I am looking for a source for a more detailed explanation of the section "Cartesian coordinates for a regular n-dimensional simplex in $\mathbb{R}^n$" in the Wikipedia page for simplexes.
Especially regarding where "A highly symmetric way to construct a regular n-simplex is to use a representation of the cyclic group..." is described.
Once fixed an arbitrary (convenient) orientation of the $n$-simplex, provided that the centroid is on the origin, my actual aim is computing the coordinates of the centers of the edges of the $n$-simplex, though it should not be difficult once the coordinates of vertices is known.
I am particularly interested in polar coordinates.
Any hint?
Consider the subgroup of $S_n$ generated by the permutation $\sigma=(1\;2\;\cdots\; n)$, which is isomorphic to $\mathbb{Z}_n$. It acts via the permutation representation on $\mathbb{R}^n$ via the permutation representation, ie $\sigma$ maps a vector $(v_1,v_2,\dots,v_n)\mapsto(v_n,v_1,\dots,v_{n-1})$. As a matrix it is given by $$P= \begin{pmatrix} 0&1&0&\cdots&0\\ 0&0&1&\cdots&0\\ 0&0&0&\cdots&0\\ \vdots&\vdots&\vdots&\ddots&\vdots&\\ 1&0&0&\cdots&0 \end{pmatrix} $$ (this not in the block matrix $Q$ form stated on Wikipedia, but is much more intuitive I think). You can think of $P$ as cyclically permuting the coordinates. Moreover the $\langle P\rangle$-orbit of the standard basis vector $e_1=(1,0,\dots,0)$ is the set of all standard basis vectors, which I will denote by $e_k=P^{k-1}e_1$ for $1\le k\le n$.
The convex hull of these points is the standard $(n-1)$-simplex, and it lives in the affine hyperplane $H$ orthogonal to the vector $(1,\dots,1)$ which contains $e_1$. The centre of this simplex is the point $\frac{1}{n}(1,\dots,1)$.
I can write $$H=\{ (v_1,\dots,v_n) \in\mathbb{R}^n \mid v_1+\cdots+v_n =1\}.$$ Now, $H$ does not contain the origin, but we can translate everything by the vector $t=-\frac{1}{n}(1,\dots,1)$ so that $t+H$ is a linear hyperplane which contains a copy of the standard $(n-1)$-simplex centred on the origin.
Incidental remark: the representation of $\mathbb{Z}_n$ on $\mathbb{R}^n$ is reducible, and decomposes as $\mathbb{R}(1,\dots,1)\oplus t+H$ where the action on the first factor is trivial. For $n\ge 4$ the action on $t+H$ is still reducible but we don't need to decompose it further. In fact all irreducible representations of $\mathbb{Z}_n$ are $1$- or $2$-dimensional, and this is where the block decomposition of $Q$ comes from.
Back to the main point. The vertices of the translated simplex are $$e_k+t=\frac{1}{n}(-1,\dots,-1,\overset{k\textrm{th term}}{\overbrace{n-1}},-1,\dots,-1),$$ where $\frac{n-1}{n}$ appears in the $k$th coordinate. The midpoint of the edge joining $e_j+t$ and $e_k+t$ for $j<k$ is $$\frac{1}{2}(e_j+t+e_k+t)=\frac{1}{n}(-1,\dots,\overset{j\textrm{th term}}{\overbrace{\frac{n-2}{2}}},\dots,\overset{k\textrm{th term}}{\overbrace{\frac{n-2}{2}}},\dots,-1)$$ where the two $\frac{n-2}{2n}$ terms appear in the $j$th and $k$th coordinates.
Here's a sketch of what's going on for $n=3$.
This is all for a regular simplex with sides of length $\sqrt{2}$, so if you want a different side length, scale everything appropriately. If you want a non-regular simplex, then first apply a suitable linear transformation $A=\mathrm{Diag}(a_1,\dots,a_n)$ for $a_k>0$ to get a simplex with vertices $\{Ae_k\}_k$ living in the affine hyperplane $AH$ centred at the point $\frac{1}{n}A(1,\dots,1)$, and translate everything by minus this vector to get a simplex centred on the origin, and compute midpoints as above.
This is all done in cartesian coordinates of course, but it should be straightforward to translate everything into polar coordinates from here.