Let $v\in\mathcal{C}^1(\mathbb{R}^n,\mathbb{R}^n)$ $(n\geq 1)$ a velocity field such that every solution $(x_t)_{t\geq 0}$ of $(d/dt)x_t=v(x_t)$ is periodic. Denote, for a non-stationary point $x\in\mathbb{R}^n$ (meaning $v(x)\neq0$), by $T(x)$ the period of such a solution $(x_t)_{t\geq0}$ such that $x(0)=x$.

Quetion: Is $T$ continuous over the set of non-stationnary points?

  • 3
    $\begingroup$ Not an answer, but probably this indicates a negative response to your question. numdam.org/item/PMIHES_1976__46__5_0 In any case, it shows that your problem will depend on the fact that the vector field is defined in Rn with those properties. $\endgroup$
    – rpotrie
    Jun 29, 2021 at 17:32
  • $\begingroup$ This seems implicit, but just to clarify: for this question, the vector field $v$ may have zeros (stationary points)? $\endgroup$ Jun 30, 2021 at 12:04

2 Answers 2


The answer is trivially negative already for $n=3$: Start with the flow following along the "long" lines of a thickened (say with disc-shaped cut) Möbius strip. You can imagine this like a thick “rope” bent to a loop, making half a rotation on its way, and the flow follows the fibres of the rope (with a fixed speed).

Obviously, the fibre in the "center" of the rope has half of the period than the nearby fibres.

To extend the flow to $\mathbb R^3$, decrease the speed smoothly near the "boundary" of the rope to $0$, and then extend the field by letting it $0$ outside of the rope.

However, this extension works only, because the question explicitly admits stationary points. If one wants to exclude stationary points, there is a topological obstacle in the extension of this particular flow to $\mathbb R^3$. After an embedding into $\mathbb R^4$ this obstacle trivially vanishes, as the Möbius strip can be "unentangled" in $\mathbb R^4$, so in $\mathbb R^4$ there is obiously a non-singular counterexample.

I conjecture that for $n=3$ there is no non-singular counterexample at all, but I do not know how to prove this.


For $n = 2$, the answer is Yes. Topological considerations (as in the proof of the Poincare-Bendixson theorem) mean that the first-return map for a transversal to a point on a periodic orbit will be the identity locally, which implies continuity of $T$.

For sufficiently high $n$, the answer is No. The paper by Sullivan, mentioned by rpotrie, describes a compact smooth 5-manifold $M$ with a $C^{\infty}$ vector field $f$ on which every orbit is periodic, the periods are unbounded, and there are no stationary points. For this flow the period map $T$ can't be continuous.

Now consider $n$ large enough that $M$ smoothly embeds in $\mathbb{R}^n$. A sufficiently small tubular neighborhood of $M$ will be diffeomorphic to $M \times D$, where $D$ is the unit $(n-5)$-ball. You can extend the flow $f$ on $M$ to a flow $g$ on $M \times D$, slowing down to zero velocity as you reach the boundary $M \times \partial D$. Define $g$ to be zero outside this tubular neighborhood, and you have a counterexample.

As for intermediate values of $n$: there is an example like Sullivan's that works on a 4-manifold (Epstein and Vogt, 1978), so that gives a counterexample for $n \geq 9$, but I can't say much else.


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