An identity involving a Killing field Does anyone know how to prove the following identity. We assume that $\Omega$ is a Killing field and $U, V$ are vector fields. Then
$[\Omega ,\nabla _UV]-\nabla _U([\Omega, V])=\nabla _{[\Omega ,U]}V.$
Reference - bottom of page 342 of Christodolou's paper on the formation of black holes and singularities.
Many thanks
 A: If you expand the first two brackets in terms of the covariant derivative you get
$$\begin{align} 
[\Omega ,\nabla _UV]-\nabla _U([\Omega, V]) - \nabla _{[\Omega ,U]}V &= \nabla_\Omega \nabla_U V - \nabla_U \nabla_\Omega V - \nabla _{[\Omega ,U]}V + \nabla_U \nabla_V \Omega - \nabla_{\nabla_U V} \Omega 
\\
&= R(\Omega, U)V+\nabla^2_{U,V} \Omega,
\end{align}$$
so this difference is tensorial in $U,V$. Naming it $Z(U,V) = Z^k{}_{ij} U^i V^j \partial_k$, we see from the curvature symmetry $R_{likj}=R_{kjli}$ that
$$\begin{align}
Z_{kij} &= R_{likj}\Omega^l+\Omega_{k;ij} \\
&= \Omega_{i;kj} -\Omega_{i;jk} +\Omega_{k;ji}.
\end{align}$$
Now using the fact that $\Omega$ is Killing; i.e. $\Omega_{i;j} = -\Omega_{j;i}$:
$$ Z_{kij} = \Omega_{i;kj} + \Omega_{j;ik} + \Omega_{k;ji}. $$
But apply the Bianchi identity to $R_{likj}\Omega^l = \Omega_{i;kj} +\Omega_{j;ik}$ and you get
$$ 0 = \Omega_{i;kj} + \Omega_{j;ik} + \Omega_{k;ji} + \Omega_{i;kj} + \Omega_{j;ik} + \Omega_{k;ji} = 2Z_{kij},$$
so $Z(U,V) = 0$.
