Conserved quantity in a spacetime with Killing vector

Question

I am trying to prove that that the expression $Q=-\frac{1}{\kappa}\int_{S_\infty} \nabla^i \xi^k \mathrm{d}\sigma_{ik}$ is a conserved quantity for a spacetime with Killing vector $\xi^i$ where $S_\infty$ is the 'boundary at infinity' of the spacelike hypersurface $\Sigma$.

Using the Gauss Theorem, I have written $Q$ as follows:

$$\begin{align*}Q&=\frac{1}{\kappa}\int_{\Sigma} \nabla_c \nabla^c \xi^k \mathrm{d}\sigma_{k}\\&=\frac{1}{\kappa}\int_{\Sigma} R^{kb} \xi_b \mathrm{d}\sigma_{k}\end{align*}$$

Is there any way of seeing why this last expression is constant?

Thanks!

Answer 1

I will just sketch this out. The Killing equation $\xi_{a;b} + \xi_{b;a} = 0$ tells you that $\nabla^i\xi^k$ is antisymmetric in indices. I am going to write this as the commutator between $U$ the vector along which we differentiate $\xi$ with $\nabla_U$ so in compact notation we have $$ Q = -\frac{1}{\kappa}\int[U, \xi]d\sigma $$ I now use Stokes' rule to get this on the volume, where now the commutator is $[V, [U, \xi]]$ and this equals $R(U, V)\xi + \nabla_{[U,V]\xi}$. This gives $$ Q = -\frac{1}{\kappa}\int \left(R(U, V)\xi + \nabla_{[U,V]}\xi\right) dVol $$ This leads to the equation you have. However, in general the $\sigma$ is an area two-form and that $Vol$ is a volume three-form. This means the indices in tensor notation on $[V, [U, \xi]]$ have permutation symmetry and the integral is then over the permutation of this commutator. Then by Jacobi theorem this is zero and all that is left is a constant of integration.