{I\!I})(Y,{e_i}),{e_0}} \right\rangle } - \sum\limits_{i = 1}^n {\left\langle {({\nabla _Y}\mathrm{I\!I})({e_i},{e_i}),{e_0}} \right\rangle } .$

We now evaluate the right hand side. In order to achieve that goal, we make use of the formula involving the covariant derivative of the second fundamental form. Indeed, we first write

$\displaystyle\begin{array}{lcl} ({\nabla _{{e_i}}}\mathrm{I\!I})(Y,{e_i}) &=& \nabla _{{e_i}}^ \bot (\mathrm{I\!I}(Y,{e_i})) - \mathrm{I\!I}(\nabla _{{e_i}}^MY,{e_i}) - \mathrm{I\!I}(Y,\nabla _{{e_i}}^M{e_i}) \hfill \\&=& - \nabla _{{e_i}}^ \bot (K(Y,{e_i}){e_0}) + K(\nabla _{{e_i}}^MY,{e_i}){e_0} + K(Y,\nabla _{{e_i}}^M{e_i}){e_0}. \end{array}$

Using the Leibniz rule and thanks to $\langle \nabla_{e_i}e_0,e_0 \rangle =0$, we can write

$\displaystyle\nabla _{{e_i}}^ \bot (K(Y,{e_i}){e_0}) = \nabla _{{e_i}}^ \bot (K(Y,{e_i})){e_0} + K(Y,{e_i})\underbrace {\nabla _{{e_i}}^ \bot {e_0}}_{ = 0} = {\nabla _{{e_i}}}(K(Y,{e_i})){e_0}.$

Thus, we get that

$\displaystyle\left\langle {({\nabla _{{e_i}}}\mathrm{I\!I})(Y,{e_i}),{e_0}} \right\rangle = {\nabla _{{e_i}}}(K(Y,{e_i})) - K(\nabla _{{e_i}}^MY,{e_i}) - K(Y,\nabla _{{e_i}}^M{e_i}).$

By summing, we obtain

$\displaystyle\sum\limits_{i = 1}^n {\left\langle {({\nabla _{{e_i}}}\mathrm{I\!I})(Y,{e_i}),{e_0}} \right\rangle } = (\text{div}_gK)(Y).$

For the second term, we do the same way as follows

$\displaystyle\begin{array}{lcl}({\nabla _Y}\mathrm{I\!I})({e_i},{e_i}) &=& \nabla _Y^ \bot (\mathrm{I\!I}({e_i},{e_i})) - \mathrm{I\!I}(\nabla _Y^M{e_i},{e_i}) - \mathrm{I\!I}({e_i},\nabla _Y^M{e_i}) \hfill \\&=& \nabla _Y^ \bot (\mathrm{I\!I}({e_i},{e_i})) - 2\mathrm{I\!I}(\nabla _Y^M{e_i},{e_i}).\end{array}$

Without loss of generality, we can assume that we are in normal coordinates at some point, say $p \in M$. This immediately implies that $\nabla _Y^M{e_i}=0$ for all $i=\overline{1,n}$. Therefore,

$\displaystyle\begin{array}{lcl}({\nabla _Y}\mathrm{I\!I})({e_i},{e_i}) &=& \nabla _Y^ \bot (\mathrm{I\!I}({e_i},{e_i})) \hfill \\&=& - \nabla _Y^ \bot (K({e_i},{e_i}){e_0}) \hfill \\&=&- \nabla _Y^ \bot (K({e_i},{e_i})){e_0} + K({e_i},{e_i})\underbrace {\nabla _Y^ \bot {e_0}}_{ = 0} \hfill \\&=&- {\nabla _Y}(K({e_i},{e_i})){e_0}. \end{array}$

Thus, by summing, we obtain

$\displaystyle\sum\limits_{i = 1}^n {\left\langle {({\nabla _Y}\mathrm{I\!I})({e_i},{e_i}),{e_0}} \right\rangle } = {\nabla _Y}\Big(\sum\limits_{i = 1}^n {K({e_i},{e_i})} \Big) = d(\text{trace}_gK)(Y).$

Thus, we have proved that

$\displaystyle\text{Ric}_{\overline g }(Y,{e_0}) = (\text{div}_gK)(Y) - d(\text{trace}_gK)(Y).$

By using the Einstein equation and thanks to $\overline g(Y,e_0)=0$, we get that

$\displaystyle \text{Ric}_{\overline g }(Y,{e_0}) = \text{Ric}_{\overline g }(Y,{e_0}) - \frac{1}{2}\underbrace {\overline g (Y,{e_0})}_{ = 0}\text{Scal}_{\overline g} = \kappa T(Y,{e_0}).$

Thus, we arrive at

$\displaystyle \text{div}_gK - d(\text{trace}_gK)=\kappa T(\cdot,{e_0}),$

known as the momentum constraint equation in general relativity.

Remark: Throughout the proof, we have used the fact that the vector $\nabla_Y e_0$ belongs to the tangent space, thus giving us $\nabla_Y^\bot e_0=0$. For a proof of this fact, see this entry.