Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $(M,\mathcal F)$ be a smooth foliated manifold. An automorphism of $(M,\mathcal F)$ is a diffeomorphism of $M$ that takes leaves of $\mathcal F$ onto leaves. Let now $L$ be a leaf of $\mathcal F$. It may happen that $L$ has an open neighborhood $U$ which is a sum of leaves and such that for every leaf $L'\subset U$ there exists an automorphism $\phi$ taking $L$ onto $L'$.

My questions are:

Does every $(M,\mathcal F)$ have a leaf $L$ with the described property?

If the answer is no, how much can we hope for instead? And is there a simple counterexample? Are there some natural classes of foliations which still have this property?

Ideally, I would like to consider a smooth foliated manifold $(M,\mathcal F)$ such that in some flat chart $\psi\colon U\to \mathbb R^n$ there exists an open set $V\subset U$ and a family $\tau_x$ of automorphisms of $(M,\mathcal F)$ indexed by a neighborhood of 0 in $\mathbb R^n$, where $\tau_x$ acts on $V$ like a translation by $x$ in the chart $\psi$. This condition is satisfied for instance when there is a neighborhood $U$ of a leaf $L$ and a foliation-preserving diffeomorphism $U\to L\times W$, where $L\times W$ has the corresponding product foliation.

share|improve this question
add comment

2 Answers

up vote 4 down vote accepted

No. There exist foliations $F$ of the plane that are rigid in the sense that no automorphism of $F$ can permute the leaves. See the following paper:

MR2407104 (2009e:54055) Gartside, Paul(1-PITT); Gauld, David(NZ-AUCK); Greenwood, Sina(NZ-AUCK) Homogeneous and inhomogeneous manifolds. (English summary) Proc. Amer. Math. Soc. 136 (2008), no. 9, 3363--3373.

share|improve this answer
add comment

this will very rarely be true! remember that the leaves without holonomy are always generic. On the other hand the union of the leaves with holonomy is often dense. In that case, since any automorphism of the foliation will preserve holonomy, the will be no automorphism of your kind.

share|improve this answer
add comment

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.