In Frankel's paper: "Manifolds with positive curvature", he proved the following theorem:
If $M^n$ is complete Riemannian manifold with positive sectional curvature, and $V^r$, $W^s$ are two compact totally geodesic submanifolds. If $r+s\ge n$ then $V$ and $W$ have a non-empty intersection.
The proof is by contradiction. If there is no intersection, then chose $\gamma$ connecting $V$ to $W$ and realized the distance between $V$ and $W$. Due to the dimension reason, he find a parallel vector fileld $X$ along $\gamma$, then apply the second variational formula.
For the boundary term $g(\nabla_X X, \dot{\gamma})$. He claimed this is zero due to the totally geodesic property. My question is: In order to take covariant derivative $\nabla_v Y$, the vector field $Y$ has to be defined at least along one curve $\sigma$ with $\dot{\sigma}=v$, right? But for $\nabla_X X$ in the proof, the vector filed $X$ is only defined along $\gamma$ not along the tangent direction $X$. (My guess is it does not depend on the extension of $X$, but I can't see why, or it's too trivial so Frankel didn't write it down?)
edit: I found that probably, we can just extend $X$ at $\gamma(0)$ and $\gamma(\ell)$ as the tengent vector of geodesic along direction $X$, and extend $X$ at $\gamma(t)$ arbitrary. So this will give the desired boundary condition. Is my claim correct?