I wanna represent the following equation with just inner products of the unit vectors $\{ \mathbf v_i : i=0, \cdots, n\}$: $$\sum _{i=0}^n \mathbf v_i = \mathbf 0.$$
I had thought following simultaneous equation is true iff sum of v_i = 0: $$ \sum_{k\le c < d \le n}\langle \mathbf v_{\sigma(c)}, \mathbf v_{\sigma(d)}\rangle -\sum_{0\le a < b \le k-1}\langle \mathbf v_{\sigma(a)}, \mathbf v_{\sigma(b)}\rangle= k-\frac{n+1}{2}\quad \left(k=0, \cdots, \left\lfloor \frac{n+1}{2} \right\rfloor\right)$$ where $\sigma$ is a permutation in symmetric group $S_n$. For example, if $n=2$; $$k=0: \quad \langle \mathbf v_0, \mathbf v_1\rangle+ \langle \mathbf v_0, \mathbf v_2\rangle+ \langle \mathbf v_1, \mathbf v_2\rangle = - \frac{3}{2};$$ $$k=1: \quad \langle \mathbf v_0, \mathbf v_1\rangle= \langle \mathbf v_0, \mathbf v_2\rangle= \langle \mathbf v_1, \mathbf v_2\rangle = - \frac{1}{2}.$$ But in higher dimension space, it seems not to be valid. Was I right? If I was wrong, what are the formulas I have to add?
Denote by $X$ your inner product space. Note that by definiteness of your inner product a vector $v \in X$ is zero iff $\def\<#1>{\left<#1\right>}\<v,v> =0$. If you apply this to $\sum_i v_i$, you get $$ \sum_i v_i = 0\iff 0 = \<\sum_i v_i, \sum_j v_j> = \sum_{i,j} \<v_i, v_j> $$