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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?

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My thought :

At $p$, $\nabla_XY =X^l \frac{\partial Y^k }{\partial x^l} \frac{\partial }{\partial x^k} + X^l Y^m \Gamma_{lm}^k \frac{\partial }{\partial x^k}$ so that $Y$ must be defined on a neighborhood of $p$.

Note that $X(t)$ is defined on $\gamma(t)$ bacause of parallel transport.

So for each $t$, on a neighborhood $U$ of $\gamma(t)$, we can define a vector field $W$ such that $W(\gamma(t))=X(t)$ and $\frac{d}{ds}_{s=a} \exp_{\gamma(t)} (sX(t)) = W( \exp_{\gamma(t)} (aX(t))$ for $0 < a < \epsilon$ where $\epsilon > 0$ is small.

Furthermore since $V$ is totally geodesic, we can define a vector field $W$, corresponded to $X(0)$, which is tangent to $V$.

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