# Tangent map between tangent spaces

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?

• Use compactness of $S^n$. – 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.
• @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 $S^n$. – student Feb 19 '12 at 19:28
• @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. – Ben Blum-Smith Feb 21 '12 at 2:53
• (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 $q$. – Ben Blum-Smith Feb 21 '12 at 3:10