---- > [!definition] Definition. ([[stereographic projection]]) > The **stereographic projections** $\sigma_{N}$ carries a point $p=(x_{1},\dots,x_{n}, x_{n+1})$ on the [[sphere]] $\mathbb{S}^{n} \subset \mathbb{R}^{n+1}$ minus its north pole $N$, onto the intersection of $\mathbb{R}^{n} \times \{ 0 \}$ with the straight [[line]] that connects $N$ to $p$. $\sigma_{S}$ is define analogously. > ![[CleanShot 2024-03-08 at 16.42.21.jpg]] If we say that the **north pole** and the **south pole** of the [[sphere]] $\ss ^{n}$ in $\rr ^{n+1}$ are the points $N=(0,\dots,0,1) \and S=(0,\dots,0,-1),$ then the **stereographic projections** are $\sigma_{N}: \ss ^{n}\cut \{ N \} \to \rrn$ and $\sigma_{S}:\ss ^{n} \cut S \to \rrn$ by $\sigma_{N}(x_{1},\dots,x_{n+1})=\frac{(x_{1},\dots,x_{n})}{1-x_{n+1}} \and \sigma_{S}(x_{1},\dots,x_{n+1})=\frac{x_{1},\dots,x_{n}}{1+x_{n+1}}.$ > [!basicexample] Example > This is how maps of the world are often created! > > [!basicexample] Basic Example. ([[stereographic projection]] in $\mathbb{R}^{3}$) > Consider the [[sphere]] $\mathbb{S}^{2}=\{ (x,y,z) \in \mathbb{R}^{2}: x^{2}+y^{2}+(z-1)^{2} \}=1$. (Note that this is shifted from definition above.) It has north pole $N=(0,0,2)$ and south pole $S=(0,0,0)$. We claim that $\pi(x,y,z)=2\frac{(x,y)}{2-z}=N_{3} \frac{(x,y)}{2-z}.$ $N_{3}=1$ in the definition above; $N_{3}=2$ in this example. To see this, recall that the equation for a line passing through $(x,y,z)$ and $N=(0,0,2)$ $\mathbb{R}^{3}$ is $\b \ell(t)=(\ell_{1}(t), \ell_{2}(t), \ell_{3}(t))=(x-0, y-0, z-2)t + (0,0,2), \ \ t \in \mathbb{R}.$ We want to find when $\ell_{3}(t)=0$. This entails solving $(z-2)t+2=0,$ for which we find $t^{*}=\frac{2}{2-z}$. We plug $t^{*}$ into the equation for $\b \ell(t)$ to find $\b \ell(t^{*})=(\frac{2x}{2-z}, \frac{2y}{2-z}, 0).$ And after dropping the final coordinate we are done. > We next claim that the [[stereographic projection]] $\pi:\mathbb{S}^2-\{ N \} \to \mathbb{R}^{2}$ has the following inverse map on $\mathbb{S}^2-\{ N \}$: $\begin{align} \pi ^{-1}(u,v)= & \big( x(u,v), y(u,v), z(u,v) \big), \text{ where } \\ x(u,v) = & \frac{4u}{u^{2}+v^{2}+4} \\ y(u,v)= & \frac{4v}{u^{2}+v^{2}+4} \\ z(u,v)= & \frac{2(u^{2}+v^{2})}{u^{2}+v^{2} + 4}. \end{align}$ To begin, $\pi(N)=0$ and thus $\pi ^{-1}(0,0)=N$ since $\pi$ is a [[bijection]]. This obviously holds when plugged into the equation above. Next assume $p=(x,y,z)$ is nonzero, and $\pi(p)=(u,v)$ (see figure). Observe that, since $p$ and $\pi(p)$ live on the same [[line]] by construction, there exists nonzero $c \in \mathbb{R}$ which scales the [[vector]] $(x,y,z)-N$ into the [[vector]] $(u,v,0)-N$ . Recalling that $N=(0,0,2)$, this implies $c(u,v,-2)=(x,y,z-2).$ And so $x=cu , y=cv, z-2=-2c$ we just need to find $c$. We haven't used yet the fact that $(x,y,z)$ is one the [[sphere]]: considering this subjects the above to the constraint $x^{2}+y^{2}+(z-1)^{2}=1$. We obtain $1=x^{2}+y^{2}+(z-1)^{2}=c^{2}u^{2}+c^{2}v^{2}+(-2c+2-1)^{2}.$ The solutions to this quadratic equation are given by Python to be $c=0, c=\frac{4}{u^{2}+v^{2}+4}.$ We discard the trivial solution because we have assumed $c \neq 0$. Substituting the nontrival solution into $(*)$, we find $\begin{align} x= & cu=\frac{4u}{u^{2}+v^{2}+4} \\ y = & cv = \frac{4v}{u^{2}+v^{2}+4} \\ \frac{z}{2}= & -c + 2 = \frac{-4}{u^{2}+v^{2}+4} + 1= \frac{ u^{2} + v^{2}}{u^{2}+v^{2}+4} \end{align}$ and this is the result. > Note that we don't need to verify $\pi \circ \pi ^{-1}=\id$; our construction of $\pi ^{-1}$ follows straight from geometric definition to be the correct inverse map. > Clearly $\pi$ and $\pi ^{-1}$ are both smooth. So $\pi$ is a [[homeomorphism]]. From Python, the [[derivative]] of $\pi ^{-1}$ is $J = \begin{bmatrix} \frac{-8u^2}{(u^2 + v^2 + 4)^2} + \frac{4}{u^2 + v^2 + 4} & \frac{-8uv}{(u^2 + v^2 + 4)^2} \\ \frac{-8uv}{(u^2 + v^2 + 4)^2} & \frac{-8v^2}{(u^2 + v^2 + 4)^2} + \frac{4}{u^2 + v^2 + 4} \\ \frac{-4u(u^2 + v^2)}{(u^2 + v^2 + 4)^2} + \frac{4u}{u^2 + v^2 + 4} & \frac{-4v(u^2 + v^2)}{(u^2 + v^2 + 4)^2} + \frac{4v}{u^2 + v^2 + 4} \end{bmatrix}$ and this has [[rank]] $2$ because clearly $J_{1,1}$ is not a scalar multiple of $J_{1,2}$. > Rotational symmetry of the [[sphere]] then implies that we can define a [[stereographic projection]] that excludes the south pole instead of the north; the two maps together cover the sphere as required. ^68968c ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```