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Let $f:X \to \mathbb{R}$ be a Morse function on some compact submanifold $X \subset \mathbb{R}^n$, and assume that $p \in X$ is not a critical point of $f$. For some $\epsilon > 0$ let $D_\epsilon(p)$ denote the Euclidean disk of radius $\epsilon$ around $p$. I'd like to claim that there are some small $\epsilon > 0$ and $\delta > 0$ so that we have a deformation-retraction (or at least a homotopy equivalence) $$ \rho:D_\epsilon(p) \cap \{f(p) \leq f \leq f(p)+\delta\} \stackrel{\simeq}{\longrightarrow} D_\epsilon(p) \cap {\{f = f(p)\}}.$$

Let's call the codomain $A$ and the domain $B$. My questions are: (a) can we impose conditions on $f$ which make $\rho: B \to A$ exist, (b) and is there a precise reference for this?


The Hope: The intuition is simply that even if the interval $[f(p),f(p)+\delta]$ contains a billion critical values of $f$, so long as none of the critical points involved are in our $\epsilon$-ball, the handle attachments will be far away from $p$ and therefore the gradient vector field of $f$ should take the $(f(p)+\delta)$-sublevelset to the $f(p)$-sublevelset without any serious incidents en route.


The Problem: Of course, there is no reason for the gradients $-\nabla f$ to point into $D_\epsilon(p)$ along the bounding upper hemisphere $$H^+ = \partial D_\epsilon(p) \cap (B-A),$$ which means that the gradient flow might be pushing points outside $D_\epsilon(p)$ laterally into $B$ rather than flowing down to $A$. I suspect that the following should suffice: if for every point $x$ in $H^+$, the gradient $-\nabla_xf$ does not lie in the tangent space $T_xH^+$, then the desired map $\rho:B \to A$ is furnished by flowing along $-\nabla f$. I could certainly try to write all of this down, but it seems like overkill and it's hard to believe that it hasn't been done before (one might expect to see it in Nicolaescu's nice Morse theory textbook for instance).

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If $p_0$ is not a critical point of $f$ then the implicit function theorem states that, there exists local coordinates $(x^1,\dotsc, x^n)$, defined in an open neighborhood $U$ of $p_0$ in $\newcommand{\bR}{\mathbb{R}}$ $\bR^n$ such that, in these coordinates we have ($m=\dim X$) $$ x^i(p_0)=0,\;\;\forall i, $$ $$ X=\{ x^{n-m+1}=\cdots =x^n=0\}, $$ $$ f(x^1,\dotsc, x^n)=f(0,\dotsc,0)+x^m. $$ If you now define the (non-Euclidean) box $\newcommand{\ve}{{\varepsilon}}$ $$ B=\big\{ |x^i|< \ve;\;\;i=1,\dotsc, m\big\}. $$ In this neighborhood, that is not an Euclidean ball, the deformation you seek is obvious.

To deal with the region $D_{\ve}(p)$ consider the smooth function $\DeclareMathOperator{\Hess}{Hess}$ $$ g:X\to\bR,\;\;g(x)=\Vert x-p\Vert^2, $$ where $\Vert-\Vert$ is the standard Euclidean norm on $\bR^n$. The Hessian of $g$ at $p$, viewed a symmetric bilinear form $T_pX\times T_pX\to\bR$ is positive definite. $\newcommand{\pa}{\partial}$

Choose local coordinates $(x^1,\dotsc, x^m)$ on $X$ near $p$ as above. In these coordinates the vector field $\pa_{x^m}$ is a gradient like vector field for $f$.

For $\ve>0$ sufficiently small the region $R_{\ve}=\{g\leq \ve\}$ is strictly convex in the above coordinates $x^i$ because the second fundamental forms along the boundary $\pa R_{\ve}$ are positive definite being small perturbations of $\Hess$.

This reduces the problem to the following situation. Suppose that $R_\ve$ is a compact, convex neighborhood of the origin in $\bR^m$ with smooth boundary. Then for $\delta>0$ sufficiently small we have a deformation retract $$ R_\ve \cap \{ 0\leq x^m\leq \delta\}\to R_{\ve}\cap\{x^m=0\}. $$ I think this is clear.

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  • $\begingroup$ Thank you for this answer (and your book, from which I learned Morse theory!) One question: I don't see how you get $$f(x^1,\ldots,x^n) = \text{stuff}$$ when $f$ is only defined on $X$ consisting of the first $m$ coordinates. Should that $n$ be an $m$, or are you using an extension? And similarly, I'm not sure what the $b$ is in the index set of your box. $\endgroup$ May 6, 2020 at 11:00
  • $\begingroup$ You are right. Use an extension of $f$. Indeed, $b$ was a typo. I meant $m$ so $B$ is a box living inside $X$. If you are interested specifically in the intersection of $X$ with a Euclidean ball then things get a bit more technical. See the new addition to my answer. $\endgroup$ May 6, 2020 at 12:15
  • $\begingroup$ Ah, wonderful. Thank you for the update, this is precisely what I was hoping for! $\endgroup$ May 6, 2020 at 13:06
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I don't see how you could exclude that the gradient lies in the tangent space near the equator.

Seems to me that you need to use the diffeomorphism provided by the Morse lemma applied to $f$ to straighten the level set of $f$ into a hyperplane locally. After that, work in the "straightened domain" and apply the Morse lemma to the distance to (the image of) $p$ to "straighten" the distorted ball into a round ball. Of course some care is needed to avoid distorting the hyperplane. Composing the two diffeomorphisms you're in a situation where an orthogonal projection gives the desired retraction.

It seems that an extension of the Morse Lemma to vector valued functions should be feasible and would solve this more directly but I don't know references on this.

EDIT: the Morse lemma for vector functions approach is probably too restrictive, as discussed in

Modification of Morse lemma with two functions

The first approach above probably can be made to work though.

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  • $\begingroup$ The space $B-A$ excludes the equator, and in general I'd expect the limiting gradient line to actually lie on the tangent at the equator (even in a hyperplane with parallel downward flow). This is fine as long as off the equator we don't have any points of tangency. $\endgroup$ May 5, 2020 at 13:12
  • $\begingroup$ yes but close to the equator higher order terms would in general make a tangency possible I think $\endgroup$
    – alesia
    May 5, 2020 at 13:15

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