# Ngô Quốc Anh

## December 26, 2012

### The evolution equations for the constraint equations

Filed under: Uncategorized — Tags: — Ngô Quốc Anh @ 1:12

Let us first recall the Gauss equation which is given by $\displaystyle\displaystyle\left\langle {\overline R (X,Y)Z,W} \right\rangle = \left\langle {R\left( {X,Y)Z} \right),W} \right\rangle - \left\langle {\mathrm{I\!I}(X,Z)} , {\mathrm{I\!I}(Y,W)} \right\rangle + \left\langle {\mathrm{I\!I}(Y,Z)} , {\mathrm{I\!I}(X,W)} \right\rangle.$

This can be proved as the following $\displaystyle\begin{array}{lcl} R\left( {X,Y)Z} \right) &=&\displaystyle {\nabla _X}{\nabla _Y}Z - {\nabla _Y}{\nabla _X}Z - {\nabla _{[X,Y]}}Z \hfill \\&=& {\nabla _X}({\overline \nabla _Y}Z - \mathrm{I\!I}(Y,Z)) - {\nabla _Y}({\overline \nabla _X}Z - \mathrm{I\!I}(X,Z)) - ({\overline \nabla _{[X,Y]}}Z - \mathrm{I\!I}([X,Y],Z)) \hfill \\&=& {\nabla _X}({\overline \nabla _Y}Z) - {\nabla _X}(\mathrm{I\!I}(Y,Z)) - {\nabla _Y}({\overline \nabla _X}Z) + {\nabla _Y}(\mathrm{I\!I}(X,Z)) - {\overline \nabla _{[X,Y]}}Z + \mathrm{I\!I}([X,Y],Z) \hfill \\&=& {\overline \nabla _X}{\overline \nabla _Y}Z - \mathrm{I\!I}(X,{\overline \nabla _Y}Z) - {\overline \nabla _Y}{\overline \nabla _X}Z + \mathrm{I\!I}(Y,{\overline \nabla _X}Z) - {\nabla _X}(\mathrm{I\!I}(Y,Z)) + {\nabla _Y}(\mathrm{I\!I}(X,Z)) \hfill \\ &&- {\overline \nabla _{[X,Y]}}Z + \mathrm{I\!I}([X,Y],Z) \hfill \\&=&\overline R (X,Y)Z - \mathrm{I\!I}(X,{\overline \nabla _Y}Z) + \mathrm{I\!I}(Y,{\overline \nabla _X}Z) - {\nabla _X}(\mathrm{I\!I}(Y,Z)) + {\nabla _Y}(\mathrm{I\!I}(X,Z)) + \mathrm{I\!I}([X,Y],Z),\hfill \\ \end{array}$

which implies $\displaystyle\begin{array}{lcl}\left\langle {\overline R (X,Y)Z,W} \right\rangle &=& \left\langle {R(X,Y)Z,W} \right\rangle + \left\langle {{\nabla _X}(\mathrm{I\!I}(Y,Z)),W} \right\rangle - \left\langle {{\nabla _Y}(\mathrm{I\!I}(X,Z)),W} \right\rangle \hfill \\&=& \left\langle {R(X,Y)Z,W} \right\rangle + \left\langle {{\nabla _X}(\mathrm{I\!I}(Y,Z)),W} \right\rangle - \left\langle {{\nabla _Y}(\mathrm{I\!I}(X,Z)),W} \right\rangle \hfill \\ &=& \left\langle {R(X,Y)Z,W} \right\rangle - \left\langle {\mathrm{I\!I}(Y,Z),{\nabla _X}W} \right\rangle + \left\langle {\mathrm{I\!I}(X,Z),{\nabla _Y}W} \right\rangle. \hfill \\ \end{array}$

By using ${\nabla _X}W = {\overline \nabla _X}W - \mathrm{I\!I}(X,W)$ and ${\nabla _Y}W = {\overline \nabla _Y}W - \mathrm{I\!I}(Y,W)$, we finally obtain $\displaystyle\left\langle {\overline R (X,Y)Z,W} \right\rangle = \left\langle {R(X,Y)Z,W} \right\rangle + \left\langle {\mathrm{I\!I}(Y,Z),\mathrm{I\!I}(X,W)} \right\rangle - \left\langle {\mathrm{I\!I}(X,Z),\mathrm{I\!I}(Y,W)} \right\rangle$

as claimed.

This identity is slightly different from that of the usual one since $\langle n, n \rangle = -1$ plays a crucial role in the above computation.

We denote by $K_{\alpha\beta}$ the components of the scalar second fundamental form $K$, i.e., $\displaystyle \mathrm{I\!I}(\cdot, \cdot) = -K(\cdot, \cdot) n.$