second derivative of product of Jacobi with geodesic is ill

Question

I have a Jacobi field $J$ along a geodesic $\gamma$ and a metric $g$. How do I show the following.

$$\frac{d^2}{dt^2}g(J_t,\dot{\gamma}(t)) = 0$$

Since the connexion is compatible, we have $$\frac{d}{dt}g(J_t,\dot{\gamma}(t)) = g(\nabla_\dot{\gamma}J,\dot{\gamma}) + g(J,\nabla_\dot{\gamma}\dot{\gamma}) = g(\nabla_\dot{\gamma}J,\dot{\gamma})$$ so second derivative is $g(\nabla_\dot{\gamma}(\nabla_\dot{\gamma}J),\dot{\gamma})$

but after that I am always turning in circles to try and find something that is zero.

Answer 1

Let $J=J(t)$ be a Jacobi vector field along $\gamma$.

Denote $J'=D_t (J)$ and $J''=D^{2}_{t}J$

Now \begin{align} D^{2}_{t} \langle J, \gamma' \rangle &= \langle J''(t), \gamma' \rangle \\ &= - \langle R(J, \gamma')\gamma',\gamma' \rangle \\ &= -R(J, \gamma', \gamma', \gamma') \\ &= 0 \end{align}