Does there exist a smooth nontrivial fiber bundle $p: F \hookrightarrow \mathbb R^n \to B$ such that $F$ and $B$ are connected manifolds with $F$ compact? "Nontrivial" here means the fiber $F$ is not a point.
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5$\begingroup$ The title says "foliation", the body of the question says "fibration": Which one do you actually mean? The question about foliations was asked earlier (one and half ears ago) at MSEhttps://math.stackexchange.com/questions/2842525/folitaion-of-mathbb-rn-by-compact-leaves $\endgroup$– Moishe KohanCommented Feb 15, 2020 at 17:38
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2 Answers
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There does not, even if you don’t require the fiber and base to be manifolds (or even connected, just that $F$ is not a single point). See
Borel, Armand; Serre, Jean-Pierre, Impossibilité de fibrer un espace euclidien par des fibres compactes, C. R. Acad. Sci. Paris 230 (1950), 2258–2260.
answered Feb 15, 2020 at 17:35
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On the other hand, if you only mean "foliation" as in your title, and not "fibration", then there is Vogt's foliation of R^3 by circles! (But it is not C^1, only differentiable). Vogt, Elmar, "A foliation of R3 and other punctured 3-manifolds by circles", Publications Mathématiques de l'IHÉS, Tome 69 (1989), p. 215-232 http://www.numdam.org/item/PMIHES_1989__69__215_0/
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2$\begingroup$ And it appears that the existence of a $C^1$-foliation by circles is an open problem. $\endgroup$ Commented Feb 16, 2020 at 15:15
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$\begingroup$ Thank you Moishe, I should have added that Vogt's foliation is not C^1. It's corrected. $\endgroup$ Commented Feb 17, 2020 at 16:09