4
$\begingroup$

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 $\gamma$?

1 Answers 1

4

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.

  • 1
    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 $[X, \gamma'] = -\partial_x$.2015-04-17