Does the Lie bracket of a certain pair of vector fields vanish?

Question

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.

Answer 1

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.