# Projection of 4D sphere to 3D - focus point

I just finished watching this interesting/funny Numberphile video on stereographic projection. In the video he demonstrated that one could draw an $n$-dimensional sphere onto an $(n-1)$-dimensional object (plane?).

In his examples he placed the light at the top of the shape, what he called the "north pole". For $\mathbb{R}^2$ and $\mathbb{R}^3$ the point was the top of the sphere. However, for obvious reasons, you can only see the output of the mapping $\mathbb{R}^4 \to \mathbb{R}^3$. So it got me wondering, how do we define the "north pole" of a sphere in $\mathbb{R}^4$, within the context of projection? In $\mathbb{R}^n$?

For 2-dimensions we use the point with the max $y$ value. For 3-dimensions we use the point with the max $z$ value. For 4-dimensions is it the maximized point along the $w$-axis? If so, what does this look like and how do we know it's a pole for stereographic projection? I don't know much advanced topology, or any for that matter, so I'm looking for an answer with a $(3 \text{ or } 2)$-dimensional analogy or simplified explanation.

For your first question: using the stereographic projection you can project an $n$-dimensional sphere onto (what is called) a hyperplane, the $(n-1)$-dimensional object of your question.

For your second question, let's start defining (in one of the many possible ways) the stereographic projection:

Consider the unit sphere in the $3$-dimensional space $\mathbb{R}^3$ and let $N = (0, 0, 1)$ be the north pole and $S=\mathbb{S}^2\setminus\{N\}$ be the rest of the sphere. The plane $z = 0$ runs through the center of the sphere and the equator is the intersection of the sphere with this plane. For any point $p$ on $S$ there is a unique line through $N$ and $p$, and this line intersects the plane $z = 0$ in exactly one point $p^\prime$. We define the stereographic projection of $p$ to be this point $p^\prime$ in the plane.

As explained in the video of your question, the stereographic projection has nice properties like the conformality, meaning that it preserves the angles, despite it does not preserve area. In such a projection, circles on the sphere that do not pass through the point of projection are projected to circles on the plane and circles on the sphere that do pass through the point of projection are projected to straight lines on the plane.

I said this is one of the many ways we can define it, and you can check here why and how to define others. Using this idea, I will try to answer your second question:

As you said in the $3$-dimensional space the notion of north pole is basically the point on the sphere with maximum $z$-coordinate for a given coordinate system and this can be extended to any dimension by just considering the point $(0,0,0\dots,0,1)$ as the distinguished north pole.

But you don't really need this point. In general, one can define a stereographic projection from any point $w\in\mathbb{S}^n$ on the $n$-sphere onto any hyperplane $\Pi\in\mathbb{R}^{n+1}$ such that

• $\Pi$ is perpendicular to the diameter through $w$
• $\Pi$ does not contain $w$.

As long as $\Pi$ meets these conditions, then for any point $p$ other than $w$ (our distinguished point now) the line through $p$ and $w$ meets $\Pi$ in exactly one point $p^\prime$, which is defined to be the stereographic projection of $p$ onto $\Pi$. I use the terminology stereographic in the previous sentence because this projection has the same essential properties as the one in the $3$-dimensional case from the north pole $N$: they are smooth bijections with smooth inverse defined everywhere except at the projection point and they are conformal and not area-preserving.

• How do we know for every point $p$ on $S$, there is a unique line thru $N$ and $p$ that intersects $z=0$? This seems rather intuitive for 2 and 3 dimensions, but I don't want to generalize based on intuition. – Dando18 Jun 16 '17 at 13:31
• @Dando18 In the notation that I used in the answer, your question would be how do we know for every point $p$ on $S$, there is a unique line thru $w$ and $p$ that intersects $\Pi$? I'm doing this remark since you ask for the generalized case. Basically we are considering $\ell\cap\Pi$, where $\ell$ is the line mentioned before. This intersection can be $\ell$ (if it's contained in $\Pi$) or something with less dimension (a point). Since $\Pi$ is perpendicular to to the diameter through $w$, $\ell$ is not contained in $\Pi$ and $\ell\cap\Pi$ is a point. – Edu Jun 16 '17 at 13:46
• @Dando18 Sorry, I am reading your question again and I think I misunderstood it: are you asking why the line joining two points is unique? That should need explanation and does not depend on the dimension of the space... – Edu Jun 16 '17 at 13:56
• no I was asking how do we know there exists such a unique line ($\exists \ell$), which your first comment answered. Thanks! – Dando18 Jun 16 '17 at 14:08