# Integral curve is periodic if and only if it is compact

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. Commented May 22, 2017 at 10:28
• its smooth actually. Commented May 22, 2017 at 11:19
• See my answer here: math.stackexchange.com/questions/391315/… Commented May 25, 2017 at 4:32
• @MoisheCohen That is very useful, you probably should make an answer either way. Commented May 25, 2017 at 12:43

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