# A smooth function $f:S^1\times S^1\to \mathbb R$ must have more than two critical points.

I am trying to show that a smooth function $f:S^1\times S^1\to \mathbb R$ must have more than two critical points. Since $f$ attains maximum and minimum, it must have at least two critical points. How would one show that they can't be two?

If one considers the gradient vector field $\nabla f$ then by the Poincare Hopf index theorem its index is equal to the Euler characteristic of the torus, i.e. it is $0$. Therefore $\nabla f$ has an even number of zeros.

This question has been posted here but it hasn't been answered.

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Are you familiar with Morse theory? If you assume the critical points are non-degenerate (which you should be able to do by taking a small perturbation), then the Morse inequalities give you the result immediately. Alternatively, you could show directly that the decomposition corresponding to a function with only two critical points must give you a sphere. – anomaly Sep 13 '14 at 21:33
@anomaly I am not familiar with Morse theory. What is this decomposition you are mentioning? – Dimitrios Nt Sep 13 '14 at 23:58
I am studying from Guillemin Pollack. An idea I have is that when we have maximum or minimum for $f$, this corresponds to a "source" or "sink" (using G-P terminology), and at those points the index of $\nabla f$ should be $+1$. Thus, in order to have Euler characteristic $0$, we must have either two saddle points that correspond to index $-1$ each, or one point with index $-2$. – Dimitrios Nt Sep 14 '14 at 0:15

The only closed manifolds which allow a function with two (maybe degenerate) critical points are spheres. In dimension 2 it is quite easy to prove.

One of these critical points is maximum and another is minimum. Any level set in between is compact and has dimension 1, so it is disjoint union of circles. Now use an argument from Morse theory: because there is no critical points between two such level sets, the gradient vector field gives diffeomorphism of them. So the whole space is the suspension over disjoint union of circles. It is not difficult to show that it is manifold only in case of one circle, and then it is 2-sphere, not torus.

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I like this argument! – Dimitrios Nt Oct 13 '14 at 15:54
I'm not sure whether this argument should require that two critical points are non-degenerate, as in the Milnor's book, *Morse Theory"? – Henry Jun 2 '15 at 15:14
@Henry, I beleive it's not necessary, as the complement of these two points is the product of $(0, 1)$ and the fiber anyway, and just some words are to be said about it being suspension. – se0808 Jun 3 '15 at 7:05

One way to prove it, is to use the Lusternik–Schnirelmann category:

Definition: Let $A$ be a closed nonempty subset of a topological space $X$. We define the Lusternik–Schnirelmann category of $A$ as the number $$\operatorname{cat}_X(A)=\min\left\{n\in\mathbb{N}:\ A\subset \bigcup_{i=1}^nA_n,\ A_n\ \mbox{is contractible to a point in}\ X,\ A_n\ \mbox{closed}\right \}.$$

The idea behind this number, is that it is related to the number of critical points of a function $f$. For instance, if $X$ is a compact manifold and $f:X\to \mathbb{R}$ is a smooth function then, the number of critical points of $f$ is bigger than or equal to $\operatorname{cat}_X(X)$.

It is an interesting problem to compute $\operatorname{cat}_X(X)$ in the case $X$ is the torus. You can show that in this case, $\operatorname{cat}_X(X)=3$.

There are plenty of references on this subject, however, as I am an analyst, I would like to refer you to the book of Drabek and Milota, in particular, chapter 6, section 6.4B.

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