A Morse function $f(x)\colon \mathbb{R}^n\to \mathbb{R}$ is a smooth function s.t. all singular points are non-degenerate. A theorem of Sard implies that for any smooth $f(x)$ and almost all $a\in \mathbb{R}^n$ the function $f(x) + ax$ is Morse.

A Morse function $f$ is called perfect Morse if in addition the singular values are distinct.

Question: Is there an analog for perfect Morse, i.e., given smooth $f(x)$ is $f(x) + ax$ perfect Morse for almost all $a$?

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    $\begingroup$ Perfect Morse function usually means something else (that the differential in the Morse chain complex vanishes). $\endgroup$ Jul 19, 2012 at 10:27

1 Answer 1


The answer to your question is negative. A counterexample is given by the Morse function

$f:\mathbb{R}\to \mathbb{R}, \;\; f(x)=\sin(x). $

Using this example it is easy to produce higher dimensional counterexamples.

(The Morse functions with at most one critical point per level set were called excellent by R. Thom. They are also known as stable functions. They form an open and dense subset in the space of smooth functions on a manifold.)

Re-Edit (Following Mark Grant's observation.) If $f:\mathbb{R}\to\mathbb{R}$ is a polynomial

$$f(x)=x^n+\sum_{j=0}^{n-1} a_j x^j, $$

then your guess is correct. Namely for all but finitely many $a$'s the function $f_a(x):=f(x)-ax$ is excellent. Here is the argument.

Consider the real algebraic set $\newcommand{\bR}{\mathbb{R}}$

$$ Z=\lbrace (x,a) \in\bR^2;\;\; f'(x)=a\rbrace. $$

We have a natural projection $\alpha: Z\to \bR$, $(x,a)\mapsto a$, whose fibers are finite sets. This is a semialgebraic map so that there exist finitely many real numbers

$$A_0< A_1< ... < A_N$$

such that the map $\alpha$ is a $C^3$-covering map over each of the intervals $(-\infty, A_0)$, $(A_0,A_1)$, ... $(A_N,\infty)$.

Let $I$ be one of these intervals. The set

$$ Z(I):=\lbrace (x,a)\in Z;\;\;a\in I\rbrace $$

consists of the graphs of finitely many twice differentiable semialgebraic functions

$$ u_1,...,u_m: I\to\mathbb{R} $$

such that $ u_i(a) < u_{i+1}(a)$, $\forall a, i$. In other words, the set of critical points of $f_a$, $a\in I$ is $u_1(a),\dotsc , u_m(a)$.

I claim that for all but finitely many $a$'s in $I$ the function $f_a$ is excellent. I argue by contradiction. If this were not the case, then the semialgebraicity of the $u_i$'s implies that there exist an open interval $ J\subset I$ and pair of indices $(i,j)$, $i\neq j$, such that

$$ f_a( u_i(a) )= f_a( u_j(a) ), \;\;f'(u_i(a))=f'(u_j(a))=a,\;\;\forall a\in J.$$

The first equality can be rewritten as

$$ f(u_i)- f(u_j)= a(u_i-u_j). $$

Differentiating the above equality with respect to $a$ and recalling that $f'(u_i)=f'(u_j)=a$ we deduce

$$ a(u_i'-u_j')= (u_i-u_j)+ a(u_i'-u_j'). $$

This clearly implies

$$ u_i(a)=u_j(a),\;\;\forall a\in J. $$


Note. The same is true for any polynomial $f$ in any number of variables. The proof in the general case is very similar to the one above in the one-dimensional case.

  • $\begingroup$ This answers my question perfectly. However, I wonder what happens when $f$ is a polynomial. $\endgroup$ Jul 19, 2012 at 11:37
  • $\begingroup$ I think that when $f$ is a polynomial your guess is correct, but I do not have a proof. $\endgroup$ Jul 19, 2012 at 12:13
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    $\begingroup$ Dear Liviu, I'm not sure I understand the edit. The condition that $f(x)-ax$ be perfect for $a\gg 0$ does not imply that it is perfect for almost all $a$. Indeed, the function $\sin(x)+ax$ is perfect for $|a|>1$ (there are no critical points). Or am I missing something? $\endgroup$
    – Mark Grant
    Jul 20, 2012 at 5:57
  • $\begingroup$ @ Mark: I guess you are right. Thanks for pointing this out. $\endgroup$ Jul 20, 2012 at 8:58
  • $\begingroup$ @ Lior I now have a proof that your guess is correct for any polynomial $f$. See the link below liviusmathblog.blogspot.com/2012/07/… $\endgroup$ Jul 20, 2012 at 13:29

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