Let $f:\mathbb{S^2} \setminus {(0,0,1)}\to\mathbb{R^2} $ be the stereographic projection of a sphere onto a plane.
Find the derivative of $f$.
Note: You should get a linear map from 2-d space to 2-d space
I'm new to the idea of differentiation of manifolds.
First I need two manifolds M and N and a smooth map between them
$M$ is the $\mathbb{S^2} \setminus {(0,0,1)} $ paramaterized as $\{(x,y,z)|x^2 + y^2 + z^2 = 1\}$
$N$ is the equatorial plane $\{z=0\}$
Now I need a smooth map between the two manifolds $f:M\to N $ $\space where \space(x,y,z)\mapsto (x,y)$
.
Now I need to associate each point $s\in M$ with a tangent space $TM_x$
This lets us define the derivative $df_x:TM_x\to TN_y$
According to my book (Topology from the Differentiable Viewpoint)
we choose a parametrization where $U$ is a subset of ${R^m}$ and M is a subset of ${R^k}$.
$g:U\to M$ of a neighborhood g(U) of x in M, with g(u) = x and the derivative is
$dg_u: {R^m} \to {R^k}$
We have our parametrization of $M$, $\{(x,y,z)|x^2 + y^2 + z^2 = 1\}$
$TM_x$ is the image of $dg_x \circ dg_u$
Now I am a little confused. I'd like a concrete example like this to work from where I can say,
This is my a manifold M, this is another manifold N, I have a smooth map between them, this is how I generate the tangent space, and this is how I calculate the derivative
Best Answer
The stereographic projection map is
$f:\mathbb{S^2} \setminus {(0,0,1)}\to\mathbb{R^2} $
$(x,y,z) \mapsto (\frac{x}{1-z},\frac{y}{1-z})$
When we try and find our Jacobian, we know we want a $2 \times2$ matrix, but we have three variable?
When we make our Jacobian, we want local coordinates
$x^2+y^2+z^2=1 \leftarrow \rightarrow z=\sqrt{1-x^2 - y^2}$
so we can define our map as
$f:\mathbb{S^2} \setminus {(0,0,1)}\to\mathbb{R^2} $
$(x,y,z) \mapsto (\frac{x}{1-\sqrt{1-x^2-y^2}},\frac{y}{1-\sqrt{1-x^2-y^2}})$
which is in two variables
The Jacobian becomes
$$ \begin{pmatrix} \frac{\partial}{\partial x} (\frac{x}{1-\sqrt{1-x^2-y^2}}) & \frac{\partial}{\partial y} (\frac{x}{1-\sqrt{1-x^2-y^2}})\\ \frac{\partial}{\partial x}(\frac{y}{1-\sqrt{1-x^2-y^2}}) & \frac{\partial}{\partial y}(\frac{y}{1-\sqrt{1-x^2-y^2}})\\ \end{pmatrix} $$
Computing this, compute the determinant, and we are finished.