Timeline for Is every vector field a gradient vector field with respect to a pseudo metric?

9 events

when toggle format	what		by	license	comment
Jun 18, 2017 at 21:42	history	edited	Ali Taghavi	CC BY-SA 3.0	added 128 characters in body
Jun 18, 2017 at 21:32	comment	added	Ali Taghavi		@RobertBryant Yes thank you very much. So I revise the question for an open manifold for example for the plane.
Jun 18, 2017 at 21:27	comment	added	Ali Taghavi		@RobertBryant for example $\nabla xy= y\partial _x -x\partial_y$ with $dx^2- dy^2$.
Jun 18, 2017 at 21:27	comment	added	Robert Bryant		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.
S Jun 18, 2017 at 21:17	history	suggested	jeq	CC BY-SA 3.0	Typo in title.
Jun 18, 2017 at 21:14	comment	added	Ali Taghavi		@RobertBryant Is not possible that $\nabla f. \nabla f=0$ along closed orbit?
Jun 18, 2017 at 21:06	comment	added	Robert Bryant		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.
Jun 18, 2017 at 20:53	review	Suggested edits
S Jun 18, 2017 at 21:17
Jun 18, 2017 at 20:32	history	asked	Ali Taghavi	CC BY-SA 3.0