If $M$ is a smooth (connected) compact manifold, then it is known that any (smooth) vector field is complete, which means that the flow exists for all $t \in \mathbb{R}$.
What about the converse? If $M$ is a smooth non-compact manifold, does there always exist some (smooth) vector field which is incomplete? I would expect the answer to be yes. I suspect one may construct a (smooth) vector field and some initial condition such that the corresponding integral curve goes to one "end" of the manifold in finite time, and cannot be continued for all $t \in \mathbb{R}$.
Am I right? Is it true that a smooth (connected) manifold $M$ is compact iff any (smooth) vector field on $M$ is complete?
Edit: I have an idea how one may prove it. By the Whitney embedding theorem, one can always embed a smooth manifold inside $\mathbb{R}^N$, for some $N$ ($N$ can be taken to be twice the dimension of the manifold, but this is irrelevant to our argument). Then one may use the well known fact (Heine-Borel) that a subset of $\mathbb{R}^N$ is compact iff it is closed and bounded. So in our case, since $M$ is noncompact, then it is either non-closed or unbounded, as a submanifold of $\mathbb{R}^N$. In case it is unbounded, choose a smooth curve on $M$ going to $\infty$, construct a vector field, tangent to $M$, which is tangent to that curve and whose parametrized integral curve for any initial point on that curve goes to infinity in finite time. The other case can be dealt with in a similar fashion.