For any $q \in \mathbb{N}_{>0}$, let $\Omega_q = \{v_1, \ldots, v_q \} \subset \mathbb{R}^{q-1}$ be the set of $q$ points belonging to the surface of a unit sphere in $\mathbb{R}^{q-1}$ and corresponding to the vertices of a $q$-simplex (or regular $q$-hedron). For each $v_i \in \Omega_q$, let $v_i^{(j)}$ be the $j-th$ cartesian coordinate of the vector $v_i \in \Omega_q$, where $j \in \{1, \ldots, q-1\}.$
Question 1: it true that, for any $q \in \mathbb{N}_{>0}$, for any coordinate $j \in \{1, \ldots, q-1\}$, the following holds?
$$ \sum\limits_{n \in \{1, \ldots, q\} } v_n^{(j)} \, \, = 0. $$
Question 2: How can I compute the following sum, for any power $m \in \mathbb{N}_{>0}$
$$ \sum\limits_{n \in \{1, \ldots, q\} } (v_n^{(j)})^m? $$
What I tried for Question 1: It is not difficult to see that the statement is true for $q = 2, 3$ (in which case the elements of $\Omega_q$ correspond to the points $\{-1,1\}$ and to the vertices of a triangle with sides of equal length when $q =1$ and $q=2$ respectively), but I would like to have the general picture and I don't know how to deal with regular $q$-hedrons for large values of $q$,which I cannot easily draw.
Additional comment: I would be happy just of receiving some suggestions about how to approach the problems.
First of all, a q-simplex in general is NOT regular. Just as in 2D a triangle needn't have to be equilateral. But as you add within parantheses the assertation "regular q-hedron", you probably mean rather the "regular q-simplex".
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Next, there is a quite easy representation of any regular q-simplex as being described by the set of q points $(1, 0, 0, …, 0)$, $(0, 1, 0, …, 0)$, …, $(0, 0, 0, …, 1)$. This simply describes the (q-1)-dimensional simplex facet of the q-dimensional regular cross-polytope (or hyperoctahedron), in fact that one of the first orthant. This affine facet plane then obviously is given by the equation that the sum of all coords equates to $1$.
The vector sum of those vertices then is the point $(1, 1, 1, …, 1)$. It clearly would be beyond that affine plane. But you could project that point down onto this plane. from the equation of that plane it is easily derived that you just have to apply the general factor 1/q. That then will be the circumcenter of your regular simplex.
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If you would prefer to deal within a true hyperplane instead (i.e. a parallel plane containing the origin as well), then you could translate every point of that affine plane by this very circumcenter vector. Up to a global scaling by q you thus could well use the vertex coordinates $(q-1, -1, -1, …, -1)$, $(-1, \ q-1, -1, …, -1)$, …, $(-1, -1, -1, …,\ q-1)$ for the regular simplex too. Obviously the equation of this hyperplane is that the sum of the coords equates to $0$.
Moreover your Question 1 becomes quite obvious in this representation.
$$$$ --- rk