coordinates of vertices of regular simplex

Question

For $d=3$, vertex coordinates of a regular simplex have a simple expression since vertices correspond to four vertices of a cube. Is there a simple expression for higher dimensions? In particular I'm interested in $d=2^n-1$, integer $n$.

Edit: by coordinates I mean points in $\mathbb{R}^d$. Every $d$-simplex has a simple expression for coordinates in $\mathbb{R}^{d+1}$, as Mariano shows below

Answer 1

It is known that there is a regular simplex of side length $\sqrt{(d+1)/2}$ whose vertices are vertices of the cube $[-1,1]^d$ in $\Bbb{R}^d$ if and only if there exists a Hadamard matrix of order $d+1$; this is a square matrix of $\pm 1$-entries with pairwise orthogonal columns.

In particular, there exist Hadamard matrices of order $2^n$, one of which can be constructed using the recursive Sylvester's construction as explained on the above linked wikipedia page:

Let $H_0=[1]$ and $H_{n+1}=\left[\array{H_n & H_n \\\\ H_n & -H_n}\right].$

Note that the first column of $H_n$ consists only of ones. Delete it to obtain $2^n$ row vectors in $\Bbb{R}^{2^n-1}$. These are the coordinates of a regular simplex.

Answer 2

One interesting problem is to determine the $n$ such that there is a regular simplex in $\mathbb{R}^n$ with rational/integer coordinates. This is a well-known old problem which I discussed in this 1998 usenet thread, and is a nice application of the Hasse-Minowski theory of rational quadratic forms. The answer is yes iff $n+1$ is the sum of one, two, four or eight odd squares.

Answer 3

The $d$ points $(0,\dots,0,1,0,\dots,0)$ are the vertices of a regular $(d-1)$-simplex. If you want it to be centered at the origin, just substract their barycenter from them.

Answer 4

Compute the full Q matrix from the QR decomposition of a column vector of ones, and drop the first column.

eg in R+

simplex <- function(n) {
    qr.Q(qr(matrix(1,nrow=n)),complete=T)[,-1]
}

Answer 5

Since this question is back on the front page, I wanted to mention the simplex code. For $b\ge 2$, and $n\ge1$, the code consists of $b^n$ code words, each of which is a $b$-ary string of length $s=(b^n-1)/(b-1)$. The Hamming distance between any two codewords is $b^{n-1}$. Therefore, if you write the $(b-1)$-simplex using your favorite coordinates in $\mathbf{R}^{b-1}$, then substituting these $b$ vertices for the $b$ digits in your codewords and concatenating the resulting words into vectors in $\mathbf{R}^{(b-1)s}$ gives you a nice set of coordinates for the $(b^n-1)$-simplex.

For example, taking $b=2$ and $n=2$ gives the 4 alternating vertices of the 3-cube.

Answer 6

The coordinates of the vertices of a regular ''n''-dimensional simplex can be obtained from these two properties,

1. For a regular simplex, the distances of its vertices to its center are equal.
2. The angle subtended by any two vertices of an ''n''-dimensional simplex through its center is $\arccos\left(\tfrac{-1}{n}\right)$

If we initialize our first vertex to be along the first dimension, we can iteratively find the subsequent vertices by following the criteria above. Here's the code in Julia:

function regular_simplex(ndims, radius)
  verts = [zeros(ndims) for i in 1:(ndims+1)];
  for i in 1:(ndims+1)
    if i > 1
      for j in i:(ndims+1)
        verts[j][i-1] = (-dot(verts[i-1], verts[j]) - 1 / ndims) / verts[i-1][i-1];
      end
    end
    if i <= ndims
      verts[i][i] = sqrt(1 - sum(verts[i].^2))
    end
  end
  return verts .* radius;
end

Answer 7

We can orient and list the n+1 n-simplex unit vectors u{0:n} in such a way that the 0th simplex unit vector u{0} is equivalent to the 0th dimensional unit vector x{0}, the 1st simplex unit vector u{1} is a linear combination of the 0th and 1st dimensional unit vectors x{0} and x{1}, the 2nd simplex unit vector u{2} is a linear combination of the 0th through 2nd dimensional unit vectors, x{0}, x{1}, and x{2}, etc.

Always using a new dimensional unit vector in describing the next simplex unit vector. The pattern is broken in describing the final simplex unit vector u{n}, because there can be only n different dimensional unit vectors, but n+1 simplex unit vectors.

n+1 total vertices for regular n-simplex with unit radius n-circumsphere given as unit vectors oriented such that the jth simplex vector u{0<=j<=n-1} is a function of dimensional unit vectors x{0:j} and the final vector u{n} a function of x{0:n-1}

for 0<=j<=n-1
u{j} = sqrt((n+1)(n-j-1)/(n(n-j))) x{j} - sum{0<=k<=j}(sqrt((n+1)/(n(n-k)(n-k-1))) x{k})

and final vector

u{n} = -sum{0<=k<=n-1}(sqrt((n+1)/(n(n-k)(n-k-1))) x{k})

n+1 simplex vectors, u, as a function of n dimensional vectors, x, oriented such that u(j) is a function of only x(k<=j)