Let $f$ be a Morse function on the compact manifold $X$. Let $f_t$ is a homotopic family function with $f_0=f$. Show every $f_t$ is Morse for $t$ is sufficiently small

Here is my argument, but my professor said it's not correct, without telling me why

Assume that $f_t$ is a homotopic family function on $R^k$ and $f_0=f$ is Morse in the compact Manifold $X$. Since $f_t\cong f_0$, there exist a function $F=〖det⁡(H)〗^2+∑_(i=1)^k(∂f/(∂x_i ))^2∶R^k×I→R$ such that $F(x,0)=f_0 (x)$ and$ F(x,t)=f_t (x)$. Note that f is smooth and continuous and $R$ is open (and closed), so $F^{-1} (R-{0})$ is open and contain $X×0$. Since $X$ is compact , $X×0$ is also compact , this guarantee that, there exits a ϵ>0 such that $X×[0,ϵ)$ also contained in $F^{-1} (R-{0})$. This allow us to conclude that for every $f_t$ such that $t<ϵ$, $f_t$ has $F(x,t)$ on $X$. Thus $f_t$ is also Morse in $X$.

I wonder if anyone would show me how to make it correct?

  • 1
    $\begingroup$ This doesn't make much sense to me. I don't understand what you are doing with $\Bbb R-0$, and nowhere have you used or indicated a definition of Morse. $\endgroup$ Oct 14, 2014 at 3:03
  • $\begingroup$ @dan Your proof is exercise #17 and this is #18. I turned in both, and only #17 is accepted. Though I feel these 2 problem are similar, but the professor disagree with that. $\endgroup$ Oct 14, 2014 at 9:19
  • $\begingroup$ Why do you want $X$ to be embedded in $R^n$? $\endgroup$ Oct 14, 2014 at 9:43
  • $\begingroup$ because it is manifold in $R^k$, isn't it? $\endgroup$ Oct 14, 2014 at 16:24
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    $\begingroup$ @hopefully, no. I've given all the keywords you need to find a more general proof in the literature. $\endgroup$ Nov 13, 2018 at 3:42

1 Answer 1


Even if you assume $X$ is in $R^k$, you'd have to work to extend functions $f_t$ to $R^k$ in Morse way, and then after you show $f_t$ is Morse on $R^k$ it does not follow it's restriction to $X$ is Morse...

You can replace $R^k$ by $X$ everywhere. Then the proof would be ok in spirit but would run into trouble because $H$ and $\frac{\partial f}{\partial x_i}$ are not defined in coordinate-independent way. You could salvage it by putting a Riemannian metric on $X$ (see below; you can use a cover by coordinate charts and a partitions of unity argument instead), but perhaps it's better to abandon this approach altogether and go with Ryan Budney's (first) suggestion.

A salvage attempt:

Put Riemannian metric on $X$ and

1) Use $|df|^2$ (aka $|\nabla f|^2$) instead of the $\sum \left(\frac{\partial f}{\partial x_i}\right)^2$

2) Use http://en.wikipedia.org/wiki/Hessian_matrix#Generalizations_to_Riemannian_manifolds instead of $H$. The entries of $H$ will depend on local coordinates, but the quadratic form is well-defined, and so its determinant does not depend on orthogonal (with respect to the Riamannian metric on $TM$) coordinate changes. Further, at critical points (where $df=0$) this is the usual Hessian quadratic form, which has non-$0$ determinant precisely when the critical point is non-degenerate.

Finally, replace all occurrences of $R^k$ in the proof by $X$; the proof becomes correct (assuming $f_t$ is a smooth, or at least $C^k$ with $k>1$, family). Perhaps the only remaining sticky point is to show that any open neighborhood $U$ of $X\times0$ in $X\times I$ indeed contains $X\times[0, \varepsilon)$. This does follow from compactness of $X$, but is perhaps not very obvious; one way is to show that $p \mapsto \{\inf t | (p,t) \in U\}$ is lower semicontinuous on $X$, hence attains a minimum.


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