Orthogonal Jacobi Fields Remain Orthogonal 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 $t$? 
 A: 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. 
A: Write $V(t)$ for the velocity field $\gamma'(t)$ and $J(t)$ as the Jacobi field in question. We can show that the function from $\mathbb{R}\to \mathbb{R}$ (or defined on some subset of $\mathbb{R}$) given by $\langle J(t),V(t)\rangle$ is of the form $at+b$ for $a,b\in \mathbb{R}$. To do this, it will suffice to show that its second derivative in $t$ vanishes identically. 
$$ \frac{d}{dt}\langle J(t), V(t)\rangle=\langle D_tJ(t),V(t)\rangle+ \langle J(t), D_tV(t)\rangle=\langle D_tJ(t),V(t)\rangle.$$
The second term in the middle expression vanishes by the geodesic condition. Then
$$ \frac{d^2}{dt^2}\langle J(t), V(t)\rangle =\frac{d}{dt}\langle D_tJ(t), V(t)\rangle=\langle D_t^2J(t), V(t)\rangle +\langle J(t), D_tV(t)\rangle=\langle R(V,J)V,V\rangle=0.$$
So, $\langle J(t),V(t)\rangle =at+b$ for $a,b\in \mathbb{R}$. When we know that $\langle D_tJ(t),V(t)\rangle=0$, this implies 
$$ a=\langle D_tJ(t),V(t)\rangle =0.$$
Now, $\langle J(t),V(t)\rangle =b$. However, we know $\langle J(0), V(0)\rangle =0$, so $b=0$ also. It follows that for all $t$
$$ \boxed{\langle J(t),V(t)\rangle=0}$$
