MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\Phi: M \times \mathbf{R} \rightarrow M$ be a smooth dynamical system having no periodic orbits, i.e. such that the canonical map $\pi:M \rightarrow \mathbf{R}\backslash M$ is a principal $\mathbf{R}$-bundle. Is $\pi$ always locally trivial? If not, are there any nice (not too contrived/complicated) counterexamples? Which conditions ensure $\pi$ to be locally trivial?

share|cite|improve this question
up vote 4 down vote accepted

The answer is negative. Take, for instance, the irrational foliation of the flat 2-torus by geodesics with the obvious ${\mathbb R}$ action via translations along leaves.

Note that every fiber bundle is locally trivial (by definition), so this should not have been one of the assumptions, only nonexistence of periodic orbits.

A necessary and sufficient condition for the ${\mathbb R}$ action to come from a fibration is that the action is proper, this follows, for instance, from the slice theorem.

share|cite|improve this answer
Nice answer, I was hoping for something along these lines. Could you explain why properness is necessary for existence of slices? I mean we only have a local condition on the action (which implies properness, locally), but a locally proper group action need not be proper. Regarding the terminology: Many people take principal $G$-bundles to be locally trivial, but many don't (e.g. Husemöller). – Nicolas Schmidt May 15 '12 at 17:30

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.