There are $binom{n}{2}$ distances between $n$ points in $mathbb{R}^d$. Not all of them can be chosen freely if $n$ exceeds the number $n_d = d + 1$. If $n = n_d$ we obviously have $binom{d+1}{2}$ distances which can be chosen (more or less) independently (restricted only by the triangle inequality).
I see two ways to count $N_n^d$, the number of independent distances between $ngeq d$ points in $mathbb{R}^d$, which is given by $nd - binom{d+1}{2}$
The first one: $nd$ coordinates minus one translation ($d$) minus one rotation ($binom{d}{2}$):
$N_n^d = nd - d - binom{d}{2} = nd - (binom{d}{2} + d) = nd - binom{d+1}{2}$
The second one: $binom{d+1}{2}$ distances between $d + 1$ base points plus $d ( n - (d + 1) )$ distances between the remaining points and $d$ of the base points (the remaining one serving to say in which half-space with respect to the $d$ base points the point is located):
$N_n^d = binom{d+1}{2} + d ( n - (d + 1) ) = nd - d (d + 1) + binom{d+1}{2} = nd - binom{d+1}{2}$. (Is this sound?)
Observation: The binomials seem to come from two very different directions (with two seemingly different interpretations), also the term $nd$. Does this tell something deep about (Euclidean) geometry? And what?
Are there further "independent" ways to compute $N_n^d$?
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