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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....

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  • $\begingroup$ What is "$y+$" in $Y=\{(y_0...y_n)\in \mathbb{R}^{n+1}|y+=0\}$? Is that a typo? Is it supposed to say $y_0 = 0$ instead? $\endgroup$ – David K Oct 19 '15 at 19:32
  • $\begingroup$ Hi David, indeed it is a typo, I'll fix that just now, thank you for pointing it out!! $\endgroup$ – Kydo Oct 19 '15 at 19:34
  • $\begingroup$ After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark ✓ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?. Furthermore, you should upvote any useful answer you get. $\endgroup$ – A.P. Oct 19 '15 at 21:36
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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.

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  • $\begingroup$ Thank you, I got the first bit now...True, I clearly shouldn't use the $y \in Y$ where the coordinates are unknown apart from $y_0=0$. I am looking at ii now but using the similar method, I get $x_0= \mu (y)$ where I have set $\mu (y)$ as a constant analogous to $\lambda (x)$. Am I wrong to set this up as $g(y)= \mu (x) (1,0...,0)+(1- \mu(x)) y$ ? $\endgroup$ – Kydo Oct 19 '15 at 20:24
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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 $$\lambda=\frac{x_0}{x_0-1}.$$

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  • $\begingroup$ Cheers! Brief but made that clear for me, thank you :) $\endgroup$ – Kydo Oct 19 '15 at 20:25
  • $\begingroup$ You're welcome! Always glad to feel helpful. $\endgroup$ – Bernard Oct 19 '15 at 20:40
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For the function $g$, we need to find a point $x=g(y)$ on the unit sphere $x^2=1$, which is also on the line between $y$ and the pole $p$. (Also $p^2=1$.) So there should be some scalar $k$ such that

$$x = p + k(y-p)$$

$$x^2 = p^2 + 2k\,p\cdot(y-p) + k^2(y-p)^2$$

Cancel $x^2=p^2$, and $k\neq0$, so

$$0 = 2p\cdot(y-p) + k(y-p)^2$$

$$k = -2\frac{p\cdot(y-p)}{(y-p)^2}$$

Substitute this $k$ in the first equation:

$$x = p - 2\Big(\frac{p\cdot(y-p)}{(y-p)\cdot(y-p)}\Big)(y-p)$$

[The second term can be recognized as the projection of $p$ parallel to $(y-p)$ :

$$x = p - 2p_\parallel = (p_\parallel + p_\perp) - 2p_\parallel = -p_\parallel + p_\perp$$

So $x=g(y)$ is the reflection of $p$ along the vector $(y-p)$. This makes sense, because a reflection of a unit vector is also a unit vector.]

We also have $p\cdot y=0$, because $y$ is in the plane normal to $p$, so

$$x = p - 2\Big(\frac{0-1}{y^2+1}\Big)(y-p)$$

$$= p + 2\frac{y-p}{y^2+1}$$

Now you can find the coordinates (though I think the vector expression is neater):

$$x_0 = x\cdot p = 1 + 2\frac{0-1}{y^2+1} = \frac{(y^2+1)+2(-1)}{y^2+1} = \frac{y^2-1}{y^2+1}$$

$$x_i = 0 + 2\frac{y_i-0}{y^2+1} = \frac{2y_i}{y^2+1},\qquad 1 \leq i \leq n$$


For the function $f$, we need to find a point $y=f(x)$ in the plane $y\cdot p=0$, which is also on the line between $x$ and $p$. So, again,

$$y = p + k(x-p)$$

$$y\cdot p = p^2 + k(x-p)\cdot p$$

$$0 = 1 + k(x-p)\cdot p$$

$$k = \frac{-1}{(x-p)\cdot p}$$

Substitute $k$ :

$$y = p - \frac{x-p}{(x-p)\cdot p}$$

$$= \frac{\big((x-p)\cdot p\big)p - (x-p)}{(x-p)\cdot p} = \frac{\big(x\cdot p\big)p - x}{(x-p)\cdot p} = \frac{x - \big(x\cdot p\big)p}{-(x-p)\cdot p}$$

[The second term can be recognized as the projection of $x$ parallel to $p$ :

$$y = \frac{x - x_\parallel}{-(x-p)\cdot p} = \frac{x_\perp}{-(x-p)\cdot p}$$]

Using that $x\cdot p=x_0$, the coordinates are

$$y_0 = \frac{x_0 - (x_0)1}{1-x_0} = 0$$

$$y_i = \frac{x_i - (x_0)0}{1-x_0} = \frac{x_i}{1-x_0},\qquad 1 \leq i \leq n$$


It can be seen that $f$ and $g$ are inverses:

$$g(f(x)) = p + 2\frac{f(x) - p}{f(x)^2+1}$$

$$= p + 2\frac{\Big(\frac{x - (x\cdot p)p}{1-x\cdot p}\Big) - p}{\Big(\frac{x - (x\cdot p)p}{1-x\cdot p}\Big)^2+1}$$

$$= p + 2\big(1-x\cdot p\big)\frac{\big(x - (x\cdot p)p\big) - p\big(1-x\cdot p\big)}{\big(x - (x\cdot p)p\big)^2+\big(1-x\cdot p\big)^2}$$

$$= p + 2\big(1-x\cdot p\big)\frac{x - p}{\big(1 - 2(x\cdot p)^2 + (x\cdot p)^2\big)+\big(1 - 2x\cdot p + (x\cdot p)^2\big)}$$

$$= p + 2\big(1-x\cdot p\big)\frac{x - p}{2 - 2x\cdot p}$$

$$= p + (x - p)$$

$$= x$$

and

$$f(g(y)) = \frac{g(y) - \big(g(y)\cdot p\big)p}{1-g(y)\cdot p}$$

$$= \frac{\big(p + 2\frac{y-p}{y^2+1}\big) - \Big(\big(p + 2\frac{y-p}{y^2+1}\big)\cdot p\Big)p}{1-\big(p + 2\frac{y-p}{y^2+1}\big)\cdot p}$$

$$= \frac{0 + 2\frac{y}{y^2+1} - 2\frac{y\cdot p}{y^2+1}p}{0 - 2\frac{y\cdot p-1}{y^2+1}}$$

$$= \frac{y - (y\cdot p)p}{1 - y\cdot p}$$

$$= \frac{y - 0}{1 - 0}$$

$$= y$$


Note that the projection functions $f$ and $g$ can easily have their domains extended to all of space (except the plane $x\cdot p=1$), but their ranges are still the sphere and the plane, so the extensions are not invertible.

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