So essentially what I'm asking is this: so let's say that I have a set of points that I arrange into a sphere. Now, with those same points I arrange them into that of a hypersphere. How do I convert the coordinates of a point in that hypersphere into coordinates of where the point would be if the points were assembled into a 3d sphere instead of its 4d hypersphere form.
Essentially project a 4-d coordinate to a 3-d one assuming that a 3-d shape retains the amount of points it has when it is shaped into its 4-d version. Essentially a sphere is rearranged into a hypersphere with the same amount of points. How do I convert a 4-d coordinate on the hypersphere into a 3-d coordinate on the sphere.
I have no formal understanding of projective geometry and I'm a bit iffy with vectors. I hope this question is understandable enough. Thank you.
Represent a point in spherical coordinates:
$$\begin{pmatrix} x\\y\\z\\w \end{pmatrix} = r \begin{pmatrix} \sin\phi \sin\psi \sin\theta \\ \sin\phi \sin\psi \cos\theta \\ \sin\phi \cos\psi \\ \cos\phi\\ \end{pmatrix}$$ then you can project on the equator by letting $\phi=\pi/2$. This gives you spherical coordinates in $(x,y,z)$:
$$\begin{pmatrix} x\\y\\z \end{pmatrix} = r \begin{pmatrix} \sin\psi \sin\theta \\ \sin\psi \cos\theta \\ \cos\psi \\ \end{pmatrix}$$
Notice however than when you are on a 3-sphere, i.e. $r=1$, then this will project to a 2-dimensional space, namely the 2-sphere.
If you want to retain more information and just project into a 3-dimensional space, you may want to use other projections like parallel projection. The easiest such projection is to just "forget" one coordinate:
$$\begin{align} \Bbb R^4 &\to \Bbb R^3 \\ (x,y,z,w) &\mapsto (x,y,z) \end{align}$$
For more flexibility, take a basis of $\Bbb R^4$ that's ortho-normal, say $e_1, \dots, e_4$ and write the points $P$ uniquely as:
$$P=\sum_{i=1}^4 p_i e_i$$ where the $p_i$ are the coordinates with respect to the $e_i$. Then do a parallel projection by taking $p_1, p_2, p_3$ as coordinates in $\Bbb R^3$, i.e. you just forget the 4th coordinate. This is then a parallel projection to the 3d-space spanned by $e_1, e_2, e_3$.