Does the Lie bracket of a certain pair of vector fields vanish? I'm trying to read section 3 in
J. Jost and Y.L. Xin [JX].
This section has to do with the geometry of Grassmannians. I have a question that comes out of the derivation at the top of page 283 in that paper. Let $M$ be a locally symmetric Riemannian manifold. Let $c$ be a geodesic in $M$ and let $J$ be a Jacobi field along $c$. Let $\cdot$ be the derivative (differential) of $c$ and assume $J$ is orthogonal to $\cdot$. Under these hypotheses is it the case that the Lie bracket,  $[\cdot, J]$, is $0$? If so, why?
[JX]            title = {{Bernstein type theorems for higher codimension.}},
year = {1999},
journal = {{Calculus of Variations}},
volume = {9},
pages = {277--296}.
I would like to include a one-page LaTeX article and a screen shot to help explain my question but I don't know how to do that.
 A: Here is the general idea: If you have a map $\Phi: (-\delta,\delta) \times (0,T) \rightarrow M$, where and $M$ is a smooth manifold, then the vector fields $S = \partial_1\Phi(0,t)$ and $T = \partial_2\Phi(0,t)$ along the curve $\gamma(t) = \Phi(0,t)$, $0 < t < T$, commute. This of course follows from the fact that, since partial derivatives commute,
$$
[S,T] = [\partial_1\Phi,\partial_2\Phi] = 0.
$$
on the entire domain $(-\delta,\delta) \times (0,T)$. In particular, if $M$ is a Riemannian manifold, then
$$
\nabla_S T = \nabla_TS.
$$
If you now restrict $S$ and $T$ to the curve $\gamma$, you can do calculations with $S$ and $T$ using the equation above, even though the equation is ill-defined along the curve.
This fact can used when doing calculations involving Jacobi fields. In particular, it is how the Jacobi equation is derived. In this specific context, $M$ would be a Riemannian manifold and $\Phi$ a $1$-parameter family of constant speed geodesics, i.e., for each $s \in (-\delta,\delta)$, the curve $t\mapsto \Phi(s,t)$ satisfies
$$
\nabla_TT = 0,
$$
where $T = \partial_t\Phi(s,t)$. On the other hand, the restriction of the vector field
$$
J = \partial_s\Phi(s,t)
$$
to $s = 0$ is a Jacobi field along each geodesic $t \mapsto \Phi(s,t)$. By the simple fact above,
$$
\nabla_JT = \nabla_TJ.
$$
Most of us do the calculation on the entire domain $(-\delta,\delta)\times [0,T]$ first and then observe that the final equation holds along the geodesic $\gamma$. However, some Riemannian geometers just use the above equation along a constant speed geodesic without comment.
