Let $\mathbb{S}^n$ be the $n$-sphere and $\mathcal{T}_s\mathbb{S}^n$ the tangent space at $\mathbf{s}\in\mathbb{S}^n$.
I am looking for a basis for the tangent $\mathcal{T}_s\mathbb{S}^n$, but I got confused.
Copying from Wikipedia, I understand that for a chart $\phi\colon\mathcal{U}\to\mathbb{R}^n$, where $\mathcal{U}$ is an open subset of $\mathbb{S}^n$, one can define an ordered basis as follows:
My questions are: a) can $\phi$ be the stereographic projection from $\mathbb{S}^n-\{N\}$ to $\mathbb{R}^n$ (where $N$ denotes the north pole of $\mathbb{S}^n$), and b) what can $f$ be in that case?
Thank you!
It's easier by example. Let us think of $\mathbb S^2$, which is a subset of $\mathbb R^3$. Denote the points of $\mathbb R^3$ as $(X,Y,Z)$.
In the extrinsic viewpoint, the tangent space at $(0,0,1)\in \mathbb S^2$ is the set $\{(X, Y, 1)\ :\ X,Y\in\mathbb R\}$.
In the intrinsic viewpoint, you need a chart. You seem to like the stereographic projection. The equations of such projection from the South Pole $(0,0,-1)$ are $$\tag{1} x=\frac{X}{1+Z},\quad y=\frac{Y}{1+Z},$$ with inverse $$\tag{2} X=\frac{2x}{1+x^2+y^2},\quad Y=\frac{2y}{1+x^2+y^2},\quad Z=\frac{1-x^2-y^2}{1+x^2+y^2}.$$ The point $(X, Y, Z)=(0,0,1)$ corresponds to $x=0, y=0$. A basis of the tangent space at this point (and any other, except for the South Pole where the chart is undefined) is $\partial_x, \partial_y$.
For a $f\in C^\infty (\mathbb S^2)$ you can compute $\partial_x f|_{x=0, y=0}$ as follows. By definition, $f=f(X, Y, Z)$. So you plug in (2), yielding $$ \frac{\partial}{\partial x} \left[ F\left( \frac{2x}{1+x^2+y^2}, \frac{2y}{1+x^2+y^2}, \frac{1-x^2-y^2}{1+x^2+y^2}\right)\right]$$ and now you can compute and finally let $x=0, y=0$ in the result.
I hope this practical example clarified what is going on here.
Final notes.
This answer is the continuation of my comment).
The original question asked about the stereographic projection from the North Pole, whereas here it is from the South Pole. There is no essential difference.