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Let $V$ be a vector field on a smooth manifold $M$.

Are there nice conditions under which there exists a (Riemannian) metric on $M$ such that $V$ is the gradient of some smooth function on $M$?

One obstruction is that gradient vector fields have no closed integral curves (since a function is increasing on integral curves of its gradient).

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    There is a theorem of Smale that says every Morse-Smale gradient-like vector field is the gradient vector field of some self-indexing Morse function with respect to some Riemannian metric. It's Theorem B in this paper: http://www.jstor.org/stable/10.2307/1970311. I've never seen any stronger results, however.2012-05-19
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    If $V$ never vanishes, then the level sets of the corresponding function will produce a transverse foliation of $M$ of codimension 1. There is considerable literature on the existence of transverse foliations for vector fields... On the other hand, if $V$ has an isolated zero, there may be a local obstruction there. For example, in 2d the index of a gradient field at an isolated zero is at most 1. Although this ought to be long known, the best reference I have is Lemma 3.1 in http://archive.numdam.org/ARCHIVE/ASNSP/ASNSP_1992_4_19_4/ASNSP_1992_4_19_4_567_0/ASNSP_1992_4_19_4_567_0.pdf2012-05-19
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    @LeonidKovalev and Henry: Thanks for the links!2012-05-19
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    To make sense of whether or not $V$ is the gradient of a specific smooth function, one needs an isomorphism between the tangent bundle and the cotangent bundle. $\:$ Does a metric necessarily give a canonical $\hspace{.4 in}$ such isomorphism? $\:$ (I know that a _Riemannian_ metric does.) $\;\;$2012-05-19
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    @RickyDemer: any non-degenerate metric will give an iso $TM \to T^* M$-- it could have indefinite signature. Actually, in my question I want to assume that the metric is Riemannian. Though it is interesting to consider the non-Riemannian case since there it seems there could be closed integral curves to gradient fields.2012-05-19
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    What would "non-degenerate" mean for a metric? $\:$ Are you just trying to emphasize that it's not only a pseudometric? $\:$ If yes, then I don't see how a general metric gives an isomorphism between those spaces.2012-05-19
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    @RickyDemer a metric $g$ is non-generate if $g(X,Y) = 0$ for all $Y \in TM$ implies that $X = 0$. This is equivalent to the map $TM \to T^* M$, $v \mapsto g(v,\cdot)$ being an isomorphism. Since Riemannian metrics satisfy $g(X,X) > 0$ for $X \ne 0$, they are non-degenerate but you can have non-Riemannian metrics (e.g. a Lorentzian metric). Though these are more subtle since e.g. not every manifold admits a Lorentzian metric though all admit Riemannian metrics.2012-05-19
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    First, that would be a metric on $M$'s tangent bundle, not on $M$. $\:$ (Your question specified $\hspace{1 in}$ that the metric was on $M$.) $\:$ Second, that would only work if the restrictions of $g$ to a tangent $\hspace{.8 in}$ space of $M$ cross itself were linear in their second arguments. $\;\;$2012-05-19
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    This is a pretty old question, so I hope it's OK for me to ask this. Have you made any further progress/found an answer since 2012?2015-05-13
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    @JoshBurby unfortunately I have not thought about this anymore.2015-05-13
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    Just a comment about the local version of the question for non-vanishing vector fields. About any point where your vector field does not vanish, there is an open neighborhood with coordinates that straighten out the integral curves, i.e. $X=\partial_z$ with $z$ one of the coordinates. $\partial_z=\nabla z$ if we take the gradient w.r.t. the pullback of the standard Euclidean metric tensor along the coordinate chart. Thus, locally around non-singular points there is always such a metric.2015-05-13

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