Here is the question:

Let M be a compact smooth manifold, $f:M\rightarrow\mathbb{R}$ be a smooth non-constant function. Show that $f$ has at least two critical points.

I am trying to show by contradiction. Suppose $f$ has less than two critical points. First assume it has none. By Inverse Function Theorem, for $x\in M$, there is an open neighbourhood $U $ containing $x$ such that $f|_U$ is a diffeomorphism. This means $f$ is an open map. Thus, $f(M)$ is open. But this is contradiction to the fact that $f(M)$ is compact (since $f$ is cont's and $M$ is compact).

Is this fine? Now I don't know how to show that there cannot be only one critical point. Only after proving this, I will be done. Some-where I should be using that $f$ is non-constant. I don't know how. I understand that if $f$ is a constant function, then every point in $M$ is trivially critical.

Intuitively I don't understand why compactness is used. I am trying to find an example of non-compact smooth manifold $M$ and a smooth map $f:M\rightarrow \mathbb{R}$ that has no critical points. I will appreciate any hint for this too.

  • $\begingroup$ I have absolutely no idea about how you apply the inverse function theorem given that $\dim M$ might be strictly larger that $1$. What you could have done was to note that if $f$ has no critical points than it must be a submersion, therefore an open map. $\endgroup$ – Alex M. Jun 9 '15 at 8:28

Since $M$ is compact and $f:M \to \Bbb R$ is continuous, there exists points $p_\sigma, p_\gamma \in M$ such that $f(p_\sigma) = \sigma$ and $f(p_\gamma) = \gamma$, where $\sigma$ and $\gamma$ are the global minimum and maximum values of $f$ on $M$. This is a basic result of elementary topology. Since $p_\sigma$ is a global minimum for $f$, we have $df_{p_\sigma} = 0$ and likewise $df_{p_\gamma} = 0$, since $p_\gamma$ is a global maximum. Since $f$ is not constant, $\sigma < \gamma$ whence $p_\sigma \ne p_\gamma$. Thus $p_\sigma$ and $p_\gamma$ are distinct critical points of $f$; we see there are at least two critical points of $f:M \to \Bbb R$. QED.

Hope this helps. Cheers,

and as always,

Fiat Lux!!!

  • $\begingroup$ quite awesome proof. I like this very much as you used very elementary ideas. $\endgroup$ – ugstudent1243 Mar 9 '14 at 17:15
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    $\begingroup$ @ ugstudent: thank you very much! It is indeed pretty "elementary"! And thanks for the "acceptance"! $\endgroup$ – Robert Lewis Mar 9 '14 at 17:17
  • $\begingroup$ you used $M$ for maximum of $f$. Perhaps a different letter would be better as it is used for manifolds already. $\endgroup$ – ugstudent1243 Mar 9 '14 at 17:20
  • $\begingroup$ @ ugstudent: you are correct! Will edit shortly. Thanks! $\endgroup$ – Robert Lewis Mar 9 '14 at 17:22
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    $\begingroup$ How to show that the property: "'p is a minimum(or max) of a function in a manifold then Dfp=0" holds to manifolds? $\endgroup$ – Luiz Feb 7 '17 at 18:49

Let's suppose that $M$ is connected. The general case then follows by considering the restriction of $f$ to any connected component of $M$. Then $f(M)$ is a connected compact subset of $\mathbb R$, and since we're assuming $f$ is nonconstant, $f(M)$ is not a single point. What must $f(M)$ look like?

For a counterexample in the case that $M$ is not compact, you could take $M = \mathbb R$ and $f(x)=x$.

  • $\begingroup$ Are you trying to give argument for why $f$ cannot have one critical point? $\endgroup$ – ugstudent1243 Mar 9 '14 at 16:56
  • $\begingroup$ I was only giving a hint. Since a complete solution has now been posted, I'll elaborate on this hint. The only compact connected subsets of $\mathbb R$ are singletons and closed intervals, so $f(M)$ must be a closed interval $[a,b]$. Choosing $p,q\in M$ with $f(p)=a$ and $f(q)=b$, you can then show that $p$ and $q$ are critical points (e.g. the inverse function theorem fails at these points). $\endgroup$ – bradhd Mar 9 '14 at 18:07

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