Wrapping Up Infinite Planes

Note: This is sort of a “part 2” to this post, which is the two-dimensional analogue to this (mainly) three-dimensional post.

It’s fairly early/late (4:39AM) and I’m pretty sleepy but I wanted to try and get this post in, so I apologize in advance if it seems “more math-y and less descriptive.”

As an expansion to the previous post, the same concept can be applied to the sphere $x^2+y^2+z^2=1$ and $\mathbb{R}^2$. And similar to before, the top of our sphere is punctured, i.e. the point $(0,0,1)$.

We now look for a function $f$ that allows us to input a point $(x_0, y_0)$ on our plane and output a point $(x_s,y_s,z_s)$ on our sphere, that is, $f: \mathbb{R}^2 \to \mathbb{S}^2$, where $\mathbb{S}^2 = \{ x \in \mathbb{R}^3 | x^2+y^2+z^2=1 \}$. The line that will assist us with this assignment is now a vector, as would be required, so here’s our new system:

$\begin{cases} x^2+y^2+z^2=1 \\ \vec{r}(t)=\left< 0,0,1 \right> + t \left< x_0,y_0,-1 \right> \ \ \ 0 \leq t \leq 1 \end{cases}$

Another, more convenient way of writing this system can be done by noting that $\vec{r}(t)$ can simply be rewritten (by adding the vectors in its definition) as

$\vec{r}(t) = \begin{cases} x(t) = x_0 t \\ y(t)=y_0t \\ z(t) = 1-t \end{cases} = \ \ \ \begin{cases} t=\frac{x}{x_0} \\ t=\frac{y}{y_0} \\ t = 1-z \end{cases}$

thus giving us

$\begin{cases} x^2+y^2+z^2=1 \\ \frac{x}{x_0}=\frac{y}{y_0}=1-z \end{cases}$

by equating the the parameter in each component. From here we just need to take turns solving for $x$, $y$, and $z$ in terms of $x_0$ and $y_0$ and we should be all set!

Solving for x:
$\displaystyle{ x^2 + \left( \frac{xy_0}{x_0} \right) ^2 + \left( 1-\frac{x}{x_0} \right) ^2 = 1}$
$\displaystyle{ \left( \frac{{x_0}^2+{y_0}^2+1}{{x_0}^2} \right)x^2 - \frac{2}{x_0} x = 0}$
$\displaystyle{ x=\frac{\frac{2}{x_0} \pm \sqrt{\frac{4}{{x_0}^2}}}{2 \left( \frac{{x_0}^2+{y_0}^2+1}{{x_0}^2} \right)} }$

$\displaystyle{ \implies x = \frac{2x_0}{{x_0}^2+{y_0}^2+1}}$

Solving for y:
$\displaystyle{ \left( \frac{yx_0}{y_0} \right) ^2 +y^2 + \left( 1-\frac{y}{y_0} \right) ^2 = 1}$
$\displaystyle{\dots}$
$\displaystyle{ \implies y = \frac{2y_0}{{x_0}^2+{y_0}^2+1} }$

Note, for $x$ and $y$, the radical was chosen to be positive or else both would’ve reduced to to being $0$ for all values of $x_0$ and $y_0$, which is not always the case, for instance at $(x_0, y_0) = (1,0)$, that should be mapped to the sphere at the point $(x_s, y_s, z_s) = (1,0,0)$ on the sphere, and similarly $(0,1) \mapsto (0,1,0)$, thus indicating we take the positive root.

Solving for z:
$\displaystyle{(x_0-x_0z)^2+(y_0-y_0z)^2+z^2=1}$
$\displaystyle{({x_0}^2+{y_0}^2+1)z^2-2({x_0}^2+{y_0}^2)z+({x_0}^2+{y_0}^2-1)=0}$
Let’s let $u ={x_0}^2+{y_0}^2$.
$\displaystyle{(u+1)z^2-2uz+(u-1)=0}$
$\displaystyle{ z = \frac{2u \pm \sqrt{4u^2-4(u+1)(u-1)}}{2(u+1)}}$
$\displaystyle{ z = \frac{u \pm 1}{u + 1}}$
Now taking the numerator that doesn’t assign $z$ to the punctured point…
$\displaystyle{ \implies z= \frac{{x_0}^2+{y_0}^2-1}{{x_0}^2+{y_0}^2+1} }$

Our Function:
Using the above information, our function $f: \mathbb{R}^2 \to \mathbb{S}^2$ is defined

$\displaystyle{ f(x_0, y_0) = \left< \frac{2x_0}{{x_0}^2+{y_0}^2+1}, \frac{2y_0}{{x_0}^2+{y_0}^2+1},\frac{{x_0}^2+{y_0}^2-1}{{x_0}^2+{y_0}^2+1} \right> }$

and there we have it!

Now, this function is actually bijective as well. Go ahead and find the explicit inverse! That is, if I give you a $(x_s, y_s, z_s)$ and you tell me what $(x_0, y_0)$ was assigned there.

On another note, doesn’t it seem like the function about really resembles the one from our previous post? Our other function $f: \mathbb{R} \to \mathbb{S}$ was defined to be $f(x_0)=\displaystyle{ \left< \frac{2x_0}{{x_0}^2+1}, \frac{{x_0}^2-1}{{x_0}^2+1} \right>}$. You know, I dare be so bold as to assert the following.

Conjecture:
The function $f: \mathbb{R}^n \to \mathbb{S}^n$ where $\mathbb{S}^n = \{ x \in \mathbb{R}^{n+1} : ||x|| = 1 \}$ defined by

$\displaystyle{ f(x_1, \dots, x_n) = \left< \frac{2x_1}{\sum_{i=1}^n x_i^2\ + 1}, \frac{2x_2}{\sum_{i=1}^n x_i^2\ + 1}, \dots, \frac{2x_{n-1}}{\sum_{i=1}^n x_i^2\ + 1}, \frac{\sum_{i=1}^n x_i^2\ - 1}{\sum_{i=1}^n x_i^2\ + 1} \right> }$

is bijective.

Proof: Too tired, but I’ll leave it up to the reader to prove/disprove.

Corollary:
The set $\mathbb{R}^n$ is isomorphic to $\mathbb{S}^n$.

Note, $||(x_1, \dots, x_n) || = \sqrt{{x_1}^2 + \dots + {x_n}^2}$.

Another way of proving the above conjecture is finding the horrendous function’s inverse (the answer is called the stereographic projection, but there’s no fun in googling first!).

Also, the conjecture implies that $\mathbb{R}^n$ and $\mathbb{S}^n$, which is really worth giving it some thought if this is your first time hearing this. That radius 1 circle from last post? It has as many points as the real number line. The sphere of radius 1 in this post? Has as many points as the infinite two-dimensional plane of real numbers. Just let that set in.

Has the existential crisis kicked in yet?

On a very final note, if you thought that this and the preceding post were interesting I highly recommend looking into stereographic projections. Also, if you want a sort of “interactive” program to mess around with spherical geometry and projections onto a plane, take a look at this java program called Sphaerica. Here’s some screenshots from it:

Well, that’s all I got for today. Probably will be a good few days until my next post so I can mainly focus on my studies. See you then!