2
$\begingroup$

In the book Geometric theory of foliations by Camacho and Neto, the following question is posed:

  1. Let $G$ be a smooth transversaly orientable foliation. Let $F$ be a compact leaf of $G$. Prove that $F$ has tubular neighborhoord diffeomorphic to a product. Hint: Prove that the normal fibration of $F$ is diffeomorphic to a product.

I tried to prove that $F$ has trivial normal bundle. But it is possible to ensure only that the submanifold $F$ possesses an atlas with transition functions $g_{ij}$ locally constant. There are nontrivial vector bundles with flat connections.

Is the exercise false?

$\endgroup$
2
$\begingroup$

This is not true, if you don't others conditions. Consider Let $A$ be ab element of $SL(2,\mathbb{Z})$ which is hyperbolic, i.e if $\lambda$ is an eigenvalue of $A$, $|\lambda|\neq 1$. Consider the suspension $M$ defined as follows: it is the quotient of $\mathbb{R}\times \mathbb{T}^2$ by the diagonal action $\gamma(x,y)=(x+1,A(y))$, the foliation is defined by the image of $\mathbb{R}\times\{y\}$ by the projection $\mathbb{R}\times \mathbb{T}^2\rightarrow M$. The projection of $\mathbb{R}\times\{0\}$ is a compact leaf $F$. The normal bundle of $F$ is not trivial, it is the bundle over $S^1$ which is the quotient of $\mathbb{R}\times \mathbb{R}^2$ by the action $g.(x,y)=x+1,A(y))$.

$\endgroup$
3
  • $\begingroup$ Every orientable vector bundle over the circle is trivial. In this case, $F$ is a circle. If your foliation is transversaly orientable, then the normal bundle to $F$ is orientable. Therefore $GL(n)-trivial$. The question is if the normal bundle is diffeomorphic to a product, in sense that exists difeomorphism $f:F\times \mathbb{R}^n \rightarrow N(F)$ such that $f(x,0)=x$. $\endgroup$ May 30 '18 at 3:46
  • $\begingroup$ en.wikipedia.org/wiki/Surface_bundle_over_the_circle $\endgroup$ May 30 '18 at 8:48
  • 1
    $\begingroup$ I believe that you have forgotten the codimension of the foliation which maybe one here. $\endgroup$ May 30 '18 at 8:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.