Integral curve is periodic if and only if it is compact

Question

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?

Do you assume that the vector field is continuous? Otherwise, I don't think this is true. — tomasz, May 22 '17 at 10:28
See my answer here: https://math.stackexchange.com/questions/391315/are-closed-orbits-of-lie-group-action-embedded/2216157#2216157 — Moishe Kohan, May 25 '17 at 04:32
@MoisheCohen That is very useful, you probably should make an answer either way. — J.E.M.S, May 25 '17 at 12:43

score 2 · Accepted Answer · answered May 25 '17 at 14:03

2

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 $V$: $$ \frac{\partial }{\partial t}f(x,t) =V(x), t\in {\mathbb R}. $$

answered May 25 '17 at 14:03

Moishe Kohan

111,854

It works, but it looks like a bit of an overkill. :-) – tomasz May 25 '17 at 14:23
@tomasz: True, but it is better to overkill than to underkill. (Some also say that it is better to overeat than to undersleep.) – Moishe Kohan May 25 '17 at 15:20

score 1 · Answer 2 · answered May 22 '17 at 10:58

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!).

hmm, actually I am more inetrested in it in non-degerate case, see my edited question. — J.E.M.S, May 22 '17 at 12:02

Integral curve is periodic if and only if it is compact

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