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.

  • $\begingroup$ 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. $\endgroup$
    – hardmath
    Nov 18 '14 at 11:43
  • $\begingroup$ 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? $\endgroup$
    – Jonathan H
    Nov 18 '14 at 11:55
  • 1
    $\begingroup$ 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 $\sum y_i = 1$. $\endgroup$
    – hardmath
    Nov 18 '14 at 12:45
  • $\begingroup$ @hardmath You're right, it's just a telescopic sum. Thank you! Well, first point checked then :) $\endgroup$
    – Jonathan H
    Nov 18 '14 at 13:31
  • $\begingroup$ 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. $\endgroup$
    – hardmath
    Nov 18 '14 at 13:37

You might check the proof of

Cartesian coordinates for vertices of a regular 16-simplex?

for a formula for the regular simplex volume.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.