A question about Killing vector and Riemann curvature tensor In Sean Carroll's Spacetime and Geometry, a formula is given as
$${\nabla _\mu }{\nabla _\sigma }{K^\rho } = {R^\rho }_{\sigma \mu \nu }{K^\nu },$$
where $K^\mu$ is a Killing vector satisfying Killing's equation ${\nabla _\mu }{K_\nu } +{\nabla _\nu }{K_\mu }=0$ and the convention of Riemann curvature tensor is
$$\left[\nabla_{\mu},\nabla_{\nu}\right]V^{\rho}={R^\rho}_{\sigma\mu\nu}V^{\sigma}.$$
So how to prove the this formula (the connection is Levi-Civita)?
 A: @C.R. This is my 'simpler proof'; I'm pretty sure it's correct, and simpler than Zhen's one as well.
From the first Bianchi identity [Carroll, (3.132)]
$$R_{\mu\nu\rho\sigma}+R_{\mu\rho\sigma\nu}+R_{\mu\sigma\nu\rho}=0$$
we have that, for every vector $V^\rho$,
$$\nabla_{[\mu}\nabla_\nu V_{\rho]}=\tfrac{1}{6}\big(R_{\rho\alpha\mu\nu}+R_{\mu\alpha\nu\rho}+R_{\rho\mu\nu\alpha}\big)V^\alpha=0,$$
where in the last equation I used the symmetry properties of the indices in the Riemann tensor to reduce it to the Bianchi identity. This is a very useful formula! I'm quite sure about the index placement thanks to the metric compatibility, $\nabla_\mu g_{\nu\rho}=0$ [Carroll, (1.32)], which implies that the metric $g_{\nu\rho}$ commutes with the covariant derivative $\nabla_\mu$.
Now, we take the Killing vector $K^\mu$ and we expand the antisymmetrization in the previous equation:
$$0=6\nabla_{[\mu}\nabla_\nu K_{\rho]}=\nabla_\mu\nabla_\nu K_\rho + 
\nabla_\nu \nabla_\rho K_\mu +\nabla_\rho\nabla_\mu K_\nu-
\nabla_\nu\nabla_\mu K_\rho-\nabla_\mu\nabla_\rho K_\nu-\nabla_\rho\nabla_\mu K_\nu.$$
Now we use the Killing's equation [Carroll, (3.174)], $\nabla_{(\nu}K_{\rho)}=0$ or $\nabla_\nu K_\rho =-\nabla_\rho K_\nu$, to simplify the previous equation:
$$\nabla_\mu\,\nabla_\nu K_\rho=-\nabla_\mu\,\nabla_\rho K_\nu\quad \Rightarrow\quad 0=2\nabla_\mu\nabla_\nu K_\rho+(\nabla_\nu\nabla_\rho-\nabla_\rho\nabla_\nu)K_\mu+\nabla_\rho\nabla_\mu K_\nu-\nabla_\nu\nabla_\mu K_\rho.$$
Again, using Killing's equations $\nabla_\rho\nabla_\mu K_\nu=-\nabla_\rho\nabla_\nu K_\mu$ and $\nabla_\nu\nabla_\mu K_\rho=-\nabla_\nu\nabla_\rho K_\mu$ we get:
$$0=2\big(\nabla_\mu\nabla_\nu K_\rho+(\nabla_\nu\nabla_\rho-\nabla_\rho\nabla_\nu)K_\mu\big)=2\big(\nabla_\mu\nabla_\nu K_\rho+R_{\mu\alpha\nu\rho}K^\alpha\big)$$
$$\nabla_\mu\nabla_\nu K_\rho=-R_{\mu\alpha\nu\rho}K^\alpha=R_{\rho\nu\mu\alpha}K^\alpha.$$
Then, we can rise the index $\rho$ multiplying with the metric $g^{\rho\sigma}$ (and using the metric compatibility):
$$\nabla_\mu\nabla_\nu K^\sigma=R^\sigma_{\phantom{\sigma}\nu\mu\alpha}K^\alpha.$$
That's all folks! I hope it would be helpful (and correct)!
A: Permit me the use of Latin indices instead of Greek indices and the convention $\nabla_a K_b=K_{b;a} $. So we wish to prove
$\newcommand{\Tud}[3]{{#1}^{#2}_{\phantom{#2}{#3}}}$
$$\Tud{K}{a}{;b c} = \Tud{R}{a}{b c d} K^d$$
where
$$\Tud{V}{a}{;b c} - \Tud{V}{a}{;c b} = \Tud{R}{a}{d c b} V^d$$
and
$$K_{a ; b} + K_{b ; a} = 0$$
Differentiating the last equation, we get
$$K_{a ; b c} + K_{b ; a c} = 0$$
so, relabelling and summing,
$$K_{a ; b c} + K_{b ; a c} - K_{b ; c a} - K_{c ; b a} + K_{c ; a b} + K_{a ; c b} = 0$$
hence,
$$K_{a; b c} + K_{a ; c b} = R_{b d a c} K^d + R_{c d a b} K^d$$
By the interchange symmetry $R_{a b c d} = R_{c d a b}$, and raising indices, we get
$$\Tud{K}{a}{;b c} - \Tud{R}{a}{b c d} K^d = -(\Tud{K}{a}{;c b} - \Tud{R}{a}{c b d} K^d)$$
On the other hand, by the first Bianchi identity and antisymmetry, we have
$$\Tud{R}{a}{d c b} = \Tud{R}{a}{b c d} + \Tud{R}{a}{c d b}$$
Hence we get
$$\Tud{K}{a}{;b c} = \Tud{K}{a}{; c b} + \Tud{R}{a}{d c b} K^d = \Tud{K}{a}{;c b} + \Tud{R}{a}{b c d} K^d + \Tud{R}{a}{c d b} K^d$$
and therefore
$$\Tud{K}{a}{;b c} - \Tud{R}{a}{b c d} K^d = \Tud{K}{a}{;c b} - \Tud{R}{a}{c b d} K^d$$
The conclusion follows.
A: A solution that is simpler still, requiring no differentiation or fancy identities other than one of the Bianchi identities, is as follows.

We know that the Riemann tensor can measure how much the covariant derivatives commute with each other, e.g
\begin{equation}
[ \nabla _{\mu } ,\nabla _{\nu }] k_{\alpha } =-R^{\beta }{}_{\alpha \mu \nu } k_{\beta }
\end{equation}
We can shuffle the indices in (4) and use the antisymmetry of the Lie bracket to get the three equations
\begin{gather}
[ \nabla _{\nu } ,\nabla _{\mu }] k_{\alpha } =R^{\beta }{}_{\alpha \mu \nu } k_{\beta } \tag{5A}\\
[ \nabla _{\mu } ,\nabla _{\alpha }] k_{\nu } =R^{\beta }{}_{\nu \alpha \mu } k_{\beta } \tag{5B}\\
[ \nabla _{\alpha } ,\nabla _{\nu }] k_{\mu } =R^{\beta }{}_{\mu \nu \alpha } k_{\beta }
\end{gather}
We now look at $\displaystyle ( 5\mathrm{C}) -( 5\mathrm{B}) -( 5\mathrm{A})$:
\begin{equation*}
[ \nabla _{\alpha } ,\nabla _{\nu }] k_{\mu } -[ \nabla _{\mu } ,\nabla _{\alpha }] k_{\nu } -[ \nabla _{\nu } ,\nabla _{\mu }] k_{\alpha } =\left( R^{\beta }{}_{\mu \nu \alpha } -R^{\beta }{}_{\nu \alpha \mu } -R^{\beta }{}_{\alpha \mu \nu }\right) k_{\beta }
\end{equation*}
Applying the first Bianchi identity,
\begin{equation*}
R^{\beta }{}_{\alpha \mu \nu } +R^{\beta }{}_{\mu \nu \alpha } +R^{\beta }{}_{\nu \alpha \mu } =0
\end{equation*}
This simplifies to
\begin{equation*}
[ \nabla _{\alpha } ,\nabla _{\nu }] k_{\mu } -[ \nabla _{\mu } ,\nabla _{\alpha }] k_{\nu } -[ \nabla _{\nu } ,\nabla _{\mu }] k_{\alpha } =2R^{\beta }{}_{\mu \nu \alpha } k_{\beta }
\end{equation*}
Expanding the Lie brackets,
\begin{gather*}
[ \nabla _{\alpha } ,\nabla _{\nu }] k_{\mu } -[ \nabla _{\mu } ,\nabla _{\alpha }] k_{\nu } -[ \nabla _{\nu } ,\nabla _{\mu }] k_{\alpha }\\
=( \nabla _{\alpha } \nabla _{\nu } -\nabla _{\nu } \nabla _{\alpha }) k_{\mu } -( \nabla _{\mu } \nabla _{\alpha } -\nabla _{\alpha } \nabla _{\mu }) k_{\nu } -( \nabla _{\nu } \nabla _{\mu } -\nabla _{\mu } \nabla _{\nu }) k_{\alpha }\\
=\nabla _{\alpha } \nabla _{\nu } k_{\mu } -\nabla _{\nu } \nabla _{\alpha } k_{\mu } +\nabla _{\alpha } \nabla _{\mu } k_{\nu } -\nabla _{\mu } \nabla _{\alpha } k_{\nu } +\nabla _{\mu } \nabla _{\nu } k_{\alpha } -\nabla _{\nu } \nabla _{\mu } k_{\alpha }
\end{gather*}
Using the linearity of $\displaystyle \nabla $,
\begin{gather*}
[ \nabla _{\alpha } ,\nabla _{\nu }] k_{\mu } -[ \nabla _{\mu } ,\nabla _{\alpha }] k_{\nu } -[ \nabla _{\nu } ,\nabla _{\mu }] k_{\alpha }\\
=\nabla _{\alpha }( \nabla _{\nu } k_{\mu } +\nabla _{\mu } k_{\nu }) +\nabla _{\mu }( \nabla _{\nu } k_{\alpha } -\nabla _{\alpha } k_{\nu }) -\nabla _{\nu }( \nabla _{\alpha } k_{\mu } +\nabla _{\mu } k_{\alpha })
\end{gather*}
Using Killing's equation $\displaystyle \nabla _{( \rho } k_{\sigma )} =0$ the first and third terms vanish, leaving
\begin{equation*}
[ \nabla _{\alpha } ,\nabla _{\nu }] k_{\mu } -[ \nabla _{\mu } ,\nabla _{\alpha }] k_{\nu } -[ \nabla _{\nu } ,\nabla _{\mu }] k_{\alpha } =\nabla _{\mu }( \nabla _{\nu } k_{\alpha } -\nabla _{\alpha } k_{\nu }) =2\nabla _{\mu } \nabla _{\nu } k_{\alpha }
\end{equation*}
Hence,
\begin{equation}
\boxed{\nabla _{\mu } \nabla _{\nu } k_{\alpha } =R^{\beta }{}_{\mu \nu \alpha } k_{\beta }}
\end{equation}
A: We could try to solve it the other way that if a vector obeys the first condition then it must be a Killing vector.
We assume,
$$
\nabla_{\mu} \nabla_{\sigma} K^{\rho} = R_{\sigma \mu \nu}^{\rho} k^{\nu}-\text { (i) }
$$
$$ where
 [\nabla_{\mu}, \nabla_{\sigma}] V^{\rho} = R_{\sigma,\mu,\nu}^{\rho} V^{\sigma}
$$
Subtracting $\nabla_{\sigma} \nabla_{\mu} K^{\rho}$ from (i)
$$
\begin{array}{l}
{\left[\nabla_{\mu}, \nabla_{\sigma}\right] K^{\rho} = R_{\sigma, \mu \nu}^{\rho} K^{\nu} - \nabla_{\sigma} \nabla_{\mu} K^{\nu}} \\
\nabla_{\sigma} \nabla_{\mu} K^{\rho}=\left(R_{\sigma, \mu \nu}^{\rho}-R_{\nu, \mu \sigma}^{\rho}\right) K^{\nu}
\end{array}
$$
$$
\begin{array}{r}
\nabla_{\sigma} \nabla_{\mu} K_{\rho}=\left(R_{\rho} _{\sigma}_{\mu \nu}- R_{\mu \sigma \rho \nu}\right) K^{\nu} \\
\left(\because R_{a b c d}=R_{c d a b}\right)
\end{array}
$$
using (1),
$$
\nabla_{\sigma} \nabla_{\mu} K_{\rho}=\nabla_{\mu} \nabla_{\sigma} K_{\rho}-\nabla_{\rho} \nabla_{\sigma} K_{\mu}
$$
Add $  \nabla_{\sigma} \nabla_{\rho} K_{\mu}$
$$\nabla_{\sigma}\left(\nabla_{\mu} K_{\rho}+\nabla_{\rho} K_{\mu}\right)=\nabla_{\mu} \nabla_{\sigma} K_{\rho}+\left[\nabla_{\sigma}, \nabla_{\rho}\right] K_{\mu}$$
$$=R_{\rho \sigma \mu \alpha} K^{\alpha}+ R_{\mu \alpha \sigma \rho} V^{\alpha}$$
$$=R_{\rho \sigma \mu \alpha} V^{\alpha}+R_{\sigma \rho \mu \alpha} V^{\alpha}$$
$$=R_{\rho \sigma \mu \alpha} V^{\alpha}-R_{\rho \sigma \mu \alpha} V^{\alpha}=0 $$
$\because \left( R_{a b c d}=R_{c d a b}\right) and $
$ \left( R_{a b c d}=-R_{b a c d}\right)$
$$
\therefore \nabla_{\sigma}\left(\nabla_{\mu} K_{\rho}+\nabla_{\rho} K_{\mu}\right)=0
$$
$ \because x^{\sigma}$ is an arbitrary direction,
$$
\nabla_{\mu} K_{\rho}+\nabla_{\rho} K_{\mu}=0$$
$ \therefore K$ is a Killing vector.
