If there exists a nontrivial vector field $V\not=0$ in Riemannian manifold $M$ and an open set $U\subset M$ such that $\nabla\_{X}V=0$ in $U$ for any vector field $X$ in $M$, then dose $U$ have to be flat?


That is, if a Riemannian maniflod exists a vector field $V$ parallel transport along any vector field, then is this maniflod flat? 



If there exists a nontrivial vector field $V$ such that $\nabla_{X}V=0$ for any vector field $X$, the manifold must be flat?
If you take a manifold like $M = \mathbb R \times M'$ with the usual metric on $\mathbb R$ and where $M'$ is some Riemannian manifold, the vector field induced from $\mathbb R$ will satisfy this condition but $M$ will not be flat in general. 


However, if you impose instead that you have $\dim M$ many independent parallel vector fields then by computing in this frame you immediately see that the curvature tensor vanishes.


