Let $M$ be a Riemannian manifold, $p\in M$ be a point in the manifold, and $\xi\in T_p M$ be a vector in its tangent space. I am wondering whether, for some small neighborhood $U\subset M$, it is possible to define a vector field $V$ such that $V(p) = \xi$ and such that $V$ is parallel. That is, the covariant derivative $\nabla V$ vanishes, $\nabla V = 0$.
The intuitive way to try to construct this is to take normal coordinates at $p$, and defining $V$ by parallel transporting it along the radial direction. I just don't know how to prove that the resulting vector field is parallel.
Suppose this procedure works to construct a parallel vector field locally. And suppose that the normal coordinates at $p$ can be extended to all of $M$ diffeomorphically (which may happen if $M$ has nonpositive curvature). Can $V$ be extended to all of $M$?
Any help with these questions would be greatly appreciated.
@John Ma's answer is a good one, but I'd also like to point out that parallel vector fields are even rarer that Killing vector fields. For example, on a round sphere, there are plenty of Killing vector fields but no nontrivial parallel fields. Basically, the existence of a parallel vector field is equivalent to the condition that the metric splits locally into a Riemannian product of a one-dimensional manifold and an $(n-1)$-dimensional one. This implies, in particular, that the sectional curvatures of planes containing $V$ are all zero.