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More precisely, is there a criterion that decides the above question?

I am particularly interested in the smooth setting: is a smooth manifold with a smooth regular foliation by circles covered by a smooth manifold with a smooth non-trivial circle action?

I am mostly interested in finite covers.

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    $\begingroup$ Do you want the action to be free? $\endgroup$ – Thomas Rot Sep 2 at 13:25
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    $\begingroup$ Do you want the leaves of the circle action to project to the leaves of the initial foliation by circles ? $\endgroup$ – BS. Sep 2 at 16:13
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    $\begingroup$ The action does not necessarily need to be free. Also the orbits of the circle action do not need to be the lifted leaves of the original foliation. Thanks a lot for your answers till now! $\endgroup$ – Caterina C. Sep 2 at 19:25
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Dennis Sullivan constructed a non-vanishing Lispschitz vector field on a closed $5$-manifold such that all orbits are periodic but they amazingly have unbounded lengths (!). An addendum by Nicholas Kuiper shows that the flow can be chosen smooth, and another by William Thurston gives another construction which is real analytic.

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    $\begingroup$ It is not clear to me how this answers the question. Why can't this manifold be covered by a manifold with a circle action? $\endgroup$ – Nick L Sep 2 at 14:34
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    $\begingroup$ @Nick L : I assumed (perhaps wrongly) that the OP asked for the circle action to have orbits the leaves of the lifted foliation. $\endgroup$ – BS. Sep 2 at 16:06
  • $\begingroup$ Indeed my assumption was wrong, according to the OP herself. The question asks precisely what was written. Sorry. $\endgroup$ – BS. Sep 3 at 5:17
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For circle foliations of compact $3$-manifolds, this is essentially answered by a theorem of Epstein: every such foliation is a Seifert fibration. Most Seifert fibrations are finitely covered by a product (surface)x(circle), and in these cases one of course has a free circle action.

One of the exceptional cases of Seifert fibrations not covered by a product is the fibration of $S^3$ obtained as follows. Take the linear foliation of slope p/q on a solid torus (completed by the core of the solid torus as another leaf), the linear foliation of slope q/p on another solid torus (again with the core as another leaf), and glue the two solid tori along their boundaries by sending one meridian to the longitude to the other and one longitude to the meridian of the other. The result is $S^3$ with a circle foliation, which does not seem to come from a free circle action if p and q are bigger than 1.

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