Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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).


share|improve this question
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: jstor.org/stable/10.2307/1970311. I've never seen any stronger results, however. –  Henry T. Horton May 19 '12 at 2:15
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 archive.numdam.org/ARCHIVE/ASNSP/ASNSP_1992_4_19_4/… –  user31373 May 19 '12 at 2:19
@LeonidKovalev and Henry: Thanks for the links! –  Eric O. Korman May 19 '12 at 2:54
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.) $\;\;$ –  Ricky Demer May 19 '12 at 3:12
@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. –  Eric O. Korman May 19 '12 at 4:54

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.