An identity involving a Killing field

Question

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

Answer 1

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 $Z(U,V) = 0$.