# Explanations about the volume of a regular simplex

I'm really sorry, this may sound ridiculous but I can't understand the Wikipedia explanation about the volume of regular n-dimensional simplices, here.

In particular, these two sentences make no sense to me:

If the coordinates of a point in a unit n-box are sorted, together with 0 and 1, and successive differences are taken, then since the results add to one, the result is a point in an n simplex spanned by the origin and the closest n vertices of the box. The taking of differences was a unimodular (volume-preserving) transformation, but sorting compressed the space by a factor of n!.

I think this might relate to the section about increasing coordinates (although I can't exactly see how), which I mostly understood but then again this sentence about volume measurement is obscure to me as well:

Indeed, the ordered simplex is a (closed) fundamental domain for the action of the symmetric group on the n-cube, meaning that the orbit of the ordered simplex under the n! elements of the symmetric group divides the n-cube into n! mostly disjoint simplices (disjoint except for boundaries), showing that this simplex has volume 1/n!.

I'm trying to read about fundamental domains to shed some light on the previous sentence, but it's not the easiest subject.. Help would be muchly appreciated.

• Yes, that is a confusing first sentence, at least taken out of context. It begins by referring to one point in the "unit box" and somehow asserts $n$ or more "vertices of the box", used (together with the origin) to define (span) an $n$-simplex. Nov 18 '14 at 11:43
• I guess the point I find non-trivial is since the results add to one. If this is true, then the point in question is indeed on the standard n-simplex in $\mathbb{R}^{n+1}$ (which is a region of the hyperplane $\sum_i x_i = 1$). But then I don't understand closest n vertices on the box either, since these vertices would always the same; they correspond to the canonical vectors. The n-box has $2^n$ vertices, why introduce such a great uncertainty in the choice of "closest vertices"? Finally, where does but sorting compressed the space by a factor of n! come from? Nov 18 '14 at 11:55
• Taking a point $(x_1,x_2,\ldots,x_n)$ "in" the unit box means $0 \le x_1,x_2,\ldots,x_n \le 1$. If the coordinates were sorted (ascending), so that $0 \le x_1 \le x_2 \;\ldots \le x_n \le 1$, then taking differences $y_i = x_{i+1} - x_i$, where for convenience we take $x_0 = 0$, $x_{n+1} = 1$, we get $y_i \in [0,1], i = 0,\ldots,n$ such that $\sum y_i = 1$. Nov 18 '14 at 12:45
• @hardmath You're right, it's just a telescopic sum. Thank you! Well, first point checked then :) Nov 18 '14 at 13:31
• It seems the idea is to present a map of the "unit box" in $\mathbb{R}^n$ onto the standard simplex, and show that it has a nice interpretation as a $n!$ to $1$ mapping that preserves volume on portions where the coordinates are already sorted. I'll have to puzzle over how to present the idea more clearly. Nov 18 '14 at 13:37