Does there exist $Y$ such that $\nabla_X Y=X$?

Question

Given a vector field $X$ on a Riemannian manifold $M$ with covariant derivative $\nabla$, does it exists a vector field $Y$ such that $\nabla_X Y=X$?

I really do not know how to start, thank you for any suggestion!

Answer 1

Let $M=S^{1}$ equipped with the standard Euclidean metric induced from $\mathbb{R}^2$, $\nabla$ be the Levi-Civita connection w.r.t this metric and $X=\dfrac{\partial}{\partial \theta}$. Suppose there exist vector field $Y$ such that $\nabla_{X}Y=X$.

We can write $Y=f\dfrac{\partial}{\partial \theta}$ for some smooth $f:S^1\to\mathbb{R}$. We compute

$\dfrac{\partial f}{\partial \theta}=X\langle Y,X\rangle=\langle\nabla_X Y,X\rangle+\langle Y,\nabla_X X\rangle=\langle X ,X\rangle+0=1$, which contradicts the fact that $f(\theta)$ is a $2\pi$ periodic function.