Jacobi field

Remarks

Example: on the unit sphere $S^2$, let $\gamma$ be a unit-speed geodesic and $E$ a parallel unit field along $\gamma$ orthogonal to $\gamma'$. For $J = \sin(t) E$, one has $D_t^2 J = -\sin(t) E$ and, since the curvature is $1$, $R(J, \gamma')\gamma' = \sin(t) E$; the sum is $0$, so $J$ is a Jacobi field.

Non-example: on the same sphere, $J = tE$ has $D_t^2 J = 0$ but $R(J, \gamma')\gamma' = tE \ne 0$ for $t \ne 0$, so $J$ is not a Jacobi field, although it is one in the flat plane.

If $\Gamma(s, t)$ is a smooth family of geodesics $t \mapsto \Gamma(s, t)$, then $J(t) = \partial_s \Gamma(0, t)$ is a Jacobi field, and every Jacobi field along a geodesic on a compact interval arises in this way. A Jacobi field is determined by $J(t_0)$ and $D_t J(t_0)$ at one time, so the Jacobi fields along $\gamma$ form a vector space of dimension $2n$, where $n = \dim M$. The fields $\gamma'$ and $t \gamma'$ are always Jacobi fields; the interesting ones are normal to $\gamma$. With the opposite sign convention for $R$, the equation reads $D_t^2 J + R(\gamma', J)\gamma' = 0$.