Vector field commutator and wave equation

Question

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 ?

Answer 1

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.