This question was not answered on math.stackexchange.

Let $M$ be a manifold (not necessarily compact) , for the sake of clearness embedded in $\mathbb{R^n}$ and $f\colon M\rightarrow \mathbb{R}$ a smooth function.

The theorem of Sard gives us that $$f+\langle\ \cdot\ ,a\rangle \colon M\rightarrow \mathbb{R}, \ x\mapsto f(x)+a_1x_1+...+a_nx_n$$ is a Morse function for almost all $a\in\mathbb{R}^n$.

Now suppose I have a finite set of regular values $c_1,...,c_n$ of $f$, so $f^{-1}(c_1),...,f^{-1}(c_n)$ do not contain critical points. Can I deform $f$ slightly to $\tilde f$, such that it becomes a Morse function, but the level sets of $c_1,..,c_n$ remain unchanged, i.e. $f^{-1}(c_i)=\tilde f^{-1}(c_i)$?

This is somehow a relative version of the density of Morse functions in the space of smooth functions.


Take any smooth nonnegative function $h\colon M\to \mathbb R$ such that all points of $h^{-1}(0)$ are regular for $f$ and $h^{-1}(0)$ includes an open neighborhood of all your level sets $f^{-1}(c_i)$.

Assume $\iota\colon M\to \mathbb R^n$ is an embedding. Note that $M\to \mathbb R^{n+1}$ defined as $$\hat\iota=x\mapsto(h(x),h(x)\cdot \iota(x))$$ is an embedding of $M\backslash h^{-1}(0)$.

Fix $a\in\mathbb R^{n+1}$ and pass to $$\tilde f=f+\langle \hat\iota,a\rangle.$$ For a generic choice of vector $a$, you get a Morse functions and $\tilde f=f$ in a neighborhood of your level sets.

(That was the first thing came to my mind, I am sure there are better solutions.)

| cite | improve this answer | |
  • $\begingroup$ I expect you mean $\tilde{f}=f+<h\cdot id_M,a>$? $\endgroup$ – Francis May 22 '15 at 9:46
  • $\begingroup$ @Francis: Now it is corrected. $\endgroup$ – Anton Petrunin May 22 '15 at 16:23

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.