$\mathbb{R}$ action on a manifold $M$ induced by a vector field $X$

Question

Suppose we have a manifold $M$ and non-vanishing vector field $X$ on it. That vector firld induces a flow $\phi^X$. If the domain of $\phi^X$ is $\mathbb{R} \times X$ than we have an action of $\mathbb{R}$ on $M$. But what happens if the flow is not global (domain is not whole $\mathbb{R} \times M$), is there any natural way to define $\mathbb{R}$ action on M than?

Answer 1

Yes, given any smooth vector field $X$ on $M$, there is an $\mathbb R$-action whose orbits are exactly the integral curves of $X$ (but with different parametrizations, of course).

Let $X\in \mathfrak X(M)$ be a smooth vector field, and let $f\colon M\to \mathbb R$ be a smooth positive exhaustion function. (To say that $f$ is an exhaustion function means that $f^{-1}(-\infty,c]$ is compact for every $c\in \mathbb R$. Such a function always exists; see, for example, Prop. 2.28 in my Introduction to Smooth Manifolds, 2nd ed.) The function $\phi\colon M\to \mathbb R$ given by $\phi(x) = \min(1,1/|Xf(x)|)$ is positive and continuous on $M$, so by the Whitney approximation theorem there is a smooth function $\psi\colon M\to\mathbb R$ such that $0<\psi(x)<\phi(x)$ for all $x\in M$. If we let $\widetilde X = \psi X$, then $|\widetilde X f|\le 1$ everywhere, which implies that $|f(\gamma(t))| \le c_0 + |t|$ along every integral curve $\gamma$ of $\widetilde X$. If $\widetilde X$ were not complete, there would be a maximal integral curve that is not defined for all time, but whose image stays in a compact set $f^{-1}(-\infty,b]$; but the escape lemma (Lemma 9.19 in my book) shows that this cannot happen. Thus $\widetilde X$ is complete, and its flow is an $\mathbb R$-action.