Describe all differentiable functions on $\mathbb{S}^n \backslash S$ (S is the south pole) [closed]

Question

Consider the sphere $\mathbb{S}^n$ embedded in $\mathbb{R}^{n+1}$. Let $N$ be the north pole of the sphere and $S$ the south pole. Every point on $\mathbb{S}^n \backslash \{N,S\}$ is defined uniquely using the exponential map by $(r,v)$, where $0<r< \pi/2$ is geodesic distance from the north pole and $v$ is a unit vector in $T_N\mathbb{S}^n$. In the following, we use these coordinates. Note that the equator is then $\{(r,v) |r = \pi/2 \}$. Let $f$ be a differentiable function on the equator of the sphere. Define $F:M \backslash \{N,S\} \to \mathbb{R}$ by $F(r,v) = f(\pi/2,v)$. Now, $rF$ is a differentiable function on $\mathbb{S}^n \backslash S$.

Claim Every differentiable function on the $\mathbb{S}^n \backslash S$ can be obtained as a function of functions of the above type.

I came across this claim on pg. 337 of the following paper (in a slightly more general context, but for me understanding it on the sphere is good enough) http://projecteuclid.org/euclid.jmsj/1261060588. My questions are: How to prove this? and What is the idea behind this claim?. The author states that "this is the idea of the normal coordinates", perhaps this is the key?

Thanks in advance for your help!

Answer 1

I understand that one must prove that every smooth function $g : \mathbb S^n\setminus \{S\} \to \mathbb R$ can be expressed as $g = G \circ h$ where $G : U \to \mathbb R$ is a smooth function on some subset $U \subset \mathbb R^m$ for some $m$ and $h : \mathbb S^n\setminus \{S\} \to U,\ p \mapsto (h_1(p), ..., h_m(p))$ is a smooth function such that its components $h_i$ are of the required form $h_i(p) = r(p)F(p)$ where $F$ is of the form that you described and $r(p)$ is the geodetic distance from N as you described.

We may identify $T_N\mathbb S^n = \mathbb R^n$ up to isometric isomorphism (which is of course also a diffeomorphism). The map $p \mapsto r(p)v(p),\ \mathbb S^n\setminus \{S\} \to T_N\mathbb S^n = \mathbb R^n$, where $r(p)$ and $v(p)$ are what you described, is then a chart of $\mathbb S^n$. The components of the vector $r(p)v(p) \in \mathbb R^n$ are called the normal coordinates of $p$ (analogously one can define normal coordinates in the neighborhood of any point of any Riemannian manifold). A chart is a diffeomorphism onto its range, thus the smooth functions on $\mathbb S^n\setminus \{S\}$ are exactly those that are smooth in the normal coordinates. Therefore every smooth function on $\mathbb S^n\setminus \{S\}$ can be expressed as a smooth function of the values of the functions $r(p)v_i(p)$ where $v_i(p)$ is the $i$-th component of $v(p)$, q. e. d.

Or for a more concrete proof, if $\phi : T_N\mathbb S^n \to \mathbb R^n$ is the isometric isomorphism between $T_N\mathbb S^n$ and $\mathbb R^n$, $\psi : p \mapsto r(p)v(p),\ \mathbb S^n\setminus \{S\} \to \psi[\mathbb S^n\setminus \{S\}] \subset T_N\mathbb S^n$ is the normal coordinate map and $g : \mathbb S^n\setminus \{S\} \to \mathbb R$ is a smooth function then $G := g \circ \psi^{-1} \circ \phi^{-1}$ is a smooth function on the image $(\phi \circ \psi)[\mathbb S^n] = $ the open disk of radius $\pi$ in $\mathbb R^n$ and thus $g$ can be expressed as $g = G \circ (\phi \circ \psi)$. The components of the function $\phi \circ \psi$ are real-valued functions of the required form.