Let $n$ points $x_1, . . . , x_n$ lying on a unit sphere $S^{d-1} \subset R^d$ (with center in the origin) are such that the angle between any two of them is the same, that is,$<x_i , x_j>$ does not depend on $i \neq j ∈ [n]$. Prove that $n ≤ d + 1$.
This the numerical I am trying to solve. But the problem I am facing is that I cannot imagine how to take n points so that they all have the same angle. I do think I can solve it using a gram matrix of points $x_1, . . . , x_n$ but somehow I am not able to implement it. Could you please help me to solve it. Thank you in advance.
Consider the Gram matrix of $d+2$ distinct unit vectors $v_1\ldots,v_{d+2}$ of $\mathbb{R}^d$:
$$G_{ij}=\langle v_i,v_j\rangle.$$
Assume $\langle v_i,v_j\rangle =\left\{\begin{array}{ll} x, \text{ if } i\neq j \\ 1, \text{ if } i=j \end{array}\right. $
Since $v_i\neq v_j$, we have $x<1$.
Notice that $G_{d+2\times d+2}=(1-x)Id_{d+2\times d+2}+xA_{d+2\times d+2}$, where $A_{ij}=1$ for every $i,j$.
So the spectrum of $G_{d+2\times d+2}$ is $\ \ (1-x)+x(d+2),\ \stackrel{d+1 \text{ times}}{\overbrace{1-x,\ldots, 1-x}}$.
The rank of $G$ is at most $d$, since $v_1,\ldots, v_{d+2}\in \mathbb{R}^d$. So the multiplicity of the eigenvalue $0$ of $G$ is at least two.
Notice that the only value of $x$ that forces $G$ to be singular satisfies $(1-x)+x(d+2)=0$, but the multiplicity of $(1-x)+x(d+2)$ is one. Absurd.
There are no $d+2$ distinct unit vectors of $\mathbb{R}^d$ with the same angle between any two of them.