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Suppose that $f$ and $g$ are two smooth functions defined on $R^n$. Assume that $(a-\epsilon, b+\epsilon)$ contains no critical point of $g$. Then $g^{-1}[a,b]$ it homomorphic to $g^{-1}(a)\times [a,b]$. Now we consider the critical points of the function $f$ restricted on $g=c$, $f|_{g=c}$, $c\in [a,b]$. My aim is to show that there is a bijection between critical points of $f|_{g=a}$ and $f|_{g=b}$.

There are counterexamples to my guess. For example, $f,g:\mathbb{R}^2\to \mathbb{R}$, $f(x,y)=x^3-xy$, $g(x,y)=y$. Then $g$ has no critical points, $f|_{g=1}$ has two critical points, but $f|_{g=0}$ has one critical point, and $f|_{g=-1}$ has no critical points.

Maybe some conditions on $f$ would be enough, like $f$ also has no critical points. Is there a theorem on this?

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    $\begingroup$ Implicit function theorem. Roughly speaking you can choose coordinates such that $g$ is one of the coordinates. Then $\mathrm{d}f|_{g^{-1}(c)}$ is a map $\mathbb{R}^n\to\mathbb{R}^{n-1}$. Its zeroes are locally one dimensional manifolds provided that the second derivatives $D^2 f$ are not degenerate. Note that the Hessian of $f$ in your example, evaluated at $x = y = 0$ is identically zero. $\endgroup$
    – Willie Wong
    Commented Nov 17, 2016 at 21:59
  • $\begingroup$ Thanks @WillieWong. To make clear, the Hessian of $f$ is computed as a function on $g^{-1}(c)$, thus $g$ can be the coordinates system of the one dimensional manifold of the critical points of $f|_g^{-1}(c)$, thus leads to the bijection. $\endgroup$
    – Simon Zhu
    Commented Nov 18, 2016 at 4:24

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Consider the Lagrange multiplier $$F(x, t, c) = f(x) + t( g(x) - c).$$ If $g$ satisfies your condition and $f$ is generic, then one can show that $$S:= \{ (x, t, c)\ |\ \nabla f(x) + t \nabla g(x) = 0,\ g(x) - c = 0 \}$$ is a 1-dimensional manifold. Its intersection with $c= a$ or $c= b$ are the critical point set of $f|_{g=a}$ and $f|_{g= b}$.

This implies that (if $a$ and $b$ are generic), then the mod 2 counting of the two critical point sets are equal.

In your counter-example, the $g = 0$ slice corresponds to a so-called "birth-death" which is not generic.

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  • $\begingroup$ so the birth-death points are where this 1-D manifold is not connected. How to test them? $\endgroup$
    – Simon Zhu
    Commented Jan 16, 2017 at 17:26
  • $\begingroup$ to Simon: the birth death is where two critical points merge to one and disappear. The 1-d manifold is still smooth and compact, although not necessarily connected; its boundary is at where g=a or g=b. $\endgroup$
    – Guangbo Xu
    Commented Jan 17, 2017 at 20:28

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