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Here's the actual question I wish to prove, along with the given hint:

Let $f: \Bbb{T}^2 \to \Bbb{R}$ be a smooth map. Show that $f$ has atleast $4$ critical points ($p$ is a critical point of $f$ if the tangent map $Tf_p: T_p\Bbb{T}^2 \to T_{f(p)}\Bbb{R}$ vanishes).

Hint: Parametrize $\Bbb{T}^2$ using angles $\theta,\varphi$, and locate the maximum and minimum points of $f(\theta, \varphi)$, for $\varphi$ fixed, say $(\theta_{\text{max}}(\varphi), \varphi)$ and $(\theta_{\text{min}}(\varphi), \varphi)$; now maximize and minimize $f$ as $\varphi$ varies.

I took a look at this question, but the answer there makes use of Morse Theory, which I do not know.


Definitions I'm working with:

The definition of torus I'm working with is $\Bbb{T}^2 = S^1 \times S^1$, and the definition of tangent space at a point $p$ of a manifold $M$ is defined to be the collection of equivalence classes of curves: \begin{align} T_pM := \{[c] : \text{$c$ is a $C^1$ curve in $M$ with $c(0) = p$} \}. \end{align} The tangent mapping is defined as $Tf_p([c]) = [f \circ c]$.


As stated, I found the problem slightly difficult, so instead I tried to solve a simpler problem in the hopes that I'd be able to generalize it. Since, $\Bbb{T}^2$ is a product manifold $S^1 \times S^1$, I tried to prove the following assertion instead:

Every smooth function $f: S^1 \to \Bbb{R}$ has atleast two critical points.

I already have a proof of this fact using the Extreme value theorem (the extrema will then be critical points), but I want to prove this assertion in a manner suggested by the hint, so that I can generalize this simple case of $f: S^1 \to \Bbb{R}$ to the general case of $f: \Bbb{T}^2 \to \Bbb{R}$.

Here's what I have done so far. In trying to follow the hint, I considered the parametrization $\alpha^{-1}: (0,2 \pi) \to S^1 \setminus \{(1,0)\}$ defined by \begin{align} \alpha^{-1}(\theta) = (\cos \theta, \sin \theta) \end{align} (I call it $\alpha^{-1}$ so that $(S^1 \setminus\{(1,0)\}, \alpha)$ is then a chart on $S^1$). Next, I managed to show that the point $\alpha^{-1}(\theta)$ is a critical point for $f$ if and only if $\theta$ is a critical point for $f \circ \alpha^{-1}: (0,2 \pi) \to \Bbb{R}$. In other words, if and only if \begin{align} (f \circ \alpha^{-1})'(\theta) = \dfrac{d}{d\theta} \big( f(\cos \theta, \sin \theta) \big) = 0 \tag{$*$} \end{align} Now, here's the problem I face: the only thing we know about $f$ is that it is a smooth function $f:S^1 \to \Bbb{R}$, but we don't have an explicit formula for it. Also, $f$ is a function whose domain is the manifold $S^1$, rather than an open set in $\Bbb{R}^2$, so I can't apply the standard chain rule to $(*)$. Because of this I'm not sure how to proceed in proving that there exist atleast two critical points for $f$. (Also, I think I might have to use a second chart so that it covers the entire circle, but I'm having trouble with the first chart itself, so I didn't bother with a second chart).


So, my questions are:

  • Is it a good idea to try to solve my general problem by considering the lower dimensional case of $f:S^1 \to \Bbb{R}$? If so, how would I prove the existence of two critical points in the $S^1$ case (without extreme value theorem), and how would I generalize this to prove existence of $4$ critical points for $f: \Bbb{T}^2 \to \Bbb{R}$?

  • If this isn't a viable approach, how would I make use of the hint given in the question to directly prove the existence of $4$ critical points?

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  • $\begingroup$ You most certainly can't ignore the periodicity involved here because there are plenty of functions defined on $\Bbb S^1\setminus\{(1,0)\}$ without critical points. $\endgroup$ Aug 1, 2019 at 3:34
  • $\begingroup$ @Fimpellizieri right, for instance, the function $f$ which assigns to each point of $S^1 \setminus \{(1,0)\}$, its angle (with the usual measurement conventions), however such a function cannot be continuous on the entire $S^1$, let alone smooth. The only reason I considered one chart is to try to simplify matters completely, and to "gain insight" into how to even start the problem. Do you have suggestions on a cleaner approach (which is still elementary) for proving existence of the $4$ critical points on the torus? $\endgroup$
    – peek-a-boo
    Aug 1, 2019 at 3:46
  • $\begingroup$ There are functions with only three critical points on the two torus, but they are not Morse. $\endgroup$
    – Thomas Rot
    Aug 2, 2019 at 9:12

2 Answers 2

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Since the hint already uses $\theta_\max$ and $\theta_\min$, coming from the extrema over the compact $S^1\times\{\phi\}$, I think you cannot really scale down the method to the one-dimensional case.

Instead, follow the hint: For fixed $\phi$, we have $S^1\to \Bbb R$, $\theta\mapsto f(\theta,\phi)$ continuous on a compact, hence tehre exist minimizers and maximizers $\theta_\min(\phi)$ and $\theta_\max(\phi)$ - but these are not continuous in $\phi$! We can still consider $f_\max\colon S^1\to\Bbb R$, $\phi\mapsto f(\theta_\max(\phi),\phi)$ and similar for $\min$. Verify (by a compactness argument) that $f_\min$ and $f_\max$ are continuous even though $\theta_\min$ and $\theta_\max$ need not be. This gives us (in general) four points: A maximizer of $f_\max$, a minimizer of $f_\max$, a maximizer of $f_\min$, a minimizer of $f_\min$. Now show that these four points are critical (because they are so in both main directions). Caveat: What happens if two of the four points coincide?

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  • $\begingroup$ your explanation clarified to me what the hint actually meant, but I'm still having trouble with (the most important) parts of the proof. Could you elaborate on why $f_{\text{max}}$ is continuous; it isn't clear to me where compactness comes into the picture. Also, I managed to show that the minimizer of $f_{\text{min}}$ and maximizer of $f_{\text{max}}$ are critical points (because they're infact global extrema of $f$). However I'm having trouble showing the "mixed" points like the maximizer of $f_{\text{min}}$ are infact critical points of $f$ (cont) $\endgroup$
    – peek-a-boo
    Aug 2, 2019 at 20:58
  • $\begingroup$ (cont) because it seems like the $\theta$ and $\phi$ are not being varied indepenently. So, could you elaborate on this as well. Thank you $\endgroup$
    – peek-a-boo
    Aug 2, 2019 at 21:00
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I would do a minimax. Here is a sketch. We already found two critical points, the maximum and the minimum. If the maximum and the minimum are not unique we are done, because we have more than two critical points.

Now in the other case for every fixed $\phi$ you can look at the curves $\gamma_\phi(\theta)=(\theta,\phi)$. Now, for such a fixed $\phi$ the function has a maximum along the curve, which might not be unique anymore. But at a point $\theta_\phi$ along which the function attains a maximum $m_\phi$ on the curve $\gamma_\phi$ at least $\frac{d}{d \theta}f(\gamma(\theta)=\frac{\partial}{\partial \theta}f(\theta,\phi)=0$.

Now argue that $m_\phi$ is a continuous function and the minimum (varying $\phi$) value $m$ exists. Argue that this value $m$ must be a critical value for the original function $f$ and that $\min f<m<\max f$.

You can do the same game, but reversing the order of the min and the max. You will also find a critical value. I think the author of the exercise assumed that these would give different critical points, but this need not be the case.

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  • $\begingroup$ A quick question on your notation: where do $\theta$ and $\phi$ live? Are you considering them to be actual points in the manifold $S^1$, or are you thinking of them as "parameters" in a certain interval of the real line, and you're merely suppressing the chart map in the notation when you write $f(\theta, \phi)$? $\endgroup$
    – peek-a-boo
    Aug 2, 2019 at 10:02
  • $\begingroup$ I think of $S^1\times S^1$ as points in $\mathbb R^2$ modulo $2\pi$. $\endgroup$
    – Thomas Rot
    Aug 2, 2019 at 10:04

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