Vector field commutator and wave equation I am studying this review on black holes https://arxiv.org/abs/0811.0354 by Dafermos and Rodnianski, and I try  to prove the proposition E.0.1. I  am currently stuck on the following equation :
$$
{\mathcal L}_X (\nabla_a \nabla_b \psi)-\nabla_a {\mathcal L}_X 
\nabla_b\psi=2 \left ((\nabla_b\, ^X\pi_{a\mu})   - 
(\nabla_\mu\, ^X\pi_{b a} ) +(\nabla_a \,^X\pi_{\mu b}) \right) \nabla^\mu\psi
$$
where
$
{}^X\pi_{\mu\nu}= =\frac12 (\mathcal{L}_X g)_{\mu\nu}.
$, $\mathcal{L}_X$ the Lie derivative of the vector field $X$, $\nabla$ the covariant derivative and $\psi$ a scalar function.
Can anyone help me to prove this formula ?
 A: Proof:  $\nabla_a\nabla_b\Psi$ is a rank-2 covariant tensor so the Lie derivative is:
$$ \mathcal{L}_X\nabla_a\nabla_b\Psi=X^c\nabla_c\nabla_a\nabla_b\Psi+(\nabla_c\nabla_b\Psi)\nabla_aX^c+(\nabla_a\nabla_c\Psi)\nabla_bX^c.$$
Similarly,
$$\mathcal{L}_X\nabla_b\Psi=X^c\nabla_c\nabla_b\Psi +(\nabla_c\Psi)\nabla_bX^c.$$
So,
$$\nabla_a\mathcal{L}_X\nabla_b\Psi=(\nabla_aX^c)\nabla_c\nabla_b\Psi+X^c\nabla_a\nabla_c\nabla_b\Psi +(\nabla_a\nabla_c\Psi)\nabla_bX^c+(\nabla_c\Psi)\nabla_a\nabla_bX^c.$$
Let's concentrate on this last term. Now, taking the convention that $2\nabla_{(a}X_{b)}=\nabla_aX_b+\nabla_bX_a$, gives
$$\nabla_a\nabla_bX_c=2\nabla_a\nabla_{(b}X_{c)}-\nabla_a\nabla_{c}X_{b}$$
Recall the Ricci identity for a 1-form $\omega$:
$$\nabla_{a}\nabla_{b}\omega_c-\nabla_{b}\nabla_{a}\omega_c=-{R^d}_{cab}\omega_d.$$
With $\omega=X_{\flat}$ one finds:
$$\nabla_a\nabla_bX_c=2\nabla_a\nabla_{(b}X_{c)}-\nabla_c\nabla_{a}X_{b}+{R^d}_{bac}X_d.$$
Using $2\nabla_{(a}X_{b)}=\nabla_aX_b+\nabla_bX_a$ again gives
$$\nabla_a\nabla_bX_c=2\nabla_a\nabla_{(b}X_{c)}-2\nabla_c\nabla_{(a}X_{b)}+\nabla_c\nabla_{b}X_{a}+{R^d}_{bac}X_d.$$
Using the Ricci identity again gives
$$\nabla_a\nabla_bX_c=2\nabla_a\nabla_{(b}X_{c)}-2\nabla_c\nabla_{(a}X_{b)}+\nabla_b\nabla_{c}X_{a}-{R^d}_{acb}X_d+{R^d}_{bac}X_d.$$
Using $2\nabla_{(a}X_{b)}=\nabla_aX_b+\nabla_bX_a$ one last time gives
$$\nabla_a\nabla_bX_c=2\nabla_a\nabla_{(b}X_{c)}-2\nabla_c\nabla_{(a}X_{b)}+2\nabla_b\nabla_{(c}X_{a)}-\nabla_b\nabla_{a}X_{c}-{R^d}_{acb}X_d+{R^d}_{bac}X_d.$$
Using the Ricci identity gives
$$\nabla_a\nabla_bX_c=2\nabla_a\nabla_{(b}X_{c)}-2\nabla_c\nabla_{(a}X_{b)}+2\nabla_b\nabla_{(c}X_{a)}-\nabla_a\nabla_{b}X_{c}+{R^{d}}_{cba}X_d-{R^d}_{acb}X_d+{R^d}_{bac}X_d.$$
So we've established
$$2\nabla_a\nabla_bX_c=2\nabla_a\nabla_{(b}X_{c)}-2\nabla_c\nabla_{(a}X_{b)}+2\nabla_b\nabla_{(c}X_{a)}+{R^{d}}_{cba}X_d-{R^d}_{acb}X_d+{R^d}_{bac}X_d.$$
Now the first (algebraic) Bianchi identity gives
$$ {R^d}_{abc}+{R^d}_{bca}+{R^d}_{cab}=0.$$
Hence,
$$2\nabla_a\nabla_bX_c=2\nabla_a\nabla_{(b}X_{c)}-2\nabla_c\nabla_{(a}X_{b)}+2\nabla_b\nabla_{(c}X_{a)}-2{R^d}_{acb}X_d.$$
So,
$$\nabla_a\mathcal{L}_X\nabla_b\Psi=(\nabla_aX^c)\nabla_c\nabla_b\Psi+X^c\nabla_a\nabla_c\nabla_b\Psi +(\nabla_a\nabla_c\Psi)\nabla_bX^c+(\nabla_c\Psi)\Big(\nabla_a\nabla_{(b}X_{c)}-\nabla_c\nabla_{(a}X_{b)}+\nabla_b\nabla_{(c}X_{a)}-{R^d}_{acb}X_d\Big).$$
Therefore, combining and canceling terms gives
$$\mathcal{L}_X\nabla_a\nabla_b\Psi-\nabla_a\mathcal{L}_X\nabla_b\Psi=X^c\nabla_c\nabla_a\nabla_b\Psi-X^c\nabla_a\nabla_c\nabla_b\Psi -(\nabla_c\Psi)\Big(\nabla_a\nabla_{(b}X_{c)}-\nabla_c\nabla_{(a}X_{b)}+\nabla_b\nabla_{(c}X_{a)}-{R^d}_{acb}X_d\Big).$$
One final use of the Ricci identity gives on the 1-form $d\Psi$ gives
$$\mathcal{L}_X\nabla_a\nabla_b\Psi-\nabla_a\mathcal{L}_X\nabla_b\Psi=-X^c{R^d}_{bca}\nabla_d\Psi+{R^d}_{acb}X_d\nabla^c\Psi-(\nabla_c\Psi)\Big(\nabla_a\nabla_{(b}X_{c)}-\nabla_c\nabla_{(a}X_{b)}+\nabla_b\nabla_{(c}X_{a)}\Big).$$
Relabelling dummy indices and using the symmetry $R_{abcd}=R_{cdab}$ gives the result.
