I have read that vector fields which are the gradients of some scalar field cannot have periodic orbits. See, e.g., (1)(2). This probably expresses the fact from vector calculus that $\nabla \times \nabla f = 0$.

In particular, this direction might also follow from De Rham cohomology because the existence of a periodic orbit might allow one to smoothly define the vector field on a non-simply connected region (i.e. by puncturing out some point contained within the periodic orbit).

Question: However, is the converse direction also true? I.e., if a vector field has no periodic orbits, then is it the gradient of some scalar field, or equivalently (contrapositive), if a vector field is not the gradient of any scalar field, then must it have at least one periodic orbit?

Context: The analogous statement is true for discrete vector fields, see for example Theorem 1.8 on p.8 of Forman's paper "Morse Theory and Evasiveness". Implicitly appealing to this equivalence, Ghrist defines, on p.150 of Elementary Applied Topology, a discrete gradient field to be a discrete vector field with no periodic orbits.

Thus, I wonder whether or not this is a new feature of discrete dynamical systems, or if this equivalence is actually the same as in the classical case of smooth dynamical systems. A pointer to a reference will suffice for an answer; a complete counterexample or proof is not needed.


2 Answers 2


This is false. Any function on a closed manifold must have critical points (for example corresponding to the maximum or the minimum), hence the corresponding gradient vector field must have equilibria.

Therefore any flow without equilibria cannot be a gradient vector field. The irrational rotation on the two torus is an example of a flow without periodic orbits or equilibria.

Any three manifold also admits a smooth flow without equilibria or periodic orbits (this is a theorem of Kuperberg).


The answer is no.

I do not know a reference offhand, but there are simple counterexamples: For example, consider the vector field $${\bf F}(x, y) = (P(x, y), Q(x, y)) := (0, x)$$ on $\{(x, y) \in \Bbb R^2 : x > 0\} \subset \Bbb R^2$.

It is not a gradient field: Since ${\bf F}$ is differentiable, if we had ${\bf F} = \nabla f$ for some $f$ and hence $(P, Q) = (f_x, f_y)$, we would have $f_{xy} = f_{yx}$ and so $P_y = Q_x$, but $P_y = 0 \neq 1 = Q_x$. On the other hand, the integral curves of $\bf F$ are the curves $t \mapsto (a, at + b)$ for constants $a, b$ (with $a > 0$); they traverse vertical lines upward and so are not periodic.

  • $\begingroup$ In your answer you assume that the gradient is chosen with respect to the standard Euclidean norm, generated by the dot product. It is quite easy to construct gradient vector fields with respect to a different scalar product. $\endgroup$
    – Artem
    Feb 22, 2017 at 14:38
  • 1
    $\begingroup$ @Artem This is a good point which I probably should have addressed. Since I didn't, I think it's fair for Travis to assume that I meant the one using the standard Euclidean norm for Euclidean space. Honestly I probably would not have understood another counterexample well anyway. $\endgroup$ Feb 22, 2017 at 20:22
  • $\begingroup$ @Artem Like you hint, the gradient operator that maps a $C^1$ function to its gradient vector depends on the choice of Riemmanina metric, so at least for me just talking about a gradient vector field entails an assumption that such a metric has been fixed in the first place, unless, like William says, this is mentioned otherwise. $\endgroup$ Feb 22, 2017 at 21:01
  • $\begingroup$ @Travis What I meant, basically, is that the condition $P_y=Q_x$ is not equivalent to the fact that vector field $(P,Q)$ is gradient. $\endgroup$
    – Artem
    Feb 22, 2017 at 21:14
  • $\begingroup$ @Artem As far as I remember, gradient flows are also quite restrictive in terms of what kind of stable and unstable manifolds intersections are allowed. If I am not mistaken (don't have a good reference at hand now) and heteroclinic cycles are also forbidden in gradient flows, something like this might be another counter example? $\endgroup$
    – Evgeny
    Feb 23, 2017 at 11:58

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