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In 1960, R. Hermann showed the following:

Theorem Let $M$ be a manifold with a foliation $F$ and a bundle-like metric, if all leaves are compact and the holonomy group of each leaf is trivial, then $M/F$ is a smooth manifold. (It is the partially result of the main theorem on Hermann, R., On the differential geometry of foliations, Ann. Math. (2) 72, 445-457 (1960). ZBL0196.54204.)

Q If we drop the condition on bundle-like and admit the trivial holonomy group, can we get the same result? That is to say:

Let $M$ be a manifold with a foliation $F$, if all leaves are compact and diffeomorphism to each other, and the holonomy group of each leaf is trivial, is it true that $M/F$ is a smooth manifold? Any reference is welcome.

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This follows from Theorem 2 in Thurston's 1974 paper "A generalization of the Reeb stability theorem", at least if $H^1(L,R)=0$.

https://core.ac.uk/download/pdf/82172971.pdf

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  • $\begingroup$ I see the local Reeb theorem. Thanks a lot. I still have another question, if $F$ is compact leaf with finite nontrivial holonomy, then there is a neighborhood $N(F)$ of $F$ such that $N(F)$ is saturated. Then, on this local saturated neighborhood $N(F)$, I think we can equip a bundle-like metric. Is this right? And for any foliated manifold $(M,F)$, suppose that each leaf is compact and with finite nontrivial holonomy, then $(M,F)$ admits a bundle-like metric. Am I right? $\endgroup$
    – DLIN
    Apr 29 '19 at 5:13
  • $\begingroup$ @DLIN, yes, from local stability, if $F$ is compact and with finite holonomy then the foliation restricted to $N(F)$ is congruent to the foliation given by the suspension of the holonomy map $\pi_1(F,x)\to\mathrm{Hol}_x(F)\subset\mathrm{Diff}(T)$, where $T$ is a small transverse at $x$, so you can choose a $\mathrm{Hol}$-invariant Riemmanian metric on $T$ (by averaging a random one over $\mathrm{Hol}$) and pull it back to $N(F)$ to obtain a transverse metric for the foliation on $N(F)$, which you can complete to a bundle-like metric by choosing a randon metric on the leaves. $\endgroup$
    – Caramello
    May 8 '19 at 6:27
  • $\begingroup$ @DLIN also, if each leaf is compact and with finite holonomy then this local analysis shows that $M/\mathcal{F}$ is naturally an orbifold. Like manifolds, any orbifold admits a riemannian metric, which you can pull-back to an $\mathcal{F}$-transverse metric on $M$ and (as before) complete to a bundle-like metric. $\endgroup$
    – Caramello
    May 8 '19 at 6:31
  • $\begingroup$ @Caramello Thank you very much. I have one more question: If the holonomy is not finite, is there an example such that even though the leaves are compact, but the quotient space $M/F$ fails to be orbifold. $\endgroup$
    – DLIN
    May 27 '19 at 4:53
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    $\begingroup$ @DLIN Take a look at numdam.org/item/PMIHES_1976__46__5_0 I think the unboundness of the lenghts of orbits implies that some orbit has to have infinite holonomy. $\endgroup$
    – Caramello
    May 28 '19 at 6:19

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