2
$\begingroup$

An exercise of the book "Introduction to smooth manifolds - John M. Lee" asks to prove that if $S$ is a closed embedded submanifold of a manifold $M$, and $X$ is a vector field on $M$ tangent to $S$, then every integral curve of $X$ that intersect $S$ is contained in $S$.

Can someone show me a counteresample in the "closed-immersed" case?

$\endgroup$
2
$\begingroup$

It's true also for closed immersed submanifolds, and I guess it's not really any harder to prove in that generality. But the closed embedded case is by far the most common and the most useful, so that's the only one that seemed to be worth stating.

$\endgroup$
0
$\begingroup$

I believe the issue here is more of a semantic nature: what exactly is the definition of an "immersed submanifold"? Ordinarily, they are defined as the image of an immersion map $f : S' \to M$, and the image is not required by definition to be a "submanifold" of $M$ (see, for example, https://en.wikipedia.org/wiki/Submanifold#Immersed_submanifolds). As an example, consider the usual self-intersecting Klein bottle model in $\mathbb{R}^3$. As per the usual definition, that will be an immersed submanifold, which will fail what you want for the problem.

$\endgroup$
  • $\begingroup$ For me an immersed submanifold is the image of an injective immersion, so for me the Klein bottle is the image $M$ of the "figure 8 immersion" en.wikipedia.org/wiki/Klein_bottle#The_figure_8_immersion , but with domain $(0,2\pi) \times (0,2\pi)$ (so that map is injective). Why is so obvious that it fails what I want? Intuitively, a vector field $X$ in $\mathbb R ^3 $ that is tangent to $M$ has to vanish on the "self-intersecting" circle $C$ (i.e. images of points with $v=0$ ), so $M$ contains every integral curve of $X$. $\endgroup$ – Ervin Jul 1 '15 at 6:26

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.