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A map of a sphere (as in, e.g. the map of the Earth) is a projection/folding out of the surface area of a 3-dimensional ball onto a 2-dimensional plane.

Do we have images that visualize the equivalent for a 4-dimensional ball?

Ideally the "texture image" of the surface of the 4D ball would have to be visible on the 3d ball, in the same way that the distribution of water and land on the Earth is visible on a 2d map.

i.e. Do we have a map of the surface area of a 4-dimensional ball onto a 3-dimensional plane?

user56834
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  • You can use the usual projections. For example take each point on the 4ball surface ($x^2+y^2+z^2+w^2=1$) and put $w=0$, in this way you can assign a value to each point of the sphere in 4d. – N74 Dec 28 '16 at 08:20
  • I actually meant if these images already exist. I wouldn't know, or have the time to make them myself (although ofcourse I'd want to know the algorithms behind them) – user56834 Dec 28 '16 at 10:03
  • Well... The image of the azimuthal projection I was suggesting is a couple of spheres (just like the azimuthal projection of the Earth that is a couple of circles). If you don't have a map to represent it's hard to draw a picture of it, and I don't know of any painted 4d ball around. – N74 Dec 28 '16 at 10:15

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Just as an example, the so called stereographic projection could be generalized for the four dimensional case.

In three dimensions, the definition is as follows:

enter image description here

Let the Southern pole of the sphere of radius one above be tangent to the plane $z=0$. The coordinates of the North pole are then $(0,0,2)$. Say, we have a point $(a,b,c)$ on the sphere with an $\color{green}{R}$ on its surface. The vector pointing from $(a,b,c)$ to the North is $(a,b,c-2)$. So the equation of the straight line through the North pole and the point with the point $\color{green}{R}$ is

$$(x(t),y(t),z(t))=t(a,b,c-2)+(0,0,2).$$

This straight punches the plane $z=0$ at the $t$ for which $z(t)=t(c-2)+2=0$. So $t=\frac{2}{2-c}$ and the coordinates of the punching point are:

$$\left(\frac{2a}{2-c},\frac{2b}{2-c},0\right).$$

By definition the point above is the stereographic image of the green $\color{green}{R}$ on the horizontal plane.

It is easy generalize this projection if our sphere is four dimensional and its Southern pole touches the hyper plane $w=0$. (Let the coordinates of the North pole be $(0,0,0,2)$.) If we have a point $\color{red}{Q}$ on the surface of the four dimensional sphere at $(a,b,c,d)$ then the vector pointing from this point to the North pole is $(a,b,c,d-2)$, that is the equation of the straight through the North pole going through the point with the $\color{red}{Q}$ is

$$(x(t),y(t),z(t),w(t))=t(a,b,c,d-2)+(0,0,0,2).$$

We can repeat the last steps of the three dimensional case and we will find the location of the $\color{red}{Q}$ in the three dimensional space given by $w=0$.

I guess that any projection based on setting up equations of straights in the four dimensional space could be generalized for higher dimensions.

zoli
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