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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 $\gamma$?

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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.

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Thanks. I see the idea, but you mentioned that the Lie derivative $[X,\gamma']$ vanishes because $X$ is a Killing field. However, $[X,\gamma']$ is a bracket of vector fields; do we not have a problem at that stage since $\gamma'$ is not defined on an open set, or since $\gamma'$ need not be extendible? – user20609 Dec 10 '11 at 9:52
You are right, but, since $\gamma$ is a geodesic it is not difficult to see that $\gamma^{\prime}$ can be smoothly extended locally (which is sufficient). In that case, if $p$ is a point on $\gamma$, let $V\subset T_p M$ the subspace orthogonal to $\gamma^{\prime}$ and let $N$ be the image of $V\cap U$ under the exponential map $\exp_p$ of $M$ in $p, U$ a neighbourhood of $0\in T_p M$. This will be a smooth $n-1$ dimensional submanifold orthogonal to $\gamma$. Then look at the geodesics orthogonal to $N$ Their tangent vectors extend $\gamma^'$ (... to be continued) – user20266 Dec 10 '11 at 11:26
(...continuation) smoothly :-) This was not your question, so I assumed this as a known construction, since you need this kind of construction to be able to work with Lie brackets of vector fields which are defined only on submanifolds. -- This procedure only works if your $M$ is smooth enough, admittedly. – user20266 Dec 10 '11 at 11:29
If you have doubts that the construction outlined in my comments works out as claimed you will need to look up the concept of the exponential map of a manifold, the fact that it's locally a diffeomorphism $T_pM\cap U \rightarrow M$ locally, and you may need to look up the key words 'tubular neighbourhood' and 'Fermi coordinates', too. – user20266 Dec 10 '11 at 11:40
It's not in general true that $[X, \gamma']$ will vanish for your extension of $\gamma'$. As a counterexample let $M = \mathbb{R}^2$, and $\gamma(t) = (0,t)$. Then your definition of $\gamma'$ gives $\gamma' = \partial_y$. Let $X = y \partial_x - x \partial_y$ be the Killing field for (clockwise) rotation about the origin. Then $[X, \gamma'] = -\partial_x$. – mollyerin Apr 17 '15 at 7:35

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