Techniques for finding the edges of a 4-dimensional convex hull I am looking for some techniques in general, but let's do it on a (for me relevent) example. I have a set of $n^2$ points
$$p_{i,j}:=\frac1{\sqrt 2} \begin{pmatrix} \cos(2\pi i/n) \\ \sin(2\pi i/n) \\ \cos(2\pi j/n) \\ \sin(2\pi j/n) \end{pmatrix}\in\Bbb R^4,\qquad\text{for $i,j=1,...,n$}.$$
Let $P:=\mathrm{conv}\{p_{i,j}\}$ be the convex hull of all these points. Since all the points are on the boundary of the unit sphere of $\Bbb R^4$, they will form the vertices of $P$. 
Now I wonder, what are the edges of $P$? My guess would be that the edges are formed by the vertex pairs
$$(p_{i,j},p_{i+1,j})\quad\text{and}\quad (p_{i,j},p_{i,j+1})$$
for all $i=1,...,n-1$, and no others. Currently I have no tools to show this easily. What would be the fastest/most elegant way to approach this problem?
 A: That special polytope in fact is the duoprism of two n-gons, as already was mentioned in the comments.
In the general case, when considering n arbitrary vertices within $\mathbb{R}^4$ and would be asking for the edges of the convex hull, you would first have to consider any subset of 4 vertices each, i.e. ${n\choose 4}$ selections, and derive the thereby spanned hyperplanes. Next you would have to determine the subset of those, which have the remaining vertices on just a single side thereof. Those are the bounding hyperplanes.
The subset of vertices contained within either of these bounding hyperplanes then reduces your problem by 1 dimension. That is, you would have some set of $m<n$ vertices in $\mathbb{R}^3$, which then defines some bounding cell by means of the same convex hull procedure. Thus you take all subsets of 3 therefrom in turn, i.e. ${m\choose 3}$ selections, providing you with a set of 2D planes. Then again you have to detect the subset of those, which have the remaining vertices (of this set of $m$ for sure) on a single side therefrom. These now are the bounding 2D planes
Finally you do the same dimensional reduction once again, in order to derive the searched for edges / bounding 1D lines.
--- rk
