In a finite simple graph $X$, for any $t\in\mathbb{R}$, a vector $t$-coloring of $G$ is a mapping $\phi_t: V(X)\longrightarrow S^m$ for some $m\in\mathbb{N}$ (where $S^m$ is the $m$-sphere in $\mathbb{R}^{m+1}$) such that for any $x, y\in V(X)$, $\langle\, \phi_t(x) \,,\, \phi_t(y) \,\rangle \leq -\dfrac{1}{t-1}$ whenever $x\sim y$. The vector chromatic number of $G$ is the infimum among all real numbers $t\in\mathbb{R}$ such that $G$ has a vector $t$-coloring. The definition and more details can be found in this link.
Further, for any $t\in \mathbb{R}$, a strict vector $t$-coloring is a mapping $\psi_t:V(X) \longrightarrow S^m$ for some $m\in\mathbb{N}$ such that $\langle\, \psi_t(x) \,,\, \psi_t(y) \,\rangle = -\dfrac{1}{t-1}$, and the strict vector chromatic number $\chi_{sv}(G)$ is defined similarly. Clearly $\chi_v(G)\leq \chi_{sv}(G)$ and in the link above it is proved that $\omega(G)$ the max clique number of $G$ is less than or equal to $\chi_v(G)$. My question is that how to show $\chi_{sv}(K_n)\leq n$?
If we place the vertices of $K_n$ at the corners of an $(n-1)$-simplex with center at the origin, then we'll see the desired inner products of $-\frac1{n-1}$.
Although this can be done in $\mathbb R^{n-1}$, it's easier to implement in $\mathbb R^n$. Put the first vertex at the point $$\frac{(1-n, 1, 1, 1, \dots, 1)}{\sqrt{n(n-1)}}$$ and all other vertices at cyclic shifts of this point. Now check that: