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Suppose we have a complete, non-singular vector field on $M$. Take an integral curve $\phi: \mathbb{R} \rightarrow M$. If it is periodic than $\phi(\mathbb{R})$ is compact as an image of a compact set. Suppose $\phi(\mathbb{R})$ is compact, I have a problem with showing it is periodic, which I suppose is trivial and that even there is no arbitrary, continuous bijection from $\mathbb{R}$ to a compact space.

EDIT: Since it may not be so trivial as I expected let me be very precise. M is compact smooth manifold, we have a smooth (necessarily complete since $M$ is compact) non-singular vector field on it. Is it true that an integral curve compact in induced topology must be periodic?

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  • $\begingroup$ Do you assume that the vector field is continuous? Otherwise, I don't think this is true. $\endgroup$
    – tomasz
    Commented May 22, 2017 at 10:28
  • $\begingroup$ its smooth actually. $\endgroup$
    – J.E.M.S
    Commented May 22, 2017 at 11:19
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    $\begingroup$ See my answer here: math.stackexchange.com/questions/391315/… $\endgroup$
    – Moishe Kohan
    Commented May 25, 2017 at 4:32
  • $\begingroup$ @MoisheCohen That is very useful, you probably should make an answer either way. $\endgroup$
    – J.E.M.S
    Commented May 25, 2017 at 12:43

2 Answers 2

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Yes, the orbit has to be closed; this is a special case of my answer here. In your case the Lie group is ${\mathbb R}$ acting via the diffeomorphisms generated by the vector field $V$: $$ \frac{\partial }{\partial t}f(x,t) =V(x), t\in {\mathbb R}. $$

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  • $\begingroup$ It works, but it looks like a bit of an overkill. :-) $\endgroup$
    – tomasz
    Commented May 25, 2017 at 14:23
  • $\begingroup$ @tomasz: True, but it is better to overkill than to underkill. (Some also say that it is better to overeat than to undersleep.) $\endgroup$
    – Moishe Kohan
    Commented May 25, 2017 at 15:20
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In general, this is not true.

Consider a curve on the plane which is $(x,\sin(1/x))$ for $\lvert x\rvert\geq 1$, with the part for $x\in (-1,1)$ "looping around", passing through the interval $\{0\}\times [-1,1]$ and connecting the two components. This can be made into a compact, bijective and smooth (non-singular) curve, and then you can extend its derivative arbitrarily to obtain a vector field on the whole ${\bf R}^2$ (but not to a continuous vector field!).

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  • $\begingroup$ hmm, actually I am more inetrested in it in non-degerate case, see my edited question. $\endgroup$
    – J.E.M.S
    Commented May 22, 2017 at 12:02

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