Foliation with trivial leaf holonomy

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.

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

• 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?
• @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. May 8 '19 at 6:27
• @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. May 8 '19 at 6:31
• @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.