Does a compact leaf of the smooth transversaly orientable foliation have trivial normal bundle?

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?

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 $g.(x,y)=x+1,A(y))$.
• 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 $f(x,0)=x$. May 30 '18 at 3:46