Proof of a solution of Klein-Gordon equation with Killing vector In this (very good) notes: http://people.physics.tamu.edu/pope/geomlec.pdf it is set as an exercise to proof that if $u_i$ solves the Klein-Gordon equation:
$$(\Box -m^2 )u_i = 0$$
then you can proof, appealing to the properties of the Killing vectors, that the Killing vector $K^\mu \partial_\mu$ also solves
$$(\Box -m^2 )K^\mu \partial_\mu u_i = 0$$
and that the key point is to prove that $\Box(K^\mu \partial_\mu u_i ) = K^\mu \partial_\mu\Box u_i$
Honestly, I don't get with the key. Any help will be appreciated.
 A: If $K$ is a Killing vector, it satisfies
$$
\nabla_\mu K_\nu  + \nabla_\nu K_\mu = 0 \quad \implies \quad \nabla^\mu K_\mu = 0 ~. 
$$
Another property we will need can be proved by acting on the above euqation with $\nabla^\nu$. We find
\begin{align}
\Box K_\mu 
&= - \nabla^\nu \nabla_\mu K_\nu \\
&= [ \nabla_\mu ,  \nabla_\nu ]K^\nu \\
&= R^\nu{}_{\lambda\mu\nu} K^\lambda \\
&= - R_{\mu\nu} K^\nu~. 
\end{align}
Now, we have
$$
( \nabla^\mu \nabla_\mu - m^2 ) K^\nu \nabla_\nu \phi = \Box K^\nu \nabla_\nu \phi +  2  \nabla^\mu   K^\nu \nabla_\mu \nabla_\nu \phi +  K^\nu  \nabla^\mu  \nabla_\mu \nabla_\nu \phi - m^2 K^\nu \nabla_\nu \phi ~. 
$$
Note
$$
 \nabla^\mu   K^\nu \nabla_\mu \nabla_\nu \phi 
=\frac{1}{2} ( \nabla^\mu   K^\nu + \nabla^\nu   K^\mu )  \nabla_\mu \nabla_\nu \phi = 0 ~,
$$
and
\begin{align}
K^\nu  \nabla^\mu  \nabla_\mu \nabla_\nu \phi 
&= K^\nu  \nabla^\mu   \nabla_\nu  \nabla_\mu \phi \\
&= K^\nu \nabla_\nu  \Box  \phi + K^\nu  [ \nabla^\mu  , \nabla_\nu ]  \nabla_\mu \phi \\
&= m^2 K^\nu \nabla_\nu \phi + K^\mu R_{\mu\nu} \nabla^\mu \phi  ~. 
\end{align}
Putting all this together, we find
\begin{align}
( \nabla^\mu \nabla_\mu - m^2 ) K^\nu \nabla_\nu \phi 
&= \Box K^\nu \nabla_\nu \phi   +  m^2 K^\nu \nabla_\nu \phi + K^\mu R_{\mu\nu} \nabla^\mu \phi  - m^2 K^\nu \nabla_\nu \phi  \\
&= 0 ~.
\end{align}
QED.
