Edit: According to the comment of Prof. Bryant we revise the question as follows:

Assume that $X$ is a smooth vector field on an open manifold $M$, for exmple $\mathbb{R}^2$. Is there a non degenerate pseudo Riemannian metric on $M$ such that $X$ is a gradient vector field? If not what type of obstructions would appear?

  • 4
    $\begingroup$ No. Let $X$ be a vector field on the $2$-sphere that has a closed orbit. It cannot be the gradient vector field of any non-degenerate metric on the $2$-sphere. Closed integral curves are clearly an obstruction, and there are many such kinds of of obstructions of a dynamial nature, probably too many to classify. $\endgroup$ – Robert Bryant Jun 18 '17 at 21:06
  • $\begingroup$ @RobertBryant Is not possible that $\nabla f. \nabla f=0$ along closed orbit? $\endgroup$ – Ali Taghavi Jun 18 '17 at 21:14
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    $\begingroup$ No. Any non-degenerate pseudo-Riemannian metric on $S^2$ has to be either positive or negative definite because the tangent bundle of $S^2$ cannot be split into the sum of two line bundles. The same argument applies to any surface with nonzero Euler characteristic and a vector field with a closed orbit. $\endgroup$ – Robert Bryant Jun 18 '17 at 21:27
  • $\begingroup$ @RobertBryant for example $\nabla xy= y\partial _x -x\partial_y$ with $dx^2- dy^2$. $\endgroup$ – Ali Taghavi Jun 18 '17 at 21:27
  • $\begingroup$ @RobertBryant Yes thank you very much. So I revise the question for an open manifold for example for the plane. $\endgroup$ – Ali Taghavi Jun 18 '17 at 21:32

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