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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

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3 Answers 3

up vote 6 down vote accepted

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.

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Actually, I was already looking at Hadamard matrices in relation to this problem because such matrix is a common design matrix for exponential families, somewhat special in that it gives a transformation that decorellates components of a uniform multinomial random variable. So now, in addition to decorrelating components, it maps the variable (which lies in $2^n-1$ simplex embedded in $d=2^n$ space since variable l_1 norm is fixed) into d=2^n-1 dimensions with nice expression for extreme points, neat! –  Yaroslav Bulatov Sep 14 '10 at 22:14
    
Do you have a reference or a place where I can read more about this simplex in cube<->hadamard connection ? –  Yaroslav Bulatov Sep 20 '10 at 2:16
    
There is also an equivalence with inscribing cross polytopes in cubes. A nice reference to this and more is Adams, Zvengrowski and Laird, Vertex embeddings of regular polytopes, Expo. Math. 21 (2003), no. 4, 339--353. I can't find an open version on the web, but the link to the published version is dx.doi.org/10.1016/S0723-0869(03)80037-3 . –  Konrad Swanepoel Sep 20 '10 at 9:17

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.

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Yes, this seems to work to get a $(d-1)$-simplex in $\mathbb{R}^d$, but the question is I think about a $d$-simplex in $\mathbb{R}^d$. –  Louigi Addario-Berry Sep 14 '10 at 19:00
    
that gives coordinates in d+1 dimensions, but it's not clear when it corresponds to easy expression in d dimensions (like it does for 3-simplex) –  Yaroslav Bulatov Sep 14 '10 at 19:01
    
You may want to add the detail that you want the points to be in some specific space to the question, Yuraslov. –  Mariano Suárez-Alvarez Sep 14 '10 at 19:04
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I've done this recently, you just add in a point at $$ (-t, -t, \ldots, -t) $$ and figure out what $t$ needs to be. Then shift if it needs to be centered at the origin. –  Will Jagy Sep 14 '10 at 19:18
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In $R^n,$ all coordinates of the final point are $$ \frac{1 - \sqrt{1 + n}}{n} \; \; = \;\; \frac{-1}{1 + \sqrt{1 + n}} $$ –  Will Jagy Sep 14 '10 at 21:08

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.

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