Stereographic projection...A more concise explanation?

Question

I have been looking at stereographic projections in books, online but they all seem...I don't know how else to put this, but very pedantic yet skipping the details of calculations.

Say, I have a problem here which asks;

Let $n \geq 1$ and put $S^n=\{(x_0,x_1...,x_n) \in \mathbb{R}^{n+1}|{x_0}^2+...+{x_n}^2=1\}$ (So I understand this is a unit sphere in $n+1$ dimensions). Let $P=\{(1,0,...,0)\}$ and consider $S^n$\ $P$ and $Y=\{(y_0...y_n)\in \mathbb{R}^{n+1}|y_0=0\}$ both with Euclidean metric. Thus $X$ is an $n$-sphere with a point removed and $Y \cong \mathbb{R}^n$.

That is the set up. The problem I don't know how to approach is,

$i$) For $x=(x_0,...,x_n) \in X$ and let $f(x)$ be a unique point of $Y$ such that $P=(1,0...0)$ and $x$, $f(x)$ are collinear. So, find $ \lambda(x)$ such that $f(x)=\lambda(x)P+(1-\lambda(x))x$ fo some $\lambda(x) \in \mathbb{R}$.

$ii$) For $y=(y_0,...y_n) \in Y$ let $g(y)$ be a unique point of $X$ such that $(1,0,...,0)$, $y$ and $g(y)$ are collinear, by similar method to $i$), find a formula for $g(y)$.

$iii$) Prove that $f : X \rightarrow Y$ and $g : Y \rightarrow X$ are inverse to each other and deduce that $X$ is homeomorphic to $Y$

Leaving aside part iii, I don't get how to do i. I have seen some examples on removing the North Pole(which is NOT my case) but then the equation for $f(x)$ appears out of nowhere in those cases. I don't know where and how they were obtained, no explanations and steps were given.

So, $\lambda(x)$ is a real number or, at least a scalar I understand. Given my conditions, I tried substituting the points to the given collinear form but found only that $\frac{y_0-x_0}{1-x_0} = \lambda(x)$ and also $\frac{x_i-y_i}{x_i} = \lambda(x)$ unless $i=0$. Which doesn't make sense to me, as I found well, 2 different $\lambda(x)$s (haven't I??)

I just don't get this stereographic projection thing "analytically". I have seen pictures and diagrams which visualises it and that's all very nice but algebraically/analytically, I cannot make sense of it.

Would anyone care helping me out at all?? Thank you so much....

Answer 1

For part i) you have to find the point of the line (Px) that lies in the hyperplane $x_0=0$, i. e. you have to solve for $$\lambda +(1-\lambda)x_0=0,\tag{1}$$ since the straight line $(Px)$ has a parametic representation: $$\lambda(1 ,0,\dots,0)+(1-\lambda)(x_0,x_1,\dots,x_n)=\bigl(\lambda +(1-\lambda)x_0, (1-\lambda)x_1, \dots,(1-\lambda)x_n\bigr).$$ Solving for $\lambda$ in $(1)$, we get $$\lambda=\frac{x_0}{x_0-1}.$$

Answer 2

The point of finding equations with $\lambda(x)$ and individual coordinates $x_k$ and $y_k$ is that you do not already know what all the values of all the coordinates are. You may have been given the coordinates $(x_0, \ldots, x_n)$ but not the coordinates $(y_0, \ldots, y_n)$, and you want to find those coordinates.

Or in other words, given the stereographic projection as defined in the question, and the coordinates of a point $x \in X$, find the coordinates of $f(x)$.

So initially, you do not know $y_i$ in general, and you cannot use $\frac{x_i - y_i}{x_i} = \lambda(x)$ to derive $\lambda(x)$. But you do know from the definition of the projection that $y_0 = 0$, so you can use $\frac{y_0 - x_0}{1 - x_0} = \lambda(x)$ to find $\lambda(x)$, and then for $i\neq 0$ you can use $\frac{x_i - y_i}{x_i} = \lambda(x)$ with the known value of $\lambda(x)$ to derive the (previously unknown) value of $y_i$.

If you did know the values of all the coordinates already, that is, if you were given then coordinates $(x_0, \ldots, x_n)$ of $x$ and the coordinates $(y_0, \ldots, y_n)$ of $f(x)$, then the $n+1$ equations of the form $\frac{y_0 - x_0}{1 - x_0} = \lambda(x)$ and $\frac{x_i - y_i}{x_i} = \lambda(x)$ must all agree on the same value of $\lambda(x)$. If solving any of these equations for $\lambda(x)$ resulted in a value of $\lambda(x)$ different from the value that solves any other equation, the points you were given would not be a correct stereographic projection.