Orthogonal Jacobi Fields Remain Orthogonal

Question

Let $\gamma(t)$ be a geodesic and suppose $\left<J(0), \gamma'(0)\right> = \left<J'(0), \gamma'(0)\right> = 0$ where $J'(0)$ indicates the covariant derivative of $J$ along $\gamma$. There exists a unique Jacobi field $J(t)$ with initial conditions $J(0), J'(0)$. How can I see that $J(t)$ and $\frac{DJ}{dt}$ remain perpendicular to $\gamma(t)$ for all $t$?

Answer 1

Since $\langle J(0), \gamma'(0) \rangle =0$, it suffices to show that $\frac{d}{dt}\langle J(t), \gamma'(t) \rangle =0$ for all $t$. To do this, first compute $\frac{d}{dt}\langle J(t), \gamma'(t) \rangle = \langle J'(t), \gamma'(t) \rangle + \langle J'(t), \gamma''(t) \rangle = \langle J'(t), \gamma'(t) \rangle, $

where we've used compatibility of the Levi-Civita connection with the metric and the fact that $\gamma$ is a geodesic.

Now, since $\langle J'(0), \gamma'(0) \rangle=0$, it suffices to show that $\frac{d}{dt}\langle J'(t), \gamma'(t) \rangle =0$ for all $t$. To this end, compute $\frac{d}{dt}\langle J'(t), \gamma'(t) \rangle = \langle J''(t), \gamma'(t) \rangle = \langle R(\gamma'(t), J(t))\gamma'(t),\gamma'(t)\rangle=0$,

where we've used that $J$ is a Jacobi field and that the Riemann tensor is antisymmetric in its last two entries.