Killing vector field along a geodesic I was trying to show that a Killing vector field satisfies the Jacobi Equation for a geodesic, just by assuming that 
\begin{equation}
\nabla_\mu X_\nu + \nabla_\nu X_\mu=0
\end{equation}
Indeed, if I take into account that $[X,\gamma']=0$ being $\gamma$ the geodesic, I can show it easily without using the Killing equation. Yet I can't understand how to prove that the commutator vanishes by using this equation.
 A: Suppose $(M,g)$ is a Riemannian manifold, $I\subseteq\mathbb R$ is an interval, and $\gamma\colon I\to M$ is a geodesic. A smooth map $F\colon I\times (-\varepsilon,\varepsilon)\to M$ is called a "variation of $\gamma$." A standard calculation shows that if $F$ is a variation through geodesics (meaning that $s\mapsto F(s,t)$ is a geodesic for each $t$), then its variation field $V(s) = \partial F(s,t)/\partial t|_{t=0}$ is a Jacobi field. (See Theorem 10.2 in my Riemannian Geometry.) 
Now if $X$ is a Killing field and $\theta$ is its flow, then for each $t\in(-\varepsilon,\varepsilon)$, the diffeomorphism $\theta_t$ takes geodesics to geodesics. Thus $F(s,t) = \theta_t(\gamma(s))$ is a variation through geodesics, so its variation field $V(s) = X(\gamma(s))$ is a Jacobi field. 
A: I continue to believe that no good sense can be made of $[X, \gamma'] = 0$. However, ignoring this, I've supplied below a proof that Killing fields are Jacobi. My method of proof was just "calculate until the desired result falls out". In particular, I offer no guarantees that there isn't a better way.
Quick conventions: My curvature tensor convention is
$$
   R(X, Y)Z = \nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z - \nabla_{[X,Y]} Z
$$
which makes the Jacobi equation
$$
   D_t^2 J + R(J, \dot{\gamma}) \dot{\gamma} = 0.
$$
The condition for $X$ being a Killing field is that
$$
  \langle \nabla_Y X, Z \rangle = - \langle Y, \nabla_Z X \rangle \tag{1}
$$
for all vector fields $Y, Z$. Setting $Y = Z$, one sees that this also implies
$$
 \langle \nabla_Y X, Y \rangle = 0 \tag{2}
$$
for any $Y$.
Proof that Killing fields are Jacobi. Let $\Gamma$ be any extension of $\gamma'$, and let $Y$ be any vector field. It suffices to show that
$$
   \langle \nabla_\Gamma \nabla_\Gamma X + R(X, \Gamma)\Gamma, Y \rangle = 0
$$
for points on the geodesic.
We have
\begin{align}
   \langle \nabla_\Gamma \nabla_\Gamma X, Y \rangle &= \Gamma \langle \nabla_\Gamma X, Y \rangle - \langle \nabla_\Gamma X, \nabla_\Gamma Y \rangle \\ &= \Gamma \langle \Gamma, \nabla_Y X \rangle - \langle \Gamma, \nabla_{\nabla_\Gamma Y} X \rangle \\ &= \langle \Gamma, \nabla_\Gamma \nabla_Y X - \nabla_{\nabla_\Gamma Y} X \rangle. \tag{3}
\end{align}
(In the last equation we've used $\nabla_\Gamma \Gamma = 0$, which is only true along the geodesic; I don't wish to clutter the notation any further by showing these expressions as evaluated at points $\gamma(t)$, but it is implied.)
On the other hand
\begin{align}
   \langle R(X, \Gamma)\Gamma, Y \rangle &= -\langle R(Y, \Gamma)X, \Gamma \rangle \\ &= \langle \nabla_Y \nabla_\Gamma X - \nabla_\Gamma \nabla_Y X - \nabla_{[Y, \Gamma]} X, \Gamma \rangle. \tag{4}
\end{align}
Adding $(3)$ and $(4)$, and using $\nabla_\Gamma Y + [Y, \Gamma] = \nabla_Y \Gamma$, we get
\begin{align}
    \langle \nabla_\Gamma \nabla_\Gamma X + R(X, \Gamma)\Gamma, Y \rangle &= \langle \Gamma, \nabla_Y \nabla_\Gamma X - \nabla_{\nabla_Y \Gamma} X \rangle. \tag{5}
\end{align}
But
$$
   \langle \Gamma, \nabla_Y \nabla_\Gamma X \rangle = Y \langle \Gamma, \nabla_\Gamma X \rangle - \langle \nabla_Y \Gamma, \nabla_\Gamma X \rangle = -\langle \nabla_Y \Gamma, \nabla_\Gamma X \rangle,
$$
the first term vanishing because of $(2)$, whereas
$$
  \langle \Gamma, \nabla_{\nabla_Y \Gamma} X \rangle = -\langle \nabla_\Gamma X, \nabla_Y \Gamma \rangle
$$
by $(1)$.
Thus $(5)$ reduces to zero, which is what we wanted to show.
