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?

  • $\begingroup$ I don't know a source for this, but would it be helpful if I described in detail the construction in the first half of that section (using $e_1,\dots,e_n$) in some of the language of the second half, or do you really want the full generality? $\endgroup$ Sep 1, 2021 at 17:45
  • $\begingroup$ That would be helpful, thank you. $\endgroup$
    – BillyJoe
    Sep 1, 2021 at 17:55

1 Answer 1


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 $n=3$.

enter image description here

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.

  • $\begingroup$ So this is for a $(n-1)$-simplex in $\mathbb{R}^n$ rather than $\mathbb{R}^{n-1}$ like in the Wikipedia page, however better than that, the coordinates are simpler. Thank you. $\endgroup$
    – BillyJoe
    Sep 1, 2021 at 21:06
  • $\begingroup$ The construction begins in $\mathbb{R}^n$, but $t+H$ is a linear subspace of dimension $n-1$ so can be identified with $\mathbb{R}^{n-1}$. One choice of basis for this space could be $\{(1,-1,0\dots,0),(0,1,-1,\dots,0),\dots,(0,\dots,1,-1)\}$ (I'm not claiming this is a particularly good or natural basis, just that it does the job). $\endgroup$ Sep 1, 2021 at 21:15
  • $\begingroup$ And putting the $(n-1)$-simplex in in $\mathbb{R}^n$ allows to avoid a possible reflection in the context of this question right? Meaning that we can solve the problem only with rotations. $\endgroup$
    – BillyJoe
    Sep 2, 2021 at 10:19
  • $\begingroup$ Yes I think that's right $\endgroup$ Sep 2, 2021 at 10:46
  • $\begingroup$ The main trick behind this representation of the n-simplex by coords of n+1 space is simply the fact that the n-simplex is a facet of the n+1-orthoplex (or cross-polytope), i.e. the generalization of the octahedron. $\endgroup$ Sep 14, 2021 at 16:42

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