2
$\begingroup$

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!

| cite | improve this question | | | | |
$\endgroup$
  • 1
    $\begingroup$ Writing out this equation in local coordinates should reduce the problem to an usual PDE I believe. $\endgroup$ – MSobak Oct 17 '19 at 8:06
3
$\begingroup$

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.

| cite | improve this answer | | | | |
$\endgroup$
  • 1
    $\begingroup$ This works actually in any nonzero vector fields $X$ since $\nabla _X Y = X$ is linear in $X$. $\endgroup$ – Arctic Char Oct 17 '19 at 18:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.