Let $X$ be a circle that with one corner (i.e. think of a triangle where we smooth out two of the vertices). Now let us consider the topological torus $M \cong \mathbb{T}^n$ which is the product of $n$ copies of $X$. Note that $M$ contains $n$ distinct circles along which it is not smooth.

Finally, suppose we are given a function $f:M \rightarrow \mathbb{R}$ which is continuous on all of $M$ and is smooth wherever $M$ is smooth. Is there any way to conclude that there exists a critical point on the smooth part of $M$? What if we replace some of the non-smooth circles with smooth ones? That is take $M$ to be the product of $k$ smooth circles, and $n -k$ non-smooth ones?

I'm not concerned whether the critical point is non-degenerate.

An easy example is the case $n = 1$: The $\min$ or $\max$ of $f$ will correspond to a critical point on the smooth part of $X$. This example tells us that generically we should expect $f$ to have a critical point on the smooth part of $M$ since, generically, the $\min$ or $\max$ should not lie on the measure zero subset of $M$ that is not smooth.

I heard of something called Stratified Morse Theory, but I'm not sure if this applies, or whether there is a more elementary way to think about the problem not using Stratified Morse Theory.

  • $\begingroup$ So... any space homeomorphic to a smooth manifold can be given a smooth structure. What do you mean by "non-smooth circle"? $\endgroup$ Jul 22, 2010 at 15:41
  • $\begingroup$ You are right in that it can be given a smooth structure. But suppose that I am given a non-smooth manifold and want to consider a function on the non-smooth manifold itself, not a smooth version of it. By a non-smooth circle I mean: something homeomorphic to a circle but has one corner (so it is smooth everywhere except at one point). $\endgroup$
    – user7807
    Jul 22, 2010 at 17:24
  • 1
    $\begingroup$ It appears to me that your $M$ is singular in codimension 1, not in dimension 1 as implied by the 3rd sentence. $\endgroup$ Jul 22, 2010 at 20:55

2 Answers 2


Answer to your question is negative if $n\ge 1$ ($n$ is a number of smooth circles and I suppose that there is non-smooth circles). Indeed, in that case your manifold is homeomorphic to torus $T^k$ and the homeomorphism $\varphi\colon M\to T^k$ could be taken smooth on the smooth part of $M$. The image of non-smooth part is a "big" subset (more than finite), denote that subset by $A$. There exists a smooth function $f$ on $T^k$, such that all its critical points are contained in $A$. (For any function on $T^k$ with a finite number of critical points there is a diffeomorphism of $T^k$ sending the set of critical points to $A$). Function $\varphi^*f$ is continuous, smooth on the smooth part of $M$ and has no smooth critical points.


Let $n=2.$ The height function $h$ for the standard embedding of the torus (as in the first pages of Milnor's Morse Theory) has four critical points $A_i, i=1,2,3,4.$ Now draw two simple circles $S_1$ and $S_2$ transversally intersecting in a single point of the torus and containing the points $A_i$ and change the smooth structure by declaring $S_1$ and $S_2$ to be singular ridges. Then $h$ has no critical points on the smooth part of the torus. By taking a product with a number of smooth circles, you get a counterexample in higher dimensions. In fact, even one non-smooth circle ($n-k=1$) is sufficient to spoil the game.

  • $\begingroup$ Your answer is a particular case of my construction. Петя Пушкарь $\endgroup$
    – Petya
    Jul 22, 2010 at 21:43
  • $\begingroup$ Hi Petya, nice to see you here! I typed my answer before I saw yours. $\endgroup$ Jul 22, 2010 at 23:27

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