Let us call the spacetime $M$ with a metric $g_{ab}$. There is a unit spacelike vector field $\eta^a$ orthogonal to a hypersurface. So that we can define the so-called gravitational electric and magnetic fields using Weyl tensor:
$E_{ab}=C_{ambn}\eta^m\eta^n$, $B_{ab}=*C_{ambn}\eta^m\eta^n$
where $*C_{ambn}=\epsilon_{ampq}C^{pq}_{\phantom{dd}bn}$ with $\epsilon_{ampq}$ the alternating tensor. This definition is not the standard one, but it does not matter in my question. Plus, I should not call $E_{ab}, B_{ab}$ gravitational electric, magnetic fields because $\eta^a$ is not timelike. But let me use these names to represent the two fields. This is just nomenclature.
The Weyl tensor satisfies the following relation:
$\nabla_{[m}C_{ab]cd}=2g_{c[m}C_{ab]pq}\eta^p+2g_{d[m}C_{ab]cp}\eta^p+\eta_{[m}C_{ab]cd}$
I want to show that
$D_{[a}E_{b]c}=0,\quad D_{[a}B_{b]c}=0$
Here, $D_a$ is the covariant derivative compatible with the induced metric $h_{ab}=g_{ab}-\eta_a\eta_b$.
It is easy to prove $D_{[a}E_{b]c}=0$, by simply contracting $\eta^q\eta^c$ with both sides of the equation of Weyl tensor and obtaining:
$\nabla_mE_{bd}-\nabla_bE_{md}=\eta_mE_{bd}-\eta_bE_{md}$
and then projecting this to the hypersurface to get the result.
But it is not so easy to get the 2nd result: $D_{[a}B_{b]c}=0$. One possible method is to contract $\eta_f\eta^b\epsilon^{cdef}$, then you will get the following complicated expression:
$\nabla_mB_{ae}-\nabla_aB_{me}=4\eta^b(\epsilon_{ab}^{\phantom{dd}ef}E_{md}-\epsilon_{mb}^{\phantom{dd}ef}E_{ad})-2\eta^b*C_{amb}^{\phantom{edd}e}-(\eta_mB_{ae}-\eta_aB_{me})$
The last bracket will disappear after projected to the hypersurface, but the remaining terms seem to survive. Do you have any idea of how to solve this?
Thanks!
