3
$\begingroup$

Assume that we have a $n-1$ dimensional integrable distribution $D $ on $\mathbb{R}^n \setminus \{0\}$ which generates a foliation $\mathcal{F}$. We fix an orientation for $D$.(For $n=2$ we assume that it is orientable, that is a vector field generates the distribution. For $n>2$ the distribution is automatically orientable then we fix an orientation for $D$. )

The spheres with radius $1$ and $2$ are denoted by $S_1$ and $S_2$, respectively.

Assume that these spheres determine two leaves of the foliation and the $D$-orientation on $S_2$ coincide the standard orientation of $S_2$ while the $D$-orientation on $S_1$ opposite the standard orientation of $S_1$.

Questions:

  1. Is there a precise example of this situation of opposite orientation for two nested spheres?
  2. Is it true to say that there exist a closed curve $\gamma \subseteq \{z\in \mathbb{R}^n \mid 1< \parallel z \parallel <2 \}$ such that the normall bundle of $\gamma$ is identical to the restriction of $D$ to $\gamma$?

The motivation for this question is the following wonderful counterexample of a nongeodesible foliation of torus determined by a non vanishing vector field .

Note that for $n=2$ the answer to the above question is affirmative.

Assume that $X=P\partial_x +Q\partial_y$ is a non vanishing vector field on the punctured plane. Assume that $C_1,C_2$ are two closed orbits of $X$ such that $C_1$ lies in the interior of $C_2$. Moreover the flow- orientation of $C_2$ is anti clockwise and the flow orientation of $C_1$ is clockwise. (the situation in the above linked counter example). Lets consider the orthogonal vector field $Y=-Q\partial_x+P\partial_y$. Then the annular region $R$ bounded by $C_1,C_2$ is invariant under the positive flow of $Y$. So the Poincare Bendixon theorem implies that there is a closed orbit $\gamma \subseteq R$ for $Y$. Since $Y\perp X$ thus the normal bundle to $\gamma$ is identical to the $X$ direction.

$\endgroup$

Your Answer

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

Browse other questions tagged or ask your own question.