4
$\begingroup$

Let $M$ be a smooth manifold and $f\in C^\infty(M)$. Let $S:=f^{-1}(\{c\})$ for some $c\in \operatorname{Im}(f)\subseteq\mathbb{R}$. When does exist a manifold $N$ with $\dim(N)<\dim(M)$ and a smooth immersion $\Phi :N\rightarrow M$ such that $\Phi (N)=S$? I guess that the interior of $S$ has to be empty.

Furthermore: when is $N$ unique up to dipheomorphism?

It's a well known result that if $c$ is a regular value then $S$ is a closed embedded hypersurface, so, in that line of thinking, probably you need to put some conditions to $c$, but I don't know which ones.

Partial answers are also welcome.

Thanks in advance.

Improve this question
$\endgroup$
12
  • $\begingroup$ I might take an issue with your question. When you say you are asking for the "next level" of generality, what do you mean? Can you describe all the levels of generality in some system? I think usually people would take your line of questioning in a different direction, i.e. have a disagreement on the form of generalization. $\endgroup$
    – Ryan Budney
    Aug 28, 2022 at 3:49
  • $\begingroup$ @RyanBudney true. I'll delete it. $\endgroup$
    – user1234567890
    Aug 28, 2022 at 6:46
  • 1
    $\begingroup$ I am fairly certain this will be true if every critical point for $c$ is Morse (i.e. for every $x$ such that $f-c$ vanishes at the second order, the Hessian is non degenerate) by blowing up the singularities at those points. This is a generic condition on $f$, so for most $f$ all the level sets will be immersed manifolds. That said, there must be someone more qualified for a complete answer on MO. $\endgroup$
    – Pierre PC
    Dec 25, 2022 at 9:47
  • $\begingroup$ @PierrePC anyway thank you for your comment. Now I have somewhere to start. $\endgroup$
    – user1234567890
    Dec 25, 2022 at 11:53
  • $\begingroup$ @PierrePC: the level sets in that case are not immersed manifolds, in general. Locally yes it's true, but there is a global problem. Take for example the standard linear height function on a torus. You need to leave the smooth category and be okay with topological immersions to make sense of your answer. I don't think there will be a non-tautological answer that is satisfactory to the OP. The main reason is that immersed submanifolds aren't level sets of any natural family of functions on manifolds that I know of. $\endgroup$
    – Ryan Budney
    Dec 25, 2022 at 21:00

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.