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Let $$ \varphi:\mathbb{R}^2 \longrightarrow \mathbb{S}^2-{N} \subset \mathbb{R}^3$$ be the (inverse) stereographic projection from the North pole on the unit sphere centred at the origin.

$$\varphi(x,y) = \left( \frac{2x}{x^2+y^2+1},\frac{2y}{x^2+y^2+1},\frac{x^2+y^2-1} {x^2+y^2+1} \right) $$ Let $$\hat{f}(x,y) = (\alpha x, \alpha y) $$ where $\alpha >0. $ $$f: \mathbb{S}^2 \longrightarrow \mathbb{S}^2 = \left\{ \begin{array}{ll} \varphi \circ \hat{f} \circ \varphi^{-1} & on \quad \mathbb{R}^2 \longrightarrow \mathbb{S}^2-{N} \\ f(N)=N & \\ \end{array} \right. $$

How would I show that the map is smooth at $N$ and compute $Df|_N$ at the north pole?

In the Jacobian representation, $Df|_N$ goes to infinity.

I tried taking $$\lim_{\epsilon=(a,b,c)\rightarrow (0,0,0)} \frac{f(N+\epsilon)-f(N)}{\epsilon} $$ But the same problem happens.

Is there a reformulation of $f$, say $h$ which is equal to $f$ but expressed entirely differently (for example in polar coordinates?)

Any hints or clues are appreciated!

ZahaMan
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    Your $f$ is the identity, so the derivative is the identity at any point. You need a chart around the north pole, e.g. stereographic projection from the south pole, to compute the derivative in the chart representation. – Daniel Fischer Nov 11 '14 at 19:05
  • Sorry I've missed a piece out - just editing it in now – ZahaMan Nov 11 '14 at 19:14
  • Okay. The second part of my comment still applies. You need a chart around $N$ and express $f$ in that chart. – Daniel Fischer Nov 11 '14 at 19:22
  • So something like -(φS o fhat o φS^-1) around and including N – ZahaMan Nov 11 '14 at 19:29
  • More something like $\varphi_S^{-1}\circ f \circ \varphi_S$. You want a function $\mathbb{R}^2 \to \mathbb{R}^2$ to compute the derivative. – Daniel Fischer Nov 11 '14 at 19:39
  • I'm gonna try and compute that now - thanks for the great explanation – ZahaMan Nov 11 '14 at 20:02
  • Ok I get the Matrix $\begin{matrix} 1 & 0 \ 0 & 1 \ \end{matrix}$ ie the plane parallel to xy at the north pole (which sounds right geometrically!) – ZahaMan Nov 11 '14 at 20:55
  • There should appear an $\alpha$ somewhere. It should be of the form $$g(\alpha)\cdot\begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix},$$ where I don't know $g(\alpha)$ without computing, but it should be similar to $\frac{1}{\alpha}$. – Daniel Fischer Nov 11 '14 at 21:00
  • By the way I'm really sorry to take so much of your time! – ZahaMan Nov 11 '14 at 21:01
  • You're not "taking" it. It's my own decision. If I wanted to, I could just ignore any comments here or turn off the computer or whatever. – Daniel Fischer Nov 11 '14 at 21:06
  • Point taken, still grateful :) I think I know what I did wrong; when I use $g=\varphi_{S}^{-1} \circ f \circ \varphi_{S}$ and compute $Dg = D\varphi_{S}^{-1} \circ Df \circ D\varphi_{S}$ I get that $Df|{N} = I{3 \times 3}$ ie the identity. I shouldn't be using that should I? It's more fruitful to use $f = \varphi_{N} \circ \hat{f} \circ \varphi_{N}^{-1}$ right? – ZahaMan Nov 11 '14 at 21:21
  • The sphere is two-dimensional. It doesn't make sense to say $Df\lvert_N$ is a $3\times 3$-matrix. You can only express $Df\lvert_N$ as a matrix if you have a basis of the tangent space. Typically, you get that via a coordinate chart. Here, the stereographic projection from the south pole is a natural coordinate chart around the north pole. Then you express $Df\lvert_N$ in that chart, as $Dg\lvert_0$. – Daniel Fischer Nov 11 '14 at 22:09
  • That seems to be where my understanding falls short; how would you express $Df|{N}$ as $Dg|{0}$; I just can't grasp how to rewrite f in the chart as g - is there a formula I should be using? :( – ZahaMan Nov 11 '14 at 23:28
  • $g = (\varphi_S^{-1}\circ\varphi) \circ \hat{f} \circ (\varphi^{-1}\circ \varphi_S)$. Compute $\varphi^{-1}\circ \varphi_S$ and its inverse, and you have an explicit representation of $g$. Computing the derivative of $g$ is standard. Then you have the representation of $Df\lvert_N$ in the chart $\varphi_S$. Since it happens to be a multiple of the identity, the representation is the same in all charts, but that's coincidental, generally the representation of the derivative is chart-dependent. – Daniel Fischer Nov 12 '14 at 10:43

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