Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\gamma$?

share|improve this question

1 Answer 1

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.

share|improve this answer
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? –  Kaloyan Marinov 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 at 7:35

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.