TITLE: How to find $n+1$ equidistant vectors on an $n$-sphere?
QUESTION [5 upvotes]: Following this question, Proving the existence of a set of vectors, I'm looking for a way to find $n+1$ equidistant vectors on a Euclidean $n$-sphere.
For $n=2$, you can pick the vertices of any equilateral triangle.
For $n=3$, pick a tetrahedron.
What about larger dimensions?

REPLY [3 votes]: This description is pretty concrete.
In $\mathbb{R}^{n+1}$, consider the standard orthonormal basis $\{e_1, e_2, \ldots, e_{n+1} \}$. Clearly all $n+1$ points have the same distance to all others. Now, they all lie in the $n$-dimensional hyperplane (affine space) given by the equation $x_1+x_2+\dots+x_{n+1}=1$ (where the $x_i$ are the coordinates in $\mathbb{R}^{n+1}$) which is geometrically just a copy of  $\mathbb{R}^n$.
The average of those $n+1$ points is of course the point $C=\left( \frac{1}{n+1}, \frac{1}{n+1}, \ldots, \frac{1}{n+1}\right)$. Inside the hyperplane mentioned, with $C$ as center, draw the sphere with radius chosen such that it passes through all the $e_i$.
The radius would be $|C - e_1|=\sqrt{\frac{n+n^2}{(n+1)^2}}=\sqrt{\frac{n}{n+1}}$ it seems, but you can easily scale to whatever radius you desire.
See also Simplex (Wikipedia).