Killing vector fields restricted to geodesics Given a Riemannian manifolds $(M,g)$, a Killing vector field $X$ on $M$, and a geodesic $\gamma: K \rightarrow M$ defined on an interval $K \subseteq \mathbb{R}$, how does one show that $X \circ \gamma$ is a Jacobi field along $\gamma$?
 A: Wlog $\gamma = \gamma(t)$ is parameterized by arclength. Then this is shown by differentiating $X\circ \gamma$ twice wRt $t$, using the defining equations for a geodesic and Killing field and the rules for interchanging covariant derivates -- these, you need to know, of course. Differentiating once results in:
$$\nabla_{\frac{\partial}{\partial t}} X\circ\gamma = \nabla_{\gamma^{\prime}} X = \nabla_X \gamma^{\prime}$$
the last equality being true cause $X$  is killing (implying that the Lie derivative $[X,\gamma^{\prime}]$ vanishes). Hence
$$\nabla_{\frac{\partial}{\partial t}}  \nabla_{\frac{\partial}{\partial t}} X\circ\gamma = 
\nabla_{\gamma^{\prime}} \nabla_X \gamma^{\prime} = \nabla_X \nabla_{\gamma^{\prime}} \gamma^{\prime} + R(\gamma^{\prime},X)\gamma^\prime $$
(depending on the sign conventions you are using for the curvature tensor the last term may appear with a different sign.) In the last expression the first term vanishes, cause $\gamma$ is a geodesic, and the term involving the Lie derivate does not appear, again because X is Killing. So you are done. 
A: A vector field $X$ is Killing if and only if it generates a flow $\phi_s$ by isometries. We obtain a geodesic variation $\Gamma(s,t)=\phi_s \gamma(t)$, and 
$$ \frac{\partial \Gamma}{\partial s}\bigg|_{s=0} = \frac{ \partial \phi_s \gamma(t) }{\partial s} \bigg|_{s=0} =X_{\gamma(t)} $$ 
is a Jacobi field. 
