# Whether a flow domain always admitted a flow?

Let $M$ be a non compact manifold, a flow domain $D\subset \mathbb R\times M$ is a open set such that its fibre at a point $D_p=\{t: (t,p)\in D\}$ is a open interval containing 0 (i.e. it's connected).

For every vector field, its maximal domain of the flow is a flow domain. I want to think about the converse. That is, given a flow domain, can we find a (smooth) vector field on $M$ such that the maximal domain of its flow is exactly $D$.

I need M noncompact just because it admits some flows that cannot be globally defined

## 1 Answer

Interesting question. It appears the answer is no. Here's a counterexample.

Let $M = \mathbb R^2$, and let $D$ be the following flow domain: $$D = \{(t,p): |t|(1-|p|)^2<1\}.$$ Thus for each point $p$ on the unit circle, $D_p$ is all of $\mathbb R$, while for any other $p$, $D_p$ is the interval $(-1/(1-|p|)^2,\ +1/(1-|p|)^2)$.

Suppose $X$ is a vector field whose flow domain is $D$. The only points whose integral curves are defined for all time are those on the unit circle. If an integral curve starting anywhere intersects the unit circle at some time $t_0$, then it must be defined for all time, because the integral curve starting at $0$ is just a time-translation of the one starting at $t_0$. Therefore, integral curves starting in the interior of the unit disk must stay in the interior, because otherwise they'd have to cross the circle and thus be defined for all time.

But this contradicts a basic lemma called the Escape Lemma, which says that any maximal integral curve whose image remains in a compact set must be defined for all time.

• Good construction! Is that holds just because there is a compact set captures (a connected compoent of) set of points with nontrivial fiber? – yaoliding Nov 8 '17 at 19:35
• @yaoliding: Yes, something like that is the basic idea behind that example. You might think about whether there could be some assumptions on $D$ that would rule out such examples. – Jack Lee Nov 8 '17 at 20:08