2-distance Set

Lemma

Let $X=\{x_1,\dots ,x_v\}$ be a finite subset of the unit sphere in ${\mathbb{R}}^f$.
We say that $X$ is a spherical two-distance set if there are two constants $\alpha ,\beta$ such that for any two distinct points $x_i,x_j\in X$,

$$⟨x_i,x_j⟩\in \{\alpha ,\beta \}.$$

Equivalently, the squared distances $\Vert x_i-x_j{\Vert }^2$ take exactly two possible values for $i\ne j$.

Then $v\le \frac{f(f+3)}{2}$.

\begin{proof} Let $X=\{x_1,\dots ,x_v\}\subseteq {\mathbb{R}}^f$.

Assume that for any $i\ne j$, the squared distance $\Vert x_i-x_j{\Vert }^2$ takes one of two values, say $\alpha$ or $\beta$.

For each point $x_i$, define a polynomial function on ${\mathbb{R}}^f$ by

$$p_i(x)=(\Vert x-x_i{\Vert }^2-\alpha )(\Vert x-x_i{\Vert }^2-\beta ).$$

Note that

$$p_i(x_j)=\{\begin{array}{ll}0& \text{if }j\ne i,\\ \alpha \beta & \text{if }j=i.\end{array}$$

Therefore, the functions $p_1,\dots ,p_v$ are linearly independent: if ${\sum }_{i=1}^v{c}_i{p}_i(x)=0$, then evaluating at $x=x_j$ gives $c_j{p}_j(x_j)=0$, so $c_j=0$ for every $j$.

Thus $v\le \dim W$, where $W$ is the vector space spanned by all polynomial functions of the form $p_i$.

Each $p_i(x)$ is a polynomial in the coordinates of $x=(x_1,\dots ,x_f)$ of degree at most $2$ after restricting to the sphere containing the two-distance set. Hence $W$ lies inside the space of polynomials of degree at most $2$ in $f$ variables, with the constant term accounted for by the spherical relation.

The dimension of this space is

$$\left(\genfrac{}{}{0ex}{}{f+2}{2}\right)-1=\frac{(f+2)(f+1)}{2}-1=\frac{f(f+3)}{2}.$$

Therefore, $v\le \frac{f(f+3)}{2}$. \end{proof}