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

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!

• Writing out this equation in local coordinates should reduce the problem to an usual PDE I believe. – MSobak Oct 17 '19 at 8:06

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.
• This works actually in any nonzero vector fields $X$ since $\nabla _X Y = X$ is linear in $X$. – Arctic Char Oct 17 '19 at 18:03