How to show that there are 2 points of $S^n$ where a smooth map $g: S^n \to\mathbb R$ has tangent map equal to zero?

  • $\begingroup$ Use compactness of $S^n$. $\endgroup$ – student Feb 19 '12 at 15:37

$S^n$ is compact, so on a purely topological level, $g$ (as a continuous function on it) must have a maximum and a minimum. Now if $g$ is also smooth, these must be critical values, by the same argument as in elementary calculus. Since the range of $g$ is 1-dimensional, anywhere $dg_x$ does not have full rank it must be zero.

  • $\begingroup$ I'm new on the subject, so I'll ask: how do you actually prove that the rank is not full? Would it be possible to consider the 2 charts of the atlas of S^n and for each chart domain find a point where the differential is zero? I tried to look at it through charts and work in R^n in local coordinates but I'm stuck...but I'd like to use this approach...any idea? $\endgroup$ – John Feb 19 '12 at 16:27
  • $\begingroup$ @John You should first prove that if $p$ is a point of maximum or minimum of a function $f : \mathbb{R}^n \rightarrow \mathbb{R}$, then $df(p)$ must be zero. Then, using charts, you can extend this result to any manifold, and in particular to $S^n$. $\endgroup$ – student Feb 19 '12 at 19:28
  • $\begingroup$ @John - good questions. It doesn't work to go to charts first, before talking about the max and min, because then you lose the ability to refer to $S^n$'s compactness, because the charts are individually not compact. In particular, there's no reason to think if you cover $S^n$ with two charts, the critical points will be split between them. (They could easily both be in the same chart and the other chart is critical point-free.) Example: Let $S^n$ be earth's surface, let $g$ be latitude. The two critical points are the north and south poles. $\endgroup$ – Ben Blum-Smith Feb 21 '12 at 2:53
  • $\begingroup$ (cont'd) Now suppose it happened that earth is parametrized by two charts, one of which covers the greater New York metropolitan area and the other covers the rest of the globe; they overlap on the outer suburbs. The big chart contains both critical points, and the New York chart contains none. The moral is that the true essential thing about a manifold is the ability to be covered by charts, which is really nothing more than the ability to be covered by a small chart in the neighborhood of any given point. Any particular atlas is adding information extrinsic to the manifold itself. $\endgroup$ – Ben Blum-Smith Feb 21 '12 at 3:00
  • $\begingroup$ (cont'd) What this means for you is that the first move is to observe: $S^n$ is a compact topological space, so any real-valued function $g$ attains a maximum and a minimum, say at $p,q\in S^n$. Now put local coordinates around $p$. You now have a function $\mathbb{R}^n\rightarrow \mathbb{R}$ that has a maximum at $p$. (More precisely, at $p$'s image in the local coordinates.) The problem has turned into the exercise in multivariable calculus that Leandro mentioned: prove a function $\mathbb{R}^n\rightarrow \mathbb{R}$ has vanishing derivative at a maximum point. Then do the same for $q$. $\endgroup$ – Ben Blum-Smith Feb 21 '12 at 3:10

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